Fluid Mechanics (Solution Manual) [6 ed.] 9780124059351, 012405935X

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Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.1. Many centuries ago, a mariner poured 100 cm3 of water into the ocean. As time passed, the action of currents, tides, and weather mixed the liquid uniformly throughout the earth’s oceans, lakes, and rivers. Ignoring salinity, estimate the probability that the next sip (5 ml) of water you drink will contain at least one water molecule that was dumped by the mariner. Assess your chances of ever drinking truly pristine water. (Consider the following facts: Mw for water is 18.0 kg per kg-mole, the radius of the earth is 6370 km, the mean depth of the oceans is approximately 3.8 km and they cover 71% of the surface of the earth.) Solution 1.1. To get started, first list or determine the volumes involved: υd = volume of water dumped = 100 cm3, υc = volume of a sip = 5 cm3, and V = volume of water in the oceans = 4 πR 2 Dγ , where, R is the radius of the earth, D is the mean depth of the oceans, and γ is the oceans' coverage fraction. Here we've ignored the ocean volume occupied by salt and have assumed that the oceans' depth is small compared to the earth's diameter. Putting in the numbers produces: € V = 4 π (6.37 ×10 6 m) 2 (3.8 ×10 3 m)(0.71) = 1.376 ×1018 m 3 . For well-mixed oceans, the probability Po that any water molecule in the ocean came from the dumped water is: (100 cm3 of water) υ d 1.0 ×10−4 m 3 Po = = = = 7.27 ×10−23 , € 18 3 (oceans' volume) V 1.376 ×10 m Denote the probability that at least one molecule from the dumped water is part of your next sip as P1 (this is the answer to the question). Without a lot of combinatorial analysis, P1 is not easy to calculate directly. It is easier to proceed by determining the probability P2 that all the € in your cup are not from the dumped water. With these definitions, P can be molecules 1 determined from: P1 = 1 – P2. Here, we can calculate P2 from: P2 = (the probability that a molecule was not in the dumped water)[number of molecules in a sip]. The number of molecules, Nc, in one sip of water is (approximately) 1.00g gmole molecules N c = 5cm 3 × × × 6.023×10 23 = 1.673×10 23 molecules 3 cm 18.0g gmole 23

€

Thus, P2 = (1− Po ) Nc = (1− 7.27 ×10 −23 )1.673×10 . Unfortunately, electronic calculators and modern computer math programs cannot evaluate this expression, so analytical techniques are required. First, take the natural log of both sides, i.e. ln(P2 ) = N c ln(1− Po ) = 1.673×10 23 ln(1− 7.27 ×10 −23 ) then expand the natural logarithm using ln(1–ε) ≈ –ε (the first term of a standard Taylor series for ε → 0 ) ln(P2 ) ≅ −N c ⋅ Po = −1.673×10 23 ⋅ 7.27 ×10 −23 = −12.16 , and exponentiate to find: P2 ≅ e−12.16 ≅ 5 ×10 −6 ... (!) Therefore, P1 = 1 – P2 is very close to unity, so there is a virtual certainty that the next sip of water you drink will have at least one molecule in it from the 100 cm3 of water dumped many years ago. So, if one considers the rate at which they themselves and everyone else on the planet uses water it is essentially impossible to enjoy a truly fresh sip.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.2. An adult human expels approximately 500 ml of air with each breath during ordinary breathing. Imagining that two people exchanged greetings (one breath each) many centuries ago, and that their breath subsequently has been mixed uniformly throughout the atmosphere, estimate the probability that the next breath you take will contain at least one air molecule from that age-old verbal exchange. Assess your chances of ever getting a truly fresh breath of air. For this problem, assume that air is composed of identical molecules having Mw = 29.0 kg per kg-mole and that the average atmospheric pressure on the surface of the earth is 100 kPa. Use 6370 km for the radius of the earth and 1.20 kg/m3 for the density of air at room temperature and pressure. Solution 1.2. To get started, first determine the masses involved. m = mass of air in one breath = density x volume = (1.20kg /m 3 )(0.5 ×10−3 m 3 ) = 0.60 ×10−3 kg ∞

M = mass of air in the atmosphere = 4 πR

2

∫ ρ(z)dz

z= 0

Here, R is the radius of the earth, z is the elevation above the surface of€the earth, and ρ(z) is the € air density as function of elevation. From the law for static pressure in a gravitational field, z= +∞

€ Ps, on the earth is determined from Ps − P∞ = dP dz = −ρg , the surface pressure,

∫ ρ(z)gdz so

z= 0

Ps − P∞ (10 5 Pa) = 4π (6.37 ×10 6 m)2 = 5.2 ×1018 kg . −2 g 9.81ms where the pressure (vacuum) in outer space = P∞ = 0, and g is€assumed constant throughout the atmosphere. For a well-mixed atmosphere, the probability Po that any molecule in the atmosphere came from the age-old verbal exchange is 2 × (mass of one breath) 2m 1.2 ×10−3 kg Po = = = = 2.31×10−22 , 18 (mass of the whole atmosphere) M 5.2 ×10 kg where the factor of two comes from one breath for each person. Denote the probability that at least one molecule from the age-old verbal exchange is part of your next breath as P1 (this is the answer to the question). Without a lot of combinatorial analysis, P1 is not easy to calculate € directly. It is easier to proceed by determining the probability P2 that all the molecules in your next breath are not from the age-old verbal exchange. With these definitions, P1 can be determined from: P1 = 1 – P2. Here, we can calculate P2 from: P2 = (the probability that a molecule was not in the verbal exchange)[number of molecules in one breath]. The number of molecules, Nb, involved in one breath is 0.6 ×10−3 kg 10 3 g molecules Nb = × × 6.023 ×10 23 = 1.25 ×10 22 molecules 29.0g /gmole kg gmole Nb −22 1.25×10 22 Thus, P2 = (1− Po ) = (1− 2.31×10 ) . Unfortunately, electronic calculators and modern computer math programs cannot evaluate this expression, so analytical techniques are required. First, € take the natural log of both sides, i.e. ln(P2 ) = N b ln(1− Po ) = 1.25 ×10 22 ln(1− 2.31×10−22 ) € then expand the natural logarithm using ln(1–ε) ≈ –ε (the first term of a standard Taylor series for ε → 0 ) ln(P2 ) ≅ −N b ⋅ Po = −1.25 ×10 22 ⋅ 2.31×10−22 = −2.89 , € and exponentiate to find: M = 4π R 2

that:

€

€ €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

P2 ≅ e−2.89 = 0.056 . Therefore, P1 = 1 – P2 = 0.944 so there is a better than 94% chance that the next breath you take will have at least one molecule in it from the age-old verbal exchange. So, if one considers how often they themselves and everyone else breathes, it is essentially impossible to get a breath of € truly fresh air.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.3. The Maxwell probability distribution, f(v) = f(v1,v2,v3), of molecular velocities in a gas flow at a point in space with average velocity u is given by (1.1). a) Verify that u is the average molecular velocity, and determine the standard deviations (σ1, 12 #1 & 2 3 σ2, σ3) of each component of u using σ i = % ∫∫∫ (vi − ui ) f (v)d v( for i = 1, 2, and 3. $ n all v ' b) Using (1.27) or (1.28), determine n = N/V at room temperature T = 295 K and atmospheric pressure p = 101.3 kPa. c) Determine N = nV = number of molecules in volumes V = (10 µm)3, 1 µm3, and (0.1 µm)3. d) For the ith velocity component, the standard deviation of the average, σa,i, over N molecules is σa,i = σ i N when N >> 1. For an airflow at u = (1.0 ms–1, 0, 0), compute the relative uncertainty, 2σ a,1 u1 , at the 95% confidence level for the average velocity for the three volumes listed in part c). e) For the conditions specified in parts b) and d), what is the smallest volume of gas that ensures a relative uncertainty in U of one percent or less? Solution 1.3. a) Use the given distribution, and the definition of an average: 32 " m % +∞ +∞ +∞ * m 1 23 (v)ave = ∫∫∫ v f (v)d v = $ v − u . d 3v . ' ∫ ∫ ∫ v exp +− n all u # 2π kBT & −∞ −∞ −∞ , 2kBT / Consider the first component of v, and separate out the integrations in the "2" and "3" directions. 32 ! m $ +∞ +∞ +∞ . m 1 *+(v1 − u1 )2 + (v2 − u2 )2 + (v3 − u3 )2 ,-2 dv1dv2 dv3 (v1 )ave = # & ∫ ∫ ∫ v1 exp /− " 2π kBT % −∞ −∞ −∞ 0 2kBT 3 3 2 +∞

! m $ =# & " 2 π k BT %

* m(v1 − u1 )2 - +∞ * m(v2 − u2 )2 - +∞ * m(v3 − u3 )2 v exp − dv exp − dv exp + . + . ∫ 1 , 2k T / 1 ∫ , 2k T / 2 ∫ +,− 2k T ./dv3 B B B −∞ −∞ −∞ 12

The integrations in the "2" and "3" directions are equal to: (2πk B T m) , so 1 2 +∞

! m $ (v1 )ave = # & " 2 π k BT %

* m(v1 − u1 )2 . dv1 2kBT /

∫ v exp +,− 1

−∞

12 € The change of integration variable to β = (v1 − u1 ) ( m 2kBT ) changes this integral to:

! ! 2k T $1 2 $ 1 ∫ ## β #" mB &% + u1 && exp {−β 2 } d β = 0 + π u1 π = u1 , % −∞ " where the first term of the integrand is an odd function integrated on an even interval so its contribution is zero. This procedure is readily repeated for the other directions to find (v2)ave = u2, and (v3)ave = u3. Thus, u = (u1, u2, u3) is the average molecular velocity. Using the same simplifications and change of integration variables produces: 32 ! m $ +∞ +∞ +∞ . m 1 2 2 *+(v1 − u1 )2 + (v2 − u2 )2 + (v3 − u3 )2 ,-2 dv1dv2 dv3 σ1 = # & ∫ ∫ ∫ (v1 − u1 ) exp /− " 2π kBT % −∞ −∞ −∞ 0 2kBT 3 1 (v1 )ave = π

1 2 +∞

! m $ =# & " 2 π k BT %

+∞

* m(v1 − u1 )2 1 ! 2kBT $ +∞ 2 ∫ (v1 − u1 ) exp +,− 2k T ./ dv1 = π #" m &% ∫ β exp {−β 2 } d β . B ±∞ ±∞ 2

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

The final integral over β is:

π 2 , so the standard deviations of molecular speed are 12 σ1 = ( k B T m) = σ 2 = σ 3 , where the second two equalities follow from repeating this calculation for the second and third directions. € b) From (1.27), n V = p kBT = (101.3kPa) [1.381×10 −23 J / K ⋅ 295K ] = 2.487 ×10 25 m −3 € n = 2.487 ×1010 for V = 103 µm3 = 10–15 m3 c) From n/V from part b): n = 2.487 ×10 7 for V = 1.0 µm3 = 10–18 m3 n = 2.487 ×10 4 for V = 0.001 µm3 = 10–21 m3 d) From (1.29), the gas constant is R = (kB/m), and R = 287 m2/s2K for air. Compute: € 12 12 12 2σ a,1 u1 = 2 ( kBT m€ n ) [1m / s ] = 2 ( RT n ) 1m / s = 2 ( 287 ⋅ 295 n ) = 582 n . Thus, for V = 10–15 m3 € : 2σ a,1 u1 = 0.00369, –18 3 V = 10 m : 2σ a,1 u1 = 0.117, and V = 10–21 m3 : 2σ a,1 u1 = 3.69. e) To achieve a relative uncertainty of 1% we need n ≈ (582/0.01)2 = 3.39 × 109, and this corresponds to a volume of 1.36 × 10-16 m3 which is a cube with side dimension ≈ 5 µm. € €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.4. Using the Maxwell molecular speed distribution given by (1.4), a) determine the most probable molecular speed, b) show that the average molecular speed is as given in (1.5), 12 #1 ∞ & c) determine the root-mean square molecular speed = vrms = % ∫ 0 v 2 f (v)dv( , $n ' d) and compare the results from parts a), b) and c) with c = speed of sound in a perfect gas under the same conditions. Solution 1.4. a) The most probable speed, vmp, occurs where f(v) is maximum. Thus, differentiate (1.4) with respect v, set this derivative equal to zero, and solve for vmp. Start from: 32 ! m $ 2 ( mv 2 + f (v) = 4π n # , , and differentiate & v exp )− " 2 π k BT % * 2kBT 32 2 , 3 (* mvmp (* mv 2 ,*2 ! m $ / * mvmp df = 4π n # exp )− mp -4 = 0 -− & 12vmp exp )− dv " 2π kBT % 10 +* 2kBT .* kBT +* 2kBT .*43 Divide out common factors to find: mv 2 2kBT . 2 − mp = 0 or vmp = m k BT b) From (1.5), the average molecular speed v is given by: 32 # m & ∞ 3 * mv 2 1∞ v = ∫ v f (v)dv = 4π % . dv . ( ∫ v exp +− n 0 $ 2 π k BT ' 0 , 2kBT / Change the integration variable to β = mv 2 2kBT to simplify the integral: 12

12 12 ! m $ k BT ∞ ! 8kBT $ ! 8kBT $ −β −β ∞ v = 4# & ∫ β exp {−β } d β = #" π m &% (−β e − e )0 = #" π m &% , " 2 π k BT % m 0 and this matches the result provided in (1.5). c) The root-mean-square molecular speed vrms is given by: 32 # m & ∞ 4 * mv 2 1∞ 2 2 vrms = ∫ v f (v)dv = 4π % . dv . ( ∫ v exp +− n 0 $ 2 π k BT ' 0 , 2kBT /

12

Change the integration variable to β = v ( m 2kBT )

to simplify the integral:

12 ∞

4 ! 2kBT $ 4 ! 2kBT $ 3 π 3kBT 4 2 . = # & ∫ β exp {−β } d β = # & m π" m % 0 π" m % 8 Thus, vrms = (3kBT/m)1/2. d) From (1.28), R = (kB/m) so vmp = 2RT , v = (8 / π )RT , and vrms = 3RT . All three speeds 2 vrms =

have the same temperature dependence the speed of sound in a perfect gas: c = γRT , but are factors of

2γ,

8 πγ and

3 γ , respectively, larger than c. €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.5. By considering the volume swept out by a moving molecule, estimate how the mean-free path, l, depends on the average molecular cross section dimension d and the molecular number density n for nominally spherical molecules. Find a formula for ln1 3 (the ratio of the mean-free path to the mean intermolecular spacing) in terms of the nominal molecular volume ( d 3 ) and the available volume per molecule (1/n). Is this ratio typically bigger or smaller than one?

€

€

Solution 1.5. The combined collision cross section for two spherical molecules having diameter d is πd 2 . The mean free path l is the average distance traveled by a molecule between collisions. Thus, the average molecule should experience one collision when sweeping a volume equal to πd 2 l . If the molecular number density is n, then the volume per molecule is n–1, and the d mean intermolecular spacing is n–1/3. Assuming that the swept volume necessary to produce one collision is proportional to the volume per € molecule produces: 2 π d l = C n or l = C ( nπ d 2 ) ,

€

where C is a dimensionless constant presumed to be of order unity. The dimensionless version of this equation is: mean free path l = −1 3 = l n1 3 mean intermolecular spacing n " n −1 % C C = 23 2 = = C $ 3' 3 23 n πd #d & nd ( )

23

23

" volume per molecule % = C$ ' , # molecular volume &

where all numerical constants like π have been combined into C. Under ordinary conditions in gases, the molecules are not tightly packed so l >> n −1 3 . In liquids, the molecules are tightly packed so l ~ n −1 3 .

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.6. Compute the average relative speed, vr , between molecules in a gas using the Maxwell speed distribution f given by (1.4) via the following steps. a) If u and v are the velocities of two molecules then their relative velocity is: vr = u – v. If the angle between u and v is θ, show that the relative speed is: vr = |vr| = u 2 + v 2 − 2uv cosθ where u = |u|, and v = |v|. b) The averaging of vr necessary to determine vr must include all possible values of the two speeds (u and v) and all possible angles θ. Therefore, start from: 1 vr = 2 ∫ vr f (u) f (v)sin θ dθ dvdu , 2n all u,v,θ and note that vr is unchanged by exchange of u and v, to reach:

1 ∞ ∞ π ∫ ∫ ∫ u2 + v2 − 2uv cosθ sinθ f (u) f (v)dθ dvdu n 2 u=0 v=u θ =0 c) Note that vr must always be positive and perform the integrations, starting with the angular one, to find: 12 # 16k T & 1 ∞ ∞ 2u3 + 6uv 2 vr = 2 ∫ ∫ f (u) f (v)dvdu = % B ( = 2 v . $ π ' 3n u=0 v=u uv vr =

Solution 1.6. a) Compute the dot produce of vr with itself: 2

v r = v r ⋅ v r = (u − v)⋅ (u − v) = u ⋅ u − 2u ⋅ v + v ⋅ v = u 2 − 2uv cosθ + v 2 . Take the square root to find: |vr| = u 2 + v 2 − 2uv cosθ . b) The average relative speed must account for all possible molecular speeds and all possible angles between the two molecules. [The coefficient 1/2 appears in the first equality below because the probability density function of for the angle θ in the interval 0 ≤ θ ≤ π is (1/2)sinθ.] 1 vr = 2 ∫ vr f (u) f (v)sin θ dθ dvdu 2n all u,v,θ

∫

u 2 + v 2 − 2uv cosθ f (u) f (v)sin θ dθ dvdu

1 ∞ ∞ π = 2 ∫ ∫ ∫ u 2 + v 2 − 2uv cosθ sin θ f (u) f (v)dθ dvdu. 2n u=0 v=0 θ =0 In u-v coordinates, the integration domain covers the first quadrant, and the integrand is unchanged when u and v are v! dv! swapped. Thus, the u-v integration can be completed above the line u = v if the final result is doubled. Thus, 1 ∞ ∞ π vr = 2 ∫ ∫ ∫ u 2 + v 2 − 2uv cosθ sin θ f (u) f (v)dθ dvdu . n u=0 v=u θ =0 Now tackle the angular integration, by setting β = u 2 + v 2 − 2uv cosθ so that d β = +2uvsin θ dθ . This leads to 2

1 ∞ ∞ (v+u) dβ vr = 2 ∫ ∫ ∫ β 1 2 f (u) f (v)dvdu , n u=0 v=u β =(v−u)2 2uv

v!

all u,v,θ

=

1 2n 2

u

=

du! u!

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

and the β-integration can be performed: 1 ∞ ∞ 2 3 2 (v+u)2 f (u) f (v) 1 ∞ ∞ f (u) f (v) vr = 2 ∫ ∫ β dvdu = 2 ∫ ∫ (v + u)3 − (v − u)3 dvdu . 2 (v−u) 2n u=0 v=u 3 u v 3n u=0 v=u u v Expand the cubic terms, simplify the integrand, and prepare to evaluate the v-integration: 1 ∞ ∞ f (v) f (u) vr = 2 ∫ ∫ 2u3 + 6uv 2 dv du 3n u=0 v=u v u

( )

(

(

)

)

32

# m & 2 * mv 2 - f (u) 1 ∞ ∞ 3 2 4π = ∫ ∫ (2u + 6uv ) v %$ 2π k T (' v exp +,− 2k T ./ dv u du. 3n u=0 v=u B B Use the variable substitution: α = mv2/2kBT so that dα = mvdv/kBT, which reduces the vintegration to: 32 1 ∞ ∞ ! 3 k BT $ ! m $ kT f (u) vr = α &4 π # du & exp {−α } B dα # 2u + 6u ∫ ∫ 3n u=0 mu2 kBT " m % " 2 π k BT % m u 12 ∞

2! m $ = # & 3n " 2π kBT %

∞

! 3 k T $ −α f (u) du # 2u + 6u B α &e dα " m % u u=0 mu2 kBT

∫ ∫

12 ∞

2! m $ = # & 3n " 2π kBT %

∞

! $ kT f (u) ∫ #"−2u3e−α + 6u mB (−α e−α − e−α )&% 2 u du mu kBT u=0

12

! mu 2 $ f (u) 2! m $ ∞! 3 k BT $ = # 8u +12u exp du. & ∫# #− & & 3n " 2π kBT % u=0 " m % " 2kBT % u The final u-integration may be completed by substituting in for f(u) and using the variable 12 substitution γ = u ( m kBT ) . 12 ∞

32 32 ! ! k T $3 2 ! k BT $ $ ! m $ ! k BT $ 3 B ∫ ##8#" m &% γ +12 #" m &% γ &&4π #" 2π k T &% #" m &%γ exp (−γ 2 ) dγ % B 0 "

1 ! 2m $ vr = # & 3 " π k BT %

12 ∞

2 !k T $ = # B & 3π " m %

12

$ 2 !k T $ ! 3 1 ∫ (8γ +12γ ) exp (−γ ) dγ = 3π #" mB &% #"8 8 π +12 4 π &% 0 4

12

2

12

2

! 16k T $ 4 ! k BT $ = # & = # B & = 2v " πm % π" m % Here, v is the mean molecular speed from (1.5).

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.7. In a gas, the molecular momentum flux (MFij) in the j-coordinate direction that crosses a flat surface of unit area with coordinate normal direction i is: 1 MFij = ∫∫∫ mvi v j f (v)d 3v where f(v) is the Maxwell velocity distribution (1.1). For a perfect V all v gas that is not moving on average (i.e., u = 0), show that MFij = p (the pressure), when i = j, and that MFij = 0, when i ≠ j. Solution 1.7. Start from the given equation using the Maxwell distribution: 32 * m 1 nm " m % +∞ +∞ +∞ 3 MFij = ∫∫∫ mvi v j f (v)d v = v12 + v22 + v32 ). dv1dv2 dv3 $ ' ∫ ∫ ∫ vi v j exp +− ( V all u V # 2π kBT & −∞ −∞ −∞ , 2kBT / and first consider i = j = 1, and recognize ρ = nm/V as the gas density, as in (1.28). 32 ! m $ +∞ +∞ +∞ 2 * m MF11 = ρ # v12 + v22 + v32 ). dv1dv2 dv3 & ∫ ∫ ∫ u1 exp +− ( " 2π kBT % −∞ −∞ −∞ , 2kBT / 3 2 +∞

! m $ = ρ# & " 2 π k BT %

2 1

∫v

−∞

( mv12 + +∞ ( mv22 + +∞ ( mv32 + exp )− , dv1 ∫ exp )− , dv2 ∫ exp )− ,dv3 * 2kBT - −∞ * 2kBT * 2kBT −∞

The first integral is equal to ( 2kBT m )

( 2 π k BT

32

(

)

π 2 while the second two integrals are each equal to

12

m ) . Thus: 32

! m $ ! 2kBT $3 2 π ! 2π kBT $1 2 ! 2π kBT $1 2 kT MF11 = ρ # & # & # & # & = ρ B = ρ RT = p 2 " m % " m % m " 2 π k BT % " m % where kB/m = R from (1.28). This analysis may be repeated with i = j = 2, and i = j = 3 to find: MF22 = MF33 = p, as well. Now consider the case i ≠ j. First note that MFij = MFji because the velocity product under the triple integral may be written in either order vivj = vjvi, so there are only three cases of interest. Start with i = 1, and j = 2 to find: 32 ! m $ +∞ +∞ +∞ * m MF12 = ρ # v12 + v22 + v32 ). dv1dv2 dv3 & ∫ ∫ ∫ v1v2 exp +− ( " 2π kBT % −∞ −∞ −∞ , 2kBT / 32

! m $ +∞ ( mv12 + +∞ ( mv22 + +∞ ( mv32 + = ρ# , dv1 ∫ v2 exp )− , dv2 ∫ exp )− ,dv3 & ∫ v1 exp )− " 2π kBT % −∞ * 2kBT - −∞ * 2kBT * 2kBT −∞ Here we need only consider the first integral. The integrand of this integral is an odd function because it is product of an odd function, v1, and an even function, exp {−mv12 2kBT } . The

integral of an odd function on an even interval [–∞,+∞] is zero, so MF12 = 0. And, this analysis may be repeated for i = 1 and j = 3, and i = 2 and j = 3 to find MF13 = MF23 = 0.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.8. Consider the viscous flow in a channel of width 2b. The channel is aligned in the x-direction, and the velocity u in the x-direction at a distance y from the channel centerline is 2 given by the parabolic distribution u(y) = U 0 1− ( y b) . Calculate the shear stress τ as a

[

]

function y, µ, b, and Uo. What is the shear stress at y = 0? * $ y '2du d y € Solution 1.8. Start from (1.3): τ = µ = µ U o ,1− & ) / = –2µU o 2 . At y = 0 (the location of dy dy + % b ( . b

maximum velocity) τ = 0. At At y = ±b (the locations of zero velocity), τ = 2µ U o b . €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.9. Hydroplaning occurs on wet roadways when sudden braking causes a moving vehicle’s tires to stop turning when the tires are separated from the road surface by a thin film of water. When hydroplaning occurs the vehicle may slide a significant distance before the film breaks down and the tires again contact the road. For simplicity, consider a hypothetical version of this scenario where the water film is somehow maintained until the vehicle comes to rest. a) Develop a formula for the friction force delivered to a vehicle of mass M and tire-contact area A that is moving at speed u on a water film with constant thickness h and viscosity µ. b) Using Newton’s second law, derive a formula for the hypothetical sliding distance D traveled by a vehicle that started hydroplaning at speed Uo c) Evaluate this hypothetical distance for M = 1200 kg, A = 0.1 m2, Uo = 20 m/s, h = 0.1 mm, and µ = 0.001 kgm–1s–1. Compare this to the dry-pavement stopping distance assuming a tire-road coefficient of kinetic friction of 0.8. Solution 1.9. a) Assume that viscous friction from the water layer transmitted to the tires is the only force on the sliding vehicle. Here viscous shear stress at any time will be µu(t)/h, where u(t) is the vehicle's speed. Thus, the friction force will be Aµu(t)/h. du u b) The friction force will oppose the motion so Newton’s second law implies: M = −Aµ . dt h This equation is readily integrated to find an exponential solution: u(t) = Uo exp (−Aµt Mh ) , where the initial condition, u(0) = Uo, has been used to evaluate the constant of integration. The distance traveled at time t can be found from integrating the velocity: t t x(t) = ∫ u(t!)dt! = Uo ∫ exp (−Aµt! Mh ) dt! = (Uo Mh Aµ )$%1− exp (−Aµt Mh )&' . o

o

The total sliding distance occurs for large times where the exponential term will be negligible so: D = Uo Mh Aµ 2 c) For M = 1200 kg, A = 0.1 m , Uo = 20 m/s, h = 0.1 mm, and µ = 0.001 kgm–1s–1, the stopping distance is: D = (20)(1200)(10–4)/(0.1)(0.001) = 24 km! This is an impressively long distance and highlights the dangers of driving quickly on water covered roads. For comparison, the friction force on dry pavement will be –0.8Mg, which leads to a vehicle velocity of: u(t) = Uo − 0.8gt , and a distance traveled of x(t) = Uot − 0.4gt 2 . The vehicle stops when u = 0, and this occurs at t = Uo/(0.8g), so the stopping distance is 2 ! Uo $ ! Uo $ Uo2 , D = Uo # − 0.4g = & # & " 0.8g % " 0.8g % 1.6g which is equal to 25.5 m for the conditions given. (This is nearly three orders of magnitude less than the estimated stopping distance for hydroplaning.)

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.10. Estimate the height to which water at 20°C will rise in a capillary glass tube 3 mm in diameter that is exposed to the atmosphere. For water in contact with glass the contact angle is nearly 0°. At 20°C, the surface tension of a water-air interface is σ = 0.073 N/m. Solution 1.10. Start from the result of Example 1.4. 2σ cos α 2(0.073N / m)cos(0°) h= = = 9.92mm 3 ρ gR (10 kg / m 3 )(9.81m / s 2 )(1.5 ×10 −3 m)

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.11. A manometer is a U-shaped tube containing mercury of density ρm. Manometers are used as pressure-measuring devices. If the fluid in tank A has a pressure p and density ρ, then show that the gauge pressure in the tank is: p − patm = ρmgh − ρga. Note that the last term on the right side is negligible if ρ « ρm. (Hint: Equate the pressures at X and Y.)

Solution 1.9. Start by equating the pressures at X and Y. pX = p + ρga = patm + ρmgh = pY. Rearrange to find: p – patm = ρmgh – ρga.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.12. Prove that if e(T, υ) = e(T) only and if h(T, p) = h(T) only, then the (thermal) equation of state is (1.28) or pυ = kT, where k is constant.

€

Solution 1.12. Start with the first equation of (1.24): de = Tds – pdυ, and rearrange it: $ ∂s ' $ ∂s ' 1 p ds = de + dυ = & ) de + & ) dυ , % ∂e (υ % ∂υ ( e T T where the second equality holds assuming the entropy depends on e and υ. Here we see that: 1 # ∂s & p $ ∂s ' = % ( , and = & ) . T $ ∂e 'υ T % ∂υ ( e € $ ∂ $ ∂s ' ' $ ∂ $ ∂ s ' ' Equality of the crossed second derivatives of s, & & ) ) = & & ) ) , implies: % ∂υ % ∂e (υ ( e % ∂e % ∂υ ( e (υ $ ∂ (1€T ) ' $ ∂ ( p T ) ' € & ) =& ) . % ∂υ ( e % ∂e (υ $ ∂ (1 T ) ' $ ∂ (1 T ) ' € However, if e depends only on T, then (∂/∂υ)e = (∂/∂υ)T, thus & ) =& ) = 0 , so % ∂υ ( e % ∂υ (T # ∂( p T) & € % ( = 0 , which can be integrated to find: p/T = f1(υ), where f1 is an undetermined function. $ ∂e 'υ Now repeat this procedure using the second equation € of (1.24), dh = Tds + υdp. % ∂s ( % ∂s ( 1 υ ds = dh − dp = ' * dh + ' * dp . & ∂h ) p T T & ∂p ) h 1 # ∂s & υ % ∂s ( Here equality of the coefficients of the differentials implies: = % ( , and − = ' * . T $ ∂h ' p T & ∂p ) h # ∂ (1 T ) & # ∂ (υ T ) & € So, equality of the crossed second derivatives implies: % ( = −% ( . Yet, if h depends $ ∂p ' h $ ∂h ' p # ∂ (1 T ) & €# ∂ (1 T ) & € % ∂ (υ T ) ( only on T, then (∂/∂p)h = (∂/∂p)T, thus % ( =% ( = 0 , so −' * = 0 , which can $ ∂p ' h $ ∂p 'T & ∂h ) p € undetermined function. be integrated to find: υ/T = f2(p), where f2 is an Collecting the two results involving f1 and f2, and solving for T produces: p υ € =T = or pf 2 ( p) =€υf1 (υ ) = k , f1 (υ ) f 2 ( p) where k must be is a constant since p and υ are independent thermodynamic variables. Eliminating f1 or f2 from either equation on the left, produces pυ = kT. And finally, using both versions € of (1.24) we can write: dh – de = υdp + pdυ = d(pυ). € When e and h only depend on T, then dh = cpdT and de = cvdT, so dh – de = (cp – cv)dT = d(pυ) = kdT , thus k = cp – cv = R, where R is the gas constant. Thus, the final result is the perfect gas law: p = kT/υ = ρRT.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.13. Starting from the property relationships (1.24) prove (1.31) and (1.32) for a reversible adiabatic process involving a perfect gas when the specific heats cp and cv are constant. Solution 1.13. For an isentropic process: de = Tds – pdυ = –pdυ, and dh = Tds + υdp = +υdp. Equations (1.31) and (1.32) apply to a perfect gas so the definition of the specific heat capacities (1.20), and (1.21) for a perfect gas, dh = cpdT, and de = cvdT , can be used to form the ratio dh/de: dυ dρ dp dh cp dT cp υ dp =γ = . or −γ = = =γ =− υ ρ p de cv dT cv pdυ The final equality integrates to: ln(p) = γln(ρ) + const which can be exponentiated to find: p = const.ργ, which is (1.31). The constant may be evaluated € at a reference condition po and ρo to find: γ p po = ( ρ ρ o ) and this may be inverted to put the density ratio on the left 1γ

€

ρ ρ o = ( p po ) , which is the second equation of (1.32). The remaining relationship involving the temperature is found by using the perfect gas law, p = ρRT, to eliminate ρ = p/RT: 1γ −1 γ " p %(γ −1) γ ρ p RT € pTo # p & T p" p% , = = = % ( or = $ ' =$ ' ρ o po RTo poT $ po ' To po # po & # po & which is the first equation of (1.32). €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.14. A cylinder contains 2 kg of air at 50°C and a pressure of 3 bars. The air is compressed until its pressure rises to 8 bars. What is the initial volume? Find the final volume for both isothermal compression and isentropic compression. Solution 1.14. Use the perfect gas law but explicitly separate the mass M of the air and the volume V it occupies via the substitution ρ = M/V: p = ρRT = (M/V)RT. Solve for V at the initial time: Vi = initial volume = MRT/pi = (2 kg)(287 m2/s2K)(273 + 50°)/(300 kPa) = 0.618 m3. For an isothermal process: Vf = final volume = MRT/pf = (2 kg)(287 m2/s2K)(273 + 50°)/(800 kPa) = 0.232 m3. For an isentropic process: 1γ

1 1.4

V f = Vi ( pi p f ) = 0.618m 3 (300kPa 800kPa )

= 0.307m 3 .

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.15. Derive (1.35) starting from Figure 1.9 and the discussion at the beginning of Section 1.10.

€

Solution 1.15. Take the z axis vertical, and consider a small fluid element δm of fluid having volume δV that starts at height z0 in a stratified fluid medium having a vertical density profile = ρ(z), and a vertical pressure profile p(z). Without any vertical displacement, the small mass and its volume are related by δm = ρ(z0)δV. If the small mass is displaced vertically a small distance ζ via an isentropic process, its density will change isentropically according to: ρ a (z0 + ζ ) = ρ (z0 ) + ( dρ a dz)ζ + ... where dρa/dz is the isentropic density gradient at z0. For a constant δm, the volume of the fluid element will be: ) δm δm δm & 1 dρ a δV =€ = = ζ + ...+ (1− ρ a ρ (z0 ) + ( dρ a dz)ζ + ... ρ(z0 ) ' ρ(z0 ) dz * The background density at z0 + ζ is: ρ(z0 + ζ ) = ρ (z0 ) + ( dρ dz)ζ + ... If g is the acceleration of gravity, the (upward) buoyant force on the element at the vertically € displaced location will be gρ(z0 + ζ)δV, while the (downward) weight of the fluid element at any vertical location is gδm. Thus, a vertical application Newton's second law implies: € ) d 2ζ δm & 1 dρ a δm 2 = +gρ (z0 + ζ )δV − gδm = g( ρ(z0 ) + ( dρ dz)ζ + ...) ζ + ...+ − gδm , (1− dt ρ(z0 ) ' ρ(z0 ) dz * where the second equality follows from substituting for ρ(z0 + ζ) and δV from the above equations. Multiplying out the terms in (,)-parentheses and dropping second order terms produces: d 2ζ gδm dρ gδm dρ a gδm ' dρ dρ a * δm 2 = gδm + ζ− ζ + ...− gδm ≅ ) − ,ζ dt ρ (z0 ) dz ρ(z0 ) dz ρ (z0 ) ( dz dz + Dividing by δm and moving all the terms to the right side of the equation produces: d 2ζ g % dρ dρ a ( − ' − *ζ = 0 2 & ) dt ρ (z ) dz dz 0 € Thus, for oscillatory motion at frequency N, we must have g $ dρ dρ a ' N2 = − & − ), ρ(z0 ) % dz dz ( € which is (1.35).

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.16. Starting with the hydrostatic pressure law (1.14), prove (1.36) without using perfect gas relationships. Solution 1.16. The adiabatic temperature gradient dTa/dz, can be written terms of the pressure gradient: # ∂T & dTa # ∂T & dp =% ( = −gρ% ( dz $ ∂p ' s dz $ ∂p ' s where the hydrostatic law dp/dz = –ρg has been used to reach the second equality. Here, the final partial derivative can be exchanged for one involving υ = 1/ρ and s, by considering: # ∂h & # ∂h & € dh = % ( ds + % ( dp = Tds + υdp . $ ∂s ' p $ ∂p ' s # ∂ # ∂h & & # ∂ # ∂ h & & Equality of the crossed second derivatives of h, % % ( ( = % % ( ( , implies: $ ∂p $ ∂s ' p ' s $ ∂s $ ∂ p ' s ' p € # ∂T & # ∂υ & # ∂υ & # ∂T & # ∂υ & # ∂s & % ( =% ( =% ( % ( =% ( % ( , $ ∂p ' s $ ∂ s ' p $ ∂T ' p $ ∂ s ' p $ ∂T ' p $ ∂T ' p where the second two equalities are mathematical manipulations that allow the introduction of € & ∂υ ) ! ∂h $ ! ∂s $ 1 & ∂ρ ) α = − ( + = ρ( + , and cp = # & = T # & . ' ∂T * p " ∂T % p " ∂T % p ρ ' ∂T * p € Thus, " ∂T % "c % " ∂υ % " ∂ s % dTa gαT . = −gρ $ ' = −gρ $ ' $ ' = −gα $ p ' = − # ∂T & p # ∂T & p dz cp #T & # ∂ p &s €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.17. Assume that the temperature of the atmosphere varies with height z as T = T0 + g KR ! T0 $ Kz where K is a constant. Show that the pressure varies with height as p = p0 # where & " T0 + Kz % g is the acceleration of gravity and R is the gas constant for the atmospheric gas. Solution 1.17. Start with the hydrostatic and perfect gas laws, dp/dz = –ρg, and p = ρRT, eliminate the density, and substitute in the given temperature profile to find: dp p p dp g dz = −ρg = − g=− g or =− . dz RT R(T0 + Kz) p R (T0 + Kz) The final form may be integrated to find: g ln p = − ln(T0 + Kz) + const. RK € At z = 0, € the pressure must be p0, therefore: g ln p0 = − ln(T0 ) + const. RK € equation above and invoking the properties of logarithms produces: Subtracting this from the " p% g " T0 + Kz % ln$ ' = − ln$ ' RK # T0 & # p0 & € Exponentiating produces: −g/KR g/KR " T0 % p "T0 + Kz % , which is the same as: p = p0 $ =$ ' ' . p0 # €T0 & # T0 + Kz &

€

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.18. Suppose the atmospheric temperature varies according to: T = 15 − 0.001z, where T is in degrees Celsius and height z is in meters. Is this atmosphere stable? Solution 1.18. Compute the temperature gradient: dT d °C °C = (15 − 0.001z) = −0.001 = −1.0 . dz dz m km For air in the earth's gravitational field, the adiabatic temperature gradient is: dTa gαT (9.81m / s 2 )(1 / T )T °C . =− = = −9.8 2 2 dz cp 1004m / s °C km € Thus, the given temperature profile is stable because the magnitude of its gradient is less than the magnitude of the adiabatic temperature gradient.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.19. A hemispherical bowl with inner radius r containing a liquid with density ρ is inverted on a smooth flat surface. Gravity with acceleration g acts downward. Determine the weight W of the bowl necessary to prevent the liquid from escaping. Consider two cases: a) the pressure around the rim of the bowl where it meets the plate is atmospheric, and b) the pressure at the highest point of the bowl’s interior is atmospheric. c) Investigate which case applies via a simple experiment. Completely fill an ordinary soup bowl with water and concentrically cover it with an ordinary dinner plate. While holding the bowl and plate together, quickly invert the water-bowl-plate combination, set it on the level surface at the bottom of a kitchen sink, and let go. Does the water escape? If no water escapes after release, hold onto the bowl only and try to lift the water-bowl-plate combination a few centimeters off the bottom of the sink. Does the plate remain in contact with the bowl? Do your answers to the first two parts of this problem help explain your observations? Solution 1.19. a) When the pressure around the inverted bowl's rim is atmospheric, the surface does not support the weight of the liquid and any increase in elevation above the rim must produce a lower pressure in the liquid. In this case the weight of the liquid is supported by a reduced pressure on the interior of the bowl. Thus the vertical force Fz that the bowl applies to the surface is negative and includes its own weight, W, and the weight of the liquid ( 2 3) π r 3ρ g : Fz = −W − ( 2 3) π r 3ρ g . So, for Fz to reach zero, the point of bowl lift-off, W must be negative! This implies that the bowl will adhere to the surface and will not separate unless it's pulled upward. This finding concerning the pressure force on the bowl can be reached via a more traditional hydrostatic calculation using spherical coordinates (see Appendix B.5). The net vertical pressure force on the inverted bowl will be: π 2

∫

Fz, pressure = 2π

π 2

( p − po ) ( e r ⋅ e z ) r 2 sin θ dθ = 2π r 2

θ =0

∫ ( p − p )cosθ sinθ dθ , o

θ =0

where p is the pressure inside the bowl and po is the pressure outside the bowl (atmospheric pressure). Here θ is the polar angle (θ = 0 defines the positive z-axis). In this coordinate system z = rcosθ, and the pressure inside and outside the bowl must match when z = 0, so p – po = –ρgz = –ρgrcosθ, so π 2 2π 3 3 Fz, pressure = −ρ g2π r ∫ cos2 θ sin θ dθ = − r ρg . 3 θ =0 So, the bowl's "weight" must be less than − ( 2 3) π r 3ρ g to keep the liquid trapped. b) In this case the pressure force on the bowl is different because atmospheric pressure is reached at the inside top of the bowl. In this case, p – po = –ρg(z – r), and the vertical pressure force calculation looks like: π 2

Fz, pressure = 2π r

2

π 2 3

∫ ( p − p )cosθ sinθ dθ = −ρ g2π r ∫ (cosθ −1)cosθ sinθ dθ o

θ =0

θ =0

1 = ρ gπ r 3. 3 So, the static force balance for the bowl is: Fz = −W + (1 3) π r 3ρ g . Thus, the bowl's weight must

be more than (1 3) π r 3ρ g to keep the liquid trapped.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

c) Interestingly, even though a little bit of water may escape from underneath the inverted bowl, nearly all the water remains trapped under the bowl when the bowl-plate combination is released, and the inverted bowl does adhere to the plate when it is lifted. These observations closely match the part a) answer, so it is the physically meaningful one.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.20. Consider the case of a pure gas planet where the hydrostatic law is:

dp dz = −ρ (z)Gm(z) z 2 , where G is the gravitational constant and m(z) = 4π

∫

z o

ρ (ζ )ζ 2 dζ is the

planetary mass up to distance z from the center of the planet. If the planetary gas is perfect with gas constant R, determine ρ(z) and p(z) if this atmosphere is isothermal at temperature T. Are these vertical profiles of ρ and p valid as z increases without bound? Solution 1.20. Start with the given relationship for m(z), differentiate it with respect to z, and use the perfect gas law, p = ρRT to replace the ρ with p. ) dm d & z p(z) = ( 4 π ∫ ρ (ζ )ζ 2 dζ + = 4 πz 2 ρ(z) = 4 πz 2 . dz dz ' 0 RT * Now use this and the hydrostatic law to obtain a differential equation for m(z), dp Gm(z) d # RT dm & # 1 dm & Gm(z) = −ρ (z) 2 → % . ( = −% ( $ 4π z 2 dz ' $ 4π z 2 dz ' z 2 dz z dz € After recognizing T as a constant, the nonlinear second-order differential equation for m(z) simplifies to: RT d " 1 dm % 1 dm . $ 2 '=− 4 m G dz # z dz & z dz This equation can be solved by assuming a power law: m(z) = Azn. When substituted in, this trial solution produces: RT d −2 RT z Anz n−1 ) = (n − 3) Anz n−4 = −z−4 A 2 nz 2n−1. ( € G dz G Matching exponents of z across the last equality produces: n – 4 = 2n – 5, and this requires n = 1. For this value of n, the remainder of the equation is: RT RT (−2) Az−3 = −z−4 A 2 z1 , which reduces to: A = 2 . € G G Thus, we have m(z) = 2RTz/G, and this leads to: 2RT 1 2R 2T 2 1 ρ(z) = , and p(z) = . 2 G 4 πz 2 € € G 4 πz Unfortunately, these profiles are not valid as z increases without bound, because this leads to an unbounded planetary mass. €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.21. Consider a gas atmosphere with pressure distribution p(z) = p0 (1− (2 / π )tan −1 (z / H )) where z is the vertical coordinate and H is a constant length scale. a) Determine the vertical profile of ρ from (1.14) b) Determine N2 from (1.35) as function of vertical distance, z. c) Near the ground where z > H. Specify the value of z/H above which this atmosphere is stable. Solution 1.21. a) Start from (1.14): dp 2 1 1 2 po 1 = −ρ g = − po so ρ (z) = . 2 dz π 1+ (z / H ) H π gH 1+ (z / H )2 b) First determine dρ/dz: " 2z H 2 % dρ 2 po 1 2z . =− = − ρ $ 2' dz π gH (1+ (z / H )2 )2 H 2 # 1+ (z / H ) & Determine the local adiabatic density gradient by expanding (1.31) for small vertical displacement ζ that leads to an adiabatic density change: γ γ ! ζ d ρa $ p(z) + ζ ( dp dz ) +... ! ρ (z) + ζ ( d ρ a dz ) +... $ ζ dp =# +... = #1+ +... & . & or 1+ p(z) ρ (z) p dz " ρ dz % " % For small ζ, expand the final term and cancel the leading 1's, to find: 1 dp γ d ρ a ρg d ρa ρ 2g , thus: . = =− =− p dz ρ dz p dz γp where the second equality follows from (1.14). Now use (1.35) g " d ρ d ρa % g " " 2z H 2 % ρ 2 g % " 2z H 2 ρg % N2 = − $ − + = g − ' ' = − $ −ρ $ ' $ ' ρ # dz dz & ρ # # 1+ (z / H )2 & γ p & # 1+ (z / H )2 γ p &

" % 2z H 2 2 1 ' = g$ − . $ 1+ (z / H )2 γπ H (1− (2 / π )tan −1 (z / H )) 1+ (z / H )2 ' # & " % 2g H $ z 1 ' = − 1+ (z / H )2 $# H γπ (1− (2 / π )tan −1 (z / H )) '& c) In the limit as z H → 0 , the terms inside the big parentheses go to –1/γπ, which is negative, so N2 is negative, and this indicates an unstable atmosphere. In the limit as z H → ∞ , both terms inside the big parentheses become large, but the first one is larger so N2 is positive, and this indicates a stable atmosphere. The boundary between the two regimes occurs when the terms inside the big parentheses sum to zero. This condition leads to an implicit equation for z/H: % z 1 z" −1 " z % − = 0 , or γ π − 2 tan $ ' $ ' =1, # H && H γπ (1− (2 / π )tan −1 (z / H )) H# which has numerical solution z/H = 0.274 when γ = 1.4.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.22. Consider a heat-insulated enclosure that is separated into two compartments of volumes V1 and V2, containing perfect gases with pressures and temperatures of p1 and p2, and T1 and T2, respectively. The compartments are separated by an impermeable membrane that conducts heat (but not mass). Calculate the final steady-state temperature assuming each gas has constant specific heats. Solution 1.22. Since no work is done and no heat is transferred out of the enclosure, the final energy Ef is the sum of the energies, E1 and E2, in the two compartments. E1 + E2 = Ef implies ρ1V1cv1T1 + ρ2V2cv2T2 = (ρ1V1cv1 + ρ2V2cv2)Tf, where the cv's are the specific heats at constant volume for the two gases. The perfect gas law can be used to find the densities: ρ1 = p1/R1T1 and ρ2 = p2/R2T2, so p1V1cv1/R1 + p2V2cv2/R2 = (p1V1cv1/R1T1 + p2V2cv2/R2T2)Tf. A little more simplification is possible, cv1/R1 = 1/(γ1 – 1) and cv2/R1 = 1/(γ2 – 1). Thus, the final temperature is: p1V1 (γ1 −1) + p2V2 (γ 2 −1) . Tf = p1V1 [(γ1 −1)T1 ] + p2V2 [(γ 2 −1)T2 ]

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Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.23. Consider the initial state of an enclosure with two compartments as described in Exercise 1.22. At t = 0, the membrane is broken and the gases are mixed. Calculate the final temperature.

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Solution 1.23. No heat is transferred out of the enclosure and the work done by either gas is delivered to the other so the total energy is unchanged. First consider the energy of either gas at temperature T, and pressure P in a container of volume V. The energy E of this gas will be: E = ρVcvT = (p/RT)VcvT = pV(cv/R) = pV/(γ – 1). where γ is the ratio of specific heats. For the problem at hand the final energy Ef will be the sum of the gas energies, E1 and E2, in the two compartments. Using the above formula: E f = p1V1 (γ1 −1) + p2V2 (γ 2 −1) . Now consider the mixture. The final volume and temperature for both gases is V1+V2, and Tf. However, from Dalton's law of partial pressures, the final pressure of the mixture pf can be considered a sum of the final partial pressures of gases "1" and "2", p1f and p2f: € pf = p1f + p2f. Thus, the final energy of the mixture is a sum involving each gases' partial pressure and the total volume: E f = p1 f (V1 + V2 ) (γ1 −1) + p2 f (V1 + V2 ) (γ 2 −1) . However, the perfect gas law implies: p1f(V1+V2) = n1RuTf, and p2f(V1+V2) = n2RuTf where n1 and n2 are the mole numbers of gases "1" and "2", and Ru is the universal gas constant. The mole numbers are obtained from: € n1 = p1V1/RuT1, and n2 = p2V2/RuT2, Thus, final energy determined from the mixture is: $pV ' T nRT n R T $ pV ' R T $ p V ' R T $ pV ' T Ef = 1 u f + 2 u f =& 1 1) u f +& 2 2 ) u f =& 1 1) f +& 2 2 ) f . γ1 −1 γ1 −1 % RuT1 ( γ1 −1 % RuT2 ( γ 2 −1 % T1 ( γ1 −1 % T2 ( γ 2 −1 Equating this and the first relationship for Ef above then produces: p1V1 (γ1 −1) + p2V2 (γ 2 −1) . Tf = p1V1 [(γ1 −1)T1 ] + p2V2 [(γ 2 −1)T2 ]

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Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.24. A heavy piston of weight W is dropped onto a thermally insulated cylinder of cross-sectional area A containing a perfect gas of constant specific heats, and initially having the external pressure p1, temperature T1, and volume V1. After some oscillations, the piston reaches an equilibrium position L meters below the equilibrium position of a weightless piston. Find L. Is there an entropy increase? Solution 1.24. From the first law of thermodynamics, with Q = 0, ΔE = Work = (W + p1A)L. For a total mass m of a perfect gas with constant specific heats, E = mcvT so ΔE = E2 – E1 = mcv(T2 – T1) = (W + p1A)L. Then T2 = T1 + (W + p1A)L/mcv. Also, for a perfect gas, PV/T = constant so p1V1/T1 = p2V2/T2. For the cylinder, V2 = V1 – AL, and p2 = p1 + W/A. Therefore: p1V1 ( p1 + W A)(V1 − AL) . = T1 T1 + (W + p1 A)L mcv Here, m = p1V1/RT1 so mcv = (p1V1/T1)(cv/R) = (p1V1/T1)(γ – 1)–1, and this leaves: p1V1 ( p1 + W A)(V1 − AL) ( p1 + W A)(V1 − AL) , or p1V1 = , = T1 T1 + (γ −1)(W + p1 A)LT1 p1V1 1+ (γ −1)(W + p1 A)L p1V1 after multiplication on both sides by T1. Solve for L. p1V1 + (γ −1)(W + p1 A)L = p1V1 + WV1 A − p1 AL − WL , W WV1 A . L [(γ −1)(W + p1 A)L + p1 A + W ] = V1 , or L = A γ (W + p1 A) The initial length Lo of the column of gas is V1/A, so this final answer can be written: L W p1 A . = Lo γ (1+ W p1 A) For an isentropic process, pVγ = constant, so for a constant cross section cylinder: 1

" %γ L 1 p1Lγo = ( p1 + W A) (Lo − L)γ , or = 1− $ ' . Lo # 1+ W p1 A & This isentropic compression distance is always greater than the "sudden compression" distance determined above. Therefore, the sudden compression does lead to an increase in entropy.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.25. Starting from 295 K and atmospheric pressure, what is the final pressure of an isentropic compression of air that raises the temperature 1, 10, and 100 K. Solution 1.25. Start from the first equation of (1.32), and solve for the pressure: γ −1

γ

7

! T $γ −1 ! 295 + ΔT $ 2 T ! p$γ = # & , or p = po # & = (101.3kPa) # & , " 295 % To " po % " To % where the final equality applies for the values specified in the problem statement when γ = 1.4. For the three given values of ΔT, the final pressures are: 102.5 kPa, 113.8 kPa, and 281.4 kPa.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.26. Compute the speed of sound in air at –40°C (very cold winter temperature), at +45°C (very hot summer temperature), at 400°C (automobile exhaust temperature), and 2000°C (nominal hydrocarbon adiabatic flame temperature) Solution 1.26. The speed of sound c of a perfect gas is given by (1.33): c = γ RT , or c = γ R[273+ (T in °C)] . Presuming that the R = 287 m2s–2K–1 applies at the four temperatures given, the four sound speeds are: 306 ms–1, 357 ms–1, 520 ms–1, and 956 ms–1, respectively.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.27. The oscillation frequency Ω of a simple pendulum depends on the acceleration of gravity g, and the length L of the pendulum. a) Using dimensional analysis, determine single dimensionless group involving Ω, g and L. b) Perform an experiment to see if the dimensionless group is constant. Using a piece of string slightly longer than 2 m and any small heavy object, attach the object to one end of the piece of string with tape or a knot. Mark distances of 0.25, 0.5, 1.0 and 2.0 m on the string from the center of gravity of the object. Hold the string at the marked locations, stand in front of a clock with a second hand or second readout, and count the number (N) of pendulum oscillations in 20 seconds to determine Ω = N/(20 s) in Hz. Evaluate the dimensionless group for these four lengths. c) Based only on the results of parts a) and b), what pendulum frequency do you predict when L = 1.0 m but g is 16.6 m/s? How confident should you be of this prediction? Solution 1.27. a) Construct the parameter & units matrix using Ω, g, and L.

M L T

Ω 0 0 -1

g 0 1 -2

L 0 1 0

This rank of this matrix is two, so there is one dimensionless group that is readily found by inspection to be Ω(L/g)1/2. b) The following table lists the results of the simple pendulum experiments: L (m) 0.25 0.50 1.00 2.00

N 20 14.5 10 7.0

Ω (s–1) 1.00 0.725 0.500 0.350

Ω(L/g)1/2 0.160 0.164 0.160 0.158

Here g = 9.81 ms–2 has been used to evaluate the final column. These results do suggest that the dimensionless group is constant within the uncertainty of these simple experiments, and that the constant is ~0.1605. (For small angular oscillations in the absence of air resistance, this constant should be 1/2π ≈ 0.159) c) Using the results of part b), Ω(L/g)1/2 = 0.1605 , so Ω = 0.1605 (g/L)1/2, which implies: Ω = 0.1605(16.6/1.0)1/2 = 0.654 s–1 for the specified conditions. Given the results of part b), the confidence in this prediction should be high; it might have ± 1 or 2% error at most.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.28. The spectrum of wind waves, S(ω), on the surface of the deep sea may depend on the wave frequency ω, gravity g, the wind speed U, and the fetch distance F (the distance from the upwind shore over which the wind blows with constant velocity). a) Using dimensional analysis, determine how S(ω) must depend on the other parameters. b) It is observed that the mean-square wave amplitude, η 2 =

∫

∞ 0

S(ω )dω , is proportional to F.

Use this fact to revise the result of part a). c) How must η2 depend on U and g? Solution 1.28. a) Construct the parameter & units matrix using S, ω, g, U, and F. The units of S are (length2)(time).

M L T

S 0 2 1

ω 0 0 -1

g 0 1 -2

U 0 1 -1

F 0 1 0

This rank of this matrix is two, so there are 5 – 2 = 3 dimensionless groups. Three suitable groups are readily found by inspection to be Π1 = SωF–2, Π2 = ωFU–1, and Π3 = U2F–1g–1. Thus, S F 2 "ωF U 2 % must depend on the other parameters as follows: S = Ψ$ , '. ω # U gF & b) Use the given equation and the result of part a): ∞ ∞ ∞ ∞ $ U 2 ' dγ F 2 $ωF U 2 ' F 2 $ωF U 2 ' 2 η 2 = ∫ Sdω = ∫ Ψ& , d ω = Ψ , d ω = F Ψ ) ∫ ω &% U gF )( ∫ &%γ, gF )( γ , ω U gF % ( 0 0 0 0 where the final equality follows from changing the integration variable to γ = ωF/U. From this equation, the only way that η2 can be proportional to F is for the undetermined function be proportional to its second argument: Ψ (γ ,U 2 gF ) = (U 2 gF ) Φ(γ ) . Combine this result with the F 2 "ωF U 2 % F 2 U 2 "ωF % U 2F "ωF % Ψ$ , Φ$ Φ$ '= '. '= ω # U gF & ω gF # U & gω # U & c) Substitute the final answer from part b) back into the integral relationships to find: ∞ U 2F ∞ 1 $ωF ' U 2F ∞ dγ U 2 F 2 η = ∫ Sdω = Φ& Φ (γ ) = ⋅ const. )dω = ∫ ∫ g 0ω %U ( g 0 γ g 0 Thus, η2 must be proportional to U2 and inversely proportional to g.

result of part a): S =

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.29. One military technology for clearing a path through a minefield is to deploy a powerfully exploding cable across the minefield that, when detonated, creates a large trench through which soldiers and vehicles may safely travel. If the expanding cylindrical blast wave from such a line-explosive has radius R at time t after detonation, use dimensional analysis to determine how R and the blast wave speed dR/dt must depend on t, r = air density, and E´ = energy released per unit length of exploding cable. Solution 1.29. Follow example 1.10 and construct the parameter & units matrix using R, t, r, and E´. The units of E´ are energy/length. Construct the parameter & units matrix.

M L T

R 0 1 0

t 0 0 1

ρ 1 -3 0

E´ 1 1 -2

This rank of this matrix is three, so there is 4 – 3 = 1 dimensionless group, and it may be found by inspection: Π1 = E´t2/ρR4. Since there is only one dimensionless group, it must be constant so: R(t) = const.[E´/ρ]1/4t1/2. To find dR/dt, differentiate this with respect to time to find: dR/dt = (const./2)[E´/ρ]1/4t–1/2 = R/(2t).

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.30. One of the triumphs of classical thermodynamics for a simple compressible substance was the identification of entropy s as a state variable along with pressure p, density ρ, and temperature T. Interestingly, this identification foreshadowed the existence of quantum physics because of the requirement that it must be possible to state all physically meaningful laws in dimensionless form. To see this foreshadowing, consider an entropic equation of state for a system of N elements each having mass m. a) Determine which thermodynamic variables amongst s, p, ρ, and T can be made dimensionless using N, m, and the non-quantum mechanical physical constants kB = Boltzmann’s constant and c = speed of light. What do these results imply about an entropic equation of state in any of the following forms: s = s(p,ρ), s = s(ρ,T), or s = s(T,p)? b) Repeat part a) including  = Planck’s constant (the fundamental constant of quantum physics). Can an entropic equation of state be stated in dimensionless form without  ? Solution 1.30. a) Select each thermodynamic variable in turn, and see if it can be made dimensionless using m, kB, and c. Here N is dimensionless and it is the only extensive variable so it need not be considered when seeking the dimensionless forms of the intensive (per unit mass) thermodynamic variables. The dimensions of the remaining parameters and constants are:

M L T θ

s 0 2 -2 -1

p 1 -1 -2 0

ρ 1 -3 0 0

T 0 0 0 1

m 1 0 0 0

kB 1 2 -2 -1

c 0 1 -1 0

Exponent algebra can be used to reach the following results: • entropy s can be made dimensionless via s/mkB; • pressure p cannot be made dimensionless using m, kB, and c; • density ρ cannot be made dimensionless using m, kB, and c; and • temperature can be dimensionless via kBT/mc2. These results imply that an entropic equation of state in any of the forms s = s(p,ρ), s = s(ρ,T), or s = s(T,p) cannot be made dimensionless using non-quantum mechanical physical constants. b) The units of  are (energy)(time) = ML2T–1. With the addition of this parameter, exponent algebra can be used to reach the following results: • entropy s can be made dimensionless via s/mkB; • pressure p can be made dimensionless via p 3 m 4 c 5 ; • density ρ can be made dimensionless via ρ 3 m 4 c 3 ; and • temperature can be dimensionless via kBT/mc2. These results imply that an entropic equation of state in any of the forms s = s(p,ρ), s = s(ρ,T), or s = s(T,p) can be made dimensionless with  .

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.31. The natural variables of the system enthalpy H are the system entropy S and the pressure p, which leads to an equation of state in the form: H = H(S, p, N), where N is the number of system elements. a) After creating ratios of extensive variables, use exponent algebra to independently render H/N, S/N, and p dimensionless using m = the mass of a system element, and the fundamental constants kB = Boltzmann’s constant,  = Planck’s constant, and c = speed of light. b) Simplify the result of part a) for non-relativistic elements by eliminating c. c) Based on the property relationship (1.24), determine the specific volume = υ = 1 ρ = (∂h ∂p)s from the result of part b). d) Use the result of part c) and (1.25) to show that the sound speed in this case is compare this result to that for a monotonic perfect gas.

5p 3ρ , and

Solution 1.31. a) Here H and S are extensive variables so they must be proportional to N. So parameter & units matrix can be constructed as: M L T θ

H/N 1 2 -2 0

S/N 1 2 -2 -1

p 1 -3 0 0

m 0 0 0 1

kB 1 2 -2 -1

 1 2 -1 0

c 0 1 -1 0

This rank of this matrix is four, so there are 7 – 4 = 3 dimensionless groups. Three suitable groups that isolate H/N, p and S/N are readily found by inspection or exponent algebra to be: Π1 = HN–1m–1c–2, Π2 = p3m −4 c −5 , and Π3 = SN −1kB−1 . The following scaling law then applies to H/N: " p3 S % H = Θ $ 4 5, ' Nmc 2 # m c NkB & where Θ is an undetermined function. b) To eliminate c, extract the second dimensionless group from the argument of Θ, raise it to the –2/5 power, and multiply on the left side with the altered group involving H: −2 5 25 ! S $ ! sm $ H ! p3 $ H ! m 4c 5 $ hm8 5 or = Θ = = Θ# & , # & 2 # 4 5& 2 # 3 & 25 65 Nmc " m c % Nmc " p % p  " NkB % " kB % where H/Nm = h, and s = S/Nm has been used for the second-to-last equality. c) Use the final equality of part b), and perform the indicated differentiation: " s % 1 " ∂h % " ∂ ) p 2 5 6 5 " sm %,% 2  6 5 υ = = $ ' = $$ + 8 5 Θ $ '.'' = Θ $ '. 85 35 ρ # ∂p &s # ∂p * m # kB &-&s 5 m p # mkB & 53

 2 ( 2 " s %+ 53 d) Use the final result of part c) get an equation for the pressure, p = 8 3 * Θ $ '- ρ , and m ) 5 # mkB &, 53

" ∂p % 5  2 ) 2 " s %, 5p −2 3 = differentiate as in (1.25) to find: c = $ ' = , which is the + Θ$ '. ρ 83 3 m * 5 # mkB &3ρ # ∂ρ &s correct relationship for the speed of sound in a monotonic perfect gas.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.32. A gas of noninteracting particles of mass m at temperature T has density ρ, and internal energy per unit volume ε. a) Using dimensional analysis, determine how ε must depend on ρ, T, and m. In your formulation use kB = Boltzmann’s constant,  = Plank’s constant, and c = speed of light to include possible quantum and relativistic effects. b) Consider the limit of slow-moving particles without quantum effects by requiring c and  to drop out of your dimensionless formulation. How does ε depend on ρ and T? What type of gas follows this thermodynamic law? c) Consider the limit of massless particles (i.e., photons) by requiring m and ρ to drop out of your dimensionless formulation of part a). How does ε depend on T in this case? What is the name of this radiation law? Solution 1.32. a) Construct the parameter & units matrix noting that kB and T must go together since they are the only parameters that involve temperature units.

M L T

€

ε 1 -1 -2

ρ 1 -3 0

kBT 1 2 -2

m 1 0 0

 1 2 -1

c 0 1 -1

This rank of this matrix is three. There are 6 parameters and 3 independent units, so there will be 3 dimensionless groups. Two of the dimensionless groups are energy ratios that are easy spot: Π1 = ε ρc 2 and Π 2 = kB T mc 2 . There is one dimensionless group left that must contain  . A ! k BT ρ  3 $ ρ 3 ε bit of work produces: Π3 = 4 3 , so = ϕ1 # 2 , 4 3 & . mc ρc 2 " mc m c % €  means dropping Π3. Eliminating c means combining Π1 and Π2 to create a new b) Dropping Π1 ε ρc 2 εm = = dimensionless group that lacks c: . However, now there is only one 2 Π 2 kB T mc ρk B T % ρk T ( dimensionless group so it must be a constant. This implies: ε = const ⋅ ' B * which is the & m ) caloric equation of state for a perfect gas. € c) Eliminating ρ means combining Π1 and Π3 to create a new dimensionless group that lacks ρ: ε ρ 3 ε 3 € dimensionless group with Π2 to eliminate Π1 ⋅ Π3 = 2 ⋅ 4 3 = 4 5 . Now combine this new ρc m c m c 4 ε 3 1 ε 3 # mc 2 & ε  3c 3 m: 4 5 ⋅ 4 = 4 5 ⋅ % . Again there is only a single dimensionless group so it ( = m c Π 2 m c $ k BT ' ( k BT ) 4

must equal a constant; therefore ε =

const 4 ⋅ ( kBT ) . This is the Stephan-Boltzmann radiation law. 3 3 c

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.33. A compression wave in a long gas-filled constant-area duct propagates to the left at speed U. To the left of the wave, the gas is quiescent with uniform density ρ1 and uniform pressure p1. To the right of the wave, the gas has uniform density ρ2 (> ρ1) and uniform pressure is p2 (> p1). Ignore the effects of viscosity in this problem. Formulate a dimensionless scaling law for U in terms of the pressures and densities. U! p1, ρ1!

p2, ρ2!

u1 = 0!

Solution 1.33. a) Construct the parameter & units matrix:

M L T

U 0 1 -1

ρ1 1 -3 0

ρ2 1 -3 0

p1 1 -1 -2

p2 1 -1 -2

This rank of this matrix is just two. There are 5 parameters but just 2 independent units, so there will be 3 dimensionless groups. These are readily found by inspection: Π1 = U ρ1 p1 , Π 2 = ρ1 ρ2 , and Π3 = p1 p2 . Thus, the scaling law is:

U ρ1 p1 = Φ ( ρ1 ρ2 , p1 p2 ) . where Φ is an undetermined function.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.34. Many flying and swimming animals – as well as human-engineered vehicles – rely on some type of repetitive motion for propulsion through air or water. For this problem, assume the average travel speed U, depends on the repetition frequency f, the characteristic length scale of the animal or vehicle L, the acceleration of gravity g, the density of the animal or vehicle ρo, the density of the fluid ρ, and the viscosity of the fluid µ. a) Formulate a dimensionless scaling law for U involving all the other parameters. b) Simplify your answer for a) for turbulent flow where µ is no longer a parameter. c) Fish and animals that swim at or near a water surface generate waves that move and propagate because of gravity, so g clearly plays a role in determining U. However, if fluctuations in the propulsive thrust are small, then f may not be important. Thus, eliminate f from your answer for b) while retaining L, and determine how U depends on L. Are successful competitive human swimmers likely to be shorter or taller than the average person? d) When the propulsive fluctuations of a surface swimmer are large, the characteristic length scale may be U/f instead of L. Therefore, drop L from your answer for b). In this case, will higher speeds be achieved at lower or higher frequencies? e) While traveling submerged, fish, marine mammals, and submarines are usually neutrally buoyant (ρo ≈ ρ) or very nearly so. Thus, simplify your answer for b) so that g drops out. For this situation, how does the speed U depend on the repetition frequency f? f) Although fully submerged, aircraft and birds are far from neutrally buoyant in air, so their travel speed is predominately set by balancing lift and weight. Ignoring frequency and viscosity, use the remaining parameters to construct dimensionally accurate surrogates for lift and weight to determine how U depends on ρo/ρ, L, and g. Solution 1.34. a) Construct the parameter & units matrix

M L T

U 0 1 -1

f 0 0 -1

L 0 1 0

g 0 1 -2

ρo 1 -3 0

ρ 1 -3 0

µ 1 -1 -1

The rank of this matrix is three. There are 7 parameters and 3 independent units, so there will be 4 dimensionless groups. First try to assemble traditional dimensionless groups, but its best to use the solution parameter U only once. Here U is used in the Froude number, so its dimensional counter part, gL , is used in place of U in the Reynolds number.

ρ gL3 U = Froude number, Π 2 = = a Reynolds number µ gL The€next two groups can be found by inspection: ρ f Π 3 = o = a density ratio , and the final group must include f: Π 4 = , and is a frequency ρ g L € ratio between f and that of simple pendulum with length L. Putting these together produces: $ ρ gL3 ρ U f ' o & ) where, throughout this problem solution, ψi , i = 1, 2, 3, … are = ψ1& , , ) µ ρ gL g L € % ( unknown functions. Π1 =

€

€

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

b) When µ is no longer a parameter, the Reynolds number drops out:

$ρ U f ' = ψ 2 && o , )) . gL % ρ g L(

c) When f is no longer a parameter, then U = gL ⋅ ψ 3 ( ρ o ρ ) , so that U is proportional to L . This scaling suggests that taller swimmers have an advantage over shorter ones. [Human swimmers best approach the necessary conditions for this part € of this problem while doing freestyle (crawl) or backstroke where the arms (and legs) are used for propulsion in an € € also applies to alternating (instead of simultaneous) fashion. Interestingly, this length advantage ships and sailboats. Aircraft carriers are the longest and fastest (non-planing) ships in any Navy, and historically the longer boat typically won the America’s Cup races under the 12-meter rule. Thus, if you bet on a swimming or sailing race where the competitors aren’t known to you but appear to be evenly matched, choose the taller swimmer or the longer boat.] d) Dropping L from the answer for b) requires the creation of a new dimensionless group from f, g, and U to replace Π1 and Π4. The new group can be obtained via a product of original $ρ ' Uf g $ρ ' U f Uf = ψ 4 & o ) , or U = ψ 4 & o ) . Here, dimensionless groups: Π1Π 4 = . Thus, = g f %ρ( g gL g L %ρ( U is inversely proportional to f which suggests that higher speeds should be obtained at lower frequencies. [Human swimmers of butterfly (and breaststroke to a lesser degree) approach the conditions required for this part of this problem. Fewer longer strokes are typically preferred € ones. Of course, the trick for € € is to properly lengthen each over many short reaching top speed stroke without losing propulsive force]. e) When g is no longer a parameter, a new dimensionless group that lacks g must be made to U gL Π U replace Π1 and Π5. This new dimensionless group is 1 = , so the overall scaling = Π 5 f g L fL %ρ ( law must be: U = fL ⋅ ψ 5 ' o * . Thus, U will be directly proportional to f. Simple observations of &ρ) swimming fish, dolphins, whales, etc. verify€that their tail oscillation frequency increases at higher swimming speeds, as does the rotation speed of a submarine or torpedo’s propeller. f) Dimensionally-accurate surrogates for weight and lift are: ρ o L3 g and ρU 2 L2 , respectively. Set € these proportional to each other, ρ o L3 g ∝ ρU 2 L2 , to find U ∝ ρ o gL ρ , which implies that larger denser flying objects must fly faster. This result is certainly reasonable when comparing similarly shaped aircraft (or birds) of different sizes. € €

€

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.35. The acoustic power W generated by a large industrial blower depends on its volume flow rate Q, the pressure rise ΔP it works against, the air density ρ, and the speed of sound c. If hired as an acoustic consultant to quiet this blower by changing its operating conditions, what is your first suggestion? Solution 1.35. The boundary condition and material parameters are: Q, ρ, ΔP, and c. The solution parameter is W. Create the parameter matrix:

Mass: Length: Time:

W Q ΔP ρ c –––––––––––––––––––––––––––– 1 0 1 1 0 2 3 -1 -3 1 -3 -1 -2 0 -1

This rank of this matrix is three. Next, determine the number of dimensionless groups: 5 parameters - 3 dimensions = 2 groups. Construct the dimensionless groups: ∏1 = W/QΔP, ∏2 = ΔP/ρc2. Now write the dimensionless law: W = QΔPΦ(ΔP/ρc2), where Φ is an unknown function. Since the sound power W must be proportional to volume flow rate Q, you can immediately suggest a decrease in Q as means of lowering W. At this point you do not know if Q must be maintained at high level, so this solution may be viable even though it may oppose many of the usual reasons for using a blower. Note that since Φ is unknown the dependence of W on ΔP cannot be determined from dimensional analysis alone.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.36. The horizontal displacement Δ of the trajectory of a spinning ball depends on the mass m and diameter d of the ball, the air density ρ and viscosity µ, the ball's rotation rate ω, the ball’s speed U, and the distance L traveled. a) Use dimensional analysis to predict how Δ can depend on the other parameters. b) Simplify your result from part a) for negligible viscous forces. c) It is experimentally observed that Δ for a spinning sphere becomes essentially independent of the rotation rate once the surface rotation speed, ωd/2, exceeds twice U. Simplify your result from part b) for this high-spin regime. d) Based on the result in part c), how does Δ depend on U?

Top view!

spinning ball trajectory ! Δ"

U!

L!

Solution 1.36. a) Create the parameter matrix using the solution parameter is Δ, and the boundary condition and material parameters are: Q, ρ, ΔP, and c.

Mass: Length: Time:

Δ 0 1 0

m 1 0 0

d 0 1 0

ρ 1 -3 0

µ 1 -1 -1

ω 0 0 -1

U 0 1 -1

L 0 1 0

This rank of this matrix is three. Next, determine the number of dimensionless groups: 8 parameters - 3 dimensions = 5 groups. Construct the dimensionless groups: Π1 = Δ/d, Π2 = m/ρd3, Π3 = ρUd/µ, Π4 = ωd/U, and Π5 = L/d. Thus, the dimensionless law for Δ is: # m ρUd ω d L & Δ = Φ% 3 , , , (, d µ U d' $ ρd where Φ is an undetermined function. b) When the viscosity is no longer a parameter, then the third dimensionless group (the Reynolds number) must drop out, so the part a) result simplifies to: # m ωd L & Δ = Ψ% 3 , , (, d $ ρd U d ' where Ψ is another undetermined function. c) When the rotation rate is no longer a parameter, the fourth dimensionless group from part a) (the Strouhal number) must drop out, so the part b) result simplifies to: # m L& Δ = Θ% 3 , ( , d $ ρd d ' where Θ is another undetermined function. d) Interestingly, the part c) result suggests that Δ does not depend on U at all!

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.37. A machine that fills peanut-butter jars must be reset to accommodate larger jars. The new jars are twice as large as the old ones but they must be filled in the same amount of time by the same machine. Fortunately, the viscosity of peanut butter decreases with increasing temperature, and this property of peanut butter can be exploited to achieve the desired results since the existing machine allows for temperature control. a) Write a dimensionless law for the jar-filling time tf based on: the density of peanut butter ρ, the jar volume V, the viscosity of peanut butter µ, the driving pressure that forces peanut butter out of the machine P, and the diameter of the peanut butter-delivery tube d. b) Assuming that the peanut butter flow is dominated by viscous forces, modify the relationship you have written for part a) to eliminate the effects of fluid inertia. c) Make a reasonable assumption concerning the relationship between tf and V when the other variables are fixed so that you can determine the viscosity ratio µnew/µold necessary for proper operation of the old machine with the new jars. d) Unfortunately, the auger mechanism that pumps the liquid peanut butter develops driving pressure through viscous forces so that P is proportional to µ. Therefore, to meet the new jarfilling requirement, what part of the machine should be changed and how much larger should it be? Solution 1.37. a) First create the parameter matrix. The solution parameter is tf. The boundary condition and material parameters are: V, ρ, P, µ, and d.

Mass: Length: Time:

tf 0 0 1

V 0 3 0

P 1 -1 -2

ρ 1 -3 0

d 0 1 0

µ 1 -1 -1

This rank of this matrix is three. Next, determine the number of dimensionless groups: 6 parameters - 3 dimensions = 3 groups. Construct the dimensionless groups: Π1 = Ptf/µ, Π2 = µ2/ρd2P, Π3 = V/d3, and write a dimensionless law: tf = (µ/P)Φ(µ2/ρd2P,V/d3), where Φ is an unknown function. b) When fluid inertia is not important the fluid's density is not a parameter. Therefore, drop ∏2 from the dimensional analysis formula: tf = (µ/P)Ψ(V/d3), where Ψ is yet another unknown function. c) One might reasonably expect that tf ∝ V (these are the two extensive variables). Therefore, we end up with tf = const⋅µV/Pd3. Now form a ratio between the old and new conditions and cancel common terms: (t f ) new (µV /Pd 3 ) new (µV ) new V µnew 1 =1= = , so new = 2 → = 3 (µV /Pd ) old (µV ) old Vold µold (t f ) old 2 d) If P is proportional to µ, then to achieve the same filling time for twice the volume using the part c) result for tf implies, Vold 2V € € € =€ old 3 € € 3 (d old ) (d new ) Thus, the machine’s nozzle diameter must be increased so that dnew = 3 2 dold .

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.38. As an idealization of fuel injection in a Diesel engine, consider a stream of highspeed fluid (called a jet) that emerges into a quiescent air reservoir at t = 0 from a small hole in an infinite plate to form a plume where the fuel and air mix. a) Develop a scaling law via dimensional analysis for the penetration distance D of the plume as a function of: Δp the pressure difference across the orifice that drives the jet, do the diameter of the jet orifice, ρo the density of the fuel, µ∞ and ρ∞ the viscosity and density of the air, and t the time since the jet was turned on. b) Simplify this scaling law for turbulent flow where air viscosity is no longer a parameter. c) For turbulent flow and D > do, only the momentum flux of the jet matters, so Δp and do are replaced by the single parameter Jo = jet momentum flux (Jo has the units of force and is approximately equal to Δpdo2 ). Recreate the dimensionless law for D using the new parameter Jo. Solution 1.38. a) The parameters are: D, t, Δp, ρo, ρ∞, µ∞, and do. With D as the solution parameter, create the parameter matrix: € D t Δp ρo ρ∞ µ∞ do –––––––––––––––––––––––––––––––––––––––– Mass: 0 0 1 1 1 1 0 Length: 1 0 -1 -3 -3 -1 1 Time: 0 1 -2 0 0 -1 0

€

Next, determine the number of dimensionless groups. This rank of this matrix is three so 7 parameters - 3 dimensions = 4 groups, and construct the groups: Π1 = D do , Π 2 = ρ o ρ∞ , Π 3 = Δpt 2 ρ∞ do2 , and Π 4 = ρ∞Δpdo2 µ∞2 . Thus, the dimensionless law is: % ρ o Δpt 2 ρ∞Δpdo2 ( D = f' , , * , where f is an unknown function. 2 do µ∞2 ) € € & ρ∞ ρ∞ do b) For high Reynolds number turbulent flow when the reservoir viscosity is no longer a € parameter, the above result becomes: % ρ Δpt 2 ( D € , = g' o , 2* do & ρ∞ ρ∞ do ) where g is an unknown function. c) When do and ρ∞ are not parameters, there is only one dimensionless group: Δpt 2 ρ∞ D2 , so the dimensionless law becomes: € D = const ⋅ t Δp ρ o . d) When Δp and do are replaced by the single parameter Jo = jet momentum flux, there are two dimensionless parameters: J o t 2 ρ∞ D4 , and ρ o ρ∞ , so the dimensionless € law becomes: 14

D = ( J o t 2 ρ∞ ) F ( ρ o ρ∞ ) , € where F is an unknown function. [The results presented € here are the€fuel-plume penetration scaling laws for fuel injection in Diesel engines where more than half of the world's petroleum ends up being burned.] €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.39. One of the simplest types of gasoline carburetors is a tube with small port for transverse injection of fuel. It is desirable to have the fuel uniformly mixed in the passing air stream as quickly as possible. A prediction of the mixing length L is sought. The parameters of this problem are: ρ = density of the flowing air, d = diameter of the tube, µ = viscosity of the flowing air, U = mean axial velocity of the flowing air, and J = momentum flux of the fuel stream. a) Write a dimensionless law for L. b) Simplify your result from part a) for turbulent flow where µ must drop out of your dimensional analysis. c) When this flow is turbulent, it is observed that mixing is essentially complete after one rotation of the counter rotating vortices driven by the injected-fuel momentum (see downstreamview of the drawing for this problem), and that the vortex rotation rate is directly proportional to J. Based on this information, assume that L ∝ (rotation time)(U) to eliminate the arbitrary function in the result of part b). The final formula for L should contain an undetermined dimensionless constant.

Solution 1.39. a) The parameters are: L, J, d, µ, ρ, and U. Use these to create the parameter matrix with L as the solution parameter: L J d µ ρ U ––––––––––––––––––––––––––––––––––– Mass: 0 1 0 1 1 0 Length: 1 1 1 -1 -3 1 Time: 0 -2 0 -1 0 -1 Next, determine the number of dimensionless groups. This rank of this matrix is three so 6 parameters - 3 dimensions = 3 groups, and construct them: Π1 = L/d, Π2 = ρUd/µ, Π3 = ρU2d2/J. And, finally write a dimensionless law: L = d Φ(ρUd/µ, ρU2d2/J), where Φ is an unknown function. b) At high Reynolds numbers, µ must not be a parameter. Therefore: L = dΨ(ρU2d2/J) where Ψ is an unknown function. c) Let Ω = vortex rotation rate. The units of Ω are 1/time and Ω must be proportional to J. Putting this statement in dimensionless terms based on the boundary condition and material J parameters of this problem means: Ω = const = (rotation time)-1 ρUd 3

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

L ρU 2 d 3 ρU 2 d 2 , or = const . Thus, for transverse d J J injection, more rapid mixing occurs (L decreases) when the injection momentum increases. Therefore: L = const (Ω-1)U = const

€

€

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.40. Consider dune formation in a large horizontal desert of deep sand. a) Develop a scaling relationship that describes how the height h of the dunes depends on the average wind speed U, the length of time the wind has been blowing Δt, the average weight and diameter of a sand grain w and d, and the air’s density ρ and kinematic viscosity ν. b) Simplify the result of part a) when the sand-air interface is fully rough and ν is no longer a parameter. c) If the sand dune height is determined to be proportional to the density of the air, how do you expect it to depend on the weight of a sand grain? Solution 1.40. a) The solution parameter is h. The boundary condition and material parameters are: U, Δt, w, d, ρ, and ν. First create the parameter matrix: h U Δt w d ρ ν ––––––––––––––––––––––––––––––––––––––– Mass: 0 0 0 1 0 1 0 Length: 1 1 0 1 1 -3 2 Time: 0 -1 1 -2 0 0 -1 Next determine the number of dimensionless groups. This rank of this matrix is three so 7 parameters - 3 dimensions = 4 groups. Construct the dimensionless groups: Π1 = h/d, Π2 = Ud/ν, Π3 = w/ρU2d2, and Π4 = UΔt/d. Thus, the dimensionless law is & Ud h w UΔt ) = Φ( , 2 2 , +, d ' ν ρU d d * where Φ is an unknown function. b) When ν is no longer a parameter, Π2 drops out: % w UΔt ( h € = Ψ' 2 2 , *, d & ρU d d ) where Ψ is another unknown function. c) When h is proportional to ρ, then h ρU 2 d 2 % UΔt ( € = Θ' *, & d ) d w where Θ is another unknown function. Under this condition, dune height will be inversely proportional to w the sand grain weight.

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.41. The rim-to-rim diameter D of the impact crater produced by a vertically-falling object depends on d = average diameter of the object, E = kinetic energy of the object lost on impact, ρ = density of the ground at the impact site, Σ = yield stress of the ground at the impact site, and g = acceleration of gravity. a) Using dimensional analysis, determine a scaling law for D. b) Simplify the result of part a) when D >> d, and d is no longer a parameter. c) Further simplify the result of part b) when the ground plastically deforms to absorb the impact energy and ρ is irrelevant. In this case, does gravity influence D? And, if E is doubled how much bigger is D? d) Alternatively, further simplify the result of part b) when the ground at the impact site is an unconsolidated material like sand where Σ is irrelevant. In this case, does gravity influence D? And, if E is doubled how much bigger is D? e) Assume the relevant constant is unity and invert the algebraic relationship found in part d) to estimate the impact energy that formed the 1.2-km-diameter Barringer Meteor Crater in Arizona using the density of Coconino sandstone, 2.3 g/cm3, at the impact site. The impact energy that formed this crater is likely between 1016 and 1017 J. How close to this range is your dimensional analysis estimate? g!

D!

Ε# ρ, Σ" d!

Solution 1.41. The solution parameter is D. The boundary condition and material parameters are: d, E, θ, ρ, Σ, and g. First create the parameter matrix:

M L T

D 0 1 0

d 0 1 0

E 1 2 –2

ρ 1 –3 0

Σ 1 –1 –2

g 0 1 –2

The rank of this matrix is 3, so there are 6 – 3 = 4 dimensionless groups. These groups are: Π1 = D d , Π 2 = E ρ gd 4 , and Π3 = Σ ρ gd . Thus, the scaling law is:

D / d = fn(E / ρ gd 4 , Σ / ρ gd) , where fn is an undetermined function. b) Use the second dimensionless group to remove d from the other two: # & −1 4 ! Σ ! E $−1 4 $ D! E $ D Σ % (. # & , or =Φ # & = Φ# # 14 14( 4& & % 3 3 d " ρ gd 4 % ρ gd ρ gd " % % ( E ρ g) " $ (ρ g E ) ' where Φ is an undetermined function. c) In this case, the two dimensionless groups in the part b) must be combined to eliminate ρ,

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Σ1 3

D 14

( E ρ g)

1 12

(ρ g E ) 3 3

=

D 13

( E Σ)

= const. ,

and this lone dimensionless group must be constant. In this case, gravity does not influence the crater diameter, and a doubling of the energy E increases D by a factor of 21 3 ≅ 1.26 . d) When S is irrelevant, then the part b) result reduces to: D = const. 14 ( E ρ g) In this case, gravity does influence the crater diameter, and a doubling of the energy E increases D a factor of 21 4 ≅ 1.19 . e) If the part d) constant is unity, then that result implies: E = ρ gD 4 . For the Barringer crater, this energy is: E = (2300 kgm–3)(9.81 ms–2)(1200m)4 ≈ 4.7 x 1016 J, an estimate that falls in the correct range, a remarkable result given the simplicity of the analysis.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.42. An isolated nominally spherical bubble with radius R undergoes shape oscillations at frequency f. It is filled with air having density ρa and resides in water with density ρw and surface tension σ. What frequency ratio should be expected between two isolated bubbles with 2 cm and 4 cm diameters undergoing geometrically similar shape oscillations? If a soluble surfactant is added to the water that lowers σ by a factor of two, by what factor should air bubble oscillation frequencies increase or decrease? Solution 1.42. The boundary condition and material parameters are: R, ρa, ρw, and σ. The solution parameter is f. First create the parameter matrix: f R ρa ρw σ –––––––––––––––––––––––––––– Mass: 0 0 1 1 1 Length: 0 1 -3 -3 0 Time: -1 0 0 0 -2 Next determine the number of dimensionless groups. This rank of this matrix is three, so 5 parameters - 3 dimensions = 2 groups. Construct the dimensionless groups: Π1 = f ρ w R 3 σ , and Π2 = ρw/ρa. Thus, the dimensionless law is %ρ ( σ f = Φ w *, 3 ' ρw R & ρa ) € where Φ is an unknown function. For a fixed density ratio, Φ(ρw/ρa) will be constant so f is proportional to R–3/2 and to σ1/2. Thus, the required frequency ratio between different sizes bubbles is: € −3 2 ( f ) 2cm " 2cm % =$ ' = 2 2 ≅ 2.83 . ( f ) 4 cm # 4cm & Similarly, if the surface tension is decreased by a factor of two, then ( f )σ 2 #1/2 &−1 2 1 =% ( = ≅ 0.707 . ( f )σ $ 1 ' 2 €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.43. In general, boundary layer skin friction, τw, depends on the fluid velocity U above the boundary layer, the fluid density ρ, the fluid viscosity µ, the nominal boundary layer thickness δ, and the surface roughness length scale ε. a) Generate a dimensionless scaling law for boundary layer skin friction. b) For laminar boundary layers, the skin friction is proportional to µ. When this is true, how must τw depend on U and ρ? c) For turbulent boundary layers, the dominant mechanisms for momentum exchange within the flow do not directly involve the viscosity µ. Reformulate your dimensional analysis without it. How must τw depend on U and ρ when µ is not a parameter? d) For turbulent boundary layers on smooth surfaces, the skin friction on a solid wall occurs in a viscous sublayer that is very thin compared to δ. In fact, because the boundary layer provides a buffer between the outer flow and this viscous sub-layer, the viscous sublayer thickness lv does not depend directly on U or δ. Determine how lv depends on the remaining parameters. e) Now consider nontrivial roughness. When ε is larger than lv a surface can no longer be considered fluid-dynamically smooth. Thus, based on the results from parts a) through d) and anything you may know about the relative friction levels in laminar and turbulent boundary layers, are high- or low-speed boundary layer flows more likely to be influenced by surface roughness? Solution 1.43. a) Construct the parameter & units matrix and recognizing that τw is a stress and has units of pressure. τw U ρ µ δ ε ––––––––––––––––––––––––––––––– M 1 0 1 1 0 0 L -1 1 -3 -1 1 1 T -2 -1 0 1 0 0

€

The rank of this matrix is three. There are 6 parameters and 3 independent units, thus there will be 6 – 3 = 3 dimensionless groups. By inspection these groups are: a skin-friction coefficient = τ ρUδ ε Π1 = w 2 , a Reynolds number = Π 2 = , and the relative roughness = Π 3 = . Thus the ρU µ δ & ρUδ ε ) τ , + where f is an undetermined function. dimensionless law is: w 2 = f ( ρU ' µ δ* €Π1 to be proportional to 1/ Π2 b) Use the result of part € a) and set τ w ∝ µ . This involves requiring τ µ &ε) g( + , where g is an so the revised form of the dimensionless law in part a) is: w 2 = ρU ρUδ ' δ * € µU % ε ( g' * . Thus, in laminar undetermined function. € Simplify this relationship to find: τ w = δ &δ ) boundary layers, τw is proportional to U and independent of ρ. € c) When µ is not a parameter the second dimensionless group from part a) must be dropped. &ε ) τ Thus, the dimensionless law becomes: w 2 = h(€ + where h is an undetermined function. Here 'δ * ρU 2 we see that τ w ∝ ρU . Thus, in turbulent boundary layers, τw is linearly proportional to ρ and €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

quadratically proportional to U. In reality, completely dropping µ from the dimensional analysis is not quite right, and the skin-friction coefficient (Π1 in the this problem) maintains a weak dependence on the Reynolds number when ε/δ 0, when q is high enough and gravity tends to push the injected gas toward the plate surface. Reformulate your answer to part a) by dropping τw and L to determine a dimensionless law for the minimum air injection rate, qc, necessary to form an air layer. c) Simplify the result of part c) when surface tension can be neglected. d) Experimental studies (Elbing et al. 2008) find that qc is proportional to U2. Using this information, determine a scaling law for qc involving the other parameters. Would an increase in g cause qc to increase or decrease?

€

€ €

Solution 1.44. a) Construct the parameter & units matrix and recognizing that τw is a stress and has units of pressure. τw L q U ρ µ σ g ––––––––––––––––––––––––––––––––––––––––––– M 1 0 0 0 1 1 1 0 L -1 1 2 1 -3 -1 0 1 T -2 0 -1 -1 0 -1 -2 -2 This rank of this matrix is three. There are 8 parameters and 3 independent units, thus there will be 8 – 3 = 5 dimensionless groups. By inspection these groups are: a skin-friction coefficient = τ ρUL U Π1 = w 2 , a Reynolds number = Π 2 = , a Froude number = Π 3 = , a capillary number ρU µ gL ρq µU = Π4 = , and flux ratio = Π 5 = . Thus the dimensionless law is: µ σ % ρUL U µU τw € ρq ( € function. = f' , , , * where f is an undetermined 2 ρU gL σ µ ) & µ € dropping Π1. Dropping L means combining Π2 and Π3 to form a new b) Dropping τw means ρUL U 3 ρU 3 = dimensionless group: Π 2Π 23 = . Thus, with Π5 as the solution parameter, the µ gL µg

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

scaling law for the minimum air injection rate, qc, necessary to form an air layer is: ρqc µ = φ ( ρU 3 µg, µU σ ) where φ is an undetermined function.

€

c) When σ is not a parameter, Π4 can be dropped leaving: ρqc µ = ϕ ( ρU 3 µg) where ϕ is an undetermined function. d) When qc is proportional to U2, then dimensional analysis requires: 23

qc = (µ ρ)const.( ρU 3 µg€) So, an increase in g would cause qc to decrease. €

13

= const.U 2 (µ ρg 2 ) .

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.45. An industrial cooling system is in the design stage. The pumping requirements are known and the drive motors have been selected. For maximum efficiency the pumps will be directly driven (no gear boxes). The number Np and type of water pumps are to be determined based on pump efficiency η (dimensionless), the total required volume flow rate Q, the required pressure rise ΔP, the motor rotation rate Ω, and the power delivered by one motor W. Use dimensional analysis and simple physical reasoning for the following items. a) Determine a formula for the number of pumps. b) Using Q, Np, ΔP, Ω, and the density (ρ) and viscosity (µ) of water, create the appropriate number of dimensionless groups using ΔP as the dependent parameter. c) Simplify the result of part b) by requiring the two extensive variables to appear as a ratio. d) Simplify the result of part c) for high Reynolds number pumping where µ is no longer a parameter. e) Manipulate the remaining dimensionless group until Ω appears to the first power in the numerator. This dimensionless group is known as the specific speed, and its value allows the most efficient type of pump to be chosen (see Sabersky et al. 1999). Solution 1.45. a) The total power that must be delivered to the fluid is QΔP. The power that one pump delivers to the fluid will be ηW. Thus, Np will be the next integer larger than QΔP/ηW. b) Construct the parameter & units matrix using ΔP as the solution parameter M L T

ΔP 1 -1 -2

Q 0 3 -1

Np 0 0 0

Ω 0 0 -1

ρ 1 -3 -0

µ 1 -1 -1

The rank of this matrix is three. There are 6 parameters and 3 independent units, thus there will be 6 – 3 = 2 dimensionless groups. By inspection these groups are: a pressure coefficient = ΔP ρQ 2 3Ω1 3 , the number of pumps = Π = N , and a Reynolds number = . Π1 = Π = 2 p 3 23 µ ρ (Ω2Q ) c) The two extensive parameters (Q & Np) must for a ratio, so defining q = Q/Np, the two ΔP ρ q 2 3Ω1 3 dimensionless groups are: , and . 23 µ ρ (Ω2 q ) d) At high Reynolds number, the second dimensionless group will not matter. Thus,

ΔP

ρ (Ω2 q )

23

alone will characterize the pump, and this group will be a constant. Ωq1 2 e) The specific speed = . Low values of the specific speed (below ~1/2 or so) 34 ΔP ρ ( ) correspond centrifugal pumps that move relatively small amounts of liquid against relativelyhigh pressure differences (a water pump for circulating a coolant through narrow passageways). High values of the specific speed (above ~2 or so) correspond to propeller pumps that move relatively high volumes of fluid against relatively-low pressure differences (a ventilation fan).

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.46. Nearly all types of fluid filtration involve pressure driven flow through a porous material. a) For a given volume flow rate per unit area = Q/A, predict how the pressure difference across the porous material = Δp, depends on the thickness of the filter material = L, the surface area per unit volume of the filter material = Ψ, and other relevant parameters using dimensional analysis. b) Often the Reynolds number of the flow in the filter pores is very much less than unity so fluid inertia becomes unimportant. Redo the dimensional analysis for this situation. c) To minimize pressure losses in heating, ventilating, and air-conditioning (HVAC) ductwork, should hot or cold air be filtered? d) If the filter material is changed and Ψ is lowered to one half its previous value, estimate the change in Δp if all other parameters are constant. (Hint: make a reasonable assumption about the dependence of Δp on L; they are both extensive variables in this situation). Porous Material!

Q/A!

Q/A!

Gage pressure ! = Δp!

Gage pressure ! = 0! L!

Solution 1.46. This question can be answered with dimensional analysis. The parameters are drawn from the problem statement and the two fluid properties ρ = density and µ = viscosity. The solution parameter is Δp, and the unit matrix is: Δp Q/A L Ψ ρ µ –––––––––––––––––––––––––––––––––––––– M 1 0 0 0 1 1 L –1 1 1 –1 –3 –1 T –2 –1 0 0 0 –1 There will be: 6 – 3 = 3 dimensionless groups. These groups are: a pressure coefficient = Π1 = Δp ρ (Q / A)2 , a dimensionless thickness = Π 2 = LΨ , and a thickness-based Reynolds number Π3 = ρ (Q / A)L µ . The dimensionless relationship must take the form: # Δp ρ (Q / A)L & = fn % LΨ, (. 2 ρ (Q / A) µ $ ' b) Dropping the density reduces the number of dimensionless groups. The product of the first and third group is independent of the density, thus the revised dimensional analysis result is: Δp ρ (Q / A)L ΔpL ⋅ = = fn ( LΨ ) . 2 ρ (Q / A) µ (Q / A)µ c) The viscosity of gases increases with increasing temperature, thus to keep Δp low for a given filter element (LΨ) the viscosity should be low. So, filter cold air.

Fluid Mechanics, 6th Ed.

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d) A reasonable assumption is that Δp will be proportional to the thickness of the porous material, and this implies: (Q / A)µ 2 Δp = ⋅ co nst. ( LΨ ) = const.(Q / A)µ LΨ 2 . L So, if Ψ is lowered by a factor of ½, then Δp will be lowered to ¼ of is previous value.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.47. A new industrial process requires a volume V of hot air with initial density ρ to be moved quickly from a spherical reaction chamber to a larger evacuated chamber using a single pipe of length L and interior diameter of d. The vacuum chamber is also spherical and has a volume of Vf. If the hot air cannot be transferred fast enough, the process fails. Thus, a prediction of the transfer time t is needed based on these parameters, the air’s ratio of specific heats γ, and initial values of the air’s speed of sound c and viscosity µ. a) Formulate a dimensionless scaling law for t, involving six dimensionless groups. b) Inexpensive small-scale tests of the air-transfer process are untaken before construction of the commercial-scale reaction facility. Can all these dimensionless groups be matched if the target size for the pipe diameter in the small-scale tests is d´ = d/10? Would lowering or raising the initial air temperature in the small-scale experiments help match the dimensionless numbers? Solution 1.47. a) This question can be answered with dimensional analysis. The parameters are drawn from the problem statement. The solution parameter is t, and the unit matrix is:

M L T

t V Vf L d γ c ρ µ ––––––––––––––––––––––––––––––––––––––––––––––––– 0 0 0 0 0 0 0 1 1 0 3 3 1 1 0 1 -3 -1 1 0 0 0 0 0 -1 0 -1

The rank of this matrix is 3, so there will be: 9 – 3 = 6 dimensionless groups. These groups are: a dimensionless time: Π1 = ct/d; a volume ratio: Π2 = V/Vf; another volume ratio: Π2 = d3/Vf; an aspect ratio: Π4 = L/d; the ratio of specific heats: Π5 = γ; and a sonic Reynolds number: Π6 = ρcd/µ. Thus, the scaling law is: ! V d 3 L ρ cd $ ct && . = φ ## , , , γ , d V V d µ " f f % b) The first dimensionless group will be matched if the other five are matched. So, let primes denote the small-scale test parameter. Matching the five independent dimensionless groups means: V V ! d 3 d !3 L L! ρ cd ρ !c!d ! , , = , γ = γ´, and . (1, 2, 3, 4, 5) = = = V f V f! V f V f! d d ! µ µ! Starting from the target pipe size ratio, d´ = d/10, (2) implies: d 3 Vf = = 10 3 or V f! = V f 10 3 , and V ! = V 10 3 . 3 d ! V f! where the finding for V´ follows from (1). Similarly, from (3), the results for the length of the pipe are: d L = = 10 , so L! = L 10 . d ! L! The next independent dimensionless group, γ, can be matched by using air in the scale model tests since it will be used in the full-scale device. The final independent dimensionless group is likely to be the most difficult to match. Using (5) and d´ = d/10, implies:

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

ρ cd ρ !c!d ρ !c! ρc or = = 10 , µ 10µ ! µ! µ and this relationship only involves material properties. The initial temperature (and pressure P) of the small scale experiments can be varied, ρ is proportional to T–1 (and P+1), while both c and µ are proportional to T1/2 (and nearly independent of P). Thus, small scale testing at reduced temperature (and elevated pressure) might make it possible to match values of this dimensionless group between the large- and small-scale experiments.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.48. Create a small passive helicopter from ordinary photocopy-machine paper (as shown) and drop it from a height of 2 m or so. Note the helicopter’s rotation and decent rates once it’s rotating steadily. Repeat this simple experiment with different sizes of paper clips to change the helicopter’s weight, and observe changes in the rotation and decent rates. a) Using the helicopter’s weight W, blade length l, and blade width (chord) c, and the air’s density ρ and viscosity µ as independent parameters, formulate two independent dimensionless scaling laws for the helicopter’s rotation rate Ω, and decent rate dz/dt. b) Simplify both scaling laws for the situation where µ is no longer a parameter. c) Do the dimensionless scaling laws correctly predict the experimental trends? d) If a new paper helicopter is made with all dimensions smaller by a factor of two. Use the scaling laws found in part b) to predict changes in the rotation and decent rates. Make the new smaller paper helicopter and see if the predictions are correct.

7 cm!

7 cm! paper clip! 2 cm! 4 cm!

Solution 1.48. The experiments clearly show that Ω and dz/dt increase with increasing W. a) This question can be answered with dimensional analysis. The parameters are drawn from the problem statement. The first solution parameter is Ω (the rotation rate) and the units matrix is:

M L T

Ω W l c ρ µ –––––––––––––––––––––––––––––––– 0 1 0 0 1 1 0 1 1 1 -3 -1 -1 -2 0 0 0 -1

The rank of this matrix is 3, so there will be: 6 – 3 = 3 dimensionless groups. These groups are: Π1 = Ωρ1/2l2/W1/2, Π2 = l/c, and Π3 = ρW/µ2. Thus, the scaling law is: " l ρW % ρl 4 Ω = φ$ , 2 ', W #c µ & where φ is an undetermined function. The second solution parameter is dz/dt (the descent rate), and it has the same units as Ωl, so it's scaling law is: ! l ρW $ ! dz $ ρl 2 =ψ# , 2 &. # & " dt % W "c µ % where ψ is an undetermined function.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

b) When µ is no longer a parameter, both scaling laws simplify #l& ! dz $ ρl 2 !l$ ρl 4 = Φ % ( and # & = Ψ# &, $c' " dt % W "c% W where Φ and Ψ are different undetermined functions. These laws imply 1 W #l& dz 1 W " l % Ω= 2 Φ % ( and = Ψ$ ' , l ρ $c' dt l ρ # c & c) These scaling laws are consistent with experimental results; the rotation and decent rates both increase when W increases. d) If the aspect ratio, l/c, is fixed, then decreasing l by a factor of two should increase Ω by a factor of 4, and dz/dt by a factor of two. When the smaller helicopter is made, the experiments do appear to confirm these predictions; both Ω and dz/dt do increase, and the increase for Ω is larger. Ω

Fluid Mechanics, 6th Ed.

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€

Kundu, Cohen, and Dowling

Exercise 2.1. For three spatial dimensions, rewrite the following expressions in index notation and evaluate or simplify them using the values or parameters given, and the definitions of δij and εijk wherever possible. In b) through e), x is the position vector, with components xi. a) b ⋅ c where b = (1, 4, 17) and c = (–4, –3, 1) b) (u ⋅ ∇ ) x where u a vector with components ui. c) ∇φ , where φ = h ⋅ x and h is a constant vector with components hi. d) ∇ × u, where u = Ω × x and Ω is a constant vector with components Ωi. "1 2 3& $ $ e) C€ ⋅ x , where C = #0 1 2' $0 0 1$ % ( Solution 2.1. a) b ⋅ c = bic i = 1(−4) + 4(−3) + 17(1) = −4 −12 + 17 = +1 +x . + % ∂ ( % ∂ ( % ∂ (.- 1 0 € ∂ b) (u ⋅ ∇ ) x = u j x i = -u1' * + u2 ' * + u3 ' *0 x 2 ∂x j ∂x1 ) ∂x 2 ) ∂x 3 )/- 0 & & & , € -,x 3 0/ ) # ∂x & # ∂x & # ∂x & , + u1% 1 ( + u2 % 1 ( + u3 % 1 (. $ ∂x 2 ' $ ∂x 3 '. )u ⋅1+ u ⋅ 0 + u ⋅ 0, + $ ∂x1 ' ) u1 , 2 3 # ∂x 2 & # ∂x 2 &. + 1 + # ∂x 2 & . + . = +u1% ( + u2 % ( + u3 % (. = +u1 ⋅ 0 + u2 ⋅1+ u3 ⋅ 0. = u jδij = +u2 . = ui $ ∂x 2 ' $ ∂x 3 '. + $ ∂x1 ' + . +*u3 .# & # & # ∂x 3 &. *u1 ⋅ 0 + u2 ⋅ 0 + u3 ⋅1∂x 3 + ∂x 3 +u1%$ ∂x (' + u2 %$ ∂x (' + u3 %$ ∂x ('. * 1 2 3 ∂φ ∂ ∂x i c) ∇φ = = ( hi x i ) = hi = hiδij = h j = h ∂x j ∂x j ∂x j € × u = ∇ × (Ω × x ) = ε ∂ (ε Ω x ) = ε ε Ω δ = (δ δ − δ δ )Ω δ = (δ δ − δ δ )Ω d) ∇ ijk klm l m ijk klm l jm il jm im jl l jm il jj ij jl l ∂x j = ( 3δil − δil )Ωl = 2δil Ωl = 2Ωl = 2Ω Here, the following identities have been used: εijkεklm = δilδ jm − δimδ jl , δijδ jk = δik , δ jj = 3 , and δij Ω j = Ωi #1 2 3'# x1 ' # x1 + 2x 2 + 3x 3 ' € % %% % % % e) C ⋅ x = Cij x j = $0 1 2($ x 2 ( = $€ x 2 + 2x 3 ( € € %0 0 1%% x % % % x3 & )& 3 ) & )

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 2.2. Starting from (2.1) and (2.3), prove (2.7). Solution 2.2. The two representations for the position vector are: x = x1e1 + x 2e 2 + x 3e 3 , or x = x1"e1" + x 2" e"2 + x 3" e"3 . Develop the dot product of x with e1 from each representation, e1 ⋅ x = e1 ⋅ ( x1e1 + x 2e 2 + x 3e 3 ) = x1e1 ⋅ e1 + x 2e1 ⋅ e 2 + x 3e1 ⋅ e 3 = x1 ⋅1+ x 2 ⋅ 0 + x 3 ⋅ 0 = x1 , and e1 ⋅ x = e1 ⋅ ( x1#e1# + x #2e#2 +€x #3e#3 ) = x1#e1 ⋅ e1# + x #2e1 ⋅ e#2 + x #3e1 ⋅ e#3 = x #iC1i , € set these equal to find: x1 = x "iC1i , € where Cij = e i ⋅ e#j is a 3 × 3 matrix of direction cosines. In an entirely parallel fashion, forming € the dot product of x with e2, and x with e2 produces: € x 2 = x "iC2i and x 3 = x "iC3i . Thus, for any component xj, where j = 1, 2, or 3, we have: € x j = x "iC ji , which is (2.7). € €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 2.3. For two three-dimensional vectors with Cartesian components ai and bi, prove the Cauchy-Schwartz inequality: (aibi)2 ≤ (ai)2(bi)2. Solution 2.3. Expand the left side term, (ai bi )2 = (a1b1 + a2 b2 + a3b3 )2 = a12 b12 + a22 b22 + a32 b32 + 2a1b1a2 b2 + 2a1b1a3b3 + 2a2 b2 a3b3 , then expand the right side term, (ai )2 (bi )2 = (a12 + a22 + a32 )(b12 + b22 + b32 ) = a12 b12 + a22 b22 + a32 b32 + (a12 b22 + a22 b12 ) + (a12 b32 + a32 b12 ) + (a32 b22 + a22 b32 ). Subtract the left side term from the right side term to find: (ai )2 (bi )2 − (ai bi )2

= (a12 b22 − 2a1b1a2 b2 + a22 b12 ) + (a12 b32 − 2a1b1a3b3 + a32 b12 ) + (a32 b22 − 2a2 b2 a3b3 + a22 b32 ) 2

= (a1b2 − a2 b1 )2 + (a1b3 − a3b1 )2 + (a3b2 − a2 b3 )2 = a × b .

Thus, the difference (ai )2 (bi )2 − (ai bi )2 is greater than zero unless a = (const.)b then the difference is zero.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 2.4. For two three-dimensional vectors with Cartesian components ai and bi, prove the triangle inequality: a + b ≥ a + b . Solution 2.4. To avoid square roots, square both side of the equation; this operation does not change the equation's meaning. The left side becomes:

(a + b)

2

2

2

= a +2 a b + b ,

and the right side becomes: 2

2

2

a + b = (a + b)⋅ (a + b) = a ⋅ a + 2a ⋅ b + b ⋅ b = a + 2a ⋅ b + b . So,

(a + b)

2

2

− a + b = 2 a b − 2a ⋅ b .

Thus, to prove the triangle equality, the right side of this last equation must be greater than or equal to zero. This requires: a b ≥ a ⋅ b or using index notation: ai2 bi2 ≥ ai bi , which can be squared to find: ai2 bi2 ≥ (ai bi )2 , and this is the Cauchy-Schwartz inequality proved in Exercise 2.3. Thus, the triangle equality is proved.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 2.5. Using Cartesian coordinates where the position vector is x = (x1, x2, x3) and the fluid velocity is u = (u1, u2, u3), write out the three components of the vector: (u ⋅ ∇)u = ui ∂u j ∂x i .

(

)

Solution 2.5.

€

€

+ % ∂u ( % ∂u ( % ∂u ( / - u1' 1 * + u2 ' 1 * + u3 ' 1 * & ∂x 2 ) & ∂x 3 ) - & ∂x1 ) % ∂u ( % ∂u ( % ∂u ( % ∂ u ( - % ∂u ( % ∂u ( % ∂u (a) (u ⋅ ∇ )u = ui ' j * = u1' j * + u2 ' j * + u3 ' j * = , u1' 2 * + u2 ' 2 * + u3 ' 2 *0 & ∂x i ) & ∂x1 ) & ∂x 2 ) & ∂x 3 ) - & ∂x1 ) & ∂x 2 ) & ∂x 3 )% ( % ( % ∂u3 (- ∂u3 ∂u3 u + u + u ' * ' * *- 1 2 3' & ∂x 2 ) & ∂x 3 )1 . & ∂x1 ) ) # ∂u & # ∂u & # ∂u & + u% ( + v% ( + w% ( + $ ∂z ' + + $ ∂x ' $ ∂y ' + # ∂v & # ∂v & # ∂v & + = * u% ( + v% ( + w% ( . $ ∂z ' + + $ ∂x ' $ ∂y ' # & + # ∂w & # ∂w &+ ∂w + u%$ (' + v% ( + w%$ ('+ ∂z / $ ∂y ' , ∂x

(

)

The vector in this exercise, (u ⋅ ∇ )u = ui ∂u j ∂x i , is an important one in fluid mechanics. As described in Ch. 3, it is the nonlinear advective acceleration. €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 2.6. Convert ∇ × ∇ρ to indicial notation and show that it is zero in Cartesian coordinates for any twice-differentiable scalar function ρ. Solution 2.6. Start with the definitions of the cross product and the gradient. ∂ ∂ 2ρ ∇ × ( ∇ρ ) = εijk (∇ρ )k = εijk ∂xj ∂ x j∂ xk Write out the vector component by component recalling that εijk = 0 if any two indices are equal. Here the "i" index is the free index. ! 2 2 # ε123 ∂ ρ + ε132 ∂ ρ ∂ x2∂ x3 ∂ x3∂ x2 # # 2 2 # ∂ ρ ∂ ρ ∂ 2ρ εijk = " ε 213 + ε 231 ∂ x j∂ xk # ∂ x1∂ x3 ∂ x3∂ x1 # 2 ∂ ρ ∂ 2ρ # ε312 + ε321 ∂ x1∂ x2 ∂ x2∂ x1 # $

% ! 2 2 # # ∂ ρ – ∂ ρ # # ∂ x2∂ x3 ∂ x3∂ x2 # # # # ∂ 2ρ ∂ 2ρ = – + & " # # ∂ x1∂ x3 ∂ x3∂ x1 # # ∂ 2ρ ∂ 2ρ # # − # ' # $ ∂ x1∂ x2 ∂ x2∂ x1

% # # # # &=0, # # # # '

where the middle equality follows from the definition of εijk (2.18), and the final equality follows ∂ 2ρ ∂ 2ρ when ρ is twice differentiable so that . = ∂x j∂x k ∂x k∂x j

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 2.7. Using indicial notation, show that a × (b × c) = (a ⋅ c)b − (a ⋅ b)c. [Hint: Call d ≡ b × c. Then (a × d)m = εpqmapdq = εpqmapεijqbicj. Using (2.19), show that (a × d)m = (a ⋅ c)bm − (a ⋅ b)cm.] Solution 2.7. Using the hint and the definition of εijk produces: (a × d)m = εpqmapdq = εpqmapεijqbicj = εpqmεijq bicjap = –εijqεqpm bicjap. Now use the identity (2.19) for the product of epsilons: (a × d)m = – (δipδjm – δimδpj) bicjap = – bpcmap + bmcpap. Each term in the final expression involves a sum over "p", and this is a dot product; therefore (a × d)m = – (a ⋅ b)cm + bm(a ⋅ c). Thus, for any component m = 1, 2, or 3, a × (b × c) = − (a ⋅ b)c + (a ⋅ c)b = (a ⋅ c)b − (a ⋅ b)c.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 2.8. Show that the condition for the vectors a, b, and c to be coplanar is εijkaibjck = 0. Solution 2.8. The vector b × c is perpendicular to b and c. Thus, a will be coplanar with b and c if it too is perpendicular to b × c. The condition for a to be perpendicular with b × c is: a ⋅ (b × c) = 0. In index notation, this is aiεijkbjck = 0 = εijkaibjck.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 2.9. Prove the following relationships: δijδij = 3, εpqrεpqr = 6, and εpqiεpqj = 2δij. Solution 2.9. (i) δijδij = δii = δ11 + δ22 + δ33 = 1 + 1 + 1 = 3. For the second two, the identity (2.19) is useful. (ii) εpqrεpqr = εpqrεrpq = δppδqq – δpqδpq = 3(3) – δpp = 9 – 3 = 6. (iii) εpqiεpqj = εipqεpqj = – εipqεqpj = – (δipδpj – δijδpp) = – δij + 3δij = 2δij.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 2.10. Show that C⋅CT = CT⋅C = δ, where C is the direction cosine matrix and δ is the matrix of the Kronecker delta. Any matrix obeying such a relationship is called an orthogonal matrix because it represents transformation of one set of orthogonal axes into another. Solution 2.10. To show that C⋅CT = CT⋅C = δ, where C is the direction cosine matrix and δ is the matrix of the Kronecker delta. Start from (2.5) and (2.7), which are x "j = x iCij and x j = x "iC ji , respectively, and change the index "i" into "m" on (2.5): x "j = x m Cmj . Substitute this into (2.7) to find: x j = x "iC ji = ( x m Cmi )C ji = CmiC ji x m . € € However, we also have xj = δjmxm, so € δ jm x m = CmiC ji x m → δ jm = CmiC ji , which can be written:€ δ jm = CmiCijT = C⋅CT, and taking the transpose € of the thisT produces: T (δ jm ) = δmj = (CmiCijT ) = CmiT Cij = CT⋅C. €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 2.11. Show that for a second-order tensor A, the following quantities are invariant under the rotation of axes: I1 = Aii , I 2 =

A11

A12

A21

A22

+

A22

A23

A32

A33

+

A11

A13

A31

A33

, and I3 = det(Aij).

[Hint: Use the result of Exercise 2.8 and the transformation rule (2.12) to show that Iʹ′1 = Aʹ′ii = Aii.= I1. Then show that AijAji and AijAjkAki are also invariants. In fact, all contracted scalars of the I 2 = 12 "# I12 − Aij A ji $% , form AijAjk ⋅⋅⋅ Ami are invariants. Finally, verify that

I 3 = 13 "# Aij A jk Aki − I1 Aij A ji + I 2 Aii $% . Because the right-hand sides are invariant, so are I2 and I3.] Solution 2.11. First prove I1 is invariant by using the second order tensor transformation rule (2.12): " = Cim C jn Aij . Amn Replace Cjn by CnjT and set n = m,

" = CimCnjT Aij → Amm " = Cim CmjT Aij . Amn T Use the result of Exercise 2.8, € δij = Cim Cmj = , to find: " = δij Aij = Aii . I1 = Amm €

Thus, the first invariant € is does not depend on a rotation of the coordinate axes. Now consider€whether or not AmnAnm is invariant under a rotation of the coordinate axes. Start with a double application of (2.12): € T " Anm " = Cim C jn Aij C pn Cqm A pq = C jn CnpT Cim Cmq Amn Aij A pq . From the result of Exercise 2.8, the factors in parentheses in the last equality are Kronecker delta functions, so " Anm " = δ jpδiq Aij A pq = Aij A ji . Amn € Thus, the matrix contraction AmnAnm does not depend on a rotation of the coordinate axes. The manipulations for AmnAnpApm are a straightforward extension of the prior efforts for Aii and AijAji. € " Anp " A"pm = Cim C jn Aij Cqn Crp Aqr Csp Ctm Ast = C jn CnqT Crp C psT CimCmtT Aij Aqr Ast . Amn Again, the factors in parentheses are Kronecker delta functions, so " Anp " A"pm = δ jqδrsδit Aij Aqr Ast = Aiq Aqs Asi , Amn which implies that the matrix contraction AijAjkAki does not depend on a rotation of the coordinate € axes. Now, for the second invariant, verify the given identity, starting from the given definition € for I2. A11 A12 A22 A23 A11 A13 I2 = + + A21 A22 A32 A33 A31 A33 = A11 A22 − A12 A21 + A22 A33 − A23 A32 + A11 A33 − A13 A31 = A11 A22 + A22 A33 + A11 A33 − ( A12 A21 + A23 A32 + A13 A31 )

(

(

€ € €

€ €

) (

)(

)(

)(

) (

)(

)

)(

)(

(

)

= 12 A112 + 12 A222 + 12 A332 + A11 A22 + A22 A33 + A11 A33 − A12 A21 + A23 A32 + A13 A31 + 12 A112 + 12 A222 + 12 A332 =

1 2

[ A11 + A22 + A33 ]

2

−

1 2

(2A

12

2 11

2 22

2 33

A21 + 2A23 A32 + 2A13 A31 + A + A + A

)

)

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

= 12 I12 − 12 ( A11 A11 + A12 A21 + A13 A31 + A12 A21 + A22 A22 + A23 A32 + A13 A31 + A23 A32 + A33 A33 )

= 12 I12 − 12 ( Aij A ji ) = 12 (I12 − Aij A ji ) Thus, since I2 only depends on I1 and AijAji, it is invariant under a rotation of the coordinate axes because I1 and AijAji are invariant under a rotation of the coordinate axes. The manipulations for the third invariant are a tedious but not remarkable. Start from the given definition for I3, and group like terms.

€

€

I3 = det ( Aij ) = A11 (A22 A33 − A23 A32 ) − A12 (A21 A33 − A23 A31) + A13 (A21 A32 − A22 A31 ) = A11 A22 A33 + A12 A23 A31 + A13 A32 A21 − ( A11 A23 A32 + A22 A13 A31 + A33 A12 A21 )

€

€

€ €

€

€ €

€

(a)

Now work from the given identity. The triple matrix product AijAjkAki has twenty-seven terms: A113 + A11 A12 A21 + A11 A13 A31 + A12 A21 A11 + A12 A22 A21 + A12 A23 A31 + A13 A31 A11 + A13 A32 A21 + A13 A33 A31 + 3 € A21 A11 A12 + A21 A12 A22 + A21 A13 A32 + A22 A21 A12 + A22 + A22 A23 A32 + A23 A31 A12 + A23 A32 A22 + A23 A33 A32 +

A31 A11 A13 + A31 A12 A23 + A31 A13 A33 + A32 A21 A13 + A32 A22 A23 + A32 A23 A33 + A33 A31 A13 + A33 A32 A23 + A333 These can be grouped as follows: Aij A jk Aki = 3(A12 A23 A31 + A13 A32 A21 ) + A11(A112 + 3A12 A21 + 3A13 A31) + 2 2 (b) A22 (3A21 A12 + A22 + 3A23 A32 ) + A33 (3A31 A13 + 3A32 A23 + A33 ) The remaining terms of the given identity are: −I1 Aij A ji + I2 Aii = I1(I2 – Aij A ji ) = I1 (I2 + 2I2 − I12 ) = 3I1I2 – I13 , where€the result for I2 has been used. Evaluating the first of these two terms leads to: 3I1I2 = 3(A11 + A22 + A33 )(A11 A22 − A12 A21 + A22 A33 − A23 A32 + A11 A33 − A13 A31 ) = 3(A11 + A22 + A33 )(A11 A22 + A22 A33 + A11 A33 ) − 3(A11 + A22 + A33 )(A12 A21 + A23 A32 + A13 A31 ) . € to (b) produces: Adding this Aij A jk Aki + 3I1I2 = 3(A12 A23 A31 + A13 A32 A21 ) + 3(A11 + A22 + A33 )(A11 A22 + A22 A33 + A11 A33 ) + 2 2 A11 (A112 − 3A23 A32 ) + A22 (A22 − 3A13 A31 ) + A33 (A33 − 3A12 A21 ) = 3(A12 A23 A31 + A13 A32 A21 − A11 A23 A32 − A22 A13 A31 − A33 A12 A21 ) + 3 3 (c) 3(A11 + A22 + A33 )(A11 A22 + A22 A33 + A11 A33 ) + A113 + A22 + A33 The last term of the given identity is: €3 3 3 2 2 2 2 2 2 I13 = A€ 11 + A22 + A33 + 3(A11 A22 + A11 A33 + A22 A11 + A22 A33 + A33 A11 + A33 A22 ) + 6A11 A22 A33 3 3 = A113 + A22 + A33 + 3(A11 + A22 + A33 )(A11 A22 + A11 A33 + A22 A33 ) – 3A11 A22 A33 € Subtracting this from (c) produces: Aij A jk Aki + 3I1I2 − I13 = 3(A12 A23 A31 + A13 A32 A21 − A11 A23 A32 − A22 A13 A31 − A33 A12 A21 + A11 A22 A33 ) = 3I3 . This verifies that the given identity for I3 is correct. Thus, since I3 only depends on I1, I2, and AijAjkAki, it is invariant under a rotation of the coordinate axes because these quantities are € invariant under a rotation of the coordinate axes as shown above. €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 2.12. If u and v are vectors, show that the products uiυj obey the transformation rule (2.12), and therefore represent a second-order tensor. Solution 2.12. Start by applying the vector transformation rule (2.5 or 2.6) to the components of u and v separately, u"m = Cim ui , and v "n = C jn v j . The product of these two equations produces: u"m v "n = Cim C jn uiv j , which is the same as (2.12) tensors. € for second order €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 2.13. Show that δij is an isotropic tensor. That is, show that δʹ′ij = δij under rotation of the coordinate system. [Hint: Use the transformation rule (2.12) and the results of Exercise 2.10.] Solution 2.13. Apply (2.12) to δij,

" = CimC jnδij = CimCin = CmiT Cin = δmn . δmn where the final equality follows from the result of Exercise 2.10. Thus, the Kronecker delta is invariant under coordinate rotations.

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 2.14. If u and v are arbitrary vectors resolved in three-dimensional Cartesian 2

coordinates, use the definition of vector magnitude, a = a ⋅ a , and the Pythagorean theorem to show that u⋅v = 0 when u and v are perpendicular.

€

Solution 2.14. Consider the magnitude of the sum u + v, 2 u + v = (u1 + v1 ) 2 + (u2 + v 2 ) 2 + (u3 + v 3 ) 2 = u12 + u22 + u32 + v12 + v 22 + v 32 + 2u1v1 + 2u2v 2 + 2u3v 3 2 2 = u + v + 2u ⋅ v , which can be rewritten: 2 2 2 u + v − u − v = 2u ⋅ v . € €When u and v are perpendicular, the Pythagorean theorem requires the left side to be zero. Thus, u ⋅ v = 0. €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 2.15. If u and v are vectors with magnitudes u and υ, use the finding of Exercise 2.14 to show that u⋅v = uυcosθ where θ is the angle between u and v. Solution 2.15. Start with two arbitrary vectors (u and v), and view them so that the plane they define is coincident with the page and v is horizontal. Consider two additional vectors, βv and w, that are perpendicular (v⋅w = 0) and can be summed together to produce u: w + βv = u. u w

θ

v

βv Compute the dot-product of u and v: u⋅v = (w + βv) ⋅v = w⋅v + βv⋅v = βυ2. where the final equality holds because v⋅w = 0. From the geometry of the figure: βv βυ u , or β = cosθ . cosθ ≡ = υ u u Insert this into the final equality for u⋅v to find: %u ( u ⋅ v = ' cos θ *υ 2 = uυ cosθ . &υ € ) €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 2.16. Determine the components of the vector w in three-dimensional Cartesian coordinates when w is defined by: u⋅w = 0, v⋅w = 0, and w⋅w = u2υ2sin2θ, where u and v are known vectors with components ui and υi and magnitudes u and υ, respectively, and θ is the angle between u and v. Choose the sign(s) of the components of w so that w = e3 when u = e1 and v = e2. Solution 2.16. The effort here is primarily algebraic. Write the three constraints in component form: u⋅w = 0, or u1w1 + u2 w 2 + u3 w 3 = 0 , (1) v⋅w = 0, or υ1w1 + υ 2 w 2 + υ 3 w 3 = 0 , and (2) The third one requires a little more effort since the angle needs to be eliminated via a dot product: € w⋅w = u2υ2sin2θ = u2υ2(1 – cos2θ) = u2υ2 – (u⋅w)2 or € w12 + w 22 + w 32 = (u12 + u22 + u32 )(υ12 + υ 22 + υ 32 ) − (u1υ1 + u2υ 2 + u3υ 3 ) 2 , which leads to (3) w12 + w 22 + w 32 = (u1υ 2 − u2υ1 ) 2 + (u1υ 3 − u3υ1 ) 2 + (u2υ 3 − u3υ 2 ) 2 . Equation (1) implies: (4) w1 = −(w 2 u2 + w 3 u3 ) u1 € Combine (2) and (4) to eliminate w1, and solve the resulting equation for w2: € $ υ ' $ υ ' −υ1 (w 2 u2 + w 3 u3 ) u1 + υ 2 w 2 + υ 3 w 3 = 0 , or &− 1 u2 + υ 2 )w 2 + & − 1 u3 + υ 3 ) w 3 = 0 . % u1 ( % u1 ( € Thus: $υ ' $ υ ' $u υ − uυ ' (5) w 2 = +w 3 & 1 u3 − υ 3 ) &− 1 u2 + υ 2 ) = w 3 & 3 1 1 3 ) . € % u1 ( € % u1 ( % u1υ 2 − u2υ1 ( Combine (4) and (5) to find: ' w $$ υ u − υ 3 u1 ' w 3 $ υ1u3 u2 − υ 3 u1u2 + υ 2 u1u3 − υ1u2 u3 ' w1 = − 3 && 1 3 +) ) u2 + u3 ) = − & u1 %% υ 2 u1 − υ1u2 ( u1 % υ 2 u1 − υ1u2 ( ( € $ u2υ 3 − u3υ 2 ' w $ −υ u u + υ 2 u1u3 ' (6) =− 3& 3 1 2 ) = w3& ). u1 % υ 2 u1 − υ1u2 ( % u1υ 2 − u2υ1 ( Put € (5) and (6) into (3) and factor out w3 on the left side, then divide out the extensive common factor that (luckily) appears on the right and as the numerator inside the big parentheses. $€ (u2υ 3 − u3υ 2 ) 2 + (u3υ1 − u1υ 3 ) 2 + (u1υ 2 − u2υ1 ) 2 ' 2 2 2 w & ) = (u1υ 2 − u2υ1 ) + (u1υ 3 − u3υ1 ) + (u2υ 3 − u3υ 2 ) 2 (u1υ 2 − 2u$2υ1 ) 1 ' % ( , so w 3 = ±(u1υ 2 − u2υ1 ) . w3 & = 1 ) 2 % (u1υ 2 − u2υ1 ) ( If u = (1,0,0), and v = (0,1,0), then using the plus sign produces w3 = +1, so w 3 = +(u1υ 2 − u2υ1 ) . Cyclic permutation of the indices allows the other components of w to be determined: w1€= u2υ 3 − u3υ 2 , € w 2 = u3υ1 − u1υ 3 , € w 3 = u1υ 2 − u2υ1 . 2 3

€

€ € €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 2.17. If a is a positive constant and b is a constant vector, determine the divergence and the curl of u = ax/x3 and u = b×(x/x2) where x = x12 + x 22 + x 32 ≡ x i x i is the length of x.

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Solution 2.17. Start with the divergence calculations, and use x = x12 + x 22 + x 32 to save writing. $ ' $ ' $ ax ' $ ∂ ∂ ∂ ' & x1€, x 2 , x 3 ) = a& ∂ , ∂ , ∂ ) ⋅ $& x1, x 2 , x 3 ') ∇ ⋅ & 3 ) = a& , , )⋅ % x ( % ∂x1 ∂x 2 ∂x 3 ( & [ x 2 + x 2 + x 2 ] 3 2 ) % ∂x1 ∂x 2 ∂x 3 ( % x 3 ( 2 3 % 1 ( € # ∂ # x1 & ∂ # x 2 & ∂ # x 3 && # 1 3 x1 & 1 3 x2 1 3 x3 = a% % 3 ( + 2x1 ) + 3 − 2x 2 ) + 3 − 2x 3 )( % 3(+ % 3 (( = a% 3 − 5 ( 5 ( 5 ( ' 2x x 2x x 2x $ ∂x1 $ x ' ∂x 2 $ x ' ∂x 3 $ x '' $ x # 3 3( x 2 + x 2 + x 2 ) & # 3 3& 1 2 3 ( = a% 3 − 3 ( = 0 . = a%% 3 − 5 ( x x ' $x ' $x Thus, the vector field ax/x3 is divergence free even though it points away from the origin everywhere. % b × x ( % ∂ ∂ ∂ ( % b2 x 3 − b3 x 2 ,b3 x1 − b1 x 3 ,b1 x 3 − b2 x1 ( ∇⋅' 2 * =' , , *⋅' * & x ) & ∂x1 ∂x 2 ∂x 3 ) & x12 + x 22 + x 32 ) $ ∂ $ b x − b x ' ∂ $ b3 x1 − b1 x 3 ' ∂ $ b1 x 2 − b2 x1 '' =& & 2 3 2 3 2)+ & )+ & )) ( ∂x 2 % ( ∂x 3 % (( x x2 x2 % ∂x1 % # 2 & # 2 & # 2 & = (b2 x 3 − b3 x 2 )%− 4 (2x1 )( + (b3 x1 − b1 x 3 )% − 4 (2x 2 )( + (b1 x 2 − b2 x1 )%− 4 (2x 3 )( $ x ' $ x ' $ x ' 4 = − 4 (b2 x 3 x1 − b3 x 2 x1 + b3 x1 x 2 − b1 x 3 x 2 + b1 x 2 x 3 − b2 x1 x 3 ) = 0 . x This field is divergence free, too. The curl calculations produce: $ ax ' $ ∂ ∂ ∂ ' $ x1, x 2 , x 3 ' $ ∂x −3 ∂x −3 ∂x −3 ∂x −3 ∂x −3 ∂x −3 ' ∇ × & 3 ) = a& , , × = a x − x , x − x , x − x ) & ) & ) 2 1 3 2 1 % x ( % ∂x1 ∂x 2 ∂x 3 ( % x 3 ( % 3 ∂x 2 ∂x 3 ∂x 3 ∂x1 ∂x1 ∂x 2 ( # 3 x3 & 3 x2 3x 3 x3 3 x2 3 x1 = a% − 2x 2 ) + 2x 3 ),− 15 (2x 3 ) + 2x1 ),− 2x1 ) + 2x 2 )( = (0,0,0) 5 ( 5 ( 5 ( 5 ( 5 ( $ 2x ' 2x 2x 2x 2x 2x 3 Thus, thus the vector field ax/x is also irrotational. $ b × x ' $ ∂ ∂ ∂ ' $ b2 x 3 − b3 x 2 ,b3 x1 − b1 x 3 ,b1 x 2 − b2 x1 ' ∇ ×& 2 ) = & , , ) ×& ). % x ( % ∂x1 ∂x 2 ∂x 3 ( % x12 + x 22 + x 32 ( There are no obvious simplifications here. Therefore, compute the first component and obtain the others by cyclic permutation of the indices. $b × x' ∂ $ b1 x 2 − b2 x1 ' ∂ $ b3 x1 − b1 x 3 ' ∇ ×& 2 ) = & )− & ) % x (1 ∂x 2 % ( ∂x 3 % ( x2 x2 # & # −2 & b −2 b = 12 + (b1 x 2 − b2 x1 )% 4 (2x 2 + 12 − (b3 x1 − b1 x 3 )% 4 (2x 3 $x ' $x ' x x 2 2 2 2b x − 4b1 x 2 + 4b2 x1 x 2 + 4b3 x1 x 3 − 4b1 x 3 2b 4 x = 1 = − 21 + 41 (b1 x1 + b2 x 2 + b3 x 3 ) 4 x x x This field is rotational. The other two components of its curl are: € €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

$b × x' $b × x' 2b 4 x 2b 4 x ∇ × & 2 ) = − 22 + 42 (b1 x1 + b2 x 2 + b3 x 3 ) , ∇ × & 2 ) = − 23 + 43 (b1 x1 + b2 x 2 + b3 x 3 ) . % x (2 % x (3 x x x x

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Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 2.18. Obtain the recipe for the gradient of a scalar function in cylindrical polar coordinates from the integral definition (2.32).

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Solution 2.18. Start from the appropriate form of (2.32), ez 1 "z z ∇Ψ = lim ∫∫ ΨndA , where Ψ is a scalar function of e! V →0 V A "! position x. Here we choose a nearly rectangular volume "R eR V = (RΔϕ)(ΔR)(Δz) centered on the point x = (R, ϕ, z) with sides aligned perpendicular to the coordinate y directions. Here the e unit vector depends on ϕ so its ! direction is slightly different at ϕ ± Δϕ/2. For small Δϕ, R x this can be handled by keeping the linear term of a simple Taylor series: [eϕ ] ≅ eϕ ± (Δϕ 2)(∂eϕ ∂ϕ ) = eϕ  (Δϕ 2)e R . Considering the drawing ϕ

ϕ ±Δϕ 2

and noting that n is an outward normal, there are six contributions to ndA: # $ ΔR & ΔR ' outside = % R + inside = −& R − (ΔϕΔze R , )ΔϕΔze R , $ % 2 ' 2 ( € % % Δϕ ( Δϕ ( e R *, eR * , close vertical side = ΔRΔz'−eϕ − more distant vertical side = ΔRΔz'eϕ − & ) & ) 2 2 top = RΔϕΔRe z , € €and bottom = −RΔϕΔRe z . Here all the unit vectors are evaluated at the center of the volume. Using a two term Taylor series approximation for Ψ on each of the six surfaces, and taking the six contributions in the same € € order, the integral definition becomes a sum of six terms representing ΨndA. € € 5.( 9 1 .( 1 ΔR ∂Ψ +( ΔR + ΔR ∂Ψ +( ΔR + -* R + -e R ΔϕΔz3 − 0* Ψ − -* R − -e R ΔϕΔz3 + 70* Ψ + 7 2 ∂R ,) 2 , 2 ∂R ,) 2 , 2 /) 2 7/) 7 7 77 . 1 . 1 ( 7( 1 Δϕ ∂Ψ +( Δϕ + Δϕ ∂Ψ +( Δϕ + ∇Ψ = lim −e − e ΔRΔz + Ψ + e − e ΔRΔz + 60* Ψ − : * * 3 0 3 - ϕ * - ϕ R R ΔR →0 RΔϕΔRΔz ) , ) , 2 ∂ϕ 2 2 ∂ϕ 2 ) , ) , / 2 / 2 7 7 Δϕ →0 Δz →0 7.( 7 1 .( 1 + + 70* Ψ + Δz ∂Ψ -e z RΔϕΔR3 − 0* Ψ − Δz ∂Ψ -e z RΔϕΔR3 + ... 7 78/) 7; 2 ∂z , 2 ∂z , 2 /) 2

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Here the mean value theorem has been used and all listings of Ψ and its derivatives above are evaluated at the center of the volume. The largest terms inside the big {,}-brackets are proportional to ΔϕΔRΔz. The remaining higher order terms vanish when the limit is taken. /( Ψ 3 ( Ψ R ∂Ψ + R ∂Ψ + e R -ΔϕΔRΔz − *− e R − e R -ΔϕΔRΔz +1 1* e R + ) 2 2 ∂R , 2 ∂R , 1) 2 1 ( eϕ ∂Ψ e R + 1(eϕ ∂Ψ e R + 1 1 ∇Ψ = lim − Ψ-ΔϕΔRΔz + * − Ψ-ΔϕΔRΔz + 0* 4 ΔR →0 RΔϕΔRΔz ) 2 ∂ϕ 2 2 ∂ϕ 2 , ) , 1 1 Δϕ →0 Δz →0 1( R ∂Ψ + 1 ( R ∂Ψ + e z -ΔϕΔRΔz − *− e z -ΔϕΔRΔz + ... 1* 1 ) 2 ∂z , 2) 2 ∂z , 5 'Ψ ∂Ψ 1 ∂Ψ Ψ ∂Ψ * ∂Ψ 1 ∂Ψ ∂Ψ ∇Ψ = ( e R + eR + eϕ − e R + ez + = eR + eϕ + ez ∂R R ∂ϕ R ∂z , ∂R R ∂ϕ ∂z )R

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 2.19. Obtain the recipe for the divergence of a vector function in cylindrical polar coordinates from the integral definition (2.32). Solution 2.19. Start from the appropriate form of (2.32), ez 1 "z z ∇ ⋅ Q = lim ∫∫ n ⋅ QdA , where Q = (QR, Q , Qz) is a vector V →0 V A "! function of position x. Here we choose a nearly rectangular "R volume V = (RΔϕ)(ΔR)(Δz) centered on the point x = (R, ϕ, z) with sides aligned perpendicular to the coordinate directions. Here the e unit vector depends on ϕ so its ! direction is slightly different at ϕ ± Δϕ/2. Considering the x R drawing and noting that n is an outward normal, there are six contributions to ndA: # $ ΔR & ΔR ' outside = % R + (ΔϕΔze R , inside = −& R − )ΔϕΔze R , close vertical side = −ΔRΔz[eϕ ]ϕ −Δϕ 2 , $ % 2 ' 2 ( ϕ

€

ϕ

more distant vertical side = ΔRΔz[eϕ ]

, top = RΔϕΔRe z , and bottom = −RΔϕΔRe z . € otherwise specified. Using Here the unit vectors are evaluated at the center of the volume unless € € a two-term Taylor series approximation for the components of Q on each of the six surfaces, and taking the six contributions to n ⋅ QdA in€the same order, the integral € definition becomes: € 5.( 1 . 1 9 ( ΔR ∂QR +( ΔR + ΔR ∂QR +( ΔR + -* R + -ΔϕΔz3 − 0* QR − -* R − -ΔϕΔz3 +7 70* QR + 2 ∂R ,) 2 , 2 ∂R ,) 2 , 2 /) 2 7 7/) 77 1 .( 1 1 € 77.( Δϕ ∂Qϕ + Δϕ ∂Qϕ + ∇ ⋅ Q = lim −ΔRΔz + Q + ΔRΔz + 60* Qϕ − : )3 0* ϕ 3 -( ΔR →0 RΔϕΔRΔz 2 ∂ϕ 2 ∂ϕ ) , ) , / 2 / 2 7 7 Δϕ →0 Δz →0 7.( 7 1 .( 1 + + 70* Qz + Δz ∂Qz -RΔϕΔR3 − 0*Qz − Δz ∂Qz - RΔϕΔR3 + ... 7 78/) 7; 2 ∂z , 2 ∂z , 2 /) 2 ϕ +Δϕ 2

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Here the mean value theorem has been used and all listings of the components of Q and their derivatives are evaluated at the center of the volume. The largest terms inside the big {,}brackets are proportional to ΔϕΔRΔz. The remaining higher order terms vanish when the limit is taken. 0(QR R ∂QR + 4 ( QR R ∂QR + Δ ϕ ΔRΔz − − − Δ ϕ ΔRΔz + 2* + 2 *) 2 2 ∂R -, 2 ∂R -, 2) 2 2 ( + ( + 2 2 1 1 ∂Qϕ 1 ∂Qϕ ∇ ⋅ Q = lim ΔϕΔRΔz + * ΔϕΔRΔz + 1−*− 5 ΔR →0 RΔϕΔRΔz )2 ∂ϕ , 2 ) 2 ∂ϕ , 2 Δϕ →0 Δz →0 2( R ∂Ψ + 2 ( R ∂Ψ + Δ ϕ ΔRΔz − − Δ ϕ ΔRΔz + ... 2* 2 *) 2 ∂z -, 3) 2 ∂z , 6 &Q ∂Q 1 ∂Qϕ ∂Qz ) 1 ∂ 1 ∂Qϕ ∂Qz ∇⋅Q =' R + R + + + *= (RQR ) + ∂R R ∂ϕ ∂z + R ∂ R R ∂ϕ ∂z (R

e! eR y

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 2.20. Obtain the recipe for the divergence of a vector function in spherical polar coordinates from the integral definition (2.32). z Solution 2.20. Start from the appropriate 1 form of (2.32), ∇ ⋅ Q = lim ∫∫ n ⋅ QdA , V →0 V A #r where Q = (Qr, Q , Q ) is a vector function of position x. Here we choose a nearly er #" rectangular volume V = (rΔ θ )(rsin θ Δ ϕ )(Δr) € " centered on the point x = (r, θ, ϕ) with sides e" aligned perpendicular to the coordinate directions. Here the unit vectors depend on y θ and ϕ so directions are slightly different at e! rsin" ! θ ± Δθ/2, and ϕ ± Δϕ/2. Considering the x #! drawing and noting that n is an outward normal, there are six contributions to ndA: # Δr & # Δr & $ Δr ' $ Δr ' outside = % r + (Δθ% r + ( sin θΔϕ (e r ) , inside = & r − )Δθ& r − ) sin θΔϕ (−e r ) , $ % 2' $ 2' 2( % 2( bottom = [ r sin(θ + Δθ 2)ΔϕΔr](eθ )θ +Δθ 2 , top = [ r sin(θ − Δθ 2)ΔϕΔr](−eθ )θ −Δθ 2 , θ

ϕ

close vertical side = rΔθΔr(−eϕ ) , and more distant vertical side = rΔθΔr(+eϕ ) . ϕ −Δϕ 2 ϕ +Δϕ 2 €Here the unit vectors are evaluated at the center € of the volume unless otherwise specified. Using a two-term Taylor series approximation € for the corresponding components of Q on each of the €

six surfaces produces: % € Δθ ∂Qθ ( $ € Δr ∂Qr ' % Δr ∂Qr ( outside: &Qr + , *(eθ ) )e r , inside: 'Qr − *e r , bottom: 'Qθ + & % & 2 ∂θ ) θ +Δθ 2 2 ∂r ( 2 ∂r ) & & Δθ ∂Qθ ) Δϕ ∂Qϕ ) top: (Qθ − +(eθ )θ −Δθ 2 , close vertical side : (Qϕ − +(−eϕ )ϕ −Δϕ 2 , and ' 2 ∂θ * 2 ∂ϕ * ' Δϕ ∂Qϕ ( € € % € more distant vertical side : 'Qϕ + *(eϕ )ϕ +Δϕ 2 . 2 ∂ϕ ) & to n ⋅ QdA , the integral definition becomes: € € Collecting and summing the six contributions 1 ∇ ⋅ Q = lim × Δr →0 (rΔθ )(r sin θΔϕ )Δr € ΔΔθϕ →0 →0 € 2 2 70* 3 0* 3; Δr ∂Qr - * Δr Δr ∂Qr - * Δr 92,Qr + /Δθ, r + / sin θΔϕ5 − 2,Qr − /Δθ, r − / sin θΔϕ5 9 2 ∂r . + 2. 2 ∂r . + 2. 91+ 4 1+ 49 9 9 3 0* 39 9 0* * * Δθ ∂Qθ Δθ Δθ ∂Qθ Δθ 8+2, Qθ + /r sin,θ + /ΔϕΔr5 − 2, Qθ − / r sin,θ − /ΔϕΔr5< + + 2 ∂θ . 2 . 2 ∂θ . 2 . 4 1+ 49 9 1+ 9 0* 9 3 0* 3 ∂Q ∂Q 9−2, Qϕ − Δϕ ϕ /rΔθΔr5 + 2,Qϕ + Δϕ ϕ / rΔθΔr5 + ... 9 2 ∂ϕ . 2 ∂ϕ . 9: 1+ 9= 4 1+ 4

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Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

The largest terms inside the big {,}-brackets are proportional to ΔθΔϕΔr. The remaining higher order terms vanish when the limit is taken. 0* ∂Q 4 2,r 2 r + 2rQr /Δθ sin θΔϕΔr 2 . 2+ ∂r 2 2 2 * 1 ∂Q ∇ ⋅ Q = lim × 1+,sin θ θ + cos θQθ /rΔθΔϕΔr5 Δr →0 (rΔθ )(r sin θΔϕ )Δr . ∂θ 2 + 2 Δθ →0 Δϕ →0 2 *∂Qϕ 2 2+, 2 /rΔθΔϕΔr + ... 3 + ∂ϕ . 6 Cancel the common factors and take the limit, to find: .' ∂Q * ' * '∂Q * 1 1 ∂Q ∇⋅Q = × /)r 2 r + 2rQr , sin θ + )sin θ θ + cosθQθ ,r + ) ϕ ,r2 + ( + ( ∂ϕ + 3 (r)(r sin θ ) 0( ∂r ∂θ € &∂ ∂Q ) 1 ∂ = 2 × ' ( r 2Qr ) sin θ + r (sin θQθ ) + r ϕ * r sin θ ( ∂r ∂θ ∂ϕ + 1 ∂ 1 ∂ 1 ∂Qϕ € = 2 ( r 2Qr ) + (sin θQθ ) + r ∂r r sin θ ∂θ r sin θ ∂ϕ €

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Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 2.21. Use the vector integral theorems to prove that ∇ ⋅ (∇ × u) = 0 for any twicedifferentiable vector function u regardless of the coordinate system. Solution 2.21. Start with the divergence theorem for a vector € function Q that depends on the spatial coordinates, ∫∫∫ ∇ ⋅ QdV = ∫∫ n ⋅ QdA V

A

n

t1 V

nc1

nc2 t2

where the arbitrary closed volume V has surface A, and the A1 outward normal is n. For this exercise, let Q = ∇ × u so that A2 C ∫∫∫ ∇ ⋅ (∇ × u)dV = ∫∫ n ⋅ (∇ × u) dA . € V A Now split V into two sub-volumes V1 and V2, where the surface of V1 is A1 and the surface of V2 € surfaces, but A1+ A2 = A so: is A2. Here A1 and A2 are not closed ∫∫∫ ∇ ⋅ (∇ × u)dV = ∫∫ n ⋅ (∇ × u) dA + ∫∫ n ⋅ (∇ × u) dA . € V A1 A2 where n is the same as when the surfaces were joined. However, the bounding curve C for A1 and A2 is the same, so Stokes theorem produces: ∫∫∫ ∇ ⋅ (∇ × u) dV = ∫ u ⋅ t1 ds + ∫ u ⋅ t 2 ds . € V

C

C

Here the tangent vectors t1 = nc1 × n and t 2 = nc 2 × n have opposite signs because nc1 and nc2, the normals to C that are tangent to surfaces A1 and A2, respectively, have opposite sign. Thus, the two terms on the right side of the last equation are equal and opposite, so (i) ∫∫∫ ∇ ⋅ (∇ × u)dV = 0 . € € V For an arbitrary closed volume of any size, shape, or location, this can only be true if ∇ ⋅ (∇ × u) = 0 . For example, if ∇ ⋅ (∇ × u) were nonzero at some location, then integration in small volume centered on this location would not be zero. Such a nonzero integral is not allowed € by (i); thus, ∇ ⋅ (∇ × u) must be zero everywhere because V is arbitrary.

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Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 2.22. Use Stokes’ theorem to prove that ∇ × (∇φ ) = 0 for any single-valued twicedifferentiable scalar φ regardless of the coordinate system. Solution 2.22. From (2.34) Stokes Theorem€is: ∫∫ (∇ × u) ⋅ ndA = ∫ u ⋅ tds . A

€

C

Let u = ∇φ , and note that ∇φ ⋅ tds = (∂φ ∂s) ds = dφ because the t vector points along the contour C that has path increment ds. Therefore: (ii) (∇ × [∇φ ]) ⋅ ndA = ∫ ∇φ ⋅ tds = ∫ dφ = 0 , €∫∫ A C C € where the final equality holds for integration on a closed contour of a single-valued function φ. For an arbitrary surface A of any size, shape, orientation, or location, this can only be true if ∇ × ( ∇φ ) = 0 . For example, if ∇ × ( ∇φ ) = 0 were nonzero at some location, then an area € integration in a small region centered on this location would not be zero. Such a nonzero integral is not allowed by (ii); thus, ∇ × ( ∇φ ) = 0 must be zero everywhere because A is arbitrary.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 3.1. The gradient operator in Cartesian coordinates (x, y, z) is: ∇ = e x (∂ ∂x ) + e y (∂ ∂y ) + e z (∂ ∂z) where e x , e y , and e z are the unit vectors. In cylindrical polar coordinates (R, ϕ, z) having the same origin, (see Figure 3.3b), coordinates and unit vectors are related by: R = x 2 + y 2 , ϕ = tan−1 ( y x ) , and z = z; and e R = e x cosϕ + e y sin ϕ , € the following in the cylindrical polar coordinate eϕ = −e x sin ϕ + e y cosϕ , and e z€= e€ z . Determine system. eϕ ∂ϕ a) ∂e R ∂ϕ and ∂€ € € b) the gradient operator ∇ € c) the divergence of the velocity field ∇⋅u d) the Laplacian operator ∇ ⋅ ∇ ≡ ∇ 2 € e) the advective acceleration term (u⋅∇)u

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€ €

Solution 3.1. The € Cartesian unit vectors do not depend on the coordinates so the unit vectors from the cylindrical coordinate system can be differentiated when they are written in terms of ex, ey, and ez. a) First work with eR, use the given unit vector definition, and proceed with straightforward differentiation. The variables R and z do not appear in the formula for eR, so ∂e R ∂R = ∂e R ∂z = 0 . However eR does depend on the angle ϕ. Thus, ∂e R ∂ = (e x cosϕ + e y sinϕ ) = −e x sinϕ + e y cosϕ = eϕ . ∂ϕ ∂ϕ Proceed to determine the derivatives of eϕ. Again note that the variables R and z do not appear in the formula for eϕ, so ∂eϕ ∂R = ∂eϕ ∂z = 0 . However, like eR, eϕ does depend on the angle ϕ. Thus, € ∂eϕ ∂ = (−e x sinϕ + e y cosϕ ) = −e x cosϕ − e y sinϕ = −e R . ∂ϕ ∂ϕ € The third unit vector, ez, is the same as the Cartesian unit vector and does not depend on the coordinates. b) Start by constructing the expressions for ex, ey, and ez in terms of eR, eϕ, and ez. This can be € $ cosϕ sin ϕ 0($e x ( $e R ( & && & & & done my inverting the linear system %−sin ϕ cosϕ 0)%e y ) = % eϕ ) to find & 0 0 1&*&'e z &* &' e z &* '

€

e x = e R cosϕ − eϕ sin ϕ , e y = e R sin ϕ + eϕ cos ϕ , and e z = e z The next step is to use the coordinate definitions: € R = x 2 + y 2 , ϕ = tan−1 ( y x ) , and z = z to transform € € the Cartesian partial derivatives. € % y( ∂ ∂ ∂R ∂ ∂ϕ ∂ ∂z ∂ x ∂ 1 ∂ = + + = − − + (0) ' * 2 2 ∂x ∂x ∂R ∂x ∂ϕ ∂x ∂z ∂z x 2 + y 2 ∂R 1+ ( y x ) & x ) ∂ϕ € € ∂ sin ϕ ∂ = cosϕ − ∂R R ∂ϕ € €

(1,2,3) (4,5,6)

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

$ 1' ∂ ∂ ∂R ∂ ∂ϕ ∂ ∂z ∂ y ∂ 1 ∂ = + + = + + (0) & ) 2 ∂y ∂y ∂R ∂y ∂ϕ ∂y ∂z ∂z x 2 + y 2 ∂R 1+ ( y x ) % x ( ∂ϕ ∂ cosϕ ∂ = sin ϕ + ∂R R ∂ϕ ∂ ∂R ∂ ∂ϕ ∂ ∂z ∂ ∂ ∂ ∂ ∂ € = + + = (0) + (0) + (1) = ∂z ∂z ∂R ∂z ∂ϕ ∂z ∂z ∂R ∂ϕ ∂z ∂z ∂ ∂ ∂ Now€reassemble the gradient operator ∇ = e x + e y + e z using the cylindrical coordinate ∂x ∂y ∂z unit vectors and differentiation definitions: € & & ∂ sinϕ ∂ ) ∂ cos ϕ ∂ ) ∂ ∇ = (e R cos ϕ − eϕ sin ϕ )( cosϕ − + e sin ϕ + e cos ϕ sin ϕ + ( ) + ( + + ez . R ϕ R ∂ϕ * ∂R R ∂ϕ * ∂z ' € ∂R ' Collect all of the terms with like unit vectors and differential operators together: ∂ 1 ∂ ∇ = e R (cos2 ϕ + sin 2 ϕ ) + e R (−cosϕ sin ϕ + sin ϕ cosϕ ) ∂R R ∂ϕ € ∂ 1 ∂ ∂ eϕ (−sin ϕ cos ϕ + cosϕ sin ϕ ) + eϕ (sin 2 ϕ + cos2 ϕ ) + ez ∂R R ∂ϕ ∂z The terms in (,)-parentheses are either +1 or 0. When evaluated they produce: € ∂ 1 ∂ ∂ ∇ = eR + eϕ + ez . ∂R R ∂ϕ ∂z € c) In cylindrical coordinates, the divergence of the velocity is: & ∂ 1 ∂ ∂) ∇ ⋅ u = (e R + eϕ + e z + ⋅ (e R uR + eϕ uϕ + e z uz ) . R ∂ϕ ∂z * € ' ∂R Further simplification requires that both the unit vectors and the u's be differentiated. Completing this task term by term produces: ∂ ∂u ∂e ∂u e R € ⋅ (e R uR ) = e R ⋅ e R R + uR e R ⋅ R = R , ∂R ∂r ∂R ∂ R ∂ u ∂ e ∂ eR ⋅ (eϕ uϕ ) = e R ⋅ eϕ ϕ + uϕ e R ⋅ ϕ = 0 , ∂R ∂R ∂R ∂uz ∂e z ∂ eR ⋅ (e z uz ) = e R ⋅ e z + uze R ⋅ = 0, € ∂R ∂R ∂r eϕ ∂ e ⋅ e ∂u u ∂e u u ⋅ (e R uR ) = ϕ R R + R eϕ ⋅ R = 0 + R eϕ ⋅ eϕ = R , € R ∂ϕ R ∂ϕ R ∂ϕ R R e e ⋅ e ∂ u u ∂ e ∂ u u ∂ 1 1 ∂uϕ ϕ ϕ € ⋅ (eϕ uϕ ) = ϕ ϕ ϕ + ϕ eϕ ⋅ ϕ = − ϕ eϕ ⋅ e R = , R ∂ϕ R ∂ϕ R ∂ϕ R ∂ϕ R R ∂ϕ eϕ ∂ e ⋅ e ∂u u ∂e € ⋅ (e z uz ) = ϕ z ϕ + z eϕ ⋅ z = 0 R ∂ϕ R ∂ϕ R ∂ϕ ∂ ∂ u ∂ e € e z ⋅ (e R uR ) = e z ⋅ e R R + uR e z ⋅ R = 0 , ∂z ∂z ∂z ∂u ∂e ∂ € e z ⋅ (eϕ uϕ ) = e z ⋅ eϕ ϕ + uϕ e z ⋅ ϕ = 0 , and ∂z ∂z ∂z ∂uz ∂e z ∂uz ∂ € e z ⋅ (e z uz ) = e z ⋅ e z + uze z ⋅ = . ∂z ∂z ∂z ∂ z €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Reassembling the equation produces:

∂uR uR 1 ∂uϕ ∂uz + + + . ∂R R R ∂ϕ ∂z The 1st & 2nd terms on the right side are commonly combined to yield: 1 ∂ 1 ∂uϕ ∂uz ∇⋅u= + . (10) (RuR ) + R ∂ R R ∂ϕ ∂ z € d) The Laplacian operator is ∇ 2 ≡ ∇ ⋅ ∇ , and its form in cylindrical coordinates can be found by evaluating the dot product. Fortunately, the results of part c) can be used via the following replacements for the second gradient operator of the dot product: € 1 ∂ ∂ ∂ € , uϕ ↔ , and uz ↔ . (7,8,9) uR ↔ R ∂ϕ ∂z ∂R Inserting the replacements (7,8,9) into (10) produces: 1 ∂ $ ∂ ' 1 ∂2 ∂2 ∇2 = &R ) + 2 2 + 2 . R ∂R % ∂R ( €R ∂ϕ ∂z € € e) Start with the answer to part b) and compute the first dot product to find: ∂ 1 ∂ ∂ u ⋅ ∇ = uR + uϕ + uz ∂R R ∂ϕ ∂z € This is the scalar operator applied to u = uR e R + uϕ eϕ + uz e z to find the advective acceleration: & ∂ 1 ∂ ∂) + uz +( uR e R + uϕ eϕ + uze z ) . (u ⋅ ∇)u = ( uR + uϕ R ∂ϕ ∂z * € ' ∂R Here the components of u and the unit vectors eR and eϕ depend on the angular coordinate. & ∂u 1 ∂ uR ∂u ) 1 ∂e R + uz R + + uR uϕ + (u ⋅ ∇)u = e R ( uR R + uϕ ∂ R R ∂ϕ ∂ z R ∂ϕ ' * € ! ∂u ! ∂u ∂u $ ∂u $ 1 ∂ uϕ 1 ∂ eϕ 1 ∂ uz eϕ # uR ϕ + uϕ + uz ϕ & + uϕ2 + e z # uR z + uϕ + uz z & R ∂ϕ ∂z % R ∂ϕ R ∂ϕ ∂z % " ∂R " ∂R Use the results of part a) to evaluate the unit vector derivatives. & ∂u e 1 ∂uR ∂u ) + uz R + + uR uϕ ϕ + (u ⋅ ∇)u = e R ( uR R + uϕ R ∂ϕ ∂z * R ' ∂R $ ∂uϕ ' $ ∂u ∂u ∂u ' 1 ∂uϕ e 1 ∂uz eϕ & uR + uϕ + uz ϕ ) − uϕ2 R + e z & uR z + uϕ + uz z ) R ∂ϕ ∂z ( R R ∂ϕ ∂z ( % ∂R % ∂R Collect components ' ∂u u2 * 1 ∂uR ∂u + uz R − ϕ , + (u ⋅ ∇)u = e R ) uR R + uϕ R ∂ϕ ∂z R+ ( ∂R € $ ∂u $ ∂u ∂u u u ' ∂u ' 1 ∂uϕ 1 ∂uz eϕ & uR ϕ + uϕ + uz ϕ + R ϕ ) + e z & uR z + uϕ + uz z ) R ∂ϕ ∂z R ( R ∂ϕ ∂z ( % ∂R % ∂R ∇⋅u=

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Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 3.2. Consider Cartesian coordinates (as given in Exercise no. 1) and spherical polar coordinates (r, θ, ϕ) having the same origin (see Figure 3.3c). Here coordinates and unit vectors are related by: r = x 2 + y 2 + z 2 , θ = tan−1

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x 2 + y 2 z , and ϕ = tan−1 ( y x ) ; and

e r = e x cosϕ sin θ + e y sin ϕ sin θ + e z cos θ , eθ = e x cosϕ cos θ + e y sin ϕ cos θ − e z sin θ , and eϕ = −e x sin ϕ + e y cosϕ . In the spherical polar coordinate system, determine the following items. a) ∂e€ r ∂θ , ∂e r ∂ϕ , ∂eθ€∂θ , ∂eθ ∂ϕ , and ∂eϕ ∂ϕ € b) the gradient operator ∇ € c) the divergence of the velocity field ∇⋅u d)€the Laplacian ∇ ⋅€∇ ≡ ∇ 2 € € e) the advective acceleration term (u⋅∇)u

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Solution € 3.2. The Cartesian unit vectors do not depend on the coordinates so the unit vectors from the spherical coordinate system can be differentiated when they are written in terms of ex, ey, and ez. a) First work with er, use the given unit vector definition, and proceed with straightforward differentiation. The variable r doesn’t even appear in the formula for er, so ∂e r ∂r = 0 . However er does depend on both angles. Thus, ∂e r ∂ = (e x cosϕ sin θ + e y sin ϕ sin θ + e z cos θ ) = e x cos ϕ cosθ + e y sin ϕ cos θ − e z sin θ = eθ ∂θ ∂θ € and, ∂e r ∂ = (e x cosϕ sinθ + e y sinϕ sinθ + e z cosθ ) = −e x sinϕ sinθ + e y cosϕ sinθ = eϕ sinθ . ∂ϕ ∂ϕ Proceed to determine the derivatives of eθ. Again note that the variable r doesn’t appear in its formula, so ∂eθ ∂r = 0 . However, like er, eθ does depend on both angles. Thus, ∂eθ ∂ = (e x cosϕ cos θ + e y sin ϕ cos θ − e z sin θ ) = −e x cosϕ sin θ − e y sin ϕ sin θ − e z cos θ = −e r ∂θ ∂θ and, € ∂e ∂ θ = (e x cosϕ cosθ + e y sinϕ cosθ − e z sinθ ) = −e x sinϕ cosθ + e y cosϕ cosθ = eϕ cosθ . ∂ϕ ∂ϕ Now consider eϕ and note that the variables r and θ don’t appear in its formula, so ∂eϕ ∂r = ∂eϕ ∂θ = 0 . However, eϕ does depend on ϕ. Thus, ∂eϕ ∂ = ($) (−e x sinϕ + e y cosϕ ) = −e x cosϕ − e y sinϕ . ∂ϕ ∂ϕ The question now is how to relate the right side of this equation back to er, and eθ [note: because eϕ ⋅ ∂eϕ ∂ϕ = 0 , ∂eϕ ∂ϕ can only be a linear combination of er and eθ]. Assuming ∂eϕ ∂ϕ = ae r + be € θ , then ($) and the unit vector definitions require: acosϕ sin θ + bcosϕ cos θ = −cosϕ , asin ϕ sin θ + bsin ϕ cosθ = −sin ϕ , and € acosθ − bsin θ = 0 . After dividing out common factors, the first two equations are the same: asin θ + bcos θ = −1. When this simplified € equation is combined with the third equation, we obtain: a = −sin θ and b = −cosθ . Thus, ∂€eϕ ∂ϕ = −e r sin θ − eθ cos θ . € € €

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Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

b) Start by constructing the expressions for ex, ey, and ez in terms of e r , eθ , and eϕ . This can be % cosϕ sin θ sin ϕ sin θ cosθ )%e x ) % e r ) ' '' ' ' ' done my inverting the linear system &cos ϕ cosθ sin ϕ cosθ −sin θ *&e y * = &eθ * to find ' −sin ϕ cosϕ € € 0 '+'(€e z '+ '(eϕ '+ (

e x = e r cosϕ sin θ + eθ cosϕ cos θ − eϕ sin ϕ e y = e r sin ϕ sin θ + eθ sin ϕ cosθ + eϕ cos ϕ e z = e r cosθ − eθ sin θ . € The next step is to€use the coordinate definitions:

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2 2 2 −1 x2 + y2 z € r = x + y + z , θ = tan € to transform the Cartesian partial derivatives. ∂ ∂r ∂ ∂θ ∂ ∂ϕ ∂ x ∂ 1 = + + = + ∂x ∂x ∂r ∂x ∂θ ∂x ∂ϕ r ∂r 1+ x 2 + y 2 z 2 2z € € € ∂ cos ϕ cosθ ∂ sin ϕ ∂ = cosϕ sin θ + − ∂r r ∂θ r sin θ ∂ϕ ∂ ∂r ∂ ∂θ ∂ ∂ϕ ∂ y ∂ 1 = + + = + ∂y ∂y ∂r ∂y ∂θ ∂y ∂ϕ r ∂r 1+ x 2 + y 2 z 2 2z

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∂ sin ϕ cos θ ∂ cos ϕ ∂ + + ∂r r ∂θ r sin θ ∂ϕ ∂ ∂r ∂ ∂θ ∂ ∂ϕ ∂ z ∂ 1 = + + = + ∂z ∂z ∂r ∂z ∂θ ∂z ∂ϕ r ∂r 1+ x 2 + y 2 z

(1,2,3)

) , and ϕ = tan

−1

( y x)

2x

∂ y ∂ − 2 2 x + y ∂θ x + y ∂ϕ 2

2

2y

∂ x ∂ + 2 2 x + y ∂θ x + y ∂ϕ 2

2

= sin ϕ sin θ

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Now reassemble the gradient operator ∇ = e x

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2

− x2 + y2 ∂ ∂ sin θ ∂ = cos θ − 2 z ∂θ ∂r r ∂θ

∂ ∂ ∂ + e y + e z using the spherical coordinate unit ∂x ∂y ∂z

vectors and differentiation definitions:

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(4,5,6)

' ∂ cos ϕ cosθ ∂ sin ϕ ∂ * ∇ = (e r cos ϕ sin θ + eθ cos ϕ cosθ − eϕ sin ϕ )) cosϕ sin θ + − , ∂r r ∂θ r sin θ ∂ϕ + ( € % ∂ sin ϕ cosθ ∂ cos ϕ ∂ ( +(e r sin ϕ sin θ + eθ sin ϕ cosθ + eϕ cos ϕ )'sin ϕ sin θ + + * ∂r r ∂θ r sinθ ∂ϕ ) & % ∂ sin θ ∂ ( +(e r cosθ − eθ sin θ )'cos θ − *. & ∂r r ∂θ ) Collect all of the terms with like unit vectors and differential operators together: ∂ ∇ = e r (cos2 ϕ sin 2 θ + sin 2 ϕ sin 2 θ + cos 2 θ ) ∂r 1 ∂ +e r (cos 2 ϕ cosθ sin θ + sin 2 ϕ cosθ sin θ − cosθ sin θ ) r ∂θ 1 ∂ +e r (−cos ϕ sin ϕ sin θ + sin ϕ cosϕ sin θ ) r sin θ ∂ϕ ∂ +eθ (cos 2 ϕ cosθ sin θ + sin 2 ϕ cosθ sin θ − sin θ cosθ ) ∂r

Fluid Mechanics, 6th Ed.

1 ∂ r ∂θ 1 ∂ +eθ (−cos ϕ sin ϕ cos θ + sin ϕ cosϕ cos θ ) r sin θ ∂ϕ ∂ +eϕ (−sin ϕ cosϕ sin θ + cos ϕ sin ϕ sin θ ) ∂r 1 ∂ +eϕ (−sin ϕ cosϕ cos θ + cosϕ sin ϕ cos θ ) r ∂θ 1 ∂ 2 2 +eϕ (sin ϕ + cos ϕ ) r sin θ ∂ϕ The terms in (,)-parentheses are either +1 or 0. When evaluated they produce: ∂ 1 ∂ 1 ∂ ∇ = e r + eθ + eϕ . ∂r r ∂θ r sin θ ∂ϕ c) In spherical coordinates, the divergence of the velocity is: ' ∂ 1 ∂ 1 ∂ * ∇ ⋅ u = )e r + eθ + eϕ , ⋅ (e r ur + eθ uθ + eϕ uϕ ) . ∂ r r ∂θ r sin θ ∂ϕ ( + € Further simplification requires that the unit vectors and the u's be differentiated. Completing this task term by term produces: ∂ ∂u ∂e ∂ u e r ⋅ (e r€ ur ) = e r ⋅ e r r + ure r ⋅ r = r , ∂r ∂r ∂r ∂ r ∂ ∂uθ ∂e e r ⋅ (eθ uθ ) = e r ⋅ eθ + uθ e r ⋅ θ = 0 , ∂r ∂r ∂r ∂uϕ ∂eϕ ∂ e r ⋅ (eϕ uϕ ) = e r ⋅ eϕ + uϕ e r ⋅ = 0, ∂r ∂r ∂r eθ ∂ e ⋅ e ∂u u ∂e u u ⋅ (e r ur ) = θ r r + r eθ ⋅ r = 0 + r eθ ⋅ eθ = r , r ∂θ r ∂θ r ∂θ r r eθ ∂ eθ ⋅ eθ ∂uθ uθ ∂eθ 1 ∂uθ uθ 1 ∂uθ , ⋅ (eθ uθ ) = + eθ ⋅ = − eθ ⋅ e r = r ∂θ r ∂θ r ∂θ r ∂θ r r ∂θ e ⋅ e ∂u u ∂e eθ ∂ ⋅ (eϕ uϕ ) = θ ϕ ϕ + ϕ eθ ⋅ ϕ = 0 r ∂θ r ∂θ r ∂θ eϕ ∂ eϕ ⋅ e r ∂ur ur ∂e ur u ⋅ (e r ur ) = + eϕ ⋅ r = eϕ ⋅ (eϕ sin θ ) = r , r sin θ ∂ϕ r sin θ ∂ϕ r sin θ ∂ϕ r sin θ r eϕ ∂ eϕ ⋅ eθ ∂uθ u ∂e u u ⋅ (eθ uθ ) = + θ eϕ ⋅ θ = θ eϕ ⋅ eϕ cosθ = θ , r sin θ ∂ϕ r sin θ ∂ϕ r sin θ ∂ϕ r sin θ r tan θ & ) eϕ ∂ ∂uϕ ∂e 1 ⋅ (eϕ uϕ ) = + uϕ eϕ ⋅ ϕ + ( eϕ ⋅ eϕ r sin θ ∂ϕ r sin θ ' ∂ϕ ∂ϕ * * 1 ' ∂uϕ 1 ∂uϕ = − uϕ eϕ ⋅ (e r sin θ + eθ cos θ ), = ) r sin θ ( ∂ϕ + r sin θ ∂ϕ Reassembling the equation produces: ∂u 2u 1 ∂uθ u 1 ∂uϕ ∇⋅u= r + r + + θ + ∂r r r ∂θ r tan θ r sin θ ∂ϕ € The 1st & 2nd terms, and the 3trd and 4th terms on the right side are commonly combined to yield: +eθ (cos 2 ϕ cos2 θ + sin 2 ϕ cos2 θ + sin 2 θ )

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Kundu, Cohen, and Dowling

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Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

1 ∂ 2 1 ∂ 1 ∂uϕ r ur ) + . (10) (sin θ uθ ) + ( 2 r ∂r r sin θ ∂θ r sin θ ∂ϕ d) The Laplacian operator is ∇ 2 ≡ ∇ ⋅ ∇ , and its form in spherical polar coordinates can be found by evaluating the dot product. Fortunately, the results of part c) can be used via the following replacements for the second gradient operator of the dot product: € 1 ∂ ∂ 1 ∂ € , uθ ↔ , and uϕ ↔ . (7,8,9) ur ↔ r sin θ ∂ϕ ∂r r ∂θ Inserting (7,8,9) into (10), the Laplacian then becomes: 1 ∂$ ∂' 1 ∂ $ ∂ ' 1 ∂2 ∇2 = 2 &r2 ) + 2 . & sin θ )+ 2 2 r € ∂r % ∂r ( r sin θ€∂θ % ∂θ ( r sin θ ∂ϕ 2 € e) Start with the answer to part b) and compute the first dot product to find: ∂ 1 ∂ 1 ∂ u ⋅ ∇ = ur + uθ + uϕ ∂r r ∂θ r sin θ ∂ϕ € This is the scalar operator applied to u = ure r + uθ eθ + uϕ eϕ to find the advective acceleration: ' ∂ 1 ∂ 1 ∂ * + uϕ (u ⋅ ∇)u = ) ur + uθ ,( ure r + uθ eθ + uϕ eϕ ) . r ∂θ r sin θ ∂ϕ + ( ∂r € Here the components of u and € the unit vectors depend on the angular coordinates. ' ∂ur 1 ∂ur 1 ∂ ur * 1 ∂e r 1 ∂e r + uϕ + ur uϕ + (u ⋅ ∇)u = e r ) ur + uθ , + ur uθ r ∂θ r sin θ ∂ϕ + r ∂θ r sin θ ∂ϕ € ( ∂r % ∂u 1 ∂uθ 1 ∂uθ ( 2 1 ∂eθ 1 ∂eθ eθ ' ur θ + uθ + uϕ + uθ uϕ + * + uθ r ∂θ r sin θ ∂ϕ ) r ∂θ r sin θ ∂ϕ & ∂r % ∂u 1 ∂uϕ 1 ∂uϕ ( 1 ∂eϕ 1 ∂eϕ eϕ ' ur ϕ + uθ + uϕ + uϕ2 * + uϕ uθ r ∂θ r sin θ ∂ϕ ) r ∂θ r sin θ ∂ϕ & ∂r Use € the results of part a) to evaluate the unit vector derivatives. ' ∂u e sin θ 1 ∂ur 1 ∂ur * eθ + uϕ + ur uϕ ϕ + (u ⋅ ∇)u = e r ) ur r + uθ , + ur uθ r ∂θ r sin θ ∂ϕ + r r sin θ ( ∂r € % ∂u eϕ cosθ 1 ∂uθ 1 ∂uθ ( 2 e r eθ ' ur θ + uθ + uϕ * − uθ + uθ uϕ r ∂θ r sin θ ∂ϕ ) r r sin θ & ∂r % ∂uϕ ( 1 ∂uϕ 1 ∂uϕ 1 2 eϕ ' ur + uθ + uϕ (−e r sin θ − eθ cosθ ) * + 0 + uϕ r ∂θ r sin θ ∂ϕ ) r sin θ & ∂r Collect components € 2 2 ( ∂u 1 ∂ur 1 ∂ur uθ + uϕ + + uϕ − -+ (u ⋅ ∇)u = e r * ur r + uθ r ∂θ r sin θ ∂ϕ r , ) ∂r € 2 & ∂u ) 1 ∂uθ 1 ∂uθ ur uθ uϕ θ eθ ( ur + uθ + uϕ + − cot θ + r ∂θ r sin θ ∂ϕ r r ' ∂r * % ∂uϕ ( 1 ∂uϕ 1 ∂uϕ ur uϕ uθ uϕ eϕ ' ur + uθ + uϕ + + cot θ * r ∂θ r sin θ ∂ϕ r r & ∂r ) € ∇⋅u=

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Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 3.3. In a steady two-dimensional flow, Cartesian-component particle trajectories are given by: x(t) = ro cos (γ (t − to ) + θ o ) and y(t) = ro sin (γ (t − to ) + θ o ) where ro = xo2 + yo2 and

θ o = tan −1 ( yo xo ) . a) From these trajectories determine the Lagrangian particle velocity components u(t) = dx/dt and v(t) = dy/dt, and convert these to Eulerian velocity components u(x,y) and v(x,y). b) Compute Cartesian particle acceleration components, ax = d2x/dt2 and ay = d2y/dt2, and show that they are equal to D/Dt of the Eulerian velocity components u(x,y) and v(x,y). Solution 3.3. a) Differentiate as suggested to find: dx(t) dy(t) u(t) = = −γ ro sin (γ (t − to ) + θ o ) and v(t) = = γ ro cos (γ (t − to ) + θ o ) . dt dt Now use the original trajectory equations to eliminate the trig-functions: u = −γ y and v = γ x . b) Again differentiate as suggested to find: d 2 x(t) ax (t) = = −γ 2 ro cos (γ (t − to ) + θ o ) = −γ 2 x and 2 dt d 2 y(t) ay (t) = = −γ 2 ro sin (γ (t − to ) + θ o ) = −γ 2 y . 2 dt Compute Du/Dt and Dv/Dt from the final two answers of part a): Du ∂u ∂u ∂u = + u + v = 0 − γ y(0) + γ x(−γ ) = −γ 2 x and Dt ∂t ∂x ∂y Dv ∂v ∂v ∂v = + u + v = 0 − γ y(γ ) + γ x(0) = −γ 2 y . Dt ∂t ∂x ∂y The final equalities match as appropriate: ax = Du/Dt, and ay = Dv/Dt.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 3.4. In a steady two-dimensional flow, polar coordinate particle trajectories are given by: r(t) = ro and θ (t) = γ (t − to ) + θ o . a) From these trajectories determine the Lagrangian particle velocity components ur(t) = dr/dt and, uθ (t) = rdθ/dt, and convert these to Eulerian velocity components ur(r,θ) and uθ (r, θ ) . 2

b) Compute polar-coordinate particle acceleration components, ar = d 2 r dt 2 − r ( dθ dt ) and aθ = rd 2θ dt 2 + 2 ( dr dt ) ( dθ dt ) , and show that they are equal to D/Dt of the Eulerian velocity

with components ur(r,θ) and uθ (r, θ ) . Solution 3.4. a) Differentiate as suggested to find: dr(t) dθ (t) ur (t) = = 0 and uθ (t) = r = rγ . dt dt These equations are readily interpreted as Eulerian velocity components: ur = 0 and uθ = γ r . b) Start with the particle accelerations: 2

ar = d 2 r dt 2 − r ( dθ dt ) = 0 − r(γ )2 = −γ 2 r and aθ = rd 2θ dt 2 + 2 ( dr dt ) ( dθ dt ) = 0 + 2(0)γ = 0 . Compute Dur/Dt and Duθ/Dt from the forms given in Appendix B and the two answers of part a): Dur ∂ur ∂u u ∂u u 2 (γ r)2 = + ur r + θ r − θ = 0 + 0 − = −γ 2 r and Dt ∂t ∂r r ∂θ r r Duθ ∂uθ ∂uθ uθ ∂uθ ur uθ = + ur + + = 0 + 0(γ ) + γ (0) + 0 = 0 . Dt ∂t ∂r r ∂θ r The final equalities match as appropriate: ar = Dur/Dt, and aθ = Duθ/Dt.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 3.5. if ds = (dx, dy, dz) is an element of arc length along a streamline (Figure 3.5) and u = (u, v, w) is the local fluid velocity vector, show that if ds is everywhere tangent to u then dx u = dy v = dz w .

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Solution 3.5. If ds = (dx, dy, dz) and u are parallel, then they must have the same unit tangent vector t: ds (dx,dy,dz) (u,v,w) u . t= = = = ds (dx) 2 + (dy) 2 + (dz) 2 u2 + v 2 + w 2 u The three components of this equation imply: dx u dy v dz w = , = , and = . ds u ds u ds u € But these can be rearranged to find: ds dx dy dz = = = . u u€ v w € €

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Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 3.6. For the two-dimensional steady flow having velocity components u = Sy and v = Sx, determine the following when S is a positive real constant having units of 1/time. a) equations for the streamlines with a sketch of the flow pattern b) the components of the strain rate tensor c) the components of the rotation tensor d) the coordinate rotation that diagonalizes the strain rate tensor, and the principal strain rates. e) How is this flow field related to that in Example 3.5. Solution 3.6. a) For steady streamlines in two dimensions: dy v Sx x dx dy = = = , which implies: or = dx u Sy y u v ydy = xdx → y 2 2 = x 2 2 + const.

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y

Solving for y(x) produces: y = ± x 2 + const . These are hyperbolae that asymptote to the lines y = ±x. € b) Compute the strain rate tensor from its definition: € 1 # ∂u ∂u & Sij = %% i + j ((€ x 2 $ ∂x j ∂x i ' 1 # ∂u ∂x ∂u ∂y + ∂v ∂ x ) & 2( = $1 ' ∂v ∂y % 2 (∂v ∂x + ∂u ∂y ) ( 1 " 0 S + S )% " 0 S % 2( = #1 &=# & 0 ' $S 0' $ 2 ( S + S) c) Compute the rotation tensor from its definition: 0 ∂u ∂ y − ∂v ∂x ' $ 0 S − S ' $0 0' ∂u ∂ u $ Rij = i − j = % (=% (=% ( 0 0 ) &0 0) ∂x j ∂x i &∂v ∂x − ∂u ∂y ) &S − S d) From Example 2.4, a θ = 45° coordinate rotation diagonalizes the strain rate tensor. The $cosθ −sin θ ' 1 $1 −1' direction cosine matrix is: Cij = % (= % ( , and the rotated strain rate matrix 2 &1 1 ) & sin θ cosθ ) S´ is: 1 % 1 1(% 0 S ( 1 %1 −1( 1 % S S (%1 −1( % S 0 ( S" = CT ⋅ S ⋅ C = & )& ) & )= & )& )=& ). 2 '−1 1*' S 0 * 2 '1 1 * 2 ' S −S *'1 1 * ' 0 −S * € e) When this flow field is rotated 45° in the clockwise direction, it is the same as the flow field in Example 3.5.

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Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 3.7. At the instant shown in Figure 3.2b, the (u,v)-velocity field in Cartesian coordinates is u = A(y 2 − x 2 ) (x 2 + y 2 ) 2 , and v = −2A xy (x 2 + y 2 ) 2 where A is a positive constant. Determine the equations for the streamlines by rearranging the first equality in (3.7) to read udy − vdx = 0 = (∂ψ ∂y ) dy + (∂ψ ∂x ) dx and then looking for a solution in the form ψ(x,y) = const. € €

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Solution 3.7. Rearrange the two-dimensional streamline condition, dx/u = dy/v, to obtain udy – vdx = 0 as the description of a streamline. Assume this differential equation is solved by the function ψ(x,y) = const, so that (∂ψ/∂x)dx + (∂ψ/∂y)dy = 0. Comparing the two equations requires: u = ∂ψ/∂y , and v = –∂ψ/∂x. Now use the given velocity field to find: (a,b) ∂ψ ∂y = A(y 2 − x 2 ) (x 2 + y 2 ) 2 , and ∂ψ ∂x = +2A xy (x 2 + y 2 ) 2 . Integrate (b) treating y as a constant: ' −1 * ∂ψ 2x 2xdx = Ay 2 → ψ − ψ o = Ay ∫ 2 = Ay) 2 2 2 2 2 2 ,, ∂ x (x + y ) (x + y ) x + y ( + € € where ψo may depend on y. Differentiate this result with respect to y to determine ψo: ∂ ∂ % −Ay ( −A (−Ay) A(y 2 − x 2 ) (ψ − ψ o ) = ' 2 2 * = 2 2 − 2 2 2 (2y) = 2 2 2 = u . ∂y ∂y & x + y ) x + y (x + y ) (x + y ) € This result and equation (a) implies ∂ψo/∂y = 0, so it is at most a constant. Thus, the streamlines are given by: Ay ψ (x, y) = const. = − 2 . € x + y2

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Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 3.8. Determine the equivalent of the first equality in (3.7) for two dimensional (r,θ)polar coordinates, and then find the equation for the streamline that passes through (ro, θo) when u = (ur, u ) = (A/r, B/r) where A and B are constants. θ

Solution 3.8. The two-dimensional streamline condition in Cartesian coordinates is dx/u = dy/v, and is obtained from considering the streamline-tangent vector t: e dx + e dy e u+e v u ds t= = x 2 y 2 = x 2 y2 = . ds u (dx) + (dy) u +v In two-dimensional polar coordinates this becomes: ds e dr + e rdθ e u +e u u t= = r 2 θ = r r 2 θ 2θ = . 2 ds u (dr) + (rdθ ) ur + uθ € Equating components produces two equations: ds dr rdθ dr ur rdθ uθ and , or . = = = = u ur uθ ds u ds u € Thus, using the last equality and the given velocity field: 1 dr ur A r A A = = = → ln(r) = θ + const. r dθ € uθ B r B B € € The initial condition allows the constant to be evaluated: "r% A $A ' A ln(ro ) = θ o + const., which leads to ln$ ' = (θ − θ o ) or r = ro exp& (θ − θ o )) . %B ( # ro & B €B

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Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 3.9. Determine the streamline, path line, and streak line that pass through the origin of coordinates at t = t´ when u = Uo + ωξocos(ωt) and v = ωξosin(ωt) in two-dimensional Cartesian coordinates where Uo is a constant horizontal velocity. Compare your results those in Example. 3.3 for U o → 0 . Solution 3.9. (i) For the streamline, time is a constant. Use the first equality of (3.7) to find: dy v ωξ o sin(ωt) = = = m(t) , € dx u U o + ωξ o cos(ωt) where m is the streamline slope. Since m does not depend on the spatial coordinate, this equation is readily integrated to find straight time-dependent streamlines: y = m(t)x + const. Thus, the streamline that passes through (0,0) at t = t´ is: € ωξ o sin(ωt $) y= x. U o + ωξ o cos(ωt $) (ii) For the path line, use both components of (3.8): dx dy = U o + ωξ o cos(ωt) and = ωξ o sin(ωt) , dt dt € and integrate in time to find: x − x o = U o t + ξ o sin(ωt) and y − y o = –ξ o cos(ωt) . Determine xo and yo by requiring the path line to pass through the origin at at t = t´: € €t #) and 0 − y = –ξ cos(ωt %) . 0 − x o = U o t # + ξ o sin(ω o o The final component equations are: € € x = U o (t − t #) + ξ o (sin(ωt) − sin(ωt #)) and y = –ξ o (cos(ωt) − cos(ωt %)) . These two parametric equations for x(t) and y(t) can be combined to eliminate some of the t€ € dependence: 2 2 x − U o (t − t #) + ξ o sin(ωt #)) + ( y − ξ o cos(ωt #)) = ξ o2 , ( € € which describes a moving circle with center located at (U o (t − t #) − ξ o sin(ωt #),ξ o cos(ωt #)) . (iii) For the streak line, use the path line results but this time evaluate the constants at t = to instead of at t = t´ to find: € x = U o (t − t o ) + ξ o (sin(ωt) − sin(ωt o )) and y = –ξ o (cos(ωt) − cos(ωt o )) . € Now evaluate these equations at t = t´ to produce two parametric equations for the streak line coordinates x(to) and y(to): x = U o ( t " − t o ) + ξ o (sin(ωt ") − sin(ωt o )) and y = –ξ o (cos(ωt $) − cos(ωt o )) . € € combining the equations: Some of the to dependence can be eliminated by 2 2 ( x − U o (t # − t o ) − ξ o sin(ωt #)) + ( y + ξ o cos(ωt #)) = ξ o2 , which describes a circle with € a to-dependent center located at € (U o (t " − t o ) + ξ o sin(ωt "),−ξ o cos(ωt ")) . These results differ from those in Example 3.3 by the uniform translation velocity Uo so € they can be put into correspondence with a Galilean transformation x´ = x – Uo(t – t´).

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 3.10. Compute and compare the streamline, path line, and streak line that pass through (1,1,0) at t = 0 for the following Cartesian velocity field u = (x, –yt, 0). Solution 3.10. (i) For the streamline, time is a constant. Use the first equality of (3.7) to find: dy v yt dy dx = =− → = −t → ln y = −t ln x + const. , or y = const.x–t. dx u x y x Evaluating at x = y = 1 and t = 0 requires the constant to be unity, so the streamline is: y = 1. (ii) For the path line, use both components of (3.8): dx dy = x and = −yt , € dt dt and integrate these in time to find: x = C1e t and y = C2 exp{−t 2 2} , where C1 and C2 are constants. € Evaluating € at x = y = 1 and t = 0 requires C1 = C2 = 1. Eliminate t from the y-equation using t = ln(x) to find the path line as: y = exp{−(ln x) 2 2} . € € (iii) For the streak line, use the path line results but this time evaluate the constants at t = to instead of at t = 0 to find: x = exp{t − t o } and y = exp{(t o2 − t 2 ) 2} . € Evaluate at t = 0, and eliminate to from the resulting equations to find: y = exp{+(ln x) 2 2} .

€

€ €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 3.11. Consider a time-dependent flow field in two-dimensional Cartesian coordinates where u = τ t 2 , v = xy τ , and  and τ are constant length and length time scales, respectively. a) Use dimensional analysis to determine the functional form of the streamline through x´ at time t´. b) Find the equation for the streamline through x´ at time t´ and put your answer in € € dimensionless form. € c) Repeat b) for the path line through x´ at time t´. d) Repeat b) for the streak line through x´ at time t´.

€

Solution 3.11. a) The streamline y(x) will depend on x, t, t´, x´= (x´,y´),  , and τ. There are eight % x x # y # t t #( y parameters and two dimensions, thus there are six dimensionless groups: = Ψ' , , , , * . &   τ τ)  Here there are too many variables and parameters for dimensional analysis to be really useful. € the remainder of the solution. However, this effort provides a reminder to check units throughout i) For the streamline, time is a constant. Use the first equality of (3.7) to find: €2 2 dy v xy τ dy t 2 xdx t x = = → = → ln y = + const. 2 2 2 dx u τ t y τ  τ 2 2 2 The initial condition requires, x = x´ and y = y´ at t = t´, and this allows the constant to be determined, yielding: # y & (t 2 x 2 − t "2 x "2 ) ln% ( = . € 2 2τ 2 $ y"' (ii) For the path line, use both components of (3.8): dx τ dy xy = 2 and = , dt t dt τ € in time and use the initial condition to find: and integrate the first of these x = −τ t + const. or x − x # = −τ ( t −1 − t #−1 ) . Use this result for x(t) in the € second equation € for y(t): ( dy y % % 1 1 ( dy % 1 1 1 ( = ' −τ ' − * + x $* or =' − + x "* dt . dt τ & & t € t $) y & t " t τ ) ) € The last expression can be integrated to find: −1 #y& t +# x "t " &# t &. # t & # x" 1 & #t& x" y #t& ln% ( = −1+ − ln% ( = % + ((t − t ") − ln% ( or = % ( exp-% + 1(% −1(0 . $ t " ' $ τ t " '€ $ t "' '$ t " '/ τ y" $ t"' ,$ τ $ y"' € t" Now use the final equations for x(t) and y(t) to eliminate t. The equation for x(t) can be rearranged to find: t" (x − x ") t " = 1− € , t τ so the equation for y becomes: −1 (. y % (x − x ") t " ( +% x "t " (%% (x − x ") t " ( = '1− + 1*'''1− * exp-' * −1**0. )&& y " & € τ ) -,& τ τ ) )0/ (iii) For the streak line, use the path line results but this time evaluate the integration constants at t = to instead of at t = t´ to find:

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

x − x # = −τ ( t − t −1

−1 o

)

−1 +# x "t &. &# t y #t& , and = % ( exp-% o + 1(% −1(0. '$ t o '/ y" $ to ' ,$ τ

Now eliminate to find: €

−1 +% x " % 1 (x − x ") (−1 (% (x − x ")t (. y % (x − x ")t ( = '1+ * exp-'' ' + * + 1**' *0. y" & τ €) τ ) -,& τ & t )& τ )0/

And evaluate at t = t´ to reach: −1 +% x " % 1 (x − x ") (−1 (% (x − x ") t " (. y % (x − x ") t " ( = '1+ * exp-'' ' + * + 1**' *0. y" & τ ) τ ) -,& τ & t " € )& τ )0/

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 3.12. The velocity components in an unsteady plane flow are given by u = x (1+ t) and v = 2y (2 + t) . Determine equations for the streamlines and path lines subject to x = x0 at t = 0.

€

Solution 3.12. i) For the streamline, time is a constant. Use the first equality of (3.7) to find: € t) dy v 2y (2 + t) dy 2(1+ t) dx 2(1+ = = → = → ln y = ln x + const. dx u x (1+ t) y (2 + t) x (2 + t) Use of the initial condition produces: 2(1+ 0) ln y 0 = ln x 0 + const., (2 + 0) € so the final answer is: 2(1+t )

" y % 2(1+ t) " x % y " x % (2+t ) ln$ ' = ln$ ' or . =$ ' € # y 0 & (2 + t) # x 0 & y0 # x0 & (ii) For the path line, use both components of (3.8): dx x dy 2y and , = = dt 1+ t dt 2 + t € and integrate the € these in time to find: ln x = ln(1+ t) + const. and ln y = 2ln(2 + t) + const. Use the initial condition to determine the two constants, and exponentiate both equations: € € 2 x = x 0 (1+ t) and y = y 0 (1+ t 2) . To determine the path line, eliminate t to find: € € 2 y = y 0 (1+ (x − x 0 ) 2x 0 ) . €

€ €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 3.13. Using the geometry and notation of Fig. 3.8, prove (3.9).

€

€

Solution 3.13. Before starting this problem, it is worthwhile to note that the acceleration of a fluid particle is invariant under the specified Galilean transformation so the components of U cannot be part of the final answer. Thus, transformation errors can be readily detected if terms are missing in the final results or extra ones have appeared. Figure 3.8 supports the following vector addition formula: x = Ut + x "o + x " . Thus, the Cartesian-coordinate transformations in this case are given by: x " = x − (e x ⋅ U) t − x o" , y " = y − (e y ⋅ U) t − y "o , z" = z − (e z ⋅ U) t − z"o , and t " = t . The transformation of the spatial derivatives between the stationary frame of reference, Oxyz, € and the steadily moving frame, O´x´y´z´ is straightforward mathematics: € ∂ ∂x # ∂ ∂y # ∂ ∂z#€∂ ∂t # ∂ ∂ + + + = , € = ∂x ∂x ∂x # ∂x ∂y # ∂x ∂z# ∂x ∂t # ∂x # ∂ ∂x # ∂ ∂y # ∂ ∂z# ∂ ∂t # ∂ ∂ = + + + = , and ∂y ∂y ∂x # ∂y ∂y # ∂y ∂z# ∂y ∂t # ∂y # ∂ ∂x # ∂ ∂y # ∂ ∂z# ∂ ∂t # ∂ ∂ € = + + + = , ∂z ∂z ∂x # ∂z ∂y # ∂z ∂z# ∂z ∂t # ∂z# where the final equality on each line follows from differentiating the definitions of the moving € coordinate variables given above. The time derivative requires more effort ∂ ∂x # ∂ ∂y # ∂ ∂z# ∂ ∂t # ∂ ∂ ∂ ∂ ∂ = + + + = −e x ⋅ U − ey ⋅ U − ez ⋅ U + . € ∂t ∂t ∂x # ∂t ∂y # ∂t ∂z# ∂t ∂t # ∂x # ∂y # ∂z# ∂t # ∂ The first three equations imply: ∇ = ∇ # and the fourth implies: −U ⋅ ∇ $ + . The velocities will ∂t $ be related by: u = U + u´. Now use these results to assemble the fluid particle acceleration € starting in the stationary coordinate system, and converting each velocity and differential operation to the moving€coordinate system. € ' ∂u ∂* + (u ⋅ ∇ )u = ) −U ⋅ ∇ & + ,[U + u&] + ([U + u&] ⋅ ∇ &)[U + u&] ( ∂t ∂t & + ∂U ∂u$ = −U ⋅ ∇ $U + − U ⋅ ∇ $u$ + + (U ⋅ ∇ $)U + (u$ ⋅ ∇ $)U + (U ⋅ ∇ $)u$ + (u$ ⋅ ∇ $)u$ ∂t $ ∂t $ ∂u$ ∂u$ = −U ⋅ ∇ $u$ + + (U ⋅ ∇ $)u$ + (u$ ⋅ ∇ $)u$ = + (u$ ⋅ ∇ $)u$ ∂t $ ∂t $ Here most of the simplifications occur because all derivatives of U are zero; it is a constant. € as expected, the form of the fluid particle acceleration is frame invariant for coordinate Thus, systems related by Galilean transformations. €

Fluid Mechanics, 6th Ed.

Exercise 3.14. Determine the unsteady, ∂u/∂t, and advective, (u⋅∇)u, fluid acceleration terms for the following flow fields specified in Cartesian coordinates. a) u = ( u(y,z,t),0,0) b) u = Ω × x where Ω = (0,0,Ωz (t)) c) u = A(t)( x,−y,0) d) u = (Uo + uosin(kx – Ωt), 0, 0) where Uo, uo, k and Ω are positive constants

€

€ €

Kundu, Cohen, and Dowling

€

Solution 3.14. a) Here there is only one component of the fluid velocity; thus ∂u ∂t = (∂ ∂t )( u(y,z,t),0,0) = (∂u ∂t,0,0) , and [u ⋅ ∇]u = [(u(y,z,t),0,0) ⋅ (∂ ∂x,∂ ∂y,∂ ∂z)]( u(y,z,t),0,0) = [u(∂ ∂x)]( u(y,z,t),0,0) = 0 . b) Here the fluid velocity has two components: u = Ω × x = (–Ωzy, +Ωzx, 0), so % dΩz dΩz ( dΩz € ∂u ∂t = (∂ ∂t )(−Ωz y,Ωz x,0) = ' −y ,x ,0* = (−y, x,0) , and & dt dt ) dt € [u ⋅ ∇]u = [(−Ωz y,Ωz x,0) ⋅ (∂ ∂x, € ∂ ∂y,∂ ∂z)](−Ωz y,Ωz x,0) = [−Ωz y(∂ ∂x) + Ωz x(∂ ∂y)](−Ωz y,Ωz x,0) = (−Ω2z x,−Ω2z y,0) = −Ω2z ( x, y,0) .

€ c) Again the fluid velocity has two components: (Ax, –Ay, 0), so $ dA dA ' dA € ∂u ∂t = (∂ ∂t )( Ax,−Ay,0) = & x ,−y ,0) = ( x,−y,0) , and % dt dt ( dt € [u ⋅ ∇]u = [( Ax,−Ay,0) ⋅ (∂ ∂x,∂ ∂y,∂ ∂z)]( Ax,−Ay,0)

= [ Ax(∂ ∂x) − Ay(∂ ∂y)]( Ax,−Ay,0) = ( A 2 x, A 2 y,0) = A 2 ( x, y,0) .

€ d) Here again there is only one component of the fluid velocity; thus ∂u ∂t = (∂ ∂t )(U o + uo sin(kx − Ωt),0,0) = (−Ωuo cos(kx − Ωt),0,0) , and € €[u ⋅ ∇ ]u = [(U o + uo sin(kx − Ωt),0,0) ⋅ (∂ ∂x,∂ ∂y,∂ ∂z)](U o + uo sin(kx − Ωt),0,0) .

= [(U o + uo sin(kx − Ωt))(∂ ∂x)](U o + uo sin(kx − Ωt),0,0) = ((U o + uo sin(kx − Ωt)) kuo cos(kx − Ωt),0,0)

€ €

(

)

= kU o uo cos(kx − Ωt) + 12 kuo2 sin[2(kx − Ωt)],0,0 . € €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 3.15. Consider the following Cartesian velocity field u = A(t)( f (x),g(y),h(z)) where A, f, g, and h are non-constant functions of only one independent variable. a) Determine ∂u/∂t, and (u⋅∇)u in terms of A, f, g, and h, and their derivatives. b) Determine A, f, g, and h when Du/Dt = 0, u = 0 at x = 0, and u is finite for t > 0. c) For the conditions in b), determine the equation for€the path line that passes through xo at time to, and show directly that the acceleration a of the fluid particle that follows this path is zero.

€

€ €

Solution 3.15. a) Here there are three components of the fluid velocity; thus # dA dA dA & dA ∂u ∂ = ( Af , Ag, Ah ) = % f ,g ,h ( = ( f ,g,h ) , and $ dt ∂t ∂t dt dt ' dt + % ∂ ∂ ∂ (. [u ⋅ ∇]u = -A( f ,g,h) ⋅ ' , , *0( Af , Ag, Ah) & ∂x ∂y ∂z )/ , ) df dg dh , # ∂ € ∂ ∂& = A 2% f + g + h (( f ,g,h ) = A 2 + f ,g ,h . , ∂y ∂z ' $ ∂x * dx dy dz € where the partial derivatives have become total derivatives because f, g, and h are only functions of one variable. b) For the given velocity field, Du/Dt = 0 implies: € dA dg dA dh dA df + A 2g = 0 , and h + A 2h = 0. f + A2 f =0 , g dt dy dt dz dt dx Start with the first equation, assume A2f is not zero, and divide by it to find: 1 dA df + = 0. A 2 dt dx € € € only depends on t while the second one only depends on x, thus, The first term in this equation each must be equal and opposite, and constant (= C). So, 1 dA dA 1 1 df = −C → − 2 = Cdt € → = C(t − t o ) → A = , and = C → f = C(x − x o ) , 2 A dt A A C(t − t o ) dx where to and xo are constants of integration. Similarly, for the other two component directions: g = C(y − y o ) and h = C(z − zo ) . Here, we presume to ≤ 0 so that A is finite for t > 0. € c) The x-component of the path line is defined by: dx C(x − x o ) dx dt = u = Af = = € → €→ ln(x − x o ) = ln(t − t o ) + const. → x − x o = U(t − t o ) , dt C(t − t o ) x − x o t − to where U is a constant. Similarly for the other Cartesian directions: y − y o = V (t − t o ) , and z − zo = W (t − t o ) , where V and W are constants. So defining the constant vector U = (U, V, W), the path line of interest is: x – xo = U(t – to). Thus, x(t) is linear function of t, so dx(t)/dt = U, and a = d€2x(t)/dt2 = 0. This exercise illustrates how the unsteady and advective acceleration terms may be equal and opposite.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 3.16 If a velocity field is given by u = ay and v = 0, compute the circulation around a circle of radius ro that is centered on at the origin. Check the result by using Stokes’ theorem. Solution 3.16. In plane polar coordinates, the vector path-length element, ds, on a circle of radius ro is ds = tds = eθ ro dθ . From Example 2.1, the radial and angular velocity component are: ur = ucosθ + vsinθ = arsinθcosθ + 0, and u = –usinθ + vcosθ = –arsin2θ , where u = ay = arsinθ and v = 0 has been used. Thus, circulation is: θ = 2π θ = 2π θ = 2π Γ = ∫ u ⋅ ds = ∫θ = 0 u ⋅ eθ ro dθ = ∫θ = 0 uθ ro dθ = − ∫θ = 0 aro2 sin 2 θdθ = −aπro2 . € Now use Stokes' theorem to reach the same result. Start by computing the vorticity: ex ey ez θ

∇ × u = ∂ ∂x ∂ ∂y ∂ ∂z = −ae z . ay 0 0 Insert this into the Stokes' theorem area integral using n = ez, and dA = rdrdθ: θ = 2π r= r Γ = ∫∫ (∇ × u) ⋅ ndA = ∫θ = 0 ∫ r= 0o (−ae z ) ⋅ e z rdrdθ = −aπro2 ,

€

A

€ of the prior circulation calculation. and this matches the result

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 3.17. Consider a plane Couette flow of a viscous fluid confined between two flat plates a distance b apart. At steady state the velocity distribution is u = Uy/b and v = w = 0, where the upper plate at y = b is moving parallel to itself at speed U, and the lower plate is held stationary. Find the rates of linear strain, the rate of shear strain, and vorticity in this flow. Solution 3.17. Here there is only one velocity component: u = Uy/b. The strain rate tensor is: 1 * U 2b∂ u ∂x ∂u ∂y + ∂v ∂x )- * 0 1 # ∂ui ∂u j & 2( Sij = %% + (( → Sij = + 1 .=+ .. 0 / 2 $ ∂x j ∂x i ' ∂ v ∂y , 2 (∂v ∂x + ∂u ∂y ) / ,U 2b Thus, the linear strain rates are both zero, and the shear strain rate is U/2b. The vorticity vector has one non-zero component: ∂v ∂u U ωz = − = − . € ∂x ∂y b

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 3.18. The steady two-dimensional flow field inside a sloping passage is given in (x,y)-

(

Cartesian coordinates by u = (u, v) = (3q 4h ) 1− ( y h )

2

) ( 1, ( y h) (dh dx)) where q is the volume

flow rate per unit length into the page, and h is the passage's half thickness. Determine the streamlines, vorticity, and strain rate tensor in this flow away from x = 0 when h = αx where α is a positive constant. Sample profiles of u(x,y) vs. y are shown at two x-locations in the figure. What are the equations of the streamlines along which the x- and y-axes are aligned with the principal axes of the flow? What is the fluid particle rotation rate along these streamlines? y!

h(x)! x!

Solution 3.18. For planar flow in Cartesian coordinates, the streamlines are determined from: dy v ( y h ) ( dh dx ) = = = ( y α x )α = y x . dx u 1 Separate the differentials and integrate to find: ln(y) = ln(x) + const. Exponentiate to find: y = Cx, where C is a constant. Thus, the streamlines are straight lines through the origin of coordinates. The vorticity ωz is determined from: ∂v ∂u ∂ )# 3q &# y 2 & y dh , ∂ )# 3q &# y 2 &, ω z = − = +% (%1− 2 ( . − +% (%1− (. ∂x ∂y ∂x *$ 4h '$ h ' h dx - ∂y *$ 4h '$ h 2 '3q ∂ ) 1 # y 2 & y , 3q ∂ ) 1 # y 2 &, = + %1− 2 2 ( α . − + %1− 2 2 (. 4 ∂x *α x $ α x ' α x - 4 ∂y *α x $ α x '3q ) 2y 4y 3 , 3q ) 2y , 3qy ) 2y 2 1, = +− 3 + 3 5 . − +− 3 3 . = −1+ + + . 4 * α x α x - 4 * α x - 2α x 3 * α 2 x2 α 2 The strain-rate tensor Sij is computed from the following velocity derivatives: ∂u ∂ )" 3q %" y 2 %, 3q ∂ ) 1 " y 2 %, 3q ) 1 3y 2 , 3q ) 3y 2 , S11 = = +$ '$1− 2 '. = −1+ + $1− 2 2 '. = +− 2 + 3 4 . = + . ∂x ∂x *# 4h &# h &- 4 ∂x *α x # α x &- 4 * α x α x - 4α x 2 * α 2 x2 ∂v ∂ )" 3q %" y 2 % y dh , 3q ∂ ) 1 " y2 % y , 3q " 3y 2 % S22 = = +$ '$1− 2 ' .= + $1− ' α. = $1− ' , and ∂y ∂y *# 4h &# h & h dx - 4 ∂y *α x # α 2 x 2 & α x - 4α x 2 # α 2 x 2 & 1 " ∂u ∂v % 1 ∂ /( 3q +( y 2 +2 1 ∂ /( 3q +( y 2 + y dh 2 3qy / 1 2y 2 2 S12 = S21 = # + & = 1− + 1− = − −1+ 1* -* 1* -* 4 -4 1 4, 2 $ ∂y ∂x ' 2 ∂y 0) 4h ,) h 2 ,3 2 ∂x 0) 4h ,) h 2 , h dx 3 4α x 3 0 α 2 α 2 x2 3 where the differentiation details for S12 are available above in the calculation of ωz. The principle axes occur then the off-diagonal strain rate components are zero. This occurs when

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

3qy " 1 2y 2 % − −1+ $ '= 0 . 4α x 3 # α 2 α 2 x2 & This equality is satisfied along the x-axis where y = 0, and when the contents of the [,]-brackets are zero: 1 2y 2 or y = ±x (1+ α 2 ) 2 +1 = 2 2 2 α α x These streamlines only occur inside flow wedge when α > 1. The fluid particle rotation rate is ωz/2, so from the results above for ωz: S12 = S21 =

!ω $ !ωz $ # & = 0 , and # z & " 2 %y=± x " 2 %y=0

=± (1+α 2 ) 2

3q (1+ α 2 ) 2 . 2α 3 x 2

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 3.19. For the flow field u = U + Ω × x , where U and Ω are constant linear- and angularvelocity vectors, use Cartesian coordinates to a) show that Sij is zero, and b) determine Rij. Solution 3.19. Since no€simplifications are given, all the components of U = (U1, U2, U3) and Ω = (Ω1, Ω2, Ω3) should be treated as being non-zero. In Cartesian coordinates, the velocity field is e1 e 2 e 3 u = U + Ω × x = U1e1 + U 2e 2 + U 3e 3 + Ω1 Ω2 Ω3 x1 x 2 x 3 = (U1 + Ω2 x 3 − Ω3 x 2 )e1 + (U 2 + Ω3 x1 − Ω1 x 3 )e 2 + (U 3 + Ω1 x 2 − Ω2 x1 )e 3 a) Use this velocity field result to compute the velocity gradient tensor, and its transpose (indicated with a superscript "T" below) to sort out which derivatives are zero and which ones € are not. € #∂u1 ∂x1 ∂u1 ∂x 2 ∂u1 ∂x 3 ' # 0 −Ω3 +Ω2 ' % % % ∂ui % = $∂u2 ∂x1 ∂u2 ∂x 2 ∂u2 ∂x 3 ( = $+Ω3 0 −Ω1 ( ∂x j % % % 0 %) &∂u3 ∂x1 ∂u3 ∂x 2 ∂u3 ∂x 3 ) &−Ω2 +Ω1 )∂u1 ∂x1 ∂u2 ∂x1 ∂u3 ∂x1 - ) 0 +Ω3 −Ω2 T + + + ∂u j # ∂ui & + = %% 0 +Ω1 . (( = *∂u1 ∂x 2 ∂u2 ∂x 2 ∂u3 ∂x 2 . = *−Ω3 ∂x i $ ∂ x j ' + + + 0 +/ € ,∂u1 ∂x 3 ∂u2 ∂x 3 ∂u3 ∂x 3 / ,+Ω2 −Ω1 This result can be used to construct the strain rate tensor Sij: 1 + 0 −Ω3 + Ω3 ) 12 (+Ω2 − Ω2 )/ T& # 2( # & # & 1 ∂ui ∂u j 1 ∂u ∂u 1 + 0 −Ω1 + Ω1 ) 0 (( = % i + %% i (( ( = , 12 (+Ω3 − Ω3 ) ( € Sij = %% 2 2 $ ∂x j ∂x i ' 2 % ∂x j $ ∂x j ' ( - 1 $ ' . (−Ω2 + Ω2 ) 1 (+Ω1 − Ω1 ) 0 1 2 2 "0 0 0& $ $ = #0 0 0' . $0 0 0$ % ( € b) Similarly for the rotation tensor: + 0 −Ω3 − Ω3 +Ω2 + Ω2 / T ∂ui ∂u j ∂ui $ ∂ui ' R = − = − = +Ω + Ω 0 −Ω − Ω & ) , 0 ij 3 3 1 & ) € ∂x j ∂ x i ∂x j % ∂ x j ( 0 . −Ω2 − Ω2 +Ω1 + Ω1 1 $ 0 −2Ω3 +2Ω2 ( & & = %+2Ω3 0 −2Ω1 ) . &−2Ω +2Ω 0 &* € ' 2 1

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 3.20. Starting with a small rectangular volume element δV = δx1δx2δx3, prove (3.14). Solution 3.20. The volumetric strain rate for a fluid element is: 1 D 1 D 1 D 1 D 1 D (δV ) = (δx1δx 2δx 3 ) = (δx1 ) + (δx 2 ) + (δx 3 ) . δV Dt δx1δx 2δx 3 Dt δx1 Dt δx 2 Dt δx 3 Dt From Section 3.4 in the text, the linear strain rate corresponding to elongation or contraction of a fluid element in the first direction is: 1 D ∂u (δx1) = S11 = 1 , € δx1 Dt ∂x1 and this can be immediately extended to the other two directions, 1 D ∂u 1 D ∂u (δx 2 ) = S22 = 2 and (δx 3 ) = S33 = 3 , δx 2 Dt ∂x 2 δx 3 Dt ∂x 3 € because its geometric derivation (see Figure 3.10) did not rely on any special properties of the first direction. Substitution of these relationships into the final version of the volumetric strain rate given above produces: € € 1 D ∂ u ∂u ∂u ∂u (δV ) = S11 + S22 + S33 = 1 + 2 + 3 = i = Sii . δV Dt ∂x1 ∂x 2 ∂x 3 ∂x i

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 3.21. Let Oxyz be a stationary frame of reference, and let the z-axis be parallel with the fluid vorticity vector in the vicinity of O so that ω = ∇ × u = ω ze z in this frame of reference. Now consider a second rotating frame of reference Ox "y "z" having the same origin that rotates about the z-axis at angular rate Ωez. Starting from the kinematic relationship, u = (Ωe z ) × x + u$ , show ω " = ∇ " × u" in the rotating frame of reference can only be that in the vicinity of O the vorticity € zero when 2Ω = ωz, where ∇ " is the € gradient operator in the primed coordinates. The following unit vector transformation rules may be of use: e"x = e x cos(Ωt) + e y sin(Ωt) , € e"y = −e x sin(Ωt) + e y cos(Ωt)€ , and e"z = e z .

€ Solution 3.21. The approach here is to compute ω " = ∇ " × u" in the stationary frame of reference € and then determine the parameter choice(s) necessary for ω´ to be zero. The vector x must have a € € consistent representation in either frame, so x = xe x + ye y = x "e"x + y "e"y = x "(e x€cos(Ωt) + e y sin(Ωt)) + y "(−e x sin(Ωt) + e y cos(Ωt)) . Equating components in the stationary frame of reference produces: x = x " cos(Ωt) − y " sin(Ωt) , and y = x " sin(Ωt) + y " cos(Ωt) . The remaining independent variables are the same in either frame: t = t ", and z = z". Thus, spatial € derivatives are related by: ∂ ∂x ∂ ∂ y ∂ ∂z ∂ ∂ t ∂ ∂ ∂ = + + = cos(Ωt) + sin(Ωt) , € € + ∂x # ∂x # ∂x ∂x # ∂y ∂x # ∂z ∂x # ∂t ∂x € ∂y € ∂ ∂x ∂ ∂ y ∂ ∂z ∂ ∂ t ∂ ∂ ∂ = + + + = −sin(Ωt) + cos(Ωt) , and ∂y # ∂y # ∂x ∂y # ∂y ∂y # ∂z ∂y # ∂t ∂x ∂y ∂ ∂x ∂ ∂ y ∂ ∂z ∂ ∂ t ∂ ∂ € = + + + = . ∂z# ∂z# ∂x ∂z# ∂y ∂z# ∂z ∂z# ∂t ∂z Using the above information, the gradient operator in the rotating coordinates can be rewritten in € terms of the stationary frame coordinates and unit vectors: ∂ ∂ ∂ ∇" = € e"x + e"y + e"z ∂x " ∂y " ∂z" $ ∂ ∂' = (e x cos(Ωt) + e y sin(Ωt))&cos(Ωt) + sin(Ωt) ) ∂x ∂y ( % % ∂ ∂( ∂ € +(−e x sin(Ωt) + e y cos(Ωt))'−sin(Ωt) + cos(Ωt) * + e z ∂x ∂y ) ∂z & & ∂ ∂ ∂ ∂) € ∇ " = (e x cos 2 (Ωt) + e x sin(Ωt)cos(Ωt) + e y sin(Ωt)cos(Ωt) + e y sin 2 (Ωt) + ∂x ∂y ∂x ∂y * ' % ∂ ∂ ∂ ∂( ∂ € +' e x sin 2 (Ωt) − e x sin(Ωt)cos(Ωt) − e y sin(Ωt)cos(Ωt) + e y cos 2 (Ωt) * + e z ∂x ∂y ∂x ∂y ) ∂z & & ) ∂ ∂ ∂ ∂ € ∇ " = (e x cos 2 (Ωt) + e x sin(Ωt)cos(Ωt) + e y sin(Ωt)cos(Ωt) + e y sin 2 (Ωt) + ∂x ∂y ∂x ∂y * ' % ∂ ∂ ∂ ∂( ∂ € +' e x sin 2 (Ωt) − e x sin(Ωt)cos(Ωt) − e y sin(Ωt)cos(Ωt) + e y cos 2 (Ωt) * + e z ∂x ∂y ∂x ∂y ) ∂z & ∂ ∂ ∂ € ∇" = e x + e y + e z . ∂x ∂y ∂z € €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

This result seems too simple, but should not be a surprising to a routine user of vector calculus. Now resolve rotating-frame velocity components in the stationary frame of reference: u" = u − (Ωe z ) × x = ue x + ve y + we z + Ωye x − Ωxe y . Using the stationary frame components and unit vectors, the vorticity in the rotating frame is: ex ey ez ex ey ez ex ey ez ω " = ∇ " × u" = ∂ ∂x ∂ ∂y ∂ ∂z = ∂ ∂ x ∂ ∂y ∂ ∂z + ∂ ∂x ∂ ∂ y ∂ ∂z € u + Ωy v − Ωx w u v w Ωy −Ωx 0 ex ey ez = ∇ × u + ∂ ∂x ∂ ∂y ∂ ∂z = ω ze z + e x (0) + e y (0) + e z (−Ω − Ω) = e z (ω z − 2Ω) Ωy −Ωx 0

€

Thus, ω´ will be zero when ωz = 2Ω. €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 3.22. Consider a plane-polar area element having dimensions dr and rdθ. For twodimensional flow in this plane, evaluate the right-hand side of Stokes’ theorem ∫ ω ⋅ ndA = ∫ u ⋅ ds and thereby show that the expression for vorticity in plane-polar coordinates 1∂ 1 ∂u is: ω z = (ruθ ) − r . r ∂r r ∂θ €

Solution 3.22. Using the element shown with angular width dθ,

€

ur + (!ur/!!)d! u!

dr

u! + (!u!/!")dr ur

!

application of Stokes' theorem provides: (

( + ∂uθ + ∂u dr-( r + dr) dθ − * ur + r dθ -dr − uθ rdθ , or ) ∂r , ∂θ , ∂u ∂u ω z rdθdr = ur dr + ruθ dθ + r θ drdθ + uθ drdθ − ur dr − r dθdr − uθ rdθ + ... ∂r ∂θ ∂uθ ∂ur =r drdθ + uθ drdθ − dθdr + ... ∂r ∂θ where + ... indicates the presence of higher order terms in dr and dθ. Division by rdθdr and € passing to the limit where dr and dθ go to zero produces: ∂u 1 1 ∂ur 1 & ∂ ∂u ) € ω z = θ + uθ − = ( ( ruθ ) − r + . ∂r r r ∂θ r ' ∂θ ∂θ *

∫ ω ⋅ ndA = ∫ u ⋅ ds → ω z rdθdr = ur dr + *) uθ +

€

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 3.23. The velocity field of a certain flow is given by u = 2xy 2 + 2xz 2 , v = x 2 y , and w = x 2 z . Consider the fluid region inside a spherical volume x2 + y2 + z2 = a2. Verify the validity of Gauss’ theorem ∫∫∫ ∇ ⋅ udV = ∫∫ u ⋅ ndA by integrating over the sphere. V

A

€ € Solution 3.23. First compute the divergence of the velocity field. ∂u ∂v ∂w ∂ ∂ ∂ ∇ ⋅€u = + + = (2xy 2 + 2xz 2 ) + ( x 2 y ) + ( x 2 z) = 2y 2 + 2z 2 + 2x 2 = 2r 2 . ∂x ∂y ∂z ∂ x ∂y ∂z The volume integral of ∇ ⋅ u is: r= a 8 ∇ ⋅ udV = ∫∫∫ ∫ (2r 2 ) 4πr 2 dr = 5 πa 5 . V r= 0 € Now work on the surface integration using spherical coordinates (see Figure 3.3d, and Appendix e r = e x cosϕ sin θ + e y sin ϕ sin θ + e x cos θ , so B). Here, n = € u ⋅ n = ucosϕ sin θ + v sin ϕ sin θ + w cos θ € = (2xy 2 + 2xz 2 ) cosϕ sin θ + ( x 2 y ) sin ϕ sin θ + ( x 2 z) cos θ

€

Unfortunately, this result is in mixed variables so convert everything to spherical polar € coordinates using € x = r cosϕ sin θ , y = r sin ϕ sin θ , z = r cosθ . € This conversion produces: u ⋅ n = 2r 3 (cos ϕ sin 2 ϕ sin 3 θ + cos ϕ sin θ cos 2 θ ) cos ϕ sin θ 3 2 € 2 θ cos θ ) cos θ +r 3 (cos 2 ϕ€ sin ϕ sin 3 θ ) sin ϕ sin € θ + r (cos ϕ sin

= 2r 3 (sin 2 ϕ sin 2 θ + cos2 θ ) cos2 ϕ sin 2 θ + r 3 (sin 2 ϕ sin 2 θ ) cos2 ϕ sin 2 θ + r 3 (cos 2 θ ) cos 2 ϕ sin 2 θ

€ € €

= r 3 ( 3sin 2 ϕ sin 2 θ + 3cos 2 θ ) cos 2 ϕ sin 2 θ = r 3

So, the surface integral produces: ϕ = 2π θ = π

∫∫ u ⋅ ndA = ∫ ∫

€

A

=a €

a3

3 4

2

3 4

2

sin (2ϕ )dϕ

2

θ=π

∫

θ=0

2

2

2

(1− cos2 θ ) sinθdθ + a5 +1

2

2

ϕ = 2π

∫

ϕ= 0

#3 & 2 = a % π ( ∫ (1− β 2 ) dβ + a 5 ( 3π ) ∫ β 2 (1− β 2 )dβ $ 4 ' −1 −1 # & # 3 16 4 12 12 & 8 5 5 € = a % π ( + a ( 3π ) = % + (πa 5 = πa 5 , $ 4 ' 15 15 $ 15 15 ' 5 €which is the same as the result of the volume integration. €

2

2

2

2

2

( sin (2ϕ)(1− cos θ) + 3cos ϕ cos θ(1− cos θ))a sinθdθdϕ

ϕ= 0 θ = 0 ϕ = 2π 3 5 2 4 ϕ= 0 +1 5

∫

( sin (2ϕ)(1− cos θ) + 3cos ϕ cos θ(1− cos θ)) 3cos2 ϕdϕ

2

θ=π

∫ cos2 θ (1− cos2 θ ) sin θdθ

θ=0

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 3.24. A flow field on the xy-plane has the velocity components u = 3x + y and v = 2x – 3y. Show that the circulation around the circle (x − 1)2 + (y − 6)2 = 4 is 4π. Solution 3.24. The circle is centered at (1,6) and its radius is 2. So, if (x,y) is a point on the circle then x = 1 + 2cosθ, and y = 6 + 2sinθ, where θ is the angle from the horizontal. The velocity component tangent to the circle will be uθ = eθ ⋅ u = (−sin θ,cos θ ) ⋅ (u,v) = −usin θ + v cos θ . Evaluate this velocity component. uθ = −(3x + y)sin θ + (2x − 3y)cos θ = (−3sin θ + 2cosθ )x − (sin θ + 3cosθ )y

= (−3sin θ + 2cos θ )(1+ 2cosθ ) − (sin θ + 3cos θ )(6 + 2sin θ ) € = −3sin θ − 6sin θ cosθ + 2cos θ + 4 cos2 θ − 6sin θ − 2sin 2 θ −18cosθ − 6cos θ sin θ € = −9sin θ −12sin θ cosθ −16cos θ + 4 cos2 θ − 2sin 2 θ Now compute the circulation: 2π

€

€ €

2π

∫ uθ rdθ = ∫ (−9sin θ −12sinθ cosθ −16cosθ + 4 cos2 θ − 2sin 2 θ )2dθ θ=0 θ=0 = (0 + 0 + 0 + 4 π − 2π )2 = 4 π

Γ=

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 3.25. Consider solid-body rotation about the origin in two dimensions: ur = 0 and uθ = ω0r. Use a polar-coordinate element of dimension rdθ and dr, and verify that the circulation is vorticity times area. (In Section 5 this was verified for a circular element surrounding the origin.) Solution 3.25. Using the element shown with angular width dθ, ur + (!ur/!!)d! u!

dr

u! + (!u!/!")dr ur

!

the circulation is around the element is the sum or four terms: ' * ' * ∂u ∂u Γ = ∫ u ⋅ ds = ur dr + ) uθ + θ dr,( r + dr) dθ − ) ur + r dθ , dr − uθ rdθ . ( ( ∂r + ∂θ + Now substitute in the velocity field: ur = 0 and uθ = ω0r. Γ = (ω 0 r + ω 0 dr)( r + dr) dθ − ω 0 r 2 dθ = 2ω 0 rdrdθ + ω 0 (dr) 2 dθ = (2ω 0 )rdrdθ , 2 where € the final equality holds in the limit as the differential elements become small and (dr) dθ is negligible compared to rdrdθ. Thus, since the voriticity is 2ω0 and rdrdθ is the area of the element, the final relationship states: circulation = (vorticity) x (area). €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

$ ' −Ay + Ax & , ,0) . Exercise 3.26. Consider the steady Cartesian velocity field u = & ( x 2 + y 2 )β ( x 2 + y 2 )β ) % ( a) Determine the streamline that passes through x = (x o , y o ,0) b) Compute Rij for this velocity field. c) For A > 0, explain the sense of rotation (i.e. clockwise or counter clockwise) for fluid elements € for β < 1, β = 1, and β > 1. € β

Ax ( x 2 + y 2 ) dy v x Solution 3.26. a) Use the definition of a streamline: = = = − . Use the β dx u −Ay ( x 2 + y 2 ) y

two ends of this extended equality to find: ydy = −xdx , and integrate the resulting differential relationship to get: y 2 2 = −x 2 2 + const . Evaluate the constant using the required condition:

€

x 2 + y 2 = x o2 + y o2 . This is a circle with radius€ x o2 + y o2 . Therefore the streamlines are circles. € " 0 ∂u ∂ y − ∂v ∂ x 0 & $ $ ∂u ∂u b) The rotation tensor is: Rij = i − j = # ∂ v ∂ x − ∂ u ∂ y 0 0 ' , where the ∂ x€j ∂ xi $ $ 0 0 0 $( $% second equality comes from putting the specified velocity field into the definition of Rij with u = (u, v, w) = (u1, u2 , u3 ) = ui as usual. Evaluating the derivatives produces: " & 0 −x 2 − y 2 + 2 β y 2 − ( x 2 + y 2 − 2 β x 2 ) 0 $ $ $ 2 2 $ A Rij = 0 0 ' # x + y − 2 β x 2 − (−x 2 − y 2 + 2 β y 2 ) β +1 $ ( x 2 + y2 ) $$ 0 0 0 $ % ( " 0 −1 0 & $ 2A(1− β ) $ = # 1 0 0 ' β ( x 2 + y2 ) $% 0 0 0 $( c) The following answers are based on A > 0. For β < 1, fluid particles rotate counter clockwise (positive ωz). For β = 1, fluid particles do not rotate. For β > 1, fluid particles rotate clockwise (negative ωz). Interestingly, the streamlines are the same (circular!) for all three possibilities.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 3.27. Using indicial notation (and no vector identities) show that the acceleration a of a fluid particle is given by: a = ∂u ∂t + ∇

( u ) + ω × u where ω is the vorticity. 1 2

2

Solution 3.27. The acceleration of a fluid particle is a = Du Dt ≡ ∂u ∂t + ( u ⋅ ∇) u . Thus, the task is to prove (u ⋅€∇ )u = ∇

( u ) + ω × u. Start from the advective acceleration written in index 1 2

2

notation and force the rotation tensor to appear: $ ∂u ∂u ' ∂u ∂u 1 ∂ 2 u j i = u j && i − j )) − u j j = u j Rij + (u j ) . ∂x j ∂x i 2 ∂x i % ∂x j ∂x i ( € From (3.15), Rij = −εijkω k , so this becomes (1 2+ ∂u 1 ∂ 2 1 ∂ 2 u j i = −εijk u jω k + u j ) = εikjω k u j + u j ) = ω × u + ∇* u - , ( ( )2 , 2 ∂x i 2 ∂x i € ∂x j where € the final equality follows from the index notation definitions of the cross product (2.21), 2 gradient (2.22), and vector magnitude ( u 2j = u ).

€ €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 3.28. Starting from (3.29), show that the maximum uθ in a Gaussian vortex occurs when 1+ 2(r 2 σ 2 ) = exp(r 2 σ 2 ) . Verify that this implies r ≈ 1.12091σ. Solution 3.28. Differentiate the uθ equation from (3.29) with respect to r and set this derivative equal to zero. € 2 2 2 2 2 2 d Γ d '1− exp(−r σ ) * Γ ' 1− exp(−r σ ) exp(−r σ ) ' 2r ** ) ,= ) − )− 2 ,,, = 0 . (uθ (r)) = , 2π )− ( σ ++ dr 2π dr )( r r2 r + ( Eliminate common factors assuming r ≠ 0. 0 = −1+ exp(−r 2 σ 2 ) + (2r 2 σ 2 ) exp(−r 2 σ 2 ) = −1+ (1+ 2r 2 σ 2 ) exp(−r 2 σ 2 ) . € This can be rearranged to: exp( r 2 σ 2 ) = 1+ 2r 2 σ 2 , which is the desired result. When r/σ ≈ 1.12091, then € exp( r 2 σ 2 ) = 3.51289 and 1+ 2r 2 σ 2 = 3.51288 , which is suitable numerical€agreement. €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 3.29. Using (3.35) in two dimensions with F = 1, show that the time-rate-of-change of the area of the parallelogram shown is hl(dθ/dt)cosθ when θ depends on time while h and l are constants.

y!

l!

θ(t)!

h! x!

Solution 3.29. With F = 1, the volume integral of ∂F/∂t on the right side of (3.35) is zero, and the integrands are simplified, so (3.35) simplifies to: d ∫ dV = + ∫ A*(t ) b ⋅ n dA . dt V*(t ) (Here the volume integration is two dimensional, and produces the parallelogram's area, hlsinθ, which can be time differentiated, d(hlsinθ)/dt, to reach the desired result. However, this is not the intended solution path for this problem.) When θ is time-dependent, the parallelogram has three moving sides (a left side of length h, a right side of length h, and a top side of length l). Thus, the simplified version of (3.35) reduces to: d ∫ dV = ∫ left si de + ∫ right si de + ∫ top si de b ⋅ ndA . dt V*(t ) Here we note that the left and right sides move identically, so b will be the same. However, n points in opposite directions on these two sides, so the contributions from these two sides cancel. The x-coordinates of points on the parallelogram's top side are hcosθ ≤ x(t) ≤ l + hcosθ. The ycoordinate of the parallelogram's top side is y(t) = hsinθ. Time differentiate the location of any point on the parallelogram's top side to find b: b = (dx/dt, dy/dt) = (–hsinθ, hcosθ)(dθ/dt). And, on the parallelogram's top side, n is ey, so b ⋅ n = ( h cosθ ) ( dθ dt ) , and in two dimensions dA is just dx. Thus, the simplified version of (3.35) is: l+h cosθ d dθ dθ , dV = b ⋅ n dA = h cosθ dx = hl cosθ ∫ ∫ ∫ V*(t ) top si de dt dt dt h cosθ and this is the desired result.

{

}

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 3.30. Using (3.35) in two dimensions with F = 1, show that the time-rate-of-change of the area of the triangle shown is 12 b ( dh dt ) when h depends on time and b is constant.

y! b! h(t)! x! Solution 3.30. With F = 1, the volume integral of ∂F/∂t on the right side of (3.35) is zero, and the integrands are simplified, so (3.35) simplifies to: d ∫ dV = + ∫ A*(t ) b ⋅ n dA . dt V*(t ) (Here the volume integration is two dimensional, and produces the triangle's area, hb/2, which can be time differentiated, (d/dt)(hb/2) to reach the desired result. However, this is not the intended solution path for this problem.) When h is time-dependent, the triangle has two moving sides (a top side of length b, and a hypotenuse of length [h2 + b2]1/2). Thus, the simplified version of (3.35) reduces to: d ∫ dV = ∫ top si de + ∫ hypotenuse b ⋅ ndA . dt V*(t ) The x-coordinates of the triangle's top side are 0 ≤ x ≤ b. The y-coordinate of the triangle's top side is y(t) = h(t). Time differentiate the location of any point on the triangles's top side to find b: b = (dx/dt, dy/dt) = (0, dh/dt). And, on the triangle's top side, n is ey, so b ⋅ n = dh dt , The x-y coordinates of the triangle's hypotenuse fall on the line y = (h/b)x for 0 ≤ x ≤ b. The motion of points on the hypotenuse can be described by a purely vertical velocity. So, for any constant x-location, differentiate the equation of this line to find b: b = (dx/dt, dy/dt) = (0, (x/b)dh/dt). Tangent and normal vectors to an x-y curve are (1, dy/dx) and (dy/dx, –1), respectively. Thus, the outward unit normal on the hypotenuse of the triangle is: (h b, −1) = (h, −b) , so b ⋅ n = − x ( dh dt ) . n= h2 + b2 (h / b)2 +1 h2 + b2 Thus, the simplified version of (3.35) can be written: b b x ( dh dt ) $ d dh dV = b ⋅ n dA + b ⋅ n dA = dx − 1+ (h / b)2 dx &' , ∫ ∫ ∫ ∫ ∫ % V*(t ) 2 2 dt h +b top si de hypotenuse 0 dt 0

{

}

where, in two dimensions, dA is just dx on the top side and dA is a path length element along the hypotenuse, ds = [1 + (dy/dx)2]1/2dx. The factor in [,]-brackets is this path length element in terms of dx. Perform the integrations to find: b d dh 1 dh # x 2 & b dh ∫ dV = b dt − b dt %$ 2 (' = 2 dt , dt V*(t ) 0 and this is the desired result.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 3.31. Using (3.35) in two dimensions with F = 1, show that the time-rate-of-change of the area of the ellipse shown is π b ( da dt ) when a depends on time and b is constant. y!

b! x! a(t)!

Solution 3.31. With F = 1, the volume integral of ∂F/∂t on the right side of (3.35) is zero, and the integrands are simplified, so (3.35) simplifies to: d ∫ dV = + ∫ A*(t ) b ⋅ n dA . dt V*(t ) (Here the volume integration is two dimensional, and produces the ellipse's area, πab, which can be time differentiated, (d/dt)( πab) to reach the desired result. However, this is not the intended solution path for this problem.) When a is time-dependent, the contour that defines the ellipse, (y/b)2 + (x/a)2 = 1, is also time dependent. However, the symmetry of the ellipse allows the analysis to completed in the first quadrant alone and then multiplied by 4. Thus, the simplified version of (3.35) reduces to: d ∫ dV = 4 ∫ b ⋅ ndA . dt V*(t ) first qua drant The equation for the ellipse in the first quadrant is: x = +a 1− ( y b)

2

2

or y = +b 1− ( x a ) .

The motion of points on this curve can be described by a purely horizontal velocity when a varies but b is constant. So, for any constant y-location, differentiate the first equation to find b: ! dx $ ! da $ ! x da $ 2 b = # , 0& = # 1− ( y b) , 0 & = # , 0& , " dt % " dt % " a dt % where the final equality comes from changing the independent variable from y to x. Tangent and normal vectors to an x-y curve are (1, dy/dx) and (–dy/dx, 1), respectively. Thus, the outward unit normal on the ellipse in the first quadrant is: (−dy / dx,1) , so b ⋅ n = x(−dy / dx) da . n= (dy / dx)2 +1 a (dy / dx)2 +1 dt Thus, the simplified version of (3.35) can be written: a d x(−dy / dx) da $ dV = 4 b ⋅ n dA = 4 1+ (dy / dx)2 dx &' , ∫ ∫ ∫ % V*(t ) 2 dt first qua drant 0 a (dy / dx) +1 dt where, in two dimensions, dA is a path length element, ds = [1 + (dy/dx)2]1/2dx, along the firstquadrant portion of the ellipse. This path length element is the factor in [,]-brackets above.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Simplify the integrand, insert –dy/dx = (bx/a2)/[1 – (x/a)2]1/2, and perform the integration using the substitution x = asinθ to find: a d x(−dy / dx) da 4 da a xb(x / a 2 ) dV = 4 ∫ dx = dx ∫ ∫ dt V*(t ) a dt a dt 0 1− (x / a)2 0 da dt and this is the desired result. = 4b

π 2

∫ 0

sin 2 θ 2

1− sin θ

cosθ dθ = 4b

da dt

π 2

∫ sin 0

2

θ dθ = 4b

da π da = πb , dt 4 dt

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 3.32. For the following time-dependent volumes V*(t) and smooth single-valued integrand functions F, choose an appropriate coordinate system and show that ( d dt ) ∫ V *(t ) FdV obtained from (3.30) is equal to that obtained from (3.35). a) V*(t) = L1(t)L2L3 is a rectangular solid defined by 0 ≤ xi ≤ Li, where L1 depends on time while L2 and L3 are constants, and the integrand function F(x1,t) depends only on the first coordinate € and time. 2 b) V*(t) = (π/4)d (t)L is a cylinder defined by 0 ≤ R ≤ d(t)/2 and 0 ≤ z ≤ L, where the cylinder’s diameter d depends on time while its length L is constant, and the integrand function F(R,t) depends only on the distance from the cylinder’s axis and time. c) V*(t) = (π/6)D3(t) is a sphere defined by 0 ≤ r ≤ D(t)/2 where the sphere’s diameter D depends on time, and the integrand function F(r,t) depends only on the radial distance from the center of the sphere and time. Solution 3.32. The two equations are (3.30)

d x =b(t ) ∫ F(x,t)dx = dt x =a(t )

b

∫ a

∂F db da dx + F (b,t ) − F ( a,t ) , ∂t dt dt

d ∂F(x,t) F(x,t)dV = ∫ dV + ∫ F(x,t)b ⋅ ndA . ∫ dt V *(t ) ∂t V *(t ) A *(t ) a) Use Cartesian coordinates with the€origin at xi = 0. The cross sectional area of the rectangular solid, L2L3, is constant, so dV = L2L3dx1. The volume integral proceeds from x1 = 0 (= a) to x1 = L1(t) (= b) so (3.30) implies: € L1 d d x =L1 (t ) ∂F dL (a1) F (x,t)dV = F(x ,t)L L dx = L2 L3 dx1 + 1 F ( L1,t ) L2 L3 . ∫ ∫ ∫ 1 2 3 1 dt V *(t ) dt x1 =0 dt 0 ∂t Now start from (3.35) using a control volume that encloses the rectangular solid. In this case the only control surface that moves lies at x1 = L1, has outward normal n = e1, and moves with velocity b = (dL1/dt)ex. First evaluate the left side term from (3.35). € d d L1 (t ) L2 L3 d L1 (t ) F(x,t)dV = F( x ,t) dx dx dx = L L ∫ ∫ ∫ ∫ 1 1 2 3 2 3 dt ∫ F( x1,t)dx1 . dt V *(t ) dt x1 = 0 x 2 = 0 x 3 = 0 x1 = 0 Now evaluate the right side terms from (3.35). L1 (t ) L 2 L 3 L2 L3 $ dL ' d ∂F( x1,t) F(x,t)dV = dx dx dx + F(L1,t)& 1 e1 ) ⋅ e1dx 2 dx 3 ∫ ∫ ∫ ∫ ∫ ∫ 1 2 3 % dt ( dt V *(t ) ∂t x1 = 0 x 2 = 0 x 3 = 0 x2 = 0 x3 = 0 €

and (3.35)

L1 (t )

∂F( x1,t) dL dx1 + L2 L3 1 F(L1,t) ∂t dt x1 = 0 Setting the left and right side terms equal produces L1 (t ) d L1 (t ) ∂F( x1,t) dL (a2) L2 L3 F( x ,t)dx = L L dx1 + L2 L3 1 F(L1,t) , ∫ 1 1 2 3 ∫ dt ∂ t dt x = 0 x = 0 1 1 € which is identical to (a1). b) Use cylindrical coordinates with the origin at R = z = 0. The length area of the cylinder, L, is constant, so dV = L(2πRdR). The volume integral proceeds from R = 0 (= a) to R = d(t)/2 (= b) so € (3.30) implies: = L2 L3

∫

d /2 d ( d 2) d d d (t )/ 2 ∂F d F (x,t)dV = 2πL F(R,t)RdR = 2πL ∫ RdR + 2πL F ( d /2,t ) . ∫ ∫ dt V *(t ) dt R =0 dt 2 0 ∂t

€

(b1)

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Now start from (3.35) using a control volume that encloses the cylinder. In this case the only control surface that moves lies at R = d/2, has outward normal n = eR, and moves with velocity b = (d(d/2)/dt)eR. First evaluate the left side term from (3.35). d d L d / 2 2π d d /2 F(x,t)dV = F(R,t)RdϕdRdz = 2πL ∫ ∫ ∫ ∫ ∫ F(R,t)RdR . dt V *(t ) dt z= 0R = 0 ϕ = 0 dt 0 Now evaluate the right side terms from (3.35). L d / 2 2π L 2π & d(d /2) ) d ∂F(R,t) d F(x,t)dV = Rd ϕ dRdz + F(d/2,t)( e R + ⋅ e R dϕdz ∫ ∫ ∫ ∫ ∫ ∫ ' dt * dt V *(t ) ∂t 2 z= 0R = 0 ϕ = 0 z= 0ϕ = 0 € d /2 ∂F(R,t) d(d /2) d = 2πL ∫ RdR + 2πL F(d/2,t) ∂t dt 2 R= 0 Setting the left and right side terms equal produces d /2 d d /2 ∂F(R,t) d(d /2) d 2πL F(R,t)RdR = 2πL ∫ RdR + 2πL F(d/2,t) , (b2) ∫ dt 0 ∂t dt 2 R= 0 € which is identical to (b1). c) Use spherical coordinates with the origin at r = 0. The sphere expands symmetrically so the volume element is dV = 4πr2dr. The volume integral proceeds from r = 0 (= a) to r = D(t)/2 (= b) € implies: so (3.30) D /2 % D (2 d ( D 2) d d D(t )/ 2 ∂F 2 2 ∫ F (x,t)dV = 4π dt ∫ F (r,t)r dr = 4π ∫ ∂t r dr + 4 π dt F (D /2,t )'& 2 *) . (c1) dt V *(t ) r =0 0 Now start from (3.35) using a control volume that encloses the sphere. In this case the moving control surface lies at r = D/2, has outward normal n = er, and moves with velocity b = (d(D/2)/dt)er. First evaluate the left side term from (3.35). € d d D / 2 π 2π d D /2 F(x,t)dV = F(R,t)r 2 dr sin θdθdϕ = 4 π ∫ ∫ ∫ ∫ ∫ F(r,t)r 2 dr . dt V *(t ) dt r= 0 θ = 0 ϕ = 0 dt 0 Now evaluate the right side terms from (3.35). D / 2 π 2π π 2π ' d(D /2) * ' D * 2 d ∂F(r,t) 2 F(x,t)dV = r dr sin θ d θ d ϕ + F(d/2,t) e R , ⋅ e R ) , sin θdθdϕ ) ∫ ∫ ∫ ∫ ∂t ∫ ∫ ( dt + (2+ dt€V *(t ) r= 0 θ = 0 ϕ = 0 θ = 0ϕ = 0

' D *2 ∂F(r,t) 2 d(D /2) = 4π ∫ r dr + 4 π F(D/2,t)) , (2+ ∂t dt r= 0 Setting the left and right side terms equal produces D /2 % D (2 d D /2 ∂F(r,t) 2 d(D /2) 2 4π ∫ F(r,t)r dr = 4π ∫ ∂t r dr + 4π dt F(D/2,t)'& 2 *) , dt 0 r= 0 which is identical to (c1). D /2

€

€

(c2)

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 3.33. Starting from (3.35), set F = 1 and derive (3.14) when b = u and V*(t) = δV → 0 . Solution 3.33. With F = 1, b = u, and V*(t) = δV with surface δA, (3.35) becomes: d € ∫ dV = 0 + ∫ u ⋅ ndA . dt δV δA The first integral is merely δV. Use Gauss' divergence theorem on the second term to convert it to volume integral. d € (δV ) = ∫ ∇ ⋅ udV . dt δV As δV → 0 the integral reduces to a product of δV and the integrand evaluated at the center point of δV. Divide both sides of the last equation by δV and take the limit as δV → 0 : 1 d 1 1 lim δV ) =€ lim ∇ ⋅ udV = lim ( [(∇ ⋅ u)δV + ...] = ∇ ⋅ u = Sii , ∫ δV →0 δV dt δV →0 δV δV →0 δV € δV € and this is (3.14). €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 3.34. For a smooth single valued function F(x) that only depends on space and an arbitrarily-shaped control volume that moves with velocity b(t) that only depends on time, show that ( d dt ) ∫ V *(t ) F(x)dV = b ⋅ ∫ V *(t ) ∇F(x)dV .

(

€

)

Solution 3.34. Start from Reynolds transport theorem (3.35): d ∂F(x,t) F(x,t)dV = ∫ dV + ∫ F(x,t)b ⋅ ndA . ∫ dt V *(t ) ∂t V *(t ) A *(t ) When F does not depend on time, the first term on the right drops out. d ∫ F(x,t)dV = ∫ F(x,t)b ⋅ ndA . dt V *(t ) A *(t ) € When b does not depend on location, it can be taken outside the surface integral. $ ' d F(x,t)dV = b ⋅ & ∫ F(x,t)ndA) . ∫ dt V *(t ) % A *(t ) ( € Apply Gauss' theorem to the integral in large parentheses, to reach the desired form: % ( d F(x,t)dV = b ⋅ ∇F(x,t)dA ' ∫ *. ∫ dt & ) V *(t ) V *(t ) €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

3.35. Show that (3.35) reduces to (3.5) when V*(t) = δV → 0 and the control surface velocity b is equal to the fluid velocity u(x,t).

€

€

Solution 3.35. When V*(t) = δV with surface€δA, δV is small, and b = u, δV represents a fluid particle. Under these conditions (3.35) becomes: d ∂F(x,t) F(x,t)dV = ∫ dV + ∫ F(x,t)u ⋅ ndA , ∫ dt δV ∂t δV δA and the time derivative is evaluated following δV. Use Gauss' divergence theorem on the final term to convert it to a volume integral, ∫ F(x,t)u ⋅ ndA = ∫ ∇ ⋅ ( F(x,t)u)dV , € δA δV so that (3.35) becomes: '∂F(x,t) * '∂F(x,t) * d F(x,t)dV = ∫ ) + ∇ ⋅ ( F(x,t)u),dV = ∫ ) + F(x,t)∇ ⋅ u + (u ⋅ ∇ ) F(x,t),dV , ∫ + + dt δV ∂t δV ( δV ( € ∂t where the second equality follows from expanding the divergence of the product Fu. As δV → 0 the various integrals reduce to a product of δV and the integrand evaluated at the center point of δV. Divide both sides of the prior equation by δV and take the limit as δV → 0 to find: + 1 d 1 (∂F(x,t) € lim F(x,t)dV = lim + F(x,t)∇ ⋅ u + (u ⋅ ∇ ) F(x,t)-dV , ∫ ∫ * δV →0 δV dt δV →0 δV ,€ ∂t δV δV ) 0 * 1 d 1 -' ∂F(x,t) lim F(x,t)δV + ...] = lim + F(x,t)∇ ⋅ u + (u ⋅ ∇ ) F(x,t),δV + ...2, or ) [ / δV →0 δV dt δV →0 δV .( + ∂t 1 d 1 d ∂F(x,t) € F(x,t) + F(x,t) lim + F(x,t)∇ ⋅ u + (u ⋅ ∇) F(x,t) , (δV ) = δ V →0 dt δV dt ∂t where the product rule for derivative has been used on product FδV in [,]-braces on the left. 1 d From (3.14) or Exercise 3.33: lim (δV ) = ∇ ⋅ u , so the second terms on both sides of the δV →0 δV dt € last equation are equal and may be subtracted out leaving: d ∂F(x,t) F(x,t) = + (u ⋅ ∇ ) F(x,t) , dt ∂t and this is (3.5) when€the identification D Dt ≡ d dt is made.

€ €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.1. Let a one-dimensional velocity field be u = u(x, t), with v = 0 and w = 0. The density varies as ρ = ρ0(2 − cos ωt). Find an expression for u(x, t) if u(0, t) = U. Solution 4.1. Here u = u(x,t)ex, and the density field is given, so a solution for u(x,t) might be found from the continuity equation: ∂ρ ∂ρ ∂ρ ∂u + ∇ ⋅ ( ρu) = 0 , or specifically for this problem: + u + ρ = 0. ∂t ∂t ∂x ∂x The given density field only depends on time so ∂ρ/∂x = 0, and this leads to: ' sin(ωt) * ∂u 1 ∂ρ ρ 0 sin(ωt) =− =− → u = −) , x + C(y,z,t) . ∂x ρ ∂t ρ 0 (2 − cos(ωt)) € € ( 2 − cos(ωt) + where C is function of integration that does not depend on x. The initial condition requires: u(0, t) = U = C(y,z,t), so the final answer for u(x, t) is € $ sin(ωt) ' u = U −& )x . % 2 − cos(ωt) (

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.2. Consider the one-dimensional Cartesian velocity field: u = (α x t, 0, 0 ) where α is a constant. a) Find a spatially uniform, time-dependent density field, ρ = ρ(t), that renders this flow field mass conserving when ρ = ρo at t = to. b) What are the unsteady (∂u/∂t), advective ( [u ⋅ ∇]u ), and particle (Du/Dt) accelerations in this flow field? What does α = 1 imply? Solution 4.2. a) Use the continuity equation and the given velocity field with ρ = ρ(t): ∂ρ dρ α + ∇ ⋅ ( ρu) = 0 implies + ρ = 0. ∂t dt t Separate variables and integrate: dρ α ∫ ρ = − ∫ t dt ––> ln ρ = −α ln t + C . € € −α Exponentiate and evaluate the integration constant at t = to to find: ρ = ρ o ( t t o ) . b) For the given flow field: € 2 ∂u αx €⋅ ∇)u = u ∂u e = $ α x ' α e = α x e , and Du = ∂u + (u ⋅ ∇)u = α (α −1)x e . = − 2 e x , (u & ) x x x x % t (t ∂x t2 ∂t t t2 € Dt ∂t When α = 1 the unsteady and advective acclerations are non-zero, but they are equal and opposite so that Du/Dt is zero.

Fluid Mechanics, 6th Ed.

€

Kundu, Cohen, and Dowling

Exercise 4.3. Find a non-zero density field ρ(x,y,z,t) that renders the following Cartesian velocity fields mass conserving. Comment on the physical significance and uniqueness of your solutions. a) u = (U sin(ωt − kx),0,0) where U, ω, k are positive constants. [Hint: exchange the independent variables x,t for a single independent variable ξ = ωt–kx] b) u = (−Ωy,+Ωx,0) with Ω = constant. [Hint: switch to cylindrical coordinates] c) u = ( A x,B y,C z) where A, B, C are constants.

Solution 4.3. a) u = (U sin(ωt − kx),0,0) where U, ω, k are positive constants. Use the expanded ∂ρ ∂u ∂ρ form of the continuity equation for a unidirectional velocity, u = ( u,0,0) : +ρ +u = 0, ∂t ∂x ∂x and plug in the given velocity field to find: € ∂ρ ∂ρ − ρkU cos(ωt − kx) + U sin(ωt − kx) = 0 . ∂t ∂x € € Now follow the hint and change from independent variables (x, t) to ξ, where ξ = ωt–kx: ∂ ∂ξ d d ∂ ∂ξ d d = =ω = = −k , , and ∂t ∂t dξ dξ ∂x ∂x dξ dξ € dρ dρ ω − kU sin ξ = ρkU cosξ that can be to find a first-order differential equation: dξ dξ dρ kU cosξ € € dξ –> ln ρ = −ln(ω − kU sin ξ ) + C(y,z) . separated and integrated: ∫ ρ = ∫ ω − kU sin ξ Here, the constant C must be used to € make the solution dimensionally sound. Noting that ω/k has units of velocity, define M = kU/ω. Since the original equation did not contain any y or z € dependence the constant of integration might depend on these variables. So, the final solution € can be obtained by exponentiating the last equation: ρ (y,z) ρ o (y,z) , ρ(x, y,z,t) = o = 1− M sin ξ 1− ( kU ω ) sin(ωt − kx) where ρo(y,z) is an undetermined function. Thus, this solution is not fully determined; it is not unique. This is the density field that corresponds to a traveling-wave disturbance in a stationary medium. In the limit as M → 0 with ρo = constant, this wave becomes an acoustic plane wave € with ω/k = the speed of sound and M = the Mach number of the fluid particle motions. b) u = (−Ωy,+Ωx,0) with Ω = constant. The change to cylindrical coordinates is straight forward using: x = Rcosϕ & y = Rsin ϕ , and uR = ucosϕ + v sin ϕ & uϕ = −usin ϕ + v cosϕ . uR = −Ωy cosϕ + Ωx sin ϕ = −ΩRsin ϕ cos ϕ + ΩRcos ϕ sin ϕ = 0 , and uϕ = +Ωy sinϕ + Ωx cosϕ = ΩRsin 2 ϕ + ΩRcos 2 ϕ = ΩR . € ∂ρ €1 ∂ ∂ρ ∂ρ € € € Thus, u = ΩRe + +Ω = 0. ( ρΩR) = 0 or ϕ , so the continuity equation is: ∂t R ∂ϕ ∂t ∂ϕ € The solution of this equation can be obtained from the method of characteristics. The idea is to € determine special or characteristic directions or paths in the r-θ-z-t space along which the € solution for ρ is easy to find. Start by postulating the existence of such a path € € ρ = ρ( R(t),ϕ (t),z(t),t ) so that the total derivative of ρ with respect to time is: dρ ∂ρ dR ∂ρ dϕ ∂ρ dz ∂ρ = + + + . dt ∂R dt ∂ϕ dt ∂z dt ∂t

€ €

€ €

Fluid Mechanics, 6th Ed.

€

€

Kundu, Cohen, and Dowling

Comparing this equation with the one above shows that the characteristic paths are defined by dR dt = 0 , dϕ dt = Ω , and dz dt = 0 . Thus, after a single time integration of each equation, the characteristic paths are determined to be: R = Ro , ϕ = Ωt + ϕ o , and z = zo . where the quantities with subscript zero are constants. Along these paths the original equation € € for ρ becomes dρ/dt = 0, which implies the density is constant along these paths. If the density at Ro, ϕo, zo and t = 0 is ρ o (Ro ,ϕ o ,zo ) , then the density at all later times can be obtained by € € € substituting for ro, θo, zo from the three equations that specify the characteristic path. Hence, the solution for the density is: ρ = ρ o (R,ϕ − Ωt,z) . Here again ρo is an undetermined function so this solution is not fully determined; it is € not unique. In this case, the velocity field corresponds to solid-body rotation about the z-axis at angular rate Ω, so conservation of mass implies that the density variations must revolve around € the z-axis at the angular rate Ω as well. c) u = ( A x,B y,C z) where A, B, C are constants. ∂ρ Use the expanded form of the continuity equation: + ρ∇ ⋅ u + u ⋅ ∇ρ = 0 , and plug in the given ∂t " A B C % A ∂ρ B ∂ρ C ∂ρ ∂ρ velocity field to find: − ρ$ 2 + 2 + 2 '+ + + = 0 . This equation is linear in ∂t y z & x ∂ x y ∂ y z ∂z #x ρ, so try a separation of variables solution: € i.e. ρ = X(x)Y (y)Z(z)T(t) . Putting in the trial solution, dividing the whole equation by the trial solution, and grouping terms yields: T " $ A X " A ' $ B Y " B ' $C Z " C ' + − +& − )+ − =0 T &% x X €x 2 )( % y Y y 2 ( &% z Z z 2 )( where ( )´ denotes derivative of ( ) with respect to is argument. Because each group of terms depends on only one of the independent coordinates, this equation can only be satisfied if each term is equal to a constant, and the 4 constants sum to zero. This means setting: € A X" A B Y" B C Z" C T" − 2 = b , and − = c where a + b + c + d = 0 = d, − 2 = a, y Y y z Z z2 T x X x The solution of the first equation is: T(t) = Toe dt where To is a constant, while that of the second can be found from: " ax 2 % X " ax € 1 ax 2 € € € &, = + → ln X = + ln x + const., or X(x) = X o x exp# X A x€ 2A $ 2A ' where Xo is a constant. The solutions of the third and fourth equations are similar to that of the second. Combining the solutions of these equations and condensing the leading product of constants to C1 = To X oYo Z o produces: € $ ax 2 €by 2 cz 2 ' ρ(x, y,z,t) = C1 xyz exp% + + − (a + b + c)t ( & 2A 2B 2C ) € where C1, a, b, c are undetermined constants, and the above restriction on a, b, c, and d has been used to eliminate d. Any particular version of this solution is acceptable as long as C1 ≠ 0. Here, the density field is zero everywhere that ∇ ⋅ u → ∞ . The constants C1, a, b, and c are not determined € so again this solution is not fully determined; it is not unique. €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Overall, conservation of mass is typically inadequate to fully specify a fluid density or velocity field; conservation of momentum and initial & boundary conditions are needed for complete flow field solutions.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.4. A proposed conservation law for ξ, a new fluid property, takes the following form: d ∫ ρξ dV + ∫ Q ⋅ n dS = 0 , where V(t) is a material volume that moves with the fluid velocity dt V (t ) A(t ) u, A(t) is the surface of V(t), ρ is the fluid density, and Q = −ργ∇ξ . a) What partial differential equation is implied by the above conservation statement? ∂ξ 1 b) Use the part-a) result and the continuity equation to show: + u ⋅ ∇ξ = ∇ ⋅ ( ργ∇ξ ) . ∂t ρ Solution 4.4. a) Start with the given material CV equation, and apply Reynolds Transport Thm. to the first term on the left side and Gauss’s Divergence Thm. to the other term: d ∂ ρξ dV + ∫ Q ⋅ n dS = 0 → ∫ ( ρξ ) dV + ∫ ρξ ( u ⋅ n) dV + ∫ ∇ ⋅ Q dV = 0 . ∫ dt V (t ) S(t ) V (t ) ∂ t A(t ) V (t ) Now apply Gauss’s divergence theorem to the remaining surface integral, subsitute in the specified relationship Q = −ργ∇ξ , and combine all the terms into one volume integral to find: $∂ ' ∫ %&∂ t ( ρξ ) + ∇ ⋅ ( ρξ u) − ∇ ⋅ ( ργ∇ξ )() dV = 0 . V (t ) Here V(t) is an arbitrary material volume, so the integrand must be zero. Thus, the partial differential equation implied by the given CV equation is: ∂ ( ρξ ) + ∇ ⋅ ( ρξ u) = ∇ ⋅ ( ργ∇ξ ) . ∂t b) Expand the left side of the part a) result and group the terms that have ξ as a coefficient: # ∂ρ & ∂ξ ∂ρ ∂ξ ρ + ξ + ξ∇ ⋅ ( ρ u) + ρ u ⋅ ∇ξ = ρ + ρ u ⋅ ∇ξ + ξ % + ∇ ⋅ ( ρ u) ( . $ ∂t ' ∂t ∂t ∂t The contents of the parentheses is zero because of the continuity equation, so dividing by ρ yields: ∂ξ 1 + u ⋅ ∇ξ = ∇ ⋅ ( ργ∇ξ ) , ∂t ρ which can be recognized as the advection-diffusion equation for a conserved passive scalar.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.5. The components of a mass flow vector ρu are ρu = 4x2y, ρv = xyz, ρw = yz2. a) Compute the net mass outflow through the closed surface formed by the planes x = 0, x = 1, y = 0, y = 1, z = 0, z = 1. b) Compute ∇ ⋅ ( ρu) and integrate over the volume bounded by the surface defined in part a) c) Explain why the results for parts a) and b) should be equal or unequal. Solution 4.5. a) The specified volume is a cube, and the mass outflow will have six contributions € for each side): (one 1

∫∫ ρu ⋅ ndA =

cube surface

+

1

∫

∫ [4 x 2 y ] x=1 dydz −

1

1

y= 0z= 0

1

1

∫ ∫ [4 x 2 y ] x= 0 dydz

y= 0z= 0 1

∫ ∫ [ xyz] y=1 dxdz − ∫ ∫ [ xyz] y= 0 dxdz

x= 0z= 0 1

+

1

∫

x= 0z= 0

1

∫ [ yz 2 ] dxdy −

x= 0y= 0

z=1

1

1

∫ ∫ [ yz 2 ] z= 0 dxdy

x= 0y= 0

The integral evaluations are straightforward and unremarkable, and lead to: 1 1 11 ∫∫ ρu ⋅ ndA = 2 + 0 + 4 + 0 + 2 + 0 = 4 . cube surface € b) First compute the divergence of the mass flow: ∂ ∂ ∂ ∇ ⋅ ( ρu) = ( ρu) + ( ρv ) + ( ρw ) = 8xy + xz + 2yz , ∂x ∂y ∂z € then integrate it in the cubical volume: 1 1 1 & 1 )& 1 ) & 1 )& 1 ) & 1 )& 1 ) 11 ∫∫∫ ∇ ⋅ (ρu)dV = ∫ ∫ ∫ (8xy + xz + 2yz)dxdydz = 8(' 2 +*(' 2 +* + (' 2 +*(' 2 +* + 2(' 2 +*(' 2 +* = 4 . cube € x= 0y= 0z= 0 c) The two answers should be the same because of Gauss' Divergence Theorem.

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.6. Consider a simple fluid mechanical model for the atmosphere of an ideal spherical star that has a surface gas density of ρo and a radius ro. The escape velocity from the surface of the star is ve. Assume that a tenuous gas leaves the star’s surface radially at speed vo uniformly over the star’s surface. Use the steady continuity equation for the gas density ρ and fluid velocity u = (ur , uθ , uϕ ) in spherical coordinates

1 ∂ 2 1 ∂ 1 ∂ r ρur ) + ( ρuθ sinθ ) + ( ρuϕ ) = 0 ( 2 r ∂r r sin θ ∂θ r sin θ ∂ϕ for the following items

(

)

a) Determine ρ when vo ≥ ve so that u = (ur , uθ , uϕ ) = vo 1− ( ve2 vo2 ) (1− ( ro r )), 0, 0 . b) Simplify the result from part a) when vo >> ve so that: u = (ur , uθ , uϕ ) = (vo , 0, 0) . c) Simplify the result from part a) when vo = ve . d) Use words, sketches, or equations to describe what happens when vo < ve . State any assumptions that you make. Solution 4.6. a) ur is the only non-zero velocity component so only the continuity equation simplifies to 1 ∂ 2 (r ρur ) = 0 , r 2 ∂r multiply by r2 and integrate to find: r 2 ρur = C(θ,ϕ ) , where C is the function of integration that cannot depend on r. Thus, −1 2 C(θ,ϕ ) ρ o ro2 , ρ =€ 2 = 2 1− (v e2 v o2 )(1− ( ro r)) r ur r € where C(θ,ϕ ) = ρ o ro2v o = const. is determined from the boundary conditions: ρ = ρo & v = vo at r = ro . 2 2 b) When v o >> v€ e , the square root factor in the result of part a) becomes unity, so ρ = ρ o ro r . −1 2 32 € c) When v o = v e , the square root factor simplifies to ( ro r) , so ρ = ρ o [ ro r] d) Here, the gas leaving the star’s surface does not escape the star’s gravitational field, so it must €fall back to star’s surface. For a tenuous gas, it might be acceptable € to ignore molecular

[

]

(

)

+ 2 2 moving €collisions so that the velocity field of the outward € € gas, ur = v o 1− v e v o (1− ( ro r)) for r < rmax = ro [1− v o2 v e2 ] , is equal and opposite to the velocity field of the inward falling gas so

€ €

that u−r = −ur+ . Both outward and inward mass fluxes must balance because the net mass flux must be zero. Thus, the resulting density field will have € equal parts contributed by inward and outward flowing stellar gas. Under these conditions, the density field will be double that from part a) with a radial limitation: % ρ o ro2 ) −1 2 2 2 '2 ' 1− v v 1− r r for r < r ( ) ( ) ( ) e o o max 2 ρ=& r * where rmax = ro [1− v o2 v e2 ] . '(0 for r ≥ rmax '+

[

]

€

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.7. Consider the three-dimensional flow field ui = β xi or equivalently u = βrer, where β is a constant with units of inverse time, xi is the position vector from the origin, r is the distance from the origin, and eˆr is the radial unit vector. Find a density field ρ that conserves mass when: a) ρ(t) depends only on time t and ρ = ρo at t = 0, and b) ρ(r) depends only on the distance r and ρ = ρ1 at r = 1 m. c) Does the sum ρ(t) + ρ(r) also conserve mass in this flow field? Explain your answer. Solution 4.7. The full continuity equation is: ∂ρ ∂t + (u ⋅ ∇)ρ + ρ∇ ⋅ u = 0 . a) When ρ depends only on t, (u ⋅ ∇)ρ = 0 so the cont. equation reduces to: d ρ dt + ρ∇ ⋅ u = 0 . For the given velocity field, ∇ ⋅ u = ∂ui ∂xi = β ∂xi ∂xi = β (∂x1 ∂x1 + ∂x2 ∂x2 + ∂x3 ∂x3 ) = 3β , so

d ρ dt = −3βρ . This equation has the exponential solution: ρ (t) = ρo e−3βt when ρ = ρo at t = 0. b) When ρ only depends on distance r, ∂ρ/∂t = 0 so the continuity equation reduces to: (β reˆr )⋅ ∇ρ + 3βρ = 0 , or r ( d ρ dr ) = −3ρ . This equation has the power law solution 3

ρ (r) = ρ1 ⋅ (1m r ) when ρ = ρ1 at r = 1 m. c) The proposed sum solution, ρ(t) + ρ(r), does conserve mass, and this is readily shown by substitution of ρ(t) + ρ(r) into the full continuity equation. This sum solution conserves mass because the continuity equation is linear in ρ when u is specified independently of ρ. When ρ when u are both dependent field variables, the continuity equation is non-linear.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.8. The definition of the stream function for two-dimensional constant-density flow in the x-y plane is: u = −e z × ∇ψ , where ez is the unit vector perpendicular to the x-y plane that determines a right-handed coordinate system. a) Verify that this vector definition is equivalent to u = ∂ψ ∂y , and v = −∂ψ ∂x in Cartesian coordinates. € b) Determine the velocity components in r-θ polar coordinates in terms of r-θ derivatives of ψ. c) Determine an equation for the z-component of the vorticity in terms of ψ. € € Solution 4.8. a) Start from the definition of the cross product: ex ey ez ' ∂ψ * ' ∂ψ * ' ∂ψ * ' ∂ψ * u = −e z × ∇ψ = − 0 0 1 = −e x )− , − e y ) , = e x ) , + e y ) − , . ( ∂x + ( ∂x + ( ∂y + ( ∂y + ∂ψ ∂x ∂ψ ∂y 0 Setting components equal from the extreme ends of this extended equality produces: u = ∂ψ ∂y , and v = −∂ψ ∂x . b) In r-θ polar coordinates u = urer + u e , e z × e r = eθ , and e z × eθ = −e r . Plus, € ∂ψ 1 ∂ψ . Therefore, using the same definition in r-θ polar coordinates produces: ∇ψ = e r + eθ ∂r r ∂θ € € ( ∂ψ 1 ∂ψ + ∂ψ 1 ∂ψ u = ure r + uθ eθ = −e z × ∇ψ = −e z × *€e r + eθ + er . €θ - = −e ) ∂r r ∂θ , ∂r r ∂θ Matching components implies: 1 ∂ψ ∂ψ and uθ = − . ur = r ∂θ ∂r c) € The z-component of the vorticity is: ∂v ∂u ∂ & ∂ψ ) ∂ & ∂ψ ) & ∂ 2ψ ∂ 2ψ ) ω z = − = ( − + − ( + = −( 2 + 2 + = −∇ 2ψ . ∂x€ ∂y ∂x ' ∂x€* ∂y ' ∂y * ' ∂x ∂y * θ

€

€

θ

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.9. A curve of ψ (x, y) = C1 (= a constant) specifies a streamline in steady twodimensional constant-density flow. If a neighboring streamline is specified by ψ (x, y) = C2 , show that the volume flux per unit depth into the page between the streamlines equals C2 – C1 when C2 > C1. € Solution 4.9. Start with a picture of the two streamlines and let Q = volume€flow rate (per unit depth) between them.

 ∇ψ

ψ=C

2

d

ψ=C

1

€

€

2

By definition, Q = ∫ | u | d where d is an increment of length that lies perpendicular to the flow 1

in the stream tube. Again by definition, ∇ψ lies in the direction perpendicular to the lines of constant ψ. Thus: € ∂ψ ∂x dx + ∂ψ ∂y dy ) ( ) = −vdx + udy = dψ , so | u | d = dψ . € d = ∇ψ ⋅ dx = ( 2 2 | ∇ψ | |u | v 2 + u2 € ∂x ) + (∂ψ ∂y ) (∂ψ 2

Q = ∫ | u | d =

This means that:

1

€ €

2

∫ dψ = C2 − C1 , 1

which completes the proof. €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.10. Consider steady two-dimensional incompressible flow in r-θ polar coordinates where u = (ur , uθ ) , ur = + ( Λ r 2 ) cosθ , and Λ is positive constant. Ignore gravity. a) Determine the simplest possible uθ . b) Show that the simplest stream function for this flow is ψ = ( Λ r ) sin θ . c) Sketch the streamline pattern. Include arrowheads to show stream direction(s). d) If the flow is frictionless and the pressure far from the origin is p∞, evaluate the pressure p(r, θ) on θ = 0 for r > 0 when the fluid density is ρ. Does the pressure increase or decrease as r increases? 1 ∂ 1 ∂u (rur ) + θ = 0 , to find: r ∂r r ∂θ # & 1 ∂ 1 ∂ Λ Λ 1 ∂uθ ∂uθ Λ Λ , or = + 2 cosθ → uθ = + 2 sin θ , % + cosθ ( = − 3 cosθ = − (rur ) = ' r ∂r r ∂r $ r r r ∂θ ∂θ r r where the final constant of integration has been dropped to produce the simplest possible uθ . b) By definition: ur = (1 r ) (∂ψ ∂θ ) , and uθ = −∂ψ ∂r . Thus,

Solution 4.10. a) Place the given ur = + ( Λ r 2 ) cosθ into ∇ ⋅ u =

1 ∂ψ Λ ∂ψ Λ Λ = + 2 cosθ so = + cosθ → ψ = sin θ + f (r) , and r ∂θ r ∂θ r r ∂ψ Λ Λ − = + 2 sin θ so ψ = sin θ + g(θ ) . ∂r r r y! The simplest stream function is recovered when f = g = 0: ψ = ( Λ r ) sin θ . c) Start from the result of b) and switch to Cartesian coordinates: Λ y , which implies x 2 + y 2 = ( Λ ψ ) y , or ψ= 2 2 2 2 x +y x +y 2

2

x 2 + ( y − Λ 2ψ ) = ( Λ 2ψ ) . Thus, the streamlines are circles centered on the y-axis that are tangent to the x-axis at the origin of coordinates. The stream direction is determined from the radial velocity ur, which must be positive when θ is near zero. d) Use the ordinary Bernoulli-equation to find: & 1 1 # Λ2 Λ2 p∞ = p(r, θ )θ =0 + ρ ( vr2 + vθ2 ) = p(r, 0) + ρ % 4 cos2 (0) + 4 sin 2 (0)( , or θ =0 2 2 $r r ' 1 Λ2 p(r, 0) = p∞ − ρ 4 . 2 r

The pressure increases as r increases.

x!

Fluid Mechanics, 6th Ed.

€

Kundu, Cohen, and Dowling

Exercise 4.11. The well-known undergraduate fluid mechanics textbook by Fox et al. (2009) provides the following statement of conservation of momentum for a constant-shape (nonrotating) control volume moving at a non-constant velocity U = U(t). d dU ρurel dV + ∫ ρurel (urel ⋅ n)dA = ∫ ρgdV + ∫ fdA − ∫ ρ dV . ∫ dt V *(t ) dt A *(t ) V *(t ) A *(t ) V *(t ) Here urel = u − U(t) is the fluid velocity observed in a frame of reference moving with the control volume while u and U are observed in a non-moving frame. Meanwhile, (4.17) states this law as € d ∫ ρudV + ∫ ρu(u − U) ⋅ ndA = ∫ ρgdV + ∫ fdA dt V *(t ) A *(t ) V *(t ) A *(t ) where the replacement b = U has been made for the velocity of the accelerating control surface A*(t). Given that the two equations above are not identical, determine if these two statements of conservation of fluid momentum are contradictory or consistent. € Solution 4.11. The goal here is to derive one result from the other. Start with the equation from the undergraduate textbook, substitute using urel = u − U , and put the forces on the left and move the extra acceleration term to the other side of the equation. dU d ∫ ρgdV + ∫ fdA = ∫ ρ dt dV + dt ∫ ρ(u − U)dV + ∫ ρ(u − U)(u − U) ⋅ ndA V *(t ) A *(t ) V *(t ) V *(t ) A *(t ) € Now expand the terms on the terms on the right side. ∫ ρgdV + ∫ fdA = V *(t )

€

€

€

A *(t )

dU d dU d dV + ρudV − ρdV − U ∫ ∫ ∫ ρdV + ∫ ρu(u − U) ⋅ ndA − U ∫ ρ(u − U) ⋅ ndA dt dt V *(t ) dt V *(t ) dt V *(t ) V *(t ) A *(t ) A *(t ) Here, U and dU/dt depend on time only – that is, they are uniform over V*(t) and A*(t) – so they can move inside or outside of the volume and surface integrals over V*(t) and A*(t), but not through time differentiations. Thus, the first and third terms on the right side are equal and opposite. Rearrange the other terms to find: &d ) d ∫ ρgdV + ∫ fdA = dt ∫ ρudV + ∫ ρu(u − U) ⋅ ndA − U( dt ∫ ρdV + ∫ ρ(u − U) ⋅ ndA+. ' V *(t ) * V *(t ) A *(t ) V *(t ) A *(t ) A *(t ) Conservation of mass requires the contents of the [,]-brackets to equal zero. After removing these terms, the remaining simplified equation, d ∫ ρudV + ∫ ρu(u − U) ⋅ ndA = ∫ ρgdV + ∫ fdA , dt V *(t ) A *(t ) V *(t ) A *(t ) is the same as (4.17). The two statements of conservation of momentum are consistent. The essential difference between the two CV formulations of conservation of momentum amounts to a transformation between an inertial and an accelerating frame of reference when mass is also conserved. €

∫

ρ

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.12. A jet of water with a diameter of 8 cm and a speed of 25 m/s impinges normally on a large stationary flat plate. Find the force required to hold the plate stationary. Compare the average pressure on the plate with the stagnation pressure if the plate is 20 times the area of the jet. Solution 4.12. Choose a rectangular CV with one side that covers and the flat stationary plate.

Ajet Ujet

x

Denote the plate area by A, and the jet velocity & area by Ujet and Ajet, respectively. In this problem only the x-component of the integral momentum equation is required, and x-direction fluid momentum only enters the CV of its left side. Assume that atmospheric pressure Po acts on the three sides of the CV that are not in contact with the plate. −ρU 2jet A jet + 0 = Po A − ∫ PdA = Fx , or A

π (0.08m) 2 = −3.142kN 4 2 3 ρ U F (10 kg /m 3 )(25m /s) 2 jet x € = = = 31.25kPa . The average pressure is: Pave = 20A jet 20 20 € P 1 1 (10 3 kg /m 3 )(25m /s) 2 The stagnation pressure is: Ps = ρU 2jet = = 312.5kPa , so ave = Ps 10 2 2 Here Ps and Pave€are reported as gauge pressures. 2 jet

Fx = −ρU A jet = −(10 kg /m )(25m /s) 2

€

3

3

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.13. Show that the thrust developed by a stationary rocket motor is F = ρAU2 + A(p − patm), where patm is the atmospheric pressure, and p, ρ, A, and U are, respectively, the pressure, density, area, and velocity of the fluid at the nozzle exit. Solution 4.13. Use the control volume shown, where F is the force that holds the rocket motor in place. patm p A !

F

U

Assuming the flow is steady and that control volume is not moving (b = 0), the horizontal component of (4.17) becomes: ρU 2 A = F + patm A − pA , or F = ρU 2 A + A( p − patm ) .

€

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.14. Consider the propeller of an airplane moving with a velocity U1. Take a reference frame in which the air is moving and the propeller [disk] is stationary. Then the effect of the propeller is to accelerate the fluid from the upstream value U1 to the downstream value U2 > U1. Assuming incompressibility, show that the thrust developed by the propeller is given by F = ρ A(U 22 −U12 ) 2, where A is the projected area of the propeller and ρ is the density (assumed constant). Show also that the velocity of the fluid at the plane of the propeller is the average value U = (U1 + U2)/2. [Hint: The flow can be idealized by a pressure jump of magnitude Δp = F/A right at the location of the propeller. Also apply Bernoulli’s equation between a section far upstream and a section immediately upstream of the propeller. Also apply the Bernoulli equation between a section immediately downstream of the propeller and a section far downstream. This will show that Δp = ρ (U 22 −U12 ) / 2 .] Solution 4.14. The CV and nominal propeller-edge streamlines are:

1

€

a b

2

where the mass flux within the curved propeller-edge streamlines is constant and equal to m˙ = ρU1 A1 = ρU 2 A2 . Apply the steady constant density Bernoulli equation from "1" to "a", and from "b" to "2": p1 + 12 ρU12 = pa + 12 ρU a2 , and pb + 12 ρU b2 = p2 + 12 ρU 22 . Since p1 = p2, and Ua = Ub, these equations become: pb − pa = + 12 ρ(U 22 − U12 ) . Now apply€the momentum principle across € the propeller plane: 2 2 −ρU a + ρU b = 0 = pa A − pb A + F , or F = A( pb − pa ) = 12 ρA(U 22 − U12 ) . where Ua = Ub, and F is the €force applied to the propeller (positive to the right) by the device that holds it in place. To find the velocity at the propeller plane, use the stationary rectangular control volume € € shown above, and start with (4.5), −ρU1 A1 + ρU 2 A2 + ρU1 (A1 − A2 ) + m˙ sides = 0 , where the four terms correspond to the inlet, fast flow at the outlet, ordinary speed flow at the outlet, and stream-wise sides of the CV. Here the inlet flow is uniform at velocity U1. Using the definition of m˙ , the conservation of mass statement becomes: € ρU1 (A1 − A2 ) + m˙ sides = 0 . where A2 is the fast-flow area on the outlet side of the CV. Now evaluate the horizontal component of (4.17) € −m˙ U1 + m˙ U 2 + ρU12 (A1 − A2 ) + U1m˙ sides = ( p1 − p2 )A1 + F . €

€

Fluid Mechanics, 6th Ed.

€

Kundu, Cohen, and Dowling

Substitute in the conservation of mass result and note that p1 = p2 to find: −m˙ U1 + m˙ U 2 = F , or F = ρUA(U 2 − U1 ) . where m˙ = ρUA , while U (= Ua = Ub) and A are flow velocity and area inside the curved lines at the propeller plane. Equate the relationship for F found from the Bernoulli equations with the one found from conserving mass and momentum to develop an equation for U: € € F = 12 ρA(U 22 − U12 ) = ρUA(U 2 − U1 ) , 1

Use the second equality, cancel common factors, and solve for U to find: U = 2 (U 2 + U1 ) . €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.15. Generalize the control volume analysis of Example 4.1 by considering the control volume geometry shown for steady two-dimensional flow past an arbitrary body in the absence of body forces. Show that the force the fluid exerts on the body is given by the Euler momentum integral, F j = − ∫ ( ρui u j − τ ij )n i dA , and that 0 = ∫ ρui n i dA . A1

A1

Solution 4.15. For steady flow through a stationary control volume (b = 0) and no body force, (4.5) and (4.17) simplify to: € ∫ ρ(u ⋅€n)dA = 0 → ∫ ρui n i dA = 0 , and A *(t )

A *(t )

∫ ρu(u ⋅ n) dA = ∫ f(n,x,t)dA

A *(t )

∫ ρu j ui n i dA = ∫ n iτ ij dA ,

→

A *(t )

A *(t )

A *(t )

where the second equality in each case is the same as first after conversion to index notation, and (2.15) or (4.20b)€has been used to write the surface force f in terms of the stress tensor τ€ij. Here the control surface A* is composed of three pieces, A* = A1 + A2 + A3, where A1 is the outer surface, A2 conforms to the body's surface, and A3 is the connection between A1 and A2. Here n (with components ni) is the outward normal on the combined surface. Thus, n points outward on A1 (shown as n1 in the figure) and points into the body on A2 (shown as n2 in the figure). Thus, the conservation laws require: ∫ ρui n i dA = 0 and ∫ (ρu j ui − τ ij )n i dA = 0 . A1 +A 2 +A 3

A1 +A 2 +A 3

Now consider the contribution of each surface individually. In general, there are no simplifications to be made on A1 and all the integrand terms must be retained. The surface A2 coincides with the surface of the solid body, thus u ⋅ n = uini = 0 on A2. This fact eliminated the € of mass equation, and causes the first integrand term to € A in the conservation contribution from 2 dropout of the conservation of momentum equation. On A3, as the width of the opening goes to zero, the magnitude of the integrands on € the upper and lower surfaces become equal but the normal on these two surfaces is opposite; therefore, A3 provides no net contribution to either conservation law. Thus, the integral laws become: ∫ ρui n i dA = 0 and − ∫ τ ij n i dA + ∫ (ρu j ui − τ ij )n i dA = 0 . A1

A2

A1

The first of these is sought as part of the solution to this exercise. The second becomes the Euler momentum integral after considering the lone integral over A2 and orientation of n on A2. If n pointed outward from the body, this integral would represent the force, F, the fluid applies to the € on A , n (= n ) € body. However 2 2 points inward so –n points outward from the body, thus F = − ∫ τ ij n i dA . A2

Therefore the final result (the Euler momentum integral) is: F = − ∫ ( ρu j ui − τ ij ) n i dA . A1

€ €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.16. The pressure rise Δp = p2 − p1 that occurs for flow through a sudden pipe-crosssectional-area expansion can depend on the average upstream flow speed Uave, the upstream pipe diameter d1, the downstream pipe diameter d2, and the fluid density ρ and viscosity µ. Here p2 is the pressure downstream of the expansion where the flow is first fully adjusted to the larger pipe € diameter. a) Find a dimensionless scaling law for Δp in terms of Uave, d1, d2, ρ and µ. b) Simplify the result of part a) for high-Reynolds-number turbulent flow where µ does not matter. c) Use a control volume analysis to determine Δp in terms of Uave, d1, d2, and ρ for the high Reynolds number limit. [Hints: i) a streamline drawing might help in determining or estimating the pressure on the vertical surfaces of the area transition, and ii) assume uniform flow profiles wherever possible.] d) Compute the ideal flow value for Δp and compare this to the result from part d) for a diameter ratio of d1/d2 = ½. What fraction of the maximum possible pressure rise does the sudden expansion achieve? d2!

d1 ! p1!

Uave!

p2!

Solution 4.16. a) This is a dimensional analysis task. Construct the units matrix. Δp Uave d1 d2 ρ µ M 1 0 0 0 1 1 L -1 1 1 1 -3 -1 T -2 -1 0 0 0 -1 The matrix has rank three so there are 6 – 3 = 3 dimensionless groups. These can be found by 2 inspection; thus: Δp ( ρU ave ) = f (d1 d2 , ρU ave d1 µ) where f is an undetermined function. b) At high Reynolds number the viscosity will not be a parameter, so the Reynolds number can 2 be dropped from the part a) result: Δp ( ρU ave ) = f (d1 d2 ) . The remainder of this exercise involves € finding f. c) Place a cylindrical CV in the duct that abuts the area change on the upstream € side and extends far enough downstream to fully enclose the flow’s reattachment zone. For high-Re turbulence, its OK to assume uniform inflow and outflow Separation profiles. Denote the outflow velocity by Reattachment U2 . π 2 π Cons. of mass implies: − d1 U ave + d22U 2 = 0 ; (a) 4 4 π π π π 2 Cons. of horizontal momentum: − d12 ρU ave (b) + d22 ρU 22 = p1 d22 − p2 d22 . 4 4 4 4

€ €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Here, the cross-stream momentum equation suggests that p1 acts on the flat surfaces of the area transition because the flow separates at the area junction and is parallel to the x-direction and this implies that ∂p ∂r = 0 . Use the above equations to determine Δp = p2 − p1. Equation (b) implies: 2 p2 − p1 = ρd12U ave d22 − ρU 22 , and equation (a) reduces to: U 2 = ( d12 d22 )U ave . Eliminate U2 to find: 2 2 2 2 € p2 − p1 = ρU ave p2 − p1 = ρd12U ave d22 − ρ( d12 d22 ) U ave , or € (d12 d22 ) 1− (d12 d22 ) .

€

(

1 2

d) For an ideal flow: p2 − p1 = ρU

2 ave

1 2

2 2

1 2

− ρU = ρ€U

(

)

2 ave

(1− d

4 1

d

4 2

) = (15 32)ρU

)

2 ave

From part d): p2 − p1 = ρU ( d d ) 1− ( d d ) = ( 3 16) ρU . Thus, the pressure rise in a €sudden expansion to four times the area with € separated flow is only 40% of that possible in an ideal flow. € Abrupt area changes are exceptionally common in automotive piston-engine exhaust € systems where they are used for low-frequency noise control. However, they lead to pumping losses that reduce the engine's horsepower, especially under wide-open-throttle conditions, because of the principles illustrated in this exercise. 2 ave

2 1

2 2

2 1

2 2

2 ave

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.17. Consider how pressure gradients and skin friction develop in an empty wind tunnel or water tunnel test section when the flow is incompressible. Here the fluid has viscosity µ and density ρ, and flows into a horizontal cylindrical pipe of length L with radius R at a uniform horizontal velocity Uo. The inlet of the pipe lies at x = 0. Boundary layer growth on the pipe’s walls induces the horizontal velocity on the pipe’s centerline to be UL at x = L; however, the pipe-wall boundary layer thickness remains much smaller than R. Here, L/R is of order 10, and ρUoR/µ >> 1. The radial coordinate from the pipe centerline is r. a) Determine the displacement thickness, δ L* , of the boundary layer at x = L in terms of Uo, UL , and R. Assume that the boundary layer displacement thickness is zero at x = 0. [The boundary layer displacement thickness, δ*, is the thickness of the zero-flow-speed layer that displaces the € the actual boundary layer. For a boundary layer velocity outer flow by the same amount as profile u(y) with y = wall-normal coordinate and U = outer flow velocity, δ* is defined by: ∞ δ* = ∫ 0 (1− ( u U ))dy .]

€

€

b) Determine the pressure difference, ΔP = PL – Po, between the ends of the pipe in terms of ρ, Uo, and UL. c) Assume the horizontal velocity profile at the outlet of the pipe can be approximated by:

(

u(r, x = L) = U L 1− ( r R)

n

) and estimate average skin friction,

τ w , on the inside of the pipe

between x = 0 and x = L in terms of ρ, Uo, UL, R, L, and n. d) Calculate the skin friction coefficient, c f = τ w 12 ρU o2 , when Uo = 20.0 m/s, UL = 20.5 m/s, R = € 1.5 m, L = 12 m, n = 80, and the fluid is water, i.e. ρ = 103 kg/m3. € r!

U o!

U L! x = 0!

x = L!

2R!

Solution 4.17. a) Apply the principle of conservation of mass. The effective flow area will be 2

2

πR 2 at the inlet (x = 0), but will be π ( R − δL* ) at the outlet (x = L): U oπR 2 = U L π ( R − δL* ) .

(

)

Solve for δL* = R 1− U o U L . €

€

b) Start from the steady Bernoulli equation (4.19) for horizontal flow (no body force): 1 1 1 1 1 ρU o2 + Po = ρU L2 + PL ,€and rearrange it to find: PL − Po€= ρU o2 − ρU L2 = ρ(U o2 − U L2 ) . 2 2 2 2 2 € c) Choose a CV that encloses all the fluid in the tube between x = 0 and x = L, and apply the CV form of conservation of momentum (4.17). Here the flow is steady and there is no net body force, thus: € R R n 2 −2πρ ∫ U o2 rdr +2πρ ∫ U L2 1− ( r R) rdr = ( Po − PL )πR 2 − 2πRLτ w r= 0

r= 0

(

)

Cancel common factors, evaluate the integrals, and substitute in the part b) result for the pressure difference: R

€

r= 0

€

(

n

−ρU o2 R 2 + 2 ρ ∫ U L2 1− 2( r R) + ( r R)

2n

)rdr = (P − P )R o

L

2

− 2RLτ w

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

$ R 2 2 R n +2 1 R 2n +2 ' 2 −ρU o2 R 2 + 2 ρU L2 & − n − 2n ) = ( Po − PL ) R − 2RLτ w 2 R n + 2 R 2n + 2 % ( $ ' 4 1 1 2 2 2 −ρU o2 R 2 + ρU L2 R 2 &1− + ) = ρ(U L − U o ) R − 2RLτ w % n + 2 n + 1( 2 2 Divide € by ρR and collect terms: # 4 1 & 1 2 Lτ 2 −U o2 + U L2 %1− + ( = (U L − U o ) − 2 w $ n + 2 n + 1' 2 ρR € #1 4 1 & 2 1 2 Lτ w + % − (U − U = −2 $ 2 n + 2 n + 1' L 2 o ρR Solve for the average shear stress: € ρR + 2 % 8 2 ( 2. τw = + *U 0 -U o − '1− & n + 2 n + 1) L / 4L , € 2τ R+ % 8 2 ( U L2 . + d) Use the result of part c) to find: c f = w2 = * 0 . Evaluate: -1− '1− ρU o 2L , & n + 2 n + 1) U o2 / 2 € 1.5m ) # 8 2 &# 20.5 & , cf = +1− %1− + (% ( .= 0.00162 2(12m) * $ 82 81'$ 20.0 ' The numbers provided here € are approximately applicable to the William B. Morgan Large Cavitation Channel in Memphis, Tennessee, the world's largest low-turbulence water tunnel. €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.18. An acid solution with density ρ flows horizontally into a mixing chamber at speed V1 at x = 0 where it meets a buffer solution with the same density moving at speed V2. The inlet flow layer thicknesses are h1 and h2 as shown, the mixer chamber height is constant at h1 + h2, and the chamber width into the page is b. Assume steady uniform flow across the two inlets and the outlet. Ignore fluid friction on the interior surfaces of the mixing chamber for parts a) and b). a) By conserving mass and momentum in a suitable control volume, determine the pressure difference, Δp = p(L) – p(0), between the outlet (x = L) and inlet (x = 0) of the mixing chamber in terms of V1, V2, h1, h2, and ρ. Do not use the Bernoulli equation. b) Is the pressure at the outlet higher or lower than that at the inlet when V1 ≠ V2? c) Explain how your answer to a) would be modified by friction on the interior surfaces of the mixing chamber.

V1!

h1!

V2!

h2!

x = 0!

x = L!

Solution 4.18. a) Choose a stationary control volume (b = 0) that captures the fluid between x = 0 and x = L. Here the inflows and outflows are presumed steady. Thus, conservation of mass from (4.5) reduces to: ∫ ρu ⋅ n dA = 0 . A*(t )

At x = 0 (the inlet CV surface), n = –ex but the velocities of both streams are positive. At x = L (the outlet CV surface), n = +ex and the velocity of the mixed stream is positive. On both surfaces dA = bdy. Thus, the reduced form of (4.5) can be written: ∫ ρu ⋅ n dA = − ∫ ρub dy + ∫ ρub dy = −ρV1bh1 − ρV2bh2 + ρV3b(h1 + h2 ) = 0 , A*(t )

inlet

outlet

where V3 is the outlet flow speed. Here, the density ρ and flow width b are constants, so this result can be simplified to: V1h1 +V2 h2 = V3 (h1 + h2 ) . (†) The same control volume should be used to conserve of momentum. Again the flow is presumed steady, so the horiztonal component of (4.17) reduces to: ∫ ρu(u ⋅ n)dA = ∫ fx (n, x, t)dA , A*(t )

A*(t )

where fx is the horizontal surface force on the control volume. The body force term is absent here because gravity acts vertially. As for conservation of mass with this control volume, only the inflow and outflow surfaces contribute. Evaluating the reduced form of (4.17) produces: ∫ ρu(u ⋅ n)dA = − ∫ ρu2b dy + ∫ ρu2b dy = −ρV12bh1 − ρV22bh2 + ρV32b(h1 + h2 ) A*(t )

inlet

=

∫

A*(t )

outlet

f x (n, x, t)dA = +

∫

inlet

p(0)b dy −

∫

p(L)b dy = ( p(0) − p(L)) b(h1 + h2 ),

outlet

where the friction terms on the upper and lower CV boundaries have been neglected. Divide out the common factor of b, and rearrange this to isolate the pressure difference on the left:

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

p(0) − p(L) = ρV32 −

ρ V12 h1 +V22 h2 ) . ( h1 + h2

Use (†) to eliminate V3, 2

" V h +V h % ρ p(0) − p(L) = ρ $ 1 1 2 2 ' − V12 h1 +V22 h2 ) , ( # h1 + h2 & h1 + h2 and – after some alegbra – this can be simplified to: h1h2 2 p(L) − p(0) = ρ V −V2 ) . 2 ( 1 (h1 + h2 ) b) Interestingly, the pressure always rises when the two streams have different speeds: p(L) ≥ p(0), and it does not matter which velocity is larger. c) If friction were included in the calculation the pressure rise would be less.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.19. Consider the situation depicted below. Wind strikes the side of a simple residential structure and is deflected up over the top of the structure. Assume the following: twodimensional steady inviscid constant-density flow, uniform upstream velocity profile, linear gradient in the downstream velocity profile (velocity U at the upper boundary and zero velocity at the lower boundary as shown), no flow through the upper boundary of the control volume, and constant pressure on the upper boundary of the control volume. Using the control volume shown: a) Determine h2 in terms of U and h1, and b) Determine the direction and magnitude of the horizontal force on the house per unit depth into the page in terms of the fluid density ρ, the upstream velocity U, and the height of the house h1. c) Evaluate the magnitude of the force for a house that is 10 m tall and 20 m long in wind of 22 m/sec (approximately 50 miles per hour).

Solution 4.19. 3. a) Use the control volume formulation of the continuity equation to find that: h2 h2 Uy U h22 Uh2 Uh1 = ∫ u(y)dy = ∫ dy = = → h2 = 2h1. h2 2 2 0 0 h2 b) The control volume form of the x-momentum equation is h1

2

h2

2

h1

h2

− ∫ ρU dy + ∫ ρu dy = ∫ p∞ dy + ∫ − p∞ dy + ∫ + p∞ (n ⋅ e x )dy + Φx . 0 0 0 0 top € where Φx is the force on the fluid per unit depth into the page. It is delivered through the foundation of the house. Here the flow is assumed inviscid so there's no friction on the lower CV surface. Evaluating the integrals yields: € 1 1 −ρU 2 h1 + ρU 2 h2 = p∞ (+h1 − h2 + (h2 − h1 )) + Φx → Φx = − ρU 2 h1 . 3 3 1 Therefore, the force on the house in the +x direction per unit depth into page = + ρU 2 h1 . 3 3 2 c) (Force on the house) = (1.2 kg/m )(22 m/sec) (10 m)(20 m)/3 = 38.7 kN ≈ 8700 pounds. The € € force attempts to push the house in the positive x-direction, that is: downwind. €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.20. A large wind turbine with diameter D extracts a fraction η of the kinetic energy from the airstream (density = ρ = constant) that impinges on it with velocity U. a) What is the diameter of the wake zone, E, downstream of the windmill? b) Determine the magnitude and direction of the force on the windmill in terms of ρ, U, D, and η. c) Does your answer approach reasonable limits as η → 0 & η → 1?

Solution 4.20. Place a control volume around the stream tube that hits the windmill with vertical inflow-outflow surfaces well upstream and downstream of the wind turbine. a) For ρ = const, the volume flux in the stream tube must be constant. Therefore: V = A1U/A2. The efficiency of the wind turbine, η, implies: (1 – η)ρU2/2 = ρV2/2, or V = U(1 – η)1/2. Eliminate V with the continuity result to find: A1U/A2 = U(1 – η)1/2 , or E = D/(1 – η)1/4 . b) For the chosen CV, cons. of horizontal momentum implies: –ρU2A1 + ρV2A2 = +Φx, where Φx is force the wind turbine applies to the fluid. Here the flow speed outside the CV is assumed to be U everywhere. Therefore, the pressure on the CV surface is P∞ everywhere because of the steady constant-density horizontal flow Bernoulli equation. Thus, the pressure integration drops out of the momentum equation because P∞ acts on the entire control surface. Substituting in the results from part a) produces: 2 π D2 π D2 2 πD # 12 Force on the windmill = −Φ x = ρ U 2 − ρ U 2 (1− η ) = ρ U $1− (1− η ) %& . 12 4 4(1− η ) 4 The force on the windmill points in the downstream direction, and is dimensionally sound (ρU2D2 is has units of force). c) This answer approaches reasonable limits. When as η → 0 the force on the wind turbine also goes to zero. So, when there is no change in airspeed, there should be no force on the wind turbine. When η → 1 the force on the windmill approaches πρU2D2/4, which precisely balances the momentum flux in the incoming stream tube.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.21. An incompressible fluid of density ρ flows through a horizontal rectangular duct of height h and width b. A uniform flat plate of length L and width b attached to the top of the duct at point A is deflected to an angle θ as shown. a) Estimate the pressure difference between the upper and lower sides of the plate in terms of x, ρ, Uo, h, L and θ when the flow separates cleanly from the tip of the plate. b) If the plate has mass M and can rotate freely about the hinge at A, determine a formula for the angle θ in terms of the other parameters. You may leave your answer in terms of an integral.

Solution 4.21. a) The question says to estimate the pressure difference. Thus, reasonable simplifying assumptions should be acceptable. The first of these is to ignore the two-dimensional character of the flow field and assume uniform horizontal flow at each x location. The second is to assume steady flow. So, for the conditions stated (incompressible), the pressure difference between the upper and lower side the plate can be estimated from the Bernoulli equation applied between a midduct point comfortably upstream of the plate, where the flow speed is Uo and the pressure is po, and a location x that is connected to the first point by a streamline, where the flow speed and pressure below the plate are U(x) and p(x). 1 1 po + ρU o2 = p(x) + ρU 2 (x) 2 2 Conservation of mass in the duct requires: U o h = U(x)h(x) = U(x)( h − x tan θ ) . Combining these two equations to eliminate U(x) produces: € ( 1 % ( 1 1 % U o2 h 2 1 2 p(x) − po = ρ U o2 − U 2 (x) = ρ'U o2 − = ρ U 1− ' *. * o 2€ 2 & (h − x tan θ ) 2 ) 2 & 1− (x /h) 2 tan 2 θ ) When the flow separates from the plate as shown, the pressure above the dotted separating streamline will be the same as that below it. This pressure is % ( 1 1 pupper = p(x = L cos θ ) − po = ρU o2 '1− € *, 2 & 1− (L /h) 2 sin 2 θ ) so that % ( 1 1 1 p(x) − pupper = ρU o2 '1− −1+ * 2 1− (L /h) 2 sin 2 θ ) & 1− (x /h) 2 tan 2 θ € % ( 1 1 1 = ρU o2 ' − . 2 2 2 2 * 2 &1− (L /h) sin θ 1− (x /h) tan θ ) b) The € angle θ can be determined requiring the net moment on the plate to be zero. For a small element of the plate of length dl = dx/cosθ located at horizontal location x, the moment around A due to gravity is: – x(M/L)g(dx/cosθ). For this same element, the moment due to fluid mechanical €

(

)

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

pressure is: +(x/cosθ)(p(x) – pupper)b(dx/cosθ). Thus, the angle that solves the problem is specified by: x= L cosθ $ M b ' = 0 = Net moment on the plate ∫ &%−g L cosθ + ( p(x) − pupper ) cos2 θ )(xdx . x= 0 Using the above relationship for p(x) – pupper, and pressing on with the integration and a little algebra yields a final transcendental equation: % ( ML cosθ €1 1 1 2 2 *, 0 = −g + ρU o2bL2 '' ln 1− L h sin θ + ( ) 2 2 2 2 * 2 4 L h sin θ 1− L h sin θ ( ) ( ) & ) which could be solved for specific values of the various parameters. Although complicated, this formula does reach the correct limits. When the fluid dynamic force is negligible, the second term can be ignored, the plate hangs straight down, and the answer θ = π/2 is provided by the €first term. When the plate's weight is negligible, the first term can be ignored so the plate's angle of deflection should be very small and this answer is recovered from the formula above when the terms inside the big parentheses go to zero, which occurs when θ → 0.

{

}

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.22 A pipe of length L and cross sectional area A is to be used as a fluid distribution manifold that expels a steady uniform volume flux per unit length of an incompressible liquid from x = 0 to x = L. The liquid has density ρ, and is to be expelled from the pipe through a slot of varying width, w(x). The goal of this problem is to determine w(x) in terms of the other parameters of the problem. The pipe-inlet pressure and liquid velocity at x = 0 are Po and Uo, respectively, and the pressure outside the pipe is Pe. If P(x) denotes the pressure on the inside of 1 the pipe, then the liquid velocity through the slot Ue is determined from: P(x) − Pe = 2 ρU e2 . For this problem assume that the expelled liquid exits the pipe perpendicular to the pipe’s axis, and note that wUe = const. = UoA/L, even though w and Ue both depend on x. a) Formulate a dimensionless scaling law for w in terms of x, L, A, ρ, Uo, Po, and Pe. € cross section of the pipe are b) Ignore the effects of viscosity, assume all profiles through the uniform, and use a suitable differential-control-volume analysis to show that: dU d dP . A + wU e = 0 , and ρ U 2 = − dx dx dx c) Use these equations and the relationships stated above to determine w(x) in terms of x, L, A, ρ, Uo, Po, and Pe. Is the slot wider at x = 0 or at x = L?

€

€

Solution 4.22. a) Create the parameter matrix. w x L A ρ Uo Po Pe ––––––––––––––––––––––––––––– M 0 0 0 0 1 0 1 1 L 1 1 1 2 -3 1 -1 -1 T 0 0 0 0 0 –1 -2 -2 There will be 8 – 3 = 5 dimensionless groups # x A ρU 2 P & w P w x A ρU o2 By inspection the groups are: , , 2 , , and o , thus: = fn% , 2 , o , o ( Po Pe L L L L $ L L Po Pe ' b) For a differential slice of the pipe located between x and x+dx, cons. of mass implies: −ρU(x)A + ρU e (x)w(x)dx + ρU(x + dx)A = 0 . Group the U-terms together divide by dx: € € €and € € ρA (U(x + dx) − U(x)) dx € + ρU e (x)w(x) = 0 . Divide by ρ and€take the limit as dx goes to zero to find: A dU dx + U e w = 0 . Using the same CV conserv horizontal momentum € −ρU 2 (x)A + ρU 2 (x + dx)A = P(x)A − P(x + dx)A Divide by A & dx, and take the limit as dx goes to zero to find: € ρ dU 2 dx = − dP dx . c) Integrate the two equations found for part b): U = −(U e w A) x + C1, and ρU 2 = −P + C2 €

€ €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

The boundary conditions for U are: U(0) = Uo, and U(L) = 0, and from the problem statement: wUe = const. = UoA/L, so U(x) = U o (1− x L) . Noting that P(0) = Po and U(0) = Uo, the integrated momentum equation becomes: P(x) + ρU 2 (x) = Po + ρU o2 . Put this together with the COMA result and the exit velocity condition, € 2 P(x) − Pe = 12 ρU e2 = 12 ρ(U o A wL) , to find: € 2

2

Pe + 12 ρ(U o A wL) + ρU o2 (1− x L) = Po + ρU o2 ,

which can be solved for € w(x) to get

[(

w(x) = ( A L) (Po − Pe )

€ at x = 0. The slot is wider €

1 2

(

ρU o2 ) + 2 1− (1− x L)

2

)]

−1 2

.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.23. The take-off mass of a Boeing 747-400 may be as high as 400,000 kg. An Airbus A380 may be even heavier. Using a control volume (CV) that comfortably encloses the aircraft, explain why such large aircraft do not crush houses or people when they fly low overhead. Of course, the aircraft’s wings generate lift but they are entirely contained within the CV and do not coincide with any of the CV’s surfaces; thus merely stating the lift balances weight is not a satisfactory explanation. Given that the CV’s vertical body-force term, −g ∫ ρdV , will exceed CV

4x106 N when the airplane and air in the CV’s interior are included, your answer should instead specify which of the CV’s surface forces or surface fluxes carries the signature of a large aircraft’s impressive weight. € Solution 4.23. Assume that the airplane is in steady level flight so that the flow is steady in the frame of reference of the airplane. Therefore, choose a frame of reference attached to the airplane and select a control volume (CV) that comfortably encloses the aircraft. In this frame of reference the CV is not moving but is instantaneously aligned above the house. The overall problem is really three dimensional, but will be simplified here. Therefore, take the CV to be a cube of volume V = L3 with z = 0 at mean roof level. Retaining the roof-peak in the solution is not essential since there is no flow through the bottom of the control volume. Denote the speed and mass of the aircraft by Uac and Mac, let the Cartesian-coordinate fluid velocity with respect  to the control volume be u = (u,v,w) , and choose the front of the control volume to coincide with the y-z plane. z y

€

x For steady flow of a perfect fluid with density ρ (viscous forces are irrelevant here because of the enormous Reynolds number), the integral momentum equation in the z-direction is: − ∫ ρwudS + ∫ ρwudS + ∫ ρw 2 dS − ∫ ρwvdS + ∫ ρwvdS front

back

top

= −g( ρ airV + M ac ) −

close side

∫ PdS +

top

far side

∫ PdS

bottom

where g is the acceleration of gravity, P is pressure, and the volume of the aircraft has been ignored € compared V. There is no flux term on the bottom of the CV because the rooftop is assumed to be solid and impenetrable. € back of the control volume u is approximately equal to U . The On the front and ac contribution from sides and top of the control volume will be small since |w|,|v| « Uac at any reasonable distance from the aircraft. For low-Mach-number subsonic flight, such as during take-off and landing, the fluid density can be treated as constant. If the length of one side of the cubical control volume is L, then momentum conservation simplifies to : L L

ρU ac

∫ ∫ (w 0 0

L L back

− w front ) dydz = −g( ρ airV + M ac ) +

∫ ∫ (P

bottom

0 0

− Ptop ) dxdy .

(a)

In the absence of a difference between the front and back vertical velocity on the CV surface, the weight of the aircraft and air in the CV would be borne by the ground through the pressure

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

difference Pbottom – Ptop. Here we can set Pbottom = Patm + Pextra, where Patm is the atmospheric pressure that supports the air in the CV and counter acts the pressure on the top of the CV, and Pextra is the extra pressure on the ground caused by the aircraft. In particular, evaluate (a) when the aircraft is absent to find a hydrostatic balance, L L

0 = −gρ airV +

∫ ∫ (Patm − Ptop )dxdy , 0 0

and subtract this from (a) to find: L L

L L

∫ (w front − w back )dydz + ∫ ∫ Pextra dxdy .

(b) 0 0 0 0 € Thus, the weight of the aircraft is borne by moving air (the first term in b) and an extra pressure on the ground (the second term in b). In reality, the pressure distribution on the aircraft's wings causes wfront – wback to be both nonzero and positive (i.e. wfront is an updraft and wback is a € this term primarily balances the impressive weight of the aircraft. For a landing downdraft), and speed of 100 m/s, L ≈ 100 m (~50% larger than the aircraft’s wingspan), gMac = 4 mega-Newton, we can estimate the vertical velocity difference necessary to support the aircraft: 4 ×10 6 N w front − w back ≈ ~ 3.3m /s. (1.2kg /m 3 )(100m /s)(10 4 m 2 ) Such wind speeds are not much of a hazard to most people, buildings, or other structures. Further consideration of the flow-field produced by a flying aircraft shows that heavy aircraft do not crush houses and people because the plane's weight ends up being spread over an € area. A flying aircraft's pressure distribution may exist for a kilometer or more in extremely large all directions, so the supporting area for a plane with a weight of 4 mega-Newtons might easily exceed 1 km2 or 106 square meters. Therefore the extra pressure on the ground will be at most a few Pa (less than 0.01% of an atmosphere). gM ac = ρU ac ∫

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.24. An inviscid incompressible liquid with density ρ flows in a wide conduit of height H and width B into the page. The inlet stream travels at a uniform speed U and fills the conduit. The depth of the outlet stream is less than H. Air with negligible density fills the gap above the outlet stream. Gravity acts downward with acceleration g. Assume the flow is steady for the following items. a) Find a dimensionless scaling law for U in terms of ρ, H, and g. b) Denote the outlet stream depth and speed by h and u, respectively, and write down a set of equations that will allow U, u, and h to be determined in terms of ρ, H, and g. c) Solve for U, u, and h in terms of ρ, H, and g. [Hint: solve for h first.] Solution 4.24. a) Density is the only %! &! '! parameter with units of mass. Thus, it cannot be part of a dimensionless law $" "#$! for U. The remaining parameters (U, g, !" #" ! H) can only be assembled into one 2 dimensionless group, Π1 = U gH , thus: ! U = const ⋅ gH . b) Place the € CV around the change in liquid level as shown above, ignore friction, set atmospheric pressure to zero, and assume uniform inflow and outflow. Cons. of mass: (1) ρUH = ρuh 1 1 2 2 2 2 € of horizontal momentum −ρU H + ρu h = P0 H + 2 ρgH − 2 ρgh Cons. (2)

P0 + 12 ρU 2 + ρgH = 0 + 0 + ρgH

Bernoulli from ‘0’ to ‘1’:

(3) 1 2 € 0 + 0 + ρgH = 0 + 2 ρu + ρgh Bernoulli from ‘1’ to ‘2’: (4) The zeros arise in the Bernoulli € equations because point ‘1’ is a stagnation point and the pressure there is atmospheric pressure. There are four unknowns (U, u, h, P0) so these four equations € should be sufficient for a complete solution. € − h) . Use this to eliminate u from (1) and (2), and use (3) to c) Use (4) to find u = 2g(H eliminate P0 from (2). The remaining equations are: ρUH = ρh 2g(H − h) , and −ρU 2 H + ρ2g(H − h)h = − 12 ρU 2 H + 12 ρg(H 2 − h 2 ) . Divide each by ρ. Square the first and divide by H, and simplify the second to get: € equation 1 1 2 2 2 2 U H = 2g(H − h)h 2 H , and − 2 U H + 2g(H − h)h = + 2 g(H − h ) . U 2 H in the second, and then divide by (H–h) to find: €Use the first of these to eliminate € − gh 2 H + 2gh = + 12 g(H + h) , or h 2 − 32 Hh + 12 H 2 = 0 = (h − H)(h − 12 H) .

€ h = H 2 so (4) implies u = €gH , and these together with (1) produce: U = gH 2 . Thus, €

€ €

€ €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.25. A hydraulic jump is the shallow-water-wave equivalent of a gas-dynamic shock wave. A steady radial hydraulic jump can be observed safely in one’s kitchen, bathroom, or backyard where a falling stream of water impacts a shallow pool of water on a flat surface. The radial location R of the jump will depend on gravity g, the depth of the water behind the jump H, the volume flow rate of the falling stream Q, and stream’s speed, U, where it impacts the plate. In your work, assume 2gh U 2 o = ( H + ho ) –> ho %2 H 2 H 2 $ gH ' € Now use the relationship ho = h(R) = Q (2πRU) to eliminate ho and solve for R: ' Q $ U2 ' Q $ U2 2 −1) = H –> R = 2 − H & & ) € 2πRU % gH (€ 2πUH 2 % € g ( € into dimensionless form just takes a little algebra. The final result is: c) Putting this result gH ' R 1 Q $ U = 2 − & ) H 2π € gH 5 % gH U ( €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.26. A fine uniform mist of an inviscid incompressible fluid with density ρ settles steadily at a total volume flow rate (per unit depth into the page) of 2q onto a flat horizontal surface of width 2s to form a liquid layer of thickness h(x) as shown. The geometry is twodimensional. a) Formulate a dimensionless scaling law for h in terms of x, s, q, ρ, and g. b) Use a suitable control volume analysis, assuming u(x) does not depend on y, to find a single cubic equation for h(x) in terms of h(0), s, q, x, and g. c) Determine h(0).

Solution 4.26. a) There appear to be six parameters, but the density is the only parameter that includes the units of mass so it can be set aside. Thus, there will only be two independent dimensions, length (L) and time (T). Therefore, the units matrix is: h x s q g L 1 1 1 2 1 T 0 0 0 -1 -2 There will be 5 – 2 = 3 dimensionless groups. By inspection: Π1 = h s , Π 2 = x s , and

(

€

2

3

)

Π 3 = q 2 gs3 , so h s = fn x s,q gs . b) Conserve mass using the differential CV: −u(x)h(x) + u(x + dx)h(x + dx) − (q€s) dx = 0 ,€or d(uh) dx = q s . Conserve horizontal momentum in same differential CV with hydrostatic pressure forces: € −ρu 2 (x)h(x) + ρu 2 (x + dx)h(x + dx) = 12 ρgh 2 (x) − 12 ρgh 2 (x + dx) , or d(u 2 h) dx = −( g 2)€dh 2 dx . € Integrate these equations and evaluate the constants with u(0) = 0 to get 2 2 2 uh = qx s , and u h = −( g 2) h − h (0) . €

(

2

2

)

(

2

2

)

2

€ two equations to find: q x hs = −( g 2) h − h (0) , which is a Eliminate u(x) between these cubic equation so h(x) cannot be put into a simple form. € of mass implies: c) Global conservation q = u(s)h(s) . If Pa = atmospheric pressure, a Bernoulli € streamline from x = 0 to x = s on the flat surface (the liquid surface is not a streamline) produces: Pa + ρgh(0) = Pa + ρgh(s) + 12 ρu€2 (s) , or u 2 (s) = 2g( h(0) − h(s)) = q 2 h 2 (s) . 2

(

2

2

)

This final equality and the result € of part b) evaluated at x = s, q h(s) = −( g 2) h (s) − h (0) , represent two equations in two unknowns, h(0) & h(s), that can be solved simultaneously to determine h(0) in terms of q and g: € € 13 h(0) = 3(q 2 4g) = 3h(s).

€

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.27. A thin-walled pipe of mass mo, length L, and cross sectional area A is free to swing in the x-y plane from a frictionless pivot at point O. Water with density ρ fills the pipe, flows into it at O perpendicular to the x-y plane, and is expelled at a right angle from the pipe’s end as shown. The pipe’s opening also has area A and gravity g acts downward. For a steady mass flow rate of m˙ , the pipe hangs at angle θ with respect to the vertical as shown. Ignore fluid viscosity. a) Develop a dimensionless scaling law for θ in terms of mo, ˙ , and g. L, A, ρ, € m b) Use a control volume analysis to determine the force components, Fx & Fy, applied to the pipe at the pivot point in ˙ , and g. € terms of θ, mo, L, A, ρ, m c) Determine θ in terms of mo, L, A, ρ, m˙ , and g. d) Above what value of m˙ will the pipe rotate without stopping? € €

Solution 4.27. a) The € units matrix is: ˙ θ mo L A ρ g m ––––––––––––––––––––––––––––––––––––– M 0 1 0 0 1 1 0 L 0 0 1 2 € -3 0 1 T 0 0 0 -1 0 -1 -2 This matrix has rank three and there are seven parameters so four dimensionless groups should be constructed. By inspection the dimensionless groups are: θ, m˙ 2 L gmo2 , L2 A , and ρAL mo ,

(

2

2

2

)

so the dimensionless scaling law is: θ = f m˙ L gmo ,L A, ρAL mo , where f is an undetermined function. b) Note that this part of the problem involves steady flow. Place€the CV around the outside of € € the pipe except at the pivot point O where the CV slices through the pipe parallel to the x-y € is in the z-direction across this control surface slice at O. Here, plane. Assume that the flow conservation of mass merely requires m˙ = const. The control surface pressure is atmospheric everywhere outside the pipe, and pressure forces can only act in the z-direction on the control surface slice at O. Thus, there is no pressure contribution to the reaction forces, Fx and Fy, which arise from the portion of the € control surface that passes through the pipe structure. The momentum flux of water at O is perpendicular to the x-y plane so only the outflow momentum flux is relevant for determining Fx and Fy. For the chosen stationary control volume, this outflow flux term can be readily evaluated because the water’s discharge velocity and the CV’s outward normal are parallel. Assume the outflow velocity is uniform too; u = ( m˙ ρA)n , so conservation of momentum implies: m˙ % m˙ ( m˙ 2 ∫ ρu(u ⋅ n)dA = ρ ρA n' ρA * A = − ρA (e x cosθ + e y sin θ ) = −( mo + ρAL) ge y + Fxe x + Fye y , & ) outlet € where n = −e x cos θ − e y sin θ . Equating x & y components produces the final answers:

Fx = −( m˙ 2 ρA) cosθ , and Fy = (mo + ρAL)g − ( m˙ 2 ρA) sinθ . €

c) This part of the problem again involves STEADY flow. A small CV placed inside the pipe at € its end shows that the force resulting from the flux of fluid discharged from the pipe primarily acts at end of the pipe. Based on this, equate the gravity- and flow-induced moments on the pipe

€

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

(

)

2 about O so that Fx and Fy are not needed: (mo + ρAL)g( L 2) sin θ = m˙ ρA L . Solve for the

(

)

2 angle: sin θ = 2 m˙ ρA (mo + ρAL)g .

(

2

)

d) When 2 m˙ ρA (mo + ρAL)g > 1 there is no answer for θ since sinθ ≤ 1. Thus, when

m˙ > ρA(mo + ρAL)g 2 , no steady€solution is possible, and the pipe will spin without stopping.

€ € €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.28. Construct a house of cards, or light a candle. Get the cardboard tube from the center of a roll of paper towels and back away from the cards or candle several feet so that by blowing you cannot knock down the cards or blow out the candle unaided. Now use the tube in two slightly different configurations. First, place the tube snugly against your face encircling your mouth, and try to blow down the house of cards or blow out the candle. Repeat the experiment while moving closer until the cards are knocked down or the candle is blown out (you may need to get closer to your target than might be expected; do not hyperventilate; do not start the cardboard tube on fire). Note the distance between the end of the tube and the card house or candle at this point. Rebuild the card house or relight the candle and repeat the experiment, except this time hold the tube a few centimeters away from your face and mouth, and blow through it. Again, determine the greatest distance from which you can knock down the cards or blow out the candle. a) Which configuration is more effective at knocking the cards down or blowing the candle out? b) Explain your findings with a suitable control-volume analysis. c) List some practical applications where this effect might be useful. Solution 4.28. a) The most effective configuration is when the tube is held a few centimeters in front of the face and mouth. b) The difference between the two cases is the greater volume flux through the tube when it is held a few centimeters in front of the mouth. Since the exit area of the tube does not change, the volume flux change results entirely from an increased exit velocity. Therefore, the control volume solution should determine the exit velocity from the tube in terms of other parameters. In the following, the subscript ‘o’ will refer to conditions at the inlet end of the tube, the subscript ‘e’ will refer to conditions at the exit or downstream end of the tube, and the subscript ‘a’ will refer to ambient (motionless) conditions some distance from the tube. Consider a control volume that encloses the air inside the tube, ignore the frictional losses between the tube and the flowing air, and align the tube with the x-axis (see picture). For the air speeds and temperature differences under consideration the flow is essentially incompressible and isothermal which means, ρ ≈ const. po

Uo

pe Tube Walls

Ao

Af

Uf

Ue Control Volume Boundary

Let Af be the cross sectional area of the fast moving air blown out of the mouth, and Uf be the velocity of this air stream where it enters the tube. When ρ = const, conservation of mass reduces to conservation of volume implying: A f U f + (Ao − A f )U o = AeU e (1)

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

The horizontal-component of the momentum equation is then −ρA f U 2f − ρ(Ao − A f )U o2 + ρAeU e2 = ∫ Po dA − Ao

∫ Pe dA

(2)

Ae

In both cases, Ao = Ae = A. Using these relationships and assuming that the inlet and exit pressure integrals can be approximated by the PoAo and PeAe, produces: A f U f + (A − A f )U o = AU e (3) € (Po − Pe )A + ρA f U 2f + ρ(A − A f )U o2 = ρAU e2 (4) These are two equations in three unknowns: (Po–Pe), Uo, and Ue. The third equation is determined by the tube configuration. € CASE I. When the tube is € held against the face, the third relationship is Uo = 0, so (3) implies: U e = A f A U f = sU f , (5) where s = Af/A. CASE II. When the tube is held away from the mouth. The remaining relationship comes from a pressure € pressure of the air drawn into the tube from the ambient condition balance outside the tube. The can be estimated from the steady Bernoulli equation: Pa + 12 ρU a2 = Pa = Po + 12 ρU o2 (6) where, of course, Ua ≈ 0. In addition, if the exit velocity is parallel to the tube axis, then the component of the momentum equation perpendicular to the tube axis suggests that Pa ≈ Pe so that Pe − Po = 12 ρU o2 . (7) € Put (7) into (4), then the equations to be solved for Ue are (3) and − 12 U o2 A + A f U 2f + (A − A f )U o2 = AU e2 . (8) where the common factor of ρ has € been divided out. Rearrange (3) and (8) with s = Af/A to find: U e − sU f 1 , and ( 2 − s)U o2 = U e2 − sU 2f . (9a,b) Uo = 1− s € for U in terms of U and s (this involves some algebra). Eliminate Uo and solve e f U e s(2s −1) + 2s(1− s) 3 (10) = Uf € (1− s) 2 + s2 € Although it is not obvious by looking at it, this result, (10), is always larger than the result for Case I, (5), especially when s is small. For example, when s = 1/10: Case I –> Ue = 0.100Uf, while Case II –> Ue = 0.368Uf, almost 4 times larger! Or, when s = 1/4: Case I –> Ue = 0.250Uf, € while Case II –> Ue = 0.535U f, more than twice as large! c) For this problem, the fast-moving air from the mouth and the cardboard tube form a crude ejector pump. Applications for ejector pumps occur wherever rapid or steady pumping without moving parts is required for inflation or evacuation of a container or chamber. Examples: air bags, emergency exit ramps on commercial aircraft, emergency life support (ventilation of the lungs), and a variety of industrial processes involving corrosive materials. In addition, if the fast-moving stream carries a fuel and the entrained fluid carries the oxidizer (for example), the shape (i.e. the area ratio) of the ejector pump can be used to mix reactants to a set the mixture ratio for combustion.

(

)

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.29. Attach a drinking straw to a 15-cm-diameter cardboard disk with a hole at the center using tape or glue. Loosely fold the corners of a standard piece of paper upward so that the paper mildly cups the cardboard disk (see drawing). Place the cardboard disk in the central section of the folded paper. Attempt to lift the loosely folded paper off a flat surface by blowing or sucking air through the straw. a) Experimentally determine if blowing or suction is more effective in lifting the folded paper. b) Explain your findings with a control volume analysis.

Solution 4.29. a) It would seem that the only way to lift the paper would be to suck on the straw. However, blowing is found to be just as successful! b) This can be explained with a control volume analysis. Choose a control volume that encloses a portion of the fluid between the paper and the cardboard disk. For simplicity assume that the flow is axisymmetric, constant density, and inviscid. Also assume that the paper and the cardboard are perfectly planar surfaces. Denote the distance between the paper and the cardboard disk by h, the diameter of the cardboard disk by R, and the radius & cross sectional area of the straw by rs & As. 2rs ur r

h Cardboard Paper

control volume p∞

R Conservation of mass implies: 2πrhur = UiAs for r >> rs, where r is the radius of the control volume, ur is the local radial component of the flow at the circular edge of the control volume, and Ui is the inlet velocity from the straw. Therefore: ur = U i As (2πhr) for r >> rs. What we need to know is the pressure distribution above the piece of paper in terms of the parameters of the problem and p∞, the pressure below the piece of paper. The Bernoulli equation in this case is: € + 1 ρu 2 = p + 1 ρU 2 p + 1 ρu 2 = p(r) ∞

2

R

2

r

i

2

i

where the 'i' subscript refers to conditions at the lower tip of the straw. For the paper to remain suspended we need the following to be true:

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

r= R

W p ≤ 2π

∫ [− p(r) + p ]rdr

(†)

∞

r= o

where Wp is the weight of the paper. This pressure integral can be estimated from the vertical momentum equation for a control volume like that shown above with radius R: r= R

r= R

r= rs

r= 0

∫ ρU i (U ie z ⋅ e€z )dA = ρU i2 As = −2π ∫ p(r)rdr − pi As + 2π ∫ p(r)rdr .

As

The first and second integrals on the right side come from the cardboard and paper surfaces, respectively. Rearrange this, use the Bernoulli equation, and then approximate the flow under the cardboard as purely radial for r > rs. € r= R r= R 2π ∫ p(r)rdr = ρU i2 As + 2π ∫ p(r)rdr + pi As r= 0

r= rs

r= R

∫ p(r)rdr = ρU

2π

r= 0

r= R 2 i

As + 2π

∫ (p

∞

)

+ 12 ρuR2 − 12 ρur2 rdr + pi As .

r= rs

Now start reconstructing the integral of the pressure difference (†): € r= R

2π

r= R

2 ∞ − p(r)] rdr = − ρU i As − 2π

∫ [p

r= 0 € Inserting ur = uR ( R r) yields:

∫(

r= rs

1 2

ρuR2 − 12 ρur2 ) rdr − pi As + p∞ As

r= R

2π

∫ [p

∞

− p(r)] rdr = −ρU i2 As − 12 πρuR2 ( R 2 − rs2 ) + πρuR2 R 2 (ln R − ln rs ) − pi As + p∞ As .

r= 0 € Collect terms and introduce Wp: €

W p ≤ ( p∞ − pi − ρU i2 ) As + 12 πρuR2 ( rs2 + 2R 2 ln( R rs ) − R 2 ) .

€

2 € Simplifying further using As = πrs , the Bernoulli relationships, and uR = U i As (2πhR) = U i rs2 (2hR) produces: % ( 1 r2 W p ≤€− 12 ρU i2 As + 12 πρuR2 R 2 (2ln( R rs ) −1) , or, W p ≤ ρU i2 As'−1+ s 2 (2ln( R rs ) −1)* . 2 4h € & ) For a straw diameter rs = 3 mm, h = 1 mm, and R = 80 mm, the factor inside the large parentheses is about +11.5. With these dimensions and Ui ≈ 20 m/sec, Wp can be as large as a € 0.08 N. This strange lifting effect is used in manufacturing processes to pick up computer chips € without touching them. Note also that Ui enters the final answer as U i2 so it doesn’t matter which ways the flow goes; suction or blowing through the straw should yield identical results.

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.30. A compression wave in a long gas-filled constant-area duct propagates to the left at speed U. To the left of the wave, the gas is quiescent with uniform density ρ1 and uniform pressure p1. To the right of the wave, the gas has uniform density ρ2 (> ρ1) and uniform pressure is p2 (> p1). Ignore the effects of viscosity in this problem. a) Formulate a dimensionless scaling law for U in terms of the pressures and densities. b) Determine U in terms of ρ1, ρ2, p1, and p2 using a control volume. c) Put your answer to part b) in dimensionless form and thereby determine the unknown function from part a). d) When the density and pressure changes are small, they are proportional: p2 − p1 = c 2 ( ρ2 − ρ1 ) for ( ρ2 − ρ1 ) ρ2 ρ 2) and does not add any mass to the airflow. (This mass flow assumption isn’t true, but it serves to keep the algebra under control in this problem.) The relevant parameters are shown in the figure. Use the steady Bernoulli equation into the inlet and from the outlet of the fire, but perform a control volume analysis across the fire. Ignore the vertical extent of A1 compared to H, and the effects of viscosity. Solution 4.60. Start from the stagnant air (Uo ≈ 0) on the suction side of the chimney and use the steady Bernoulli equation and a CV analysis to follow the flow all the way to the chimney exit. The effects of viscosity are ignored throughout this problem. Po = P1 + 12 ρ oU12 Suction flow into the fireplace opening: (1) Cons. of mass through the fireplace: (2,3) ρ oU1 A1 = ρ 2U1a A1 = ρ 2U 2 A2 2 2 Cons. of horiz. mom through the fire: (4) −ρ oU1 A1 + ρ 2U1a A1 = P1 A1 − P1a A1 1 1 2 2 Outflow from the fire: P1a + 2 ρ 2U1a = P2 + 2 ρ 2U 2 + ρ 2 gH (5) € Static pressure outside the chimney: (6) P2 + ρ o gH = Po € All of the velocities and pressures subscripted “1” and “1a” need to be eliminated along with Po € and P2 to find a single equation for U2. First, divide (4) by A1 and use (2,3) to substitute for the € unwanted velocities. € $ ρ 2U 2 A2 ' 2 $ ρ 2U 2 A2 ' 2 ρ 2U 22 A22 $ ρ 2 ' , or (7) −ρ o & + ρ = P − P &1− ) = P1 − P1a ) ) 2& 1 1a A12 % ρ o ( % ρ o A1 ( % ρ 2 A1 ( Use (1), (2,3), and (7) to find: 2 ρ 2U 22 A22 $ ρ2 ' ρ 2U 22 A22 $ ρ 2 ' 1 $ ρ 2U 2 A2 ' P − P = 1− , or (8) 1− = P − ρ − P & ). & ) ) o 1a o o& 1a A12 % 2 ρ o ( A12 % ρ o ( 2 % ρ o A1 ( € € Combine (2,3), (5), (6), and (8): ρ 2U 22 A22 $ ρ 2 ' ρ 2U 22 A22 Po − = Po − ρ o gH + 12 ρ 2U 22 + ρ 2 gH &1− )+ 2 2 A1 % 2 ρ o ( 2A1 € € Cancel Po’s, divide by ρ2, and collect all the terms with U2 to one side of the equation. $ A2 $ ρ ' A 2 1 ' ρ − ρo U 22 & − 22 &1− 2 ) + 22 − ) = 2 gH ρ2 % A1 % 2 ρ o ( 2A1 2 ( € Change signs on both sides of the equation, simplify terms involving the area ratio, and solve for U2 to find: 2( ρ o ρ 2 −1) gH € U2 = 1+ ( A22 A12 )(1− ρ 2 ρ o ) Thus, for inviscid flow, U2 increases with increasing g, H, and A1, and with decreasing ρ2 and A2.

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.61. A hemispherical vessel of radius R containing an inviscid constant-density liquid has a small rounded orifice of area A at the bottom. Show that the time required to lower the level from h1 to h2 is given by 2π $ 2R 3 2 1 52 32 52 ' t 2 − t1 = & ( h1 − h2 ) − ( h1 − h2 )) ( 5 A 2g % 3

€

€

Solution 4.61. Let z denote the height of the R liquid in the vessel. Although there is slight € 2 unsteadiness even when πR >> A, use the r h1 steady Bernoulli equation between the free surface and the orifice at the bottom of the dz h2 tank: z 2 1 # dz & 1 A patm + ρ% ( + ρgz = patm + ρU e2 , (1) $ ' 2 dt 2 where Ue is the flow speed exiting the tank. For an incompressible liquid, conservation of mass reduces to conservation of volume: (2) πr 2 ( dz dt ) = AU e . Eliminate Ue from (1) and (2) to find: $ π 2 r 4 '$ dz ' 2 $ π 2 r 4 '1 2 $ dz ' 2gz = & 2 −1)& ) , or −& 2 −1) & ) = 2gz , % A (% dt ( % A ( % dt ( € where the minus sign arises because z decreases as the tank empties. When πr2 >> A, this relationship can be simplified to −πr 2 ( dz dt ) = A 2gz . From the geometry shown, (R − z) 2 + r 2 = R 2 , so r 2 = 2Rz − z€2 . Thus, the simplified conservation of mass relationship becomes: € 2 −π (2Rz − z )( dz dt ) = A 2gz . Separate the € € differentials and integrate. h2 t2 h2 π π & z3 2 z5 2 ) 12 32 − ∫ dt = − ∫ (2Rz − z )dz → t2 − t1 = − ( 2R + . A €2g h1 A 2g ' 3 2 5 2 * h1 t1 Evaluate and simplify. 2π $ 2R 3 2 1 52 32 52 ' t 2 − t1 = & ( h1 − h2 ) − ( h1 − h2 )) . ( 5 A 2g % 3 €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.62. Water flows through a pipe in a gravitational field as shown in the accompanying figure. Neglect the effects of viscosity and surface tension. Solve the appropriate conservation equations for the variation of the cross-sectional area of the fluid column A(z) after the water has left the pipe at z = 0. The velocity of the fluid at z = 0 is uniform at Vo and the cross-sectional area is A0. Solution 4.62. Using the z coordinate shown and the steady Bernoulli equation between the pipe exit and the vertical location z: 1 1 (1) patm + ρVo2 = patm + ρu 2 (z) + ρgz , 2 2 where u(z) is the flow speed at vertical location z. For an incompressible liquid, conservation of mass reduces to conservation of volume: (2) Vo Ao = u(z)A(z) , € u(z) from (1) and (2) to find: Eliminate 2 Vo2 = (Vo Ao A(z)) + 2gz . Solve for A(z) to find: € Vo Ao . A(z) = 2 V − 2gz o €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.63. Redo the solution for the orifice-in-a-tank problem allowing for the fact that in Fig. 4.16, h = h(t) but ignoring fluid acceleration. Estimate how long it takes for the tank take to empty. Solution 4.63. Let h(t) denote the height of the liquid in the vessel. Let the tank and orifice cross sectional areas be At and Ao, respectively. Although there is slight unsteadiness even when At >> Ao, use the steady Bernoulli equation between the free surface and the orifice at the bottom of the tank: 2 1 $ dh ' 1 patm + ρ&− ) + ρgh = patm + ρU o2 (1) 2 % dt ( 2 where Uo is the flow speed exiting the tank throught the orifice, and the minus sign appears because the water level is decreasing. For an incompressible liquid, conservation of mass reduces to conservation of volume: € (2) At (−dh dt ) = AoU o . Eliminate Uo from (1) and (2) to find: # At2 &# dh & 2 # At2 &−1 2 1 dh 2gh = % 2 −1(%− ( , or 2g % 2 −1( = − , h dt $ A€o '$ dt ' $ Ao ' which is a differential equation for h(t). The left side is merely a constant. Integrate this equation from t = 0 when h = h0 to t = tf when h = 0 to find: # At2 &−1 2 # h1 2 & 0 12 € 2g % 2€−1( t f = −% ( = 2h0 , $ Ao ' $ 1 2 ' h0 which can be rearranged to yield: 12 2h0 # At2 & tf = % −1( . g $ Ao2 ' €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.64. Consider the planar flow of Example 3.5, u = (Ax,–Ay), but allow A = A(t) to depend on time. Here the fluid density is ρ, the pressure at the origin or coordinates is po, and there are no body forces. a) If the fluid is inviscid, determine the pressure on the x-axis, p(x,0,t) as a function of time from the unsteady Bernoulli equation. b) If the fluid has viscosities µ and µυ, determine the pressure throughout the flow field, p(x,y,t), from the x-direction and y-direction differential momentum equations. c) Are the results for parts a) and b) consistent with each other? Explain your findings. p& #1 2 p& + gz + ( = % u + gz + ( , and choose the ρ '2 $ 2 ρ '1 1 streamline along the x-axis from the origin to the location x: x x "1 2 p% "1 2 p% "1 ∂u dA p(x, 0, t) % p or dx + u + = u + x dx + $ A 2 x 2 + ' $ ' '= 0+ o . ∫ ∂ t $# 2 ∫ ρ &x # 2 ρ &0 ρ & ρ #2 0 0 dt The integral is elementary, so after some rearrangement the last equation implies: " dA % x2 p(x, 0, t) − po = −$ + A2 ' . # dt &2 ρ b) For this velocity field ∇ ⋅ u = 0 , so the flow is incompressible. Thus, the relevant version of the Navier-Stokes momentum equation is (4.39b) with g = 0: Du ρ = −∇p + µ∇ 2 u . Dt Using u = (Ax,–Ay) in this equation leads to: (! dA $ ! dA $ + ∂p ∂p ρ *# x + A 2 x & e x + # − y + A 2 y & e y - = − e x − e y + 0 . % " dt % , ∂x ∂y )" dt Separating this into two component equations and integrating yields: " dA % x2 " dA % y2 p = −ρ $ + A 2 ' + B(y, t) and p = −ρ $ − + A 2 ' + C(x, t) , # dt &2 # dt &2 where B and C are functions of integration. These two equations are consistent and match the origin condition p(0,0,t) = po when " dA % x 2 " dA % y2 p(x, y, t) − po = −$ + A2 ' − $− + A2 ' . # dt & 2 # dt &2 ρ c) The results of parts b) and c) are consistent. Fluid viscosity does not influence the pressure in this flow field because there is no net viscous force on fluid particles. 2

Solution 4.64. a) Use (4.82):

∂u

#1

∫ ∂ t ⋅ ds + %$ 2 u

2

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.65. A circular plate is forced down at a steady velocity Uo against a flat surface. Frictionless incompressible fluid of density ρ fills the gap h(t). Assume that h 0, FD opposes the sphere’s downward motion. At first the sphere is moving slowly so its Reynolds number is low, but Re D = ρuz D µ increases with time as the sphere’s velocity increases. To account for this variation in ReD, the sphere’s coefficient of drag may be approximated as: CD ≅ 12 + 24 Re D . For the following items, provide answers in terms of m, ρ, µ, g and D; do not use z, uz, FB, or FD. a) Assume the sphere's vertical equation of motion will be solved by a computer after being put into dimensionless form. Therefore, use the information provided and the definition t * ≡ ρ gtD µ d Re D to show that this equation may be rewritten: = A Re 2D + B Re D + C , and determine the * dt coefficients A, B, and C. b) Solve the part a) equation for ReD analytically in terms of A, B, and C for a sphere that is initially at rest. c) Undo the dimensionless scaling to determine the terminal velocity of the sphere from the part c) answer as t → ∞ .

g!

z! D! m!

µ, ρ! vz! Solution 4.72. a) The buoyant and drag forces on the sphere will be: π 1 π 1 π ! 1 24 $ FB = ρ D 3g and FD = ρuz2 D 2CD = ρuz2 D 2 # + &. 6 2 4 2 4 " 2 Re D % Thus, the approximate equation for the sphere’s velocity is: " 1 24 % du π 1 π m z = mg − ρ D 3g − ρuz2 D 2 $ + '. dt 6 2 4 # 2 Re D & Now introduce the dimensionless scalings one at a time. First use uz = µ Re D ρ D to eliminate uz. 2 duz µ d Re D π 3 1 " µ Re D % π 2 " 1 24 % m =m = mg − ρ D g − ρ $ '. ' D $ + dt ρ D dt 6 2 # ρ D & 4 # 2 Re D & Simplify the right side:

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

" 1 24 % mµ d Re D " π % π = $ m − ρ D 3 ' g − µ 2 Re 2D $ + '. # ρ D dt 6 & 8ρ # 2 Re D &

Switch to the dimensionless time t* using the chain rule for differentiation and t * = ρ gtD µ : mµ d Re D mµ ! d Re D dt * $ mµ ! d Re D ρ gD $ = # &= # & ρ D dt ρ D " dt * dt % ρ D " dt * µ % ! 1 24 $ d Re D ! π $ π = mg = # m − ρ D 3 & g − µ 2 Re 2D # + &. * " dt 6 % 8ρ " 2 Re D % Divide both sides of the final equality by mg, and rearrange the right side: " µ2 % " π ρ D3 % d Re D π " µ2 % 2 = − Re − 3 π Re + $ ' $ ' '. D D $1− dt * 16 # ρ mg & # ρ mg & # 6 m & " µ2 % π ρ D3 π " µ2 % Thus: , , and . C = 1− A=− $ B = −3 π ' $ ' 6 m 16 # ρ mg & # ρ mg & b) Assuming C is positive, the scaled equation found for part a) can be separated and integrated: # 2A Re + B & 2 d Re D D tanh −1 $ = ∫ dt * , or t ∗ + const. = − '. ∫ A Re2 + B Re 2 2 % B − 4AC ( B − 4AC D D+ C This integrated result can be algebraically inverted: # 1 &, 1 ) Re D = B 2 − 4AC (t ∗ + const.)(- . *−B + B 2 − 4AC tanh %− $ 2 '. 2A + The constant can be evaluated with the initial condition ReD = 0 at t = 0 to find: / ) B 2 − 4AC # &,31 1 1 B Re D = t ∗ + tanh −1 % 0−B + B 2 − 4AC tanh +− (.4 , 2A 12 2 +* $ B 2 − 4AC '.-15 which is a formal solution to the problem. c) Terminal velocity will occur when dReD/dt* = 0. Use this fact and the result of part a) to find: " µ2 % " π ρ D3 % π " µ2 % 2 0=− $ Re − 3 π Re + ' T $ ' T $1− ', 16 # ρ mg & # ρ mg & # 6 m & where ReT is the Reynolds number at terminal velocity. Divide by the leading coefficient: 16 " ρ mg %" π ρ D 3 % 0 = ReT2 + 48ReT − $ 2 '$1− '. π # µ &# 6 m & Thus, using the quadratic formula and choosing the appropriate sign to ensure ReT > 0, yields: " " 16 " ρ mg %" π ρ D 3 %% % 1$ 16 " ρ mg %" π ρ D 3 % 2 ReT = −48 + 48 − 4(1) $ − $ 2 '$1− ' ' = −24 + 24 2 + $ 2 '$1− ' '. 2 $# π # µ &# 6 m & # π # µ &# 6 m && '& 2 " % % " B % C µ $ −B µ "$ 16 " ρ mg %" π ρ D 3 % ' (uz )T = + $ ' − '= −24 + 24 2 + $ 2 '$1− so: ' . # 2A & A '& ρ D $# ρ D $# 2A π # µ &# 6 m & '&

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.73. a) From (4.101), what is the dimensional differential momentum equation for steady incompressible viscous flow as Re → ∞ when g = 0. b) Repeat part a) for Re → 0 . Does this equation include the pressure gradient? c) Given that pressure gradients are important for fluid mechanics at low Re, revise the pressure scaling in (4.100) to obtain a more satisfactory low-Re limit for (4.39b) with g = 0. Solution 4.73. Equation (4.101) with g = 0 is a scaled equation that involves the Strouhal number St = Ωl/U, and the Reynolds number Re = ρUl/µ. ∂ u∗ 1 St ∗ + u∗ ⋅ ∇* u∗ = −∇* p∗ + ∇∗2 u∗ . ∂t Re a) When Re → ∞ , the viscous terms drop out. In dimensional terms, the result is the Euler equation: ∂u 1 + ( u ⋅ ∇) u = − ∇p , ∂t ρ which does apply to high-speed flows with weak viscous effects. b) Multiply (4.101) by Re to find: ∂ u∗ ReSt ∗ + Re u∗ ⋅ ∇* u∗ = −Re ∇* p∗ + ∇∗2 u∗ . ∂t Taking the limit as Re → 0 leads to: ∇ 2 u = 0 , which does not include the pressure gradient and is not correct. c) The problem is the assumed scaling for the pressure that went into creating (4.101). As an alternative, assume that pressure p should be scaled by a viscous stress ~ µU/l; p* = pl/µU. In this case, the dimensionless version of (4.39b) with g = 0 becomes: " µ % * ∗ " µ % *2 ∗ " Ωl % ∂ u∗ ∗ * ∗ '∇ p + $ ' ∇ u , or $# '& ∗ + u ⋅ ∇ u = −$ U ∂t # ρUl & # ρUl &

(

)

(

(

St

)

)

∂ u∗ 1 1 + ( u∗ ⋅ ∇* ) u∗ = − ∇* p∗ + ∇*2 u∗ , ∗ ∂t Re Re

where the scalings for all the other dependent variables (and not the pressure) are given by (4.100). As Re → 0 with this pressure scaling, the dimensional momentum equation becomes: ∇p = µ∇ 2 u , which is the correct equation for low-Reynolds number creeping flow.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 4.74. a) Simplify (4.45) for motion of a constant-density inviscid fluid observed in a frame of reference that does not translate but does rotate at a constant rate Ω = Ωez. b) Use length, velocity, acceleration, rotation, and density scales of L, U, g, Ω, and ρ to determine the dimensionless parameters for this flow when g = –gez and x = (x, y, z). (Hint: subtract out the static pressure distribution.) c) The Rossby number, Ro, in this situation is U/ΩL. What are the simplified equations of motion for horizontal constant-density inviscid flow, u = (u, v, 0), observed in the rotating frame of reference when Ro 0. Use the method of images for the following. a) Based on symmetry arguments, determine the horizontal velocity u of the vortex when h = H/2. b) Show that for 0 < h < H the horizontal velocity of the vortex is: ∞ * Γ ' 1 u(0,h) = )1− 2∑ ,, 2 4 πh ( n=1 (nH /h) −1+ and evaluate the sum when h = H/2 to verify your answer to part a). Solution 5.14. There are two flat solid walls at y = 0 yt € and y = H. The combined effect of each requires many y = 3H image vortices. t y = 2H a) When the vortex is located at x = (0, H/2), the t induced velocities from the two closest image vortices y=H t x cancel, and the induced velocities from the next closest y=0 t image vortices cancel as well. In fact, the total induced y = –H velocity from the whole array of image vortices is zero t because vortex contributions cancel in pairs. Therefore, y = –2H u(0, H/2) = 0. b) Construct the induced velocity by adding the contributions from each image vortex. When 0 < h < H, the first image vortex below the lower solid wall must be located at y = –h and it must have a strength of –Γ to mimic the effect of the wall at y = 0. Based on the drawing, this induced velocity will be in the positive x-direction, so ' Γ Γ $1 u(0,h) = + ... = + ...) & ( 4 πh 2π % 2h Now consider the first image vortex above the upper wall. It must be located at y = 2H – h and it must have a strength of –Γ to mimic the effect of the wall at y = H. Based on the drawing, this induced velocity will be in the negative x-direction, so € ( Γ %1 1 u(0,h) = − + ...* ' ) 2π & 2h 2H − 2h However, the first upper image vortex spoils the lower wall boundary condition, so another image vortex must be added at y = –2H + h and it must have a strength of +Γ to mimic the effect of the wall at y = 0. Based on the drawing, this induced velocity will be in the negative x€ direction, so ( Γ %1 1 1 u(0,h) = − − + ...* ' ) 2π & 2h 2H − 2h 2H But, now another vortex with strength +Γ needs to added at y = 2H + h to preserve the upper wall boundary condition, so ( Γ %1 1 1 1 € u(0,h) = − − + + ...*. ' ) 2π & 2h 2H − 2h 2H 2H Continuing this construction leads to: ( Γ %1 1 1 1 1 1 1 1 1 u(0,h) = − − + + − − + + ...*, ' 2π€& 2h 2H − 2h 2H 2H 2H + 2h 4H − 2h 4H 4H 4H + 2h )

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

which can be simplified by removing terms that are equal and opposite: ( Γ %1 1 1 1 1 1 1 u(0,h) = − + − + − + ...*. ' 2π & 2h 2H − 2h 2H + 2h 4H − 2h 4H + 2h 6H − 2h 6H + 2h ) The first term is unique while all the others follow in pairs, so 0 ∞ ∞ % (0 Γ -1 1 1 Γ 1 / 2. u(0,h) = + − + = 1− 2 ' * ∑ ∑ / 2 2 2 2π . 2h n=1& 2nH − 2h 2nH + 2h )1 4 πh /. € n=1 n ( H h ) −12 1 where the second equality follows from algebraic manipulations within the big parentheses. When h = H/2, this formula becomes: ∞ Γ ' 1 * u(0,H /2) = 1− 2 € ∑ ) ,. 2 4 πh ( n=1 4n −1+ The sum is 1/2, and this can be found by looking in an appropriate mathematical reference (see Gradshteyn, I. S., and Ryzhik, I. M., Tables of Integrals, Series, and Products [Academic Press, New York, 1980], p. 8), or by considering the terms in the sum: ∞ . 1€ 1 ∞% 1 1 ( 1 +% 1 ( % 1 1 ( % 1 1 ( − * = -'1− * + ' − * + ' − * + ...0 ∑ 2 = ∑'& 2 n=1 2n −1 2n + 1) 2 ,& 3 ) & 3 5 ) & 5 7 ) / n=1 4n −1 . 1 1 + % 1 1( % 1 1( % 1 1( = -1+ '− + * + '− + * + '− + * + ...0 = . 2 , & 3 3) & 5 5 ) & 7 7 ) / 2 The pair-cancellation of terms is the mathematical signature of the symmetry discussed in part a), and series of this type are sometimes known as telescoping series. With either evaluation approach, the net induced velocity is € % 1 (. Γ + u(0,H /2) = -1− 2' *0 = 0 , & 2 )/ 4 πh , which matches the part a) result.

€

Fluid Mechanics, 6th Ed.

€

Kundu, Cohen, and Dowling

Exercise 5.15. The axis of an infinite solid circular cylinder with radius a coincides with the zaxis. The cylinder is stationary and immersed in an incompressible inviscid fluid, and the net circulation around it is zero. An ideal line vortex parallel to the cylinder with circulation Γ passes through the x-y plane at x = L > a and y = 0. Here two image vortices are needed to satisfy the boundary condition on the cylinder’s surface. If one of these is located at x = y = 0 and has strength Γ. Determine the strength and location of the second image vortex. y Solution 5.15. If the net circulation around the cylinder is zero, then the second image vortex must (xs,ys) have strength –Γ. Therefore, the fluid velocity u at any point will be a sum of velocities induced by the vortex x ! ! at (0, a), the first image vortex at (0, 0), and second (0,0) (0,L) image vortex at (x´, y´). This velocity must be tangent ! (x´,y´) to the surface of the cylinder. Therefore, determine the fluid velocity at the cylinder-surface point (xs, ys). Using Cartesian unit vectors and the diagram to the right, with the circulation directions of the vortices shown, leads to: % ( Γ % ys x ( Γ ys (L − x s ) '− * u(x s, y s ) = e − e '− e x + s e y * + x y 2πa & a a ) 2π (L − x s ) 2 + y s2 '& (L − x s ) 2 + y s2 (L − x s ) 2 + y s2 *) % ( Γ ys − y+ xs − x+ ' * + e − e x y 2π (x s − x +) 2 + (y s − y +) 2 '& (x s − x +) 2 + (y s − y +) 2 (x s − x +) 2 + (y s − y +) 2 *) This can be simplified to: ) Γ & ys ys ys − y% u(x s, y s ) = + e (− 2 − 2 2 2 2+ x 2π ' a (L − x s ) + y s (x s − x %) + (y s − y %) * ) Γ & xs (L − x s ) xs − x% − e . ( 2− 2 2 2 2+ y 2π ' a (L − x s ) + y s (x s − x %) + (y s − y %) * Here the outward normal to the surface of the cylinder is e r = ( x s a)e x + ( y s a)e y . Therefore to find (x´, y´), set the dot product of u and er to zero, * Γ ' xsys xsys x s (y s − y &) € u(x s, y s ) ⋅ e r = + )− 2 − , 2πa ( a (L − x s ) 2 +€y s2 (x s − x &) 2 + (y s − y &) 2 + +

* Γ ' xsys (L − x s )y s (x s − x &)y s − = 0. ) 2 − 2 2 2 2, 2πa ( a (L − x s ) + y s (x s − x &) + (y s − y &) + After dividing out common factors and cancelling equal and opposite terms this reduces to: Ly s x #y s − x s y # − + = 0, 2 2 (L − x s ) + y s (x s − x #) 2 + (y s − y #) 2 € which must be true for all values of (xs, ys). So, at xs = a and ys = 0, this equation becomes: −ay # =0 , (a − x #) 2 + y #2 € which can only be true when y´ = 0, and this leaves a single equation for x´: +

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Ly s x #y s + = 0. 2 2 (L − x s ) + y s (x s − x #) 2 + y s2 Expand the denominators, divide by ys, and use x s2 + y s2 = a 2 , to find: L x# = 2 . 2 2 L − 2Lx s + a a − 2x s x # + x #2 € Cross multiply and cancel common terms € to reach: 2 2 2 Lx " − (L + a ) x " + La 2 = L( x " − L)( x " − a 2 L) = 0 . The root that leads to € a vortex location inside the cylinder is: x´ = a2/L. So, the final answers are: • the second image vortex has strength –Γ, and • its location is (a2/L, 0). € −

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 5.16. Consider the interaction of two vortex rings of equal strength and similar sense of rotation. Argue that they go through each other, as described near the end of section 5.6. VC Solution 5.16. Consider two vortex rings with a similar sense of rotation as shown in the figure to the right. The radii and speeds of C A VB the two vortices are equal. The motion at A is the resultant of VB, VB VD VC, and VD, while the motion at C is the resultant of VA, VB, and VD VD. Comparing the velocity components on A and C, it is clear that the net resultant is an enlargement of the vortex on the right, and a VA contraction of the vortex on the left, as indicated by the vertical arrows in the lower figure (a). Parts (b) and (c) of the lower figure show that left-side vortex accelerates to the right and passes through the right-side vortex. Part c) show that the new right side vortex grows and slows while the new left-side vortex contracts and accelerates. This results in part d), which is identical to the starting condition shown in the first figure except that the ring vortices have exchanged places. Thus, the entire process starts D B over.

(a)

(b)

(c)

(d)

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 5.17. A constant density irrotational flow in a rectangular torus has a circulation Γ and volumetric flow rate Q. The inner radius is r1, the outer radius is r2, and the height is h. Compute the total kinetic energy of this flow in terms of only ρ, Γ, and Q.

h

will be:

€

Solution 5.17. For an irrotational vortical flow with circulation Γ. The velocity in the angular direction will be: Γ . uθ = r1 2πr r2 The volumetric flow rate will be: r2 h r2 h Γ hΓ & r2 ) Q = ∫ ∫ uθ dzdr = ∫ ∫ dzdr = ln( + . 2π ' r1 * r1 € 0 r1 0 2πr The total kinetic energy, KE, of the fluid in the in the torus r2 h 2 π 1 2 ρΓ 2 r2 h 1 ρΓ 2 h ' r2 * 1 KE = ∫ ∫ ∫ ρuθ dzrdrdθ = ∫ ∫ dzdr = 4π ln) r , = 2 ρΓQ . 4 π r1 0 r ( 1+ r1 0 0 2 €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 5.18. Consider a cylindrical tank of radius R filled with a viscous fluid spinning steadily about its axis with constant angular velocity Ω. Assume that the flow is in a steady state. (a) Find ∫ A ω ⋅ ndA where A is a horizontal plane surface through the fluid normal to the axis of rotation and bounded by the wall of the tank, and n is the normal on A. (b) The tank then stops spinning. Find again the value of ∫ A ω ⋅ ndA . € Solution 5.18. a) At steady state the flow inside the tank will be in solid body rotation, and the vorticity will be twice the rotation rate: ω = 2Ω = constant. Thus, ∫ A ω ⋅ ndA = 2ΩπR 2 . € b) Let C be the bounding curve around A, and let ds be an element of C. Then, 2π ∫ A ω ⋅ dA = ∫ C u ⋅ ds = ∫ 0 [uθ ] r= R Rdθ = 0 , since uθ = 0 at r = R because of the no-slip boundary condition. €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 5.19. Using Figure 5.13, prove (5.32) assuming that: (i) the two vortices travel in circles, (ii) each vortex's speed along it's circular trajectory is constant, and (iii) the period of the motion is the same for both vortices.

Solution 5.19. When the two vortices travel in circles, with radii h1 and h2, at constant speeds with the same period T, the circumference of each circular trajectory divided by the induced speed of the vortex on that trajectory must match: 2π h1 2π h2 , =T = Γ 2 2π h Γ1 2π h where the term on the left applies to the vortex on the left (1), and the term on the right applies to the vortex on the right (2). This relationship can be simplified to: (#) h1 Γ 2 = h2 Γ1 . Now use h = h1 + h2 to substitute for h2 using h2 = h – h1, and solve for h1 to find: −1 "1 1% h h"1 1% Γ2h h1 h − h1 , or h1 $ + ' = , which implies h1 = $ + ' = . = Γ1 # Γ 2 Γ1 & Γ 2 + Γ1 Γ2 Γ1 # Γ 2 Γ1 & Γ1 Γh The relationship (#) then implies: h2 = 1 for the other vortex. Γ1 + Γ2 Thus, G is located at the place where the diagonal line connecting the tips of the induced velocity vectors crosses the line segment that connects the vortices. So, G is halfway between the two vortices when they are of equal strength, but it lies closer to the stronger vortex when the vortices are of unequal strength. € Here G is also the "center of vorticity". For example, when G is located at the origin of coordinates: 2

center of vorticity ≡ ∑ rn Γ n n=1

% −Γ hΓ + Γ hΓ ( = (−h1, 0)Γ1 + (h2 , 0)Γ 2 = (−h1Γ1 + h2 Γ 2 , 0 ) = ' 2 1 1 2 , 0 * = (0, 0). Γ1 + Γ 2 & )

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 5.20. Consider two-dimensional steady flow in the x-y plane outside of a long circular cylinder of radius a that is centered on and rotating about the z-axis at a constant angular rate of Ωz. Show that the fluid velocity on the x-axis is u(x,0) = (Ωza2/x)ey for x > a when the cylinder is replaced by a) a circular vortex sheet of radius a with strength γ = Ωza, and b) a circular region of uniform vorticity ω = 2Ωzez with radius a. c) Describe the flow for x2 + y2 < a2 for parts a) and b). y Solution 5.20. a) For a circular vortex sheet, an element of the circle of length adθ located at angle θ will have a circulation of dΓ = γadθ = Ωza2dθ. The distance between the vortex element and the location

)

& Ωz a e y 1 ( = θ+ 2π 2x (( ' 2

€

!

x (x, 0) is (x − acos θ ) 2 + a 2 sin 2 θ . Thus, the magnitude and direction of the induced velocity increment du at (x, 0) are (asin θ )e x + (x − acos θ )e y . dΓ 2π (x − acosθ ) 2 + a 2 sin 2 θ and € (x − acosθ ) 2 + a 2 sin 2 θ The total induced velocity at (x, 0) is determined from integrating over all vortex elements. +π Ωz a 2 ' (asin θ )e x + (x − acosθ )e y * Ωz a 2 + π ' (asin θ )e x + (x − acos θ )e y * u(x,0) = ∫ d θ = ∫ ) x 2 + a 2 − 2ax cosθ ,dθ . ) , 2 2 2 € θ + 2π − π ( + θ =− π 2π ( (x − acos θ ) + a sin € Separate the components and consider them individually. Ωz a 2e y π (x − acos θ )dθ Ωz a 2e x π asin θdθ u(x,0) = + . ∫ ∫ 2π − π x 2 + a 2 − 2ax cosθ 2π − π x 2 + a 2 − 2ax cosθ The x-component of u is zero because it is specified as an integration of an odd function on an even interval. The remaining y-component integration can be rewritten to reach a tabulated integral form, and then evaluated: € ) Ω a 2e 1 π & x 2 − a2 u(x,0) = z y 1+ ∫ ( +dθ 2 2 2π 2x − π ' x + a − 2ax cosθ *

(

€

(acos!,asin!) du !

& 2(x − a ) −1( tan ( 2 ( x 2 + a2 ) − 4a2 x 2 (' 2

2

2

( x 2 + a2 ) − 4a2 x 2 x 2 + a 2 − 2ax

+π )) θ ++ tan ++ 2 ++ **− π

+π 2 & 2 Ωz a 2e y 1 & Ωz a 2e y 1 & & π )) Ωz a 2e y θ )) π −1 x − a = tan = 2 π + 2 − 2 . (θ + 2tan ( + (− ++ = + ( 2 ' 2 ** 2π 2x ' 2 **− π 2π 2x ' 2 x ' (x − a) +π (x − acos θ )dθ 2π = [Thus, u(x,0) = ∫ 2 and is independent of a!] 2 x − π x + a − 2ax cos θ b) For a uniform distribution of vorticity within the circle specified by x2 + y2 < a2, an element of area dA = rdθdr will have a circulation dΓ = ωzdA = 2Ωzrdθdr. The distance between the vortex

€ element and the location (x, 0) is the induced velocity at (x, 0) are

€

(x − r cos θ ) 2 + r 2 sin 2 θ . Thus, the magnitude and direction of

Fluid Mechanics, 6th Ed.

(

dΓ 2π (x − r cosθ ) 2 + r 2 sin 2 θ

€

Kundu, Cohen, and Dowling

) and

(r sin θ )e x + (x − r cos θ )e y

. (x − r cosθ ) 2 + r 2 sin 2 θ The total induced velocity at (x, 0) is determined from integrating over all vortex sheet elements. a +π 2Ωz ' (r sin θ )e x + (x − r cosθ )e y * Ωz a + π ' (r sin θ )e x + (x − r cosθ )e y * u(x,0) = ∫ ∫ rd θ dr = ∫ ∫ ) x 2 + r 2 − 2rx cosθ ,dθrdr . ) , € 0 − π 2π ( (x − r cos θ ) 2 + r 2 sin 2 θ + π 0 −π ( + € The angular integration is the same as that completed in part a) with a replaced by r, so Ωze y a 2π Ωze y 2π a 2 Ωz a 2e y u(x,0) = ∫ rdr = π x 2 = x . π 0 x c) In part a) the fluid is stationary for x2 + y2 < a2. In part b), the fluid is in solid body rotation for x2 + y2 < a2.

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 5.21. An ideal line vortex in a half space filled with an inviscid constant-density fluid has circulation Γ, lies parallel to the z-axis, and passes through the x-y plane at x = 0 and y = h. The plane defined by y = 0 is a solid surface. a) Use the method of images to find u(x,y) for y > 0 and show that the fluid velocity on y = 0 is u(x,0) = Γhe x [π (x 2 + h 2 )] . b) Show that u(0,y) is unchanged for y > 0 if the image vortex is replaced by a vortex sheet of strength γ (x) = −u(x,0) on y = 0. c) (If you have the patience) Repeat part b) for u(x,y) when y > 0. € Solution 5.21. a) If the actual vortex is at (0, h) and has circulation Γ, then an image vortex of € with circulation –Γ at (0, –h) produces a no-through-flow boundary condition on y = 0. In Cartesian coordinates, the velocity u at any location (x, y) is the sum of the two induced velocities: % −(y − h)e + xe ( % (y + h)e − xe ( Γ Γ x y x y ' *+ ' *, u(x, y) = 2 2 ' 2 2 * 2 2 ' 2 2 * 2π x + (y − h) & x + (y − h) ) 2π x + (y + h) & x + (y + h) ) and this form can be simplified to: Γ % −(y − h)e x + xe y (y + h)e x − xe y ( u(x, y) = + 2 (†) ' *. 2π & x 2 + (y − h) 2 x + (y + h) 2 ) € When evaluated on y = 0, the velocity is: Γ % he x + xe y he x − xe y ( Γhe x u(x,0) = + 2 = . ' 2 2 2 * 2π & x + h x + h ) π (x 2 + h 2 ) € b) First of all, determine u(0,y) from the results of part a) Γ % −e x e ( u(0, y) = + x *. (%) ' 2π & y − h y + h ) € For a flat vortex sheet lying on y = 0, an element of the length dx´ located at x´ will have a circulation of dΓ = γ(x´)dx´. The distance between this vortex element and the location (x, y) is and direction of the induced velocity increment du at (x, y) are (x − x #) 2 + y 2 . The magnitude € −ye x + (x − x #)e y . dΓ 2π (x − x %) 2 + y 2 and (x − x #) 2 + y 2 When the image vortex is replaced by a vortex sheet, u(x,y) for y > 0 will be the sum of the induced velocity from the actual vortex and that from the vortex sheet, which is an integral. € Γ / −(y − h)e x + xe y +∞& −ye x + (x − x %)e y ) γ ( x %)dx % 2 € + ∫( (£) u(x, y) = 0 3. 2 2 + 2π 1 x 2 + (y − h) 2 * Γ 4 −∞ ' (x − x %) + y For this part of this exercise, finding u(0,y) from this formula is sufficient, so put x = 0 to find: Γ / −e x +∞& −ye x − x %e y ) γ ( x %)dx % 2 u(0, y) = + ∫( 0 3. + 2π 1 y − h −∞' x %2 + y 2 * Γ 4 € Comparing this equation with (%), sets the remaining task as showing the equality of +∞$ −ye − x #e ' γ ( x #)dx # e ∫ & x #2x + y 2 y ) Γ = y +x h . ( −∞ % € To do this set, γ(x´) = – u(x´, 0) from part a), and separate the left side into components:

(

€

€

)

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

$ −ye x − x #e y ' he y +∞ −h hye x +∞ dx # x #dx # # ∫ & x #2 + y 2 ) π ( x #2 + h 2 ) dx = π ∫ ( x #2 + y 2 )( x #2 + h 2 ) + π ∫ ( x #2 + y 2 )( x #2 + h 2 ) . ( −∞ % −∞ −∞ The y-component integral is zero because it involves an odd integrand and an even interval. This leaves the x-component: hye x +∞ dx # hye x 1 +∞' 1 1 * = − 2 ∫ ,dx # 2 2 2 2 2 2 ∫) 2 2 π −∞ ( x # + y )( x # + h ) π h − y −∞( x # + y x# + h 2 + hye x π - 1 1 0 hye - h − y 0 ex = − 2= 2 x2/ , 2= 2 2/ π h − y . y h 1 h − y . hy 1 y + h where first equality follows from a partial fractions decomposition of the integrand on the left, +∞

€

+∞

−1

the second equality follows from: I(a) = ∫ −∞ ( x 2 + a 2 ) dx = π a , and the final equality achieves the € desired form. Therefore, the vortex sheet on y = 0 appropriately mimics an image vortex. c) The effort here is nearly a repeat of part b). This time start from (£) and compare it to (†). The task here is to show that the following equality is holds: €+∞$ −ye + (x − x #)e ' γ ( x #)dx # (y + h)e − xe (&) ∫ & (xx− x #) 2 + y 2 y ) Γ = x 2 + (y x+ h) 2 y . ( −∞ % First set, γ(x´) = – u(x´, 0) from part a), and begin working on L, the left side of (&): +∞$ −ye + (x − x #)e ' −hdx # h +∞ ( ye x − (x − x #)e y ) dx # x y . L = Lx e x + Ly e y = ∫ & = ∫ 2 2 ) 2 2 € ( π ( x # + h ) π −∞ ((x − x #) 2 + y 2 )( x #2 + h 2 ) −∞ % (x − x #) + y Consider the x-component integration first: hy +∞ dx # hy +∞ ' A( x # − x) + B Cx # + D * Lx = = ∫ ∫ ( ( x # − x) 2 + y 2 + x #2 + h 2 +dx #, 2 2 2 2 π π # # (x − x ) + y ( x + h ) ) , −∞ −∞ ( ) € where A, B, C, and D are constants. Some algebra is required to find: A = −2x E , B = (h 2 + x 2 − y 2 ) E , C = +2x E , and D = (x 2 + y 2 − h 2 ) E , where E = (x 2 + y 2 − h 2 ) 2 + 4 x 2 h 2 . The integrations that include A and C are zero because they € involve odd integrands and even intervals. The integrals involving B and D can be evaluated using the I(a) formula above, to determine: € € € € hy # Bπ Dπ & h 3 + hx 2 − hy 2 + yx 2 + y 3 − yh 2 Lx = % + . € ( = Bh + yD = π $ y h ' (x 2 + y 2 − h 2 ) 2 + 4 x 2 h 2 The numerator and denominator in can be rearranged: (h + y)[ h(h − y) + x 2 + y(y − h)] h(h 2 − y 2 ) + x 2 (h + y) + y(y 2 − h 2 ) . Lx = = (x 2 + y 2 + h 2 ) 2 − 4 y 2 h 2 € ( x 2 + y 2 + h 2 + 2yh)[ x 2 + y 2 + h 2 − 2yh] Further rearrangement shows that the factors in square brackets are equal, so (h + y)[ x 2 + (y − h) 2 ] h+ y , Lx = 2 = 2 2 2 2 2 € ( x + (y + h) )[ x + y + h − 2yh] x + (y + h)2 which matches the x-component on the right side of (&). Now consider the y-component integration: h +∞ ( x # − x)dx # h +∞ ' F( x # − x) + G Hx # + J * = + + dx #, € Ly = ∫ ∫( π −∞ ((x − x #) 2 + y 2 )( x #2 + h 2 ) π −∞ ) ( x # − x) 2 + y 2 x #2 + h 2 , where F, G, H, and J are constants. Some algebra is required to find: €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

F = (h 2 + x 2 − y 2 ) E , G = 2xy 2 E , H = (−h 2 − x 2 + y 2 ) E , and J = −x(x 2 + y 2 + h 2 ) E , The integrations that include F and H are zero because they involve odd integrands and even intervals. The integrals involving G and J can be evaluated using the I(a) formula above, to determine: € € € x [ x 2 + y 2 − 2yh + h 2 ] h # Gπ Jπ & h 2xyh − x 3 − xy 2 − xh 2 Ly = % + =− 2 (=G +J = 2 π$ y h' y (x + y 2 − h 2 ) 2 + 4 x 2 h 2 (x + y 2 + h 2 ) 2 − 4 y 2 h 2

€

=−

x [ x 2 + y 2 − 2yh + h 2 ]

=−

x , x + (y + h) 2

+ y + h + 2yh )[ x + y + h − 2yh ] and the final form matches the y-component on the right side of (&). Thus, the equivalence of a variable strength vortex sheet and an image vortex is established. €

(x

2

2

2

2

2

2

2

1

Chapter 6 Excercises Problem 1 Show, by Taylor expansion, that fj+2 − 2fj+1 + 2fj−1 − fj−2 d3 f ≈ . 3 dx 2∆x3 What is order of this approximation? Solution Expand around xj : fj+1 = fj +

∂f ∂ 2 f ∆x2 ∂ 3 f ∆x3 ∂ 4 f ∆x4 ∆x + + + + O(∆x5 ) ∂x ∂x2 2 ∂x3 6 ∂x4 24

(1)

fj−1 = fj −

∂f ∂ 2 f ∆x2 ∂ 3 f ∆x3 ∂ 4 f ∆x4 ∆x + − + + O(∆x5 ) 2 3 ∂x ∂x 2 ∂x 6 ∂x4 24

(2)

fj+2 = fj +

∂f ∂ 2 f 4∆x2 ∂ 3 f 8∆x3 ∂ 4 f (2∆x)4 2∆x + + + + O(∆x5 ) 2 3 ∂x ∂x 2 ∂x 6 ∂x4 24

∂f ∂ 2 f 4∆x2 ∂ 3 f 8∆x3 ∂ 4 f (2∆x)4 2∆x + − + + O(∆x5 ) ∂x ∂x2 2 ∂x3 6 ∂x4 24 Subtract equation (2) from equation (1): fj−2 = fj −

fj+1 − fj−1 =

∂f ∂ 3 f ∆x3 2∆x + + O(∆x5 ) ∂x ∂x3 3

(3) (4)

(5)

and equation (4) from equation (3) fj+2 − fj−2 =

∂f ∂ 3 f 8∆x3 4∆x + + O(∆x5 ). ∂x ∂x3 3

(6)

Multiplying equation (5) by 2 and subtracting it from equation (6) yields fj+2 − 2fl+1 + 2fj−1 − fj−2 =

∂ 3 f 6∆x3 + O(∆x5 ), ∂x3 3

which can be rearranged to give d3 f fj+2 − 2fj+1 + 2fj−1 − fj−2 = + O(∆x2 ). 3 dx 2∆x3 Thus, this is a second order approximation to the third derivative.

(7)

2 Problem 2 Consider the following “backward in time” approximation for the diffusion equation:  ∆tD  n+1 n+1 n+1 f + f − 2f fjn+1 = fjn + j+1 j−1 j ∆x2 (a) Determine the accuracy of this scheme. (b) Find its stability properties by von Neumann’s method. How does it compare with the forward in time, centered in space approximation considered earlier? Solution Expand around fjn+1 fjn = fjn+1 −

∂f ∂ 2 f ∆t2 ∆t + 2 + O(∆t3 ) ∂t ∂t 2

n+1 fj+1 = fjn+1 +

∂f ∂ 2 f ∆x2 ∂ 3 f ∆x3 ∆x + + + O(∆x4 ) 2 ∂x ∂x 2 ∂x3 6

n+1 fj−1 = fjn+1 −

∂f ∂ 2 f ∆x2 ∂ 3 f ∆x3 ∆x + − + O(∆x4 ) 2 ∂x ∂x 2 ∂x3 6

Substituting into: fjn+1 = fjn +

 ∆tD  n+1 n+1 n+1 f + f − 2f j−1 j ∆x2 j+1

and rearranging the terms gives: ∂2f ∂ 2 f ∆t ∂ 4 f D∆x2 ∂f −D 2 = 2 + 4 + O(∆t2 , ∆x4 ). ∂t ∂x ∂t 2 ∂t 12

(1)

The right hand side is the error, so the accuracy is O(∆t, ∆x2 ). To find stability, write fjn = n eikxj and substitute into the equation given in the problem statement, using that eik(xj ±∆x) = eikxj e±ik∆x . The result is n+1 = n + (

∆tD n+1 ik∆x ) (e − 2 + e−ik∆x ), ∆x2

or n+1 1 1 = = n 1 − (cos k∆x − 1)2 (∆tD/∆x2 ) 1 + 4(∆tD/∆x2 ) sin2 Thus, we always have n+1 n ≤ 1,  and the scheme is unconditionally stable.

k∆x 2

.

3 Problem 3 Approximate the linear advection equation ∂f ∂f +U =0 U >0 ∂t ∂x by the backward in time method from problem 2. Use the standard second order centered difference approximation for the spatial derivative. (a) Write down the finite difference equation. (b) Write down the modified equation (c) Find the accuracy of the scheme (d) Use the von Neuman’s method to determine the stability of the scheme. Solution (a) The scheme is: n+1 n+1 fj−1 − fj−1 fjn+1 − fjn +U =0 ∆t 2∆x (b) First expand around fjn+1 :

fjn = fjn+1 −

∂f ∂ 2 f ∆t2 ∆t + 2 + O(∆t3 ) ∂t ∂t 2

Rearrange to get fjn+1 − fjn ∂f ∂ 2 f ∆t = − 2 + O(∆t2 ) ∆t ∂t ∂t 2 Then expand n+1 fj+1 = fjn+1 +

∂f ∂ 2 f ∆x2 ∂ 3 f ∆x3 ∆x + + + O(∆x4 ) 2 ∂x ∂x 2 ∂x3 6

∂f ∂ 2 f ∆x2 ∂ 3 f ∆x3 ∆x + − + O(∆x4 ) ∂x ∂x2 2 ∂x3 6 subtract and divide by 2∆x to get: n+1 fj−1 = fjn+1 −

n+1 n+1 fj+1 − fj−1 ∂f ∂ 3 f ∆x2 = + 3 + O(∆x4 ) 2∆x ∂x ∂t 6 The modified equation is therefore

∂f ∂ 2 f ∆t ∂ 3 f ∆x2 ∂f +U = 2 −U 3 ∂t ∂x ∂t 2 ∂x 6 (c) From the modified equation the error is O(∆t, ∆x2 ) (d) Substituting fjn = n eikxj into the discrete equation and using that eik(xj ±∆x) = eikxj e±ik∆x , we get n+1 − n U n+1 n+1 ∆tU = n+1 (eik∆x − e−ik∆x ) ⇒ − 1 + i i sin k∆x = 0 ∆t 2∆x n n ∆x n+1 1  = ⇒ ∆tU n  1 + i ∆x i sin k∆x

4 n+1 Thus, the method is unconditionally stable since  n ≤ 1.

5 Problem 4 Consider the following finite difference approximation to the diffusion equation: fjn+1 = fjn + 2

 ∆tD  n n+1 n−1 n f − f − f + f . j+1 j−1 j j ∆x2

This is the so-called Dufort-Frankel scheme, where the time integration is the ”Leapfrog” method, and the spatial derivative is the usual center difference approximation, except that we have replaced fjn by (1/2)(fjn+1 +fjn−1 ) . Derive the modified equation and determine the accuracy of the scheme. Are there any surprises? Solution Writing ∂ 2 f ∆t2 ∂ 3 f ∆t3 ∂ 4 f ∆t4 ∂f ∆t + 2 + 3 + 4 + O(∆t5 ) ∂t ∂t 2 ∂t 6 ∂t 24 ∂f ∂ 2 f ∆t2 ∂ 3 f ∆t3 ∂ 4 f ∆t4 fjn−1 = fjn − ∆t + 2 − 3 + 4 + O(∆t5 ) ∂t ∂t 2 ∂t 6 ∂t 24 ∂ 2 f ∆x2 ∂ 3 f ∆x3 ∂ 4 f ∆x4 ∂f n ∆x + + + + O(∆x5 ) fj+1 = fjn + 2 3 ∂x ∂x 2 ∂x 6 ∂x4 24 ∂f ∂ 2 f ∆x2 ∂ 3 f ∆x3 ∂ 4 f ∆x4 n fj−1 = fjn − ∆x + − + + O(∆x5 ) ∂x ∂x2 2 ∂x3 6 ∂x4 24 fjn+1 = fjn +

and substituting into the finite difference equation gives ∂2f ∂ 4 f D∆x2 ∂ 3 f ∆t ∂ 2 f D∆t2 ∂f −D 2 = 4 − 3 − 2 + O(∆t3 , ∆x3 ) ∂t ∂x ∂t 12 ∂t 6 ∂t ∆x2 2

2

∆t ∆t The accuracy is O(∆t2 , ∆x2 , ∆x 2 ). Notice the ∆x2 term, which must go to zero as ∆x → 0 and ∆t → 0 for the results to converge. This, the scheme is only conditionally consistent.

6 Problem 5 The following finite difference approximation is given fjn+1 =

 1 n ∆tU  n n n fj+1 − fj−1 (fj+1 + fj−1 )− 2 2∆x

(a) Write down the modified equation (b) What equation is being approximated? (c) Determine the accuracy of the scheme (d) Use the von Neuman’s method to examine under which conditions this scheme is stable. Solution (a) Start by expanding ∂f ∂ 2 f ∆t2 ∆t + 2 + O(∆t3 ) ∂t ∂t 2 ∂f ∂ 2 f ∆x2 ∂ 3 f ∆x3 = fjn + ∆x + + + O(∆x4 ) ∂x ∂x2 2 ∂x3 6 ∂f ∂ 2 f ∆x2 ∂ 3 f ∆x3 = fjn − ∆x + − + O(∆x4 ) 2 ∂x ∂x 2 ∂x3 6 fjn+1 = fjn +

n fj+1 n fj−1

Substitute to get fjn +

∂ 2 f ∆t2 1 ∂2f U ∆t  ∂f ∂ 3 f ∆x3  ∂f 2 ∆t + 2 = (2fjn + ∆x ) − 2∆x + ∂t ∂t 2 2 ∂x2 2∆x ∂x ∂x3 3

Or:

∂f ∂f ∂ 2 f ∆t ∂ 2 f ∆x2 ∂ 3 f U ∆x2 +U =− 2 + − 2 ∂t ∂x ∂t 2 ∂x 2∆t ∂x3 6

(b) The equation being approximated is ∂f ∂f +U =0 ∂t ∂x 2

2 2 (c) The accuracy is O(∆t, ∆x ∆t , ∆x ). If ∆x ∼ ∆t, then ∆x /∆t ∼ ∆x and the method is first order in time and space. (d) To determine the stability, we substitute fjn = n eikxj into the discrete equation and use that eik(xj ±∆x) = eikxj e±ik∆x , giving:

1 U ∆t ik∆x n+1 = n (eik∆x + e−ik∆x ) − n (e − e−ik∆x ). 2 2∆x Rearrange n+1 U ∆t = cos k∆x − i sin k∆x. n  h

7 The absolute value is n+1 2  U ∆t 2 sin2 k∆x. n = cos2 k∆x +  ∆x So the scheme is stable if

U ∆t ≤ 1. ∆x

8 Problem 6 Consider the equation ∂f = g(f ), ∂t and the second-order predictor-corrector method: fj∗ = fjn + ∆tg(f n ) ∆t fjn+1 = fjn + (g(f n ) + g(f ∗ )). 2 Show that this method can also be written as: fj∗ = fjn + ∆tg(f n ) fj∗∗ = fj∗ + ∆tg(f ∗ ) fjn+1 = (1/2)(f n + f ∗∗ ). That is, you simply take two explicit Euler steps and then average the solution at the beginning of the time step and the end. This makes it particularly simple to extend a first order explicit time integration scheme to second order. Solution The first equations in each formulation are the same. To show that the second two equations in the second formulation are the same as the final equation of the first formulation, substitute the second and first equation into the third one: fjn+1 = fjn+1 =

1 n (f + fj∗∗ ) 2 j

1 n (f + fj∗ + ∆tg(fj∗ )) 2 j

1 n (f + fjn + ∆tg(fjn ) + ∆tg(fj∗ ) 2 j 1 fjn+1 = fjn + (∆tg(fjn ) + ∆tg(fj∗ )) 2

fjn+1 =

9 Problem 7 Modify the code used to solve the one-dimensional linear advection equation (Code 1) to solve the Burgers equation:

∂2f ∂ f2  ∂f =D 2 + ∂t ∂x 2 ∂x

using the same initial conditions. What happens? Refine the grid. How does the solution change if we add a constant (say 1) to the initial conditions?

Solution The modified code is:

% one-dimensional NONLINEAR advection-diffusion % by the FTCS scheme N=21; nstep=10; L=2.0; dt=0.05;D=0.05; k=1; dx=L/(N-1); for j=1:N, x(j)=dx*(j-1);end f=zeros(N,1); fo=zeros(N,1); time=0.0; for j=1:N, f(j)=0.5*sin(2*pi*k*x(j)); end; for m=1:nstep, m, time plot(x,f,’linewidt’,2); axis([0 L -1.5, 1.5]); pause fo=f; for j=2:N-1 f(j)=fo(j)-(0.25*dt/dx)*((fo(j+1)+fo(j))^2-... (fo(j)+fo(j-1))^2)+... D*(dt/dx^2)*(fo(j+1)-2*fo(j)+fo(j-1)); end; f(N)=fo(N)-(0.25*dt/dx)*((fo(2)+fo(N))^2-... (fo(N)+fo(N-1))^2)+... D*(dt/dx^2)*(fo(2)-2*fo(N)+fo(N-1)); f(1)=f(N); time=time+dt; end;

10 1.5 1 0.5 0 -0.5 -1 -1.5

0

0.5

1

1.5

2

Running the code for for three different resolutions N = 21, 41, 81 with dt = 0.05, 0.025, 0.0125 up to time 0.25 (for 5, 10, 20) steps, produces the figure above. Even for the coarsest resolution, the solution is essentially fully converged, so we have used thicker lines for the low resolution results, since otherwise those would not be visible. For the second part of the problem, replace the initial conditions by for j=1:N, f(j)=1.0+0.5*sin(2*pi*k*x(j)); end; Taking dt=0.0125, gives a solution that travels to the right.

11 Problem 8 Modify the code used to solve the two-dimensional linear advection equation (Code 2) to simulate the advection of an initially square blob with f = 1 diagonally across a square domain by setting u = v = 1. The dimension of the blob is 0.2 × 0.2 and it is initially located near the origin. Refine the grid and show that the solution converges by comparing the results before the blob flows out of the domain. Solution The modified code is:

% Two-dimensional unsteady diffusion by the FTCS scheme Nx=32;Ny=32;nstep=20;D=0.025;Lx=2.0;Ly=2.0; dx=Lx/(Nx-1); dy=Ly/(Ny-1); dt=0.02; f=zeros(Nx,Ny);fo=zeros(Nx,Ny);time=0.0; for i=1:Nx,for j=1:Ny,x(i,j)=dx*(i-1);y(i,j)=dy*(j-1);end,end; u=1.0; v=1.0; x0=0.2;y0=0.2; for i=2:Nx-1, for j=2:Ny-1 if x(i,j)>x0-0.5*0.2 & x(i,j)y0-0.5*0.2 & y(i,j) 0. 2π a) Determine x a (t) and x b (t) when x a (0) = (−L,0) and x b (0) = (+L,0) b) If the pressure far from the origin is p∞ and the fluid density is ρ, determine the pressure p at x = y = 0 as function of p∞, ρ, qs, and xa(t).

€

€ € € Solution 7.33.€a) All the action takes place in the x-direction so ignore the vertical (v) component of velocity. $ ! ∂φ $ qs !# x − xa (t) x − xb (t) & u(x, y, t) = # & = − # + "∂ x % 2π " ( x − xa (t))2 + y 2 ( x − xb (t))2 + y 2 &% % " % q " xa − xb ' = − qs $ 1 ' = dxa , and For x = x a (t) and y = 0: u(xa , 0, t) = − s $ 2 2π $# ( xa − xb ) + y 2 '& 2π # xa − xb & dt y=0 €

for x = x b (t) and y = 0: u(xb , 0, t) = −

% qs "$ xb − xa ' 2π $# ( xb − xa )2 + y 2 '&

=− y=0

qs " 1 % dxb = −u(xa , 0, t) . $ '= 2π # xb − xa & dt

Thus, dx a dt + dx b dt = 0 , or x a + x b = const. = 0 where final equality comes from the initial €

conditions.

So, dxa dt = −qs ( 4π xa ) or xa = ± const. − qs t 2π .

The xa-initial condition

2

2 x (t) = −xb (t) = − L − qst 2π . € requires the minus sign € and const. = L ; thus, a

1

2

b) Use the unsteady Bernoulli equation: ∂φ ∂t + 2 u + p ρ = p∞ ρ . From the symmetry of the flow, x = y = 0 is a stagnation point (u = 0), so p = p∞ − ρ∂φ ∂t . ! ∂φ $ qs ( − ( x − xa (t)) ( dxa dt ) ( x − xb (t)) ( dxb dt ) + qs ! dxa dt dxb dt $ * = − − = − + # & # & 2 2 " ∂ t %x=y=0 2π *) ( x − xa (t)) € xb % + y2 ( x − xb (t)) + y 2 -,x=y=0 2π " xa €

! ∂φ $ q ! dx dt $ q 2 ! 1 $ ρq2 = − s # a & = s 2 # 2 & , so p = p∞ − 2 s 2 # & " ∂ t %x=y=0 π " xa % 4π " xa % 4π xa Thus, the pressure at the origin becomes very negative as the two sinks approach each other.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 7.34. Consider the unsteady potential flow of an ideal source and sink located at x1 (t) = ( x1 (t),0) and x 2 (t) = ( x 2 (t),0) that are free to move along the x-axis in an ideal fluid that is stationary far from the origin. Assume that the source and sink will move in the velocity field induced by the other. qs " 2 2 % ln ( x − x1 (t)) + y 2 − ln ( x − x2 (t)) + y 2 ' , with qs > 0. € φ (x, y, t) = $ # & 2π a) Determine x1 (t) and x 2 (t) when x1 (0) = (−, 0 ) and x 2 (0) = (+, 0 ) .

€

b) If the pressure far from the origin is p∞ and the fluid density is ρ, determine the pressure p at x = y = 0 as function of p∞, ρ, qs, and x1(t).

€

€

Solution 7.34. a) All the action takes place in the x-direction so ignore the vertical (v) component of velocity. $ ! ∂φ $ q ! x − x1 (t) x − x2 (t) & u(x, y, t) = # & = s ## − " ∂ x % 2π " ( x − x1 (t))2 + y 2 ( x − x2 (t))2 + y 2 &% % " % q " x1 − x2 ' = − qs $ 1 ' = dx1 , and For x = x1 (t) and y = 0: u(x1, 0, t) = s $ − 2 2 2π $# ( x1 − x2 ) + y '& 2π # x1 − x2 & dt y=0 for x = x 2 (t) and y = 0: u(x2 , 0, t) =

€

% qs "$ x2 − x1 ' 2π $# ( x2 − x1 )2 + y 2 '&

= y=0

qs " 1 % dx2 = u(x1, 0, t) . $ '= 2π # x2 − x1 & dt

Thus, dx1 dt = dx 2 dt , or x1 − x2 = const. = −2 where final equality comes from the initial conditions. So, dx1 dt = qs ( 4π ) or x1 = ( qs 4π ) t + const. The x1-initial condition requires

€ €

const. = –  ; thus, x1 (t) = ( qs 4π ) t −  , and x2 (t) = ( qs 4π ) t +  . 1

2

b) Use the unsteady Bernoulli equation: ∂φ ∂t + 2 u + p ρ = p∞ ρ . From the symmetry of the 1

flow, the vertical velocity component will be zero at the origin, so p = p∞ − ρ∂φ ∂t − 2 ρu 2 . ! ∂φ $ q ( − ( x − x1 (t)) ( dx1 dt ) − ( x − x2 (t)) ( dx2 dt ) + qs ! dx1 dt dx2 dt $ = s* − = − # & # & 2 " ∂ t %x=y=0 2π *) ( x − x1 (t))2 + € x2 % y2 ( x − x2 (t)) + y 2 -,x=y=0 2π " x1

€

q ! qs 4π  qs 4π  $ qs ! 2qs $ 1 && = = s ## − & # 2π " ( qs 4π ) t −  ( qs 4π ) t +  % 2π " 4π  % ( qs 4π )2 t 2 −  2 =

qs2 1 2 4π ( qs 4π )2 t 2 −  2

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

$ ! ∂φ $ qs !# x − x1 (t) x − x2 (t) qs ! −1 1 $ & = u(0, 0) = − = + # & # & 2 2 " ∂ x %x=y=0 2π #" ( x − x1 (t)) + y 2 ( x − x2 (t)) + y 2 &% 2π " x1 (t) x2 (t) % x=y=0 =

$ q qs ! −1 1 1 ## && = s (−2) + 2 2π " ( qs 4π ) t −  ( qs 4π ) t +  % 2π (qs 4π ) t 2 − 2

=−

so p = p∞ − ρ

qs  1 π ( qs 4π )2 t 2 −  2

qs2 1 qs2  2 1 − ρ . 2 2 2 4π ( qs 4π ) t −  2 2π # q 4π  2 t 2 −  2 %2 ( ) $ s &

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 7.35. Consider a free ideal line vortex oriented parallel to the z-axis in a 90° corner defined by the solid walls θ = 0 and θ = 90°. If the vortex passes the through the plane of the flow at (x, y), show that the vortex path is given by: x–2 + y–2 = constant. [Hint: Three image vortices are needed at points (−x, −y), (−x, y) and (x, −y). Carefully choose the directions of rotation of these image vortices, show that dy/dx = v/u = −y3/x3, and integrate to produce the desired result.] Solution 7.35. The induced velocity at point (x, y) will include contributions from the three image vortices. If the first vortex at 2 1 (x, y) has strength +Γ, then the second vortex at (−x, y) and the third vortex at (x, −y) will have strength –Γ. The fourth vortex at (−x, −y) will have strength +Γ. Therefore: # # −y && Γ % 1 1 1 x 4 3 % (( u= − ey + ex + e + e x y( % 2 2 2 2 2 2 % ( 2π $ 2x 2y 2 x +y $ x +y x +y '' which is the contribution of the second, third, and fourth vortices, respectively. This velocity can be rewritten: + x2 % 1 Γ +% 1 y ( x ( . Γ y2 . u= e + − + e = e − e y 0 = ue x + ve y -' − 2 0 * x ' x 2 2* y 4 π ,& y x + y 2 ) x / & x x + y ) / 4π ( x 2 + y 2 ) , y The path lines for the location x = (x, y) of the first vortex are: dx Γ x2 dy Γ y2 , and . =u= = v = − dt dt 4π ( x 2 + y 2 ) y 4π ( x 2 + y 2 ) x € Divide the second equation by the first to find: dy dy dx Γ y2 Γ x2 y3 . = =− = − 3 2 2 2 2 dx dt dt x y x 4 π x + y 4 π x + y ( ) ( ) € € Using the two ends of this equality, separate and integrate the equation: dy dx 1 1 − 3= 3 → + 2 = const. 2 y x y x € This is the desired result. €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 7.36. In ideal flow, streamlines are defined by dψ = 0, and potential lines are defined by dφ = 0. Starting from these relationships, show that streamlines and potential lines are perpendicular. a) in plane flow where x and y are the independent spatial coordinates, and b) in axisymmetric flow where R and z are the independent spatial coordinates. [Hint: For any two independent coordinates x1 and x2, the unit tangent to the curve x2 = f(x1) is

t = (e1 + (df dx1 )e 2 )

2

1+ (df dx1) ; thus, for a) and b) it is sufficient to show

(t )ψ = const ⋅ (t)φ = const = 0 ] € €

∂ψ ∂ψ dx + dy = −vdx + udy. Thus a curve ∂x ∂y defined by dψ = 0, has a slope dy dx = v u . Using the hint, the tangent vector to this curve is: Solution 7.36. a) In two-dimensional ideal flow, dψ =

(t )ψ = const = (e x + (v

€

2

uz )e R ) 1+ (uR uz ) = ( uze z + uR e R ) uz2 + uR2 . € ∂φ ∂φ dR + dz = u Similarly, dφ = R dR + uz dz , so a curve defined by dφ = 0, has a slope ∂R € ∂z dR dz = − uz uR . Using the hint, the tangent vector to this curve is: € (t )φ = const = (e z + (−uz uR )e y ) 1+ (uz uR ) 2 = ( uR e z − uze R ) uR2 + uz2 . € Forming the dot product of the two unit vectors produces: (u e + u e ) (u e − u e ) u u − u u (t )ψ = const ⋅ (t)φ = const = z z 2 R 2 R ⋅ R z2 z 2 R = z R2 R2 z = 0 . uR + uz uz + uR uR + uz €

(t )ψ = const = (e z + (uR

€

2

u)e y ) 1+ (v u) = ( ue x + ve y ) u 2 + v 2 . € ∂φ ∂φ dx dy = udx + vdy , so a curve defined by dφ = 0, has a slope Similarly, dφ = €+ ∂x ∂y dy dx = − u v . Using the hint, the tangent vector to this curve is: € (t )φ = const = (e x + (−u v)e y ) 1+ (u v) 2 = (ve x − ue y ) v 2 + u 2 . € Forming the dot product of the two unit vectors produces: (ue x + ve y ) ⋅ (ve x − ue y ) = uv − vu = 0 . (t )ψ = const ⋅ (t)φ = const = v 2 + u2 u2 + v 2 v 2 + u2 € ∂ψ ∂ψ dR + dz = Ruz dR − RuR dz. Thus a curve b) In three-dimensional axisymmetric flow, dψ = ∂R ∂z defined by€dψ = 0, has a slope dR dz = uR uz . Using the hint, the tangent vector to this curve is:

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 7.37. Consider a three-dimensional point source of strength Q (m3/s). Use a spherical control volume and the principle of conservation of mass to argue that the velocity components in spherical coordinates are u = 0 and ur = Q/4πr2 and that the velocity potential and stream function must be of the form φ = φ(r) and ψ = ψ(θ). Integrate the velocity, to show that φ = −Q/4πr and ψ = −Qcosθ /4π . θ

Solution 7.37. For a point source of strength Q (m3/s), the tangential velocity is zero because of symmetry. The radial velocity times the area 4πr2 equals Q so ur = Q/ 4πr2. Therefore, ∂φ 1 ∂ψ Q 1 ∂φ 1 ∂ψ , and uθ = ur = = 2 = =− = 0. 2 ∂r r sin θ ∂θ 4 πr r ∂θ r sin θ ∂r The uθ equations, requires φ = φ(r) and ψ = ψ(θ). Using these results and integrating the ur equations produces: Q Qcos θ . φ (r) = − € and ψ (θ ) = − € 4 πr 4π

€

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 7.38. Solve the Poisson equation ∇ 2φ = Qδ (x − x &) in a uniform unbounded threedimensional domain to obtain the velocity potential φ = –Q/4π|x – x´| for an ideal point source located at x´.

€

€ to Exercise 5.9. First apply a simple shift transformation Solution 7.38. This exercise is similar that places x´ at the origin of coordinates. Define these new coordinates by: X = x − x #, Y = y − y #, Z = z − z# , and set r = x − x # = X 2 + Y 2 + Z 2 . The gradient operator ∇ XYZ in the shifted coordinates X = (X, Y, Z) is the same as ∇ in the unshifted coordinates (x, y, z), so the field equation for φ becomes: € ∇ 2XYZ φ = Qδ (X) = Qδ (x − x &) . € € € Integrate this equation inside a sphere€of radius r:

∫∫∫ ∇

2 2 2 2 2 z= +r y= + r −z x=−+ r −y −z

2 XYZ

φdV = ∫∫ ∇ XYZ φ ⋅ ndA = Q ∫ ∫ ∫ δ(X)dXdYdZ , € spherical surface z=−r y=− r 2 −z 2 x=− r 2 −y 2 −z 2 where the first equality follows from Gauss' divergence theorem, and the triple integral on the right side includes the location X = 0 (aka x = x´) so a contribution is collected from the threedimensional delta function. Thus, the right side of this equation is Q (times unity). € The dot product in the middle portion of the above equation simplifies to ∂φ/∂r because n = er on the spherical surface and e r ⋅ ∇ XYZ = ∂ ∂r . Plus, in an unbounded uniform environment, there are no preferred directions so φ = φ(r) alone (no angular dependence). Thus, the integrated field equation simplifies to π 2π π 2π $ ∂φ ' 2 ∂φ 2 ∂φ r sin θ d θ d ϕ = r sin θdθdϕ = 4 πr 2 = Q, & ) ∫ ∫€% ∂r ( ∫ ∫ ∂r θ = 0 ϕ = 0 ∂r θ = 0ϕ = 0 which implies Q Q ∂φ Q =− or φ = − . = 2 4 πr 4π x − x % ∂r 4 πr € sphere

€

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 7.39. Using (R,ϕ,z) -cylindrical coordinates, consider steady three-dimensional potential flow for a point source of strength Q at the origin in a free stream flowing along the zaxis at speed U: Q € φ (R,ϕ,z) = Uz − . 4π R 2 + z 2 a) Sketch the streamlines for this flow in any R-z half-plane. b) Find the coordinates of the stagnation point that occurs in this flow. c) Determine the pressure gradient, ∇p , at the stagnation point found in part b). € stream surface that encloses the fluid that emerges from the source, d) If R = a(z) defines the determine a(z) for z → +∞ . e) Use Stokes’ stream function to determine an equation for a(z) that is valid for any value of z. f) Use the control-volume momentum equation, ∫ ρu(u ⋅ n) dS = − ∫ pndS + F where n is the €

S

S

outward normal from the control volume, to determine the force F applied to the point source to hold it stationary. g) If the fluid expelled from the source is replaced by a solid body having the same shape, what € is the drag on the front of this body? Solution 7.39. a) A simple sketch appears to the right. b) From symmetry, the stagnation point will lie on R = 0, thus only the axial velocity needs to be considered to find its axial location zs . $ ∂φ ' Qz s . Solve & ) = uz (zs,0) = U + 32 % ∂z ( r= 0 4π (z 2 )

R

z

s

2 32 s

( )

€

3

= zs to find zs = − Q 4 πU . for zs using z c) Take the gradient of the Bernoulli equation or consider the steady ideal-flow momentum 1 equation to find: (u ⋅ ∇ )u = − ∇p . The fluid velocity is zero at a stagnation point, so ∇p = 0 at ρ € € a stagnation point. d) Far downstream of the source, all of the fluid will be moving horizontally at speed U. Thus, the fluid that comes from the source will emerge from a stream tube of radius € a∞ determined €2 from: πa∞U = Q , which implies: a∞ = Q πU . e) Set the velocity relationships for the Stokes stream function ψS equal to those from the potential: 1 ∂ψ S QR ∂φ 1 ∂ψ S Qz ∂φ uR = − =€ =U + , and uz = . € 32 = 32 = 2 2 2 2 R ∂z ∂R R ∂R ∂z 4π (R + z ) 4π (R + z ) Use variable substitutions tan β = R z in the first equation and γ = R 2 + z 2 in the second equation to integrate these relationships to find: €

ψ S = −(Qz 4π )( R 2 + z 2 ) €

€

−1 2

2 2 2 + f (R) € , and ψ S = UR 2 − (Qz 4 π )( R + z )

€

€

−1 2

+ g(z)

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

(

So, ψ S = UR 2 2 − (Q 4 π ) 1+ z

)

R 2 + z 2 where the two functions f(r) and g(z) have been chosen

so that ψs = 0 on R = 0 when z < 0. The equation for a(z) is found by setting R = a & ψs = 0:

(

0 = Ua 2 2 − (Q 4 π ) 1+ z € €

a(z) =

)

a 2 + z 2 . This implicit equation for a can be solved to find: 1 2

(a − z + z 2 ∞

2

z 2 + 2a∞2

12

)

where a∞ = Q πU .

f) From the symmetry of the flow, only a z-direction force is anticipated and the simplest possible control volume is a sphere of radius a. Thus, using spherical coordinates where r = R 2 + z 2 denotes the distance from the source and € θ is the polar angle, the vector momentum € equation can be simplified: Fz = ∫ ρuz (u ⋅ e r ) dA + ∫ pe r ⋅ e z dA , S

S

€

where the replacement n = e r has been made. For the given potential, φ = Ur cosθ − Q 4 πr , the radial and angular velocities are ∂φ Q 1 ∂φ and uθ = = U cosθ + = −U sin θ . €ur = 2 ∂r 4 πr r ∂θ € The pressure can be obtained from the Bernoulli equation: € ( 1 1 % QU cosθ Q2 1 p(r,θ ) + ρ ur2 + uθ2 = p(r,θ ) + ρ'U 2 cos 2 θ + + + U 2 sin 2 θ * = p∞ + ρU 2 . 2 2 4 2€ 2 & 2πR 16π R 2 ) € 2 1 ' QU cosθ Q * − Thus, p(r,θ ) = p∞ + ρ) − , , so the pressure integral becomes: 2 2 ( 2πr 16π 2 r 4 + π / π € 1 ) QU cos θ Q2 ,2 1 ∫ pe r ⋅ e z dS = ∫ 1 p∞ + 2 ρ+− 2πr 2 − 16π 2 r 4 .4(cosθ )2πr 2 sin θdθ = − 2 ρQU ∫ cos2 θ sinθdθ * -3 S θ = 00 θ=0 1 € = − ρQU 3 The terms with odd-powers of the cosine do not contribute to the angular integration on the € interval θ = 0 to π. To calculate the flux integral, first find uz and change it into the spherical coordinate system € ∂φ Qz Qcosθ uz = =U + =U + . 3 2 ∂z 4 πr 2 4π [R 2 + z 2 ]

(

)

Then proceed with: π

' Qcos θ *' Q * 2 ρ)U + U cos θ + 2πr sin θdθ 2 ,) 2, +( + 4 π r 4 π r S θ=0 ( € −1 QU π QU ( β3 + 4 2 =ρ 1+ cos θ ) sin θdθ = − ρ *β + - = ρQU ( ∫ 2 θ=0 2 ) 3 ,1 3 $ 4 1' Thus, F = ρQU& − )e z = ρQUe z ; again, the odd-powers of the cosine do not contribute. This % 3 3( force€points downstream, so if it were not applied the source would move upstream! This result matches the finding for a source in a free stream in two dimensions. g) Here, consider a spherical control volume that is very large so that the velocity is nearly Ue z € on the surface of the control volume. In this case, the pressure integral will be unchanged as

∫ ρuz (u ⋅ e r )dS = ∫

€

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

long as the pressure within the body is p∞. However, the flux integral will not include the contribution from the fluid that emerges from the source. Thus, the force in this case will be equal to that obtained for part d) minus the momentum flux that is no longer present:

F = (F ) d ) − ( ρUQ)e z = 0 Therefore, there will be no drag on the front of the body when the body is long compared to its diameter.

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 7.40. In (R, ϕ, z) cylindrical coordinates, the three-dimensional potential for a point −1 2 source at (0,0,s) is given by: φ = −(Q 4 π )[ R 2 + (z − s) 2 ] . a) By combining a source of strength +Q at (0,0,–b), a sink of strength –Q at (0,0,+b), and a uniform stream with velocity Uez, derive the potential (7.85) for flow around a sphere of radius a by taking the limit as Q → ∞, and b → 0, such that d = −2bQe z = –2πa3Uez = constant. Put your € final answer in spherical coordinates in terms of U, r, θ, and a. b) Repeat part a) for the Stokes stream function starting from −1 2 ψ = −(Q 4 π )(z − s)[ R 2 + (z − s) 2 ] . € Solution 7.40. In cylindrical coordinates, the three-dimensional potential for a uniform stream of U = (0,0,U) , a point source of strength +Q at x = (0,0,−b) , and a point-sink of strength –Q at x = (0,0,+b) is: Q Q . φ = Uz − + 2 2 2 4π 4π R + (z − b) € R + (z + b) Expand the square roots for b → 0, letting r = R 2 + z 2 1 1 1 1 ≈ ≈ ≈ (1  (zb r 2 )) . 2 2 2 2 € R + (z ± b) r 1± (2zb r ) r(1± (zb r )) r

€

€ €

Make replacements in the equation € for φ: % Q zb zb( Qzb d⋅ x φ = Uz − 1− 2 −1− 2 * = Uz + = U⋅ x − . ' 3 4πr & r r ) 2πr 4π | x |3 € where the last equality was obtained by using the given definition d = −2bQe z . Using z = rcosθ (where θ = polar angle) and | d |= 2πUa 3 , the potential becomes: # $ a3 & |d | ' € φ = Ur 1+ cosθ . φ = &Ur + cos θ , or ) % 3( % 4πr 2 ( €$ 2r ' € b) In cylindrical coordinates, the Stokes stream function for a uniform stream of U = (0,0,U) , a point source of strength +Q at x = (0,0,−b) , and a point-sink of strength –Q at x = (0,0,+b) is: €

1 Q ψ = UR 2 − 2 4π

(z + b)

Q

+ € R 2 + (z + b) 2 4 π

(z − b) R 2 + (z − b) 2

.

€ As in part a), expand€the square roots for b → 0, letting r = R 2 + z 2€ 1 1 1 1 ≈ ≈ ≈ (1  (zb r 2 )) . 2 € R 2 + (z ± b) 2 r 1± (2zb r 2 ) r(1± (zb r )) r

Make replacements in the equation for ψ: € % ( % % ( 1 Q z + b zb Q z − b % zb (( ψ = UR 2 − '1− 2 ** + '1+ 2 ** ' ' & ) & r )) 2 4 π r r 4 π r & ) & € 1 Q % z b z 2b zb 2 z b z 2b zb 2 ( 1 Q % 2b 2z 2b ( 2 = UR 2 + ' − − + 3 + 3 + − + 3 + 3 * = UR + '− + 3 *. 2 4π & r r r r r r r r ) 2 4π & r r ) 3 Using R = rsinθ and z = rcosθ (where θ = polar angle), and | d |= 2πUa , as given in the problem statement, the stream function becomes: €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

1 2 2 2Qb & 1 r 2 cos2 θ ) 1 2 & a 3 ) 2 ψ = Ur sin θ + (− + + = Ur (1− 3 + sin θ . 2 4π ' r r3 * 2 ' r *

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 7.41. a) Determine the locus of points in uniform ideal flow past a circular cylinder of radius a without circulation where the velocity perturbation produced by the presence of the cylinder is 1% of the free stream value. b) Repeat for uniform ideal flow past a sphere. c) Explain the physical reason(s) for the differences between the answers for a) and b). Solution 7.41. a) From (7.34), the velocity components for ideal flow about a cylinder with radius a are: # a2 & $ a2 ' ur = U%1− 2 ( cos θ , and uθ = −U&1+ 2 ) sinθ . $ r ' % r ( Clearly, the velocity magnitude when the cylinder is absent = U. Therefore, the magnitude of the perturbation is: 2

2

2 2 2 2 2 2 2 2 2 2 € u − Ue x = U ( a r ) €cos θ + U ( a r ) sin θ = U ( a r ) . u − Ue x a 2 Therefore, a 1% perturbation implies: 0.01 = = 2 → r = 10a . U r b) From€(7.86), the velocity components for ideal flow about a sphere with radius a is: # a3 & $ a3 ' ur = U%1− 3 ( cos θ , and uθ = −U&1+ 3 ) sin θ . % 2r ( €$ r ' Clearly, the velocity magnitude when the cylinder is absent = U. Therefore, the magnitude of the perturbation is: 2

2

2 3 3 2 u − Ue U 2 ( a 3 2r 3 ) sin 2 θ = U ( a 3 r 3 ) cos 2 θ + 14 sin 2 θ . €x = U ( a r ) cos θ +€

Therefore, a 1% perturbation implies: 16 u −Ue x 0.01 = = ( a 3 r 3 ) cos2 θ + 14 sin 2 θ → r = 4.64a ( 43 cos2 θ + 14 ) . U € c) For the sphere, the 1%-perturbation distance is smaller because the sphere's projected area in the direction of the flow (πa2) is smaller than that of the cylinder (2a x span); the sphere's blockage is smaller. Thus, the flow has an easier time getting around the sphere. This effect also decreases the 1%-perturbation distance at θ = π/2 in the case of the sphere.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 7.42. Using the figure for Exercise 7.29 with A3 → 0 and r → ∞ , expand the threedimensional potential for a stationary arbitraryshape closed body in inverse powers of the distance r and € prove that ideal flow theory predicts € zero drag on the body. Solution 7.42. The figure is reproduced here and a portion of this solution follows that given for Exercise 7.29. However, this time the geometry is three-dimensional and the figure must be interpreted as a slice through three dimensional control volume so that A1 is a spherical surface. In ideal flow, the only surface forces are pressure forces. Therefore, the hydrodynamic force F on the body is defined by: F = − ∫ ( p − p∞ )n2 dA = ∫ ( p − p∞ )ndA , (a) A2

A2

where the free stream velocity is U = Uex, and n is the outward normal on the total control volume; thus it points into the body on surface A2 so the usual minus sign is missing from the final equality in (a) because n = –n2. Note that the constant p∞ can be included in the pressure integration because€ ∫ p∞ndA = 0 . closed surface

The starting point for the derivation is the integral form of the steady ideal flow momentum equation applied to the entire clam-shell control volume: ∫ ρu(u ⋅ n)dA = − ∫ ( p − p∞ )ndA € A +A +A A +A +A 1

2

3

1

2

3

with the implied limit that the tube denoted by A3 goes to zero. When this limit is taken, the net contribution of A3 is zero because the surface area of A3 goes to zero. This leaves: ∫ ρu(u€⋅ n)dA + ∫ ρu(u ⋅ n)dA = ∫ ρu(u ⋅ n)dA + 0 = − ∫ ( p − p∞ )ndA , A1 A2 A1 A1 +A 2 where the first equality follows because u ⋅ n = 0 on the solid surface A2. Now substitute in (a) for the pressure integration over A2, and rearrange: ∫ ρu(u ⋅ n)dA = − ∫ ( p − p∞ )ndA − F or F = − ∫ ( p − p∞ )ndA − ∫ ρu(u ⋅ n)dA . € A1 A1 A1 A1 € Thus, the force on the body can be obtained from integrals over a spherical surface (A1) that is distant from the body. The pressure can be written in term of the velocity using the Bernoulli equation: € € 1 1 1 2 p∞ + ρU 2 = p + ρ u = p + ρ(u ⋅ u) , 2 2 2 and this allows the pressure integral to be written: 1 1 − ∫ ( p − p∞ )ndA = − ρ ∫ U 2 − u ⋅ u ndA = ρ ∫ (u ⋅ u)ndA , 2 A1 2 A1 € A1 2 where the final equality follows because U is a constant so ∫ U 2ndA = 0 . So, the force on the

(

)

closed surface

body can be obtained entirely from considerations of the velocity field: € F = ρ ∫ 12 (u ⋅ u)n − u(u ⋅ n) dA . A1

€

(

)

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Far from the body, the fluid velocity will primarily be U = Uex = constant, so it can be written u = U + u´, where u´ is the velocity perturbation that comes from the presence of the body. Inserting this decomposition into the last equation produces: F = ρ ∫ 12 U 2n + (U ⋅ u$)n + 12 (u$ ⋅ u$)n − U(U ⋅ n) − U(u$ ⋅ n) − u$(U ⋅ n) − u$(u$ ⋅ n) dA A1

(

(

= ρ ∫ (U ⋅ u$)n + A1

)

1 2

(u$ ⋅ u$)n − U(u$ ⋅ n) − u$(U ⋅ n) − u$(u$ ⋅ n))dA.

As before the second equality follows by dropping the terms that involve the quadratic factors of the constant vector U. The five remaining integrand terms all involve the perturbation velocity. An arbitrary shape closed body may be represented in potential flow by a collection of €sources and sinks. Thus, for the limit as r → ∞ , the potential may be expanded in inverse powers of r: φ φ φ = U ⋅ x + φ $ = U ⋅ x + 1 + 22 + ... € r r where the coefficients φm may be functions of the angular variables. The first term in this expansion corresponds to the free stream. The remaining term produce u´. However, the body is closed so the φ1 term must be zero because it represents the net source strength of the potential. € The fluid velocity is the gradient of the potential, therefore the largest term representing the velocity perturbation will involve another factor of 1/r. φ u" ~ 32 + ... r Using this scaling of the velocity perturbation with radial distance, the magnitude of the five integrand terms above can be assessed using dA = r2sinθdθdϕ, & Uφ φ 2 Uφ Uφ φ2 ) const F ~ lim ρ ∫ ( 3 2 +€ 26 + 3 2 + 3 2 + + 26 + r 2 sinθdθdφ ∝ lim = 0. r →∞ r →∞ 2r r r 2r * r all angles' r Thus, pure potential flow cannot predict drag. This another statement of D'Alembert's paradox. Interestingly, if a "φ0" term is added to the expanded φ above, non-zero lift and drag are possible. However, this involves the introduction of vorticity in three-dimensions and elementary € treatment of this topic for aircraft wings is postponed to Ch. 14.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 7.43. Consider steady ideal flow over a hemisphere of constant radius a lying on the y-z plane. For the spherical coordinate system shown, the potential for this flow is: φ (r,θ,ϕ ) = Ur(1+ a 3 2r 3 )cosθ where U is the flow velocity far from the hemisphere. Assume gravity acts downward along the x-axis. Ignore fluid viscosity in this problem. a) Determine all three components of the fluid velocity on the surface of the hemisphere, r = a, in ' ∂φ 1 ∂φ 1 ∂φ * , spherical polar coordinates: (ur ,uθ ,uϕ ) = ∇φ = ) , ,. ( ∂r r ∂θ r sin θ ∂ϕ + b) Determine the pressure, p, on r = a. x! c) Determine the hydrodynamic force, Rx, on the hemisphere assuming stagnation pressure is felt everywhere € underneath the hemisphere. r! π [Hints: e r ⋅ e x = sin θ cos ϕ , ∫ 0 sin 2 θdθ = π 2 ,

€

and

€

π

∫ 0 sin4 θdθ = 3π

! ϕ

8 ].

θ!

d) For the conditions of part c) what density ρh have to remain on the € must the hemisphere € surface.

z!

y!

(

3

3

)

Solution 7.43. a) Differentiate φ (r,θ,ϕ ) = Ur 1+ a 2r cosθ and evaluate on r = a. % a3 ( # a3 & ∂φ ur = = U'1− 3 * cos θ ur (r = a) = U%1− 3 ( cosθ = 0 –> ∂r & r ) $ a ' €& 1 ∂φ a3 ) 3U uθ = = −U(1+ 3 + sin θ –> uθ (r = a) = − sin θ r ∂θ 2 ' 2r * 1 ∂φ € € –> uϕ (r = a) = 0 uϕ = =0 r sin θ ∂ϕ  2 1 1 2 € € b) The steady Bernoulli equation applies here: p∞ + 2 ρU = p(r = a) + 2 ρ u(r = a) . Use the p(r = a) − p∞ 1 2 1  2 = 2 U − 2 ρ u(r = a) = 12 U 2 − 98 U 2 sin 2 θ , or results of part a) to find: € € ρ

(

)

(

2 9 p(r = a) − p∞ = 12 ρU 2 1− 49 sin2 θ = 12 ρU€ cos 2 θ − 45 4

)

c) The hydrodynamic force will be determined by the pressure-force difference between the top 1 2 and bottom of the hemisphere. € Here the stagnation pressure is p∞ + 2 ρU , thus: Rx = (pressure force pushing up on the flat side) – (pressure force pushing down on the curved side) = p∞ + 12 ρU 2 πa 2 − ∫ p(r = a)(e r ⋅ e x ) dS

€

(

)

(

)

hemisphere π +π 2

= p∞ + 12 ρU 2 πa 2 −

€ € €

(

1 2

2

)

∫ ∫

[p

θ = 0ϕ =− π 2

π 2

= p∞ + ρU πa − a

2

∫ [p

∞

θ=0

€

(

1 9 2 2 ∞ + 2 ρU 1− 4 sin θ

1 2

2

(

9 4

2

)]

)](cosϕ sinθ )a

2

sin θdϕdθ

+π 2 2

+ ρU 1− sin θ sin θdθ

∫ cosϕdϕ

ϕ =− π 2

Fluid Mechanics, 6th Ed.

(

1 2

2

)

Kundu, Cohen, and Dowling

π 2

= p∞ + ρU πa − 2a

(

)

2

∫ [p

∞

θ=0

[

= p∞ + 12 ρU 2 πa 2 − 2a 2 p∞ + 12 ρU 2 € € €

)]

(

+ 12 ρU 2 1− 49 sin 2 θ sin 2 θdθ π

] ∫ sin θdθ − 2a ( 2

θ=0

2 1 2

π

ρU 2 )(− 49 ) ∫ sin 4 θdθ θ=0

&π ) 9 & 3π ) 27π = p∞ + 12 ρU 2 πa 2 − 2a 2 p∞ + 12 ρU 2 ( + + a 2 ρU 2 ( + = ρU 2 a 2 '2* ' * 4 8 32 where the second-to-last equality has required use of the two integral hints. d) The hemisphere will remain on the surface when its weight is greater than or equal to the pressure force and the buoyant force: $ 81 U 2 ' $2 ' $2 ' 27π ρ h & πa 3 )g ≥ ρU 2 a 2 + ρ& πa 3 ) g , and this implies: ρ h ≥ &1+ )ρ . %3 ( %3 ( 32 % 64 ga (

(

€

)

[

]

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 7.44. The flow-field produced by suction flow into a round vacuum cleaner nozzle held above a large flat surface can be easily investigated with a simple experiment, and analyzed via potential flow in (R, ϕ, z)-cylindrical coordinates with the method of images. a) Do the experiment first. Obtain a vacuum cleaner that has a hose for attachments. Remove any cleaning attachments (brush, wand, etc) or unplug the hose from the cleaning head, and attach an extension hose or something with a round opening (~4 cm diameter is recommended). Find a smooth dry flat horizontal surface that is a ~0.5 meter or more in diameter. Sprinkle the central 1/3 of the surface with a light granular material that is easy to see (granulated sugar, dry coffee grounds, salt, flour, talcum powder, etc. should work well). The grains should be 1/2 to 1 mm apart on average. Turn on the vacuum cleaner and lower the vacuum hose opening from ~0.25 meter above the surface toward the surface with the vacuum opening facing toward the surface. When the hose gets to within about one opening diameter of the surface or so, the granular material should start to move. Once the granular material starts moving, hold the hose opening at the same height or lift the hose slightly so that grains are not sucked into it. If many grains are vacuumed up, distribute new ones in the bare spot(s) and start over. Once the correct hose-opening-to-surface distance is achieved, hold the hose steady and let the suction airflow of the vacuum cleaner scour a pattern into the distributed granular material. Describe the shape of the final pattern, and measure any relevant dimensions. Now see if ideal flow theory can explain the pattern observed in part a). As a first approximation, the flow field near the hose inlet can be modeled as a sink (a source with strength –Q) above an infinite flat boundary since the vacuum cleaner outlet (a source with strength +Q) is likely to be far enough away to be ignored. Denote the fluid density by ρ, the pressure far away by p∞, and the pressure on the flat surface by p(R). The potential for this flow field will be the sum of two terms: +Q φ (R,z) = + K(R,z) 4 π R 2 + (z − h) 2 b) Sketch the streamlines in the y-z plane for z > 0. c) Determine K(R,z). d) Use dimensional analysis to determine how p(R) – p∞ must depend on ρ, Q, R, and h. e) Compute p(R) – p€ ∞ from the steady Bernoulli equation. Is this pressure distribution consistent with the results of part a)? Where is the lowest pressure? (This is also the location of the highest speed surface flow). Is a grain at the origin of coordinates the one most likely to be picked up by the vacuum cleaner? Solution 7.44. a) The pattern scoured in the granular material is axisymmetric. The scoured region is a ring that is a little larger than the vacuum cleaner opening. A pile of the granular material is left directly below the center of the vacuum cleaner opening.

Fluid Mechanics, 6th Ed.

1. a) b)

Kundu, Cohen, and Dowling

z

y c) Based on the method of images, K must represent a sink at z = –h, so: +Q K(R,z) = 2 4 π R + (z + h) 2 d) 5 parameters - 3 dimensions = 2 groups. Clearly Π1 = R h . A little more effort yields:

(

2

Π 2 = ( P(R) − P∞ ) h 4 ρQ 2 . Therefore: P(R) − P∞ = ρQ h

€

4

) f ( R h)

e) Use the potential and the € Bernoulli equ. There is no velocity very far from the origin and 1 there will be no vertical velocity on z = 0, so: P∞ = P(R) + ρ [ uR2 ] . Here: z= 0 2 € ∂φ −QR −QR −QR uR = = + , so [ uR ] z= 0 = 3 2 3 2 32 , ∂R 4 π [ R 2 + (z − h) 2 ] 4 π [ R 2 + (z + h) 2 ] 2π [ R 2 + h 2 ] 2

−€ρQ2 R 2

1 ρQ2 (R h ) . and P(R) − P∞ = = − 3 2 4 8π h 1+ ( R h ) 2 3 8π 2 [ R 2 + h 2 ] € € This pressure distribution is consistent with results of part a). The radial location of the minimum pressure (maximum surface flow speed) is found from: & * 2 € ( 2R R h) d 1 ρQ2 ( 1 ( (P(R) − P∞ ) = 0 = − 2 4 ' 3 −3 4+ 2 . 2 dR 8π h ( 1+ ( R h ) 2 1+ ( R h ) (, h )

[

[

]

]

[

(

]

2

)

Dividing out all the non-zero factors and parameters yields: R 1− 2( R h ) = 0 , which implies: R = 0, or R = h 2 . The second answer matches (within experimental accuracy) the radius of the ring € on the surface where the scouring was most complete in the experiment. The first answer, R = 0, is a stagnation point (a pressure maximum) so a grain located there is not likely to be picked € is held close enough to the surface, even up. In reality, when the vacuum cleaner hose opening € the r = 0 grains are picked up because of the finite size of the grains and because of unsteady flow processes. Overall, when the hose opening is comfortably above the surface, the primary difference between the experiment and the analysis is the size of the sink. In the experiment, air suction takes place across a finite area whereas the potential flow sink is a point in space. Thus, some minor differences between theoretical and experimental results are expected, but such differences might not be easily detected given the simplicity and qualitative nature of the experiments conducted for part a).

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 7.45. There is a point source of strength Q (m3/s) at the origin, and a uniform line sink of strength k = Q/a extending from z = 0 to z = a. The two are combined with a uniform stream U parallel to the z-axis. Show that the combination represents the flow past a closed surface of revolution of airship shape, whose total length is the difference of the roots of: z2 " z % Q ± 1' = 2$ a # a & 4 πUa 2

€

Solution 7.45. From the symmetry of the flow, the U stagnation points that define the length of the body must d! lie on the z-axis. Consider first the stagnation point, P, to ! P the right of the source and line sink. The fluid velocity, Q a uz, on the z-axis will be that due to the free stream, the source, and the line sink. a Q k uz (0,z) = U + − dζ . ∫ 2 2 4 πz 0 4 π (z − ζ ) Substitute in for the value of k, perform the integration and set the result equal to zero to get: z2 # z & Q Q $1 1 ' 0 =U + − . ), which implies: 2 %$ −1(' = 2& 2 a a 4 πUa 2 4 πz % z z(z − a) ( € When P is to the left of the source and line sink, the on-axis fluid velocity is: a Q k uz (0,z) = U − + dζ . ∫ 2 2 4 πz 0 4 π (z − ζ ) € € Repeating the prior steps produces: z2 " z % Q + 1' = . 2$ a # a & 4 πUa 2 €

€

z

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercsie 7.46. Using a computer, determine the surface contour of an axisymmetric half-body formed by a line source of strength k (m2/s) distributed uniformly along the z-axis from z = 0 to z = a and a uniform stream. The nose of this body is more pointed than that formed by the combination of a point source and a uniform stream. From a mass balance, show that far downstream the radius of the half-body is r = ak πU . Solution 7.46. The Stokes stream function for this flow is given by: 1 k a (z − ζ )dζ 1 k ψ = UR 2 − = UR 2 + (z − a) 2 + R 2 − z 2 + R 2 , ∫ € 2 2 2 4 π 0 (z − ζ ) + R 2 4π where the final equality follows from evaluating the integral. The body contour will follow the dividing streamline that starts from the stagnation point located at R = 0 on the negative part of the z-axis. On the negative z-axis, the difference of square roots above is +a, so the body contour € follow the R-z trajectory implicitly specified by: will ka 1 k = UR 2 + (z − a) 2 + R 2 − z 2 + R 2 . 4π 2 4π For computer evaluation dimensionless variable are best. Therefore, define z* = z/a and R*= R/a, and find:

(

)

(

)

2

2

2

2

1 = ΩR* + (z* −1) 2 + R* − z* + R* , € where Ω = 2πaU/k is the dimensionless flow speed. This implicit relationship is plotted below for Ω = 2.0. The flow is from left to right, the R* axis is vertical, and the z* axis is horizontal. €

%"

!#$"

!" &%"

&!#$"

!"

!#$"

%"

%#$"

'"

The flow becomes uniform with speed U far downstream of the body. If r is the asymptotic radius of the half body, then a mass balance gives: Uπr2 = flow out of the line source = ka, which implies: r = ka πU

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 7.47. Consider the radial flow induced by the collapse of a spherical cavitation bubble of radius R(t) in a large quiescent bath of incompressible inviscid fluid of density ρ. The pressure far from the bubble is p∞. Ignore gravity. a) Determine the velocity potential φ(r,t) for the radial flow outside the bubble. b) Determine the pressure p(R(t), t) on the surface of the bubble. c) Suppose that at t = 0 the pressure on the surface of the bubble is p∞, the bubble radius is Ro, and its initial velocity is −R˙ o (i.e. the bubble is shrinking), how long will it take for the bubble to completely collapse if its surface pressure remains constant? 1 ∂ % 2 ∂φ ( 'r * = 0. r 2 ∂r & ∂ r ) Away from r = 0, this equation can be integrated directly to find: φ = −A / r , where A may be a function of time. To evaluate A, match the radial fluid velocity at the surface of the sphere to the $ ∂φ ' R 2 R˙ A(t) € for A to get: φ = − surface velocity of the sphere: & ) . = R˙ (t) = 2 . Solve % ∂r ( r= R(t ) r R € ˙˙ R 4 R˙ 2 ∂φ 1 p − p∞ 2RR˙ 2 R 2 R 2 b) Use the Bernoulli equation: ρ + ρ u + p = p∞ , to get: . =+ + − ρ r r 2r 4 ∂t 2 p(R,t) −€p∞ 3 ˙ 2 € ˙˙ = R + RR Evaluate this on the surface of the bubble r = R(t) to get: ρ 2 € 0 = 3 R˙ 2 + RR ˙˙ = R−1/ 2 d (R 3€/ 2 R˙ ) , and integrate: R 3 / 2 R˙ = C . Integrate c) Set p(R,t) = p∞ to get: 2 dt again to find: 2R 5 / 2 5 = Ct + D . Now evaluate the constants from the given information: € 25 5/2 3/2 −Ro3 / 2 R˙ o = C , and 2Ro5 / 2 5 = D . Thus, R(t) = Ro − (5 2) Ro R˙€o t , so the collapse time is: t = 2Ro (5 R˙ o ) .€ €

Solution 7.47.€a) All of the flow will be in the radial direction. Thus: ∇ 2φ =

[

€ €

€

€

]

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 7.48. Derive the apparent mass per unit depth into the page of a cylinder of radius a that travels at speed U c (t) = dx c dt along the x-axis in a large reservoir of an ideal quiescent fluid with density ρ. Use an appropriate Bernoulli equation and the following time-dependent two-dimensional € a 2U c (x − x c ) potential: φ (x, y,t) = − , where xc(t) is location of the center of the cylinder, and the (x − x c ) 2 + y 2 Cartesian coordinates are x and y. [Hint: steady cylinder motion does not contribute to the cylinder’s apparent mass; keep only the term (or terms) from the Bernoulli equation necessary to determine apparent mass]. € Solution 7.48. In an ideal incompressible fluid, the hydrodynamic loads on the moving cylinder will occur through pressure forces and the pressure will be set by the unsteady Bernoulli equation (UBE). ∂φ 1 ∂φ 1 2 2 ρ + ρ ∇φ + ρgy + p = p∞ or p∞ − p = ρ + ρ ∇φ + ρgy ∂t 2 ∂t 2 The force that the fluid applies to the cylinder, Fc, is θ = 2π Fc = ∫ surface ( p∞ − p)ndS = ∫θ = 0 ( p∞ − p)e r Badθ € its axis, e r = e x cosθ + e y sin θ , and θ is an angle where€B is the length of the cylinder along measured from the positive x-axis with its apex at the center of the cylinder. In addition, define 2

2 the distance from € the center of the cylinder as R(t) = ( x − x c (t)) + y so that € can be written: φ (R,θ ) = − a 2U c R cosθ where R, θ, cosθ = ( x − x c (t)) R(t) . Now the potential and Uc are all functions of time for a fixed (x,y) location. Only the unsteady term in the UBE ˙ will contribute a term that involves dU € c dt ≡ U c , thus:

(

)

p∞ − p = −ρ ( a 2U˙ c R) cos θ + (terms leading to no net force) . € Evaluate this one unsteady term on R = a, ( p∞ − p) R= a = −ρaU˙ c cosθ + ..., and this leads to: θ = 2π F B = − ρa 2U˙ € cos θ (e cosθ + e sin θ )dθ = − ρa 2U˙ (πe + 0e ) = −πa 2 ρU˙ e

€

c

c

∫θ = 0

x

y

c

x

y

c x

So, the apparent mass of the cylinder is equal to the mass of the fluid displaced by the cylinder. € The streamlines for this flow € when Uc < 0 are shown in Figure 3.2b.

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 7.49. A stationary sphere of radius a and mass m resides in inviscid fluid with constant density ρ. a) Determine the buoyancy force on the sphere when gravity g acts downward. b) At t = 0, the sphere is released from rest. What is its initial acceleration? c) What is the sphere’s initial acceleration if it is a bubble in a heavy fluid (i.e. when m → 0)? Solution 7.49. a) The buoyancy force arises from the static pressure acting on the surface of the sphere. Choose the spherical coordinate system so that ez points opposite gravity and the hydrostatic pressure is p = po − ρgz . Thus the buoyancy force on sphere will be: 2π π

FB = − ∫

∫ ( p − po )e r a 2 sin θdθdϕ =

0 0

2π π

∫ ∫ ( po − p)(e x cosϕ sin θ + e y sin ϕ sin θ + e z cosθ )a 2 sin θdθdϕ 0 0

Performing the € ϕ integration eliminates two of the coordinate directions, so €

π

π

0

0

FB = 2π ∫ (+ ρgz)e z a 2 cos θ sin θdθ = 2πρga 3e z ∫ cos 2 θ sin θdθ ,

where p − po = −ρgz , and z = acosθ have been used. The θ-integration is completed via a straightforward change of variable: π −1 4 3 2 3 € FB = +2πρga e z ∫ cos θ sin θdθ = −2πρga e z ∫ β 2 dβ = + πρga 3e z . 3 € 0 1 du b) Newton’s second law for the sphere will be: FB − mge z = (m + madded ) s , where madded is 1/2 dt 3 dus FB − mge z (4 /3) ρa g − mg (4 /3) ρa 3 − m € of the displaced fluid. Thus, the mass = = e = ge z dt m + madded m + (2 /3) ρa 3 (2 /3)ρa 3 + m z dus € c) If m → 0, then = 2ge z . dt

€ €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 7.50. A sphere of mass m and volume V is attached to the end of a light thin flexible cable of length L. In vacuum, with gravity g acting, the natural frequencies for small longitudinal (bouncing) and transverse (pendulum) oscillations of the sphere are ωb and ωp. Ignore the effects of viscosity, and estimate these natural frequencies when the same sphere and cable are submerged in water with density ρw. What is ωp when m ρ wV ). Here, the differential€equation governing small d 2θ dθ g 1 M + ρ V +γ + (M − ρwV ) θ = 0 , changes in θ is 2 2 w dt dt L where again γ is the viscous € damping coefficient. Therefore, the inviscid natural frequency # g & M − ρ wV 1− ρ wV M estimate is . ω p,w = % ( = ωp 1 $ L ' M + 2 ρ wV 1+ ρ wV 2M € If M < ρ wV , pendulum oscillations will still occur but the sphere is now a float and the cable restrains it from rising to the water surface, thus, the governing differential equation is: d 2θ dθ g € M + 12 ρ w V +γ + (−M + ρ w V ) θ = 0 2 dt dt L # g & −M + ρ wV −1+ ρ wV M so the natural frequency is ω p,w = % ( . = ωp $ L ' M + 12 ρ wV 1+ ρ wV 2M

[

€

1+ ρ wV 2 M ,

[

]

]

€ Combining the cases produces: ω p,w = ω p

1− ρ wV M , so ω p,w = ω p 2 when M 0. Similarly, the unsteady term's contribution to the ideal-flow force is: d a π 2π € − ∫ ∫ ∫ ρuz r 2 sin θ dϕ dθ dr dt r=0 θ =0 ϕ =0

& & a 2 da d a π # # a3 cosθ (sin θ dθ r 2 dr %U %1+ 3 (1− 3cos2 θ ) ( + 2 ∫ ∫ dt r=0 θ =0 $ $ 2r ' r dt ' & a 2 da & d a +1 # # a3 = −2πρ β (d βr 2 dr %U %1+ 3 (1− 3β 2 ) ( + 2 ∫ ∫ dt r=0 β =−1 $ $ 2r r dt ' ' = −2πρ

+1

& a 2 da β 2 , 2 d a) # a3 d a 2 3 = −2πρ U β + β − β + r dr = −2 πρ U )(' r 2 dt 2 .∫ + % 2r 3 ( ∫ 2r dr dt r=0 * $ dt r=0 −1 4π d da ρU ( a 3 ) = −4πρUa 2 . 3 dt dt This term represents momentum change that occurs within the sphere. Combining the pressure and internal-momentum-change forces leads to: da (FIF ) z = −2πρUa 2 (sphere internal flow included). dt which is a thrust force when da/dt > 0; the sphere is pushed upstream when its radius increases. In this case, momentum change in the sphere's interior provides the decisive contribution to the ideal flow force. However, if the sphere's interior is considered to be massless or motionless, or to have zero€or constant momentum, then the unsteady term's contribution is zero, and the idealflow force is just the drag that results from the pressure distribution on the sphere's surface: da (FIF ) z = +2πρUa 2 (sphere-internal flow excluded). dt b) In this case the sphere's geometry doesn't change, thus the potential is: # a3 & φ = U(t)r%1+ 3 ( cosθ , $ 2r ' € and this leads to a ∂φ/∂t term in the pressure integral that is not present in the steady flow case. The requisite time derivative and the velocity components are: % a3 ( & ∂φ dU $ a3 ' ∂φ 1 ∂φ a3 ) € = r&1+ = ur = U'1− 3 * cos θ and = uθ = −U(1+ 3 + sin θ . ) cosθ , ∂t dt % 2r 3 ( ∂r r ∂θ & r ) ' 2r * This time consider a fixed-size CV that encloses the sphere. Use the integral momentum equation (4.17) to find: =−

€

€

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

d a π 2π ρuz r 2 sin θdϕdθdr . ∫ ∫ ∫ ∫ dt r= 0 θ = 0 ϕ = 0 θ = 0ϕ = 0 The flux term considered in part a) is zero here because the sphere doesn't change size; its surface is motionless (ur = 0 at r = a). The unsteady term's contribution to the ideal-flow force is € d a π 2π 4 π 3 dU , − ρuz r 2 sin θdϕdθdr = − ρa ∫ ∫ ∫ dt r= 0 θ = 0 ϕ = 0 3 dt and can be obtained by repeating the part a) analysis of this term with a = const. and U = U(t). To evaluate the pressure term, use the Bernoulli equation ∂φ 1 2 ρ + ρ ∇φ + p = const. € ∂t 2 from the stagnation point at r = a and θ = π to any other point on the surface of the sphere to find the pressure p(a,θ) on the surface of the sphere: $ 3 dU ' 1 $ ' 9 3 dU 1 ρ& a cosθ ) +€ ρ&U 2 − U 2 sin 2 θ ) + p(a,θ ) = −ρ a + ρU 2 + p(a, π ) , or % 2 dt ( 2 % ( 4 2 dt 2 '9 * 3 dU 1 [ p] r= a ≡ p(a,θ ) = p(a, π ) − ρ a (1+ cosθ ) + ρU 2) U 2 sin2 θ , . (4 + 2 dt 2 Only the term involving the cosine leads to a net force: € π 2π π ( 3 dU + dU − ∫ ∫ [ p] r = a cos θa 2 sinθdϕdθ = −2π ∫ ρ* − a cosθ - cosθa 2 sin θdθ = 2πρa 3 , ) , 2 dt dt θ = 0 ϕ = 0 θ = 0 € which is a drag force; the flow pushes the sphere downstream when dU/dt > 0. Thus, as in part a), there are two possible answers: 2 dU (sphere internal flow included), and (FIF ) z = + πρa 3 € 3 dt dU (sphere-internal flow excluded). (FIF ) z = +2πρa 3 dt When sphere's internal flow is neglected, the resulting drag arises from two sources. € First, the pressure gradient in the accelerating uniform stream (∂p/∂z = –ρdU/dt) leads to a force similar to buoyancy with dU/dt replacing the acceleration of gravity, (4/3)πρa3dU/dt, that pushes € the sphere toward lower pressure (downstream). Second, the accelerating free stream must go around the sphere and this leads to an apparent mass effect, (2/3)πρa3dU/dt, that again pushes the sphere toward lower pressure. The sum of these two effects correctly recovers the result above, 2πρa3dU/dt. c) Although it is tempting, the switch to a rotating coordinate frame with same origin and z-axis, such a switch is not necessarily helpful because of the kinematic relationship [u] inertial frame = Ω × x + [u] rotating frame , where Ω is the angular rotation rate of the rotating frame. π

(FIF ) z = −

2π

∫ [ p] r = a cosθa 2 sin θdϕdθ −

With z-axis vertical, Ω = Ωez, x = (x, y, z), and an inertial-frame velocity of u = U(excosΩt+ eysinΩt), this kinematic relationship becomes: [u] inertial frame = U(e x cosΩt + e y sinΩt) = Ω × x + [u] rotating frame = Ω(− y %e%x + x %e%y ) + u% ,

€

€

where the primes denote coordinates, velocities, and unit vectors in the rotating frame; and the cross product has been computed in the rotating frame. Here, the x´-direction can be chosen to simplify the inertial-frame velocity: e"x = e x cosΩt + e y sinΩt , and this leaves:

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

u" = [u] inertial frame − Ω(− y "e"x + x "e"y ) = Ue"x − Ω(− y "e"x + x "e"y ) ,

which is only a clear simplification of [u] inertial frame = U(e x cosΩt + e y sinΩt) along the z-axis in the rotating frame where (x´, y´) = (0, 0). Therefore, use an inertial frame of reference with the z-axis vertical, so the flow far from € sphere lies in the x-y plane. In this case, the potential can be determined starting from (7.88): € % d(t) ( * φ = ''U(t) − 3 * ⋅ x. 4 π x & ) where the free stream velocity changes direction, U = U(e x cosΩt + e y sinΩt) , and the dipole strength opposes the free stream and sets the sphere's diameter, d = −2πUa 3 (e x cosΩt + e y sinΩt) . Using spherical coordinates, € x = e x r cosϕ sin θ + e y r sin ϕ sinθ + e z r cos θ , performing the dot products, and using the two-angle sum formula € for the cosine function allows the potential to be written entirely in terms of scalars: € # a3 & € φ = U% r + 2 ( cosϕ ) sin θ , $ 2r ' where ϕ´ = ϕ – Ωt. The radial and angular velocities, and ∂φ/∂t are: % a3 ( % ∂φ 1 ∂φ a3 ( = ur = U'1− 3 * cos ϕ + sin θ , = uθ = U'1+ 3 * cosϕ + cos θ , ∂r r ∂θ & 2r ) €& r ) 3 ' % 1 ∂φ a * ∂φ a3 ( = uϕ = −U)1+ 3 , sin ϕ - , and = ΩU' r + 2 * sin ϕ + sin θ . r sinθ ∂ϕ ∂t ( 2r + & 2r ) € As in parts a) and b), the ideal € flow force on the sphere, FIF, can be obtained from a surface integral of the pressure, and a volume integral over the sphere's interior. π 2π d a π 2π 2 F = − p e a sin θ d ϕ d θ − [ ] ∫ ∫ ∫ ∫ ∫ ρur 2 sin θdϕdθdr . € € IF r=a r dt θ = 0ϕ = 0 r= 0 θ = 0 ϕ = 0 To evaluate the pressure term, use the Bernoulli equation, ∂φ 1 2 ρ + ρ ∇φ + p = const., ∂t 2 € from the stagnation point at r = a, θ = π/2, and ϕ´ = π to any other point on the surface of the sphere to find the pressure p(a,θ,ϕ #) ≡ [ p] r= a on the surface of the sphere: 3 € θ + 1 ρU 2 9 (cos 2 ϕ $ cos2 θ + sin 2 ϕ $) + p(a,θ,ϕ $) = p(a, π /2, π ) . ρΩUasin ϕ $ sin 2 2 4 Rearrange this and denote p(a, π/2, π) by ps. € 9 3 p(a,θ,ϕ #) = ps − ρU 2 (1− cos2 ϕ # sin 2 θ ) − ρΩUasin ϕ # sin θ . 8 2 € Only the final term on the right leads to a net pressure force: π

2π

− ∫ ∫ [ p] r = a e r a 2 sin θdϕdθ θ = 0ϕ = 0 € π 2π 3 = ρΩUa 2 ∫ ∫ sin ϕ ) sin 2 θ (e x cosϕ sin θ + e y sin ϕ sin θ + e z cosθ ) dϕdθ . 2 θ = 0ϕ = 0

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

The θ-integration renders the z-component zero. For the other two terms, the θ-integration is π ∫ 0 sin3 θdθ = 4 3 , so the integrated pressure force simplifies to: 2π

π

−

∫ [ p] r = a e r a 2 sinθdϕdθ = 2 ρΩUa 2

∫

θ = 0ϕ = 0

€

2π

∫ sin ϕ )(e x cosϕ + e y sinϕ )dϕ .

ϕ= 0

Expand sin ϕ " = sin ϕ cos(Ωt) − cos ϕ sin(Ωt) , and integrate to find: π

−

2π

∫ ∫ [ p] r = a e r a 2 sinθdϕdθ = 2πρΩUa3 (−e x sin(Ωt) + e y cos(Ωt)) .

θ = 0ϕ = 0 € This force pushes the sphere perpendicular to the instantaneous free stream direction. € To evaluate the sphere's internal-flow contribution to the force, use the volume integral results from part a) but adjust them for the direction of the free-stream in this problem: € a π 2π 4π ∫ ∫ ∫ ρur 2 sin θdϕdθdr = 3 ρa 3U (e x cos(Ωt) + e y sin(Ωt)) , so that: r= 0 θ = 0 ϕ = 0 d a π 2π 4π 3 − ρur 2 sin θdϕdθdr = − ρa ΩU (−e x sin(Ωt) + e y cos(Ωt)) . ∫ ∫ ∫ dt r= 0 θ = 0 ϕ = 0 3 This€is the force that results from sphere's internal flow contribution. So, as in parts a) and b), there are two possible answers: 2 FIF = πρΩUa 3 (−e x sin(Ωt) + e y cos(Ωt)) (sphere internal flow included), and € 3 FIF = 2πρΩUa 3 (−e x sin(Ωt) + e y cos(Ωt)) (sphere-internal flow excluded). The direction of the force can be partially explained by the pressure gradient in the accelerating €uniform stream, ∇p = −ρ dU = − ρΩU (−e sin(Ωt) + e cos(Ωt)) , which leads to a force similar to x y dt € buoyancy with ΩU (−e x sin(Ωt) + e y cos(Ωt)) replacing the acceleration of gravity. This buoyancy-mimicking force pushes the sphere toward lower pressure. €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 7.52. Consider the flow field produced by a sphere of radius a that moves in the xdirection at constant speed U along the x-axis in an unbounded environment of a quiescent ideal fluid with density ρ. The pressure far from the sphere is p∞ and there is no body force. The velocity potential for this flow field is: a 3U x −Ut . φ (x, y, z, t) = − 2 "(x −Ut)2 + y 2 + z 2 $3 2 # %   a) If u = (u, v, w) , what is u(x, 0, 0, t) , the velocity along the x-axis as a function of time? [Hint: consider the symmetry of the situation before differentiating in all directions.] b) What is p(x,0,0,t), the pressure along the x-axis as a z! function of time? c) What is the pressure on the x-axis at x = Ut ± a? h! d) If the plane defined by z = h is an impenetrable flat 2a ! y! surface and the sphere executes the same motion, what additional term should be added to the given potential? U! x! e) Compared to the sphere's apparent mass in an unbounded environment, is the sphere's apparent mass x = Ut ! larger, the same, or smaller when the impenetrable flat surface is present? 12

Solution 7.52. a) To save writing let, r(t) = "#(x −Ut)2 + y 2 + z 2 $% , so φ = −

a 3U (x −Ut) . 2 r 3 (t)

 ∂φ a 3U # r 2 (t) − 3(x −Ut)2 & Compute velocity components from ∇φ = u . Start with: u = =− % (, ∂x 2 $ r 5 (t) ' evaluate this on the x-axis (y = z = 0) where r(t) = x −Ut to find: u(x, 0, 0, t) =

a 3U x −Ut

3

. Here φ

∂φ ∂φ ∂y 2 ∂φ = 2 = 2y 2 ; thus, v(x,0,0,t) = 0. Similarly, φ depends on z2 so ∂y ∂y ∂y ∂y w(x,0,0,t) = 0. Thus, the fluid velocity on the x-axis has only an x-component:  3 u(x, 0, 0, t) = a 3U x −Ut e x . depends on y2 so: v =

(

)

∂φ 1  2 ∂φ 1  2 + ρ u + p = p∞ , so p − p∞ = −ρ − ρu . ∂t 2 ∂t 2 Time differentiate the potential but utilize the chain-rule and the part a) results: ∂φ ∂φ ∂(x −Ut) ∂φ = = (−U ) = −Uu . ∂t ∂(x −Ut) ∂t ∂x 3 2 " ∂φ % ρa U Evaluate on the x-axis: $ρ ' , so: =− 3 # ∂t &( x,0,0,t ) x −Ut

b) The appropriate Bernoulli equation is: ρ

p(x, 0, 0, t) − p∞ = +

# 2a 3 ρ a 3U 2 1 a 6U 2 1 a 6 &( 2 % − ρ = ρ U − 3 % x −Ut 3 x −Ut 6 ( . 2 x −Ut 6 2 x −Ut $ '

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

c) At x = Ut ± a the factor in large (,)-parentheses in the part b) result is unity, so p = p∞ + (1/2)ρU2 at these locations. This answer can also be reached by noting that the specified locations are stagnation points on the sphere when it is stationary and the flow moves past it. d) The method of images applies to this situation, so an image sphere must move along the line defined by z = 2h and y = 0. Thus, the additional term for the potential is: a 3U x −Ut . − 2 "(x −Ut)2 + y 2 + (z − 2h)2 $3 2 # % e) The impenetrable flat surface requires the fluid to move faster above the real sphere to get out of its way as it travels so the sphere's apparent mass is larger when the impenetrable flat surface is present. Here it should be noted that apparent mass is a property of the sphere's geometry, the geometry of its environment, and the fluid density. It exists independently of the sphere's motion. When the sphere accelerates both its actual and apparent mass play a role. If the sphere does not accelerate then neither its actual nor its apparent mass plays a role. However, the sphere does not loose its actual mass when moving at a constant velocity, so it does not lose its apparent mass either.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 7.53. In three dimensions, consider a solid object moving with velocity U near the origin of coordinates in an unbounded quiescent bath of inviscid incompressible fluid with density ρ. The kinetic energy of the moving fluid in this situation is: 1 2 KE = ρ ∫ ∇φ dV 2 V where φ is the velocity potential and V is a control volume that contains all of the moving fluid but excludes the object. (Such a control volume is shown in the figure for Exercise 7.29 when A3 → 0 and U = 0.) € 1 a) Show that KE = − ρ ∫ φ (∇φ ⋅ n)dA where A encloses the body and is coincident with its 2 A surface, and n is the outward normal on A.

€

1

2

b) The apparent mass, M, of the moving body may be defined by KE = 2 M U . Using this definition, the result of a), and (7.97) with xs = 0, show that M = 2πa3ρ/3 for a sphere. € Solution 7.53. Start from the given equation, use the recommended CV, add φ∇ 2φ = 0 to the integrand, and apply€ Green's divergence theorem. ρ ρ 2 KE = lim ∫ ∇φ dV = lim ∫ (∇φ ⋅ ∇φ )dV 2 A 3 →0 V 2 A 3 →0 V € ρ ρ = lim ∫ (∇φ ⋅ ∇φ + φ∇ 2φ )dV = lim ∫ ∇ ⋅ (φ∇φ )dV 2 A 3 →0 V 2 A 3 →0 V 3 ρ lim ∑ ∫ φ∇φ ⋅ ni dA 2 A 3 →0 i=1 A i where n1 is the normal on A1, n2 is the normal on A2, and n3 is the normal on A3. When the limit is taken, the net contribution of A3 is zero because the surface area of A3 goes to zero. This leaves: ρ ρ KE = ∫ φ∇φ ⋅ n1dA + ∫ φ∇φ ⋅ n2 dA , 2 A1 2 A2 where n1 points away from the origin on A1 and n2 points inward on A2. By definition, the volume V enclosed by A1 contains all the moving fluid, thus ∇φ ⋅ n1 = u ⋅ n1 = 0 on A1, so the first integral in the last equation is equal to zero. Now define the outward normal on A2 as n = – n2, € from the body-conforming surface A2 to reach: and drop the subscript ρ KE = − ∫ φ€ ∇φ ⋅ ndA . 2A b) The potential for a sphere of radius a moving at velocity U that is instantaneously centered at the origin of a spherical coordinate system is: a3 a3 φ = − 3 U ⋅ x = − 2 U cosθ , € 2r 2r where the second equality follows when the direction of the z-axis is chosen to coincide with the sphere's velocity at the moment of interest. For a sphere, n = er and surface area element is dA = a2sinθdθdϕ, so the kinetic energy integral becomes: €

=

€

Fluid Mechanics, 6th Ed.

KE = −

Kundu, Cohen, and Dowling

ρ π 2π & ∂φ ) ρ π 2π & a 2 2 ) 2 φ a sin θ d θ d ϕ = − − U cos θ+ a 2 sin θdθdϕ ∫ ∫ ∫ ∫ ( + ( *r= a 2 θ = 0 ϕ = 0 ' ∂r *r= a 2 θ = 0ϕ = 0' 2

ρa 3 2 π ρa 3 2 −1 2 ρa 3 21 2 4 2 =π U ∫ cos θ sin θdθ = −π U ∫ β dβ =π U 3 6 2 35 2 2 2 θ=0 1 1 1 2πa 3 4 2 = 3ρ 6U . 22 3 5 The parentheses in the final equality contain the apparent mass M. €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 8.1. Starting from (8.5) and working in (x,y,z) Cartesian coordinates, determine an equation that specifies the locus of points that defines a wave crest. Verify that the travel speed of the crests in the direction of K = (k, l, m) is c = ω/|K|. Can anything be determined about the wave crest travel speed in other directions?

€

Solution 8.1. The phase of the waveform in (8.5) is K ⋅ x − ωt . Thus wave crests are specified by: K ⋅ x crest − ωt = 2nπ , since cos(2nπ) = 1. In (x,y,z) Cartesian coordinates with K = (k, l, m), this is the equation for a € series of planes: (*) kx crest + ly crest + mzcrest = 2nπ + ωt , € so (8.5) describes plane waves. The unit vector perpendicular to these planes, e ⊥ , is determined from the gradient of the phase: ∇ (K ⋅ x − ωt ) K € e⊥ = = = eK . ∇ (K ⋅ x − ωt ) K € To determine the travel speed of the crests, time differentiate equation (*) to find: d d d k x crest + l y crest + m zcrest = ω , dt dt dt € since k, l, m, and 2nπ are constants. In terms of a dot product, this last equation is: d K d d ω K ⋅ x crest = ω , or ⋅ x crest = e K ⋅ x crest = = c. dt K dt dt K € Thus, the travel speed of wave crests, dxcrest/dt, in the direction of K is c. And, the phase relationship (8.5) does not provide any information about the travel speed in other directions. However, it is not possible to determine the wave speed other directions. € € Consider the following hypothesis: d x crest = ce ⊥ + de|| , dt where d the crest-parallel wave speed, and e|| is a unit vector that is parallel to the wave crests with e ⊥ ⋅ e|| = 0 = e K ⋅ e|| . However, e ⋅ ∇(K ⋅ x − ωt ) = K e|| ⋅ e K = 0 ; € || thus, there are no changes in the phase of the wave (as represented by ∇(K ⋅ x − ωt ) ) in the crestparallel direction(s). Hence it is impossible to determine anything about the crest-parallel wave speed d by observing the wave. € €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 8.2. For ka ρ1) trapped between two horizontal surfaces at z = h1 and z = –h2, and having an average interface location of z = 0. For traveling waves on the interface, assume that the interface deflection from z = 0 is ξ = ξ o Re{exp(i(ωt − kx))} . The phase speed of the waves is c = ω/k. gk( ρ 2 − ρ1 ) a) Show that dispersion relationship is: ω 2 = where g is the ρ 2 coth(kh2 ) + ρ1 coth(kh1 ) € acceleration of gravity. b) Determine the limiting form of c for short (i.e. unconfined) waves, kh1 and kh2 → ∞. c) Determine the limiting form of c for long (i.e. confined) waves, kh1 and kh2 → 0. € d) At fixed wavelength λ (or fixed k = 2π/λ), do confined waves go faster or slower than unconfined waves? e) At a fixed frequency, what happens to the wavelength and phase speed as ρ2 – ρ1 → 0? f) What happens if ρ2 < ρ1?

Solution 8.22. a) For this problem there at two velocity potentials that must be matched at the moving interface. Based on the form of the interface wave, ξ = ξ o Re{exp(i(ωt − kx))} and the development given in the chapter, the form of the two potentials can be set: φ1 (x,z,t) = Z1 (z)e i(ωt−kx ) , and φ 2 (x,z,t) = Z 2 (z)e i(ωt−kx ) Here the boundary conditions and the field equation are: ∇ 2φ = 0 : Z1 (z) = A1+e +kz + A1−€ e−kz , and Z 2 (z) = A2+e +kz + A2−e−kz

€

€

∂φ ∂ξ = on €z = 0: kA1+ − kA1– € = iωξ o = kA2+ − kA2– ∂z ∂t ∂φ ∂φ € ρ1 1 + ρ1gξ = ρ 2 €2 + ρ 2 gξ on z = 0: ∂t ∂t € iωρ1 ( A1+ + A1− ) + gρ1ξ o = iωρ 2 ( A2+ + A2− ) + gρ 2ξ o $ ∂φ1 ' $ ∂φ ' kA1+e +kh − kA1– e−kh = 0 , and kA1+e−kh − kA1– e +kh = 0 =& 2) =0 & ) % ∂z ( z= h % ∂z ( z=−h 1

€

€

1

1

2

2

(1,2)

(3) (4,5)

2

The three boundary € conditions yield five equations, enough to determine all four A’s and the dispersion relationship. First use the (1) & (4), and (2) & (5) to find: € € ωξ e  kh1 ωξ e ±kh2 , and A2,± = +i o A1,± = −i o 2k sinh( kh1 ) 2k sinh( kh2 ) Plug these into (3), to find: % ωξ cosh(kh ) ( % ωξ cosh(kh ) ( 1 2 ωρ1' o * + gρ1ξ o = −ωρ 2 ' o * + gρ 2ξ o k sinh kh k sinh kh ( 1) ) € ( 2) ) & & €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Cancel the common factor of ξo, simplify and rearrange: gk( ρ 2 − ρ1 ) ω2 = . ρ1 coth(kh1 ) + ρ 2 coth(kh2 ) ω g( ρ 2 − ρ1 ) k b) Using the results of part a), the phase speed is c = = ± . k ρ1 coth(kh1 ) + ρ 2 coth(kh2 ) For short waves, kh1 and € kh2 → ∞, and the hyperbolic co-tangent functions both approach unity so: ω g( ρ 2 − ρ1 ) c€ = =± k k( ρ1 + ρ 2 ) c) For long waves, kh1 and kh2 → 0, and the hyperbolic co-tangent functions both approach the ω g( ρ 2 − ρ1 ) k g( ρ 2 − ρ1 )h1h2 inverse of their arguments: c = = ± =± ρ1 (1/kh1 ) + ρ 2 (1/kh2 ) ρ1h2 + ρ 2 h1 € k Note that when ρ1 → 0, the answers to parts b) and c) both recover the ordinary linear water wave results. d) For simplicity, take h1 = h2 = h, then for short (unconfined) waves: € g( ρ2 − ρ1 ) 1 , c=± ( ρ1 + ρ2 ) k and for long (confined) waves: g( ρ 2 − ρ1 ) g( ρ 2 − ρ1 ) kh . c =± h =± ρ1 + ρ 2 ρ1 + ρ 2 k For k = const., as kh → 0 , the confined waves travel slower than the unconfined waves with the same wavelength (same wave number k). e) As ρ2 – ρ1 → 0, the phase speed and wavelength both go to zero. € f) If ρ2 € < ρ1, then the two fluid layers are not stably stratified. The dispersion relationship then gk( ρ1 − ρ 2 ) requires ω to be imaginary, i.e. ω = ±i . This means that there is an ρ 2 coth(kh2 ) + ρ1 coth(kh1 ) interface wave solution that is growing exponentially with increasing time. The situation is unstable, but not much more can be determined from the linearized theory. In reality, the two fluids will switch places but the linearized theory considered here is not valid throughout this € process.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 8.23. Consider the long-wavelength limit of surface and interface waves with amplitudes a and b, respectively, that occur when two ideal fluids with densities ρ1 and ρ2 (> ρ1) are layered as shown in Figure 8.28. Here, the velocity potentials – accurate through second order in kH1 and KH2 – in the two fluids are: φ1 ≅ A1 (1+ 12 k 2 (z − z1 )2 ) ei(kx−ωt ) and φ2 ≅ A2 (1+ 12 k 2 (z − z2 )2 ) ei(kx−ωt ) for kH1, KH2 b for the barotropic mode, and out-of-phase with b > a for the baroclinic mode. For this case, what are the phase speeds of the two modes? Which mode travels faster? What happens to the baroclinic mode's phase speed and amplitude as ρ2 − ρ1 → 0 ? z!

a!

z!

a! x!

x!

ρ1! ρ2!

H1! H2!

b!

b!

Barotropic!

Baroclinic!

Figure 8.28. Solution 8.23. The form of the two potentials are given in the problem statement, and the surface and interface elevations from equilibrium can be taken from (8.101) and (8.102): η = aei(kx−ωt ) and ζ = bei(kx−ωt ) . The boundary conditions are: ∂φ2/∂z = 0 on z = –(H1 + H2) (i) ∂φ1/∂z = ∂η/∂t on z = 0 (ii) ∂φ1/∂t + gη = 0 on z = 0 (iii) ∂φ1/∂z = ∂φ2/∂z = ∂ζ/∂t on z = –H1 (iv) ρ1∂φ1/∂t + ρ1gζ = ρ2∂φ2/∂t + ρ2gζ on z = –H1 (v) Apply these in turn produces the following algebraic equations: (i): A2k2(z – z2) = 0 on z = –(H1 + H2), or A2k2(–H1 – H2 – z2) = 0 (ii): A1k2(z – z1) = –iωa on z = 0, or A1k2z1 = iωa (iii) iω A1 ≅ ga −iω A1 (1+ 12 k 2 (z − z1 )2 ) + ga = 0 on z = 0, or (iv) (v)

A1k2(z – z1) = A2k2(z – z2) = –iωb on z = –H1, or A1k2(H1 + z1) = A2k2(H1 + z2) = iωb −iωρ1 A1 (1+ 12 k 2 (z − z1 )2 ) + ρ1gb = −iωρ2 A2 (1+ 12 k 2 (z − z2 )2 ) + ρ2 gb on z = –H1, or

−iωρ1 A1 + ρ1gb ≅ −iωρ2 A2 + ρ2 gb where the second order terms involving [k.(vertical distance)] appearing in (iii) and (v) have been dropped compared to unity in the equations on the right. Noting that (iv) provides two equations, the first five equations on the right form a non-linear algebraic system that can be solved to find: " gk 2 H1 % iaω " 1 gkH % z2 = –(H1 + H2), z1 = –ω2/k2g, A1 = –iga/ω, A2 = − − 2 1 ' , and b = a $1− $ '. k # kH 2 ω H 2 & ω2 & # 2

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Substitute these into the final approximate equation provided by (v): " gk 2 H1 % " gk 2 H1 % " −iga % " −iaω %" 1 gkH1 % −iωρ1 $ + ρ ga 1− ≅ −i ωρ − + ρ ga $ ' ' 1 $ ' ' $1− '. 2$ 2 # ω & # k &# kH 2 ω 2 H 2 & ω2 & ω2 & # # Divide out the common factor of a, and simplify: g 2 k 2 H1 −ρ2ω 2 # 1 gkH1 & g 2 k 2 H1 , or −ρ1 ≅ − + ρ g − ρ % ( 2 2 ω2 k $ kH 2 ω 2 H 2 ' ω2 # H & ρ ω2 g 2 k 2 H1 0 ≅ − 22 + ρ2 g %1+ 1 ( − ( ρ2 − ρ1 ) k H2 ω2 $ H2 ' Multiply –ω2k2H2/ρ2 and rearrange to find:

( ρ2 − ρ1 ) 2 2 k g kH1kH 2 . ρ2 This is a quadratic equation for ω2. The two solutions are: " " ρ − ρ % 4H1H 2 % k 2 g(H1 + H 2 ) $ '. ω2 = 1± 1− $ 2 1 ' 2 ' $ 2 ρ (H + H ) # & 2 1 2 # & Set H1 = H2 = H/2 to reach: k 2 gH ! ρ $ ω2 = ##1± 1 && . 2 " ρ2 % Consider the "+" sign for the barotropic mode. For H1 = H2 = H/2, the relationship between the surface deflection amplitude a and the interface deflection amplitude b is: % " " gk 2 H1 % "$ ρ1 ρ2 % gk 2 H ' $ '. b = a $1− = a 1− = a ' ω 2 & $ k 2 gH 1+ ρ1 ρ2 ' $# 1+ ρ1 ρ2 '& # # & The factor in parentheses is positive with a value near 1/2 when the densities are nearly equal. Thus, a and b are in phase. Now consider the "–" sign for the baroclinic mode. % " % " gk 2 H1 % "$ gk 2 H ' = a $ − ρ1 ρ2 ' . b = a $1− = a 1− ' ω 2 & $ k 2 gH 1− ρ1 ρ2 ' $# 1− ρ1 ρ2 '& # # & The factor in parentheses is negative with a magnitude much larger than unity when the densities are nearly equal. Thus, a and b are out of phase. The phase speeds of the two modes when H1 = H2 = H/2 are obtained from the dispersion relationships above: 12 ω ' gH ! ρ1 $* cp = = ) #1± &, , k )( 2 #" ρ2 &%,+ with barotropic mode (corresponding to the "+" sign) traveling faster than the baroclinic mode (corresponding to the "–" sign). As ρ2 − ρ1 → 0 , the baroclinic mode's phase speed approaches zero, and its amplitude becomes unbounded. This mode amplitude result is inconsistent with the long-wavelength & shallow-water approximations, so a more refined theory is needed to truly explain the baroclinic mode amplitude in this limit. 0 ≅ ω 4 − gk ( kH 2 + kH1 )ω 2 +

(

)

(

)

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 9.1. a) Write out the three components of (9.1) in x-y-z Cartesian coordinates. b) Set u = (u(y), 0, 0), and show that the x- and y-momentum equations reduce to: 1 ∂p 1 ∂p d 2u . 0=− + ν 2 , and 0 = − ρ ∂y ρ ∂x dy Solution 9.1. a) Equation (9.1) is the constant-viscosity Navier-Stokes' momentum equation for incompressible flow: Du 1 = − ∇p + ν∇ 2 u , Dt ρ where ν is the kinematic viscosity of the flow. Using u = (u, v, w) and ∇ = (∂ ∂x,∂ ∂y,∂ ∂z) , the three components of this equation become: & ∂ 2u ∂ 2u ∂ 2u ) ∂u ∂u ∂u ∂u 1 ∂p x: +u +v +w =− + ν( 2 + 2 + 2 + , ∂t ∂x ∂y ∂z ρ ∂x ∂y€ ∂z * ' ∂x & ∂ 2v ∂ 2v ∂ 2v ) ∂v ∂v ∂v ∂v 1 ∂p y: +u +v +w =− + ν ( 2 + 2 + 2 + , and ∂t ∂x ∂y ∂z ρ ∂y ∂y ∂z * ' ∂x & ∂ 2w ∂ 2w ∂ 2w ) € ∂w ∂w ∂w ∂w 1 ∂p z: +u +v +w =− + ν( 2 + 2 + 2 +. ∂t ∂x ∂y ∂z ρ ∂z ∂y ∂z * ' ∂x b) When € u = (u(y), 0, 0), all the terms involving v and w disappear, so the part a) equations simplify to: & ∂ 2u ∂ 2u ∂ 2u ) ∂u ∂u 1 ∂p €x: + u +0+0=− + ν( 2 + 2 + 2 +, ∂t ∂x ρ ∂x ∂y ∂z * ' ∂x 1 ∂p y: 0+0+0+0 =− + ν ( 0 + 0 + 0 ) , and ρ ∂y 1 ∂p € 0+0+0+0=− + ν (0 + 0 + 0) . z: ρ ∂z And, when u depends only on y, then ∂u/∂t = ∂u/∂x = ∂u/∂z = 0 so the part a) equations simplify further: & ∂ 2u ) 1 ∂p € x: 0=− + ν( 2 + , ρ ∂x ' ∂y * 1 ∂p 0=− y: , and ρ ∂y 1 ∂p € 0=− z: . ρ ∂z These x- and y-direction equations match those in the problem statement. € €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 9.2. For steady pressure driven flow between parallel plates (see Figure 8.3), there are 7 parameters: u(y), U, y, h, ρ, µ, and dp/dx. Determine a dimensionless scaling law for u(y), and rewrite the flow-field solution (8.5) in dimensionless form. Solution 9.2. The parameters are: u(y), U, y, h, ρ, µ, and dp/dx. First, create the parameter matrix: u U y h ρ µ dp/dx –––––––––––––––––––––––––––––––––––––––– Mass: 0 0 0 0 1 1 1 Length: 1 1 1 1 -3 -1 -2 Time: -1 -1 0 0 0 -1 -2

€

Next, determine the number of dimensionless groups. This rank of this matrix is three so 7 parameters - 3 dimensions = 4 groups, and construct the groups: Π1 = u U , Π 2 = y h , 2 Π 3 = ρUh µ , and Π 4 = h (dp /dx) Uµ . Now write a dimensionless law: # y ρUh h 2 dp & u = f% , , ( U µ dx ' $ h µ U€ € where f is€an unknown function. When rewritten in dimensionless form, (8.5) is: u y h 2 dp # y &# y & 1 = −€ % (%1− ( or Π1 = Π 2 − Π 4 Π 2 (1− Π 2 ) . U h 2Uµ dx $ h '$ h ' 2 In this flow, there is no fluid acceleration so the Reynolds number, Π3, does not appear.

€

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 9.3. An incompressible viscous liquid with density ρ fills the gap between two large smooth parallel walls that are both stationary. The upper and lower walls are located at x2 = ±h, respectively. An additive in the liquid causes its viscosity to vary in the x2 direction. Here the flow is driven by a constant non-zero pressure gradient: ∂p ∂x1 = const. a) Assume steady flow, ignore the body force, set u = ( u1 (x 2 ),0,0) and use ∂u ∂u ∂ρ ∂ ∂p ∂ + % ∂u ∂u (. ∂ +% 2 ( ∂u . + ( ρui ) = 0 , ρ j + ρui j = − + ρg j + -µ'' i + j **0 + -'µv − µ* i 0 ∂t ∂x i ∂t ∂x i ∂x j € ∂x i -, & ∂x j ∂x i )0/ ∂x j ,& 3 ) ∂x i / € 2 to determine u1(x2) when µ = µo 1+ γ ( x 2 h ) .

(

€

)

b) What shear stress is felt on the lower wall? c) What is the€volume flow rate (per unit depth into the page) in the gap when γ = 0? d) If –1 < γ < 0, will the volume flux be higher or lower than the case when γ = 0? €

Solution 9.3. a) The continuity equation is satisfied by the form of the velocity field. The j =1component of momentum equation simplifies to: 0 = −(∂p ∂x1) + (∂ ∂x 2 )[µ(∂u1 ∂x 2 )] . Integrate once with ∂p ∂x1 = const. to find: µ(∂u1 ∂x 2 ) = (∂p ∂x1 ) x 2 + C . Divide by µ and integrate again: & 1 # ∂p (∂p ∂x1 ) x 2 + C dx €u1 = ∫ % x 2 + C (dx 2 = ∫ 2 2 µ $ ∂x1 € ' µo 1+ γ ( x 2 h ) €

(

)

#x γ& # x 2 &2 & h 2 ∂p # Ch = ln%%1+ γ % ( (( + tan−1% 2 ( + D. $ h ' ' µo γ 2γµo ∂x1 $ $ h ' The boundary conditions, u1 (±h) = 0 , determine the values of the constants: C = 0, and

D = −( h 2 2γµo )(∂p ∂x1) ln(1+ γ ) , thus: 2 h 2 ∂p % 1+ γ ( x 2 h ) ( ** . u1 (x 2 ) = − ln' € 2γµo ∂x1 '& 1+ γ ) b) From the solution of part a) with C = 0: τ w = (µ∂u1 ∂x 2 ) y=−h = −h (∂p ∂x1 )

€

€

+h h 2 ∂p +h % x 22 ( 2h 3 ∂p c) When γ = 0, the flow profile is parabolic: q = ∫ u1 (x 2 )dx 2 = − 1− dx = − * 2 ∫' € 2µo ∂x1 −h & h 2 ) 3µo ∂x1 −h d) The volume flux will be higher€because the viscosity will be reduced at the wall. Manipulation of the near-wall viscosity with additives is sometimes used in long piping systems to reduce pumping power requirements. €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 9.4. An incompressible viscous liquid with density ρ fills the gap between two large smooth parallel plates. The upper plate at x2 = h moves in the positive x1-direction at speed U. The lower plate at x2 = 0 is stationary. An additive in the liquid causes its viscosity to vary in the x2 direction. a) Assume steady flow, ignore the body force, set u = ( u1 (x 2 ),0,0) and ∂p ∂x1 = 0 , and use the equations specified in Exercise 8.3 to determine u1(x2) when µ = µo (1+ γ x 2 h ) . b) What shear stress is felt on the lower plate? c) Are there any physical limits on γ? If, so specify them. € € €

€

Solution 9.4. a) For u = ( u1 (x 2 ),0,0) , no body force, and ∂p ∂x1 = 0 in steady incompressible flow, the continuity equation is automatically satisfied, and the momentum equation for j = 1 simplifies to: 0 = +(∂ ∂x 2 )[µ(∂u1 ∂x 2 )] , or, after integrating once: C = µ(∂u1 ∂x 2 ) , € € where C is a constant. Now use the specified relationship for the viscosity and integrate to find: C Cdx 2 Ch u1 (x 2 ) = ∫ dx 2 = ∫ = ln(1+ γ x 2 h ) + D µ µo (1+ γ x 2 h ) € µoγ € where D is another constant. The boundary conditions u1(0) = 0 and u1(h) = U allow C = Uγµo ( h ln(1+ γ )) and D = 0 to be determined yielding: u1 (x 2 ) = U ln(1+ γ x 2 h ) ln(1+ γ ) . € b) From part a), the shear stress is constant: τ w = µ(∂u1 ∂x 2 ) = C = Uγµo ( h ln(1+ γ )) . c) Negative viscosities violate the second law of thermodynamics, thus γ > –1 is required.

€ €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 9.5. Planar Couette flow is generated by placing a viscous fluid between two infinite parallel plates and moving one plate (say, the upper one) at a velocity U with respect to the other one. The plates are a distance h apart. Two immiscible viscous liquids are placed between the plates as shown in the diagram. The lower fluid layer has thickness d. Solve for the velocity distributions in the two fluids.

Solution 9.5. For steady viscous flow between infinite parallel plates, the fluid velocity will be unidirectional: u = (u, 0, 0). For this problem, no pressure gradient is specified so assume it to be zero. Thus, the horizontal (x1-direction) momentum equation reduces to: ∂τ 21 ∂ x2 = (∂ ∂ x2 ) (µ∂ u ∂ x2 ) = µ∂ 2 u ∂ y 2 = 0 , where the last equality follows when the viscosity is constant and x2 = y. Here, the viscosity is assumed constant within each fluid. This means that the flow profile in each fluid will be piecewise linear: # A + B1 y for 0 ≤ y ≤ d & u(y) = $ 1 ', % A2 + B2 y for d ≤ y ≤ h ( where the A’s & B’s are constants and ‘1’ implies the upper fluid layer with viscosity µ1, and ‘2’ implies the lower fluid layer with viscosity µ2. The four constants can be determined from the four boundary conditions: € i) u(0) = 0 (match the speed of the lower boundary) ii) u(h) = U (match the speed of the upper boundary) – + iii) u(d ) = u(d ), and (match flow speeds at the internal fluid-fluid interface) iv) τ(d–) = τ(d+) (match shear stress at the internal fluid-fluid interface) where τ is the shear stress in the fluid. These four boundary conditions imply: A2 = 0, A1 + B1h = U , A2 + B2 d = A1 + B1d , and µ2 B2 = µ1B1 Use the first two equations to eliminate A1 and A2 from the second two equations to find: B2 d = U − B1 (h − d) , and µ2 B2 = µ1B1. Eliminate B2 and solve for B1: € € € (µ1 µ2 ) B1d = U − B1(h − d) –> B1 = µ2U [µ2 h + (µ1 − µ2 ) d] . So, B2 = µ1U [µ2 h +€(µ1 − µ2 ) d ] , and A1 = U (µ€1 − µ2 ) d [µ2 h + (µ1 − µ2 ) d ] with A2 = 0. Thus: $ ' µ1 y for 0 ≤ y ≤ d U u(y) = % ( € µ − µ2 ) d + µ2 y for d ≤ y ≤ h ) µ2 h + (µ1 − µ€ 2 ) d &( 1

€

€ €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 9.6. Consider plane Poiseuille flow of a non-Newtonian power-law fluid using the coordinate system and geometry shown in Figure 9.3. Here the fluid's constitutive relationship is n

given by (4.37): τ xy = m (∂u ∂y) , where m is the power law coefficient, and n is the power law exponent. a) Determine the velocity profile u(y) in the lower half of the channel, 0 ≤ y ≤ h/2, using the boundary condition u(0) = 0. b) Given that the maximum velocity occurs at y = h/2 and that the flow profile is symmetric about this location, plot u(y) u(h/2) vs. y (h/2) for n = 2 (a shear thickening fluid), n = 1 (a Newtonian fluid), and n = 0.4 (a shear thinning fluid). c) Explain in physical terms why the shear-thinning velocity profile is the bluntest. Solution 9.6. In plane Poiseuille flow, the upper channel wall does not move. In the lower half of the channel, the ∂u/∂y > 0, so fractional powers are readily managed. a) For fully developed constant density flow (so that ∂p/∂x is independent of x) the horizontal momentum equation reduces to: ∂p ∂τ 0 = − + xy . ∂x ∂y The vertical momentum equation is simply, ∂p/∂y = 0, so ∂p/∂x does not depend on x or y. Thus, this horizontal momentum equation can be integrated to find: ∂p y = τ xy + C , ∂x where C is a constant. However, when y = h/2, τxy = 0 so C = (h/2)∂p/∂x, therefore: n 1n 1n " ∂u % " h %" ∂p % ∂u # 1 ∂p & # h & = %− ( % − y( , $ − y '$ − ' = τ xy = m $ ' , or # 2 &# ∂x & ∂y $ m ∂x ' $ 2 ' # ∂y & and both equations are written this way because –∂p/∂x is positive. The second equation can be integrated to find: 1n 1+1 n # 1 ∂p & 1 #h & u(y) = − % − y ( % ( +D, $ m ∂x ' 1+1 n $ 2 ' where D is another constant of integration. The horizontal velocity must go to zero on the lower channel wall at y = 0, so 1n 1+1 n " 1 ∂p % 1 "h% . D =$ ' $ ' # m ∂x & 1+1 n # 2 & Thus, the velocity profile is: 1+1 n 1n 1+1 n 1+1 n 1n % , " h % , " n % h " h ∂p % ) " n " 1 ∂p % )" h % y u(y) = ' .. $ ' +$ ' − $ − y ' . = $ ' $ ' +1− $1− # 2 & .- # n +1 & 2 # 2m ∂x & +* # h 2 & .n +1 # m ∂x & +*# 2 & b) Evaluate the result of part a) at y = h/2, and form the ratio: 1+1 n 1n $ , ! n $ h ! h ∂p $ ) ! y # & # & +1− #1− & . " % " % n +1 2 2m ∂x h 2 + " % .- ) u(y) 1+1 n * = = *1− (1− η ) ,- , 1n u(h / 2) ! n $ h ! h ∂p $ # & # & " n +1 % 2 " 2m ∂x %

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

where η = y (h/2) . The three curves for n = 2, 1, and 0.4 are plotted here:

2.00# 1.80# 1.60# 1.40#

η

1.20# 1.00# 0.80# 0.60#

n=2# n=1# n=0.4#

0.40# 0.20# 0.00# 0.000#

0.200#

0.400#

0.600#

0.800#

1.000#

u(η)/u(η = 1) c) From the part a) solution, the shear stress is:

∂p # h & % − y( . ∂x $ 2 ' The highest magnitude shear stress occurs at the upper and lower wall of the channel. In a shear thinning fluid, the local viscosity decreases when the shear stress increases. Thus, to maintain this shear stress profile, the velocity gradient magnitude must increase in regions where the shear stress magnitude is also large. This leads to a blunter profile for a shear thinning fluid compared to a Newtonian (or a shear thickening) fluid.

τ xy = −

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 9.7. Consider the laminar flow of a fluid layer falling down a plane inclined at an angle θ with respect to the horizontal. If h is the thickness of the layer in the fully developed stage, 2 2 show that the velocity distribution is u(y) = ( g 2ν ) h − y sin θ , where the x-axis points in the direction of flow along the free surface, and the y-axis points toward the plane. Show that the 3 volume flow rate per unit width is Q = gh 3ν sin θ , and that and the frictional stress on the

(

(

wall is τw = ρghsinθ.

)

)

€

x Solution 9.7. In the given coordinates, there is a y stress component of gravity, gsin€θ, acting in the x-direction, but distribution h the presence of atmospheric pressure on the liquid surface ensures that ∂p/∂x = 0. Thus, the steady flow x-direction momentum equation is: u(y) ∂ 2u ! 0 = +gsin θ + ν 2 . ∂y Integrating twice produces: gsin θ 2 u(y) = − y + Ay + B , 2ν € B are constants. The boundary condition on the liquid surface (y = 0) is zero shear where A and stress (µ∂u/∂y = 0), and this allows A to be evaluated: #∂u & # gsin θ & € 0 = % ( = %− y + A( = A . 'y= 0 ν $∂y 'y= 0 $ The boundary condition, on the solid surface (y = h) is zero velocity (u = 0), and this allows B to be evaluated: gsin θ 2 gsin θ 2 0 = u(h) = − h + B , or B = h . € 2ν 2ν so the velocity profile is: gsin θ 2 u(y) = − (y − h 2 ). 2ν € flow rate per unit width is: € The volume ) gsin θ 3 h gsin θ h 2 gsin θ & h 3 2 3 Q = ∫ 0 u(y)dy = − y − h dy = − − h h . ∫ ( += ( ) 0 2 ν 2 ν 3 3 ν ' * € The magnitude of the shear stress is: & gsin θ ) ∂u τ = µ = µ( + y = ρgy sin θ , ' ν * ∂y € and the maximum shear stress, τw = ρghsinθ, occurs at the solid surface. €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 9.8. In two-dimensional (x,y)-coordinates, the Navier-Stokes equations for the fluid velocity, u = (u, v) , in a constant-viscosity constant-density flow are: ∂u ∂x + ∂v ∂y = 0 , # ∂2 u ∂2 u & # ∂2 v ∂2 v & ∂u ∂u ∂u 1 ∂p ∂v ∂v ∂v 1 ∂p +u +v =− + ν % 2 + 2 ( , and +u +v =− +ν % 2 + 2 ( . ∂t ∂x ∂y ρ ∂x ∂t ∂x ∂y ρ ∂y $ ∂x ∂y ' $ ∂x ∂y ' a) Cross differentiate and sum the two momentum equations to reach the following equation for ω z = ∂v ∂x − ∂u ∂y , the vorticity normal to the x-y plane: " ∂2ω ∂2ω % ∂ω z ∂ω ∂ω + u z + v z = ν $ 2z + 2z ' . ∂t ∂x ∂y ∂y & # ∂x b) The simplest nontrivial solution of this equation is uniform shear or solid body rotation (ωz = constant). The next simplest solution is a linear function of the independent coordinates: ωz = ax + by, where a and b are constants. Starting from this vorticity field, derive the following velocity field: b " (ax + by)2 % a ! (ax + by)2 $ u = − $ 2 2 + c' and v = # 2 2 + c& . 2# a +b 2" a +b & % where c is an undetermined constant. c) For the part b) flow, sketch the streamlines. State any assumptions you make about a, b, and c. d) For the part b) flow when a = 0, b > 0, and u = (Uo , 0) at the origin of coordinates with Uo > 0, sketch the velocity profile along a line x = constant, and determine ∇p . Solution 9.8. a) Apply –∂/∂y to the x-direction momentum equation, and ∂/∂x to the y-direction momentum equation to reach: # ∂2 ∂ # ∂u & ∂u ∂u ∂ # ∂u & ∂v ∂u ∂ # ∂u & 1 ∂2 p ∂2 &# ∂u & − % (− −u % (− −v % (= − ν % 2 + 2 (% ( , and ∂t $ ∂y ' ∂y ∂x ∂x $ ∂y ' ∂y ∂y ∂y $ ∂y ' ρ ∂y∂x $ ∂x ∂y '$ ∂y ' " ∂2 ∂ " ∂v % ∂u ∂v ∂ " ∂v % ∂v ∂v ∂ " ∂v % 1 ∂2 p ∂2 %" ∂v % + u $ '+ +v $ '=− + ν $ 2 + 2 '$ ' . $ '+ ∂t # ∂x & ∂x ∂x ∂x # ∂x & ∂x ∂y ∂x # ∂y & ρ ∂x∂y # ∂x ∂y &# ∂x & Add these two equations together noting that pressure terms cancel, and that the second and fourth terms in each equation sum to zero because of the continuity equation, ∂u ∂x + ∂v ∂y = 0 . # ∂2 ∂ # ∂v ∂u & ∂ # ∂v ∂u & ∂ # ∂v ∂u & ∂2 &# ∂v ∂u & % − ( + u % − ( + v % − ( = ν % 2 + 2 (% − ( . ∂t $ ∂x ∂y ' ∂x $ ∂x ∂y ' ∂y $ ∂x ∂y ' $ ∂x ∂y '$ ∂x ∂y ' Substituting in the definition ω z = ∂v ∂x − ∂u ∂y leads to the final form: " ∂2ω ∂2ω % ∂ω z ∂ω ∂ω + u z + v z = ν $ 2z + 2z ' . ∂t ∂x ∂y ∂y & # ∂x b) To find the velocity field start with: ωz = ax + by = ∂v/∂x – ∂u/∂y. (1) The part a) equation implies: ua + vb = 0 or v = –(a/b)u. (2) The continuity equation is also needed: ∂u/∂x + ∂v/∂y = 0. (3) Use (2) to eliminate v, from (1) and (3) to find: a ∂u ∂u ∂u a ∂u , and ax + by = − − − = 0. b ∂x ∂y ∂x b ∂y Combine these twice, first to eliminate ∂u/∂y, then to eliminate ∂u/∂x to reach:

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

" a 2 % ∂u " a b % ∂u ax + by = − $ 2 +1' , and ax + by = − $ + ' . # b a & ∂x #b & ∂y " a2 % "a b% x2 y2 Integrate to find: axy + b = − $ 2 +1' u + f (x) , and a + bxy = − $ + ' u + g(y) , where f(x) #b a& 2 2 #b & and g(y) are unknown functions. Solve both equations for u and absorb multiplicative constants into f(x) and g(y). b " f (x) + 2abxy + b 2 y 2 % b " a 2 x 2 + ab 2 xy + g(y) % , and u=− $ u = − ' $ '. 2# a2 + b2 2# a2 + b2 & & 2 2 For consistency, the unknown functions must be: f(x) = a x + const. and g(y) = b2y2 + const. so % b " (ax + by)2 u = − $ 2 2 + c ' , and then from (2) 2# a +b & y! % a a " (ax + by)2 v = − u = $ 2 2 + c' . b 2# a +b & b! a! where c is another undetermined constant. c) dysl/dxsl = v/u = –a/b. Thus, if a and b are both x! positive constants, the streamlines are negativelyslopped straight lines. In this case, u < 0 and v > 0, so the arrows point upward and to the left. d) Evaluate the constants to find: u = Uo − ( b 2 ) y 2 and v = 0, so the y! U o! horizontal velocity profile is parabolic. To get the pressure gradient, evaluate the 2D Navier-Stokes equations using the part b) velocity field:  1 ∂p 1 ∂p 0=− + ν (−b) , and 0 = − + 0 , so ∇p = (−µ b, 0 ) . ρ ∂x ρ ∂y  

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 9.9. Consider circular Couette flow as described by (9.10) in the limit of a thin gap between the cylinders. Use the definitions: R = (R1 + R2 ) 2 , h = R2 − R1 , Ω = (Ω1 + Ω2 ) 2 , ΔΩ = Ω2 − Ω1 , and R = R + y to complete the following items. a) Show that uϕ (y) ≅ ΩR + ΔΩR ( y h ) when y > h and one that acts to provide the correct boundary condition at y = 0: € z# z& € u = U %1− ( + υ (y,z) , (i) h $ h' where υ = 0 when z = 0 and z = h, υ → 0 when y >> h, and u → 0 when y → 0 . Using –∂p/∂x = 2µU/h2, the part b) field equation becomes: ∂ 2υ ∂ 2υ 1 2µU € % 2U ( % ∂ 2υ ∂ 2υ ( + = 0. , or + ν − + ν + = 0 ' * ' * ∂€ y 2 ∂z 2 ρ h 2€ & h 2 ) & ∂y 2 ∂z€2 ) Thus, υ solves a two dimensional Laplace equation. Here, a separation of variable solution, υ = Y(y)Z(z) is appropriate. Placing this solution form into the two dimensional Laplace equation and dividing by υ, produces: € € 1 d 2Y 1 d 2 Z + = 0. Y dy 2 Z dz 2 The first term depends only on y and the second term depends only on z, so each must be constant for the equation to hold throughout the domain. Let the first term equal +k2 and the second term = –k2, so that the individual Y, and Z equations become: € d 2Y d 2Z 2 − k Y = 0 , and + k 2Z = 0. 2 2 dy dz The solutions to these two equations are: Y(y) = A± exp{±ky} , and Z(z) = Bsin(kz) + C cos(kz) . To satisfy the boundary conditions υ = 0 when z = 0 and z = h, implies C = 0 and Bsin(kh) = 0 . The second of these is satisfied € by the (useless) trivial solution B = 0, or by sin(kh) € = 0. The later condition leads to k = kn = nπ/h, where € n is a positive integer so that € Z(z) = Bsin( nπz h ) . Thus, the separation constant k may take on discrete values. To satisfy the boundary condition υ → 0 when y >> h, means setting A+ = 0. Therefore, υ(y, z) takes the following form: ∞

€

υ (y,z) = ∑ An sin( k n z) exp{−kn y} . (ii) n=1 € to allow for all possible values of k. To satisfy the final boundary condition ( u → 0 when y → 0 ), (i) and (ii) imply: z# z& z# z& ∞ €U %1− ( + υ (0,z) = 0 = U %1− ( + ∑ An sin( k n z) , h $ h' h $ h ' n=1 € €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

and this equation can be used to determine the coefficients An. Following the usual approach for a Fourier series, multiply it by sin(kmz) and integrate in z from 0 to h. The result is: ∞ ∞ h h z $ z' h sin(k z) U 1− dz = − A sin(k z) sin k z dz = − An δmn , & ) ( ) ∑ ∑ ∫ m n ∫ m n h % h( 2 0 0 n=1 n=1 where δmn is the Kronecker delta function, so the Am's are given by: 2U h z$ z' Am = − sin(k m z) &1− )dz . (iii) ∫ h 0 h % h( € Thus, the final velocity field is: # mπ & * mπ z# z& ∞ u(y,z) = U %1− ( + ∑ An sin% z( exp+− y. , h / $ h ' m=1 $ h ' h € where An is given by (iii). The integral is tedious but can be evaluated by parts to find: An = −8U n 3π 3 for n = odd, and An = 0 for n = even. The final € answer includes a sum of terms that decay exponentially with increasing y, and the first of these will have the slowest decay. At a distance of y = h, this slowest decaying term will be a factor of e– ≈ 0.04 smaller than it is at y = 0. π

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 9.45. Using the velocity field (8.49), determine the drag on Stokes’ sphere from the surface pressure and the viscous surface stresses σrr and σr . θ

Solution 9.45. There are pressure and shear stress contributions to the drag on a moving sphere at low Reynolds number. The pressure distribution is given by (8.50): 3µaU p(r,θ ) − p∞ = − cos θ . 2r 2 The pressure drag can be obtained by integrating this result: θ=π & 3U ) Fpressure = − ∫ p(r = a,θ )e r ⋅ e z dS = −2πa 2 ∫ µ( + cos 2 θ sin θdθ = 2πµUa surface θ = 0 ' 2a * € The viscous drag can be obtained from surface integrals of the viscous stresses: Fvicous = − ∫ σ rθ (r = a,θ )eθ ⋅ e z dS + ∫ σ rr (r = a,θ )e r ⋅ e z dS surface

€

= −2πa 2

surface

θ=π

θ=π

θ=0

θ=0

∫ σ rθ (r = a,θ )sin 2 θdθ + 2πa 2 ∫ σ rr (r = a,θ )cosθ sin θdθ,

& 3a 3a 3 ) & 1 ∂ur ∂uθ uθ ) µU sinθ & 3a 3 ) ∂ur where σ rθ = µ( + − +=− = 2µU cosθ( 2 − 4 + . ( 3 + , and σ rr = 2µ ' r ∂θ ∂r r* r ∂r 2r * ' 2r * ' 2r Thus, € at r = a, σr ≠ 0, but σrr = 0, so θ

θ=π

€

Fvicous = 3πµUa

3

∫ sin θdθ = 4πµUa .

θ =€ 0 Thus, one third of the drag comes from pressure forces and two thirds come from the shear stress. The total drag is the sum of these two contributions: Fdrag = Fpressure + Fvicous = 2πµUa + 4 πµUa = 6πµUa . €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 9.46. Calculate the drag on a spherical droplet of radius r = a, density ρʹ′ and viscosity µʹ′ moving with velocity U in an infinite fluid of density ρ and viscosity µ. Assume Re = ρUa/µ a. To recover a uniform flow far from the sphere, the first two constants must be A = 0 and B = 1/2. Similarly, the flow inside the sphere can be represented by ψ i = ( Ai r 4 + Bi r 2 + Ci r + Di r) sin 2 θ for r < a. € Here, Di = 0 to avoid a singularity at r = 0. Compute the velocities inside and outside the sphere: , 0 &1 C D ) −2cosθ( + o + 3o + for r > 1. € . . 1 ∂ψ . '2 r r * ur = − 2 =1 , and & 2 Ci ) r sin θ ∂θ . . ./ −2cosθ(' Ai r + Bi + r +* for r < 1.2 , 0 & C D ) sin θ(1+ o − 3o + for r > 1. . . 1 ∂ψ . ' r r * uθ = =1. & Ci ) r sin θ ∂r . 2 . € sin θ 4 A r + 2B + for r < 1 ( i + i ./ .2 ' r* For a bounded velocity inside the sphere, Ci must be 0. There are four remaining boundary and matching conditions at r = 1: 1 ur(1+) = 0 and ur(1–) = 0, and these imply + Co + Do = 0 , and Ai + Bi = 0 ; (1,2) € 2 u (1+) = u (1–), and this implies 1+ Co − Do = 4 Ai + 2Bi ; and (3) the stress continuity condition: * ∂ $ u ' 1 ∂ur * $ ' € €/ = µ0,r ∂ & uθ ) + 1 ∂ur / , and this implies σr (a+) = σr (a–) or µ,r & θ ) + + ∂r % r ( r ∂θ€.r=1+ + ∂r % r ( r ∂θ .r=1– $ ' 1 µ sinθ&−1− 2Co + 4Do + 2 + 2Co + 2Do ) = µ* sin θ [ 4 Ai − 2Bi + 2Ai + 22Bi ] . % ( 2 Fortunately, there € are many canceling terms in this last condition and it simplifies to: (4) 6µ Do = 6µ" Ai a . Simultaneous solution of the four algebraic equations produces: € 1 3µ# + 2µ 1 µ" 1 µ 1 µ Co = − , Do = , Ai = , and Bi = − . 4 µ# + µ 4 µ" + µ 4 µ" + µ 4 µ# + µ € The force on the sphere is determined from a surface intergral: Fi = − ∫ ( pδij − σ ij )n j dA . θ

θ

θ

€

θ

θ

€

€

€

sphere' s surface

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

In the following, force is made dimensionless by µUa, area is made dimensionless with a2, and stress is made dimensionless with µU/a. The normal and surface elements are: n = e r , and dA = [ r 2 sin θdθdϕ ] r=1 . The dot product of the shear stress and the directed area element is: , ∂u & ∂u 1 ∂uθ uθ )/ τ ij n j dA = -e r 2 r + eθ ( θ + − +0 sin θdθdϕ . ' ∂ r r ∂r r *1 . € ∂r € The θ-component of surface shear stress 3 µ$ eθ 6(Do /a 3 )sin 2 θdθdϕ = eθ sin θdθdϕ . $ 2 µ + µ € The r-component includes a pressure contribution which must be found by integrating Stokes' creeping-flow momentum equation: % r 2 1 µ# 1 1 3µ# + 2µ ( 2 ψ o =€' + − r* sin θ , and −∇p = ∇ × ∇ × u = −∇ 2u . # # 2 4 µ + µ r 4 µ + µ & ) In this dimensionless formulation, these give: / & 1 µ% 1 ) & 1 1 µ% 1 1 3µ% + 2µ 1 ) ∂p 1 , − = 2 − + − .sin θ cosθ(1+ +4 sin θ cos θ1 3 + (€ 3 ∂r r sin θ 4 µ% + µ r * ' 2 µ% + µ r * ' 2 4 µ% + µ r 0 € 3µ% + 2µ cosθ = µ% + µ r 3 1 ∂p sin 2 θ & 3 µ% 1 ) 2sin θ , ∂ & 1 1 µ% 1 1 3µ% + 2µ 1 )/ − = − . ( + (− +− +1 , or r ∂θ r sin θ ' 2 µ% + µ r 4 * r -∂r ' 2 4 µ% + µ r 3 4 µ% + µ r *0 ∂p 3µ% + 2µ sin θ € − = . ∂θ µ% + µ 2r 2 Integrating for p and setting p∞ = 0 produces: € 3µ" + 2µ cosθ p= . µ" + µ 2r 2 € Thus, the r-component of the force integrand is: ,$ / ∂ur ' 3 µ2 + 2µ 2 e r sin θdθdϕ . )e r [ r sin θdθdϕ ] r=11 = − cos θ .& − p + 2 ( ∂r € 2 µ2 + µ -% 0 So, the force on the liquid sphere is: π 2π % 3 µ$ + 2µ 3 µ$ ( F = ∫ ∫ '−e r cosθ + eθ * sin θdθdϕ . $ $ 2 µ + µ 2 µ + µ & ) θ = 0 ϕ = 0 € The unit vectors on the surface of the sphere are functions of the polar and azimuthal angles, θ, ϕ, respectively. They must be put in terms of the Cartesian unit vectors that do not vary with position. The necessary relationships are: € e r = e x cosϕ sin θ + e y sin ϕ sin θ + e z cos θ , and eθ = e x cosϕ cos θ + e y sin ϕ cos θ − e z sin θ . Integrating first over j, the x- and y-direction contributions to the force are zero, a result that can also be deduced from the axial symmetry of the flow. This leaves: π & 3 µ$ + 2µ ) 3 µ$ F = 2πe z ∫ −( sin 3 θ+dθ . € € cos2 θ sin θ − 2 µ + µ$ * θ = 0 ' 2 µ$ + µ Carrying out the trigonometric integrals leads to:

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

%µ$ + 2µ 2µ$ ( 3µ$ + 2µ F = −2πe z ' + . * = −2πe z µ$ + µ & µ$ + µ µ + µ$) The force on the sphere is a drag force, and in dimensionless form this drag becomes: 3µ# + 2µ 3µ# + 2µ D = 2πµUa = Ds . µ# + µ 3µ# + 3µ € where Ds = 6πµUa is the Stokes' drag on a solid sphere. €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 9.47. Consider a very low Reynolds number flow over a circular cylinder of radius r = a. For r/a = O(1) in the Re = Ua/ν → 0 limit, find the equation governing the stream function ψ(r, θ) and solve for ψ with the least singular behavior for large r. There will be one remaining constant of integration to be determined by asymptotic matching with the large r solution (which is not part of this problem). Find the domain of validity of your solution. Solution 9.47. The Stokes' flow field equation is: ∇p = µ∇ 2u , which can be rewritten: ∇p = −µ∇ × ∇ × u. Applying one more curl, eliminates the pressure gradient: € 0 = ∇ × ∇ × ∇ × u. In two-dimensional (r,θ)-polar coordinates, the relationship between the stream function ψ(r,θ) and the velocity field is u = −e€z × ∇ψ , so the field equation for the stream function is given by: € 0 = ∇ × ∇ × ∇ × (−e z × ∇ψ ) . This relationship can be converted to coordinate specific derivatives in steps. Start with the stream function definition. € 1 ∂ψ ∂ψ −e z × ∇ψ = er − eθ € r ∂θ ∂r Apply the first curl operation, using the formulae in Appendix B: .1 ∂ ' ∂ψ * 1 ∂2ψ 1 ∇ × (−e z × ∇ψ ) = −e z 0 )r , + 2 2 3. / r ∂r ( ∂r + r ∂θ 2 € Apply the second curl operation, 1 ∂ .1 ∂ ( ∂ψ + 1 ∂2ψ 1 ∂ .1 ∂ ( ∂ψ + 1 ∂2ψ 1 ∇ × ∇ × (−e z × ∇ψ ) = −e r r + + e * *r - + 2 2 3. 0 0 θ 2 23 ) , ) , r ∂θ 2 r ∂θ r ∂r ∂r r ∂ θ ∂ r r ∂r ∂r / 2 / € Apply third and final curl operation, 0 = ∇ × ∇ × ∇ × (−e z × ∇ψ ) = 46 1 ∂ ' ∂ .1 ∂ ' ∂ψ * 1 ∂2ψ 1* 1 ∂ .1 ∂ ' ∂ψ * 1 ∂2ψ 186 ez 5 r) 0 )r ,+ )r ,+ 3, + 0 39. 67 r ∂r ( ∂r / r ∂r ( ∂r + r 2 ∂θ 2 2+ r 2 ∂θ 2 / r ∂r ( ∂r + r 2 ∂θ 2 26: For large r, ur ~ cosθ, and u ~ –sinθ, so it is expected that ψ ~ sinθ. Therefore, a trial solution in the form ψ(r,θ) = f(r)sinθ may be sufficient. The resulting equation for f looks like: /' sin θ ∂ $ ∂f ' f 1 ∂ $ ∂ , 1 ∂ $ ∂ψ ' f € r& . &r ) − 2 sin θ1) − 3 & r ) + sin θ = 0 . r ∂r % ∂r - r ∂r % ∂r ( r 0( r ∂r % ∂r ( r 4 Divide by sinθ and recognize that this equation is equi-dimensional in r so it has power-law solutions, f(r) ~ (r/a)m, where the factor of a has been added for later scaling analysis. Substituting this into the equation for f(r) and dividing by (r/a)m – 4 leaves: € m 2 (m − 2) 2 − (m − 2) 2 − m 2 + 1 = 0 . This can be factored: (m −1) 2 (m + 1)(m − 3) = 0 . Thus, m = 1 is a double root which implies that the solutions for f may be proportional to r/a and € (r/a)ln(r/a). Therefore:

€

θ

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

f (r) = c1 (r /a) + c 2 (r /a)ln(r /a) + c 3 (r /a) 3 + c 4 (r /a) –1 . The growth of r3 is too fast to match the flow at large r, so c3 must be zero. This leaves: ψ (r,θ ) = (c1 (r /a) + c 2 (r /a)ln(r /a) + c 4 (r /a) –1 ) sinθ . The boundary € conditions at r = a are: %1 ∂ψ ( ur = ' = (c1 + c 4 )cosθ = 0 or c4 = –c1, and & r ∂θ *)r= a € & ∂ψ ) 1 uθ = (− + = (c1 + c 2 − c 4 )sin θ = 0 or c2 = –2c1. ' ∂r *r= a a Therefore: € %r r r a( ψ (r,θ ) = c1' − 2 ln − * sinθ . &a a a r) € The final constant is determined by matching this solution to the free-stream at large r/a. However, for large r/a, ψ ~ (r/a)ln(r/a)sinθ so ur ~ ln(r/a)cosθ and u ~ ln(r/a)sinθ. The advective acceleration u ⋅ ∇u ~ (a /r)ln(r /a) and the viscous stresses ~ ∇ 2 u ~ (a /r) 2 . The terms € neglected ~ terms retained when Re(a/r)ln(r/a) ~ (a/r)2 or (r/a)ln(r/a) ~ 1/Re. So, this solution is valid for (r/a)ln(r/a) U2 πνt ≈ 0.664 2 π (0.664) U U ReL Interestingly, this result is independent of the fluid viscosity!

€

€

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 10.2. Solve the Blasius equations (10.27) through (10.29) with a computer, using the Runge–Kutta scheme of numerical integration, and plot the results. What value of f !! at η = 0 leads to a successful profile? Solution 10.2. The Blasius equation is a non-linear third-order differential equation for f(η): d3 f 1 d2 f + f = 0. dη 3 2 dη 2 The boundary conditions are: df dη →1 as η → ∞, df/dη = f = 0 at η = 0. For a computer solution using a Runge-Kutta integration scheme, this equation must be reduced to a set of three first-order differential equations by defining: € g(η) = df dη , and h(η) = d 2 f dη 2 . € € The above equation can then be written as three equations: (A) df dη = g(η) (B) dg dη = h(η) € €dh dη = − fh 2 (C) subject to: € f(0) = g(0) = 0 and g(∞) = 1. € are readily integrated via the Runge-Kutta method starting The set of three equations € from η = 0 where the initial values of f and g are known. The initial value of h (= f !! at η = 0) is not known but it can be found by trial and error (0.33205) by looking for the value of h(0) that produces g = 1 at some suitably large value of η, perhaps 10 or 20. A simple MatlabTM code that does this based on a trial & error value of h(0) is: %Compute the Blasius Boundary Layer Profile clear; clc; f0 = 0; %The first known boundary condition g0 = 0; %The second known boundary condition h0 = 0.33205; %The value adjusted by trial & error eta_start = 0; %Starting point for eta eta_end = 10; %End point for eta %The next command invokes a Runga-Kutta integration scheme [eta,f] = ode45(@Blasius,[eta_start eta_end],[f0 g0 h0]);

with the function function df = blasius(eta,f) % Solve the Blasius LBL profile equation. % f(1)=stream function, f(2)=velocity, f(3)=velocity gradient df = zeros(3,1); df(1) = f(2); %The equivalent of (A) df(2) = f(3); %The equivalent of (B) df(3) = -0.500*f(1)*f(3); %The equivalent of (C) end

The resulting plot of df/dη vs. η is:

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

$" !#," !#+" !#*" !#)"

df u = dη U

!#(" !#'" !#&"

€

!#%" !#$" !" !"

%"

'"

)"

η = y U νx

€

+"

$!"

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 10.3. A flat plate 4 m wide and 1 m long (in the direction of flow) is immersed in kerosene at 20°C, (v = 2.29 × 10 6 m2/s, ρ = 800 kg/m3) flowing with an undisturbed velocity of 0.5 m/s. Verify that the Reynolds number is less than critical everywhere, so that the flow is laminar. Show that the thickness of the boundary layer and the shear stress at the center of the plate are δ = 0.74 cm and τ0 = 0.2 N/m2, and those at the trailing edge are δ = 1.05 cm and τ0 = 0.14 N/m2. Show also that the total frictional drag on one side of the plate is 1.14 N. Assume that the similarity solution holds for the entire plate. −

Solution 10.3. ReL = UL/ν = (0.5m/s)(1m)/(2.29x106 m2/s) = 2.18x105 < Recr ~ 106. Thus, the flow is expected to be laminar everywhere. At x = 0.5 m, Rex = 1.09x105 so the 99% thickness from (10.30) and the shear stress from (10.31) are:

δ99 = 4.9x Re1x 2 = 4.9(0.5)

1.09 ×10 5 = 0.742cm , and

τ 0 = 0.332 ρU 2 Re1x 2 = 0.332(800)(0.5) 2 1.09 ×10 5 = 0.201N /m 2 . At x = 1.0 m, Rex = 2.18x105 so δ99 = 4.9x Re1x 2 = 4.9(1.0) 2.18 ×10 5 = 1.05cm , and € τ 0 = 0.332 ρU 2 Re1x 2 = 0.332(800)(0.5) 2 2.18 ×10 5 = 0.142N /m 2 . € The total drag can be obtained from (10.33): 12 −3 CD = 1.33 Re € x = 2.85 ×10 , so 1 D = ρU 2 (Area)CD = 0.5(800)(0.5) 2 (4)(1)(0.00285) = 1.14N . € 2 € €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 10.4. A fluid with constant density and viscosity flows with a constant horizontal speed U∞ over an infinite flat porous plate placed at y = 0 through which fluid is drawn with a constant velocity Vs. For this flow the steady two-dimensional zero-pressure-gradient boundary layer equations are (7.2) and (10.18) and the boundary conditions are u(y = 0) = 0, v(y = 0) = –Vs, and u = U∞ for y → ∞ . a) Assuming u depends only on y, determine u(y) in terms of ν, Vs, U∞, and y. b) What is the wall shear stress τw? How does it depend on µ? c)€What parametric change(s) decrease the boundary layer thickness? Solution 10.4. a) If u = u(y), then ∂u/∂x = 0 = ∂v/∂y and this means that v is a most a function of x. However, the boundary condition on v at y = 0 does not depend on x, thus v = const. = –Vs. ∂u ∂ 2u Therefore, the horizontal momentum equation simplifies to: −Vs = ν 2 , which can be ∂y ∂y integrated to find: −Vs y ν = ln(∂u ∂y ) +C , where C is a constant. This can be rearranged and integrated again to find: u(y) = D + E exp(−Vs y ν ) . The boundary condition, u(0) = 0 requires D + E = 0, while u(∞) = U∞ sets D = U∞. Thus, the velocity€profile is: u(y) = U∞ [1− exp(−Vs y ν )] b) The wall € shear stress is τ w = µ(∂u ∂y ) y= 0 = −µU∞ (−Vs ν ) = ρU∞Vs . It does not depend on µ!

€ layer thickness decreases when Vs increases, and ν decreases. The horizontal c) The boundary flow speed does not influence the thickness of this boundary € layer. €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 10.5. A square-duct wind tunnel test section of length L = 1 m is being designed to operate at room temperature and atmospheric conditions. A uniform air flow at U = 1 m/s enters through an opening of D = 20 cm. Due to the viscosity of air, it is necessary to design a variable cross-sectional area if a constant velocity is to be maintained in the middle part of the crosssection throughout the wind tunnel. a) Determine the duct size, D(x), as a function of x. b) How will the result be affected if U = 20 m/s? At a given value of x, will D(x) be larger or smaller than (or the same as) the value obtained in a)? Explain. c) How will the result be affected if the wind tunnel is to be operated at 10 atm (and U = 1 m/s)? At a given value of x, will D(x) be larger or smaller than (or the same as) the value obtained in a)? Explain. [Hint: the dynamic viscosity of air (µ [N·s/m2]) is largely unaffected by pressure.] d) Does the airflow apply a net force to the wind tunnel test section? If so, indicate the direction of the force.

Solution 10.5. a) The duct length (1 m) and the flow speed (1 m/s) imply a Reynolds number of: ReL = UL/ν = (1 m/s)(1 m)/(1.5x10–5m2/s) = 67,000, which much larger than unity but still in the laminar boundary layer range. If δ* is the displacement thickness of the boundary layer at position x, then conservation of mass between the inlet of the duct and location x implies: 2

ρUD2 (0) = ρU ( D(x) − δ * ) . For constant density and velocity in the duct, the pressure gradient must be zero. Therefore, the Blasius solution for a flat plate laminar boundary layer can be used, and the above equation reduces to: € D(x) = D(0) + δ * = D(0) + 1.72 νx U = 0.20m + (1.5 ×10−5 m 2 /s)x (1m /s)

€

€

= 0.20 + 0.00387 x, where the final numbers provide D in meters when x is in meters. b) If U is 20 m/s instead of 1 m/s and the flow remains laminar, the displacement thickness of the boundary layer will be reduced by a factor of 20 ≈ 4.47. Thus, the required D(x) will be € D(x) = D(0) + 1.72 νx U = 0.20m + (1.5 ×10−5 m 2 /s)x (20m /s) = 0.20 + 0.000866 x . Here, the requisite D(x) is smaller because the boundary layer will be thinner at the higher speed. c) The factor of ten increase in pressure € causes a factor of 10 increase in density. Thus for constant µ, ν = µ/ρ will decrease by a factor of 10. This implies: D(x) = D(0) + 1.72 νx U = 0.20m + (1.5 ×10−6 m 2 /s)x (1m /s) = 0.20 + 0.00122 x .

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

So, the requisite D(x) will be smaller than the part a) result because the boundary layer will be thinner at the higher pressure because of the drop in kinematic viscosity (or equivalently, the increase in Reynolds number). d) Yes, the airflow applies a force to the wind tunnel via wall shear stress. This force points to the right.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 10.6. Use the control volume shown to derive the definition of the momentum thickness, θ, for flow over a flat plate: h u" u% Drag force on the plate from zero to x x ρU 2θ = ρU 2 ∫ $1− 'dy = = ∫ τ w dx # U& unit depth into the page 0 U 0 The words in the figure describe the upper and lower control volume boundaries. U!

y! U!

zero shear stress, ! constant pressure, no through flow! h > δ99!

ho! no slip, τw ≠ 0! x!

Solution 10.6. The CV is stationary, so the integral form of the continuity equation is: h

ρUho = ρ ∫ u(y)dy , 0

where h is defined in the figure to be greater than the boundary layer thickness. The pressure is the same everywhere so the integral form of the x-momentum equation is: h

x

€ −ρU 2 ho + ρ ∫ u 2 (y)dy = − ∫ τ w dx . 0

0

Here it is assumed that τw is positive (since µ and ∂u/∂y will be positive for the boundary layer flow inside the CV), so the right-side integral represents the force the plate applies to the fluid in the CV. Using the continuity equation, eliminate ho from the x-momentum equation: h

h

x

−ρU ∫ u(y)dy + ρ ∫ u 2 (y)dy = − ∫ τ w dx . 0

0

0

Combine the two integrals, multiply by minus one, factor out ρU2, and recognize the definition of the momentum thickness θ: h ∞ x u(y) # u(y) & 2 2 ρ ∫ u(y) (U − u(y)) dy = ρU ∫ %1− ( dy = ρU θ = ∫ τ w dx . $ U ' 0 0 U 0 Since u(y) = U for y > h, it is OK to extend the upper limit of the y-integration to +∞. The integral of the shear stress on the right side of the last equation corresponds to the force the flow exerts on the plate (per unit depth into the page). This force is positive and points in the downstream direction (increasing x) and is therefore a drag force.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 10.7. Estimate the 99% boundary layer thickness on: a) a paper airplane wing (length = 0.25 m, U = 1 m/sec), b) the underside of a super tanker (length = 300 m, U = 5 m/sec), and c) an airport run way on a blustery day (length = 5 km, U = 10 m/sec). d) Will these estimates be accurate in each case? Explain. Solution 10.7. Estimate the various thicknesses from the laminar zero-pressure-gradient (Blasius) boundary layer solution: δ99 = 4.9 νx U . a) δ99 = 4.9 (1.5 ×10−5 m 2 s–1 )(0.25m) (1ms−1 ) = 9.5mm b) δ99 = 4.9 (1.0 ×10−6 m 2 s–1 )(300m) (5ms−1 ) = 3.8cm € € €

€

2 –1 c) δ99 = 4.9 (1.5 ×10−5 m€ s )(5000m) (10ms−1 ) = 0.42m d) The estimate for part a) may be fairly accurate, but the estimates for b) and c) will not be accurate. The downstream-distance-based Reynolds numbers in a), b), and c) are 1.7 ×10 4 , 1.5 ×10 9 , and 3.3 ×10 9 respectively. At the larger two Reynolds numbers, the boundary layer will be turbulent and its 99% thickness will be much greater than the Blasius boundary layer estimate given above. €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 10.8. Air at 20°C and 100 kPa (ρ = 1.167 kg/m3, ν = 1.5 × 10 5 m2/s) flows over a thin plate with a free-stream velocity of 6 m/s. At a point 15 cm from the leading edge, determine the value of y at which u/U = 0.456. Also calculate v and ∂u/∂y at this point. [Answer: y = 0.857 mm, v = 0.384 cm/s, ∂u/∂y = 3012 s–1.] −

Solution 10.8. From Table 10.1, η = y U νx = 1.4 when u/U = 0.456. Therefore,

y = 1.4 ν x U = 1.4 (1.5 ×10 −5 )(0.15) 6 = 0.857 mm. At this wall normal distance, the Blasuis BL formula for v and Table 10.1 produce: 1 νU " € df % 1 1.5 ×10 −5 (6) v= (−0.325 +1.4 ⋅ 0.456) = 0.00384 m/s. $− f + η ' = 2 x # dη & 2 0.15 Again using Table 10.1, the slope of the velocity profile is: ∂u ∂ (u /U) ∂η =U = Uf !! U ν x = (6)(0.3074) 6 (1.5×10 −5 )(0.15) = 3012 s–1. ∂y ∂η ∂ y

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 10.9. An incompressible fluid (density ρ, viscosity µ) flows steadily from a large reservoir into a long pipe with diameter D. Assume the pipe-wall boundary layer thickness is zero at x = 0. The Reynolds number based on D, ReD, is greater than 104. a) Estimate the necessary pipe length for establishing a parabolic velocity profile in the pipe. b) Will the pressure drop in this entry length be larger or smaller than an equivalent pipe length in which the flow has a parabolic profile? Why?

Solution 10.9. a) Fortunately, the exercise asks for an estimate. The flow in the entrance length of a round pipe will accelerate on the pipe's centerline, and the inner wall of the pipe is curved. Both of these features will cause the boundary layer growth inside the pipe to differ from that of a Blasius boundary. However, the Blasius solution does account for diffusive boundary layer growth, so it should be fine for producing an estimate of the entrance length L. The presence of the pipe walls is felt on the pipe centerline when the wall boundary layer has attained a thickness of D/2. Therefore, set δ99 ≈ D/2 and estimate δ99 from the Blasius solution: D/2 ≈ 5(νL/U)1/2 = 5(LD/ReD)1/2 , and solve for L to find: L ≈ 10–2D(ReD). b) The pressure drop in the entry section will be larger because: 1) wall shear stresses are larger when BL's are thin, and 2) the central portion of the flow must accelerate to accommodate the displacement effect of the BL so the velocity at the edge of the boundary layer increases from the x = 0 to x = L.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 10.10. A variety of different dimensionless groups have been used to characterize the importance of a pressure gradient in boundary layer flows. Develop an expression for each of the following parameters for the Falkner-Skan boundary layer solutions in terms of the exponent n in Ue(x) = axn, Rex = Uex/ν, integrals involving the profile function f ! , and f !!(0) , the profile slope at y = 0. Here u(x, y) = Ue (x) f ! ( y δ (x)) = Ue f !(η ) and the wall shear stress τ w = µ (∂ u ∂ y ) y=0 = (µUe δ (x)) f !!(0) . What value does each parameter take in a Blasius boundary layer. What value does each parameter achieve at the separation condition? a) (ν Ue2 ) ( dUe dx ) , an inverse Reynolds number b) (θ2/ν)(dUe/dx), the Holstein and Bohlen correlation parameter

(

c) µ

)

ρτ w3 ( dp dx ) , Patel's parameter

d) (δ * τ w ) ( dp dx ) , Clauser’s parameter Solution 10.10. a) Here Ue(x) = axn, so dU e dx = nax n−1; thus ν dU e ν nax n−1 ν n = =n = . 2 n U e dx U e ax U e x Re x The profile function does not€enter here. This parameter is zero for the Blasius boundary layer, and is –0.0904/Rex at separation. b) This time the profile function enters through the definition of the momentum thickness. 2 € ( 2 dU e θ 2 dU e 1 % ∞ u % u ( ( dU e δ 2 % ∞ = '∫ = ' ∫ f .(η)(1− f .(η))dη* '1− *dy * ν dx ν & 0 U e & U e ) ) dx ν &0 ) dx

( 2 dU e 1 % νx (% ∞ ( 2 nU e δ2 % ∞ = ' ∫ f .(η)(1− f .(η))dη* = ' *' ∫ f .(1− f .)dη* ν &0 ) dx ν & U e )& 0 ) x %∞ (2 = n' ∫ f .(1− f .)dη* . &0 )

'∞ *2 This parameter is zero for the Blasius boundary layer, and is –0.0904 ) ∫ f "(1− f ")dη, at (0 + separation. € c) This parameter involves the shear stress and pressure gradient directly. Differentiating the steady Bernoulli equation without a body force produces dp dx + ρU e dU e dx = 0 , so € 32 # µ dp µ 1 dUe & µ # ν x Ue & nUe2 % ( = − ρ U = − ρ % ( e 32 dx ' x ρ (µUe f !!(0) δ (x)) $ ρ %$ µUe f !!(0) (' ρτ w3 dx € µρν 3 4 1 x3 4 # 1 & 2 µ1 4 1 1 1 = −n U = −n % ( e 32 32 14 32 34 14 32 x $ Ue Ue ' ρ [ f !!(0)] x Ue1 4 ρµ [ f !!(0)] = −n

1

[ f !!(0)]

32

1 Re1x 4

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

This parameter is zero for the Blasius boundary layer, and it goes to +∞ at separation since f "" = 0 at separation and for the Falkner-Skan laminar boundary layer solutions –n = +0.0904 > 0. d) This parameter involves the profile function through the displacement thickness and through the shear stress. Using the differentiated Bernoulli relation specified in part c) leads to: ' δ (x) $ $∞ ' δ 2 (x) € nUe2 δ * dp $ ∞ dUe ' = & ∫ (1− u U ) dy ) ρ & −ρUe ) = − & ∫ (1− f *) dη ) τ w dx % 0 dx ( x ( µUe f **(0) % %0 ( µUe f **(0) $∞ ' ν x Ue ' Ue2 n $∞ * = −n & ∫ (1− f ) dη ) ρ =− & ∫ (1− f *) dη ). f **(0) % 0 %0 ( µUe f **(0) x ( This parameter is zero for the Blasius boundary layer, and it goes to +∞ at separation since f "" = 0 at separation and for the Falkner-Skan laminar boundary layer solutions –n = +0.0904 > 0.

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Solution 10.11. Consider the boundary layer that develops as a constant density viscous fluid is drawn to a point sink at x = 0 on and infinite flat plate in two dimensions (x, y). Here Ue(x) = – UoLo/x, so set η = y νx |U e | and ψ = − νx |U e | f (η) and redo the steps leading to (10.36) to find f """ − f "2 + 1 = 0 . Solve this equation and utilize appropriate boundary conditions to find 2 &1− αe− 2η ) 3− 2 . α= f " =€ 3( + − 2 where − 2η € 3+ 2 '1+ αe *

€ €

Exercise 10.11. Use the specified Ue to develop the appropriate expressions for ψ and h. €ψ = νx |U | f (η) = νx U o Lo f (η) = νU L f (η) , and e o o x y y U o Lo y . η= = = 2 ν x νx |U e | νx U o Lo

The Cartesian € velocity components are: ∂ψ df (η) ∂η UL 1 UL u= = − νU o Lo = − νU o Lo f ' o o = − o o f ' , and ∂y dη ∂ y ν x x € νU o L o ∂ψ df (η) ∂η UL ( y+ v =− = + νU o Lo = νU o L o f ' o o * − 2 - = − ηf ' . ∂x dη ∂x ν ) x , x The€boundary conditions are: f " →1 as η → ∞ (far from the wall the boundary layer velocity matches the free-stream), f " → 0 as η → 0 (no slip at the wall), and € ηf # → 0 as η → 0 (no flow through the wall). ∂u ∂u 1 ∂p ∂ 2u € € + ν 2 . The The horizontal boundary layer momentum equation is (10.9), u + v = − ∂x ∂y ρ ∂x ∂y € € differentiated Bernoulli equation, , allows this to be simplified: dp dx + ρ U dU dx = 0 € € e e ∂u ∂u dU e ∂ 2u ∂u ∂u U 2 L2 ∂ 2u u + v = Ue + ν 2 or u + v = − o 3 o + ν 2 . ∂x ∂y dx ∂y ∂€ x ∂y x ∂y where the second version of the equation follows when U (x) = –U L e o o/x. Substitute in the € velocity components in the final equation to find: UL ∂ % U L ( νU o L o ∂% UL ( U 2 L2 ∂2 % U L ( €− o o f # '− o o f #* − €ηf # '− o o f #* = − o 3 o + ν 2 ' − o o f #* . ) ) ) x ∂x & x x ∂y & x x ∂y & x Perform the differentiations and simplify: νU o L o UL % UL UL η( UL UL 1 U 2 L2 UL UL 1 − o o f #' + o 2 o f # + o o f ## * + ηf # o o f ## o o = − o 3 o − ν o o o o 2 f ### , & x x x x) x x ν x x x ν x € − f #(+ f # + ηf ##) + ηf f# ## = −1− f ### or − f #(+ f # + ηf ##) + ηf f# ## = − f #2 = −1− f ###. With a little rearrangement, the final equality then leads to the desired equation: f """ − f "2 + 1 = 0 . This equation can be solved for f´ as follows. Multiply by f "" and integrate to find: € 1€ 2 1 3 € ( f "") − f " + f " + A = 0 . 2 3 € boundary condition The constant A can be evaluated by considering the limit η → ∞ . The first € provides f "(∞) = 1, and if f´ is constant then f ""(∞) = 0 . Thus, the last equation becomes €

€ €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

0 – 1/3 + 1 + A = 0 as η → ∞ , which requires A = –2/3. Therefore, 12 2 1 2 1 2 ( f "") = − f " + f "3 , or f "" = ± [2 − 3 f " + f "3 ] , 3 2 3 3 € which can be separated: df " df " df " 2 = = = dη . 1 2 1 2 1 2 3 (1− f ")[2 + f "] [(2 + f "€)(1− f ")(1− f ")] €[2 − 3 f " + f "3 ] The possibility of the minus sign in front of the square root is dropped because η is always positive along with 1− f # and 2 + f ". The final equality can be integrated with the substitution 2 F€ = 2 + f ": 2FdF 2dF 2 = = dη 2 2 (3 − F )F 3 − F 3 € € Now set F = 3 tanh(γ ) , so that dF = 3 (1− tanh 2 (γ )) dγ , thus

€

2 3 (1− tanh 2 (γ )) dγ % F ( 2dF 2 2 2 = ∫ 3 − F 2 €∫ 3 1− tanh2 (γ ) = 3 γ = 3 tanh−1'& 3 *) = 3η + C ( ) € € % 2 + f $( 2 2 Backtracking all the way to f " produces: η+C = tanh−1' *. 3 3 3 ) & # 2& € 2 tanh−1% ( The constant C can be evaluated from f "(0) = 0 : C = 3 $ 3' € $ € $ η 3C ' 2' 2 η + + tanh−1 Inverting for f " yields: f " = 3tanh 2 & ) − 2 = 3tanh & ) − 2. 2 3 2 2 % ( % ( € To reach the desired form, use the sum€formula for the hyperbolic tangent, and the definition of the hyperbolic tangent. Then manipulate the exponentials and the square roots. 2 2 € $ η 2 −η 2 ' $ ' − 2η € −e 2 ) 2 ) 2 & e & 1− e $ tanh η 2 + 2 3 ' + + η 2 − η 2 − 2 η 3 ) − 2 = 3& 1+ e 3 ) −2 +e ) − 2 = 3& e f " = 3& & ) & ) η 2 − η 2 − 2 η + . + . &1+ tanh η 2 2 3 ) −e 2) 2) % ( &1+ - e &1+ - 1− e 0 0 &% , eη 2 + e−η 2 / 3 )( &% ,1+ e− 2η / 3 )(

(

)

(

$ − &1− e = 3& & − &%1+ e

2η

2η

)

2 + 1+ e− 3

(

(

+ 1− e−

$ 3+ 2− = 3& & 3+ 2+ %

( (

2η

)

'2 $ 3 − 3e− ) ) − 2 = 3& & 3 + 3e− 2) % 3 )(

2η

) 2 )e

)

3 − 2 e− 3−

−

2

' ) 2η ) (

2η

The final equality is the desired form.

€

$ + &1− , − 2 = 3& & + &1+ &% ,

( ) 2 (1− e )

2η

+ 2 1+ e−

2η

+

3− 2. − 0e 3+ 2/ 3− 2. − 0e 3+ 2/

2η

− 2η

2

' ) ) − 2. ) 2η ) )(

2η

2

' ) −2 ) (

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 10.12. Start from the boundary layer equations, (7.2), (10.9), and (10.10), and ψ = Ue (x)δ (x) f (η ) , where η = y δ (x) , with δ(x) unspecified, to complete the following items. a) Show that the boundary-layer profile equation can be written: f !!! + α ff !! + β (1− f !2 ) = 0 , where α = (δ ν ) d(Ueδ ) dx , and β = (δ 2 ν ) dUe dx . b) The part a) equation will yield similarity solutions when α and β do not depend on x. Therefore, assume α and β are constants, set Ue = axn, and show that n = β/(2α – β). c) Deduce the values of α and β that allow the profile equation to simplify to the Falkner-Skan profile equation (10.36). Solution 10.12. a) The given stream function automatically satisfies the continuity equation, so the next step is to substitute it into (10.9): ∂u ∂u dUe ∂2 u u + v = Ue +ν 2 . ∂x ∂y dx ∂y Here, the surface-normal boundary-layer momentum equation (10.10), –∂p/∂y = 0, and the Bernoulli equation in the free stream have been used to replace the pressure gradient; − (1 ρ ) (∂p ∂x ) = Ue ( dUe dx ) . For the given form of the stream function the various velocity components and derivatives are: ∂ψ y y ∂ψ 1 = −Ue"δ f −Ueδ ! f +Ueδ f ! 2 = −Ue"δ f −Ueδ ! f +Ue f ! , u= = Ueδ f " = Ue f " , v = − ∂x δ δ ∂y δ ∂u y ∂u 1 ∂2 u 1 = Ue! f ! −Ue f !! 2 δ ! , = Uee f !! , and = Ue f """ 2 . 2 ∂x δ ∂y δ ∂y δ where a prime denotes differentiation of a function with respect to its argument. Equation (10.10) is then reconstructed: # y & # y &# 1& 1 Ue f !%Ue! f ! −Ue f !! 2 δ !( + % −Ue!δ f −Ueδ ! f +Ue f ! (%Uee f !! ( = UeUe! + νUe f !!! 2 , $ δ ' $ δ '$ δ' δ and this can be simplified by performing the indicated multiplications on the left side, and canceling equal and opposite terms: δ! 1 UeUe! f !2 −UeUe! ff !! −Ue2 ff !! = UeUe! + νUe f !!! 2 . δ δ Multiply through by δ2/νUe to reach: δ2 δ2 δδ ! δ2 Ue! f !2 − Ue! ff !! −Ue ff !! = Ue! + f !!! , ν ν ν ν collect all the terms to one side of the equation, δ2 δ !!! f + Ue! (1− f !2 ) + (δUe! +Ueδ !) ff !! = 0 , ν ν and combine the terms in the last set of parentheses: δ 2 dUe δ d f !!! + 1− f !2 ) + (δUe ) ff !! = 0 , or f !!! + β (1− f !2 ) + α ff !! = 0 , ( ν dx ν dx where the given definitions, α = (δ ν ) d(Ueδ ) dx & β = (δ 2 ν ) dUe dx , have been used to reach

the final equation.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

b) When α and β are constants, their defining equations may be used to eliminate δ. Start with the α-equation: δ d(Ueδ ) δUe dδ δ 2 dUe Ue dδ 2 α= = + = +β . ν dx ν dx ν dx 2ν dx Using the two ends of this extended equality, substitute for δ2 from the definition of β to find a non-linear equation for Ue: U dδ 2 U d ! νβ $ U β d 2Ue α= e +β = e +β . # &+ β = − e 2ν dx 2ν dx " dUe dx % 2 ( dUe dx )2 dx 2 Again using the two ends of this extended equality and simplifying leads to: 2 α − β " dUe % Ue d 2Ue , = − $ ' β # dx & 2 dx 2 and this equation has power-law solutions: Ue = axn. Substituting this trial solution into this equation produces: α −β 1 α − β 2 2 2n−2 ax n n = − (n −1) . an x =− an(n −1)x n−2 which simplifies to: β 2 β 2 Solving this final algebraic equation for n produces the desired relationship: n = β/(2α – β). β ! n +1 $ c) The part b) result can be rearranged to find: α = # & . Substitute this into the profile 2" n % equation from part a) to reach: β # n +1 & f !!! + β (1− f !2 ) + % ( ff !! = 0 , 2$ n ' and this equation will match (10.36) when β = n. Thus, the values of α and β that produce the Falkner-Skan profile equation are: α = (n + 1)/2 and β = n.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 10.13. Solve the Falkner-Skan profile equation (10.36) numerically for n = –0.0904, – 0.654, 0, 1/9, 1/3, and 1 using boundary conditions (10.28) and (10.29) and the Runge–Kutta scheme of numerical integration. Plot the results and compare to Figure 10.8. What values of f !! at η = 0 lead to successful profiles at these six values of n? Solution 10.13. The Falkner-Skan profile equation is a non-linear third-order differential equation for f(η): 2 " df % d 3 f n +1 d 2 f + f − n$ ' + n = 0 . dη 3 2 dη 2 # dη & The boundary conditions are: df dη →1 as η → ∞, df/dη = f = 0 at η = 0. For a computer solution using a Runge-Kutta integration scheme, this equation must be reduced to a set of three first-order differential equations by defining: f (η ) = f (η ) , f2 (η ) = df dη , and f3 (η ) = d 2 f dη 2 . €1 € The above equation can then be written as three equations: (A) df1 dη = f2 (η ) df2 dη = f3 (η ) (B) df3 dη = −(1 2)(n +1) f1 f3 + nf22 − n

(C)

subject to: f1(0) = f2(0) = 0 and f2(∞) = 1. The set of three equations are readily integrated via the Runge-Kutta method starting from η = 0 where the initial values of f and g are known. The initial value of f3 (= f !! at η = 0) is not known but it can be found by trial and error by looking for the value of f3 that produces f2 = 1 at some suitably large value of η, perhaps 10 or 20. These initial values of f !! are : 0.00, 0.1640, 0.33205, 0.51181, 0.75741, 1.232533. A simple MatlabTM code that does this based on a trial & error value of h(0) is: %Compute the Falkner-Skan Boundary Layer Profile clear; clc; eta_start = 0; eta_end = 10; f0 = 0; %The first known boundary condition g0 = 0; %The second known boundary condition n = [-0.0904, -0.0654, 0, 1/9, 1/3, 1]; %Values of n h0 = [0.00, 0.1640, 0.33205, 0.51181, 0.75741, 1.232533]; %Values adjusted by trial & error figure; hold all; for i = 1:length(n); %The next command invokes a Runga-Kutta integration scheme [eta, f] = ode45(@(eta, f) FS(eta, f, n(i)), [eta_start eta_end],[f0 g0 h0(i)]); xlswrite(['FS_data_n=' num2str(n(i)) '.xlsx'],[eta f])

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

x=(.5*(n(i)+1)).^.5*eta; plot(x,f(:,2)); end

df u = dη U with the function function df = FS( eta, f, n ) %UNTITLED Falkner-Skan LBL Profile %f(1)=stream function, f(2)=velocity, f(3)=velocity gradient %n=1; df = zeros(3,1); df(1) = f(2); %same as (A) df(2) = f(3); %same as (B) df(3) = -0.500*(n+1)*f(1)*f(3) + n*f(2)^2 - n; %same as (C) end

η =ηyis:U νx The resulting plot of df/dη vs. 1"

€

0.9" 0.8" 0.7" 0.6" 0.5" 0.4" 0.3" 0.2" 0.1" 0"

0"

2"

4"

6"

8"

10"

The left most curve corresponds to n = 1, and the right-most curve corresponds to n = –0.0904 with the other monotonically arrayed in between. And – except for the aspect ratio and extent of the horizontal axis – this plot is identical to Figure 10.8.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 10.14. By completing the steps below, show that it is possible to derive von Karman's boundary layer integral equation without integrating to infinity in the surface-normal direction using the three boundary layer thicknesses commonly defined for laminar and turbulent boundary layers: i) δ (or δ99) = the full boundary layer thickness that encompasses all (or 99%) of the region of viscous influence, ii) δ* = the displacement thickness of the boundary layer, and iii) θ = momentum thickness of the boundary layer. Here, the definitions of the later two involve the y=δ $ y=δ u(x, y) ' u(x, y) $ u(x, y) ' * first: δ (x) = ∫ &1− )dy and θ (x) = ∫ &1− )dy , where Ue(x) is the flow U e (x) ( U e (x) ( y= 0 % y= 0 U e (x) % speed parallel to the wall outside the boundary layer, and δ is presumed to depend on x too. a) Integrate the two-dimensional continuity equation from y = 0 to δ to show that the vertical d dU e layer is: v(x, y = δ ) = (U e (x)δ * (x)) − δ . € velocity at the edge of the boundary € dx dx b) Integrate the steady two-dimensional x-direction boundary layer momentum equation from y = τ0 d 2 δ * (x) dU e2 (x) = U e (x)θ (x) + 0 to δ to show that: . ρ dx 2 dx € # d b(x ) db & # da & b(x ) ∂f (x, y) [Hint: Use Leibnitz’s rule f (x, y)dy = % f (x,b) ( − % f (x,a) ( + ∫ dy to handle ∫ $ dx a(x ) dx ' $ dx ' a(x ) ∂x the fact that € δ = δ(x)]

(

)

∂u ∂v + = 0 , and integrate from y = 0 to δ to get: ∂x ∂y δ ∂u ∫ ∂x dy +v(x, y = δ ) = 0 0 where v(x,y=0) =€0. Use Leibnitz’s rule to get the differentiation outside the integral: δ ∂u d δ dδ v(x, y = δ ) = − ∫ dy = − udy + U e (x) ∫ dx 0 dx 0 ∂x € Add and subtract Ue(x) within the integral and rearrange the result: d δ dδ v(x, y = δ ) = − (u − U e (x) + U e (x))dy + U e (x) . ∫ dx dx € 0 δ d d dδ =− (u − U e (x)) dy − (U eδ ) + U e (x) ∫ dx 0 dx dx Solution 10.14. Start € with

€

=

δ # d# u & & dU e d dU e = (U eδ * ) − δ %U e (x) ∫ %1− (dy ( − δ dx $ U e (x) ' ' dx dx dx 0 $

€ ∂u ∂u 1 dp ∂ 2u + ν 2 , multiply the continuity equation by u and add it to this b) Start with u + v = − ∂x ∂y ρ dx ∂y 2 ∂u ∂uv 1 dp ∂ 2u € + =− + ν 2 . Use the Bernoulli equation substitution for the equation to get: ∂x ∂y ρ dx ∂y pressure and integrate this equation from y = 0 to δ to get: € δ ∂u 2 dU τ ∫ ∂x dy + U e (x)v(x, y = δ ) = U eδ dxe − ρw 0 €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Use Leibnitz’s rule to get the differentiation outside the integral, substitute from part a) for the vertical velocity at the edge of the boundary layer, and add and subtract uUe within the integral: %d d δ 2 dδ dU e ( dU e τ w u − uU e + uU e ) dy − U e2 + U e (x)' (U eδ * ) − δ − * = U eδ ( ∫ & dx dx 0 dx dx ) dx ρ Invoke the definition of the momentum thickness θ and then the definition of the displacement thickness δ*: & δ ) € − d U 2θ + d (U ∫ ( u − U + U ) dy + − U 2 dδ + U (x)&( d U δ * − δ dU e )+ = U δ dU e − τ w e e e e e e e ' dx e dx dx ' 0 dx dx * dx ρ * %d d d dδ dU e ( dU e τ w − (U e2θ ) + (−U e2δ * + U e2δ ) − U e2 + U e (x)' (U eδ * ) − δ − * = U eδ & dx dx dx dx dx ) dx ρ Now rearrange the result, and group like terms together: € τw d 2 d 2 * d dU e2δ dδ dU dU = Ueθ + Ueδ − Ue U eδ * − + U e2 + U eδ e + U eδ e ρ dx dx dx dx dx dx dx € * The final four terms sum to zero, and the terms containing δ can be expanded to obtain: 2 * * τw d 2 dU e d dU e * dU e 2 dδ 2 dδ = Ueθ + δ + Ue − Ue − δ *U e = U e2θ + U eδ * , ρ dx dx dx dx dx dx dx € and the final equality is identical to (10.43), von Karman's boundary layer integral equation.

(

)

(

( ) (

€

)

(

)

(

)

)

(

)

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 10.15. Derive the von Karman boundary layer integral equation by conserving mass and momentum in a control volume (C.V.) of width dx and height h that moves at the exterior flow speed Ue(x) as shown. Here h is a constant distance that is comfortably greater than the overall boundary layer thickness δ.

y = h!

Ue!

!!

u(x,y)! p(x)!

Solution 10.15. Use the integral laws as applied to a moving control volume. Conservation of mass implies: ' h d$ h dx ρ Bdy & ∫ ) + ∫ ρ ([ u] x−dx / 2 − U e )e x ⋅ (−e x B)dy dt % 0 ( 0

C.V.! x!

Ue!

u(x+dx,y)! p(x+dx)! x+dx!

h

+

∫ ρ([u] x +dx / 2 − U e )e x ⋅ (e x B)dy + ρ[(u − U e )e x + ve y ] y= h ⋅ e y Bdx = 0 0

The four terms above correspond to unsteady mass addition to the CV, mass flux on the left side of the CV, mass flux on the right side of the CV, and mass flux on the top of the CV. The CV is a constant size and the density is presumed to be constant, so the first term is € zero (the time derivative of a constant). Combine terms, perform the dot products, presume dx is small, and divide by ρBdx to reach: h ( ∂u d %h 1 h v = − dy = − u − u dy + v = 0 , or (*) [ ] [ ] [ ] [ ] ∫ ∫ ' ∫ udy * , ( y= h x +dx / 2 x−dx / 2 ) y= h dx & 0 dx 0 ) 0 ∂x where the final equality follows from the fact that h is constant. Now conserve horizontal momentum using the same CV. ' h d$ h € dx ρ uBdy € / 2 − U e )e x ⋅ (−e x B)dy & ∫ ) + ∫ ρ [ u] x−dx / 2 ([ u] x−dx dt % 0 ( 0 h

+

∫ ρ[ u] x +dx / 2 ([ u] x +dx / 2 − U e )e x ⋅ (e x B)dy + ρu[(u − U e )e x + ve y ] y= h ⋅ e y Bdx 0

= −τ 0 Bdx + ([ p] x−dx / 2 − [ p] x +dx / 2 ) hB

where τ0 is the wall shear stress, and p is the pressure. Simplify and combine terms, and divide by ρBdx to reach: & 1 h d #h 1 h 2 2 € [ u] x +dx / 2 − [ u] x−dx / 2 dy − U e ∫ {[u] x +dx / 2 − [u] x−dx / 2}dy + [uv ] y= h ∫ % ∫ udy ( + dt $ 0 dx 0 ' dx 0

{

=−

}

τ0 1 + ([ p]x−dx / 2 − [ p]x +dx / 2 )h ρ ρdx

d dx d d , noting = = Ue dt dt dx dx that horizontal velocity of the CV (= dx/dt) is Ue; and presume dx is small: € & d h 2 d #h d h τ 1 ∂p U e % ∫ udy ( + u dy − U udy + [ uv ] y= h = − 0 − h. ∫ ∫ e dx $ 0 dx 0 ρ ρ ∂ x ' dx 0 €

Use the chain rule on the total time derivative of the unsteady term,

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Two terms on the left cancel, and at y = h, u = Ue and v is given by (*). In addition, the steady Bernoulli equation implies: dp dx = − ρU e dU e dx . So, with these substitutions, the last equation can be rewritten: ' d h 2 d $h τ0 dU e τ0 dU e h u dy − U udy = − + U h = − + U ∫ ∫ dy . &∫ ) e e e dx€0 dx % 0 ρ dx ρ dx 0 ( Rearrange this equation so the skin friction term appears by itself with positive sign. Then manipulate the integrals to form the definitions of the momentum and displacement thicknesses. ) τ0 d h 2 d &h dU e h =− u dy + U e ( ∫ udy + + U e € ∫ ∫ dy ρ dx 0 dx ' 0 dx 0 * d & h u2 ) d & h u ) dU e h u dU e h = − (U e2 ∫ 2 dy + + (U e2 ∫ dy + − U e dy + U e ∫ ∫ dy dx ' 0 U e * dx ' 0 U e * dx 0 U e dx 0 d& h u & u) ) dU e h & u) d 2 * dU e = (U e2 ∫ . ∫ (1− +dy + + U e (1− +dy = (U e θ ) + U eδ dx ' 0 U e ' U e * * dx 0 ' U e * dx dx The final equality is identical to (10.43), von Karman's boundary layer integral equation. €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 10.16. For the following approximate flat-plate boundary layer profile: u %sin(πy 2δ ) for 0 ≤ y ≤ δ ( =& ) , where δ is the generic boundary layer thickness, determine: U '1 for y > δ* a) the displacement thickness δ*, the momentum thicknesses θ, and the shape factor H = δ * θ . b) Use the zero-pressure gradient boundary layer integral equation to find: (δ x ) Re1x 2 , (δ * x )Re1x 2 , (θ x )Re1x 2 , c f Re1x 2 , and CD Re1L 2 for the approximate profile. c) Compare these results to their equivalent Blasius boundary layer values.€

€

€

€

€ boundary layer velocity profile, u = sin$& πy ') = sin$& π ζ ') where € Solution 10.16.€a) The sinusoid

€

y , is a reasonably accurate approximate laminar boundary layer profile. The momentum δ thickness for this profile is: € π 2 δ 1 $ π '$ $ π '' u$ u' 2δ θ = ∫ &1− )dy = δ ∫ sin& ζ )&1− sin& ζ ))dζ = ∫ sin(ψ )(1− sin(ψ ))dζ % U( % 2 (% % 2 (( π 0 0 U 0

U

% 2δ (

%2 (

ζ=

The integration is not too complicated,

∫

π 2 0

sin(ψ )dψ = 1, and

∫

π 2 0

sin 2 (ψ )dψ = π 4 , so

2δ & π ) & 4 − π ) (1− + = δ( + = 0.1366δ € π ' 4 * ' 2π * The displacement thickness for profile € theπ approximate € is: 2 δ $ u' 2δ 2δ $ π ' $ π − 2 ' δ * = ∫ &1− )dy = 1− sin(ψ )) dψ = & −1) = δ& ) = 0.3634δ , ( ∫ % € U( π 0 π %2 ( % π ( 0 so that the shape factor H = δ * θ = 2.66 . Substitution into the zero-pressure gradient (ZPG) Von Karman boundary layer integral relationship produces: 2(πµU 2δ ) πν 2τ dθ ) 4 − π , dδ τw dθ € = =2 =+ = U2 , or c f = w2 = . 2 * - dx ρ U ρ U U δ dx π ρ dx € where τ w = µ(∂u ∂y ) y= 0 = πµU 2δ has been obtained directly from the approximate sinusoidal

θ=

profile. Now use the fourth and sixth terms in the extended equality to form a differential equation € for δ. € % 2π 2 ( ν & ) dδ % π 2 ( ν ν πν 4 − π d δ € =' =( –> δ 2 = ' + , or δ * * x + const = 23.0 x + const . dx & 4 − π ) U U Uδ ' π * dx &4 − π )U Here the implicit assumption is that the boundary layer thickness is zero at x = 0 so the constant ν 2 drops out. Thus: δ = 23 x = 4.80 ⋅ x ⋅ Re−1 . Using the results above for the other length x U € € € scales and parameters produces: θ 12 b) (0.664) Rex = 0.654 x €* δ Re1x 2 = 1.743 (1.721) x δ 12 € (5.0) Rex = 4.80 x

€ €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

πν 1 2 πν Rex = Re1x 2 = 0.654 (0.664) 2 Uδ U ⋅ 4.80xRe−1 x 1 L 0.654 0.654 ν L dx 2 ⋅ 0.654 1 2 12 CD Re1L 2 = ∫ dx ⋅ Re = ⋅ Re1L 2 = ReL = 1.309 (1.328) ∫ L 12 L 0 Rex L U 0 x Re1L 2 € c) The Blasius boundary layer results are listed above at the right in parentheses. The sinusoid velocity profile results are all within 4% of the Blasius values. c f Re1x 2 =

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 10.17. An incompressible viscous fluid with kinematic viscosity ν flows steadily in a long two dimensional horn with cross sectional area A(x) = Aoexp{βx}. At x = 0, the fluid velocity in the horn is uniform and equal to Uo. The boundary layer momentum thickness is zero at x = 0. a) Assuming no separation, determine the boundary layer momentum thickness, θ(x), on the lower horn boundary using Thwaites method. b) Determine the condition on β that makes the no-separation assumption valid for 0 < x < L. c) If θ(x = 0) was nonzero and positive, would the flow in the horn be more or less likely to separate than the θ(x = 0) = 0 case with the same horn geometry?

Solution 10.17. a) As stated above, use Thwaites method to estimate the boundary layer momentum thickness. Start with conservation of mass to determine the U(x): U(x)A(x) = U o Ao , so U(x) = U o Ao A(x) = U o exp{−βx} . Thus, the Thwaites equation becomes: 0.45ν x 5 0.45ν +6 β x x 5 −5 β x! ! ! U ( x )d x = e ∫ ∫ Uo e € dx! U 6 (x) 0 Uo6 0 −5 β x $ ' 0.45ν +6 β x e 1 0.09ν +6 β x = e & + )= e 1− e−5 β x ) ( Uo % −5β 5β ( βUo

θ 2 (x) = €

θ 2 dU 0.09 +6 βx = e (1− e−5 βx )(−βU oe− βx ) = −0.09(e +5 βx −1) . Separation will b) First compute λ = ν dx βU o have occurred if λ < λseparation = –0.090. Therefore, λ > λseparation at x = L is required to avoid separation. This means: −0.09(e +5 βL −1) > λseparation . € Convert this requirement to a condition on β:

λ 1 % λseparation ( 0.139 e +5 βL −1 < − separation , or β < ln&1− )≈ 5L ' 0.09 * L 0.09 € c) If the boundary layer starts with a non-zero momentum thickness it is thicker and therefore more likely to separate in an adverse pressure gradient. €

€

y

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

b Exercise 10.18. The steady two-dimensional velocity potential for a source of strength q located a distance b above a large flat surface located at y = 0 is: x q xsep 2 2 2 2 φ (x, y) = ln x + (y − b) + ln x + (y + b) 2π a) Determine U(x), the horizontal fluid velocity on y = 0. b) Use this U(x) and Thwaites method to estimate the momentum thickness, θ(x), of the laminar boundary layer that develops on the flat surface when the initial momentum thickness θo is zero. x ξ 5 dξ x 6 (x 2 + 4b 2 ) = [Potentially useful information: ∫ 0 2 ] (ξ + b 2 ) 5 24b 4 (x 2 + b 2 ) 4 c) Will boundary layer separation occur in this flow? If so, at what value of x/b does Thwaites method predict zero wall shear stress? d) Using solid lines, sketch the streamlines for the ideal flow specified by the velocity potential € given above. For comparison, on the same sketch, indicate with dashed lines the streamlines you expect for the flow of a real fluid in the same geometry at the same flow rate.

(

)

$ ! ∂φ $ q ! x x q! x $ + 2 = # 2 2& Solution 10.18. a) U(x) = # & = # 2 2 2& " ∂ x %y=0 2π " x + (y − b) x + (y + b) %y=0 π " x + b % b) Use the Thwaites integral without the θo-term. 6

6

0.45νπ 6 ( x 2 + b 2 ) x q 5 " ξ %5 0.45νπ ( x 2 + b 2 ) x ξ 5dξ 0.45ν x 5 2 θ (x) = 6 ∫ U (ξ )dξ = ∫ π 5 $# ξ 2 + b2 '& dξ = ∫ (ξ 2 + b2 )5 U (x) 0 q6 x 6 qx 6 0 0 Evaluate the integral using the given information: 0.45νπ (x 2 + b 2 )6 x 6 (x 2 + 4b 2 ) 0.45νπ (x 2 + b 2 )2 2 θ 2 (x) = = (x + 4b 2 ) 6 4 2 2 4 4 qx 24b (x + b ) 24b q c) Compute the correlation parameter λ from the results of part b): " b 2 − x 2 % 0.45 " θ 2 dU 0.45π (x 2 + b 2 )2 2 x 2 %" x 2 % 2 q λ= = (x + 4b ) = 4 + $ ' $ '$1− ' . ν dx 24b 4 q π # (x 2 + b 2 )2 & 24 # b 2 &# b 2 & Clearly when x > b, λ becomes negative and its magnitude increases with increasing x, therefore at some x > b boundary layer separation will occur. Within Thwaites method, λ = λsep = –0.090 predicts the point, xzs, where the wall shear stress is zero; therefore, use the prior result and the quadratic formula to find x zs2 &# x zs2 & 0.45 # λsep = % 4 + 2 (%1− 2 ( → 24 $ b '$ b ' 12

€

9 + 4(4 − 24 λsep 0.45) ' x zs $ 3 ) = 1.35 . € = &− + b &% 2 2 )( € d) The streamlines for the ideal flow correspond to a point source above a flat surface. The real flow streamlines must include boundary layer separation at a value of x that is of order b. Furthermore, the thickness of the separated flow region will cause upward shifts in the ideal flow streamlines.

Fluid Mechanics, 6th Ed.

€ €

Kundu, Cohen, and Dowling

Exercise 10.19. A fluid-mediated particle-deposition process requires a laminar boundary layer flow with a constant shear stress, τw, on a smooth flat surface. The fluid has viscosity µ and density ρ (both constant). The flow is steady, incompressible, and two-dimensional, and the flat surface extends from 0 < x < L. The flow speed above the boundary layer is U(x). Ignore body forces. a) Assume the boundary layer thickness is zero at x = 0, and use Thwaites’ formulation for the shear stress, τ w = (µU θ ) l( λ) with λ = (θ 2 ν )( dU dx ) , to determine θ(x) and U(x) in terms of λ, ν = µ ρ , x, and τ w µ = constant. [Hint: assume that U θ = A and l( λ) are both constants so that τ w µ = Al( λ) .] b) Using the Thwaites integral (10.50) and the results of part a), determine λ. € € c) Is boundary layer separation a concern in this flow? Explain with words or equations. € € €

Solution 10.19. a) For Thwaites method, the shear stress is provided by the correlation τ w = (µU θ ) l( λ) . Thus, the simplest way to achieve constant τw is for U to be proportional to θ, i.e. U = Aθ where A is a constant, and l(λ) = constant, where λ = (θ 2 ν )( dU dx ) = constant. Eliminating U from this system of equations leaves:

€

€

τw = Al( λ ) , and λ = A(θ 2 ν )( dθ dx ) . (1,2) µ € Integrate (2) to find: λνx = Aθ 3 3 where the constant of integration is zero because θ = 0 at x = 0. Now use equation (1) to eliminate A in the relationship for θ. Thus, the solution falls into an € implicit form involving λ and l(λ). € 13 13 2 2 2 € θ (x) = ( 3λl( λ)νµx τ w ) and U(x) = Aθ (x) = ( 3τ w λνx µ l ( λ)) b) Put the part a) results into the Thwaites integral: x 53 0.45ν 23 3τ w2 λνξ µ 2 l 2 ( λ )) dξ . (3λl(λ)νµx τ w ) = 2 ∫ ( 6 3 € (3τ w λνx µ 2l 2 (λ)) 0 € Perform the integration: 23 (3λl(λ)νµx τ w ) =

0.45ν 63

(3τ

2 w

λν µ 2 l 2 ( λ ))

53

(3τ w2 λνx µ 2l 2 (λ)) € Use the two ends of the equality and simplify to find:

0.45 0.45 3νx , or λ = = 0.05625 8 8 c) Boundary layer separation is not a concern here because λ and τw are both positive and constant, so the separation condition λ = –0.090 is not approached. 13

€

x 8 3 0.45 3νx . = 83 8 ( 3τ w2 λνx µ 2 l 2 ( λ))1 3

23 (3λl(λ)νµx τ w ) (3τ w2 λνx µ 2 l 2 (λ))

€

= 3νxλ =

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 10.20. The steady two-dimensional potential for incompressible flow at nominal horizontal speed U over a stationary but mildly wavy wall is: φ (x, y) = Ux −Uε exp (−ky) cos ( kx ) , where kε > d. Write a simple laminar-flow scaling law for τ assuming that the velocity perturbation produced by the no-slip condition on the container’s sidewall must travel inward a distance d/2 via diffusion. b) Case II: h 0 , and unstable flow when (g k )(ρ2 − ρ1 ) + σk < 0 . Thus, even when the fluid densities are not stably arranged, non-zero surface tension prevents high wave number fluctuations from being unstable. In addition, note that when U1 = U2 = ρ1 = 0, we recover the € 12 dispersion relation for water waves on € pool of finite depth: ω = ck = ( gk + σk 3 ρ 2 ) tanh(kh) . f) Given that tanh(kh) →1 for kh >> 1, the thickness h will only matter when kh co. The membrane is less unstable as the fluid density increases when U < co. The influence of the membrane mass is more easily deciphered with an algebraic revision 2Uρ ± Tk 2 ρ m + 2kρ(T − ρ mU 2 ) € of the result of part b): c = . Thus, we see that the membrane is 2 ρ + kρ m

€

Fluid Mechanics, 6th Ed.

more unstable with increasing membrane mass when U >

Kundu, Cohen, and Dowling

[( kρm

2 ρ ) −1](T ρ m ) , but that the

membrane is less unstable with increasing membrane mass when U < [( kρ m 2 ρ ) −1](T ρ m ) . co d) When U = 0: c = ± . The waves move slower with a quiescent fluid present than € 1+ 2 ρ kρ m in a vacuum because of the added mass (inertial loading) € provided by the liquid.

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 11.6. Prove that σr > 0 for the thermal instability discussed in Section 11.4 via the following steps that include integration by parts and use of the boundary conditions (11.38). a) Multiply (11.36) by Tˆ * and integrate the result from z = –1/2 to z = +1/2, where z is the 2 dimensionless vertical coordinate) to find: σI + I = ∫ Tˆ *Wdz where I ≡ ∫ Tˆ dz , 1

€

€ €

2

1

2 2& # I2 ≡ ∫ % dTˆ /dz€+ K 2 Tˆ (dz , and the limits of the integrations have been suppressed for clarity. $ ' b) Multiply (11.37) by W* and integrate from z = –1/2 to z = +1/2 to find: € € 2 2 σ J1 + J 2 = RaK 2 ∫ W *Tˆdz where J1 ≡ ∫ dW /dz + K 2 W dz , Pr 2 2 2 J 2 ≡ ∫ d 2W /dz 2 + 2K 2 dW /dz + K 4 W dz , and again the limits of the integrations have been

[

[

]

]

suppressed. c) Combine the results of€ a) and b) to eliminate the mixed integral of W & Tˆ , and use the result of this combination to show that σi = 0 for Ra > 0. [Note: the integrals I1, I2, J1, and J2 are all positive definite].

€ $ d2 ' ˆ 2 Solution 11.6. a) Equation (11.36) is &σ + K − 2 )T = W . As directed, multiply by Tˆ * and dz ( % integrate through the thickness of the layer. The term-by-term results on the left side are: 2 2ˆ $ * dTˆ '1 2 2 dTˆ * d€ Tˆ dTˆ 2 *ˆ 2 * d T ˆ ˆ ˆ ˆ (σ + K ) ∫ T Tdz = (σ + K ) ∫ T dz , and − ∫ T dz 2 dz = −&%T dz )( + ∫ dz dz dz = ∫ dz dz , € −1 2 where the boundary terms in the integration by parts are zero because of the boundary conditions on Tˆ , and all the integrals are from –1/2 to +1/2. Thus, the remnant of (11.36) is: 2 € 2 dTˆ € 2 ˆ (σ + K ) ∫ T dz + ∫ dz dz = ∫ Tˆ *Wdz , € which is readily rearranged to σI1 + I2 = ∫ Tˆ *Wdz , 2 2 2& # where I1 ≡ ∫ Tˆ dz € , and I2 ≡ ∫ % dTˆ /dz + K 2 Tˆ (dz . $ ' 2 2 $σ ' d '$ d b) Equation (11.37) is & + K €2 − 2 )& 2 − K 2 )W = −RaK 2Tˆ , which expands to: dz (% dz % Pr ( 2 € * $σ € $σ d4 2' 2 2' d 2 ,−& + K )K + & + 2K ) 2 − 4 /W = −RaK Tˆ . ( % Pr ( dz dz . + % Pr * As directed, €multiply by W and integrate through the thickness of the layer. The term-by-term results on the left side are: $σ ' $σ ' 2 −& + K 2 )K 2 ∫ W *W dz = −& + K 2 )K 2 ∫ W dz , € % Pr ( % Pr (

€

Fluid Mechanics, 6th Ed.

€

Kundu, Cohen, and Dowling

12 2 2 1 #σ #σ dW * dW 4 # σ dW 2& * d W 2& * * dW 2& dz = % + 2K (2,W −∫ dz5 = −% + 2K ( ∫ dz , % + 2K ( ∫ W $ Pr ' $ Pr '3+ ' dz dz 2 dz /.−1 2 dz dz 6 $ Pr and 4 $ * d 3W '1 2 dW * d 3W * d W −∫ W dz = − W + dz ∫ & ) dz 4 dz 3 (−1 2 dz dz 3 %

2 $ dW * d 2W '1 2 d 2W * d 2W d 2W = +& −∫ dz = − ∫ dz. 2 ) dz 2 dz 2 dz 2 % dz dz (−1 2 where all the boundary terms in the various integrations by parts are zero because of the boundary conditions on W, and all the integrals are from –1/2 to +1/2. Thus, the remnant of (11.37) multiplied by –1 is: € 2 2 #σ #σ dW d 2W 2 2& 2 2& dz + ∫ dz = RaK 2 ∫ W *Tˆdz , % + K (K ∫ W dz + % + 2K ( ∫ $ Pr ' $ Pr ' dz dz 2 which is readily rearranged to σ J1 + J 2 = RaK 2 ∫ W *Tˆdz Pr € 2 2 2 2 2 2 where J1 ≡ ∫ dW /dz + K W dz , and J 2 ≡ ∫ d 2W /dz 2 + 2K 2 dW /dz + K 4 W dz .

[

[

]

]

c) The complex conjugate of the part a) result is: € σ * I1 + I2 = ∫ TˆW * dz , integral in the result of part b) to find: € Use this to eliminate the mixed € σ J1 + J 2 = RaK 2 (σ * I1 + I2 ) . Pr € The imaginary part of this equation is: # J1 & #J & σ i % ( = −RaK 2 I1σ i , or σ i % 1 + RaK 2 I1 ( = 0 . $ Pr ' $ Pr ' € When Ra > 0 (the flow is top heavy), the contents of the final parentheses is positive definite so the final equation requires σi = 0. €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 11.7. Consider the thermal instability of a fluid confined between two rigid plates, as discussed in Section 11.4. It was stated there without proof that the minimum critical Rayleigh number of Racr = 1708 is obtained for the gravest even mode. To verify this, consider the gravest odd mode for which (1) W = Asin(q0 z) + Bsinh(qz) + C sinh(q* z) . (Compare this with the gravest even mode structure: W = Acosq0z + Bcoshqz + Ccoshq*z.) Following Chandrasekhar (1961, p. 39), show that the minimum Rayleigh number is now 17,610, reached at the wave number Kcr = 5.365. € Solution 11.7. From Section 12.4 the six roots for the vertical wave number q are:

[ (

12

) ]

±iqo = ±iK(s −1)1 2 , ±q = ±K 1+ s 1+ i 3 2

[ (

12

) ]

, and ±q* = ±K 1+ s 1− i 3 2

,

(2)

13

with s ≡ ( Ra K 4 ) . The boundary conditions on W are: &2 dW # d 2 2 € (3) W = =% −K ( € W = 0 at z = ± 1/2. € dz $ dz 2 ' € From the mode shape specification, equation (1), the first, second, and fourth derivatives of the mode shape are: dW = Aq0 cos(q0 z) + Bqcosh(qz) + Cq* cosh(q* z) , (4) € dz d 2W (5) = −Aq02 sin(q0 z) + Bq 2 sinh(qz) + Cq*2 sinh(q* z) , and 2 dz d 4W € (6) = Aq04 sin(q0 z) + Bq 4 sinh(qz) + Cq*4 sinh(q* z) . 4 dz From (5) and € (6), 2 2 2 #d & # d4 & 2 2 d 4 + K % 2 − K ( W = % 4 − 2K (W dz 2 $ dz $ dz ' €' 2

2

2

= A(q02 + K 2 ) sin(q0 z) + B(q 2 − K 2 ) sinh(qz) + C (q*2 − K 2 ) sinh(q* z). Application of all three boundary conditions (3) at either z = +1/2 or z = –1/2 leads to: # &#A& sin(q0 /2) sinh(q /z) sinh(q* /2) % (% ( qcosh(q /2) q* cosh(q* /2) € % q0 cos(q0 /2) (%B( = 0 . 2 2 2 % q 2 + K 2 sin(q /2) q 2 − K 2 sinh(q /2) q*2 − K 2 sinh(q* /2)(%$C(' ) ( ) ( ) 0 $( 0 ' For nontrivial solutions, the determinant of the 3x3 matrix has to be zero. Dividing by the first row of the matrix produces: # &#A& 1 1 1 € % (% ( * * (7) %q0 cot(q0 /2) qcoth(q /2) q coth(q /2)(%B( = 0 . 2 2 2 2 2 2 2 *2 2 % q +K ) (q − K ) (q − K ) ('%$C(' $( 0 From (2) and definition of s given above: q02 = K 2 (s −1) , 2

2

(q02 + K 2 ) = (K 2 (s −1) − K 2 ) = K 4 s2 € € €

Fluid Mechanics, 6th Ed.

(q (q

2

*2

Kundu, Cohen, and Dowling

2

( [

]

− K 2 ) = K 2 1+ 12 s(1+ i 3) − K 2 2

[

]

− K 2 ) = 14 K 4 s(1− i 3)

)

2

[

]

2

= 14 K 4 s(1+ i 3) , and

2

So, (6) becomes €

# &# & 1 1 1 % (% A( * * € (8) %q0 cot(q0 /2) qcoth(q /2) q coth(q /2)(%B( = 0 . 2 2 ( % 1 1 1 1+ i 3 1− i 3 ('%$C(' %$ 4 4 after cancelling the common factor of K4s2 from the bottom row. Expanding the determinant, and using (2) and the definition of s given above produces a relationship between Ra and K which is sketched below. € 105

[

]

[

]

Ra

K 4 2 6 This relationship has a minimum of Racr = 17,610 at K = 5.36. The critical Ra for the gravest even mode is 1708, signifying that it is more unstable.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 11.8. Consider the centrifugal instability problem of Section 11.6. Making the narrowgap approximation, work out the algebra of going from (11.50) to (11.51). Solution 11.8. The perturbation equations (11.50) are: $ U ' u ' ( ∂uR 2Uϕ uϕ 1 ∂p u + ∂u $ dU − =− + ν * ∇ 2 uR − R2 - , ϕ + & ϕ + ϕ )uR = ν &∇ 2 uϕ − ϕ2 ) , ) R( R ( ∂t R ρ ∂R R , ∂t % dR % ∂uz ∂ u 1 ∂p ∂ =− + ν∇ 2 uz , and (1) (RuR ) + z = 0 . ∂t ρ ∂z ∂R ∂z Substituting in € € uR = u(R)eσt cos kz , uϕ = v(R)eσt cos kz , uz = w(R)eσt sin kz , and p ρ = pˆ (R)eσt cos kz , the equation set€(1) becomes € # d # d 1& 2 σ& U dpˆ ν % % + ( − k − (u + 2 ϕ v = , $ ' dR dR R ν R dR $ ' € € € € # d # d 1& σ & # dU U & ν % % + ( − k 2 − (v − % ϕ + ϕ (u = 0 , ν ' $ dR R' $ dR $ dR R ' ## d 1 & d σ& € ν %% + ( − k 2 − ( w = −kpˆ , and ν' $$ dR R ' dR " d 1% € + 'u = −kw . (2) $ # dR R & Eliminating w between € the third and fourth equations of set (2) produces: ν ## d 1 & d σ &# d 1 & + ( − k 2 − (% + ( u = pˆ . 2 %% $ dR R ' dR k $€ ν '$ dR R ' Inserting this equation for pˆ into the first equation of set (2), and working on the algebra eventually leads to: U ν # d # d 1 & 2 σ &# d # d 1 & 2 & € 2% % + ( − k − (% % + ( − k (u = 2 ϕ v . (3) ν '$ dR $ dR R ' R ' € k $ dR $ dR R ' The second equation of set (2) and equation (3) are a pair of equations relating u and v. Using the radius of the outer cylinder R2, define new dimensionless variables and parameters: € r = R/R2, k 2 = K 2 R22 , and ω = σR22 ν , so that the relevant equation pair becomes: " d " d 1% %" d " d 1 % % BK 2 " 1 AR22 % 2 2 + − K − ω + − K u = 2 ' ' $ + 'v , and $ $ '$ $ ' ν # r2 B & # dr # dr r & € &# dr # dr r€& & " d " d 1% % A 2 2 $ $ + ' − K − ω 'v = 2 R2 u , ν # dr # dr r & & where € U = Ar + B/r. It is convenient to make the transformation AR22 2 u → u, ν € so that the equations take the more convenient forms: " d " d 1% %" d " d 1 % % % 2 2 2" 1 $ $ + ' − K − ω '$ $ + ' − K ' u = −TaK $ 2 − κ 'v , and #r & # dr # dr r & € &# dr # dr r & & ϕ

€

Fluid Mechanics, 6th Ed.

€

Kundu, Cohen, and Dowling

" d " d 1% % 2 $ $ + ' − K − ω 'v = u , # dr # dr r & & 2 A 2 1− µ η 2 AB Ω2 R 4 (1− µ)(1− µ η ) Ω R κ = − R2 = where Ta = −4 2 R22 = 4 1 2 1 , , µ= 2, η= 1, 2 2 Ω1 R2 B 1− µ ν ν (1− η ) € 1− µ η 2 1− µ A = −Ω1η 2 , and B = Ω1R12 . 2 1− η 1− η 2 € € The no-slip and boundary conditions€at the walls require: € u = v = 0 and (d/dr)u = 0 at r = η and 1, where the last of the three conditions is equivalent to w = 0 (see the final equation of set (1)). € Now consider the narrow gap approximation that is valid when R2 − R1 > 1/r so % B r − R1 ( d 1 d + ≅ , and A + 2 ≅ Ω1'1− (1− µ) *. r R2 − R1 ) dr €r dr & Now convert the independent radial coordinate (R) to one (x) that starts on the inner cylinder using the gap dimension, d = R2 − R1, as the length scale, and let k = K/d, and w = σd2/ν to find: $ d2 '$ d 2 ' 2Ω1d 4 2 € 2 2 K (1− (1− µ)x )v , and & 2 − K − ω€)& 2 − K ) u = ν % dx (% dx ( € $ d2 ' 2Ad 4 2 − K − ω u, & 2 )v = ν % dx (

€ 2Ω d 2K 4 as the relevant equation set. By the further transformation u → 1 u , these equations ν become: € 2 $d '$ d 2 ' 2 2 & 2 − K − ω )& 2 − K ) u = (1+ αx )v , and % dx (% dx (€ $ d2 ' 2 2 & 2 − K − ω )v = −TaK u , dx % ( where € 4 AΩ Ta = − 2 1 d 4 , and α = –(1 – µ). ν € as (11. 51). These are the same equations €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 11.9. Consider the centrifugal instability problem of Section 11.6. From (11.51) and (11.53), the eigenvalue problem for determining the marginal state (σ = 0) is 2 2 (11.92,93) (d 2 dR 2 − k 2 ) uˆ R = (1+ αx)uˆϕ , (d 2 dR 2 − k 2 ) uˆϕ = −Tak 2uˆ R , with uˆ R = duˆ R dR = uˆϕ = 0 at x = 0 and 1. Conditions on uˆϕ are satisfied by assuming solutions of the form ∞

€

€

uˆϕ€= ∑ Cm sin(mπx) . (11.94) m=1 € Inserting this into (11.92), obtain an equation for uˆ R , and arrange so that the solution satisfies the four remaining conditions on uˆ R . With uˆ R determined in this manner and uˆϕ given by (11.94), (11.93) leads to an eigenvalue problem for Ta(k). Following Chandrasekhar (1961, p. 300), show € that the minimum Taylor number is given € by (11.54) and is reached at kcr = 3.12. € € € into (11.92) to find: Solution 11.9. Continuing the effort from Exercise 11.8, insert (11.94) 2

∞

(d 2 dx 2 − K 2 ) u = (1+ αx)v = (1+ αx) ∑ Cm sin(mπx) ,

(5)

m=1

€

where a switch has been made to the dimensionless variables of Exercise 11.8. Now arrange that the solution satisfies the four remaining boundary conditions on u. With u determined in this fashion and v given by (11.94), (11.93) will lead to an equation for Ta. € The solution of (5) is straightforward. The general solution can be written in the form: $ A1(m ) coshKx + B1(m ) sinhKx + A2(m ) x coshKx + B2(m ) x sinhKx ( ∞ & & Cm (6) u= ∑ 2 2 % ), 4αmπ 2 2 + (1+ αx)sin(mπx) + 2 2 cos(m π x) m=1 (m π + K ) & & ' * m π + K2 where the constants are determined by the boundary conditions u = (d/dx)u = 0 at x = 0 and 1. These conditions lead to the four equations: 4αmπ , KB1(m ) + A2(m ) = −mπ , A1(m ) = − 2 2 € m π + K2 4αmπ , (7) A1(m ) coshK + B1(m ) sinhK + A2(m ) coshK + B2(m ) sinhK = (−1) m +1 2 2 m π + K2 ) € + K sinhK ) + B(m ) (sinhK + K coshK ) = (−1) m +1 (1+ α )mπ A1(m )K sinhK + B1(m€ K coshK + A2(m ) (coshK 2 The solution of these equations is: 4αmπ € A1(m ) = − 2 2 m π + K2 mπ B1(m ) = {K + β m (sinhK + K coshK ) − γ m sinhK} Δ mπ A2(m ) = − € {sinh 2 K + β m K (sinhK + K coshK ) − γ m K sinhK } Δ m π (8) B2(m ) = € {(sinhK coshK − K ) + β mK 2 sinhK − γ m (K coshK − sinhK )} Δ where: € 4α Δ = sinh 2 K − K 2 , β m = 2 2 (−1) m +1 + coshK }, and 2{ m π +K € 4α γ m = (−1) m +1 (1− α ) + 2 2 K sinhK . m π + K2 € €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Substituting u from (6) and v from (11.94) into (11.93) produces: ∞

∑ Cn ( n 2π 2 + K 2 ) sin nπx n=1

& A1(m ) coshKx + B1(m ) sinhKx + A2(m ) x coshKx + B2(m ) x sinhKx * (9) ( ( C = TaK 2 ∑ 2 2 m 2 2 ' +. 4αmπ + (1+ αx)sin(mπx) + 2 2 cos(mπx) m=1 (m π + K ) ( (, 2 ) m π +K Multiply this equation by sin(nπx) and integrate from x = 0 to x = 1, to obtain a system of linear homogeneous equations for the constants Cm. The requirement that these constants are not all zero leads to the equation: ∞

€

€

€

* . ' 2K n +1 (m ) n +1 (m ) n +1$ sinhK ) A2(m ) , ,,[1+ (−1) coshK ] A1 + [(−1) sinhK ] B1 + (−1) &coshK − 2 2 2 , % ( n π +K nπ + / 2 2 2 $ ' (m ) n π +K , 2K n +1 n +1 , + &(−1) sinhK − 2 2 1+ (−1) coshK })B2 2{ , ,0 % ( where n π +K % ( 3 δ 1 1 +αx nm + δnm − ( n 2π 2 +0K 2 ) 2nm = 0 (10) if m + n is even and m ≠ n + 2 2 K Ta + + x nm = & 1/4 if m = n ). + 4nm % + ( 2 1 − if m + n is odd & ) + 2 + 2 2 2 2 π 2 (n 2 − m 2 ) * 'n − m 'm π + K * On using the first two equations of (7), equation (10) simplifies to: % 4mπα ( nπ 2K (−1) m +n −1] − 2 2 (−1) n +1 A2(m ) sinhK + (−1) n +1 B2(m ) coshK + B2(m ) ]) & 2 2 2 2 2 2 [ 2 [ ' * n π +K n π +K € m π +K 3 δ 1 1 +αx nm + δnm − ( n 2π 2 + K 2 ) 2nm = 0 (11) 2 2 K Ta After substituting for the constants A2(m ) and B2(m ) given by (8), (11) becomes: 4mnπ 2α (−1) m +n −1] 2 2 2 2 2 2 [ (n π + K )(m π + K )

€

€

% ) '(sinhK coshK − K)[1+ (1+ α )(−1) m +n ] ' 2 ' ' 2Kmnπ − 2 2 &+(sinhK − K coshK)[(−1) n +1 + (1+ α )(−1) m +1 ]* 2 2 2 (n π + K )(sinh K − K ) ' 4kα sinhK ' m +1 m +n '− sinhK + K(−1) ][(−1) −1] ' ( m 2π 2 + K 2 [ + 3 δ 1 1 +αx nm + δnm − ( n 2π 2 + K 2 ) 2nm = 0 (12) 2 2 K Ta A first approximation to the solution of (12) is obtained by setting the (1,1)-element of the determinant zero. This implies: €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

1 2 1 1 2Kπ 2 (2 + α ) 2 3 δ nm π + K = α + − ( ) K 2Ta 4 2 π 2 + K 2 sinh2 K − K 2 [(sinhK coshK − K) + (sinhK − K coshK)] 2 ( )( )

and this simplifies to: 3

(π 2 + K 2 ) 2 Ta = 2 + α K 2 1−16Kπ 2 cosh 2 (K /2) π 2 + K 2 2 (sinhK + K) ( )

{

€

[

]}

A plot of Ta(2 + α) as a function of K from this solution shows that the minimum value of the Taylor number is: Tacr = 3430 (2 + α ) , and is reached at Kcr = 3.12. €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 11.10. For a Kelvin–Helmholtz instability in a continuously stratified ocean, obtain a globally integrated energy equation in the form 1 d ∂U u 2 + w 2 + g 2 ρ 2 ρ 02 N 2 ) dV = − ∫ uw dV . ∫ ( 2 dt ∂z (As in Figure 11.25, the integration in x takes place over an integer number of wavelengths.) Discuss the physical meaning of each term and the mechanism of instability.

€

€ From (11.57), multiplying the perturbation equations for the Kelvin-Helmholtz Solution 11.10. instability by u, w, and g 2 ρ ρ 02 N 2 and add them together $ ∂u $ ∂w ∂u ∂U 1 ∂ p ' ∂w ρ 1 ∂p ' g 2 ρ $ ∂ρ ∂ρ N2 ' u& + U + w + + w + U + g + + + U − ρ w ) & ) & ) = 0. 0 ∂x ∂z ρ 0 ∂ x ( ∂x ρ 0 ρ 0 ∂z ( ρ 02 N 2 % ∂t ∂x g ( % ∂t % ∂t This produces: € #∂ ∂ &# u 2 w 2 g2ρ 2 & ∂U 1 # ∂p ∂p & + 2 2 ( + uw + % + (=0 % + U (% + $ ∂t ∂x '$ 2 2 2ρ0 N ' ∂z ρ 0 $ ∂x ∂z ' The final term can be rewritten: #∂ ∂ &# u 2 w 2 g2ρ 2 & ∂U 1 # ∂ ( pu) ∂ ( pw) & # ∂u ∂w & + 2 2 ( + uw + % + % + U (% + ( − p% + ( = 0 . $€∂t ∂x '$ 2 2 2ρ0 N ' ∂z ρ 0 $ ∂x ∂z ' $ ∂ x ∂z ' After global integration over an integer number of x-direction wavelengths, the terms involving U(∂/∂x) will be zero, and the third term, which is equal to ρ −1 0 ∫ ∇ ⋅ ( pu)dV , transforms into a −1 €surface integral, ρ 0 ∫ pu ⋅ ndA , which vanishes because no flow crosses the channel walls and an there is an equal in-flux and out-flux across the open boundaries on the CV (see Figure 11.25). The last term vanishes because of the continuity equation. Thus, the resulting CV equation is: € 1 d ∂U u 2 + w 2 + g 2 ρ 2 ρ 02 N 2 ) dV = − ∫ uw dV . ∫ ( € 2 dt ∂z This equation shows that the rate of change of the sum of the kinetic and potential energies is equal to the energy extracted from the correlation of the stream-wise (u) and vertical (w) velocity fluctuations with the average shear (∂U/∂z). However, in order for the perturbations € from the mean shear, they must be anti-correlated so that uw is negative on to extract energy average when ∂U/∂z is positive. This is typical of shear instabilities. And, when u and w represent turbulent fluctuations the average correlation of u and w is called a Reynolds shear stress (see Ch. 12).

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 11.11. In two-dimensional (x,y)-Cartesian coordinates, consider the inviscid stability of $ S + y for y > 0' (, horizontal parallel shear flow defined by two linear velocity gradients: U(y) = % − & S y for y ≤ 0) where S+ and S– are real constants. Assume an infinitesimal velocity perturbation with vertical component v = f (y)exp {ik(x − ct)} , where k is positive real but ω may be complex. fU !! = 0 with f (y) →€0 as y → ∞ to find f(y). U −c b) Require the pressure perturbation associated with v to be continuous across y = 0, and determine a single equation for the disturbance phase speed c in terms of the other parameters. c) For what values of S+, S–, and k, is this flow€stable, unstable, € or neutrally stable? + – d) What is special about the case S = S ?

a) Use the Rayleigh equation f !! − k 2 f −

y!

U(y)!

x!

Solution 11.11. a) Here d2U/dy2 = 0, so the Rayleigh equation simplifies to d2f/dy2 – k2f = 0 which has solutions A±exp(±ky). To satisfy the given boundary conditions for k positive & real, the decaying exponential must be chosen for y > 0 and y < 0, so f(y) = Aexp(–k|y|). ∂ ∂v # b) The continuity equation is (U(y) + u#) + = 0 . For v = A exp {ik(x − ct) − k y } , this ∂x ∂x u = − ∫ (∂ v ∂ y) dx = (−1 ik ) ( df dy) exp {ik(x − ct)} = (−A ik ) ( ∓k ) exp {ik(x − ct) − k y } implies: where the upper & lower signs go with y > 0 & y < 0, respectively. The linearized horizontal ∂u ∂ u ∂U 1 ∂ p" € momentum equation for the disturbance is: . Evaluate the +U(y) + v =− ∂t ∂x ∂y ρ ∂x derivatives on the left side and integrate in x to find: ( −ikc + " ik % p/ ( ∓k ) + S ± y $# '& ( ∓k ) + S ± - ∫ A exp {ik(x − ct) − k y } dx = − , or * −ik ρ ) −ik , "#c ( ∓k ) − S ± y ( ∓k ) + S ± $% ∫ A exp {ik(x − ct) − k y } dx = − p' . ρ Evaluate this equation above and below y = 0 and set the two results equal to find: −ck + S + = +ck + S − or c = (S+ – S–)/2k. c) Here c is always real; this flow is neutrally stable in all cases. d) When S+ = S–, then c = 0, and this implies zero phase speed for the disturbance.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 11.12. Consider the inviscid stability of a constant vorticity layer of thickness h between uniform streams with flow speeds U1 and U3. Region 1 lies above the layer, y > h/2 with U(y) = U1. Region 2 lies within the layer, |y| ≤ h/2, U(y) = 12 (U1 + U 3 ) + (U1 − U 3 )( y h ) . Region 3 lies below the layer, y < –h/2 with U(y) = U3. fU "" a) Solve the Rayleigh equation, f "" − k 2 f − = 0 , in each region, then use appropriate U −c € boundary and matching conditions to obtain: f1 (y) = ( Acosh( kh 2) + Bsinh( kh 2))e−k ( y−h 2) for y > +h/2, for |y| ≤ h/2, f 2 (y) = € Acosh( ky ) + Bsinh( ky )

f 3 (y) = ( Acosh( kh 2) − Bsinh( kh 2))e +k ( y +h 2) for y < –h/2. where f defines the spatial extent of the disturbance: v " = f (y)e ik(x−ct ) and u" = −( f " ik )e ik(x−ct ) , € and A and€B are undetermined constants. ∂u# ∂u# ∂U 1 ∂p# . +U + v# =− ∂t ∂x ∂y ρ ∂x € the pressure to be€continuous at y = ± h/2, and Integrate this equation with respect to x, require simplify your results to find two additional constraint equations: U −U (c − U1 ) f1#(+h /2) = (c − U€1 ) f 2#(+h /2) + 1 3 f 2 (+h /2) , and h U1 − U 3 f 2 (−h /2) (c − U 3 ) f 3#(−h /2) = (c − U 3 ) f 2#(−h /2) + h c) Define c o = c − 12 (U1 + U 3 ) (this is the phase speed of the disturbance waves in a frame of € reference moving at the average speed), and use the results of parts a) and b) to determine a single equation € for co: # U1 − U 3 & 2 2 2 −2kh € co = % ( ( kh −1) − e $ 2kh ' [This part of this problem requires patience and algebraic skill.] d) From the result of part c), co will be real for kh >> 1 (short wave disturbances), so the € flow is stable or neutrally stable. However, for kh > τ when u(t) is uncorrelated with itself at large time lags, τ >> Λt. This leaves: €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

1 τ = +T ( τ + 1 +∞ 1− - R11 (τ )e−iωτ dτ = ∫ ∫ R11 (τ )e−iωτ dτ , * T →∞ 2π T 2 π , τ =−T ) −∞ where the final equality holds when T >> τ and R11 (τ ) → 0 for τ >> Λt. Se (ω ) = lim

€ €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 12.9. Derive the formula for the temporal Taylor microscale λt by expanding the definition of the temporal correlation function (12.17) into a two term Taylor series, and determining the time shift, τ = λt, where this two term expansion equals zero. Solution 12.9. Expand the temporal autocorrelation: # dR (τ ) & τ 2 # d 2 R11 (τ ) & R11 (τ ) = R11 (0) + τ % 11 ( + % ( + ... $ dτ 'τ = 0 2 $ dτ 2 'τ = 0 The autocorrelation function is even, so the terms involving odd derivatives are zero. Thus, the appropriate two-term expansion is: τ 2 # d 2 R11 (τ ) & R11 (τ ) = R11 (0) + % € ( + ... 2 $ dτ 2 'τ = 0 Truncate the expansion, evaluate it at t = λt, and set it equal to zero to find: λ2 $ d 2 R11(τ ) ' R11 ( λt ) = 0 = R11 (0) + t & ) + ... 2 % dτ 2 (τ = 0 € Now solve for λt to reach: % 1 d 2 R11 (τ ) ( % d 2 R11 (τ ) ( % d 2 r11 (τ ) ( λ2t = −2R11(0) ' = −2 = −2 ' * * ' * , 2 2 2 & dτ )τ = 0 & dτ )τ = 0 & R11 (0) dτ )τ = 0 € which duplicates (12.19).

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 12.10. When x, r, and k1 all lie in the stream-wise direction, the wave number spectrum S11 (k1 ) of the stream-wise velocity fluctuation u1 (x) defined by (12.45) can be interpreted as a distribution function for energy across stream-wise wave number k1. Show that the energyweighted mean-square value of the stream-wise wave number is: ) 1 +∞ 1 & d2 2 k12 ≡ 2 ∫ k12 S11 (k1 )dk1€= − 2 ( 2 R11 (r)+ , and that λ f = 2 k1 . *r= 0 u −∞ u ' dr

€

Solution 12.10. The relationship between the stream-wise autocorrelation function and the stream-wise spatial power spectrum is the Fourier inverse of € (12.45): € +∞ R11 (r) = ∫ S11 (k1 )e +ik1 r dk1 . −∞

Differentiate this twice with respect to r, and evaluate at r = 0: +∞ " +∞ " d 2 R11 (r) % " d 2 +ik r % % " +∞ 2 % +ik1 r 1 = S (k ) e dk = −k S (k )e dk = −k12 S11 (k1 )dk1 . $ ' ∫ ∫ ∫ $ ' $ ' $ ' 11 1 1 1 11 1 1 2 2 & r= 0 −∞ # dr & r= 0 # −∞ € # dr & & r= 0 # −∞

€

Divide by the mean-square fluctuation u"2 and multiply by –1 to get the desired form: 1 # d 2 R11 (r) & 1 +∞ 2 − 2% = k S (k )dk1 = k12 ( 2 2 ∫ 1 11 1 u $ dr ' r= 0 u −∞ € For, the specified geometry, R11 is the longitudinal autocorrelation function, so from (12.39) # d 2 f (r) & 1 # d 2 R11 (r) & 2 − 2% = = 2, ( % 2 2 ( u $ dr ' r= 0 $ dr ' r= 0 λ f € and this can be combined with the mean-square wave number result to find

λ f = 2 k12 .

€ €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 12.11. In many situations, measurements are only possible of one velocity component at one point in a turbulent flow, but the flow has a non-zero mean velocity and moves past the measurement point. Thus, the experimenter obtains a time history of u1 (t) at fixed point. In order to estimate spatial velocity gradients, Taylor’s frozen-turbulence hypothesis can be ∂u 1 ∂u1 invoked to estimate a spatial gradient from a time derivative: 1 ≈ − where the “1”-axis ∂x1 U1 ∂t € must be aligned with the direction of the average flow, i.e. Ui = (U1, 0, 0). Show that this approximate relationship is true when the influence of viscosity.

€

€

€

ui ui U1 > 1 in turbulent flow, the fluctuating-flow energy dissipation is more important. c) Start with the definition of the dissipation rate and the strain rate tensor: € 2 2 2) & ν & ∂ui ∂u j ) ν (& ∂ui ) ∂ui ∂u j & ∂u j ) + ε = 2ν Sij$ Sij$ = (( + +( + = ( + +2 + 2 ' ∂x j ∂x i +* 2 ((' ∂x j +* ∂x j ∂x i ' ∂x i * + ' * The first and last terms under the final overbar are the same, so % ∂u ( 2 ∂u ∂u ε = ν '' i ** + ν i j . € ∂x j ∂x i & ∂x j ) The second term can be manipulated using the requirement of fluctuating flow incompressibility: €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

∂ 2u j ∂ui ∂u j ∂ $ ∂u j ' ∂ $ ∂u j ' + =ν u − u =ν & i ) i & ui ) ∂x j ∂ x i ∂x j % ∂x i ( ∂x j ∂x i ∂x j % ∂ x i ( ∂2 ∂ % ∂ui ( ∂2 ∂2 =ν ui u j ) − ui u j ) = ui u j . 'u j *= ( ( ∂x j ∂ x i ∂x j & ∂ x i ) ∂x j ∂ x i ∂x j ∂x i +%€ ( 2 . ∂ui ∂2 Thus, ε = ν '' u u 0. * + -& ∂x j *) ∂x i∂x j i j 0 , / € d) Use the simple scaling ideas for the ratio to find: ∂2 ui u j 2 2 ∂ x ∂ x ΔU L) & ΔUL ) η 4 ( i j 2 −3 / 4 4 −1 € =( + 4 = ReL ( ReL ) = ReL . 2 ~ 2 2 (ν η ) ' ν * L ∂u ∂x ν

(

i

j

)

e) This implies that anisotropy will not be important at high Reynolds number. Therefore, an isotropic dissipation model should be suitable for engineering purposes. €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 12.26. Determine the self-preserving form of the average stream-wise velocity Uz(z,R) of a round turbulent jet using cylindrical coordinates where z increases along the jet axis and R is the radial coordinate. Ignore gravity in your work. Denote the density of the nominally-quiescent reservoir fluid by ρ. a) Place a stationary cylindrical control volume around the jet's cone of turbulence so that circular control surfaces slice all the way through the jet flow at its origin and at a distance z downstream where the fluid density is ρ. Assuming that the fluid outside the jet is nearly stationary so that pressure does not vary in the axial direction and that the fluid entrained into the volume has negligible x-direction momentum, show J0 ≡

∫

d/2 0

ρ0U 02 2π RdR =

∫

D/2 0

ρU z2 (z, R)2π RdR ,

where J0 is the jet's momentum flux, ρ0 is the density of the jet fluid, and U0 is the jet exit velocity. b) Simplify the exact mean-flow equations ∂U z 1 ∂ " % + ( RU R ) = 0 , and U z ∂U z +U R ∂U z = − 1 ∂ P + ν ∂ $ R ∂U z ' − 1 ∂ RuzuR − ∂ Ruz2 , ∂z R ∂R ∂z ∂R ρ ∂z R ∂R # ∂R & R ∂R ∂z when ∂P/∂z ≈ 0, the jet is slender enough for the boundary layer approximation ∂/∂R >> ∂/∂z to be valid, and the flow is at high Reynolds number so that the viscous terms are negligible. c) Eliminate the average radial velocity from the simplified equations to find: & ∂U ∂U # 1 R ∂U 1 ∂ U z z − $ ∫ R z dR ' z = − Ruz uR ∂z % R 0 ∂z R ∂R ( ∂R where R is just an integration variable. d) Assume a similarity form: U z (z, R) = UCL (z) f (ξ ) , −uz uR = Ψ(z)g(ξ ) , where ξ = R δ (z) and f and g are undetermined functions, use the results of parts a) and c), and choose constant values

(

(

)

( )

)

12

appropriately to find U z (z, R) = const. ( J 0 ρ ) z −1 f ( R z ) . e) Determine a formula for the volume flux in the jet. Will the jet fluid from the nozzle be diluted with increasing z? R! Uz(z,R)!

d! z!

UCL(z)!

Solution 12.26. a) Use the stationary CV shown above, consider only the steady mean flow, and ignore turbulent fluctuations. In this case the CV momentum equation is: ∫ ρUz (U ⋅ n)dA = − ∫ Pn ⋅ e z dA , Surface

Surface

since there are no shear stresses on any of the CV boundaries. When the fluid entrained into the volume has negligible z-direction momentum, only the inlet (z = 0) and outlet (z) surfaces

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

provide a contribution, and if the pressure does not vary in the z-direction, then the above equation reduces to: d 2

D2

− ∫ "#ρU z2 $% 2π R dR + ∫ "#ρU z2 $% 2π R dR ≅ 0 . 0 z 0

0

where D is a radial location that is comfortably outside the jet's cone of turbulence. [Here it must be noted that this equation is approximate. The jet entrains reservoir fluid and induces the reservoir fluid to move. Thus, the assumption of negligible pressure gradient is not precisely correct.] Noting that the first integral on the left is the jet's momentum flux and that !"ρU z2 #$ = ρ0U 02 , the above equation simplifies to: 0

d 2

∫ 0

D2

ρ0U 02 2π R dR ≡ J 0 ≅ ∫ ρU z2 2π R dR , 0

where J0 is the jet's momentum flux. b) The mean-flow continuity equation must be retained as is. However, the mean-flow zdirection momentum equation can be simplified. ∂U ∂U 1 ∂ P ν ∂ " ∂U z % 1 ∂ ∂ U z z +U R z = − + Ruz uR − Ruz2 . $R '− ∂z ∂R ρ ∂z R ∂R # ∂R & R ∂R ∂z When ∂P/∂z ≈ 0, then ∂U ∂U ν ∂ " ∂U z % 1 ∂ ∂ U z z +U R z ≅ + Ruz uR − Ruz2 . $R '− ∂z ∂R R ∂r # ∂ R & R ∂ R ∂z If the jet is slender enough for the boundary layer approximation ∂/∂R >> ∂/∂z to be valid, then ∂U ∂U ν ∂ " ∂U z % 1 ∂ U z z +U R z ≅ + Ruz uR . $R '− ∂z ∂R R ∂R # ∂R & R ∂R For high Reynolds number flow, the viscous terms are negligible, so ∂U ∂U 1 ∂ U z z +U R z ≅ − Ruz uR . ∂z ∂R R ∂R c) Use the continuity equation to eliminate UR from the problem: ∂U z 1 ∂ (RU R ) 1 R ∂U + = 0 → U R = − ∫ R z dR ∂z R ∂R R 0 ∂z so that the boundary layer RANS equation becomes: & ∂U ∂U # 1 R ∂U 1 ∂ (3*) U z z − $ ∫ R z dR ' z = − Ruz uR ∂z % R 0 ∂z R ∂R ( ∂R where R is an integration variable. d) Use the similarity solution: U z (z, R) = UCL (z) f (ξ ) , −uz uR = Ψ(z)g(ξ ) , and ξ = R δ (z) , and evaluate each term in (3*), converting R to ξ wherever possible $ ∂U ∂ −Rδ ! ' δ! 2 ! f +UCL f ! ⋅ 2 ) = UCLUCL ! f 2 −UCL U z z = UCL f (UCL f ) = UCL f &UCL ff !ξ ; % ∂z ∂z δ ( δ " 1 R ∂U z % ∂U " δ 2 R R ∂U z ( R +% % 1 " 1 ξ ∂U z dR & R = # ∫ d * -&UCL f . = # ∫ ξ d ξ &UCL f . # ∫ R ∂z δ $ξ 0 ∂ z $R 0 ' ∂ R $ r 0 δ ∂ z ) δ ,' '

(

(

( )

)

(

(

)

)

(

( )

)

)

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

'U " ξ * U δ" ξ U U" f " ξ U 2 f " δ" ξ = ( CL ∫ fξdξ − CL ∫ f ξ" 2 dξ +UCL f " = CL CL ∫ fξdξ − CL ∫ f ξ" 2 dξ ξ ξ δ η η δ ) , 0 0 0 0 ξ ξ 2 ( + U U" f " U f " δ" 2 ξ = CL CL ∫ fξdξ − CL *( fξ ) 0 − 2 ∫ fξdξ ξ η δ) , 0 0 ξ 2 %U U " f " € U f " δ "( δ" 2 = ' CL CL + 2 CL * ∫ fξdξ − UCLξff " ; and ξ ξ δ)0 δ & € 1 ∂ 1 ∂ 1 ∂ RS ∂ Ruz uR = (R δ )uz uR = ξ uz uR = − CL (ξ g) , R ∂r δ (R δ) ∂ (R δ) δξ ∂ξ δξ ∂ξ € Now reassemble (3*) and cancel terms. δ " &U U " f " U 2 f " δ ") ξ δ " RSCL ∂ 2 2 " f 2 − UCL UCLUCL ff ξ" − ( CL CL + 2 CL (ξg) + ∫ fξdξ + UCLξff " = δ ' ξ ξ δ*0 δ δξ ∂ξ &U U " f " U 2 f " δ ") ξ RSCL ∂ " f 2 − ( CL CL + 2 CL UCLUCL (ξg) + ∫ fξdξ = ξ ξ δ*0 δξ ∂ξ ' 2 to find: € Multiply by δ and divide by UCL $δUCL ' f#ξ $ RS ' 1 ∂ # ' 2 $δUCL # + 2δ #) ∫ fξdξ = & 2CL ) (ξg) . & ) f −& % UCL ( % UCL (ξ 0 % UCL ( ξ ∂ξ € € The three functions outside the [,]-brackets only depend on the similarity variable ξ. Thus, the flow will show (the simplest) self-similarity when: # # δUCL δUCL RSCL = C1 , + 2δ # = C2 , and = C3 , (a,b,c) € 2 UCL UCL UCL where C1, C2, and C3 are constants. Equations (a) and (b) require: " C − C1 % 2δ " = C2 − C1 , or δ = $ 2 ' (z − zo ) . # & 2 € € € It is easiest to choose C2 – C1 = 2, and to assume zo = 0, so that δ = z. Thus (a) implies: ! UCL C =€ 1 , or lnUCL = C1 ln z + C4 which is the same as UCL = C5 z C1 . UCL z The constraint found in part a) requires: ∞ ∞ D/2 2 R $R' 2 2 2 2 2 J 0 ≡ 2πρ ∫ U z (z, R)RdR = 2πρδ ∫ UCL f (ξ ) d & ) = 2πρδ UCL ∫ f 2 (ξ )ξ dξ . 0 δ %δ ( 0 0 Substituting in the required similarity forms for δ and UCL leads to:

(

)

(

)

(

J 0 ≡ 2πρx 2C52 x 2C1

)

∞

∫

f 2 (ξ )ξdξ .

0

2 5

Here the integral is just a number and 2πρC is just a product of constants, therefore 2+2C1, the exponent of x, must be zero, and this implies C1 = –1 and C5 = const. J 0 ρ , where "const." is a dimensionless constant.€When all this is drawn together, the final form for the self-similar round jet velocity field is:

€

12

U z (R, z) = const. (€ J 0 ρ ) z −1 f ( R z ) . e) The volume flux Q in the jet will be:

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

D2

12

12

# J 0 & −1 2 ∞ #J & Q = 2π ∫ U z R dR = 2πUCLδ ∫ f ξ d ξ = 2π const. % ( z z ∫ f ξ d ξ ∝ % 0 ( z . $ρ' $ρ' 0 0 0 Thus, the volume flux in the jet increases linearly with downstream distance, so the jet fluid will be diluted with increasing x. ∞

2

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 12.27. Consider the turbulent wake far from a two-dimensional body placed perpendicular to the direction of a uniform flow. Using the notation defined in the Figure, the result of Example 12.5 may be written:

+∞ ( FD l U(x, y) " U(x, y) % v 2 − u 2 + * = θ = ∫ U $#1− U '& + U 2 -- dy , ρUo2 o o o −∞ * ) ,

where θ is the momentum thickness of the wake flow (a constant), and U(x,y) is the average horizontal velocity profile a distance x downstream of the body. a) When ΔU 103, the outer flow should be fully turbulent so a distinct log-law region is expected. f) Fitting is always better when there is more data to fit, and the extent of the log-law increases with increasing Reynolds number. Thus, data from the highest Reynolds number, Reτ = 104, should produce the most accurate log-law constants.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 12.36. A horizontal smooth pipe 20 cm in diameter carries water at a temperature of 20 °C. The drop of pressure is dp/dx = –8 N/m2 per meter. Assuming turbulent flow, verify that the thickness of the viscous sublayer is ≈ 0.25 mm. [Hint: Use dp/dx as given by (12.97) to find τw = 0.4 N/m2, and therefore u* =0.02 m/s.] Solution 12.36. From (12.96), ∂p/∂x = –4τw/d, therefore: d ∂p (0.2m) τ 0.4Pa τw = − =− (−8Pa / m) = 0.4Pa , so u* = w = = 0.02ms −1 . 3 −3 4 ∂x 4 ρ 10 kgm From Section 12.9, the viscous sublayer thickness is: 5lν = 5(ν u* ) = 5(1.0 ×10−6 m 2 s−1 0.02ms−1 ) = 2.5 ×10−4 m = 0.25mm .

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 12.37. The cross-section averaged flow speed Uav in a round pipe of radius a may be written: volume flux 1 a 2 a U av ≡ = ∫ U(y)2π r dr = a 2 ∫ U(y)(a − y)dy , area π a2 0 0 where r is the radial distance from the pipe's centerline, and y = a – r is the distance inward from the pipe's wall. Turbulent pipe flow has very little wake, and the viscous sublayer is very thin at high Reynolds number; therefore assume the log-law profile, U(y) = (u* κ ) ln ( yu* ν ) + B , holds throughout the pipe to find

U av ≅ u* #$(1 κ ) ln ( au* ν ) + B − 3 2κ %& . Now use the definitions: C f = τ w 12 ρU av2 , Re d = 2U av a ν , f = 4C f = the Darcy friction factor, κ = 0.41, B = 5.0, and switch to base-10 logarithms to reach (12.105). Solution 12.37. Start with given relationships and integrate: 0 2 a 2 a - u ' yu * U av = 2 ∫ U(y)(a − y)dy ≅ 2 ∫ / * ln) * , + u* B2(a − y)dy a 0 a 0 .κ ( ν + 1 =

2ν a - u* a ' yu* * u* y ' yu* *0 2u B ln) ln) ,− ,2dy + *2 ∫ / 2 a κ 0 . ν ( ν + ν ( ν +1 a

a

∫ [a − y ]dy 0

a 0 2ν 2 au* ν - au* 2u* B y20 = 2 ln(β ) − β ln(β )2dβ + 2 /ay − 2 ∫ 1 a κu* 0 /. ν a . 2 10 au* ν 2ν 2 - au* β2 β20 2u B a2 0 = 2 / (β ln β − β ) − ln(β ) + 2 + *2 /a 2 − 2 a κu* . ν 2 4 10 a . 21

=

2 2ν 2 - au* ' au* au* au* * 1 ' au* * ' au* * ln − − ln) / ) , ) , ,+ a 2κu* . ν ( ν ν ν + 2( ν + ( ν +

2 1 ' au* * 0 ) , 2 + u* B 4( ν + 1

2 2 ' 1 ' au * 3 * 2ν 2 - 1 ' au* * ' au* * 3 ' au* * 0 = 2 / ) + B,. , ln) ,− ) , 2 + u* B = u* ) ln) * , − a κu* . 2 ( ν + ( ν + 4 ( ν + 1 ( κ ( ν + 2κ + The second part of this exercise is primarily algebraic. Start by rewriting the ratio Uav/u* in terms of f , and rewriting the ratio u*a/ν in terms of Red and f :

€

U av U av = = u* τw ρ

U av

=

1

=

8 12

au 2aU av , and * = ν 2ν

1 8

f

=

1 12 f Re d . 32

f € Put these into the result of part a) and recall that ln(...) = ln(10)log10(...). $ 1 12 ' 3 U av 1 $ au* ' 3 8 ln(10) = ln& +B → = log10 & f Re d ) − + B. )− 12 u* κ % ν ( 2κ f€ κ % 32 ( 2κ Rearrange to reach the desired form, and evaluate using the numbers provided above. 1 ln(10) 1 $ 3 ln 32 ' 12 12 = log f Re + − + B − & ) = 1.986log10 ( f Re d ) −1.020 . 10 ( d) 12 f κ ( 8κ 8 % 2κ € This is the desired result.

€

€

1 2

C f U av2

1 8

f

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 12.38. The cross-section averaged flow speed Uav in a wide channel of full height b may be written: b2 2 U av ≡ ∫ U(y)dy , b 0 where y is the vertical distance from the channel's lower wall. Turbulent channel flow has very little wake, and the viscous sublayer is very thin at high Reynolds number; therefore assume the log-law profile, U(y) = (u* κ ) ln ( yu* ν ) + B , holds throughout the channel to find

U av ≅ u* #$(1 κ ) ln ( bu* 2ν ) + B −1 κ %& . Now use the definitions: C f = τ w

1 2

ρU av2 , Re b = U av b ν , f = 4C f = the Darcy friction factor, κ

= 0.41, B = 5.0, and switch to base-10 logarithms to reach: f −1 2 = 2.0 log10 ( Re b f 1 2 ) − 0.59 . Solution 12.38. Start with given relationships and integrate: b2 b2 , 2 2 ) u* # yu* & U av = ∫ U(y)dy ≅ ∫ + ln % ( + u* B. dy b 0 b 0 *κ $ ν ' -

=

2ν bκ

b2

∫ 0

b2 bu 2 ν u* # yu* & 2u B 2ν * 2u B b ln % ln ( β ) d β + * ( dy + * ∫ dy = ∫ ν $ ν ' b 0 bκ 0 b 2

2ν 2ν # bu # bu & bu & bu 2 ν [ β ln β − β ]0 * + u* B = % * ln % * ( − * ( + u* B bκ bκ $ 2ν $ 2ν ' 2ν ' # 1 # bu & 1 & = u* % ln % * ( − + B (. $ κ $ 2ν ' κ ' The second part of this exercise is primarily algebraic. Start by rewriting the ratio Uav/u* in terms of f , and rewriting the ratio u*b/2ν in terms of Reb and f : =

€

U av U av = = u* τw ρ

U av

=

1

=

8

1 bu* bU av 8 f 1 12 = = f Re b . , and 2ν 2ν 32

f12 € Put these into the result of part a) and recall that ln(...) = ln(10)log10(...). ! 1 12 $ 1 U av 1 ! bu* $ 1 8 ln(10) = ln # = log10 # f Re b & − + B . &− + B → 12 " 32 % κ u* κ " 2ν % κ f κ Rearrange to reach the desired form, and evaluate using the numbers provided above. 1 ln(10) 1 " 1 ln 32 % 12 = log f Re + − + B − $ ' = 1.986 log10 f 1 2 Re d − 0.589 . 10 b 12 f κ & 8κ 8# κ This is the desired result. 1 2

(

C f U av2

)

1 8

f

(

)

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 12.39. For laminar flow, the hydraulic diameter concept is successful when the ratio is near unity. Show that this ratio is 1.5 when the duct is a ( f ⋅Uav dh ν ) ( f ⋅Uav d ν ) duct

round pipe

wide channel. Solution 12.39. For a channel with height b and width w, the hydraulic diameter from (12.103) is: bw b , and this simplifies to: (dh)channel ≈ 2b when w >> b. =2 ( dh )channel = 4 2b + 2w 1+ b w 1 L The friction factor is appears in (12.102): Pu − Pd = ρU av2 ⋅ ⋅ f , so: 2 dh U d 2(Pu − Pd )dh2 2dh2 # dp & f ⋅ av h = = %− ( . ν µU av L µU av $ dx ' Evaluate this for laminar channel and round-pipe flows using the results from section 9.2: " U av dh % 2(2b)2 " dp % = $f ⋅ ' $ − ' = 12 ⋅ 8 = 96 , and " b 2 dp % # dx & # ν &channel µ $− ' # 12µ dx & " U av dh % " dp % 2d 2 $f ⋅ ' = $ − ' = 32 ⋅ 2 = 64 . 2 " d dp % # dx & # ν & pipe µ $− ' # 32µ dx & Thus, the ratio is: ( f ⋅U av dh ν )

channel

( f ⋅U

av

d ν)

round pipe

= 96/64 = 1.5.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 12.40. a) Rewrite the final friction factor equation in Exercise 12.38 in terms of the channel's hydraulic diameter instead of its height b. b) Using the friction factor-Reynolds number ratio given in Exercise 12.39, evaluate (12.107) for a wide channel. c) Are the results of parts a) and b) in good agreement? Solution 12.40. a) For a channel with height b and width w, the hydraulic diameter from (12.103) is: bw b , and this simplifies to: (dh)channel ≈ 2b when w >> b. =2 ( dh )channel = 4 2b + 2w 1+ b w The final result of Exercise 12.38 is f −1 2 = 2.0 log10 ( Re b f 1 2 ) − 0.59 , which can be rewritten:

"1 % f −1 2 = 2.0 log10 $ Re 2b f 1 2 ' − 0.59 = 2.0 log10 ( Re dh f 1 2 ) −1.19 . #2 & b) For this problem, (12.107) implies $$ f ⋅ Re ' ' pipe d −1 2 −1 2 & & ) fchannel,turb ≅ 2.0 log10 & ⋅ Re dh ⋅ fchannel,turb ) − 0.80 ) & f ) ⋅ Re dh (laminar %% channel ( $ 1 ' −1 2 = 2.0 log10 & ⋅ Re dh ⋅ fchannel,turb ) − 0.80 % 1.5 ( −1 2 = 2.0 log10 ( Re dh ⋅ fchannel,turb ) −1.15.

c) The only difference between the answers for parts a) and b) is 0.04 in the final subtracted constant. Thus, the agreement is good and this suggests that (12.107) provides a useful empirical correction for estimating friction factors in non-circular ducts.

Fluid Mechanics, 6th Ed.

CfR

Kundu, Cohen, and Dowling

Exercise 12.41. a) Simplify (12.114) when the roughness Reynolds number is large Reks >> 1 to show that CfR is independent of ν in the fully rough regime. b) Reconcile the finding of part a) with the results in Figure 12.25 which appear to show that CfR depends on Rex for all values of Reks. c) For this fully rough regime, compare CfR computed from (12.114) with the empirical formula −2.5 provided in Schlichting (1979): C fR = ( 2.87 +1.58⋅ log10 (x / ks )) . Solution 12.41. a) Eq. (12.114) is: 2 1 " ( H 3.5) Re x C fR 2 %' 2Π , ≅ ln $$ R +B+ ' C fR κ # 1+ 0.26 Re ks C fR 2 & κ log10(x/ks) where Rex = Uex/ν and Reks = Ueks/ν. When Reks >> 1, the '1' in the denominator may be ignored, and this allows the argument of the natural log function to be simplified: 2 1 " ( H 3.5) Re x C fR 2 %' 2Π 1 "$ ( H R 3.5) C fR 2 x %' 2Π , ≅ ln $$ R + B + = ln $ +B+ ' ' C fR κ # 0.26 Re ks C fR 2 & κ κ # 0.26 ks & κ where Rex/Reks = x/ks. b) Figure (12.25) provides curves of CfR at constant Reks vs. Rex. The ratio x/ks is not held constant on any of the plotted curves. Thus, the reader must examine points on different curves having the same value of x/ks. For example, consider x/ks = 102, where CfR = 1.16 for Rex/Reks = 105/103, 106/104, and 107/105. c) A plot of (12.114) (solid line) and Schlichting's formula (dashed line) vs. log10(x/ks) looks like:

0.1$

0.01$

0.001$

2$

3$

4$

5$

6$

Here, as in Example 12.10, HR = 1.3. The agreement between the two empirical formulae is generally good. At x/ks = 102, (12.114) is 3% higher than Schlichting's formula. At x/ks = 106, (12.114) is 6% lower Schlichting's formula.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 12.42. Perhaps the simplest way to model turbulent flow is to develop an eddy viscosity from dimensional analysis and physical reasoning. Consider turbulent Couette flow with wall spacing h. Assume that eddies of size l produce velocity fluctuations of size l(∂U ∂y ) 2

so that the turbulent shear stress correlation can modeled as: −uv ∝ l 2 (∂U ∂y ) . Unfortunately, l cannot be a constant because it must disappear near the walls. Thus, more educated guessing is needed, so for this problem assume ∂U/∂y will have some symmetry about the€channel centerline (as shown) and try: l = Cy for 0 ≤ y ≤ h/2 where C is a positive dimensionless constant and y is € With this turbulence model, the RANS the vertical distance measured from the lower wall. equation for 0 ≤ y ≤ h/2 becomes: € % (2 ∂U ∂U ∂U 1 dp 1 ∂τ xy 2 2 ∂U where τ xy = µ + ρC y ' * U +V =− + ∂y ∂x ∂y ρ dx ρ ∂y & ∂y ) Determine an analytic form for U(y) after making appropriate simplifications of the RANS equation for fully developed flow assuming the pressure gradient is zero. Check to see that your final answer recovers the appropriate forms as y → 0 and C → 0. Use the fact that € € U(y = h /2) = U o 2 in your work if necessary.

€

Solution 12.42. In Couette flow V will be zero because the fluid is confined and ∂/∂x = 0 because the flow is homogeneous in this direction. Thus, with zero pressure gradient, the RANS BL 1 ∂τ xy equation becomes very simple: 0 = . Integrate this equation from lower wall to find: ρ ∂y τ xy (y) − τ w = 0 where τw is the wall shear stress. Now substitute in the model equation for the shear stress to find a quadratic equation for the velocity gradient: 2 ∂U −µ ± µ 2 + 4 ρτ wC 2 y 2 €2 2 $ ∂U ' ∂U = µ + ρC y & ) − τ w = 0 , which implies . ∂y 2 ρC 2 y 2 ∂y % ∂y ( Here, a positive velocity gradient is expected for the lower half of the channel so the “+” sign should be chosen. A bit of work or use of a table of integrals produces: 2 2 2 2 5 % y( -+ 1 τ w 41− 1+ 4 ρτ wC y µ€ 4 ρτ wC 2 y 2 -/7 € U(y) = + ln,2C ρτ w ' * + 1+ 0 -. -176 C ρ 43 µ2 2C ρτ w ( y µ) &µ) Now check to see that this model outcome has the appropriate behavior. As y → 0, the velocity should be linear in y. To find this behavior in the equation above, expand the square roots using the Taylor series 1+ ε = 1+ 12 ε + ... for ε 0. % 0 1(

∞

€

(

1 9

3

)

€

) ∂Y ∂Y ∂ & ∂Y +Uj = − u jY %++ , but not all the terms ((κ m ∂t ∂x j ∂x j ' ∂x j * are needed at high Reynolds number. Using the length, time, and mass-fraction scales, set * * * Y * = Y Yo , U j = U j T L , t * = t T , x j = x j L , and u jY " = u jY " ( LYo /T ) , and insert these into (12.34) to find: € * * Yo ∂Y L Yo * ∂Y * Yo ∂ 2Y * 1L ∂ + U j * = κm 2 * * − Yo * u jY % . * T ∂t T L ∂x€ L ∂x j ∂ x j L T ∂x j j € € € Divide by Yo and multiply by T to find that the terms on the left side are of order unity along with the second term on the right. * * ∂Y * T ∂ 2Y * ∂ * ∂Y + U = κ − * u jY % € j m 2 * * * * ∂t ∂x j L ∂x j ∂ x j ∂x j Given that κm is at most as large as ν, the coefficient of the first term on the right side is at most as large as 1/Re, where Re is the Reynolds number = L2 νT ; thus, this term can be dropped compared to the three that are of order unity. € b) For steady mean flow, the approximate Reynolds-averaged scalar transport equation becomes: ∂Y ∂ Uj €=− u jY $ . ∂x j ∂x j

Solution 12.48. a) Start with (12.34):

€

(

€

)

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Insert the given mean velocity U1 = Sx2, and turbulent diffusion model −u2Y # = ΔUL(∂Y ∂x 2 ) :

∂Y ∂ 2Y = ΔUL 2 . (@) ∂x1 ∂x 2 $ S '1 3 € Introduce the similarity variable ξ = x 2 & ) , and note that: % ΔULx1 ( 13 23 ∂Y dY ∂ξ ξ dY€ ∂Y dY ∂ξ % S ( dY ∂ 2Y $ S ' d 2Y = =− , , and . = =' =& * ) ∂x1 dξ ∂x1 3x1 dξ ∂x 2 dξ ∂x 2 & ΔULx1 ) dξ ∂x 22 % ΔULx1 ( dξ 2 Thus, the approximate € scalar transport equation becomes: % S ( 2 3 d 2Y 1 2 dY d 2Y Sx 2 dY = , which simplifies to − ξ − ξ = ΔUL' * 2 3 dξ dξ 2 3x1 dξ€ & ΔULx1 ) dξ € € after some manipulation of the factors. Integrate and exponentiate this last equation to find: $ 1 ' ξ dY = Aexp& − ξ 3 ) , and integrate again to reach Y (ξ ) = A ∫ 0 exp(− 19 ζ 3 ) dζ + B , % 9 ( dξ € € where ζ is just an integration variable. The boundary conditions are: (i) Y (x j ) = 0 for x1 < 0, (ii) Y (x j ) = 1 at x2€= 0 for x1 > 0, and € (iii) Y (x j ) → 0 as x 2 → ∞ . Condition (i) can be set€immediately since Y = 0 is a solution of the field equation (@). 0 Condition (ii) implies: Y€(0) = A ∫ 0 exp(− 19 ζ 3 ) dζ + B = 1 , or 0+B = 1 for x1 > 0. ∞ Condition (iii) implies: Y€(∞) = A ∫ 0 exp€(− 19 ζ 3 ) dζ + B = 0 , or 0+B = 1 for x1 > 0. € −1 ∞ Therefore, B = 1 and A = − ∫ 0 exp(− 19 ζ 3 ) dζ , so € ∞ ξ ∞ ξ 1 − 19 ζ 3 ) dζ ∫ 0 exp(− 19 ζ 3 ) dζ − ∫ 0 exp(− 19 ζ 3 ) dζ ∫ξ exp(− 9 ζ 3 ) dζ ∫ 0 exp ( . Y (ξ ) = 1− ∞ € = = ∞ ∞ ∫ 0 exp(− 19 ζ 3 )dζ ∫ 0 exp(− 19 ζ 3 )dζ ∫ 0 exp(− 19 ζ 3 ) dζ € form is: Thus, the final '0 for x1 < 0, x 2 > 0 + $ S '1 3 )∞ ) ∞ where ξ = x 2 & Y (x1, x 2 , x 3 ) = ( ) . exp(− 19 ζ 3 )dζ ∫ exp(− 19 ζ 3 )dζ for x1, x 2 > 0, % ΔULx1 ( )* ξ∫ ) 0 Sx 2

(

€

€

)

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 12.49. Estimate the Monin–Obukhov length in the atmospheric boundary layer if the surface stress is 0.1 N/m2 and the upward heat flux is 200 W/m2. Solution 12.49. First compute the friction velocity: τ 0.1Pa u* = w = = 0.289ms −1 . 3 −3 ρ 1.2 kgm The heat flux determines wT " Q 200Wm−2 wT " = = = 0.166ms−1K . −3 2 −2 −1 ρCp 1.2kgm (1004m s K ) Thus, the the Monin–Obukhov length is: € 3 u* (0.289ms−1 ) 3 LM = = = 10.6m . καgwT $ (0.41)(3.4 ×10−3 K −1 )(9.81ms−2 )(0.166ms−1K) €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 12.50. Consider one-dimensional turbulent diffusion of particles issuing from a point source. Assume a Gaussian Lagrangian correlation function of particle velocity r(τ ) = exp{− τ 2 t c2 } , where tc is a constant. By integrating the correlation function from τ = 0 to ∞, find the integral time scale Λt in terms of tc. Using the Taylor theory, estimate the eddy diffusivity at large times t/Λt >> 1, given that the rms € fluctuating velocity is 1 m/s and tc = 1 s. Solution 12.50. Start from the definition of the integral time scale: ∞ ∞ ∞ ∞ ' τ2* ' τ2* π Λ t = ∫ r(τ )dτ = ∫ exp)− 2 ,dτ = ∫ exp) − 2 , dτ = t c ∫ exp −β 2 dβ = t c = 0.886t c . 2 ( tc + ( tc + 0 0 0 0 From (12.129), DT ≅ u 2 Λ t = (1ms−1 ) 2 ⋅ 0.886(1s) = 0.886m 2 s−1 .

( )

€ €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 13.1. The Gulf Stream flows northward along the east coast of the United States with a surface current of average magnitude 2 m/s. If the flow is assumed to be in geostrophic balance, find the average slope of the sea surface across the current at a latitude of 45°N. [Answer: 2.1 cm per km] Solution 13.1. Given v = 2 m/s, and f = 2Ωsin45° = 2(0.73x10–4 s–1)(0.7071) = 1.03 x10–4 s–1. For geostrophic balance: 1 ∂p ∂η , fv = =g ρ ∂x ∂x so the sea-surface slope is: ∂η fv (1.03×10 −4 s−1 )(2ms−1 ) = = = 2.1×10 −5 , ∂x g 9.81ms−2 which is equivalent to an eastward surface rise of 2.1 cm per km.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 13.2. A plate with water (ν = 10−6 m2/s) above it rotates at a rate of 10 revolutions per minute. Find the depth of the Ekman layer, assuming that the flow is laminar. Solution 13.2. Given Ω = 10 rpm = 20π rad./min. = 1.047 s–1, the Coriolis frequency is f = 2Ω = 2.09 s–1. From (13.28), the Ekman layer thickness is:

δ = 2ν f = 2(10 −6 m 2s−1 ) (2.09s−1 ) = 0.978 ×10 −3 m ≅ 1 mm .

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 13.3. Assume that the atmospheric Ekman layer over the earth’s surface at a latitude of 45ºN can be approximated by an eddy viscosity of ν V = 10 m2/s. If the geostrophic velocity above the Ekman layer is 10 m/s, what is the Ekman transport across isobars? [Answer: 2203 m2/s] Solution 13.3. Given ν V = 10 m2/s, f = 2Ωsin45° = 2(0.73x10–4 s–1)(0.7071) = 1.03 x10–4 s–1, and U = 10 m/s, the Ekman layer height from (13.28) is

δ = 2ν f = 2(10m 2s−1 ) (1.03×10 −4 s−1 ) = 440.7m . Using the equation at the top of page 720, the net transport perpendicular to the geostrophic stream is: 1 Uδ = 0.5(10ms –1 )(440.7m) = 2, 203m 2s –1 . 2

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 13.4. a) From the set (13.45) – (13.47), develop the following equation for the water surface elevation η(x,y,t): # ∂2 ∂ ) ∂2 ∂2 &, * 2 + f 2 − gH % 2 + 2 (-η (x, y, t) = 0 ∂t + ∂t $ ∂x ∂y '. b) Using η (x, y, t) = ηˆ exp {i(kx + ly − ω t)} show that that the dispersion relationship reduces to ω = 0 or (13.82). c) What type of flows have ω = 0? Solution 13.4. The starting-point equation set is: ! ∂u ∂v $ ∂u ∂η ∂ v ∂η ∂η − fv = −g , + fu = −g . (13.45, 13.46, 13.47) +H# + &=0, ∂t ∂ x ∂t ∂y ∂t "∂ x ∂ y % Apply ∂/∂t to (13.46), multiply (13.47) by f, and add the results to find: " ∂2 % " ∂2η ∂η % 2 (a) +f $ 2 + f ' u = −g $ '. ∂y & # ∂t & # ∂x∂t Multiply (13.46) by –f, apply ∂/∂t to (13.47), and add the results to find: " ∂2 % " ∂2η ∂η % 2 (b) −f $ 2 + f ' v = −g $ '. ∂x & # ∂t & # ∂y∂t Apply ∂/∂x to (a) and ∂/∂y to (b), and add the results to reach: " ∂2 %" ∂u ∂v % " ∂2 % 1 ∂η % " ∂2 ∂2 % ∂η 2 2 " , (c) + f + = + f − = −g + ' $ 2 '$ '$ $ 2 ' $ 2 2' # ∂t &# ∂x ∂y & # ∂t &# H ∂t & # ∂x ∂y & ∂t where the first equality follows from (13.45). Using the final equality, multiply by –H, and collect all the terms to left side of the equation to find: " ∂2 % " ∂2 ∂2 % ∂η 2 " ∂η % (d) + f − gH + =0. $ 2 '$ ' $ 2 2' # ∂t &# ∂t & # ∂x ∂y & ∂t Factor out the common differentiation operation ∂/∂t to reach: # ∂2 ∂ ) ∂2 ∂2 &, * 2 + f 2 − gH % 2 + 2 (-η (x, y, t) = 0 . ∂t + ∂t $ ∂x ∂y '. b) Substitute the trial solution η (x, y, t) = ηˆ exp {i(kx + ly − ω t)} into the part a) differential equation to find:

{

}

−iω −ω 2 + f 2 − gH (−k 2 − l 2 ) ηˆ = 0 . For nonzero ηˆ , this equation has solutions

ω = 0 and ω 2 = f 2 + gH ( k 2 + l 2 ) = f 2 + gHK 2 ,

where K2 = k2 + l2, and the final equation is (13.82). c) The specification ω = 0 implies steady flow, and the steady flow solutions of (13.45) – (13.47) are geostrophic flows.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 13.5. Find the axis ratio of a hodograph plot for a semidiurnal tide in the middle of the ocean at a latitude of 45°N. Assume that the mid-ocean tides are rotational surface gravity waves of long wavelength and are unaffected by the proximity of coastal boundaries. If the depth of the ocean is 4 km, find the wavelength, the phase velocity, and the group velocity. Note, however, that the wavelength is comparable to the width of the ocean, so that the neglect of coastal boundaries is not very realistic. Solution 13.5. Given f = 2Ωsin45° = 2(0.73x10–4 s–1)(0.7071) = 1.03 x10–4 s–1, ω = 2 rev./day = 1.45 x10–4 s–1, and H = 4 km, the axis ratio (as given at the top of page 733) is axis ratio = ω/f = 1.45/1.03 = 1.414. At the given depth the phase speed is: c = [gH]1/2 = [(9.81 m2/s)(4,000m)]1/2 = 198 m/s, and the group speed is the same because it is a shallow water wave. The wavelength is: λ = c(period) = (198 m/s)(12 hrs)(3600 s/hr) = 8554 km. This wavelength is long enough so that the neglect of coast boundaries is an unrealistic approximation.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 13.6. An internal Kelvin wave on the thermocline of the ocean propagates along the west coast of Australia. The thermocline has a depth of 50 m and has a nearly discontinuous density change of 2 kg/m3 across it. The layer below the thermocline is deep. At a latitude of 30°S, find the direction and magnitude of the propagation speed and the decay scale perpendicular to the coast. Solution 13.6. The given information is: f = –2Ωsin30° = –0.73x10–4 s–1, H = 50 m, and Δρ = 2 kg/m3. This is the case of a shallow layer of lighter water, overlying a deep sea. From equation (8.115), the internal gravity wave speed is:

Δρ 2kgm −3 −1 c = g!H = g H = (9.81ms ) 3 (50m) = 0.99ms−1 . −3 ρo 10 kgm These waves propagate southward along the west coast of Australia. The decay scale perpendicular to the coast is the Rossby radius: c 0.99ms−1 Λ= = = 1.36 ×10 4 m = 13.6km . −4 −1 f 0.73×10 s

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 13.7. Derive (13.96) for the vertical velocity w from (4.10), (13.48), (13.49), (13.51), (13.95) by eliminating all other dependent variables. Solution 13.7. The five starting-point equations are: ∂u ∂v ∂w + + =0, ∂x ∂y ∂z ∂u 1 ∂p# ∂v 1 ∂p# , , − fv = − + fu = − ∂t ρ0 ∂x ∂t ρ0 ∂y

(4.10) (13.48, 13.49)

∂ρ " N2 (13.51) − ρ0 w = 0 , and ∂t g ∂w 1 ∂p# gρ # . (13.95) =− − ∂t ρ0 ∂z ρ0 Apply ∂/∂t to (13.48), multiply (13.49) by f, and add the results to find: " ∂2 % 1 " ∂2 p) ∂p) % 2 (a) + f u = − +f $ 2 ' $ '. ρ0 # ∂x∂t ∂y & # ∂t & Multiply (13.48) by –f, apply ∂/∂t to (13.49), and add the results to find: " ∂2 % 1 " ∂2 p) ∂p) % 2 (b) −f $ 2 + f 'v = − $ '. ρ0 # ∂y∂t ∂x & # ∂t & Apply ∂/∂x to (a) and ∂/∂y to (b), and add the results to reach: " ∂2 %" ∂u ∂v % " ∂2 % 1 " ∂2 ∂2 % ∂p) 2 2 " ∂w % , (c) $ 2 + f '$ + ' = $ 2 + f '$ − ' = − $ 2 + 2 ' ρ0 # ∂x ∂y & ∂t # ∂t &# ∂x ∂y & # ∂t &# ∂z & where the first equality follows from (4.10). Now apply ∂/∂t to (13.95), and use (13.51) to substitute for ∂ρ´/∂t: ∂2 w 1 ∂2 p# g ∂ρ # 1 ∂2 p# g $ N 2 ' 1 ∂2 p# (d) = − − = − − ρ w = − − N 2w . & ) 0 ∂t 2 ρ0 ∂z∂t ρ0 ∂t ρ0 ∂z∂t ρ0 % g ( ρ0 ∂z∂t Use the definitions ∇ 2H ≡ ∂2 ∂x 2 + ∂2 ∂y 2 to rewrite the final equality of (c), and rewrite the ends of the extended equality (d): " ∂2 % " ∂2 % 1 2 ∂p) 1 ∂2 p( 2 ∂w 2 , and (e,f) + f = ∇ + N w = $ 2 ' $ 2 ' H ∂t ρ0 ∂z∂t # ∂t & ∂z ρ0 # ∂t & Apply ∂/∂z to (e) and ∇ 2H to (f), and add the results to find: " ∂2 % 2 " ∂2 % 1 2 ∂2 p) 1 2 ∂2 p) 2 ∂ w 2 2 − ∇H =0. $ 2 + f ' 2 + ∇H $ 2 + N ' w = ∇H ρ0 ∂z∂t ρ0 ∂z∂t # ∂t & ∂z # ∂t & Rearrange the terms noting that ∇ 2H + ∂2 ∂z 2 ≡ ∇ 2 : ∂2 w ∂2 w ∇ 2 2 + f 2 2 + N 2 ∇ 2H w = 0 , ∂t ∂z which is (13.96).

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 13.8. Using the dispersion relation m2 = k2(N2 − ω2)/(ω2 − f2) for internal waves, show (N 2 − f 2 )km that the group velocity vector is given by !"cgx , cgz #$ = 2 2 3 2 2 2 2 2 1 2 [ m, −k ] . (m + k ) (m f + k N ) [Hint: Differentiate the dispersion relation partially with respect to k and m.] Show that cg and c are perpendicular and have oppositely directed vertical components. Verify that cg is parallel to u.] Solution 13.8. Start with m2 = k2(N2 − ω2)/(ω2 − f2), and algebraically rearrange it to find:

ω2 = f2 + k2(N2 − ω2)/m2. Here we note that c = (cx, cz) =

(1)

ω (k, m) , so the components of the group velocity must be k + m2 2

found next. To find cgx, differentiate (1) with respect to the horizontal wave number k: ∂ω 2k 2 k 2 # ∂ω & 2ω = 2 ( N − ω 2 ) − 2 % 2ω (. ∂k m m $ ∂k ' Set cgx = ∂ω/∂k, to find: 2k k2 2ω cgx = 2 ( N 2 − ω 2 ) − 2 ( 2ω cgx ) , m m 2 2 or ω cgx ( m + k ) = k ( N 2 − ω 2 ) , or

cgx =

k (N 2 −ω2 )

ω (m2 + k 2 )

.

(2)

From here ω can be eliminated from (2) using (1) to find: km 2 ( N 2 − f 2 ) . cgx = 2 2 32 2 2 2 2 12 (m + k ) (m f + k N ) To find cgy, differentiate (1) with respect to the vertical wave number m: ∂ω k 2 # ∂ω & 1 2 2 2 2ω = 2 % −2ω ( − 2k ( N − ω ) 3 . ∂m m $ ∂m ' m Set cgz = ∂ω/∂m, and rearrange the last equation to find: k2 (N 2 −ω2 ) cgz = − . mω ( m 2 + k 2 ) Here again, ω can be eliminated from (3) using (1) to find: k 2m ( N 2 − f 2 ) . cgz = − 2 2 32 2 2 2 2 12 m + k m f + k N ( ) ( ) Thus we can write: cg = ( cgx , cgz ) =

(N 2 − f 2 )km ( m, −k ) . (m 2 + k 2 )3 2 (m 2 f 2 + k 2 N 2 )1 2

(3)

Fluid Mechanics, 6th Ed.

zero:

Kundu, Cohen, and Dowling

To show that cg and c are perpendicular, it is sufficient to show that their dot product is

(N 2 − f 2 )km ω m, −k ) ⋅ 2 (k, m) = 0 . 2 2 32 2 2 2 2 12 ( (m + k ) (m f + k N ) k + m2 Here the sign of cz is opposite that of cgz , so it follows that the vertical components of c and cg are oppositely directed. Finally, cg and u are parallel because: cgx m u =− = , cgz k w where (13.111) has been used to reach the final equality. cg ⋅ c =

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 13.9. Suppose the atmosphere at a latitude of 45°N is idealized by a uniformly stratified layer of height 10 km, across which the potential temperature increases by 50°C. a) What is the value of the buoyancy frequency N? b) Find the speed of a long gravity wave corresponding to the n = 1 baroclinic mode. c) For the n = 1 mode, find the westward speed of nondispersive (i.e., very large wavelength) Rossby waves. [Answer: N = 0.01279 s−1; c1 = 40.71 m/s; cx = −3.12 m/s] Solution 13.9. The given info. is f = 2Ωsin45° = 2(0.73x10–4 s–1)(0.7071) = 1.03 x10–4 s–1, H = 10 km, and ΔT = 50 °C. At absolute temperature T, the coefficient of thermal expansion for a perfect gas is 1/T. a) Therefore the buoyancy frequency is given by: g dρ dT g dT 9.81ms−2 50° N2 = − = gα = = = 1.635 ×10 −4 s−2 , so N = 0.01279 s–1. ρo dz dz T dz 300K 10 4 m b) From equation (13.71), the internal gravity wave speed corresponding to the nth normal mode is cn = NH/nπ, so c1 = NH/π = (0.01279 s–1)(104 m)/π = 40.71 m/s. c) From Section 13.13, the westward speed of a non-dispersive Rossby wave (corresponding to the n = 1 mode) is: cx ≅ − β c12 f 2 = −(2 ×10 −11 s−1m −1 )(40.71ms−1 )2 (1.03×10 −4 s−1 )2 = −3.12ms−1 .

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 13.10. Consider a steady flow rotating between plane parallel boundaries a distance L apart. The angular velocity is Ω and a small rectilinear velocity U is superposed. There is a protuberance of height h 1( & & C = 0 for n = 1 % ). € p € &−∞ for n < 1& ' * This result shows that there is a stagnation point at a foil's trailing edge when the trailing-edge included angle is non-zero. €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 14.4. Consider an airfoil section in the xy-plane, the x-axis being aligned with the chord line. Examine the pressure forces on an element ds = (dx, dy) on the surface, and show that the net force (per unit span) in the y-direction is c c Fy = − ∫ 0 pu dx + ∫ 0 pl dx where pu and p1 are the pressure on the upper and the lower surfaces and c is the chord length. Show that this relation can be rearranged in the form Fy # x& Cy = = ∫ C p d% ( , € 2 $c' (1/2) ρU c 1 2 where C p = ( p0 − p∞ ) 2 ρU , and the integral represents the area enclosed in a Cp vs x/c diagram, such as Figure 14.8. Neglect shear stresses. [Note that Cy is not exactly the lift coefficient, since the airstream is inclined at a small angle α with respect to the x-axis.] €

(

)

€ Solution 14.4. Here's a nominal drawing of a foil with the x-axis along the chord line of length c. y c ! ds x

dx For a small element ds = (dx, dy) = |ds|(cosθ, sinθ) of the foil's upper surface, the y-direction force per unit span is dFy = −( pu | ds |)cosθ = − pu dx . Thus, if pu and p1 are the pressure on the upper and the lower surfaces and c is the chord length, then c

Fy = ∫€dFy = − ∫ pu dx + 0

c

c

c

∫ pl dx = − ∫ ( pu − p∞ )dx + ∫ ( pl − p∞ )dx . 0

0

0

This can be recast in terms of a vertical force coefficient by dividing by 12 ρU 2c : c c c F & x) c & x) & x) (p − p ) (p − p ) Cy = 1 y 2 = − ∫ 1 u 2 ∞ dx + ∫ 1 l 2∞ dx = − ∫ C p,u d( + + ∫ C p,l d( + = ∫ C p d( + . € 2 ρU c 'c* 0 'c* 'c* ρU c 0 0 2 ρU c 0 2 €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 14.5. The measured pressure distribution over a section of a two-dimensional airfoil at 4° incidence has the following form: Upper Surface: Cp is constant at −0.8 from the leading edge to a distance equal to 60% of chord and then increases linearly to 0.1 at the trailing edge. Lower Surface: Cp is constant at −0.4 from the leading edge to a distance equal to 60% of chord and then increases linearly to 0.1 at the trailing edge. Using the results of Exercise 14.4, show that the lift coefficient is nearly 0.32. Solution 14.5.The distribution of the pressure coefficient is sketched below. Cp E !"% x –!"#

B

A

–!"$ D

C 0.6c

0.4c

Using the result of Exercise 14.3, # x& 1 Cy = ∫ C p d% ( = ABCD + BCE = (0.4)(0.6) + (0.4)(0.4) = 0.32 . $c' 2 The coefficient of lift is: CL = Cy cosα = Cy cos 4° = Cy (0.997564) = 0.3192 , which€ is nearly 0.32.

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 14.6. Zhukhovsky transformation z = ζ + b2/ζ transforms a circle of radius b, centered at the origin of the ζ-plane, into a flat plate of length 4b in the z-plane. The circulation around the cylinder is such that the Kutta condition is satisfied at the trailing edge of the flat plate. If the plate is inclined at an angle α to a uniform stream U, show that " −iα b 2 +iα % iΓ ln ζ e−iα , where Γ = (i) The complex potential in the ζ-plane is w = U $ζ e + e ' + ζ # & 2π

(

)

4πUbsinα. Note that this represents flow over a circular cylinder with circulation, in which the oncoming velocity is oriented at an angle α. (ii) The velocity components at point P (−2b, 0) in the ζ-plane are !" 43 U cos α, 94 U sin α #$ (iii) The coordinates of the transformed point Pʹ′ in the xy-plane are [−5b/2, 0]. (iv) The velocity components at [−5b/2, 0] in the xy-plane are [Ucosα, 3Usinα]. Solution 14.6. The three complex planes for this problem appear as: !´-plane

z-plane

!-plane ! = !´ei"

U

z = !#+ b2/!

P (-2b,0)

"

P´

(-5b/2,0)

"

U

(i) From equation (14.11), the circulation will be Γ = 4 πUbsin α . In the ζ´-plane choose U to flow parallel to the real axis. Then, the transformation ζ = ζ #e iα rotates the ζ´-plane counterclockwise to produce the ζ-plane. The complex potential is: w = U (ζ " + b 2 ζ€") + (iΓ 2π ) ln ζ " = U (ζe−iα + b 2e +iα ζ€) + (iΓ 2π ) ln ζe−iα = U (ζe−iα + b 2e +iα ζ ) + (iΓ 2π )[ln ζ − iα ]

(ii) The complex velocity is:

At point P,

dw = U (e−iα − b 2e +iα ζ 2 ) + iΓ 2π ζ . dζ

€

# dw & −iα 2 +iα 2 % ( = U (e − b e (−2b) ) + iΓ 2π (−2b) $ dζ 'P€ 0 1 = U/ cosα − isin α − (cosα − isin α )2 − i(4 πUbsin α ) 4 πb . 1 4 3 5 3 9 = U cosα − i sin α − iU sin α = U cos α − i sin α = u − iv 4 4 4 4 3 9 so: u = U cosα , and v = U sin α . 4 4 2 (iii) Based on the transformation, z = ζ + b ζ , the transformed point is € €

€

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

zP" = −2b + b 2 (−2b) = −5b /2 . (iv) The complex velocity at (–5b/2, 0) in the z-plane is: −1 −1 dw dw dζ dw # dz & dw # b 2 & dw # ζ 2 & = = % ( = %1− 2 ( = % ( dz dζ€dz dζ $ dζ ' dζ $ ζ ' dζ $ ζ 2 − b 2 ' #3 &# (−2b) 2 & 9 = % U cos α − i sin α (% ( = U cos α − i3sin α = u − iv. $4 '$ (−2b) 2 − b 2 ' 4 where ζ = (–2b, 0) has been used as the corresponding point in the ζ-plane. €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 14.7. In Figure 14.13, the angle at Aʹ′ has been marked 2β. Prove this. [Hint: Locate the center of the circular arc in the z-plane.] Solution 14.7. From the discussion following Figure 14.12, point C is located on the imaginary axis in the z-plane at –2bcot2β. z-plane

!-plane

A a

z = !"+ b2/!

A´

B

#

–2b

B´

$ &

D

b % C –2bcot2#

From triangle A´DC tan γ =

2b = tan2β , so γ = 2β. 2bcot 2β

Plus, the full angle at A´ is π/2 so δ +θ =

and A´DC is a right € triangle so: δ+γ =

Comparing (i) and (ii), requires θ € = 2β . €

€

π , 2

π = δ + 2β . 2

(i) (ii)

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 14.8. Ideal flow past a flat plate inclined at angle α with respect to a horizontal free stream produces lift but no drag when the Kutta condition is applied at the plate's trailing edge. However, pressure forces can only act in the plate-normal direction and this direction is not perpendicular to the flow. Therefore, to achieve zero drag, another force must act on the plate. This extra force is known as leading-edge suction and its existence can be assessed from the potential for flow around the tip of a flat plate that is coincident with the x-axis for x > 0. In twodimensional polar coordinates, this velocity potential is φ = 2U o ar cos(θ 2) where Uo and a are velocity and length scales, respectively, that characterize the flow. a) Determine ur and u , the radial and angular-directed velocity components, respectively. b) If the pressure far from the origin is p∞, determine the pressure p at any location (r, θ). € c) Use the given potential, a circular control volume of radius ε centered at the origin of coordinates, and the control volume version of the ideal flow momentum equation, ∫ ρu(u ⋅ n)dξ = − ∫ pndξ + F , to determine the force F (per unit depth into the page) that holds θ

C

€

C

the plate stationary when ε → 0 . Here, n is the outward unit normal vector to the control volume surface, and dξ is the length increment of the circular control surface. d) If the plate is released from rest, in what direction will it initially accelerate? €

Solution 14.8. a) Directly differentiate the potential φ = 2U o ar cos(θ 2) . ur = ∂φ ∂r = 2U o a cos(θ 2) ⋅ r−1 2 2 = U o ( a r)

12

cos(θ 2) , and

uθ = (1 r)(∂φ ∂θ ) = (1 r)2U o ar [−sin(θ 2) ⋅ (1 2)] = −U o ( a r)

12

sin(θ 2)

b) Use the Bernoulli equation to determine€p = p(r, θ) : 1 1 1 2 2 2 2 2 2 € p∞ = p + 2 ρ( ur + uθ ) = p + 2 ρU o ( a r)(cos (θ 2) + sin (θ 2)) = p + 2 ρU o ( a r) . € the pressure does not depend on the angle θ: Thus,

p − p∞ = − 12 ρU o2 ( a r) , but it does become very negative as r approaches zero. € c) To evaluate the Cartesian components of F, the Cartesian components of the velocity field are needed. Convert the part a) results using: ux = ur cosθ − uθ sin θ , and uy = ur sin θ + uθ cos θ , to € find: 12 ux = U o ( a r) [cos(θ 2) cos θ + sin(θ 2) sin θ ] , and uy = U o (€a r) € €

12

[cos(θ 2) sinθ − sin(θ€2) cosθ ] .

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

These can be simplified to: u x = U o ( a r)1 2 cos(θ 2) , and uy = U o ( a r)1 2 sin(θ 2) , using trigonometric identities. Here n = e r = e x cosθ + e x sin θ , so u ⋅ n = u ⋅ e r = ur . Thus, the CV equation can be written in terms of x- and y-components: 2π 2π ∫ 0 ρ(uxe€x + uye y )urεdθ = − ∫ 0 p(e x cosθ€+ e x sinθ )εdθ + Fxe x + Fye y . where dξ = εdθ. The pressure does not depend on θ, and € € the integrals of cosθ and sinθ from 0 to 2π are both zero so the pressure integration drops out leaving: 2π

∫0

€

(

)

12

12

ρ U o (a ε) cos(θ 2) U o ( a ε) cos(θ 2)εdθ = Fx , and 2π

∫0

(

)

12

12

ρ U o (a ε) sin(θ 2) U o ( a ε) cos(θ 2)εdθ = Fy .

The ε contour-radius factors cancel, so the integrands can be simplified to: € ρU 2 a 2 π cos 2 θ 2 dθ = F , and ρU 2 a 2 π sin θ 2 cos θ 2 dθ = F . ( ) ( ) ( ) o ∫0 x o ∫0 y Use the double-angle trigonometric identities to evaluate the integrals to find: € 2π 1 2

ρU o2 a ∫ 0

(cosθ + 1)dθ = πρU o2 a = Fx ,

2π 1 sinθ 2

and ρU o2 a ∫ 0

(

)dθ = 0 = F . y

€ final answer suggests that the horizontal force, known as The€fact that ε does not appear in either leading edge suction, found in part d) will exist for any ε, even ε → 0 . This force is applied to the tip of the plate and arises from the singularity in the potential at r = 0. € Interestingly, the real-world effects of €this singularity are exploited for both natural and anthropogenic flight. The leading edges of real airfoils are rounded so that the flow remains € attached to the foil's surface and does not have a singularity. Attached flow passing around a finite-radius-of-curvature leading edge produces a leading-edge suction force of the size predicted here that increases the lift and reduces the drag of real airfoils. The only superfluous part of the above discussion and analysis is the limit ε → 0 . d) Fx holds the plate stationary by pushing it to the right. Therefore, if Fx is not applied, the plate will accelerate to the left. €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 14.9. Consider a cambered Zhukhovsky airfoil determined by the following parameters: a = 1.1, b = 1.0, and β = 0.1. Using a computer, plot its contour by evaluating the Zhukhovsky transformation. Also plot a few streamlines, assuming an angle of attack of 5°. Solution 14.9. In the case the point Q at the center of the circle in the ζ-plane lies at: (Qx ,Qy ) = (b − cos β ,asin β ) ≅ (−0.0945,0.1098) . Thus, horizontal and vertical coordinates of the circle are given by: (ζ x ,ζ y ) = (Qx + acos θ,Qy + asin θ ) , or as a complex € variable: ζ = ζ x + iζ y = Qx + acosθ + i(Qy + asin θ ) where 0 ≤ θ ≤ 360°. Using the Zhukhovsky transformation: € b2 z = x + iy = ζ + ζ € then leads to: b 2ζ b 2ζ x = ζ x + 2 x 2 and y = ζ y − 2 y 2 . ζx + ζy ζx + ζy € where (x, y) specifies the coordinates of the foils surface. Use of a spread sheet program for evaluating and plotting with a = 1.1, b = 1.0, and β = 0.1 (radians), for 1° steps of θ produces the following plot for the foil shape: € € &%

"#

%$"% !'#

%$!'%

!&#

%$!&%

!"#

%$"% !"#

%$in the z-plane.

"#

%$&%

&#

%$'%

'#

%$ Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 14.10. A thin Zhukhovsky airfoil has a lift coefficient of 0.3 at zero incidence. What is the lift coefficient at 5° incidence? Solution 14.10. From (14.12), the lift coefficient of a Zhukhovsky airfoil is: CL = 2π (α + β ) . If CL = 0.3 when α = 0, then β = 0.3/2π. Therefore when α = 5°, # 5° 0.3 & CL = 2π % π+ ( = 0.848 . $180° 2π ' €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 14.11. The simplest representation of a three-dimensional aircraft wing in flight is the rectangular horseshoe vortex. a) Calculate the induced downwash at the center of the wing. b) Assuming the result of part a) applies along the entire wingspan, estimate CDi , the lift-induced coefficient of drag, in terms of the wing’s aspect ratio: AR = s2/A, and the wing’s coefficient of lift CL = L 12 ρU 2 A , where A is the planform area of the wing. c) Explain why the result of part b) appears to surpass the performance € of the optimal elliptic lift distribution.

(

)

€

Solution 14.11. a) Use the result of exercise 14.10. For the rectangular horseshoe vortex, each of Γ Γ the trailing vortices induces a vertical velocity of w = − . (cos90° − cos180°) = − 4 π (b /2) 2πb Γ Thus, the total downwash velocity will be twice this: w total = − . πb b) If this downwash is constant along the whole span, the wing’s lift will be decreased because € −w total Γ the incoming stream will be rotated downward through the small angle ε = , so the = U πbU € Γ Γ Γ2 lift-induced drag will be: Di ≅ Lε = L = ρUΓb = ρ . The coefficient of drag will be: πbU πbU π 2 2 2 2 2 % Di ρΓ π ρUΓ ρUΓb CL2 € ( S CD,i = 1 = = = = ' * ρU 2 S 12 ρU 2 S 12 πρ 2U 4 S & 12 ρU 2 S ) 2πb 2 2πAR 2 c) The induced € drag coefficient of part b) is half as large as that obtained from the optimal elliptic lift distribution on a finite wing because the variation of the induced downwash along the wingspan was not taken into account. The two trailing vortices induce larger downwash € near the wing tips than the estimate produced from the arithmetic in part a). In fact, velocities the answer to part a) is the minimum downwash along the wing.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 14.12. The circulation across the span of a wing follows the parabolic law

(

2

)

Γ = Γ0 1− (2y s) . Calculate the induced velocity w at midspan, and compare the value with

that obtained when the distribution is elliptic. €

€

dΓ = −8Γ0 y s2 . Then, from (14.13) dy +s 2 +s 2 & y1 − s 2 )/ +s 2 1 dΓ dy 8Γ0 ydy 2Γ0 2Γ0 , w(y1 ) = = = y + y ln y − y = s + y ln . [ 1 ( 1)]−s 2 πs2 ∫ ∫ +1. 1 ( 4 π −s 2 dy y1 − y 4 πs2 −s 2 y − y1 πs2 ' y1 + s 2 *0 € distribution varies along the€span and at midspan, y = 0, it is: This downwash 1 2Γ0 . w(0) = πs In contrast, the constant downwash for an elliptic distribution, Γ 2s , is lower.

(

2

)

Solution 14.12. From Γ = Γ0 1− (2y s) , determine

€

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 14.13. An untwisted elliptic wing of 20-m span supports a weight of 80,000 N in a level flight at 300 km/hr. Assuming sea level conditions, find (i) the induced drag and (ii) the circulation around sections halfway along each wing.

€

Solution 14.13. (i) An untwisted wing with an elliptic area is expected to have an elliptic circulation distribution. The induce drag in this case is given by (14.25): 2L2 2(8 ×10 4 N) 2 Di = = = 1.22kN , πρU 2 s2 π (1.2kgm −3 )(83.3ms−1 ) 2 (20m) 2 where 300 km/hr has been converted to 83.3 m/s. (ii) Also from (14.25), the maximum circulation Γ1 is π Di = € ρΓ12 and this implies Γ1 = 8Di ρπ = 8(1.22 ×10 3 N) π (1.2kgm −3 ) = 50.9m 2 s−1 . 8 Halfway along each wing: 2 12 $ $ 2(±s /4) ' 2 '1 2 m2 $ $ 1' ' m2 . Γ(±s /4) = Γ 1− = 50.9 1− = 44.1 & ) & ) & )) € 1& & &% 2 )( ) % ( s s s % ( % (

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 14.14 . A wing with a rectangular planform (span = s, chord = c) and uniform airfoil section without camber is twisted so that its geometrical angle, αw, decreases from αr at the root 2

(y = 0) to zero at the wing tips (y = ± s/2) according to the distribution: α w (y) = α r 1− (2y s) . a) At what global angle of attack, αt, should this wing be flown so that it has an elliptical lift distribution? The local angle of attack at any location along the span will be αt + αw. Assume the two-dimensional lift curve slope of the foil section is K. € at the angle of attack determined b) Evaluate the lift and the lift-induced drag forces on the wing in part a) when: αr = 2°, K = 5.8 rad.–1, c = 1.5 m, s = 9 m, the air density is 1 kg/m3, and the airspeed is 150 m/s Solution 14.14. a) For an elliptical lift distribution without camber (β = 0), only the first two terms of the finite wing lifting line equation (14.20) are needed: ∞ $ K cKn ' Ucα = ∑&1+ )Γn sin(nγ ) . 2 4ssin γ ( n=1% The first two terms of this non-traditional Fourier series “solution” are obtained by matching the left side of the equation to the right side evaluated at n = 1. $ K cK ' cK Ucα = &1+ Γ1 € )Γ1 sin(γ ) = Γ1 sin(γ ) + 2 4s % 4ssin γ ( 2

€

For the wing twist specified in the problem statement, α = α t + α w = α t + α r 1− (2y s) . Now, switch to the angle coordinate, y = −( s 2) cos γ , of the lifting line equation so that α = α t + α r sin€ γ , and plug this into the last equation: K cK Uc (α t + α r sin( Γ1. € γ )) = Γ1 sin(γ ) + 2 4s € Require equality of the constant and sinγ terms on both sides of the equation: K cK K Ucα t = + Γ1 , and Ucα r = Γ1. 2 4s 2 € Eliminate Γ1 and solve for αt; α t = Kcα r (4s) . Note that this angle is positive so the wing must be pitched slightly upward. For the parameters given in part b) the value of αt is 0.48°. πs πsc b) The lift will be: € L = ρUΓ1 = ρU 2Kα € r , so 8 € 4 π (9m)(1.5m) L= (1kg /m 3 )(150m /s) 2 (5.8 /rad.)(2°⋅ π /180°) = 24,150 N 8 π π The lift-induced drag will be: Di = ρΓ12 = ρU 2c 2K 2α r2 , so € 8 32 π 2 3 Di = (1kg /m )[(150m /s)(1.5m)(5.8 /rad.)(2° ⋅ π /180°)] = 204 N € 32 At a flight speed of 150 m/s, overcoming this drag force requires ~30 kW (40 hp). €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 14.15. Consider the wing shown in Figure 14.25. If the foil section is uniform along the span and the wing is not twisted, show that the three-dimensional lift coefficient, CL,3D is related to the two-dimensional lift coefficient of the foil section, CL,2D, by: CL,3D = CL,2D (1+ 2 Λ) , where Λ = s2/A is the aspect ratio of the wing.

€

Solution 14.15. The wing shown in Figure 14.24 has an elliptical planform. Thus, with a uniform € and constant downwash w. The foil section and no twist, it will have an elliptical lift distribution lift force L3D of the three-dimensional wing can be calculated in terms of an integral of the circulation Γ(y) at each span location y. Starting from (14.15) and (14.17) with β = 0, this is: +s 2 +s 2 ' ' K w* K w *+s 2 K w* 2 2' L3D = ρU ∫ Γ(y)dy = ρU ∫ c(y))α − , dy = ρU )α − , ∫ c(y)dy = ρU 2 )α − , A . ( ( ( 2 U+ 2 U +−s 2 2 U+ −s 2 −s 2 The final two equalities follow because the downwash velocity w is constant for an elliptical lift distribution, and the integral of the cord c(y) over the span is the planform area A. The two ends of this extended equality imply: % L3D w( (&) CL,3D = = K 'α − * . 2 (1 2)ρU A & U ) Here, the downwash velocity is given by (14.24), w = Γ1/2s, and Γ1 is related to the total lift force L3D of the three-dimensional wing by (14.21): πs L3D = ρUΓ1 . € 4 Therefore (&) becomes: & ) & & ) 1 4L3D ) A L3D 1 CL,3D = K (α − = K (α − CL,3D + , + = K (α − 2 2 + ' * 2sU€πsρU * πs (1 2) ρU A * πΛ ' ' where Λ = s2/A is the wing's aspect ratio. Solving for CL,3D produces: CL,2D CL,2D Kα , CL,3D = ≅ = 1+ K π Λ 1+ 2 π π Λ 1+ 2 Λ ( ) ( ) ( ) € where the angle of attack α is measured from the zero-lift orientation of the wing, and the second equality is approximate because the foil section's two-dimensional lift-curve slope K in viscous flow may be a little smaller than its inviscid value K = 2π. €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 14.16. The wing-tip vortices from large heavy aircraft can cause a disruptive rolling torque on smaller lighter ones. Lifting line theory allows the roll torque to be estimated when the small airplane’s wing is modeled as a single linear vortex with strength Γ(y) that resides at x = 0 between y = –s/2 and y = +s/2. Here, the small airplane’s wing will be presumed rectangular (span s, chord c) with constant foil-shape, and the trailing vortex from the heavy airplane’s wing € will be assumed to lie along the x-axis and produce a vertical velocity distribution at x = 0 given Γ" by: w(y) = [1− exp(− y )] . To simplify your work for the following items, ignore the 2πy trailing vortices (shown as dashed lines) from the small airplane’s wing and assume U >> w. a) Determine a formula for the rolling moment, M = €

+s 2

∫−s 2 ρUyΓ(y)dy , on the small aircraft’s

wing in terms of Γ´, s, c,  , the air density ρ, the flight speed of the small aircraft U, and the liftcurve slope of the small aircraft’s wing section K = dCL,2D dα , where α is the small-aircraftwing angle of attack. € b) Calculate M when ρ = 1.2 kg/m3, U = 150 m/s, K = 6.0/rad, b = 9 m, c = 1.5 m, Γ´ = 50 m2/s, € and s/(2  ) = 1. Comment on the magnitude of this torque. € z!

w(y)!

y!

€

!´!

U x!

Solution 14.16. a) Use the second lifting line equation, Γ(y) = 12 UcK (α + w(y) U ) , and the Γ" specified vertical velocity, w(y) = [1− exp(− y )] to determine the roll torque on the small 2πy aircraft’s wing: € , +s 2 +s 2 ) 1 Γ' 2 M = ∫ −s 2 ρUyΓ(y)dy = ρU ca2D ∫ −s 2 +α + 1− exp(− y ) . ydy . 2 2πUy * € The α-term inside the integral has the wrong symmetry and doesn’t contribute to the moment so: 0 +s 2 ρUcKΓ# +s 2 ρUcKΓ# M= 1− exp(− y ) dy = y − e +y  ]−s 2 + [ y + e−y  ] 0 ∫ [ −s 2 4π € 4π ' * ρUcKΓ# ρUcKΓ# s s −s 2 = + + e−s 2 − , = )− + + e (s − 2 + 2e−s 2 ) + 4π ( 2 2 4π ρUcsΓ# ' 2 −s 2 * = )1− [1− e ],+ 4π ( s (1.2kg /m 3 )(150m /s)(1.5m)(6.0 /rad)(9m)(50m 2 /s) b) B = 1− [1− e−1 ] = 21.3 kN-m 4π This roll torque is large enough to cause loss of pilot control for a small aircraft, even if it is felt for only a short period of time.

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[

€ € €

€

]

]

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Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 14.17. Consider the ideal rectilinear horseshoe vortex of a simple wing having span s. Use the (x, y, z) coordinates shown for the following items. a) Determine a formula for the induced vertical velocity w at (x, y, 0) for x > 0 and y > 0. b) Using the results of part a), evaluate the induced vertical velocity at the following three locations (s, 0, 0), (0, s, 0), and (s, s, 0) c) Imagine that you are an efficiency-minded migrating bird and that the rectilinear horseshoe vortex shown is produced by another member of your flock. Describe where you would choose to center your own wings. List the coordinates of the part b) location that is closest to your chosen location. Solution 14.17. The induced vertical velocity w at (x, y, 0) for x > 0 and y > 0 will be the result of the sum of the induced velocities from (1) the port-side vortex located at y = –s/2, (2) the bound vortex of the wing located at x = 0, and (3) the starboard-side vortex located at y = +s/2. All three vortices have strength Γ. Using the angles in the planform drawing provided. The sum of induced velocity from vortices (1) and (2) and (3) is: Γ[cosψ11 − cosψ11 ] Γ[cosψ 21 − cosψ 22 ] Γ[cosψ 31 − cosψ 32 ] , w(x, y,0) = − − + 4 π ( y + s 2) 4 πx 4 π ( y − s 2) where the signs are determined by the y! sense of rotation of each vortex. The (x, y)! !22! various angles can be specified in terms of € !32! the coordinates: !31! (3)! +s/2! cosψ11 = (X − x) (X − x) 2 + (y + s /2) 2 , (2) !

!21!

!12! –s/2!

x!

!11!

(1)! X!

cosψ12 = −x

€ €

cosψ 21 = (y + s /2)

x 2 + (y + s /2) 2 ,

cosψ 22 = (y − s /2)

x 2 + (y − s /2) 2 ,

cosψ 31 = x

€

x 2 + (y + s /2) 2 ,

x 2 + (y − s /2) 2 , and

cosψ 32 = −(X − x)

(X − x) 2 + (y − s /2) 2 .

In the limit that the trailing vortices are long,€ X → ∞ , the first and last cosines above simplify to +1 and –1. Thus, € w=

Γ % −1− x ' 4 π '&

x 2 + (y + s /2) 2 1+ x y + s€ /2 y − s /2 − + + y + s /2 x €x 2 + (y + s /2) 2 x x 2 + (y − s /2) 2

x 2 + (y − s /2) 2 ( * y − s /2 *)

b) Evaluate w at (s, 0, 0) €

€ €

Γ % −1− s s2 + (s /2) 2 s /2 −s /2 1+ s s2 + (s /2) 2 ( ' * − + + 4 π '& s /2 −s /2 *) s s2 + (s /2) 2 s s2 + (s /2) 2 Γ %−1−1 1+ (1/2) 2 1/2 1/2 1+ 1 12 + (1/2) 2 ( ' * w= − − + 4 πs '& 1/2 −1/2 *) 1+ (1/2) 2 1+ (1/2) 2 Γ % 4 1 1 4 ( Γ Γ w= −2 − − − −2− =− 1+ 5 = − [ 3.236] ' * 4 πs & πs πs 5 5 5 5)

w=

[

Evaluate w at (0, s, 0); here ψ22 = ψ21 = 0 so there is no contribution from vortex (2). w=

€

€

]

Γ % −1 1 ( Γ % −2 ( Γ %1( + = + 2* = + ' * ' * ' ) 4 π & s + s /2 s − s /2 ) 4 πs & 3 πs & 3)

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Evaluate w at (s, s, 0) Γ % −1− s s2 + (3s /2) 2 s + s /2 s /2 1+ s s2 + (s /2) 2 ( ' * − + + 4 π '& 3s /2 s /2 *) s s2 + (3s /2) 2 s s2 + (s /2) 2 Γ %−1−1 1+ (3/2) 2 3/2 1/2 1+ 1 1+ (1/2) 2 ( ' * w= − + + 4 πs '& 3/2 1/2 *) 1+ (3/2) 2 1+ (1/2) 2 Γ % 2 4 3 1 4 ( Γ %1 5 13 ( Γ w= − − − + + 2 + = + − ' * = + [1.151] ' * 4 πs & 3 3 13 12 ) πs 13 5 5 ) πs & 3 4 w=

€ € €

c) An efficiency minded bird would like to place its wings in a region of upward induced velocity so that its lift-induced drag is minimized by a beneficial interaction with the vortex system of a companion migrating bird. From the results of part b), the rectilinear horseshoe vortex does produce upward induced velocities for (0, s, 0), and (s, s, 0), but the magnitude is greater for (s, s, 0). Thus, the coordinate location (s, s, 0) is the preferred one, and this choice is verified by observations of real migrating birds that fly in Λ-shaped formations.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 14.18. As an airplane lands, the presence of the ground changes the plane’s aerodynamic performance. To address the essential features of this situation, consider uniform flow past a horseshoe vortex (heavy solid lines below) with wingspan b located a distance h above a large flat boundary defined by z = 0. From the method of images, the presence of the boundary can be accounted for by an image horseshoe vortex (heavy dashed lines below) of opposite strength located a distance h below the boundary. a) Determine the direction and the magnitude of the induced velocity at x = (0, 0, h), the center of the wing. b) Assuming the result of part a) applies along the entire wingspan, estimate L and Di , the lift and lift-induced drag, respectively, in terms of b, h, Γ, and ρ = fluid density. c) Compare the result of part b) to that obtained for the horseshoe vortex without a large flat € surface: L = ρUΓb and Di = ρΓ 2 π . Which configuration has more lift? Which one has less drag? Why? €

€

Solution 14.18. a) Five vortices will contribute to the induced velocity at x = (0,0,h) . Number these as follows: 1 = starboard wingtip vortex 2 = starboard wingtip image vortex € 3 = port wingtip vortex 4 = port wingtip image vortex 5 = main wing image vortex The geometrical layout ensures that the induced velocities will not point in the same direction, so work through each one individually using the Biot-Savart induced velocity law from exercise 4.10: e × eRΓ ui = ω (cosθ1 − cosθ 2 ) 4π where i is the vortex number (not a coordinate index), eω points along the vortex direction, e R points along a line perpendicular to the vortex that intersects the point of interest, and θ1 & θ2 are the polar angles between the eω and the line connecting the ends of the vortex to the point of € interest. Here the first four vortices extend from x = 0 to +∞, so the cosine terms above are: € € cos(π/2) – cos(π) = +1 u1 = −e zΓ ( 4 π (b /2)) = −e zΓ (2πb) €

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

$ ' b /2 2h &e z ) + e y 4 π (b /2) 2 + 4h 2 &% (b /2) 2 + 4h 2 (b /2) 2 + 4h 2 )( u3 = −e zΓ ( 4 π (b /2)) = −e zΓ (2πb) % ( Γ b /2 2h 'e z * u4 = − ey € 4 π (b /2) 2 + 4h 2 '& (b /2) 2 + 4h 2 (b /2) 2 + 4h 2 *) € Here the extra factor in [,] brackets for the second and fourth vortex is merely a unit vector rotated to the induced velocity direction. The fifth, vortex induces a velocity that opposes the on-coming free stream and its angular extent is different from the four other vortices. € ( −Γe −Γe x % b /2 −b /2 b /2 x ' *= u5 = − 4 π (2h) '& (b /2) 2 + 4h 2 (b /2) 2 + 4h 2 *) 2π (2h) (b /2) 2 + 4h 2 The sum of these five terms is: % ( Γe Γ b /2 Γe x b /2 'e z *− utotal = − z + πb 2π (b /2) 2 + 4h 2 '& (b /2) 2 + 4h 2 *) 2π (2h) (b /2) 2 + 4h 2 € u2 =

Γ

( Γ % 4h 2 Γ b /2 e x = −uze z − ux e x & )e z − 2 2 πb ' (b /2) + 4h * 4 πh (b /2) 2 + 4h 2 € b) The lift force will be: L = ρ(U − w x )Γb , and the induced drag force will be: utotal = −

* wz Γ2 ' 4h 2 Di = Lε = ρ(U − w x )Γb = ρΓbw z = ρ ( + € U − wx π ) (b /2) 2 + 4h 2 , € → 0, the induced drag disappears. Note, that as h/b c) For constant speed flight close to the ground surface, there is less lift and less drag for a fixed value of Γ. The lowering of the lift occurs because the induced velocity from the main wing’s € image vortex slows the on-coming stream at the location of the real wing. The induced drag is lower because the image tip vortices produce upwash that partially counter acts the downwash from the actual wingtip vortices. For actual aircraft, there are two effects that more than offset the apparent reduction in lift found in part b) and mentioned here. (i) Wing circulation Γ increases as the aircraft approaches the ground because the downwash decreases, and less downwash means an increase in the wing’s angle of attack. (ii) With a constant engine-throttle setting, the loss of induced drag as h/b → 0 causes the aircraft to mildly accelerate, thereby increasing U. The aircraft’s passengers may even feel this mild acceleration, and it may lead to a steady-state speed that prevents the aircraft from descending further and touching down. When this happens the aircraft is said to be flying with or in ground effect. Once the aircraft is low enough for ground effect to be apparent, the pilot must typically reduce the wing's CL by using its spoilers or by lowering the engine throttle setting in a controlled fashion to achieve a safe smooth landing. Observant airline passengers will notice that commercial airliners sometimes fly at very low altitudes over the landing runway for an unexpectedly long period of time before touching down. This apparent delay in touching down is merely the time necessary for pilot to adjust the aircraft's trim to continue its decent while flying in ground effect. [Spoilers are flaps on the top of the wing that spoil the airflow on the suction side of the wing; they are used to increase form drag and reduce lift in a controlled manner.]

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 14.19. Before modifications, an ordinary commercial airliner with wingspan s = 30 m generates two tip vortices of equal and opposite circulation having Rankine velocity profiles (see (3.28)) and a core size σo = 0.5 m for test-flight conditions. The addition of wing-tip treatments (sometimes known as winglets) to both of the aircraft's wing tips doubles the tip vortex core size at the test condition. If the aircraft's weight is negligibly affected by the change, has the liftinduced drag of the aircraft been increased or decreased? Justify your answer. Estimate the percentage change in the induced drag. Solution 14.19. For the conditions stated, the lift-induced drag of the aircraft has decreased by the addition of winglets. This contention is supported by an approximate analysis of the kinetic energy of the vortex flow behind the aircraft. In an unbounded nominally-quiescent fluid medium, an increase in length dl of a single semi-infinite line vortex increases the kinetic energy of the fluid by, dKE, where: ∞ 1 dKE = ρ ∫ 0 uθ2 2πrdr dl . 2 A Rankine vortex is defined by: ' Γ 2πσ 2 ) r for r ≤ σ * uθ (r) = (( + for r > σ , ) Γ 2πr € where σ is the core size, and r is the distance perpendicular to the vortex axis. Thus, the kinetic energy increment for Rankine vortex is: 2 R 2 & Γ )2 1 /σ & Γ ) 2 € dKE = ρ1 ∫ ( r 2 π rdr + lim 2 π rdr 4dl . + ( + ∫ R →∞ ' 2πr * 2 0 0 ' 2πσ 2 * σ 3 The upper limit of the second integral is problematic for a single vortex. However, the airliner will have two wing-tip vortices of equal and opposite sign so evaluation of the limit is not necessary at this point. € Performing the two integrations yields: 2 ) σ R 1 Γ 1 1 , Γ2 ) 1 σ 4 R, 3 dKE = ρ + 4 ∫ r dr + lim ∫ dr.dl = ρ + lim ln r dl ( ) + σ. R →∞ 2 2π *σ 0 4 π *σ 4 4 R →∞ σ r . , Γ2 )1 =ρ + lim (ln R) − ln σ .dl 4 π +* 4 R →∞ Use the final part of this equality, divide by the time increment dt, and recognize dl/dt = U = the aircraft's velocity, to find: , dKE Γ 2U ) 1 =ρ + lim (ln R) − ln σ .. + dt 4 π * 4 R →∞ For an aircraft that produces two counter rotating tip vortices, we can write the following approximate equation + dKE Γ 2U ( 1 ≅ 2ρ + ln Rc − ln σ - ≅ UDi € * , dt 4π ) 4 where Rc is the distance perpendicular to the aircraft's flight path that is large enough for the induced flow from the aircraft's tip vortices to have effectively cancelled out. The final equality here follows because the power necessary to overcome the aircraft's induced drag is that € the aircraft's tip-vortex velocity field. necessary to produce

[

€

]

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Thus, for comparable flight speed, air density, and aircraft weight, an estimate of the induced drag ratio can be obtained from: $1 ' + ln Rc − ln σ ) & [Di ] modified % 4 (modified 0.25 + ln(90) − ln(1) 4.75 = ≈ = = 0.87 , 0.25 + ln(90) − ln(0.5) 5.44 [ Di ] original $ 1 + ln R − ln σ ' c &% 4 )( original where Rc has been estimated as three times the wing-span of 30 m. Thus, a ~10% (or more) reduction in the induced drag is estimated for a doubling of the tip-vortex core radius from 0.5 m to 1.0 m. €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 14.20. Determine a formula for the range, R, of a long-haul jet-engine aircraft in steady level flight at speed U in terms of: MF = the initial mass of usable fuel; MA = the mass of the airframe, crew, passengers, cargo, and reserve fuel; CL/CD = the aircraft's lift-to-drag ratio; g = the acceleration of gravity; and η = the aircraft's propulsion system thrust-specific fuel consumption (with units of time/length) defined by: dMF/dt = –ηD, where D = the aircraft's aerodynamic drag. For simplicity, assume that U, the ratio CL/CD, and η are constants. [Hints. If M(t) is the instantaneous mass of the flying aircraft, then L = Lift = Mg, M = MF + MA, and dM/dt = dMF/dt. The final formula is known as the Breguet range equation.] Solution 14.20. For steady level flight, an aircraft's engine thrust T will equal its drag D. Thus, the beginning equation is: dM dM f = = −η D . dt dt Use the two ends of this extended equality, and divide on the left by M and on the right by L/g. This leads to: 1 dM g 1 = −η D = − gη . M dt L CL CD Separate the differential on the left and recognize that Udt = ds = an element of path length along the flight: MA 1 1 gη t 1 gη R dM = − U dt = − ∫ ∫ ∫ ds , CL CD U 0 CL CD U 0 M F +M A M where t is the time of steady level flight and R is the range of the flight. Perform the integrations to find: ! MA $ 1 gη ln # R. &=− CL CD U " MA + MF % Solve for R to reach the final form: U (CL CD ) ! M F $ R= ln #1+ &. gη " MA % Interestingly, air density does not explicitly enter this formula. Plus, the apparent proportionality between R and U may be somewhat misleading. The aircraft's lift must balance its weight so higher U must be paired with lower CL, and lower U must be paired with higher CL. Plus, CD tends to increase with increasing CL, too (see Figure 14.28), so maximizing the range of a jet engine aircraft means achieving a high lift-to-drag ratio at high speed.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 15.1. Use (15.4), (15.5), and (15.6) to derive (15.7) when the body force is spatially uniform and the effects of viscosity are negligible. Solution 15.1. The goal is develop an equation with an inhomogeneous-medium convected-wave operator acting on p on the left side with source terms involving q, fi, and ∂ui/∂xj on the right. Dp Dρ First, use (15.6), , to eliminate Dρ/Dt from (15.4): = c2 Dt Dt 1 Dρ ∂ui 1 Dp ∂ui 1 Dp ∂u , or + =q= 2 + = q− i , 2 ρ Dt ∂xi ρ c Dt ∂xi ρ c Dt ∂xi where the final equation is just a rearrangement of the first. Apply D/Dt to the second equation to find: D ! 1 Dp $ Dq D ∂u j . (a) − # &= Dt " ρ c 2 Dt % Dt Dt ∂x j Here, the summed-over index in the final term has been switched from 'i' to 'j'. Next, apply –∂/∂xj to (15.5) to produce: ∂g ∂ Du j ∂ # 1 ∂p & ∂ # 1 ∂τ ij & ∂fi %% (( = − j − − − . % (− ∂x j Dt ∂x j $ ρ ∂x j ' ∂x j ∂x j $ ρ ∂xi ' ∂x j Note that ∂gj/∂xj = 0 because the body force is spatially uniform, drop the viscous stress term, and move the remaining term that explicitly involves uj to the right side: ∂ # 1 ∂p & ∂f ∂ Du j %% (( = − i + − . (b) ∂x j $ ρ ∂x j ' ∂x j ∂x j Dt Add equations (a) and (b): D ! 1 Dp $ ∂ ! 1 ∂p $ Dq ∂fi ! ∂ D D ∂ $ (c) # &= &uj , − +# − # &− Dt " ρ c 2 Dt % ∂x j #" ρ ∂x j &% Dt ∂x j #" ∂x j Dt Dt ∂x j &% The remaining steps involve simplifying the final term. The first operator inside the final large parentheses may be rewritten: ∂ D ∂ "∂ ∂ % ∂2 ∂u ∂ ∂2 ∂u ∂ D ∂ . = + i + ui = i + $ + ui '= ∂x j Dt ∂x j # ∂t ∂xi & ∂t∂x j ∂x j ∂xi ∂xi∂x j ∂x j ∂xi Dt ∂x j # ∂ D D ∂ & ∂u ∂u (( u j = i j , so (c) reduces to (15.7) Thus: %% − ∂x j ∂xi $ ∂x j Dt Dt ∂x j ' D ! 1 Dp $ ∂ ! 1 ∂p $ Dq ∂f j ∂ui ∂u j # &= . − + # &− Dt " ρ c 2 Dt % ∂x j #" ρ ∂x j &% Dt ∂x j ∂x j ∂xi

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 15.2. Derive (15.12) through the following substitution and linearization steps. Set q and fi to zero in (15.7) and insert the decompositions (15.9). Treat Ui, p0, ρ0 and T0 as timeinvariant and spatially uniform, and drop quadratic and higher order terms involving the fluctuations ui! , p´, ρ´, and T´. Solution 15.2. Start with (15.7) and set q and fi to zero. This leaves: D ! 1 Dp $ ∂ ! 1 ∂p $ ∂ui ∂u j # &= . # &− Dt " ρ c 2 Dt % ∂x j #" ρ ∂x j &% ∂x j ∂xi The decomposition equations are: ui = Ui + ui! , p = p0 + p´ , ρ = ρ0 + ρ´ , and T = T0 + T´ , and all derivatives of Ui, p0, ρ0, and T0 are zero. Consider the first term: Dp # ∂ ∂ & ∂p" ∂p" ∂p" . = % + (Ui + ui") ( ( p0 + p") = +Ui + ui" Dt $ ∂t ∂xi ' ∂t ∂xi ∂xi Here the final term is quadratic in the small fluctuations, so Dp # ∂ ∂ & ≅ % +Ui ( p) , Dt $ ∂t ∂xi ' and this term is linear in the fluctuation quantity p´. Thus, only the leading-order portion of the operator coefficient of Dp/Dt need be retained because any extra first-order contributions from this coefficient will wind up being second order when multiplied with Dp/Dt. Therefore, the first term of (15.7) reduces to: 2 !∂ D ! 1 Dp $ ! ∂ ∂ $ 1 !∂ ∂ $ ∂ $ & # +Ui & p) = # +Ui & p) . # & ≅ # +Ui Dt " ρ c 2 Dt % " ∂t ∂xi % ρo c 2 " ∂t ∂xi % ∂xi % " ∂t where c is the speed of sound at pressure p0 and temperature T0. The second term simplifies in a similar manner: ∂p ∂ ∂p" = ( p0 + p") = . ∂x j ∂x j ∂x j Again this term is linear in the fluctuation quantity p´, so its coefficient cannot contain any firstorder contributions from fluctuation quantities ∂ " 1 ∂p % ∂ " 1 ∂p) % 1 ∂2 p) $ '≅ $ '= . ∂x j $# ρ ∂x j '& ∂x j $# ρ0 ∂x j '& ρo ∂x 2j The lone remaining right-side term is quadratic in the fluctuation velocity, ∂ui ∂u j ∂ (Ui + ui") ∂ (U j + u"j ) ∂ui" ∂u"j , = = ∂x j ∂xi ∂x j ∂xi ∂x j ∂xi and therefore should be dropped. So, term-by-term replacements in (15.7) using the results above lead to: 2 1 !∂ ∂ $ ∂ 2 p' ' +U p − ≅0, # & i c2 " ∂t ∂ xi % ∂ xi∂ xi which is (15.2)

Fluid Mechanics, 6th Ed.

Exercise 15.3. The field equation for acoustic pressure fluctuations in an ideal compressible fluid is (15.13). Consider one-dimensional solutions where p = p(x,t) and x = x1. a) Drop the x2 and x3 dependence in (15.13), and change the independent variables x and t to ∂ 2 p# ξ = x − ct and ζ = x + ct to simplify (15.13) to = 0. ∂ξ∂ζ b) Use the simplified equation in part a) to find the general solution to the original field equation: p"(x,t) = f (x − ct) + g(x + ct) where f and g are undetermined functions. c) When € the initial conditions are: p´ = F(x) and ∂p´/∂t = G(x) at t = 0, show that: € ' & 1$ 1 x 1# 1 x f (x) = &F(x) − ∫ 0 G(x )dx) , and g(x) = %F(x) + ∫ 0 G(x )dx(, ( ' 2% c 2$ c where x is just an integration variable.

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Kundu, Cohen, and Dowling

Solution 15.3. a) The starting point is (15.13) simplified for no x2 and x3 dependence. This is the € one-dimensional acoustic wave equation, € classical 1 ∂ 2 p# ∂ 2 p# − = 0. c 2 ∂t 2 ∂x 2 Use ξ = x − ct and ζ = x + ct , and convert the partial derivatives with respect to x and t into partial derivatives with respect to ξ and ζ. ∂ ∂ξ ∂ ∂ζ ∂ ∂ (x − ct) ∂ ∂ (x + ct) ∂ ∂ ∂ = +€ = + = + , and ∂ x ∂ x ∂ξ ∂ x ∂ζ ∂ x ∂ξ ∂ x ∂ζ ∂ξ ∂ζ € ∂ ∂ξ ∂ ∂ζ ∂ ∂ (x − ct) ∂ ∂ (x + ct) ∂ ∂ ∂ = + = + = −c +c . ∂t ∂t ∂ξ ∂t ∂ζ ∂t ∂ξ ∂t ∂ζ ∂ξ ∂ζ 2 2 2 2 %∂ ∂ ∂ (% ∂ ∂ ( ∂ ∂ ∂ = ' + *' + * = 2 + 2 + 2 , and Thus: € 2 ∂x & ∂ξ ∂ζ )& ∂ξ ∂ζ ) ∂ξ ∂ξ∂ζ ∂ζ 2 2 & ∂ ∂ )& ∂ ∂ ) 2& ∂ ∂2 ∂2 ) € 2 ∂ = c − − = c − 2 + ( 2 + ( +( + ∂t 2 ∂ξ∂ζ ∂ζ 2 * ' ∂ξ ∂ζ *' ∂ξ ∂ζ * ' ∂ξ ∂ 2 p# 1€∂ 2 p# ∂ 2 p$ ∂ 2 p$ ∂ 2 p$ ∂ 2 p$ 1 2 ) ∂ 2 p$ ∂ 2 p$ ∂ 2 p$ , 1 ∂ 2 p$ So, . = → = + 2 + = ⋅ c − 2 + + 2 .= ∂x 2 c 2 ∂t 2 ∂x 2 ∂ξ 2 ∂ξ∂ζ ∂ζ 2 c 2 ∂ξ∂ζ ∂ζ 2 - c 2 ∂t 2 * ∂ξ Cancel common € terms across the middle equality to find: ∂ 2 p# ∂ 2 p# ∂ 2 p# 2 = −2 = 0. which implies ∂ξ∂ζ ∂ξ∂ζ ∂ξ∂ζ € ∂p# = C (ζ ) where C(ζ) is a b) Use the result of part a) and integrate with respect to ξ to find: ∂ζ function of integration that cannot depend on ξ. Now € € integrate this result with respect to ζ to find: p" = ∫ C (ζ )dζ + f (ξ ) where f(ξ) is a second function of integration that cannot depend on

ζ. Since C is undetermined at this point, a new function€ g can be defined to make the last equation look a little nicer: g(ζ ) = ∫ C (ζ )dζ . Thus, the general solution of the field equation for p"(x,t) is: p"(x,t) = f (ξ ) + g(ζ ) = f (x − ct) + g(x + ct) . c) Use the result of part b) and the condition p´ = F(x) at t = 0 to find: € p"(x,0) = f (x) + g(x) = F(x) . (1)

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Fluid Mechanics, 6th Ed.

€

Kundu, Cohen, and Dowling

Differentiate the result of part b) with respect to time and use the ∂p´/∂t = G(x) at t = 0 to find: $ df + df (x) dg(x) . $∂p#' ∂ (x − ct) df ∂ (x + ct) ' + + 0 = G(x) . ) = c-, − &% ∂t )( = & d(x − ct) ∂t d(x + ct) ∂t (t= 0 dx dx / % t= 0 Use the last equality, divide by c, and integrate in x to find: 1 x (2) − f (x) + g(x) = ∫ 0 G(x )dx + D , c € where D is a constant. Adding equations (1) and (2) together produces: & 1# 1x g(x) = %F(x) + ∫ G(x )dx + D( , 2$ c 0 ' € while subtracting (2) from (1) produces: ' 1$ 1x f (x) = &F(x) − ∫ G(x )dx − D) . 2% c 0 ( € Given that D drops out of the general solution, ( 1% 1 x−ct 1 x +ct p"(x,t) = f (x − ct) + g(x + ct) = 'F(x − ct) + F(x + ct) − ∫ G(x )dx + ∫ G(x )dx − D + D* 2& c 0 c 0 ) € ( 1% 1 x +ct = 'F(x − ct) + F(x + ct) + ∫ G(x )dx*, 2& c x−ct ) it may be ignored without loss of generality, a situation that is also true for the lower limit of the x-integral. Therefore: ' & 1$ 1x 1# 1x f (x) = &F(x) − ∫ G(x )dx) and g(x) = %F(x) + ∫ G(x )dx( , 2% c 0 2$ c 0 ( ' where the integration lower limit, 0, is chosen for convenience.

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Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 15.4. Starting from (15.15) use (15.14) to prove (15.16) Solution 15.4. Equation (15.15) is

p"(x,t) = f (x − ct) + g(x + ct) , and (15.14) is the linearized integrated Euler equation without the body force: 1 ∂p" u1"(x,t) = − ∫ dt . ρ o ∂x Substitute (15.15) into€(15.14), define the variables ξ = x − ct and ζ = x + ct , and note that ∂ξ ∂x = ∂ζ ∂x = 1 : 1 ∂ 1 ) df ∂ξ dg ∂ζ , 1 ) df dg , u1"(x,t) = − ∫ ( f (x − € ct) + g(x + ct)) dt = − ∫ + + .dt = − ∫ + + . dt . ρ o ∂x ρ o * dξ dζ € ρ o * dξ ∂x €dζ ∂x Separate the two terms of the integration, change integration variables to ξ and ζ, and integrate to reach (15.16). 1 df 1 dg 1 df ( dξ + 1 dg ( dζ + 1 1 u1"(x,t) = − ∫ dt − dt = − ∫ * - − f (ξ ) − g(ζ ) * -= ∫ ∫ ρ o dξ ρ o dζ ρ o dξ ) −c , ρ o dζ ) c , ρ oc ρ oc

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1 [ f (x − ct) − g(x + ct)] ρ oc From a strictly mathematical point of view, the integrations involved in these final steps should produce constants of integration. However, the average of an acoustic fluctuation – such as u´– is zero, so the constants of integration are zero. =

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Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 15.5. Consider two approaches to determining the upper Mach number limit for incompressible flow. a) First consider pressure errors in the simplest-possible steady flow Bernoulli equation. Expand (15.29) for small Mach number to determine the next term in the expansion: p0 = p + 12 ρu 2 + ... and determine the Mach number at which this next term is 5% of p when γ = 1.4. b) Second consider changes to the density. Expand (15.30) for small Mach number and determine the Mach number at which the density ratio ρ0/ρ differs from unity by 5% when γ = € 1.4. c) Which criterion is correct? Explain why the criteria for incompressibility determined in a) and b) differ, and reconcile them if you can. Solution 15.5. a) Start with (15.29),

$ γ −1 2 'γ (γ −1) p0 = p&1+ M ) , % ( 2 and expand using the Taylor series (1+ ε) β = 1+ βε + 12 β (β −1)ε 2 + ... for ε 1 , show:

ρ2 γ +1 T2 γ −1 p2 γ +1 p2 , , and u1 = M1c1 ≅ . ≅ ≅ ρ1 γ −1 T1 γ +1 p1 2 ρ1

b) For a perfect gas with internal energy per unit mass e, the internal energy per unit volume is ρe. For a hemispherical blast wave, the volume inside the blast wave will be 23 π r 3 . Thus, set

ρ2 e2 = E 23 π r 3 , determine p2, set u1 = dr/dt, and integrate the resulting first-order differential equation to show that r(t) = K(E/ρ1)1/5t2/5 when r(0) = 0 and K is a constant that depends on γ. c) Evaluate K for γ = 1.4. A full similarity solution of the non-linear gas-dynamic equations in spherical coordinates produces K = 1.033 for γ = 1.4 (see Thompson 1972, p. 501). What is the percentage error in this exercise's approximate analysis?

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Solution 15.14. Start from the normal shock jump conditions (15.39) - (15.42): $ γ −1 2 ' $ 2 γ −1' p2 2γ = 1+ M12 −1) , M 22 = &1+ M1 ) & γM1 − ), ( % ( % p1 γ +1 2 2 ( T2 2(γ −1) $ γM12 + 1' 2 ρ2 (γ + 1)M12 , and = 1+ = & )( M1 −1) . ρ1 (γ −1)M12 + 2 T1 (γ + 1) 2 % M12 ( Simplify€these four equations for€M12 >> 1 , p2 2γ γ −1 2 γ −1 ≅ M12 , M 22 ≅ M1 γM12 = , p1 γ +€1 2 2γ € T2 2(γ −1) % γM12 ( 2 2γ (γ −1) 2 ρ 2 (γ +€ 1)M12 γ + 1 , and ≅ M1 . ≅ = ' * M1 = ρ1 (γ −1)M12 γ −1 T1 (γ + 1) 2 & M12 ) (γ + 1) 2 The third equation € form so it does not need further manipulation. Use the first € is in the correct and fourth equation to eliminate M12 from the expression for the temperature ratio: T2 2γ (γ −1) 2 2γ (γ −1) γ + 1 p2 γ −1 p2 € ≅ = € M1 = , T1 (γ + 1) 2 (γ + 1) 2 2γ p1 γ + 1 p1 which is another result€in the correct form. To reach the final part a) result, invert the strongshock pressure relationship to find and expression for M1: 12 12 12 12 $ γ + 1 p2 ' " γ +1 p2 % " γ +1 p2 % p1 " γ +1 p2 % € M1 ≅ & ' γ RT1 = $ ' γ =$ ' , ) , so that u1 = M1c1 ≅ $ ρ1 # 2 ρ1 & % 2γ p1 ( # 2γ p1 & # 2γ p1 & where R is the gas constant, and the final equality completes the effort for part a). b) As stated in the question, set E p p E ρ2 e2 = 2 3 = ρ2 cvT2 = cv 2 = 2 , and solve for p2 = (γ −1) 2 3 . πr R γ −1 πr 3 3

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Put this into the final result of part a) and set u1 = dr/dt to find: 12 12 12 ! γ +1 " 2 3E % dr ! γ +1 p2 $ E $ 3 2 dr & , which implies r u1 = = # = $ γ −1 & = ## ' . (γ −1) 2 3& dt " 2 ρ1 % 2 ρ π r dt 4 ρ π # " % 1 & 3 1 This non-linear first-order differential equation integrates to: 12 15 12 25 (5" r5 2 " 2 3E % 3 % + "E% 25 2 = $ γ −1 ' t + const , or r(t) = * $ (γ −1) ' - $ ' t , 52 # 4 ρ1π & 4π & -, # ρ1 & *) 2 # where the initial condition, r(0) = 0, has been used to determine const = 0. 12 25 +5 % 3 ( . 2 c) For γ = 1.40, K = - ' (γ −1) * 0 = 1.07, a difference of less than 4% from a similarity 4π ) / ,2 & solution of the non-linear gas-dynamic equations. Thus, this simpler analysis has certainly been worthwhile.

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Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 15.15. Starting from the set (15.45) with q = 0, derive (15.47) by letting station (2) be a differential distance downstream of station (1). Solution 15.15. The equation set (15.45) with q = 0 and the second location a differential distance dx downstream of the first location is: d ( ρu) = 0 , d ( p + ρu 2 ) = − p1df , and d h + 12 u 2 = 0 .

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Use thermodynamic relationships to determine h in terms of p and ρ. p γ p . h = cpT = cp = ρ R γ −1 ρ € € € Expand all three equations: γ dp γ p ρdu + udρ = 0 , dp + ρdu 2 + u 2 dρ = − p1df , and − dρ + 12 du 2 = 0 . γ −1 ρ γ −1 ρ 2 Solve the first equation for dρ = − ρdu u , and eliminate dρ from the other two equations: γ dp γ p du 1 2 + + 2 du = 0 . dp + ρdu 2 − ρudu = − p1df , and € € € γ −1 ρ γ −1 ρ u Simplify these equations and combine terms: € γ % dp p du ( dp + ρ udu = − p df , and ' + * + udu = 0 . 1 € γ −1 & ρ ρ u ) € Solve the first of these for du = −(1 ρu)( p1df + dp) and substitute this into the second one: ( 1 γ % dp p 1 p df + dp)* − ( p1df + dp) = 0 . ' − 2 ( 1 € γ −1 & ρ ρ € ρu ) ρ Multiply through€by ρ, and collect terms: % γ ( % γ ( γ p p − −1* dp − ' + 1* p1df = 0 . ' 2 2 & γ −1 γ −1 ρu ) & γ −1 ρu ) € 2 Separate the terms across the equals sign, use γp/ρ = c , where c = sound speed, and introduce the Mach number M = u/c: $ γ $ 1 c2 ' $ ' $ 1 ' 1€ c 2 ' 1 − −1 dp = + 1) p1df , or &γ − 2 − γ + 1)dp = & 2 + γ −1) p1df , so & ) & 2 2 % ( %M ( M % γ −1 γ −1 u ( % γ −1 u (

dp 1+ (γ −1)M 2 = df . p1 1− M 2 € To reach the second equation of (15.47), multiply by p1 subtract p1df from both sides and divide by ρu to reconstruct du = −(1 ρu)( p1df + dp) on the left: 2 € ( 1 1 % 1+ (γ −1)M 1 % γM 2 ( € du = −1* p1df = (−dp − p1df ) = ' ' * p1df . ρu ρu & 1− M 2 ρu &1− M 2 ) ) Divide by u€and rearrange the right side: du p % γM 2 ( p1 1 % M 2 ( p1 1 p1 = 2' df = df = df . * 2 2' 2* u ρu & 1− M ) p M & 1− M ) p 1− M 2 p € Thus, when M < 1, positive friction (df > 0) causes the flow to accelerate.

( M 2 −1)dp = ((γ −1)M 2 + 1) p1df , which implies: −

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Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 15.16. Starting from the set (15.45) with f = 0, derive (15.48) by letting station (2) be a differential distance downstream of station (1). Solution 15.16. The equation set (15.45) with f = 0 and the second location a differential distance dx downstream of the first location is: d ( ρu) = 0 , d ( p + ρu 2 ) = 0 , and d h + 12 u 2 = h1dq .

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Use h = cpT, expand all three equations, and use the perfect gas law (p = ρRT) to find: ρdu + udρ = 0 , dp + ρdu 2 + u 2 dρ = 0 , cp dT + 12 du2 = h1dq , and dp = ρRdT + RTdρ . Solve the first€equation for € dρ = − ρdu u , and€eliminate dρ from the second and last equations to reach: dp dT du € − , dp + ρdu 2 − ρudu = dp + ρudu = 0 , and€ = € p T u € where the final form of the last equation is obtained by dividing by p or ρRT. Solve the first of these for dp = –ρudu, and use this to eliminate dp from the second: € ρudu γ dT du − = − 2 udu€= − . p c T u 2 where γp/ρ = c , where c = sound speed, has been used for the first equality. Solve for udu and use M = u/c: γ du du dT u 2 dT , or udu = 12 du 2 = − . − 2 udu +€ = −(γM 2 −1) = c u u T T (γM 2 −1) Substitute this into the differential energy equation (the one that involves cp) and use h1 = cpT1, u 2 dT cp dT − = h1dq = cpT1dq . € T (γ M 2€−1) Divide this equation by γR, recognize the factor of M2, and use cp γ R = 1 (γ −1) : $ 1 cp cpT1 M2 ' 1 u 2 dT & )dT = , or − T1dq . dT − = dq 2 2 γ −1 γR γ RT (γ M −1) γ R &%γ −1 (γM −1) )( Solve for dT/T1. $ (γ −1)M 2 ' dT $ −1+ M 2 ' dT dT 1− γM 2 &1− ) = dq . = = dq , or & ) 2 € 2 −1 ( T1 T1 1− M 2 &% (γM −1) )( T1 % γM The final equation is the first part of (15.48) which shows that heat addition leads to cooling of the gas when 1 γ < M < 1. To reach the second part of (15.48), multiply the result for dT/T1 by T1 and substitute this into the differential energy equation: € € " 1− γ M 2 % cp dT + udu = cpT1 $ dq ' + udu = h1dq . 2 1− M # & € Recognize h1 = cpT1, and collect terms: udu $ 1− γM 2 ' (γ −1)M 2 = &1− dq = dq . ) h1 % 1− M 2 ( 1− M 2 This is the second equation of (15.48). €

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 15.17. For flow of a perfect gas entering a constant area duct at Mach number M1, calculate the maximum admissible values of f and q for the same mass flow rate. Case (a) f = 0; case (b) q = 0. Solution 15.17. From Section 15.6, 12

2 M2 1+ γM 22 $1+ ((γ −1) 2) M1 + q ' = & ) . M1 1+ γM12 − f % 1+ ((γ −1) 2) M 22 ( For maximum values of f or q, the flow is choked at the duct exit so M2 = 1. a) f = 0; set M2 = 1 to find: 12 $1+ ((γ −1) 2) M12 + q ' 1 1+ γ € = & ) . M1 1+ γM12 % (γ + 1) 2 ( Solve for q by first clearing the square root. 2 2 # 1+ γM12 & 1+ ((γ −1) 2) M12 + q γ + 1 # 1+ γM12 & * γ −1 2 , so q = M1 / . % ( = % ( − ,1+ . (γ + 1) 2 2 $(1+ γ )M1 ' + 2 $(1+ γ )M1€ ' (b) q = 0; set M2 = 1 to find: 12 $1+ ((γ −1) 2) M12 ' 1 1+ γ = & ) . M1 1+ γM€12 − f % (γ + 1) 2 € ( Solve for f. 12 $1+ ((γ −1) 2) M12 ' $ γ −1 2 '1 2 2 2 M1 ) . 1+ γM1 − f = (1+ γ )M1& ) , or f = 1+ γM1 − 2(1+ γ )M1&1+ % ( 2 (γ + 1) 2 € % ( Note, f = 0 when M1 = 1.

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Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 15.18. Show that the accelerating portion of the piston trajectory (0 ≤ xp(t) ≤ cot1) shown in Figure 15.18 is: 2

1+γ

" γ +1 % " t %γ +1 2cot t " 2 %1−γ for 1 ≤ ≤ $ x p (t) = $ ' cot1 $ ' − ' . γ −1 t1 # γ +1 & # γ −1 & # t1 &

Solution 15.18. As described in Example 15.7, the C+ characteristics emanate from the origin during the time that the piston is accelerating. And, as described in the solution to this same example, the C– characteristics originate on the x-axis where u = 0 and c = co, so the I– invariant implies: γ −1 2c(x, t) 2c(x, 0) 2c u. I − = u(x, t) − = u(x, 0) − = − o or c = co + 2 γ −1 γ −1 γ −1 Thus, on the C+ characteristics, ! dx $ ! ! !x$ γ −1 $ γ +1 $ u & = # co + u& = # & # & = (u + c)C+ = # u + co + " dt %C+ " 2 %C+ " 2 %C+ " t %C+ where the final equality ensures that the C+ characteristics pass through the origin. To determine the piston trajectory xp(t), require the final equality to hold where any C+ characteristic crosses the piston path. If the time of this crossing is τ, then the last equality implies: x (τ ) γ +1 . co + x! p (τ ) = p 2 τ Replace τ with t, and rearrange to find a first-order ordinary differential equation: 2 x p (t) 2c x! p (t) − =− o . γ +1 t γ +1 The solution is the sum of homogeneous and a particular solutions, 2 2c t γ +1 x p = At − o , γ −1 where A is an undetermined constant. The initial condition, xp = cot1 at t = t1, allows A to be determined from: 2 " γ +1 % −γ 2+1 2cot1 γ +1 , or A = cot1 $ cot1 = At1 − ' t1 . γ −1 # γ −1 & Thus, 2

1+γ

" γ +1 % " t %γ +1 2cot t " 2 %1−γ for 1 ≤ ≤ $ x p (t) = $ ' cot1 $ ' − ' , γ −1 t1 # γ +1 & # γ −1 & # t1 & which is the desired result. In dimensionless form this is: 2

x p (t) " γ +1 %" t %γ +1 2 "t% =$ $ '. '$ ' − cot1 # γ −1 &# t1 & γ −1 # t1 &

–––– T = 2 –––T=3 ----T=4

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

P Exercise 15.19. –––– For the flow conditions of Figure 15.18, plot u/co and p/po as functions of x/cot1 T=2 for xp(t) < x < cot–at– t/t – 1T==2,3 3, and 4 for γ = 1.4, where co and po are the sound speed and pressure of the quiescent gas - - - upstream - T = 4 of any disturbance from the moving piston. Does the progression of these waveforms indicate expansion wave steepening or spreading as t increases? Solution 15.19. Start from the results of Example 15.7, X X % 2 γ −1 x 2 "x , co + u(x, t) = $ − co ' and c(x, t) = & γ +1 γ +1 t γ +1 # t and rearrange these equations to introduce dimensionless variables U = u/co, C = c/co, X = x/cot1, and T = t/t1: % u(x, t) 2 " x 1 2 "X % $$ (1) = −1'' or U = $ −1' , co γ +1 # cot1 (t t1 ) & γ +1 # T & and c(x, t) 2 γ −1 x 1 2 γ −1 X or C = . = + + co γ +1 γ +1 cot1 (t t1 ) γ +1 γ +1 T The final equation can be switched to the dimensionless pressure P = p/po via (15.52): 2γ

" 2 γ −1 X %γ −1 P =$ + ' . # γ +1 γ +1 T & For the given parameters, the plots of (1) and (2) look like:

(2)

1.5"

0.5%

1" !0.5%

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0"

!1.5% !1%

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1%

2%

3%

4%

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1"

2"

3"

4"

Examination of both plots suggests that the expansion wave spreads out as time increases.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 15.20. Consider the field properties in Figure 15.19 before the formation of the shock wave. a) Using the piston trajectory from Exercise 15.18, show that the time at which the piston reaches (1+γ ) (1−γ ) speed co is −t1 ((γ +1) 2 ) = –0.3349t1 for γ = 1.4. b) Plot u/co and p/po as functions of x/cot1 for xp(t) < x < cot1 at: t/t1 = –1/3, –1/6, and –1/25 for γ = 1.4, where co and po are the sound speed and pressure of the quiescent gas upstream of any disturbance from the moving piston. Does the progression of these waveforms indicate compression wave steepening or spreading as t → 0 ? Solution 15.20. a) For the situation shown in Figure 15.19, the initial piston location is –cot1, so the sign of xp must be changed in the formula given in Exercise 15.18, and the piston starts moving at t = –t1, where t1 is presumed to be positive. Thus, the starting point for this Exercise is: 2

" γ +1 % " t %γ +1 2cot1 " t % x p (t) = − $ $ ' ' cot1 $ ' + γ −1 # −t1 & # γ −1 & # −t1 & where the extra divisor factors of –t1 have been introduced to facilitate the evaluation of xp when t < 0. Time differentiate this formula, 2

−1

" γ +1 % " 2 %" t %γ +1 " 1 % 2co dx p (t) . = −$ $ '− ' cot1 $ '$ ' dt # γ −1 & # γ +1 &# −t1 & # −t1 & γ −1 Simplify and set dxp/dt = co: 2 1−γ ( , −1 " % " 2 % " t %γ +1 * dx p (t) 2co 2co t 1+γ * = co = $ − = )$ ' −1- . ' co $ ' dt γ −1 γ −1 *# −t1 & # γ −1 & # −t1 & * + . Divide out the common factor of co and solve for t: 1+γ

" γ +1 %1−γ t = −t1 $ ' = –0.3349t1, # 2 & where the numerical value applies when γ = 1.4. b) The first or lowest C+ characteristic shown on Figure 15.19 has a slope unity. To the right of this characteristic, u = 0 and p = po. Using the part a) result, and equation for the piston's trajectory, the location where the piston speed reaches co can be found: !x t$ location = # p , & = (−0.7368, −0.3349 ) . " cot1 t1 % The line from this location to the origin is the last or uppermost C+ characteristic shown on Figure 15.19. To the left of this characteristic, the flow state will be uniform with u = co and p = constant. The region in between the two characteristics must provide the appropriate transition between the left- and right-side flow states. To specify this transition, recognize that x and t are both negative, and start from the results of Example 15.7:

P

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

% 2 γ −1 x 2 "x andT c(x, . co + u(x, t) = $ − co '–––– –––– T = 2 = 2 t) = #t & γ +1 γ +1 t γ +1 –––T=3 –––T=3 Rearrange and introduce dimensionless variables ----T=4 - - - - TU==4 u/co, C = c/co, X = x/cot1, and T = t/t1. % u(x, t) 2 " x 1 2 "X % $$ = −1'' or U = $ −1' co γ +1 # cot1 (t t1 ) & γ +1 # T & X X c(x, t) 2 γ −1 x 1 2 γ −1 X or C = . = + + co γ +1 γ +1 cot1 (t t1 ) γ +1 γ +1 T The final equation can be switched to the dimensionless pressure P = p/po via (15.52):

(1)

2γ

" 2 γ −1 X %γ −1 P =$ + ' . # γ +1 γ +1 T &

(2)

2γ

p ( 2 γ −1 " −0.7368 %+γ −1 =* + The peak pressure then is: P = $ '- = 3.583, when γ = 1.40. po ) γ +1 γ +1 # −0.3349 &, For the given parameters, the plots of (1) and (2) look like: 1.25&

4"

1&

3.5" 3"

0.75&

2.5"

0.5&

2" 1.5"

0.25&

1"

0&

0.5" 0"

!0.25& !1&

!0.8&

!0.6&

!0.4&

!0.2&

0&

)1"

)0.8"

)0.6"

)0.4"

)0.2"

0"

Examination of both plots suggests that the compression wave steepens as time increases.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 15.21. For the flow conditions of Figure 15.19, assume the flow speed downstream of the shock wave is co and determine the shock Mach number, its x-t location, and the pressure, temperature and density ratios across the shock. Are these results well matched to the isentropic compression that occurred for t < 0? What additional adjustment is needed? Solution 15.21. The shock wave first forms at the origin in Figure 15.19, and the velocity difference across the shock wave is co. This is enough to determine the shock Mach number. Start with the normal shock velocity ratio condition (15.41) for a stationary normal shock wave: u1 (γ +1)M12 . (15.41) = u2 (γ −1)M12 + 2 When the situation in Figure 15.19 is subject to a Galilean transformation that creates a shockfixed coordinate system, u1 = –M1co and u2 = (1 – M1)co. Thus, (15.41) becomes: −M1co (γ +1)M12 or −(γ −1)M12 − 2 = (γ +1)M1 (1− M1 ) , = 2 (1− M1 )co (γ −1)M1 + 2 which reduces to the quadratic equation: 2 ! $ 1 # γ +1 ! γ +1 $ γ +1 2 ± # M1 − M1 −1 = 0 , so M1 = & + 4 &. & " 2 % 2 #" 2 2 % The physically meaningful root comes from the '+' sign and is M1 = 1.766 for γ = 1.4. Thus, the x-t location of the shock is: xshock = 1.7662cot, and the pressure, temperature, and density ratios are: p2 2γ " 2 $ (3.583) = 1+ # M1 −1% = 3.473 , p1 γ +1

ρ2 (γ +1)M12 (2.488) = = 2.305 , and ρ1 (γ −1)M12 + 2 T2 2(γ −1) γ M12 +1 2 (1.440) = 1+ ( M1 −1) = 1.506 . T1 (γ +1)2 M12 Interestingly, these ratios are all slightly different than the ratios produced by the isentropic compression that occurs before the shock wave forms at t = 0 (provided in parenthesis at the right). [These numbers are obtained from the solution of Exercise 15.20]. Thus, there will be an adjustment region near the origin in Fig. 15.19 that will allow the pressure behind the shock to equilibrate with the adiabatically compressed gas that is between the piston and the shock wave. The net effect of this adjustment will be to shift the shock wave location slightly farther ahead – in the positive x-direction – compared to what is calculated above because the adiabatic compression reaches a higher pressure.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 15.22. Write momentum conservation for the volume of the small rectangular control volume shown in Figure 4.20 where the interface is a shock with flow from side 1 to side 2. Let the two end faces approach each other as the shock thickness → 0 and assume viscous stresses may be neglected on these end faces (outside the structure). Show that the n component of momentum conservation yields (15.36) and the t component gives u⋅t is conserved or v is continuous across the shock. (2)! us!

u2!

+n!

dA! l!

(1)!

u1!

–n!

dA!

Solution 15.22. For a stationary control volume V containing only fluid particles bounded by a surface A, conservation of mass and momentum may be written: d d ρdV = − ∫ ρu j n j dA , and ∫ ∫ ρui dV = − ∫ [ρui u j + pδij − σ ij ]n j dA + ∫ ρgi dV . dt V dt V A A V When applied to a small rectangular control volume like that shown above (a reproduction of Fig. 4.20) in a steady flow, these simplify to: 0 = − ∫ ρu j n j dA , and 0 = − ∫ [ ρui u j + pδij − σ ij ]n j dA + ∫ ρgi dV . € € A A V For the current situation, let the interface define the location of a shockwave, and let the thickness dimension l → 0 while requiring the upper rectangular surface to remain in region 2 and the lower rectangular surface to remain in region 1. This means V → 0 so the body force € the flux through the sides of the CV may be ignored because the € be dropped, and that term may curved-side surface area goes to zero as l → 0 . For this specialized limit, the equations reduce € to: € (ρu j )1 n j = (ρu j )2 n j , and [ρuiu j + pδij − σ ij ]1 n j = [ρuiu j + pδij − σ ij ]2 n j , where nj are the components of n, € the primarily-upward-pointing unit vector shown in the figure. Here, the two larger CV surfaces are chosen to lie outside the shock structure (which is presumed to be very thin), so that σij = 0. [Inside the shock wave, the viscous stress may be large]. Thus, € the€momentum equation becomes: (ui )1 ( ρu j )1 n j + p1n i = (ui ) 2 ( ρu j ) 2 n j + p2 n i . (†) For the shock-normal component, take the dot product of (†) with n (which has components ni). The resulting equation is: ρ (u n )2 + p = ρ (u n )2 + p2 , or p1 − p2 = −ρ1u12 + ρ2 u22 , €( i i )1 1 ( i i )2 where (uini)1 = u1 and (uini)2 = u2, and the second equation is (15.36). For the shock tangent component, take the dot product (†) with either tangent unit vector, generically represented here by t (which has components ti):

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

t i (ui )1 ( ρu j )1 n j + p1t i n i = t i (ui ) 2 ( ρu j ) 2 n j + p2 t i n i , or t i (ui )1 ( ρu j )1 n j = t i (ui ) 2 ( ρu j ) 2 n j , where by definition: t ⋅ n = t i n i = 0 . Now use the final form of the mass conservation equation to divide out the terms in larger parenthesis that contain ρ; this leaves: t i (ui )1 = t i (ui ) 2 , or in vector notation: t ⋅ u1 = t ⋅ u2 . € € Thus, the tangential components of u are unchanged across a shockwave since t represents either tangent unit€vector.

€

€

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 15.23. A wedge has a half-angle of 50°. Moving through air, can it ever have an attached shock? What if the half-angle were 40°? [Hint: The argument is based entirely on Figure 15.22.] Solution 15.23. From Figure 15.22, no attached shock can produce a deflection angle of 50°. However, an attached shock with a deflection angle of 40° is possible.

Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 15.24. Air at standard atmospheric conditions is flowing over a surface at a Mach number of M1 = 2. At a downstream location, the surface takes a sharp inward turn by an angle of 20°. Find the wave angle σ and the downstream Mach number. Repeat the calculation by using the weak shock assumption and determine its accuracy by comparison with the first method.

Solution 15.24. Using the geometry shown, and assuming M2 > 1, the shock angle σ is approximately 53° based on Figure 15.22. Therefore: Mn1 = M1sinσ = 2sin(53°) = 1.6, and Mn2 = 0.668 (from Table 15.2). From the geometry: Mn2 = M2sin(σ – δ), so M 2 = M n 2 sin(σ − δ ) = 0.668 sin(53° − 20°) = 1.227 . To develop equivalent results from weak shock theory, start from (15.53) and simplify for δ