Gas Dynamics (Solution Manual) [4 ed.]

Citation preview

Solutions Mannual for the fourth edition of

Gas Dynamics Ethirajan Rathakrishnan

Preface This manual gives the detailed solution for all the problems given at the end of different chapters of the 3rd edition of Gas Dynamics. My sincere thanks to my doctoral and masters students who helped me in checking and keying in the solutions of this manual. My sincere thanks to the Continuing Education Centre of Indian Institute of Technology Kanpur for the financial support to prepare this manual. E Rathakrishnan

i

ii

Contents 1 Some Preliminary Thoughts

1

2 Basic Equations of Compressible Flow

3

3 Wave Propagation

23

4 One-Dimensional Flow

25

5 Normal Shock Waves

79

6 Oblique Shock and Expansion Waves

119

7 Potential Equation for Compressible Flow

157

8 Similarity Rules

161

9 Two Dimensional Compressible Flows

165

10 Prandtl-Meyer Flow

169

11 Flow with Friction and Heat Transfer

173

12 MOC

205

13 Measurements in Compressible Flow

207

iii

Chapter 1

Some Preliminary Thoughts

1

2

Some Preliminary Thoughts

Chapter 2

Basic Equations of Compressible Flow 2.1 In the reservoir, the air is at stagnation state. So, the entropy relation would be ! " ! " T02 p02 s2 − s1 = cp ln − R ln T01 p01 But, T01 = T02 for adiabatic process. Therefore, ! " p01 ∆s = R ln p02 ! " p01 = R ln 1 = R ln 2 2 p01 =

198.933 J/(kg K)

Note: It should be noted that, for entropy only subscripts 2 and 1 are used; since entropy is not defined like static or stagnation entropy. 2.2 Let the initial state be denoted by subscript 1 and expanded state by subscript 2. (a) Since the cylinder is insulated, preventing any heat transfer what-so-ever, the process is adiabatic. The governing equation for this process is given by p1 Vγ1 = p2 Vγ2 = constant

(1)

Also, from ideal gas state equation p1 V1 p2 V2 = =R T1 T2 3

(2)

4

Basic Equations of Compressible Flow

From Eqs. (1) and (2), we get p1 = p2

!

!

V2 V1

"γ

T1

# 10(γ−1) = 557.35 K

=

T1 T2

"γ/(γ−1)

Therefore, T2

=

∆T

=

(b) Work = Also,

$

− 842.65 K

pdv =

$

pv γ = constant

dh −

$

du −

$

vdp

(3)

from equation (1)

Differentiating equation (1), we have, pγv γ−1 dv + v γ dp = 0 Dividing throughout by v γ−1 and integrating, we get $ $ pγdv + vdp = 0 $

vdp = −γw

(4)

Substituting equation (4) in equation (3) and simplifying, we get (1 − γ) w

=

R∆T

w

=

R∆T 287 × (−842.65) = 1−γ (−0.4)

=

6.04 × 105 J/kg

Note: Since the process undergone is expansion from a high pressure, the work removed is positive, i.e, work is done by the gas. (c) Also, from equation (1) p1 = p2 Therefore,

!

V2 V1

"γ

= 101.4 = 25.1189

The pressure ratio = 25.1189

5 2.3 p1 v1γ = p2 v2γ , where v is specific volume, i.e. volume per unit mass = V/m. Therefore, ! "γ ! "γ V1 V2 p1 = p2 m1 m2

Also, V1 = V2 = V = volume of the tank. ! "γ m2 p 2 = p1 m1 ! "1.4 1 = 5 × 105 × 2 1.8946 × 105 Pa

=

From equation of state for a calorically perfect gas, p1 p2

=

T2

=

ρ1 T1 ρ2 T2 ! "! " p2 m1 T1 p1 m2 ! " 1.8946 × 2 × 500 5

=

378.92 K

= 2.4

p1 = p2 (a) Therefore, T2

!

T1 T2

"γ/(γ−1)

=

!

=

61/3.5 × 290 = 483.868 K

p2 p1

"(γ−1)/γ

T1

The change in the temperature is ∆T

= =

T2 − T1 = 483.868 − 290 193.868 K

(b) By first law of thermodynamics, we have du + d(pe) + d(ke) = dq + dw

6

Basic Equations of Compressible Flow

Here, velocity changes are neglected. Therefore, d(ke) = 0 Also, assuming d(pe) = 0 The first law of thermodynamics reduces to du = dq + dw But the process is isentropic, thus dq = 0. Therefore, du

= =

dw = cv ∆T = 717.5 × 193.868 1.39 × 105 J/kg

(c) The work done is negative, i.e. work is done on the gas. It has been computed in (b) above. 2.5 Work done by the weight on the piston goes towards increasing the internal energy of the gas. From the first law of thermodynamics E2 − E 1 = Q + W where, E, Q, and W are respectively the internal energy, heat transfered, and work done. Since no heat is transfered, Q = 0. Therefore, $ E2 − E1 = W = F . ds where, F is force and ds is distance. At the new equilibrium position, the force acting on the piston face is F = p2 Ap , Ap is the area of the piston face. The distance traveled by the piston is ds = (V1 − V2 )/Ap , V1 and V2 are the initial and final volumes. Thus we have, E2 − E1

=

p2 . Ap (V1 − V2 )/Ap

=

−p2 (V2 − V1 )

For unit mass, e2 − e1 = −p2 (V2 − V1 )

For calorically perfect gas, e = cv T . Therefore ! " RT2 RT1 cv (T2 − T1 ) = −p2 − p2 p1 " ! c v T2 T2 p2 −1 = − + R T1 T1 p1 cv & T2 % cv 1+ = + λ (where λ = p2 /p1 ) T1 R R

7 But

1 cv = . Thus, R γ−1 γ T2 γ − 1 T1

=

T2 T1

=

λ+

1 γ−1

1 + (γ − 1)λ γ

Entropy change for a perfect gas can be written as ! " ! " T2 p2 ∆s = cp ln − R ln T1 p1 ' ( ) γ * 1 1 + (γ − 1)λ γ−1 ∆s = ln R λ γ ' ( ) γ * 1 1 + (γ − 1)λ γ−1 s2 − s1 = R ln λ γ Let λ = 1 + $, where $ # 1. Therefore, + , 1 + (γ − 1)(1 + $) ∆s γ = ln − ln(1 + $) R γ−1 γ + , 1 + γ − 1 + $(γ − 1) γ ln = − ln(1 + $) γ−1 γ + , γ−1 γ ln 1 + $ − ln(1 + $) = γ−1 γ Expanding the RHS, and retaining only upto second order terms, we get, , ! + " ∆s γ γ−1 (γ − 1)2 2 $2 = $− $ − $ − R γ−1 γ 2γ 2 2 =

$−

=

$2 2γ

$2 $2 $2 γ−1 2 $2 $ −$+ =− + + 2γ 2 2 2γ 2

Note: Work has been done by the weight which is equal to p2 Ap on the gas. The weight has moved by a distance of ds. Therefore, ∆E = W . ds = p2 Ap . ds. 2.6 Since it is an open system, Work done

= −cp (T2 − T1 )

8

Basic Equations of Compressible Flow

−cp

=

!

" T2 − 1 T1 T1

-(

p2 p1

) γ−1 γ

.

=

−cp

− 1 T1

=

% & −1004.5 × 21/3.5 − 1 × 303 − 66.66 kJ/kg

=

2.7 Work done is given by W

= =

p (V2 − V1 ) = 101325 × 6 (2 − 0.3) 1.0335 MJ

since 1 atm = 101325 Pa. 2.8 The compression process is given as isentropic. Let subscripts 1 and 2 refer to initial and final states, respectively. By isentropic process relation, we have p1 ργ1 ρ2

=

p2 ργ2

=

!

=

p2 p1

"1/γ

ρ1 =

!

690 150

"1/1.3

× 1.5

4.85 kg/m3

2.9 As we know, the relation between temperature and pressure for isentropic change of state may be written as T2 = T1

!

p2 p1

"(γ−1)/γ

where subscripts 1 and 2 refer to the initial and final states, respectively.

T2

= =

T1

!

p2 p1

"0.4/1.4

519.9 K

! "0.286 7 = 298 1

9 2.10 By isentropic relation, T2 = T1

!

v1 v2

"(γ−1)

where subscripts 1 and 2 refer to the initial and final states and v is specific volume. For air γ = 1.4. Therefore, T2

!

"0.4

v1 v2

=

T1

=

(30 + 273.15) (30)

=

0.4

= 1181.7 K

908.55◦ C

γ−1 γ R, therefore, the gas constant R = cp γ−1 γ 0.4 × 1000 = 285.7 J/(kg K) R= 1.4

2.11 (a) We have cp =

Also,

Ru 8314 = M M where M is the molecular weight and Ru is universal gas constant. Thus, R=

M=

8314 = 29.1 285.7

(b) By ideal gas state equation, we have p1 V1 = mRT1 p2 V2 = mRT2 where subscripts 1 and 2 refer to initial and final states, respectively. But p1 = p2 and therefore, V2 V1

=

323.15 T2 50 + 273.15 = = T1 200 + 273.15 473.15

=

0.683

2.12 For an ideal gas, the speed of sound a may be expressed as / a = γRT

10

Basic Equations of Compressible Flow

where γ is the ratio of specific heats and R is the gas constant. For the given gas, R

=

Ru /M = 8314/29 = 286.7 J/(kg K)

Therefore, 400

=

γ

=

/

γ × 286.7 × 373.15

4002 = 1.5 286.7 × 373.15

The specific heat cp and cv can be written as cp

=

γ R γ−1

cv

=

R γ−1

=

(1.5/0.5) × 286.7

=

860.1 J/(kg K)

Therefore, cp

cv

=

286.7/0.5

=

573.4 J/(kg K)

Note: The ratio of specific heats γ = cp /cv . For the present case γ = 1.5 = 860.1/573.4 is correct. This way the answer obtained for cp and cv may be checked. 2.13 At the nozzle exit, V = 390 m/s and T = 28 + 273.15 = 301.15 K. The corresponding speed of sound is / √ a = γRT = 1.4 × 287 × 301.15 =

347.85 m/s

Thus, M

=

V 390 = a 347.85

=

1.12

11 By isentropic relation, we have T0 T

T0

γ−1 2 M 2

=

1+

=

1 + 0.2 × 1.122 = 1.25

=

1.25 × 301.15 = 376.44 K 103.29◦ C

=

For the flow, the stagnation temperature is T0 = 376.44 K. The static temperature

T = 92.5◦ C = 92.5 + 273.15 = 365.65 K

376.44 T0 = = 1.03 = 1 + 0.2 M 2 T 365.65 Thus, M2

=

0.03 = 0.15 0.2

M

=

0.387

This is the Mach number at the station where temperature is 92.5◦ C. 2.14 For hydrogen, the gas constant R = 8314/2.016 = 4124 J/(kg K). By isentropic relation, we have T2 T1

=

=

!

"(γ−1)/γ p2 p1 ! "0.286 1 = 0.573 7

Therefore, T2

=

0.573 × T1

=

0.573 × 300 = 171.9 K

By energy equation, we have h2 +

V22 = h1 , since V1 = 0 2 V22 = 2 (h1 − h2 )

12 But

Basic Equations of Compressible Flow h = cp T ,

therefore, V2 =

0

cp

2cp (T1 − T2 )

=

1.4 γ R= × 4124 γ−1 0.4

= 14434 J/(kg K) Thus, V2

=

/

=

2 × 14434 (300 − 171.9)

1923 m/s

The speed of sound is given by / √ a2 = γRT2 = 1.4 × 4124 × 171.9 =

996.23 m/s

Thus, M2 =

V2 1923 = 1.93 = a2 996.23

The mass flow rate is given by m ˙ = ρ2 A2 V2 ρ2

=

p2 , by state equation RT2

ρ2

=

101325 , since 1 atm =101325 Pa 4124 × 171.9

=

0.143 kg/m3

Thus, m ˙

= =

0.143 × 10 × 10−4 × 1923 0.275 kg/s

2.15 The given process is a polytropic process with index n = 1.32. Since air is given as an ideal gas with constant specific heats, we have from isentropic relations the change of entropy as ! " ! " T2 p2 s2 − s1 = cp ln − R ln T1 p1

13 For a polytropic process, T2 = T1

!

p2 p1

" n−1 n

Combining these two equations, we obtain + ! " , ! " n−1 p2 s2 − s1 = cp − R ln n p1 since cp =

γ R for an ideal gas, the above equation may be written as γ−1 ! " p2 (n − γ) R s2 − s1 = ln n (γ − 1) p1

With the given data, R=

γ−1 0.4 cp = × 1004 = 287 J/(kg K) γ 1.4

Therefore, s2 − s1

=

(1.32 − 1.4) 287 1100 ln 1.32 × 0.4 101

=

−103.8 J/(kg K)

Note: Since the entropy of the gas decreases for this internally reversible process, heat must be removed from the gas. This is why the cylinder used for such a compression process is usually water jacketed. Also, we know that the entropy of the system and surrounding cannot decrease. But in this problem, the entropy decreases. It should be noted that, what decreases is entropy of the system alone and not the combined entropy of the system and surrounding. 2.16 For oxygen, molecular weight M = 32. The gas constant R = 8314/32 = 259.8 J/(kg K). Therefore, cp

=

cv

=

γ R = 909.3 J/(kg K) γ−1 cp = 649.5 J/(kg K) γ

The increase in internal energy is %u

= =

cv (T2 − T1 ) = 649.5 (125 − 25) 64950 J/(kg K)

14

Basic Equations of Compressible Flow

The increase in enthalpy is %h

= cp (T2 − T1 ) = 909.3 (125 − 25) =

90930 J/(kg K)

2.17 Let the subscripts 1 and 2 refer to the inlet and exit states. At state 1, p1 = 100 kPa, ρ1 = 1.175 kg/m3 . Therefore, T1 = At state 2,

p1 100 × 103 = 296.5 K = Rρ1 287 × 1.175

p2 = 500 kPa, ρ2 = 5.875 kg/m3 . Therefore, T2 =

p2 500 × 103 = 296.5 K = Rρ2 287 × 5.875

Assuming air to be a perfect gas, the enthalpy difference is given by h2 − h1

= =

cp (T2 − T1 ) = cp (296.5 − 296.5) 0

2.18 The entropy change is given by ds =

δq T

where δq is the reversible heat addition per unit mass. For an ideal gas δq = dh − vdp where dh is the enthalpy change and dh = cp dT δq = cp dT − vdp Using the above relation, we get ds = cp By state equation

v dT − dp T T

pv = RT , R v = T p

15 Therefore, ds = cp

dT R − dp T p

Taking log of state equation and differentiating, we get dT dp dv = + T p v Substituting for

dT is ds expression, we obtain T ! " dp dv dp ds = cp + −R p v p ds = cp

But

cp − cv = R,

dv dp + (cp − R) v p

cp − R = cv . Thus, ds = cp

dv dp + cv v p

For an isentropic change of state, we have 0 = dv v cp dv cv v cp

But

cp

dv dp + cv v p

=

−cv

=

−

dp p

dp p

cp = γ. Therefore, cv γ

dp dv =− v p

Integrating both sides, we get γ ln v

=

− ln p + constant

ln p + ln v γ

=

constant

ln (pv γ )

=

constant

or pv γ = constant

16

Basic Equations of Compressible Flow

2.19 For an isothermal process, the change in entropy %s is given by ! " p1 %s = s2 − s1 = R ln p2 where the subscripts 1 and 2 stand for the initial and final states, respectively. By state equation p1 V1

=

mRT1

p2 V2

=

mRT2

But T1 = T2 , therefore, p1 V1 = p2 V2 = mRT2 . Hence, p2

=

p1 V1 0.7 × 106 × 0.014 = V2 0.084

=

0.117 MPa

Thus, the change of entropy is %s

=

R ln

!

p1 p2

"

= 287 ln

!

0.7 0.117

"

513.4 J/(kg K)

=

2.20 Let subscripts 1 and 2 refer to initial and final states, respectively. For process 1

p2 = 700 kPa.

By state equation, we have p1 V

=

mRT1

p2 V

=

mRT2

Therefore, T2

=

p2 700 × 308.15 = 616.3 K × T1 = p1 350

The change in entropy is given by %s

=

cp ln

!

T2 T1

"

− R ln

!

p2 p1

"

For air cp = 1004.5 J/(kg K) and R = 287 J/(kg K). Thus, ! " ! " 616.3 700 %s = 1004.5 ln − 287 ln 308.15 350 =

497.33 J/(kg K)

17 For 0.3 kg of air, %s = 0.3 × 497.33 = 149.2 J/K For process 2 pV1

=

mRT1

pV2 = mRT2 Therefore, T2

=

V2 × T1 V1

Initial volume is given by V1

=

0.3 × 287 × 308.15 mRT1 = = 0.0758 m3 p 350 × 103

T2

=

0.2289 × 308.15 = 930.55 K 0.0758

The entropy change is given by %s

!

T2 T1

"

=

cp ln

=

1110.163 J/(kg K)

= 1004.5 ln

!

930.55 308.15

"

For 0.3 kg of air, %S = 0.3 × 1110.163 = 333 J/K 2.21 Given flow is adiabatic and frictionless and therefore, isentropic. By energy equation, we have h1 +

V12 V2 = h2 + 2 2 2

For perfect gas h = cp T , thus, c p T1 +

V12 V2 = cp T2 + 2 2 2

For air cp = 1004.5 J/(kg K). Therefore, 1004.5 × 273.15 +

9002 3002 = 1004.5 T2 + 2 2

18

Basic Equations of Compressible Flow

Solving this, we get T2 = 631.54 K. Therefore, the temperature increase becomes = T2 − T1 = 631.54 − 273.15

%T

=

358.39 K

By isentropic relation, we have p2 = p1

!

T2 T1

γ " γ−1

Thus, p2

=

p1

= %p

T2 T1

"3.5

= 140

!

631.54 273.15

"3.5

= 2631 kPa

2.631 MPa p2 − p1

= =

2.22 For nitrogen,

!

2.491 MPa

R = Ru /M = 8314/28 = 297 J/(kg K).

For reversible process, the change in entropy may be expressed as ! " ! " T2 p2 %s = cp ln − R ln T1 p1 But

T2 = T1 , therefore, %s

= =

− R ln

!

p2 p1

"

= − 297 ln

!

300 100

"

− 0.3263 kJ/(kg K)

Note: We know that the entropy of the system and surrounding cannot decrease. But in this problem, the entropy decreases. It should be noted that, what decreases is entropy of the system alone and not the combined entropy of the system and surrounding. 2.23 By isentropic relation, p2 = p1

!

T2 T1

γ " γ−1

19 where subscripts 1 and 2 refer to initial and final states. That is, T2 T1 T2

=

=

=

!

p2 p1

" γ−1 γ

!

" 0.4 p2 1.4 T1 p1 ! "0.286 550 300 = 475.4 K 110

The change in enthalpy is (h2 − h1 ), and h = cp T . Therefore, h2 − h1 = cp (T2 − T1 ) Also, cp =

γ R γ−1

and

R = 287 J/(kg K), for air.

cp =

1.4 × 287 = 1004.5 J/(kg K) 0.4

Thus, h2 − h1

= =

1004.5 (475.5 − 300) 176.19 kJ/kg

2.24 Let the initial and final states of air are designated by subscripts 1 and 2, respectively. By perfect gas state equation, we have

Dividing one by other, we get

p1 V =

mRT1

p2 V =

mRT2

p1 T1 = p2 T2

The change in entropy for a perfect gas is given by ! " ! " T2 p2 %s = cp ln − R ln T1 p1 ! " T2 = (cp − R) ln T1

20

Basic Equations of Compressible Flow

Given, T1

=

50 + 273.15 = 323.15 K

T2

=

125 + 273.15 = 398.15 K

cp

=

γ R = 1004.5 J/(kg K) γ−1

%s

=

(1004.5 − 287) ln

=

149.75 J/(kg K)

Thus, !

398.15 323.15

"

2.25 By energy equation, we have h0 = h +

V2 2

where subscript 0 refers to stagnation condition. Assuming air to be a perfect gas, we can express h = cp T . Therefore, c p T0

=

cp T +

T0 − T

=

V2 2 cp

V2 2

From standard atmosphere table, at 10000 m, T = 223.15 K. Therefore, the speed of sound becomes / √ a = γ R T = 1.4 × 287 × 223.15 =

299.436 m/s

The flight speed is V = M a = 2 × 299.436 = 598.872 m/s Thus, the temperature becomes ∆T

=

T0 − T =

=

178.52

598.8722 2 × 1004.5

21 2.26 Let subscripts 1 and 2 refer to the initial and final states. Given, T1 = 15◦ C = 288.15 K,

v1 = 0.06 m3 ,

v2 = 0.12 m3 ,

By perfect gas state equation, pv = RT Therefore,

T1 T2 = v1 v2

since p1 = p2 . Thus, T2

=

T1 288.15 × 0.12 = 576.3 K v2 = v1 0.06

=

303.15◦ C

p 1 = p2

22

Basic Equations of Compressible Flow

Chapter 3

Wave Propagation

23

24

Wave Propagation

Chapter 4

One-Dimensional Flow 4.1 Total temperature at “1” is T01 = 300 K. This much temperature is required as static temperature at the test-section. Therefore, TT

=

300 K

T0T TT

=

1+

γ−1 2 M 2

T0T 300

=

1+

0.4 × 2.52 2

T0T

=

675 K

Hence, the temperature rise required, ∆T

= =

T0T − TT = 675 − 300 375 K

4.2 Let the test-section conditions be denoted by subscript 2, and the sonic conditions by superscript *. γ−1 2 M 2

T02 T2

=

1+

373 T2

=

1 + 0.2 × 22

T2

=

207.2 K 25

26

One-Dimensional Flow

Further, since we know that T∗ T0

=

0.8333

T∗

=

373 × 0.8333 = 310.82 K

a∗

=

√

p0

=

3.12 × 101325 = 3.16 × 105

ρ0

=

p0 RT0

=

3.16 × 105 287 × 373

=

2.95 kg/m3

ρ∗ ρ0

=

0.634

ρ∗

=

1.8703 kg/m3

m ˙

= ρ∗ A∗ V ∗ = =

1.4 × 287 × 310.82 = 353.4 m/s = V ∗

1.8703 × 80 × 10−4 × 353.4 5.288 kg/s

4.3 Let subscripts 0, 1, and e refer to stagnation state and states at the nozzle entrance and exit, respectively. We know that, Te = 0.8333 T0 This gives, T0 =

293 = 351.6 K 0.8333

Also, T0 T1

=

1+

γ−1 2 M 2

27

T1

=

351.6 1 + 0.2 × 9

=

125.57 K

pe p0

= 0.5283

p0

=

0.8 = 1.5143 atm 0.5283

=

!

p0 p1

= p1

1

γ−1 2 M 1+ 2

1 + 0.2 × 32

=

1.5143 36.73

=

0.04123 atm

γ " γ−1

23.5

√ 1.4 × 287 × 293 = 343.11 m/s

Ve

= ae =

Ae

= 40 × 10−4 m2

ρe

=

pe RTe

=

0.8 × 101325 = 0.964 kg/m3 287 × 293

m ˙

= ρe Ae Ve = 0.964 × 40 × 10−4 × 343.11 =

1.323 kg/s

Note: It should be noted that the calculation made with equations for pressure, density, and temperature in the problem can also be done using gas tables. In fact that procedure will result in considerable time saving. 4.4 Given, A1

=

0.6 × 0.4 = 0.24 m2

28

One-Dimensional Flow

A∗

=

0.3 × 0.4 = 0.12 m2

A2

=

h2 × 0.4 = 0.4 h2 m2

M2

=

2.5

T2

=

− 10◦ C

p2

=

0.15 atm

For M2 = 2.5, isentropic table gives A2 = 2.6367 A∗

and

T2 = 0.4444 T02

This gives, Thus,

A2 = 2.6367 × 0.12 = 0.3164 m2 h2

T02

=

0.3164 0.4

=

0.791 m

=

263 0.4444

=

592 K

A1 0.24 =2 = A∗ 0.12 A1 From subsonic part of isentropic table, for A ∗ = 2, we get M2 = 0.3

and

T1 = 0.9823 T01

For isentropic flow, T01 = T02 . Therefore, T1

= =

V1

= =

0.9823 × 592 581.5 K / M1 a1 = 0.3 γ R T 145 m/s

29 4.5 For M1 = 0.5, from isentropic table, we have p1 p0

=

0.84

Therefore, p2 p0

=

p2 p1 1.0 × 0.84 × = p1 p0 6.5

=

0.129

For p/p0 = 0.129, from isentropic table, we have M2

=

2.0

A∗ A2

=

0.6

Mass flow rate is given by m ˙

=

= =

ρ1 V1 A1 = p1 M1 A1

3

γ RT1

6.5 × 101325 × 0.5 × 0.016 ×

3

1.4 287.4 × 440

17.54 kg/s

4.6 If p0 is the stagnation pressure and A∗ is the critical throat area, p∗ p0

=

A A∗

=

1

1 + 0.2M 2

1 M

!

2−3.5

5 + M2 6

"3

With these relation we obtain, p1 p2

=

A1 A2

=

!

"−3.5 1 + 0.2M12 1 + 0.2M22 ! "3 M2 5 + M12 M1 5 + M22

where subscripts 1 and 2 refer to entrance and throat of venturi. Substituting proper values into the above two relations, we get

30

One-Dimensional Flow

1.5 1.2

=

4 3

=

!

"−3.5 1 + 0.2M12 1 + 0.2M22 ! "3 M2 5 + M12 M1 5 + M22 ! "3.0 M2 1 + 0.2M12 M1 1 + 0.2M22

=

These two simultaneous equations can be solved to get M1 and M2 !

4 3 Therefore,

!

1.5 1.2 1.5 1.2

"3.0/3.5 "3.0/3.5

M2 M1

=

5 4

=

From above two equations, we get,

=

!

=

M1 M2

4 3 !

1 + 0.2M12 1 + 0.2M22

"−3.0

! "6/7 5 = 1.61 4 "−3.5 1 + 0.2M12 1 + 0.2M22

M1

=

0.46

M2

=

0.74

4.7 p0 is the total pressure and p∞ is the static pressure. 1 2 ρV q∞ = 2 p0 − p∞

=

490 mm of Hg = 0.653 × 105 Pa

p∞

=

(0.35 + 1.0132) × 105 = 1.3632 × 105 Pa

p0

=

(0.653 + 1.3632) × 105 = 2.0162 × 105 Pa

T0

=

25◦ C = 298 K

31 Therefore, T∞

!

"0.286 p∞ T0 p0 ! "0.286 1.3632 298 × = 266 K 2.0162

=

= But,

1 2 2 T∞ 1 + 0.2M∞

T0

=

2 M∞

=

T0 − T∞ 298 − 266 = 0.602 = 0.2T∞ 0.2 × 266

M∞

=

0.776

a∞

=

/ √ γRT∞ = 1.4 × 287 × 266 = 326 m/s

V∞

=

253 m/s

4.8 pi

=

1.0 atm

pf

=

6.0 atm

T1 = 290 K

For isentropic compression, we have γ ! "γ ! " γ−1 ρ2 T2 p2 = = p1 ρ1 T1 (a) Tf = Ti

!

pf pi

γ " γ−1

0.4

= (6) 1.4 = 1.67

Tf = 290 × 1.67 = 484 K

∆T = 484 − 290 = 194 K (b) Change in the internal energy is given by, de = 1.39 × 105 (N m)/kg

=

cv dT =

287 × 194 0.4

32

One-Dimensional Flow

(c) Since the process is isentropic, from first law of thermodynamics, we have de = δq + δw, and δq = 0. Hence, work imparted to the air becomes, δw = δe = 1.39 × 105 N m/kg 4.9 a

=

M

=

/

γRT = 347.21 m/s

180 = 0.518 347.19

The maximum pressure that can be achieved is the isentropic stagnation pressure. Therefore, p0 p

=

!

p0

=

1.013 × 105 × 1.201 Pa

γ−1 2 1+ M 2

γ " γ−1

= 1.201

1.217 × 105 pa

=

4.10 (a) The critical pressure ratio for the sonic condition at the nozzle exit is p∗ = 0.528 p0 If the nozzle is choked, then p2

= =

p∗ = 0.528 × 6.895 × 105 3.64 × 105 Pa

since, p2 > patm , the flow is choked. Hence, the pressure in the exit plane is p2 = p∗ (b) The minimum stagnation pressure for choked flow occurs when p2 = p∗ = patm Thus, p0min =

patm p∗ = = 1.92 × 105 Pa 0.528 0.528

33 (c) For p0 = 1.724 × 105 Pa, p∗ = 0.528 × p0 = 0.91 × 105 Pa. Since p∗ < patm flow will not be choked. Further, subsonic flow is always correctly expanded, hence, p2 = patm = 1.014 × 105 Pa 4.11 Let the reaction force acting on the diffuser be R. Then, R + p1 A1 − p2 A2

=

mV ˙ 2 − mV ˙ 1

R

=

m(V ˙ 2 − V1 ) − (p1 A1 − p2 A2 )

where subscripts 1 and 2 refer to diffuser inlet and exit, respectively. Given, ˙ = 25 kg/s p1 = 0.35 × 105 Pa, V1 = 200 m/s, T1 = 230 K, m From the given data, we get ρ1

=

p1 0.35 × 105 = 0.53 kg/m3 = RT1 287 × 230

A1

=

m ˙ 25 = 0.236 m2 = ρ1 V1 0.53 × 200

200 = 0.66 1.4 × 287 × 230 From isentropic table for M1 = 0.66, we have M1

=

√

p1 T1 = 0.7465, = 0.9199 p01 T01 Thus, p01

=

0.35 × 105 = 0.468 × 105 Pa 0.7465

230 = 250 K 0.9199 For M2 = 0.2, from isentropic table, we have T2 p2 = 0.9725, = 0.9921 p02 T02 T01

p2

=

=

0.9725 p01 (since for isentropic flow p01 = p02 )

=

0.45 × 105 Pa

T2

=

0.9921 T01 = 248 K

V2

=

√ M2 a2 = 0.2 1.4 × 287 × 248 = 63 m/s

34

One-Dimensional Flow

Therefore, R

= =

25 × (63 − 200) − (0.35 × 105 × 0.236 − 0.45 × 105 × 0.5) −10.815 kN

4.12 Let subscripts 1 and 2 refer to entrance and exit of the tank. By energy equation we have, 1 cp T1 + u21 2

=

1 cp T2 + u22 2

T1 − T2

=

u22 − u21 4 × 104 − 1 × 104 = 2cp 2 × 1004

≈

15◦

T2

=

T1 − 15◦

4.13 Work

u1

u2

Figure S4.13 Schematic of the work delivering machine.

Work delivered per unit mass is given by u21 − u22 + cp (T1 − T2 ) = work delivered 2 Given, T1

=

373 K

u1 = 200 m/s

T2

=

288 K

cp = 1004 (N m)/(kg K)

Using this we get, 20000 −

u22 + 1004 × 85 = 2

100000

35

20000 −

u22 + 85340 2

=

u2

=

100000 103.3 m/s

When the machine is idling, u22 2

=

u2idling

=

20000 + 85340 = 105350 459 m/s

4.14 Let jets “1” and “2” be denoted by the subscripts 1 and 2, and let T0 denote the temperature in the reservoir. For q = 0, adiabatic energy equation gives, ! 2 " ! 2 " u2 u1 + c p T2 + m + cp T1 = 2mcp T0 m 2 2 This gives, T0

=

u2 + u22 T1 + T2 + 1 2 4cp

=

300 +

=

(1 + 9) × 104 4 × 1004

324.9 K

4.15 Let T0 denote the temperature in the tyre. Since the process is adiabatic, we have u2 2

=

c p T0

cp (T0 − T )

=

u2 2

u2

=

2cp (T0 − T )

cp T +

u = =

√

2 × 1004 × 37

272.57 m/s

36

One-Dimensional Flow

4.16 At 15000 m, we have

Speed of sound

T

=

−56.5◦ C = 216.5 K

p

=

1.206 × 104 Pa

a

=

/

γRT

=

√

=

294.94 m/s

1.4 × 287 × 216.5

Speed of the airplane is, u

=

800 km/hr = 800 ×

=

222.22 m/s

1000 3600

This gives the Mach number as M

=

222.22 u = a 294.86

=

0.753

(a) Maximum possible temperature of the airplane skin will be the stagnation temperature, at the nose of the airplane. Thus, it is the total temperature of the air, T0 T

=

1.1134

T0

=

1.1134 × 216.5

=

241.05 K

(b) Maximum possible pressure that can be felt by the airplane cannot exceed the stagnation pressure. This will be felt at the place where air comes to complete rest, i.e., at the nose of the airplane and other similar places. Thus, p0 = 1/0.6866 p p0

=

1.206 × 104 0.6866

=

1.756 × 104 Pa

37 (c) Critical velocity of air relative to the airplane is 3 2 a0 a∗ = γ+1 √ √ γRT0 γRT0 ∗ √ = √ a = 1.2 1.2 =

284.1 m/s

(d) Vmax

= =

Vmax

= =

3

3

2 ao γ−1 γ+1 ∗ a γ−1

√ 6a∗

695.9 m/s

4.17

M = 0.6 Area = A

Figure S4.17 Schematic of convergent channel.

The mass flow rate is given by m ˙

= =

=

ρAV / p A M γRT RT 3 3 γ p T0 1 √ MA p0 R p0 T T0

From the problem and theory we know that,

38

One-Dimensional Flow

T0 = 550 K

p p0

=

0.7840

p0 = 2 × 105 Pa

T T0

=

0.9328

=

1.0354

=

4

3

T0 T 3 γ R

m ˙ = 29.188 kg/s M = 0.6

= =

/ √

γ cp γ−1 γ γ cp (γ − 1)

1.4 = 0.0694 1017 × 0.4

This gives A =

=

/γ

p R p 0 p0

m ˙ 0

T0 √1 M T T0

0.10125 m2

4.18

1 m2 2m2

Ae = 4 m 2

Figure S4.18 Schematic of convergent channel.

At the mouth of the duct p

=

patm = 1.013 × 105 Pa

T0

=

288 K

(at sea level)

39 (a)

Maximum mass flow rate is given by m ˙ max

=

p 1 √ 0 A∗ 24.741 T0

=

1 1.013 × 105 √ ×1 24.741 288

=

241.26 kg/s

(b) Ae A∗

=

4,

A∗ = 0.25 Ae

For this area ratio, from isentropic table, we get pe = 0.0298 p0

Me

=

2.94,

pe

=

0.0298 × 1.013 × 105 Pa

=

3018.7 Pa

4.19

A∗ 1 m2

p, T, ρ M =2

Figure S4.19 Schematic of convergent channel.

Here we have to find m, ˙ p∗ , T ∗ , ρ∗ , A, p, T , and ρ. ρ0

m ˙

=

p0 7 × 105 kg/m3 = RT0 287 × 313

=

7.8 kg/m3

=

0.6847 × p0 A∗ √ RT0

=

1599.133 kg/s

40

One-Dimensional Flow

Fluid properties at the throat T∗ = 0.8333 T0

=⇒

T ∗ = 261 K

p∗ = 0.5283 p0

=⇒

p∗ = 3.7 × 105 Pa

ρ∗ = 0.6339 ρ0

=⇒

ρ∗ = 4.944 kg/m3

For the test-section Mach number 2, from isentropic table, we have T = 0.5556 T0

=⇒

T = 173.9 K

p = 0.1278 p0

=⇒

p = 8.946 × 104 Pa

ρ = 0.2300 ρ0

=⇒

ρ = 1.794 kg/m3

A 1 = A∗ 0.5926

=⇒

A = 1.6875 m2

4.20

60 m/s

245 m/s 1 2

Figure S4.20 Schematic of divergent channel.

V1 p1 m ˙

= = =

245 m/s 1 × 105 Pa 13.6 kg/s

(a) ρ1

= =

a1

=

p1 1 × 105 = RT1 287 × 300 1.1614 kg/m3 / γRT1 = 347.2 m/s

V2 T1

= =

60 m/s 300 K

41

M1

=

245 = 0.7056 ≈ 0.706 347.2

p1 p0

= 0.7171

=⇒

T1 T0

= 0.909

=⇒

A1

=

m ˙ 13.6 m2 = 0.0478 m2 = ρ1 V1 1.614 × 245

A1

=

1 πD12 4

=

0.2467 m

=⇒

p0 = 1.39 × 105 Pa T0 = 330 K

D1

(b) At the outlet, the energy equation is c p T2 +

V22 = c p T0 2

Therefore, T2

=

T0 −

a2

=

/

M2

=

V22 602 = 328.2 K = 330 − 2cp 2 × 1004

γRT2 = 363.1 m/s

V2 = 0.165 a2

For M1 = 0.706, A1 /A∗ = 1.09 and for M2 = 0.165, A2 /A∗ = 3.57. Thus gives A2 = 0.1566 m2 and D2 = 0.4465 m (c) The rise in static temperature is T2 − T1 = 28.2 K 4.21 p02

=

2.0 × 105 Pa

p1

=

0.15 × 105 Pa

By Rayleigh supersonic pitot formula, &1/(γ−1) % 2γ γ−1 2 M − 1 γ+1 γ+1 p1 = 1 γ+1 2 2γ/(γ+1) p02 M 2

1

42

One-Dimensional Flow

p02 p1

=

=

!

γ+1 2 Mi 2

γ+1 2 M1 2

"γ/(γ−1)

'

%

1 2γ 2 γ+1 M1

γ+1 2 2 M1 2γ γ−1 2 γ+1 M1 − γ+1

−

γ−1 γ+1

1 * γ−1

Select M1 and calculate p02 for given p1 . Check that

&1/(γ−1)

p1 p01

is less than

p1 p02 .

Trial and error method gives M1 = 3.16 . Also, check p02 /p01 calculated with normal shock relation and isentropic relation agree exactly. For this problem p02 /p01 = 0.286. 4.22 m ˙ max

=

241 kg/s

A = 3 A∗ For this area ratio from isentropic table, M = 2.64. This is the Mach number just upstream of the shock. Let the conditions just upstream and downstream of the shock be represented by subscripts 1 and 2. From Normal shock table, for M1 = 2.64, we have p02 p2 M2 = 0.5 = 0.4452 = 7.9645 p01 p1 Therefore, p02

=

0.4452 × p01 = 0.4452 × 1.0133 × 105 = 45110 Pa

For M2 = 0.5, from isentropic table, we have A2 = 1.3398 A∗2 Therefore, A∗2

=

3 A2 = = 2.239 m2 1.3398 1.3398

Ae A∗2

=

4 = 1.786 2.239

From isentropic table, for

Ae = 1.786, we have A∗2 Me

=

0.35

pe p0e

=

0.9187

43 But p0e = p02 . Therefore, pe

0.9187 × 45110

= =

41442.5 Pa

4.23 p1 = 0.7 × 105 Pa T1 = 300 K A1 = 0.15 m

V1 = 240 m/s 2

V2 = 120 m/s

(a) m ˙

p1 A1 V1 RT1

=

ρ1 A1 V1 =

=

0.7 × 105 × 0.15 × 240 287 × 300

=

29.26 kg/s

(b) a1

=

M1

=

/

γRT1 = 347.5 m/s

V1 = 0.69 A1

Thus, p1 p01

=

0.72735

T1 T01

=

0.9131

p01

=

0.9624 × 105 Pa

T01

=

328.6 K

Thus, stagnation pressure at the exit p02 = p01 = 0.9624 × 105 Pa. (c) From (b) above, the stagnation temperature at the exit is T02 = T01 = 328.6 K

44

One-Dimensional Flow

(d) 1 cp T2 + V22 2

=

c p T0

T2

=

T0 −

A2

=

/

M2

=

V2 = 0.334 A2

p2 p02

=

0.9257

V22 1202 ) = 321.4 K = 328.6 − 2cp 2 × 1004

γRT2 = 359 m/s

Therefore, the static pressure at the exit is p2 = 8.91 × 105 Pa (e) Entropy change across the diffuser is zero since the flow is isentropic. (f ) For M1 = 0.69, the exit area,

A1 A∗

= 1.1018, and for M2 = 0.334, A2 =

A2 A∗

= 1.850. This gives

A2 A∗ A1 = 0.252 m2 A∗ A1

4.24 The pressure, density, and temperature in the settling chamber can be taken as stagnation quantities. Therefore, p0 = 1.014 × 105 Pa,

ρ0 = 1.44 kg/m3 ,

T0 = 35 + 273 = 308 K

Since the effects of viscosity is neglected, the flow in the working section can be treated as isentropic. For M = 0.8, from isentropic table, we have p = 0.656, p0

ρ = 0.740, ρ0

T = 0.886 T0

Hence, p

= =

ρ = =

0.656 p0 = 0.656 × 1.014 × 105 0.665 × 105 Pa 0.740 ρ0 = 0.740 × 1.144 0.847 kg/m2

45

T

= =

0.886 T0 = 0.886 × 308 273 K

4.25 Shock 1 2

M1 = 3

Ath

Ae

Figure S4.25 Schematic of convergent-divergent channel.

Given,

Ae = 11.91, M1 = 3.0 At

M1 = 3.0 gives, M2 = 0.4752 downstream of the normal shock, and pp02 = 01 0.3281. This gives, p02 = 2.2981 × 105 Pa. Since the flow is adiabatic, T02 = T01 = 500 K. And since the flow is isentropic between the downstream of the normal shock and the nozzle exit, T0e = T02 , and p0e = p02 . A1 A1 A2 = 4.2346, and for M2 = 0.4752, A For M1 = 3.0, A ∗ = A ∗ = 1.390(from t 1 2 isentropic table). Also, since the shock is very thin, the area before and after the shock can be taken as the same, i.e. A1 = A2 . Therefore, ! " A Ae = ∗ A e A∗2

=

Ae At A2 At A1 A∗2

=

3.9094

For this area ratio, from isentropic table, we get Me

=

0.15

Te T0e

=

0.9955

pe p0e

=

0.9844

46

One-Dimensional Flow

Thus, pe

=

2.2623 × 105 Pa

Te

=

497.8 K

4.26 M

=

2.5

p0

=

Ats

=

1 m2

T0

=

7 × 105 Pa

27 + 273 = 300 K

(a) For M = 2.5 we have, A A∗

=

A∗

=

2.637 0.38 m2

(b) T∗ T0

=

0.833

T∗

=

0.833 × 300 = 249.9 K

=

− 23.1 ◦ C

(c) M

=

a

=

V

=

V a /

γRT = 20.04 ×

√ T = 231.5 m/s

M a = 578.75 m/s

(d) m ˙ A

=

p0 1 √ × A 24.743 T0 A∗

=

1 7 × 105 √ × 2.637 24.743 × 300

47 = 620 m ˙

=

620 kg/s

4.27 By definition, we have the pressure coefficient as CP

=

CP∗

=

= =

=

=

p − p∞ q∞ " ! ∗ p −1 p∞ p∞ q∞ " ! ∗ p 2p∞ −1 p∞ 2 γp∞ M∞ " ! ∗ p p0 2 −1 2 γM∞ p0 p∞ '! * 1 2 γ γ "− γ−1 2 γ−1 2 + (γ − 1)M∞ γ+1 2 −1 γ 2 γM∞ 2 2 γ−1 1  2 γ 2 γ−1 2  2 + (γ − 1)M∞ − 1 γ 1 γ+1 2 γ−1 γ 2 γM∞ × 2 γ−1 2

=

91

2 : γ 2 2 + (γ − 1)M∞ /(γ + 1) γ−1 − 1 2 /2 γM∞

4.28

p0 = 500 kPa, T0 = 30◦ C = 30 + 273.15 = 303.15 K 101325 The nozzle has to choke since pe /p0 = = 0.203, which is well below 500 × 103 the critical pressure ratio of 0.528. Therefore, m ˙ =m ˙ max =

0.6847 p0 ∗ √ A RT0

where R = 287 m2 /(s2 K) for air. Thus, m ˙

=

0.6847 × 500 × 103 √ × 0.5 × 10−4 287 × 303.15

=

0.058 kg/s

48

One-Dimensional Flow

4.29 The pressure ratio 0.1429, we get

pe pe 1 = = 0.1429. From isentropic table, for = p0 7 p0 Me = 1.93,

Te = 0.57307 T0

T0 = 180 + 273.15 = 453.15 K Therefore, Te = 259.7 K. √ The speed of sound ae = γ RTe = 323 m/s. The exit velocity is Ve

=

Me ae = 1.93 × 323

=

623.4 m/s

4.30 The pressure ratio p/p0 across the nozzle is 1/5. This is well below the critical pressure ratio and therefore the flow is chocked at the exit. The flow is adiabatic and frictionless, therefore, the maximum mass flow rate is given by m ˙

=

0.6847 × 5 × 105 √ × 6.5 × 10−4 287 × 288.15

=

0.774 kg/s

Thus, from isentropic table, for M = 1. we get T T0

=

T

=

0.83333 240.12 K

4.31 The mass flow rate may be expressed as m ˙ =

0.6847 p0 √ Ath RT0

where p0 and T0 are the stagnation pressure and temperature, respectively. The pressure ratio we get

pe pe 91.4 = 0.90495. From isentropic table, for = = 0.905, p0 101 p0 Me = 0.38 and

Ae = 1.6587 Ath

49 Therefore, Ath =

0.033 = 0.0199 m2 . Thus, 1.6587 m ˙

=

0.6847 × 101 × 103 × 0.0199 √ kg/s 287 × 293.15

=

4.74 kg/s

At the section with A = 0.022 m2 , the area ratio is A 0.022 = 1.1055 = Ath 0.0199 The corresponding Mach number is M = 0.69 and p = 0.72735 p0 Thus, the pressure at A = 0.022 m2 is

p = 73.5 kPa

4.32 Nozzle is adapted, therefore the exit pressure pe = p16000 at 16000 m altitude. From standard atmospheric table, we get p16000 = 10.299 kPa = pe Therefore, pe p0

=

10299 , since 1 atm = 101325 Pa 15 × 101325

=

0.0067762

By isentropic relation, p0 = pe 1 + 0.2

!

γ−1 Me2 1+ 2

Me2

=

!

Me

=

3.98

γ " γ−1

1 0.0067762

"0.286

For Me = 3.98, from isentropic table, we get Ae Te = 10.53 , = 0.23992 ∗ A T0 Te

= =

0.23992 × (2600 + 273.15) 689.33 K

50

One-Dimensional Flow

The exit velocity becomes Ve

= M e a e = Me

/

γRTe

√ = 3.98 1.4 × 287 × 689.33 = Thrust

2094.6 m/s

= m ˙ e Ve = (ρe Ae Ve ) Ve = ρe Ae Ve2

9000 By state equation, we have

pe 10299 = 0.052 kg/m3 = RTe 287 × 689.33

ρe = Therefore,

Ae = Thus,

9000 0.052 × (2094.6)

A∗ = Ath =

2

= 0.0394 m2

0.0394 = 0.00374 m2 10.53

4.33 Given, V = 200 m/s and T = 15 + 273.15 = 288.15 K. The speed of sound is given by / √ a = γRT = 1.4 × 287 × 288.15 =

340.3 m/s

The Mach number becomes M= From isentropic table, for p p0

=

V = 0.59 a

M = 0.59, we get

0.79013,

T ρ = 0.93491, = 0.84514 T0 ρ0

Thus, p0

=

101 0.79013

51

T0

ρ0

=

127.8 kPa

=

288.15 0.93491

=

308.2 K

=

p 1 ρ = × 0.84514 RT 0.84514

=

101 × 103 287 × 288.15 × 0.84514

=

1.445 kg/m3

ρ0 is also given by ρ0

=

p0 127.8 × 103 = R T0 287 × 308.2

=

1.445 kg/m3

Note: This problem may also be solved using the isentropic relations directly, instead of table. 4.34 The sound speed at station 1 is / √ γRT1 = 1.4 × 287 × 303.15 = 349 m/s a1 = M1

=

V1 90.5 = 0.26 = a1 349

For this flow with M1 = 0.26 to choke, the area ratio tropic table) is

A1 required (from isenA∗

A1 = 2.317 A∗ The present area ratio 2.317 = 1.60.

6.9 A2 = 0.69. Therefore, = A1 10

A2 A2 A1 = = 0.69 × ∗ A A1 A∗

For this area ratio, from isentropic table, we get M2 = 0.4 ,

p2 T2 = 0.89561, = 0.96899 p02 T02

For M1 = 0.26, from isentropic table, we get p1 T1 = 0.95408, = 0.98666 p01 T01

52

One-Dimensional Flow p01

=

T01

=

For isentropic flow, T01 = T02

100 = 104.8 kPa 0.95408

303.15 = 307.25 K 0.98666 and p01 = p02 . Therefore,

p2

=

0.89561 × 104.8 = 93.86 kPa

T2

=

0.96888 × 307.25 = 297.72 K

4.35 For CO2 the molecular weight is 44 and γ = 1.3. The gas constant for CO2 is 8314 = 189 J/(kg K) R= 44 The stagnation pressure p0 = 6 atm and the back pressure pa = 1 atm. Therefore, the pressure ratio pa 1 = = 0.167 p0 6 which is well below the critical pressure ratio of 0.54573, required for the flow to choke. Hence, the flow is choked at the orifice. That is, M = 1 at the orifice. From the isentropic table, for M = 1, we get T = 0.86957 T0 Thus, T

= =

0.86957 × (30 + 273.15) = 263.6 K −9.55◦ C

This is the temperature with which CO2 comes out of the orifice. The mass flow rate m ˙ = ρ AV . By state equation, ρ

=

0.54573 × 6 × 101325 p = RT 189 × 263.6

=

6.66 kg/m3

A

=

V

= =

1 22 π 1 × 10−3 π = × 10−6 m2 4 4 / √ M a = M γ R T = 1 × 1.3 × 189 × 263.6 254.5 m/s

53 Thus, m ˙

= =

6.66 ×

π × 10−6 × 254.5 4

0.00133 kg/s

Aliter: The mass flow rate (for the gas with γ = 1.3) can also be expressed as m ˙

=

0.6672 × p0 × Ath √ R T0

=

0.6672 × 6 × 101325 × √ 189 × 303

=

0.00133 kg/s

π × 10−6 4

4.36 Let subscripts 1 and 2 refer to inlet and exit of the nozzle. V1

=

m ˙ 0.7 = ρ1 A1 8 × 12 × 10−4

= 72.92 m/s M1

=

V1 V1 =√ a1 γ RT1

=

√

72.92 = 0.18 1.4 × 287 × 400

p1 = ρ1 RT1 = 8 × 287 × 400 = 918.4 kPa From isentropic table, for M1 = 0.18, we get p1 p01

=

0.97765

T1 T01

=

0.99356

p01

=

918.4 = 939.4 kPa 0.97765

T01

=

400 = 402.59 K 0.99356

54

One-Dimensional Flow

p2

ρ2 RT2 = 4 × 287 × 300

= =

p2 p2 = , p02 p01

344.4 kPa

since the flow is isentropic. Therefore, p2 344.4 = 0.3666 = p01 939.4 p2 = 0.3666, we get p01

From isentropic table, for

M2 = 1.29 Therefore, V2

/

=

M2 a 2 = M2

=

√ 1.29 1.4 × 287 × 300

=

γ R T2

447.87 m/s

Mass flow rate is m ˙

=

ρ2 A2 V2

A2

=

m ˙ 0.7 = ρ2 V2 4 × 447.87

=

3.9 cm2

4.37 From energy equation, we have T0

=

T1 +

V12 2 cp

where the subscripts 0 and 1 refer to stagnation and inlet conditions. T0

=

400 +

=

405 K

1002 2 × 1004.5

55 By isentropic relation, we have γ ! " γ−1 T0 p0 = p1 T1 3

!

405 400

p0

=

200 × 10

pe p0

=

150 = 0.718 209

"3.5

= 209 kPa

By isentropic relation, we have p0 pe

=

1 + 0.2Me2

=

1

1 + 0.2 Me2 1

23.5

(1.393) 3.5

Solving we get,

The speed of sound

Me

=

0.7

T0 Te

=

1+

Te

=

405 = 368.85 K 1.098

ae =

γ−1 2 Me = 1.098 2

√ γRTe = 385 m/s.

Thus, the exit velocity is Ve

= =

Me ae = 0.7 × 385 269.5 m/s

By continuity, ρ1 A1 V1 = ρe Ae Ve . or p1 pe A1 V1 = Ae Ve RT1 RTe Ae

=

p1 T e V 1 A1 pe T 1 V e

=

1 22 π 75 × 10−3 100 200 368.85 × × × 150 400 269.5 4

=

0.00202 m2

56

One-Dimensional Flow

Thus, de

3

= =

4 × 0.00202 = 0.0507 m π

50.7 mm

The mass flow rate is m ˙

=

ρ1 A1 V1

=

22 1 p1 π 75 × 10−3 × 100 RT1 4

=

π × 752 × 10−6 200 × 103 × × 100 287 × 400 4

=

0.77 kg/s

With the exit conditions, m ˙

=

ρe Ae Ve

=

150 × 103 × 0.00202 × 269.5 287 × 368.85

=

0.77 kg/s

4.38 The flow process is described by the relation pVγ = constant. Per unit mass of air, pv γ = constant. Therefore,

! By state equation, we have v1

p1 V1γ

=

p2 V2γ

"γ

=

p1 p2

v2 v1

pv = RT . Therefore,

=

R T1 287 × (400 + 273.15) = p1 3 × 106

=

0.0644 m3 /kg

Thus, v2

=

v1

!

p1 p2

" γ1

= 0.0644

!

" 1 3 1.4 0.5

57

T2

=

0.2316 m3 /kg

=

0.5 × 106 × 0.2316 p2 v 2 = R 287

=

403.4 K

(a) By energy equation h2 +

V22 2

=

h1 +

V12 2

where V1 and V2 are the velocity at the entrance and exit of the nozzle. Also, it is reasonable to assume that V1 is very small and hence can be taken as zero. Thus, V2 h2 + 2 = h1 2 Further, for air being a perfect gas, h = cp T . Therefore, V22

=

2 (h1 − h2 ) = 2cp (T1 − T2 )

=

2 × 1004.5 (673.15 − 403.4)

V2

=

736 m/s

a2

=

/

=

402.6 m/s

γRT2 =

√ 1.4 × 287 × 403.4

Therefore, M2

=

V2 736 = a2 402.6

=

1.83

(b) Since the flow at the nozzle exit is supersonic, at the throat the flow is choked. That is, at the throat Mth = 1. From isentropic table, for M = 1, we get p∗ p0

=

0.52828

58

One-Dimensional Flow T∗ T0

=

0.83333

p∗

=

0.52828 × 3 × 106 = 1.58484 MPa

T∗

=

0.83333 × 673.15 = 560.96 K

a∗

=

/

m ˙

=

ρ∗ Ath a∗

ρ∗

=

p∗ 1.584 × 106 = ∗ RT 287 × 560.96

=

9.8 kg/m3

γRT ∗ = 474.76 m/s

Thus, Ath

(c) The mass flow rate is

=

15.89 9.8 × 474.76

=

3415 mm2

m ˙ = ρ2 A2 V2 . ρ2 =

1 = 4.318 kg/m3 v2

Therefore, m ˙ = =

4.318 × 5000 × 10−6 × 736 15.89 kg/s

4.39 First of all, let us check whether the flow is isentropic. The change in entropy (assuming air to be calorically perfect gas) is ! " ! " T2 p2 s2 − s1 = cp ln − R ln T1 p1 ! " ! " 290 101 = 1004.5 ln − 287 ln 300 200 =

162 J/(kg K)

59 Hence, the flow is nonisentropic. Consider the inner surface of the nozzle as the control volume. Let F be the force acting on the nozzle, due to the flow. By momentum balance, we have p1 A1 − p2 A2 − F

=

m ˙ (V2 − V1 )

V1

=

m ˙ ρ1 A1

ρ1

=

p1 200 × 103 = RT1 287 × 300

=

2.32 kg/m3

=

5 = 10.78 m/s 2.32 × 0.2

V1 By energy equation, we have

h1 − h2

=

V2 V22 − 1 2 2

2 cp (T1 − T2 )

=

V22 − V12 = V22 − 10.782

2 × 1004.5 (300 − 290) + 10.782

=

V22

V2

=

142.15 m/s

The exit area is given by A2 =

m ˙ ρ2 V2

where ρ2

A2

=

p2 101 × 103 = RT2 287 × 290

=

1.214 kg/m3

=

5 = 0.029 m2 1.214 × 142.15

Using these values in the momentum equation, we get

200 × 103 × 0.2 − 101 × 103 × 0.029 − F = 5 (142.15 − 10.78) Thus, F

=

40000 − 2929 − 656.85

60

One-Dimensional Flow

=

36414.15 N

4.40 Let us solve the problem by using relations and also by using gas tables. The speed of sound at the flight altitude is given by / / a = γRT = 1.4 × 287 × (−15 + 273.15) =

322.1 m/s

The flight speed V = M a = 0.8 × 322.1 = 257.68 m/s. The stagnation enthalpy h0 is given by h0 = h + For an ideal gas h0 = cp T +

V2 2

V2 and therefore, 2 c p T0 = c p T +

or T0 = T +

V2 2

V2 2cp

Thus, 2

T0

=

258.15 +

=

291.2 K

(257.68) 2 × 1004.5

By isentropic relation, we have p0 p

=

p0

= =

Using gas tables

" γ T0 γ − 1 T ! "3.5 291.2 44 258.15

!

67.08 kPa

61 For M = 0.8, from isentropic table, we get p T = 0.65602 = 0.88652 p0 T0 Thus, p0

=

44 = 67.07 kPa 0.65602

T0

=

258.15 = 291.2 K 0.88652

Note: From the above solution it is seen that, use of the gas tables has tremendous advantage over using equations directly. 4.41 Static temperature T = 216 K. Total temperature T0 = 40 + 273.15 = 313.15 K. From isentropic relation, we have T0 T 1.45

γ−1 2 M 2 = 1 + 0.2 M 2

= 1+

M2

=

0.45 = 2.25 0.2

M

=

1.5

The pressure ratio p0 /p given by isentropic relation is p0 p

= =

p0

=

! 9

γ−1 2 1+ M 2

1 + 0.2 × 1.52

γ " γ−1

:3.5

= 3.671

(3.671)(0.55) = 2.019 atm

Note: This problem can be solved by using gas tables instead of solving the isentropic relations, as follows. For

T = 0.689655, from isentropic tables, we get T0 M = 1.5 and p/p0 = 0.2724

Therefore, p0

=

0.55 = 2.019 atm 0.2724

62

One-Dimensional Flow

4.42 By energy equation, we have T01

V12 2cp

=

T02 = T1 +

=

(70 + 273.15) +

=

348.13 K

1002 2 × 1004.5

Similarly, T2

T02 −

=

4002 = 268.49 K 2 × 1004.5 / √ γRT2 = 1.4 × 287 × 268.49 348.13 −

= a2

V22 2cp

= =

328.45 m/s

=

V2 400 = a2 328.45

=

1.218

Therefore, M2

Vmax

/

= =

2 cp T01 =

√

2 × 1004.5 × 348.13

836.3 m/s

By isentropic relation,

Thus,

p01

=

M1

=

p01

=

p02

=

1 23.5 p1 1 + 0.2 M12

V1 = 0.269 γRT1 23.5 1 2.5 1 + 0.2 × 0.2692 = 2.63 atm √

% &3.5 2 p2 1 + 0.2 × (1.218) = 1.24 atm p02 1.24 = 0.471 = p01 2.63

63 4.43 Given, p0 = 350 kPa, T0 = 420 K Ae = 0.22 m2 , Ve = 525 m/s By energy equation we have T0 = Te +

Te

= T0 − =

Ve2 2cp

Ve2 5252 = 420 − 2 cp 2 × 1004.5

282.8 K

√ Therefore, the speed of sound at nozzle exit ae = γRTe . That is, √ ae = 1.4 × 287 × 282.8 = 337.01 m/s Thus, Me

pe

ρe

m ˙

=

Ve 525 = ae 337.01

=

1.56

=

p0 (1 +

3.5 0.2Me2 )

=

350 4.007

=

87.35 kPa

=

pe 87.35 × 103 = RTe 287 × 282.8

=

1.076 kg/m3

= =

ρe Ae Ve = 1.076 × 0.22 × 525 124.3 kg/s

From isentropic table, for Me = 1.56, we get Ae = 1.219 A∗ Thus, A∗ =

0.22 = 0.18 m2 1.219

64

One-Dimensional Flow

4.44 The speed of sound at 1 is / √ a1 = γRT1 = 1.4 × 287 × 315 = 355.76 m/s The Mach number at 1 is M1 =

V1 150 = 0.42. = a1 355.76

Now, for M1 = 0.42, the corresponding area ratio A1 /A∗ may be calculated using the area–Mach number relation for isentropic flow, or the value of A1 /A∗ may be read directly from isentropic table. From isentropic table for M1 = 0.42, we get A1 /A∗ = 1.52891 Thus, A1 = 1.52891 × 25 = 38.22 cm2

The mass flow rate m ˙ = ρ1 A1 V1 ρ1

=

p1 152 × 103 = RT1 287 × 315

=

1.681 kg/m3

Therefore, m ˙ = 1.681 × 38.22 × 10−4 × 150 = 0.9637 kg/s 4.45

T0 = 40◦ C = 40 + 273.15 = 313.15 K, √ The exit velocity Ve = 200 m/s = Me γRTe .

γ = 1.4

By isentropic relation we have 23.5 1 p0 = 1 + 0.2 Me2 pe

By energy relation, we have

Ve2 2cp

T0

=

Te +

Te

=

313.15 −

=

/

=

343.25 m/s

The speed of sound is ae

2002 = 293.24 K 2 × 1004.5

γRTe =

√ 1.4 × 287 × 293.24

65 The Mach number at the exit is Me

p0 pe

=

Ve 200 = ae 343.25

=

0.584 1

=

1 + 0.2 × 0.5832

23.5

= 1.259

For the convergent nozzle, pexit has to be equal to the atmospheric pressure, since the subsonic flow exiting a convergent nozzle will always be correctly expanded. pe

Thus,

=

ρgh = 13.6 × 103 × 9.81 × 22.7 × 10−3

=

3028.54 Pa (gauge)

=

101325 + 3028.54 = 104353.54 Pa (absolute)

pa = 104353.54 Pa

The storage pressure is given by p0

=

1.259 × 104353.54

=

131381.11 Pa

=

224.7 mm of Hg (gauge)

Note: The standard atmospheric pressure is patm = 101325 Pa = 760 mm of Hg. 4.46 Let the first and second sections are represented by subscripts 1 and 2, respectively. At section 1, M1

V1 365 =√ γRT1 1.4 × 287 × 305.15

=

√

=

1.04

From isentropic table, for

M1 = 1.041, we get p1 p0

=

0.50389

T1 T0

=

0.82215

66

One-Dimensional Flow

Therefore, p0 = 158.76 kPa,

T0 = 371.16 K.

At station 2, p2 p0 From isentropic table, for

120 = 0.7559 158.76

=

p2 /p0 = 0.7559, we get T2 = 0.92428 T0

M2

=

T2

=

343.06 K = 69.91◦ C

a2

=

/

Therefore,

0.64

γRT2 = 371.27 m/s

V2 = M2 a2 = 237.6 m/s 4.47 For M = 3.0, from isentropic table, we get p = 0.027224, p0

T = 0.35714, T0

A = 4.23456 A∗

Thus, A∗

=

0.05 = 0.0118 m2 4.23456

p0

=

0.2 × 101325 = 744.4 kPa 0.027224

T0

=

300 = 840 K 0.35714

The mass flow rate is given by m ˙

=

0.6847 p0 A∗ √ RT0

=

0.6847 × 7.444 × 105 × 0.0118 √ 287 × 840

=

12.25 kg/s

4.48 The velocity at nozzle entrance is very low. Hence, the pressure and temperature at the entrance can be taken as the stagnation pressure and stagnation temperature. That is, p0 = 1 MPa,

T0 = 300 K

67 From isentropic table, for M = 2, we have p = 0.1278, p0

T = 0.55556, T0

ρ = 0.23005 ρ0

Therefore, p

=

0.1278 × 1 × 106 = 127.8 kPa

T

=

0.55556 × 300 = 166.67 K

ρ =

0.23005 × ρ0 = (0.23005) (p0 /RT0 )

" 106 = 2.672 kg/m3 287 × 300 √ √ a = γRT = 1.4 × 287 × 166.67 = 258.78 m/s. There0.23005

V

=

M a = 517.56 m/s

m ˙

=

ρAV = 2.672 × 0.15 × 517.56

The speed of sound fore,

=

4.49 Given,

!

=

207.43 kg/s

p0 = 3.5 MPa, T0 = 500◦ C = 773.15 K pe = 0.7 MPa,

m ˙ = 1.3 kg/s

At the nozzle exit, the pressure ratio is pe 0.7 = 0.2 = p0 3.5 From isentropic table, for this pressure ratio, we get Me = 1.71 ,

Te Ae = 0.63099, = 1.3471 T0 Ath

Therefore, Te = 0.63099 × 773.15 = 487.85 K The corresponding speed of sound ae is / √ ae = γRTe = 1.4 × 287 × 487.85 = 442.74 m/s

68

One-Dimensional Flow Ve = Me ae = 757 m/s

The exit velocity

ρe =

pe , by state equation RTe

Therefore, 0.7 × 107 = 4.9 kg/m3 287 × 487.85

ρe = The mass flow rate

m ˙ = ρe Ae Ve . Thus, Ae

Ath

4.50 Given,

=

m ˙ 1.3 = ρe Ve 4.9 × 757

=

3.5 cm2

=

3.5 Ae = 1.3471 1.3471

=

2.6 cm2

T0 = 25◦ C = 298.15 K, Ae = 15 cm2 .

Applying momentum analysis to the control volume considered, we get F = 100 N = ρe Ae Ve2 Assuming air to be an ideal gas, Ae ρe Ve2 =

Ae pe 2 Ae γpe 2 Ve = V = Ae γpe Me2 RTe γRTe e

Also, γ Ae pe Me2 = 100 N The nozzle exit flow is subsonic and hence has to be correctly expanded with pe = patm = 101325 Pa. Thus, Me2

=

100 = 0.47 1.4 × 15 × 10−4 × 101325

Me

=

0.69

From isentropic table, for Me = 0.69, we get Te = 0.91306, T0

pe = 0.72735 p0

69 Thus, Te

p0

Ve

=

298.15 × 0.91306 = 272.23 K

=

−0.92◦ C

=

101325 = 139.3 kPa 0.72735

=

1.37 atm

=

Me a e = Me

=

0.69 ×

=

/

γRTe

√ 1.4 × 287 × 272.23

228.2 m/s

4.51 Given, m ˙ = 1 kg/s, pe = 101.325 kPa. The mass flow rate may be expressed as m ˙ =

0.6847 p0 ∗ √ A RT0

Thus, ∗

A

= = =

At the exit,

√ RT0 0.6847 p0 √ 287 × 325 0.6847 × 700 × 103 6.37 cm2

pe 101325 = = 0.14475 p0 700 × 103

From isentropic table, the corresponding Mach number Me and temperature ratio are the following. Me

=

1.92

Te T0

=

0.57561

Te

=

0.57561 × 325 = 187 K

70

One-Dimensional Flow

The speed of sound ae = Thus,

√

Ve

γ RTe = 274.1 m/s. Me ae = 1.92 × 274.1

= =

526.27 m/s

4.52 Assuming air to be a perfect gas, we have γ = 1.4 and R = 287 J/(kg K). The speed of sound at nozzle inlet is / √ γRT1 = 1.4 × 287 × 300 a1 = = 347.2 m/s The inlet Mach number is M1 =

V1 100 = 0.29 = a1 347.2

From isentropic table, for M1 = 0.29, we get p1 p01

=

0.94329

T1 T01

=

0.98346

p01

=

101325 = 107416.6 Pa 0.94329

T01

=

300 = 305 K 0.98346

The flow is supersonic at the exit, therefore at the throat M ∗ = 1. For M = 1, from isentropic table we have p∗ p0

=

0.52828

T∗ T0

=

0.8333

Thus, p∗

= =

0.52828 × 107416.6, 56746 Pa

since p∗0 = p01

71

= T∗

0.56 atm 0.8333 × 305

= =

254 K

Since the flow is isentropic, T02

=

T01 = 305, K

p02

=

p01 = 107416.6 Pa

=

1.06 atm

The mass flow rate is m ˙

=

ρ1 A1 V1 = ρt At Vt

=

p1 101325 × 5 × 10−4 × 100 A1 V1 = RT1 287 × 300

=

0.0588 kg/s

At

=

0.0588 ρt Vt

ρt

=

pt 56746 = RTt 287 × 254

=

0.78 kg/m3

=

at =

=

319.5 m/s

Thus,

Vt

Therefore, At =

/

γRTt =

√ 1.4 × 287 × 254

0.0588 = 2.36 cm2 0.78 × 319.5

4.53 Let subscripts 01 and 02 refer to properties at reservoirs 1 and 2, respectively. Pressure p02 is the back pressure for the nozzle. Therefore, the nozzle pressure ratio (NPR) becomes p02 3 = = 0.5 p01 6

72

One-Dimensional Flow

Also,

!

p∗ = p01

2 γ+1

"(γ)γ−1

= 0.528

where p∗ is the pressure corresponding to sonic speed. The NPR is less than the critical pressure ratio of 0.528. Therefore, the nozzle must experience sonic condition at the exit. Thus, the mass flow through the nozzle m ˙ is m ˙ = ρ∗ V ∗ A∗ where ρ∗ and V ∗ are the density and velocity at M = 1. From isentropic table, for M = 1, we get T∗ T01

=

0.8333

Thus, T∗

=

0.8333(30 + 273.15) = 252.6 K

V∗

=

a∗ =

=

318.58 m/s

/

γRT ∗ =

√

1.4 × 287 × 252.6

From state equation, we have ρ∗

=

p∗ 6 × 101325 × 0.528 = ∗ RT 287 × 252.6

=

4.428 kg/m3

=

m ˙ 0.5 = ρ∗ V ∗ 4.428 × 318.58

=

0.0003544 m2 = 3.544 cm2

Thus, A∗

Aliter: This problem can also be solved without going into the details at the nozzle exit, as follows. We know that in terms of reservoir pressure and temperature, the mass flow rate m ˙ is given for a gas with γ = 1.4 as 0.6847 p01 A∗ m ˙ =√ RT01

73 Thus, A∗

=

/ m ˙ RT01 0.6847 p01

=

0.0003543 m2 = 3.543 cm2

4.54 The overall pressure ratio

pe /p0 =

100 × 103 = 0.1. 106

This is well below the critical pressure ratio of 0.528. Therefore, the flow is a choked flow. The area ratio

Ae /Ath = 13.5/8 = 1.6875.

From isentropic table, for Ae /Ath = 1.6875, the corresponding exit Mach number Me = 2.0 4.55 Given, p0 = pa = 101325 Pa and T0 = 288.15 K. For the flow to choke, pb /p0 ≤ 0.528. Therefore, the pressure in the tank pb has to be pb ≤ 0.528 × 101325 ≤ 53.5 kPa This implies that pbmax = 53.5 kPa. The mass flow rate is given by m ˙

=

0.6847 p0 Ath √ RT0

=

0.6847 × 101325 π √ × (0.04)2 4 287 × 288.15

=

0.303 kg/s

4.56 Given, Ae /Ath = 4 and Me > 1. Therefore, the supersonic solution of the area Mach number relation is the answer for the problem. That is, the supersonic value of M given by the following equation is the exit Mach number. Ae 1 = Ath M

!

5 + M2 6

"3

This equation may be solved directly to get Me or the value of Me corresponding to Ae /Ath = 4 can be read directly from isentropic table. From isentropic table, we get

74

One-Dimensional Flow

Me = 2.94 and

pe = 0.029795 p0

This is the pressure ratio required. 4.57 (a) For Me = 1.63, from isentropic table, we have Ae pe = 1.275 = 0.22501 A∗ p0 For correct expansion, pe = pb . Therefore, pb = 0.22510 × p0 = 0.22501 × 10 × 101325 Pa since 1 atm = 101325 Pa. Thus,

pb = 228 kPa

(b) The flow will remain supersonic at the exit for all back pressures below 228 kPa (c) For choking, M = 1 at the throat. After choking the flow can expand as a subsonic flow at the divergent portion of the nozzle. From isentropic table, for Ae /A∗ = 1.275, we have Me = 0.54 p/p0 = 0.82 (approx) Therefore, the nozzle will remain choked for all back pressures below pb = 0.82 × 10 × 101325 = 830.87 kPa 4.58 From normal shock theory, we have 1 2 2 M12 − 1 ρ2 = 1+ 2 ρ1 M1 (γ − 1) + 2 M2

=

'1 2 1 2 *1/2 γM12 + 1 − M12 − 1 2 (γM12 + 1) − (γ + 1)

where subscripts 1 and 2 refer to states upstream and downstream of the shock, respectively. Given, M1 is very large. Also, γ = 1.4 for air. The density ratio can be written as & % 1 2 1 − ρ2 M12 =1+ ρ1 (γ − 1) + M22 1

when M1 → ∞, we get

75

ρ2 ρ1

Similarly,

=

1+

1 2 =1+ γ−1 1.4 − 1

=

1+

2 =1+5 0.4

=

6

%

& % & 1/2 γ + M12 − 1 − M12 1  & % 1 M2 =  γ+1 1 2 γ + M2 − M2 1

In the limit M1 → ∞, it reduces to M2

= =

3

γ−1 = 2γ

1

3

0.4 2.8

0.378

4.59 From isentropic table, for M = 2.0, we get p = 0.1278 p0 Therefore, p

0.1278 × 3 × 101325 = 38848 Pa

=

since, 1 atm = 101325 Pa. Given, 101325 Pa = 760 mm Hg, therefore, p

=

38848 × 760 101325

=

291.384 mm of Hg

This is absolute pressure. The gauge pressure, shown by the manometer, will be pg

= =

pabs − patm = 291.384 − 760 − 468.61 mm

The negative sign indicates that the measured pressure is below the atmospheric pressure or subatmospheric.

76

One-Dimensional Flow

4.60 The difference between the measured pressures is 500 mbar. That is, ∆p = 0.5 bar We know that, 1 bar = 105 Pa Therefore, ∆p = 0.5 × 105 Pa

Also,

∆p = ρair g h where h is the vertical height climbed. Therefore, h =

0.5 × 105 9.81 × 1.1

=

4633.49 m

4.61 Let subscript ‘0& refer to stagnation state. Given that the total pressure of air is p0 . For maximum velocity, the limiting pressure is zero. Assuming the flow as incompressible, by Bernoulli equation, we have 1 p + ρV 2 = p0 2 For Vmax , p = 0, thus,

1 2 ρV = p0 2 Also, for incompressible flow, ρ = ρ0 , thus, 3 p0 Vmax, incomp. = 2 ρ0 For compressible flow, the Bernoulli equation, we have

Therefore,

γ p0 γ p V2 + = γ−1ρ 2 γ − 1 ρ0 Vmax, comp. =

That is, Vmax, comp.

=

= =

3 3

3

2

γ p0 γ − 1 ρ0

γ Vmax, incomp. γ−1 1.4 Vmax, incomp. 1.4 − 1

1.87 Vmax, incomp.

77 That is, the error in treating this compressible flow as incompressible is 87 percent. 4.62 Let the inlet and the exit of the nozzle be denoted by subscripts 1 and 2, respectively. (a) By energy equation, we have h1 +

V12 2

= h2 +

V22 2

3025 × 103 +

602 2

= 2790 × 103 +

V22

= 473600

V2

=

=

ρAV

=

A1 V1 v1

=

0.1 × 60 0.19

=

31.58 kg/s

688.2 m/s

(b) The mass flow rate is m ˙

(c) Mass flow rate is also given by m ˙ = ρ2 A2 V2 =

A2 V2 v2

Thus, A2

=

m ˙ v2 V2

=

31.58 × 0.5 688.2

=

0.0229 m2

V22 2

78

One-Dimensional Flow

Chapter 5

Normal Shock Waves 5.1

Vg

500 − Vg

500 m/s

(a) Stationary observer

(b) Observer moving with the shock

Figure S5.1

Mach number of the air stream, M1 is given by 500 M1 = √ = 1.51 1.4 × 287 × 273

From Normal shock table, for M1 = 1.51, we have p2 p1

=

500 m/s

2.493,

T2 = 1.327 T1

Therefore, p2

=

1.745 atm

T2

=

362.27 K

V2 V1

=

1 500 − Vg = 500 1.879

500 − Vg

=

266.1 m/s

Vg

=

Also,

233.9 m/s 79

80

Normal Shock Waves

Since the velocity of the observer does not affect the static properties, pb

=

p2 = 1.745 atm

Tb

=

T2 = 362.3 K

The Mach number of the flow behind the shock wave is Mb

Vg 233.9 = 381.5 γRTb

=

√

=

0.613

From isentropic table, for M = 0.613, we have p pt

=

0.7760,

T = 0.9301 Tt

Therefore, after passage of shock the stagnation pressure is p tb

T tb

=

1.745 0.7760

=

2.25 atm

=

362.3 0.9301

=

389.5 K

Note: • For a stationary observer the stagnation temperature after passage of the wave is greater than that before passage of the wave. • For an observer sitting on the wave, however, there is no change of stagnation temperature across the wave.

81 5.2

up

u2 = C s − up

Cs 2

u1 = C s

1

Figure S5.2 Shock wave motion in a tube.

/

a1

=

γRT1 = 347 m/s

u1 u2

=

Cs (γ + 1)M12 = Cs − u p (γ − 1)M12 + 2

M1

=

Cs a1

Therefore, Cs Cs − u p

=

Cs2 +2 a21

% &2 s (γ + 1) C a1 % &2 s (γ − 1) C +2 a1

=

(γ + 1)

(γ − 1)Cs2 + 2a21

=

(γ + 1)Cs2 − (γ + 1)up Cs

2Cs2 − (γ + 1)up Cs − 2a21

=

0

up Cs − a21

=

0

γ + 1 up M1 − 1 2 a1

=

0

(γ − 1)

Cs2 −

!

γ+1 2

M12 −

"

M1

= =



Cs (Cs − up ) a21

1  γ + 1 up ± 2 2 a1

1.19

4!

γ + 1 up 2 a1

"2



+ 4

Positive sign is taken here, since M1 cannot be less than 1. Hence, Cs = M1 a1 = 413 m/s . From Normal shock table, for M1 = 1.19, pp21 = 1.485. Thus, the pressure on the face of the piston is p2 = 1.485 × 1.0133 × 105 = 1.505 × 105 Pa

82

Normal Shock Waves

5.3 up

2

1

Figure S5.3a The flow field.

(a) p2 p1

=

a1

=

!

" 2γ γ − 1 |up | γ−1 1− 2 a1 / √ γRT1 = 20.04 300

=

347 m/s

Therefore, p2 p1

=

!

p2

=

0.606p1

=

120 1 − 0.2 347

"7

= 0.606

0.606 atm

(b)

up

u2 = C s − up

Cs 2

1

Figure S5.3b Schematic of the flow field.

u1 u2

=

Cs Cs − u p

=

(γ + 1)M12 (γ − 1)M12 + 2

u1 = C s

83

M12 −

(γ − 1)M12 + 2

=

(γ + 1)M12 − (γ + 1)

γ + 1 up M1 − 1 2 a1

=

0

γ + 1 up 2 a1

=

1.2 ×

up M1 a1

120 = 0.415 347

Therefore, M12 − 0.415 M1 − 1 = 0. Solving for positive value of M1 , we get M1

=

1.228

p2 p1

=

1+

=

1.595

2 2γ 1 2 M1 − 1 γ+1

Therefore, p2 = 1.595 atm 5.4 Cs

up

Cs

u1 = C s + up

Gas at rest

u2 = 0

u2 = C s

Figure S5.4 Schematic of the flow field.

The velocity of the wave relative to the pipe = Cs . Velocity of air entering the normal shock wave relative to the shock wave is u1

= Cs + up

M1

=

Cs + u p a1

u1 u2

=

(γ + 1)M12 2 + (γ + 1)M12

84

Normal Shock Waves

Cs

=

Cs + u Cs

=

M1 a 1 − u p

From this we get (γ + 1)M12 2 + (γ − 1)M12

M12

γ+1 − 2

!

up a1

"

=

M1 a 1 M1 a 1 − u p

=

M1 u M1 − a1p

M1 − 1

=

0

a1

=

/

γRT1 = 347 m/s

Solving for M1 , we get M1 = 1.2924, taking only the positive sign, since M1 is supersonic. Hence, Cs

=

M1 a1 − up = 1.2924 × 347 − 150

=

298.5 m/s

From shock tables, for M1 = 1.294, we have p2 p1

T2 = 1.185 T1

=

1.775,

=

1.775 × 1.5 × 105

Therefore, p2

= T2

= =

2.66 × 105 Pa 1.185 × 300 355.5 K

Also, since the gas is at rest,

5.5

p02

=

p2

T02

=

T2

85

Vp 1

u2 = C s

u1 = Cs + Vp

Cs 2

Figure S5.5 Schematic of the flow field.

(γ + 1)M12 2 + (γ − 1)M12

u1 u2

=

u1

= Cs

M1

Cs a1

=

= Cs − Vp

u2 Therefore, (γ + 1)

%

2 + (γ − 1)

Cs a1

%

&2

Cs a1

=

&2

= + , Cs Cs Vp (γ + 1) − a1 a1 a1 Solving for

Cs a1 ,

Cs a1 Cs a1

−

Vp a1

2 + (γ − 1)

!

Cs a1

"2

we get,

Cs = a1 In the limit as

=

Cs Cs − u p

Vp a1

!

γ+1 4

→ ∞,

Cs a1

"

' ! "2 ! "2 *1/2 Vp Vp 1 γ+1 + 1+ a1 4 2 a1

→ ∞. In the limit as

Vp a1

5.6 a4

Cs 3

up = 300 m/s

4

→ 0,

Cs a1

→ 1.

86

Normal Shock Waves

Figure S5.6a Schematic of the flow field.

(a) up

=

p3 p4

=

300 m/s ! " 2γ γ − 1 |up | γ−1 1− 2 a4 !

" 2.8 300 0.4 1 − 0.2 × 360 ! "7 5 6 ! "7 5 × 1 = 0.2791 atm 6 + , γ − 1 |up | 5 1− = 2 a4 6

=

=

Therefore, a3 =

5 6

p3

=

a3 a4

=

× 360 = 300 m/s. Slope of the terminating characteristic is dx dt

=

C3 = a4 −

=

360 −

γ+1 |up | 2

2.4 × 300 = 0 2

For the pressure on the face of the piston, !

. 12 "2γ p2 γ+1 −1 p2 γ−1 p3 p3 + γ+1

up

=

a3 γ

up

=

a3 = 300 m/s

Therefore,

!

p2 p3

"2

!

"2 p2 −1 p3

=

γ

− 3.68

p2 + 0.72 p3

=

0

2

Therefore, p2 p3

=

3.473

p3

=

0.2791 atm

' p2 p3

+

γ−1 γ+1

2γ γ+1

*

87 Therefore, pressure at the piston face p2 = 3.473 × 0.2791 = 0.969 atm (b) The velocity of the shock with respect to a stationary observer is C s = a3

+

γ − 1 γ + 1 p2 + 2γ 2γ p3

, 12

= 529.88 m/s

Therefore, time for the shock to hit the terminating characteristic after the piston has stopped is 30 = 0.0566 s t1 = 529.88 (c) t 0

Piston path

300 m/s

dx/dt = 0

t

a4

ock

300 m/s

Pis ton dx /dt pa th =| up |

Terminating characteristic

dx /d t=

Sh

Cs

x 30 m

Figure S5.6c

5.7 Given, MS = 5.0, T1 = 27c ircC = 300 K, p1 = 0.01 atm, T4 = 300 K. From normal shock table, for MS = 5, we have p2 p1 p4 p1

=

29.0

=

4 ", −2γ + ! γ4 −1 2γ1 MS2 − (γ1 − 1) 1 γ4 − 1 1− MS − γ1 + 1 γ4 + 1 MS

=

2.26 × 106

Cs 0

p2

p3

88

Normal Shock Waves

p4

=

2.26 × p1 = 2.26 × 104 atm

Note: Instead of above equation for p4 /p1 , equation (5.57) also can be used for solving this problem. 5.8 (a) p2 /p1 = 29. Therefore, static pressure behind the shock is, p2 = 29p1 = 0.29 atm. From normal shock table for M1 = 5, we get T2 = 5.8 T1 Therefore, static temperature behind the shock T2 is, T2 = 5.8 × 300 = 1740 K (b) Particle speed u2 is given by, u2

a2

M2

=

! " 2 1 a 1 MS − γ1 + 1 MS

=

1388.71 m/s

=

20.04 ×

=

836 m/s

=

1388 836

=

1.66

/

T2

From isentropic tables, for M2 = 1.66, we have T2 p02

=

0.6447

p2 p02

=

0.215

Thus, T02 = 2699 K and p02 = 1.349 atm (c) Testing time available is ∆t

=

t2 − t1 =

8 8 − u2 Cs

89

u2

Cs

=

1388.71 m/s

Ms a1 = 5 × 347 = 1735.9 m/s

=

t

dx /d t

=

u

2

t2

t1

dt = dx/

Cs

x

8m

Figure S5.8c

Therefore, ∆t

=

8 8 − 1388 1735

=

1.15 × 10−3 s

(d) µ

sin−1

= =

1 1 = sin−1 M2 1.66

37 deg

5.9 up

2

up

2

2

Cs

1

uR

up + uR

5

uR

5

90

Normal Shock Waves

Figure S5.9 Schematic of the flow field.

(a) uR Cs

=

uR up up Cs

up + uR uR

=

ρ5 , ρ2

=

ρ5 ρ1 × ρ1 ρ2

=

ζ η

from continuity

Therefore, up uR

ζ −η η

=

Also, (refer to Example 5.5) up Cs

=

1 (1 − ) η

From this, we get uR Cs

=

! " 1 η × 1− ζ −η η

=

η−1 ζ −η

(b) p2 p1

= =

" ρ1 1+ 1− ρ2 ! " 1 1 + γ1 MS2 1 − η γ1 MS2

!

Similarly, p5 p2

=

=

! " 2 (up + uR ) ρ2 × 1 − a22 ρ5 ! "2 ! " a2 u p + u R η 1 + γ2 12 1− a2 a1 ζ

1 + γ2

91 "2 ! " up uR Cs η + 1− a1 Cs a1 ζ ! "2 ! " η−1 ζ −η γ1 p1 ρ2 η−1 MS + MS = 1 + γ2 ρ1 γ2 p2 ζ − η η ζ a21 a22

= 1 + γ1

p1 (η − 1)2 ζ M2 η p2 η2 ζ −η S

Therefore, p5 p1

= = = =

p5 p2 p2 p1 (η − 1)2 ζ p2 + γ1 MS2 p1 η ζ −η ! " (η − 1)2 ζ 1 1 + γ1 MS2 1 − + γ1 MS2 η η ζ −η ! " η−1 (η − 1)ζ 1 + γ1 MS2 1+ η (ζ − η)

=

1 + γ1 MS2

η − 1 η(ζ − 1) η ζ −η

=

1 + γ1 MS2

(η − 1)(ζ − 1) (ζ − η)

(c)

Therefore,

!

= 1 + γ2

h2 h1

=

h5 h2

=

! " γ1 − 1 2 1 MS 1 − 2 1+ 2 η "2 ! " ! γ2 − 1 up + uR ρ2 1+ 1 − 22 2 a2 ρ5

=

γ2 − 1 a21 2 (η − 1)2 ζ2 1+ M S 2 a22 η2 (ζ − η)2

h2

=

a22 γ2 − 1

h1

=

a21 γ1 − 1 a21 a22

=

γ 1 − 1 h2 γ 2 − 1 h1

!

η2 1− 2 ζ

"

92

Normal Shock Waves

Therefore, γ1 − 1 h1 2 (η − 1)2 ζ + η M 2 h2 S η 2 ζ −η

h5 h2

=

1+

h5 h1

=

h5 h2 h2 h1

= = =

h2 γ1 − 1 2 (η − 1)2 ζ + η MS + h1 2 η2 ζ −η ! " γ1 − 1 2 η − 1 ζ +η 1+ MS 2 (η + 1) + (η − 1) 2 η ζ −η 1 + (γ1 − 1)MS2

(η − 1)(ζ − 1) η(ζ − η)

5.10 Given that, the stagnation pressure and temperature are p0 = 1 atm = 101325 Pa,

T0 = 15◦ C = 288.15 K

From isentropic table, for M = 3.0, we have p = 0.02722 p0 Therefore, the static pressure at the test-section is p

=

0.02722 p0

=

0.02722 × 101325

=

2758 Pa

The pitot pressure measured by a pitot tube placed in the test-section is the pressure behind a normal shock. From normal shock table, for M = 3.0, we have p02 = 0.3283 p01 Thus, the pressure that a pitot tube at the test-section will measure is p02

= 0.3283 p01 = 0.3283 × 101325 =

33.265 kPa

93 5.11 From normal shock table, for M1 = 2.5, we have p2 = 7.125, p1

T2 = 2.1375, T1

ρ2 = 3.3333, ρ1

p02 = 8.5261 p1

Therefore, M2

=

p2

=

(7.125)(1) = 7.125 atm

ρ2

=

(3.3333)(1.225) = 4.083 kg/m3

p02

=

(8.5261)(1) = 8.5261 atm

T2

=

(2.1375)(T1 )

0.51299

From state equation, we have T1

=

101325 p1 = ρ1 R 1.225 × 287

=

288.2 K

Therefore, T2

=

(2.1375)(288.2) = 616.03 K

a2

=

/

V2

=

M2 a2 = (0.51299)(497.51)

=

γ R T2 = 497.51 m/s

255.22

From isentropic table, for M2 = 0.51299, we have T2 = 0.950 T02 Therefore, T02

=

616.03 T2 = 0.950 0.950

=

648.45 K

94

Normal Shock Waves

5.12 For nitrogen, molecular weight is 28.02 and γ = 1.4. Thus, the gas constant is γ R = 1.4 R = 8314/28.02 = 297 J/(kg K) and cp = γ−1 0.4 × 297 = 1039.5 J/(kg K). The speed of sound upstream of the shock is / √ a = γ R T = 1.4 × 297 × 303 =

354.95 m/s

Therefore, the upstream Mach number M1 is M1

V1 923 = a1 354.95

=

= 2.6 Also, since the flow process across a normal shock is adiabatic, T01 = T02 . Now, from normal shock table, for M1 = 2.6, we have p2 = 7.72 p1

M2 = 0.504

T2 = 2.2383 T1

Therefore, p2 = 7.72 × 300 = 2.316 MPa a2

V2

=

/

=

531.03 m/s

=

M2 a2 = 0.504 × 531.03

=

γRT =

√ 1.4 × 297 × 2.2383 × 303

267.64 m/s

When the flow slows down isentropically from V1 to V2 , by energy relation, we have T2

= 678.32 K

V12 − V22 2cp

=

T1 +

=

303 +

9232 − 267.642 2 × 1039.5

95 By isentropic relation, we have p2 p1

=

!

T2 T1

γ " γ−1

=

!

=

16.786

678.32 303

"3.5

Thus, p2 = 5.036 MPa 5.13 For blunt nosed model at Mach 3, there will a detached bow-shock standing in front of the nose. This shock can be approximated to a normal shock at the nose of the model around the stagnation point. Therefore, pressure at the stagnation point is the total pressure behind the normal shock. From normal shock table, for M1 = 3, we have p02 = 0.32834 p01 Thus, p02

=

0.32834 p01 = 0.32834 × 10

=

3.2834 atm

=

332.69 kPa

The flow process across the shock is adiabatic. Hence, T02 = T01 . Therefore, T02 = 315 K After the normal shock, the flow decelerates isentropically to stagnation condition at the nose. Hence, the stagnation density ρ02 can be expressed as ρ02

=

p02 332690 = R T02 287 × 315

=

3.68 kg/m3

5.14 Let us make the shock stationary and look at the field. The flow field with stationary shock will look like that shown in Fig. s5.14.

96

Normal Shock Waves T1 = 300 K p1 = 101 kPa

p2 = 5000 kPa V2

V1

Shock Figure S5.14

p2 p1

=

49.5

=

1+

=

1 + 1.167 M12 − 1.167

M12

=

49.5 + 0.167 = 42.55 1.167

M1

=

6.52

2 2γ 1 2 M1 − 1 γ+1

Thus,

The velocity becomes V1

=

M1 a1 = 6.52

=

2374.16 m/s

√

1.4 × 287 × 330

By normal shock relation, we have M2

=

'

=

0.4

γ−1 2 2 M1 M12 − γ−1 2

1+ γ

*1/2

1 2 2 2(γ − 1) γ M12 + 1 1 2 48.41 × 41.5 = 9.205 M1 − 1 = 1 + 2 2 (γ + 1) M1 244.86

T2 T1

=

1+

T2

=

9.205 × 330 = 3037.65 K

The speed of sound downstream of the shock is / √ γ R T2 = 1.4 × 287 × 3038.34 a2 = =

1104.9 m/s

97 Thus, the velocity V2 is V2

Vg

=

M2 a2 = 0.4 × 1104.8

=

441.92 m/s

=

Cs − Vg

=

2374.16 − 441.92

=

1932.24 m/s

Aliter The above problem can also be solved using gas tables, as follows. From normal shock tables, for pp21 = 49.5, we have M1

≈

6.5

M2

≈

0.4

T2 T1

≈

9.2

T2

≈

3036 K

a2

=

1104.47 m/s

V2

=

M2 a2 = 0.4 × 1104.47

=

441.8 m/s

=

√ 1.4 × 287 × 330

=

364.13 m/s

=

M1 a1 = 6.5 × 364.13

=

2366.84 m/s

a1

V1

98

Normal Shock Waves = 2366.84 − 441.8

Vg

= 1925.04 m/s

5.15 Upstream of the shock, the speed of sound a1 is / √ γ RT1 = 1.4 × 287 × 300 a1 = = The Mach number M1 =

347.2 m/s V1 412 = 1.19. = a1 347.2

From isentropic table, for M1 = 1.19, we get p1 T1 = 0.41778, = 0.77929 p01 T01 Therefore, p01

T01

=

92 p1 = 0.41778 0.41778

=

220.21 kPa

=

300 T1 = 0.77929 0.77929

=

384.96 K

From normal shock table, for M1 = 1.19, we get p2 T2 p02 = 1.4854, = 1.1217, = 0.99372, M2 = 0.84846. p1 T1 p01 Therefore, p2

T2

=

1.4854 × 92

=

136.66 kPa

=

1.1217 × 300

=

336.51 K

99

T02

=

T01 = 384.96 K , since flow process across a shock is adiabatic

p02

=

0.99372 × 220.21

=

218.83 kPa

The speed of sound is a2 = fore, V2

√ = =

γRT =

√ 1.4 × 287 × 336.51 = 367.7 m/s. There-

M2 a2 = 0.84846 × 367.7 311.98 m/s

The entropy rise across a normal shock is given by ! " p01 ∆s = R ln p02 ! " 220.21 = 287 ln 218.83 =

1.8042 J/(kg K)

5.16 Let the subscripts 1 and 2 refer to conditions upstream and downstream of the normal shock, respectively. From normal shock table, for M1 = 2.5, we get p2 p02 = 7.1250, = 8.5261. p1 p1 The static pressure in the flow just downstream of the shock is, p2 = 7.125 × 1.0 = 7.125 atm If a normal shock has to be positioned at the nozzle exit, the back pressure to which the nozzle discharges has to be equal to the total pressure downstream of the shock. The total pressure downstream of the shock is p02

= =

8.5261 × 1.0 = 8.5261 atm 863.91 kPa

i.e. the back pressure has to be 863.91 kPa to position a normal shock at the nozzle exit.

100

Normal Shock Waves

5.17 From isentropic table, for Me = 2.5, we have pe = 0.058528, p0e

Ae = 2.637 A∗

where subscripts 0, e, and ∗ refer to stagnation, exit and throat, respectively. The throat area is A∗

=

4 Ae = 2.637 2.637

1.517 cm2 1 = 17.09 atm p0e = 0.058528 For the normal shock, the upstream Mach number is 1.5. From isentropic table, for M1 = 1.5, =

A1 = 1.176, A∗

p1 = 0.2724, p01

T1 = 0.68966 T01

Thus, A1

=

1.176 × 1.517 = 1.784 cm2

p1

=

0.2724 × 17.09 = 4.655 atm

T1

=

0.68966 × 500 = 344.83 K

From normal shock table, for M1 = 1.5, p2 = 2.40583, p1

T2 = 1.3202 T1

p02 = 3.4133, p1

M2 = 0.70109

The back pressure required is p02 , thus, p02 = 3.4133 × 4.655 = 15.89 atm Downstream of the shock, the flow is isentropic upto the nozzle exit. However, for this flow the effective throat area is not the same as A∗ , since p02 < p01 . Let the effective throat area downstream of shock to be A∗2 . From isentropic table, for M2 = 0.70, A2 = 1.09437 A∗2 where A2 is the area at the shock location, which is same as A1 . Thus, A∗2

=

1.784 = 1.63 cm2 1.09437

Ae A∗2

=

4 = 2.454 1.63

101 From isentropic table for

Ae A2 ∗

= 2.454, we have

T2 = 0.98, Te

Me = 0.245

Thus, Te = 0.98 × 500 = 490 K The speed of sound is ae =

√ 1.4 × 287 × 490 = 443.71 m/s

Ve

= Me × ae = 0.245 × 443.71

The flow velocity is

=

108.71 m/s

5.18 Since the Mach number upstream of the shock is 2.32, the area ratio corresponding to this Mach number will give the area at the shock location. (a) For M1 = 2.32, from isentropic table, A1 = 2.23 A∗ Thus, area at shock location is A1 = 2.23 × 5 = 11.15 cm2 (b) The Mach number downstream of the shock M2 given by normal shock table for M1 = 2.32 is M2 = 0.53 For M2 = 0.53, from isentropic table, we have A2 = 1.29 A∗ Since A1 = A2 = area at the shock location, we have A∗2 =

11.15 A2 = = 8.64 cm2 1.29 1.29

Therefore, 12.5 Ae = 1.447 = A∗2 8.64 From isentropic table, for

Ae = 1.447, the exit Mach number Me = 0.45 A∗2

102

Normal Shock Waves

(c) For the given nozzle, the area ratio

Ae is Ath

Ae Ae 12.5 = 2.5 = ∗ = Ath A 5 Ae = 2.5, we have A∗ p2 = 0.064261 Me = 2.44 and p02

From isentropic table, for

For complete isentropic flow, p02 = p0 = 700 kPa. Thus, 44.98 kPa.

p2 = 0.064261×700 =

The back pressure range for the flow to be completely isentropic is pb ≤ 44.98 kPa . 5.19 Given, T1 = 22 K and T01 = 400 K, where subscripts 1 and 2 refer to conditions ahead of and behind the shock, respectively. For

T1 22 = 0.055, from isentropic table, = T01 400 M1 = 9.2

From normal shock table, for M1 = 9.2, we have M2

=

T2 T1

=

T2

=

0.3893 17.4 382.8 K

5.20 Upstream of the normal shock, the Mach number M1 is M1

=

V1 500 =√ a1 1.4 × 287 × 300

=

1.44

From normal shock table for M1 = 1.44, we get M2 = 0.72345,

p2 = 2.2525, p1

T2 = 1.2807, T1

Thus, p2

=

2.2525 × 100 = 225.25 kPa

p02 = 0.9476 p01

103

T2

=

1.2807 × 300 = 384.21 K

V2

=

M2 a2 = 0.72345 ×

=

√

1.4 × 287 × 384.21

284.25 m/s

The entropy increase is ∆s

=

R ln

!

p01 p02

"

= 287 ln (1.0552)

15.432 J/(kg K)

=

5.21 The flow velocity at nozzle entrance is low. Therefore, the pressure and temperature of the flow at the entry can be treated as the stagnation quantities. Thus, T01 = 300 K p01 = 1 MPa, From isentropic table, for M1 = 2, we have p1 = 0.1278 p01

and

T1 = 0.55556 T01

Therefore, p1

=

(0.1278)(1) = 0.1278 MPa

T1

=

(0.55556)(300) = 166.67 K

From normal shock table, for M1 = 2, we get p2 T2 p02 = 4.5, = 1.6875, = 0.72087 p1 T1 p01

and

M2 = 0.57735

Thus, p2

=

4.5 × 0.1278 = 0.5751 MPa

T2

=

1.6875 × 166.67 = 281.3 K

p02

=

0.72087 × 1 = 0.72087 MPa

a2

=

√ 1.4 × 287 × 281.3 = 336.19 m/s

104

Normal Shock Waves

V2

M2 a2 = 0.57735 × 336.19

= =

194.1 m/s

5.22 Given, p0 = 101 kPa and T0 = 30 + 273.15 K. A normal shock at the nozzle exit implies that the entire nozzle flow is isentropic and also the flow is choked at the throat. The area ratio is Ae 0.0724 = 2.896 = ∗ A 0.025 From isentropic table, for

Ae A∗

= 2.896, we have p1 = 0.050115, p0

M1 = 2.6,

T1 = 0.42517 T0

This is the Mach number upstream of the shock. Thus, M1

=

p1

=

0.050115 × 101 = 5.06 kPa

T1

=

0.42517 × 303.15 = 128.9 K

p01

=

2.6

101 kPa

Let the conditions downstream of the shock be referred to by subscript 2. From normal shock table, for M1 = 2.6, we get M2 p2 p1

=

0.504 T2 = 2.2383, T1

= 7.72,

p2

=

39.06 kPa

T2

=

288.5 K

p02

=

46.47 kPa

p02 = 0.46012 p01

105 5.23 Given, p0 = 200 kPa and T0 = 350 K. Let subscripts 1 2 and 3 refer to locations upstream and downstream of the shock wave and the nozzle exit, respectively. Ath = A∗1 = 0.2 m2 At the shock location, A1 = 0.6 m2 , thus, A1 0.6 = 3.0 = ∗ A1 0.2 From isentropic table, for

A1 = 3.0, we get A∗1 M1 = 2.64,

p1 = 0.04711 p01

Up to the shock the flow in the nozzle is isentropic and therefore, p01 = p0 . Thus, p01

=

p1

=

200 kPa 200 × 103 × 0.04711

=

9.422 kPa

where, p1 and p01 are the static and total pressures, respectively, ahead of the shock. Let subscript 2 refer to condition behind the shock. Now, from normal shock table, for M1 = 2.64, we have p2 = 7.9645, p1

M2 = 0.50,

p02 = 0.44522 p01

Thus, p2

= =

p02

= =

7.9645 × 9.422 75.04 kPa 0.44522 × 200 89.04 kPa

Also, from normal shock theory we have (Section 4.6) p01 A2 ∗ = = 2.25 p02 A1 ∗

106

Normal Shock Waves

Thus,

A∗2 = 2.25 × 0.20 = 0.45 m2

A∗2 may also be obtained from M2 . From isentropic table, for M2 = 0.5, A2 = 1.34 A∗2 Thus,

0.6 = 0.448 m2 1.34 This A∗2 is the equivalent throat area for the flow downstream of the shock. Therefore, at the nozzle exit, A2 ∗ =

A3 0.8 = 1.786 = ∗ A3 0.448 Now, from isentropic table, for M3 = 0.35,

A3 = 1.786, A∗3

p3 = 0.91877, p03

T3 = 0.97609 T03

Here, p03 = p02 and T03 = T02 = T0 . Thus, p3

=

p03

=

T3

=

0.91877 × 89.04 = 81.81 kPa 89.04 kPa 0.97609 × 350 = 341.63 K

For area ratio A3 /A∗3 the subsonic solution from the isentropic table was used since after a normal shock the flow becomes subsonic and this flow is further decelerated in the divergent portion of the duct. 5.24 Given, p01 = 5 atm and p02 = 3.6 atm. Therefore, p02 3.6 = 0.72 = p01 5 Assuming γ = 1.4, from normal shock table, for M1 = 2.0,

p02 = 0.72, we have p01

p2 = 4.5 p1

Now, from isentropic table, for M1 = 2.0, we get p1 = 0.1278 p01

107 Hence, the pressure just behind the normal shock at the nozzle exit is p2

= =

4.5 p1 = 4.5 × 0.1278 × 5 2.8755 atm

5.25 The pitot tube will read the actual total pressure in a subsonic stream. But in a supersonic flow, the pressure measured by a pitot probe is the total pressure downstream of a detached shock which stands at the nose of the pitot tube. Therefore, it is essential to find out whether the flow is subsonic or supersonic. It can be easily seen from the isentropic relations that for M = 1, the pressure 0.95 p = = 1.8 atm. ratio pp0 = 0.528. Hence, p0 = 0.528 0.528 Thus, when p0 < 1.8 atm, the flow is subsonic, and when p0 > 1.8 atm, the flow is supersonic. (i) p0 = 1.1 atm. The flow is subsonic and hence the pitot tube is measuring the actual total pressure of the flow. p 0.95 = 0.8636 = p0 1.1 From isentropic table, for

p = 0.8636, we get p0 M = 0.465

(ii) p0 = 2.5 atm. The flow is supersonic and the pitot tube measures p02 behind a normal (detached) shock. p02 2.5 = 2.63 = p1 0.95 p02 From normal shock table, for = 2.63, we get p1 M = 1.275 (iii) p0 = 10 atm. The flow is supersonic. p02 10 = 10.526 = p1 0.95 p02 From normal shock table, for = 10.526, we have p1 M = 2.79

108

Normal Shock Waves

5.26 The shock will be only at the divergent portion of the nozzle, since only after the throat the flow becomes supersonic. The Mach number M1 just upstream of the shock will be given by the area ratio Ashock . Ath Ashock 2000 =2 = Ath 1000 From isentropic table, for area ratio 2, we have M1 = 2.2 ,

p1 = 0.093522 p01

Up to the shock, the stagnation pressure does not change and therefore, p01

=

200 kPa

p1

=

200 × 0.093522 = 18.7 kPa

Now, from normal shock table, for M1 = 2.2, we get p2 = 5.48, p1

M2 = 0.55,

p02 = 0.62814 p01

where subscript 2 refer to condition just downstream of the shock. Therefore, p2

=

5.48 × 18.7 = 102.48 kPa

p02

=

0.62814 × 200 = 125.63 kPa

For M2 = 0.55, from isentropic table, we get A2 = 1.255 Ath2 where Ath2 is the throat area required for the flow downstream of the shock to choke. A2 Ath2 = 1.255 2 But A2 = A1 = Ashock = 2000mm , therefore, Ath2 = Thus,

2000 = 1593.63 mm2 1.255

Ae 3000 = 1.8825 = Ath2 1593.63

For this area ratio, from isentropic table (subsonic part), we get Me = 0.325

109 Therefore, 1

p02 pe

=

pe

=

125.63 p02 = 1.076 1.076

=

116.76

1 + 0.2 × 0.3252

23.5

= 1.076

The pressure loss occurs only across the shock and the loss of pressure ∆p0 is ∆p0

= =

p01 − p02 = (200 − 125.63) 74.37 kPa

5.27 Let the subscripts i, 1, 2 and e refer to the inlet, just upstream and just downstream, and the nozzle exit, respectively. It can be shown for this flow that, (1) p01 A∗1 = p02 A∗2 For Mi = 2.0, from isentropic table, we get Ai = 1.6875 A∗i Therefore, A1 A∗1

=

A1 Ai Ai A∗1

=

A1 Ai , Ai A∗i

=

2 × 1.6875 = 3.375

since A∗1 = A∗i

For this area ratio, from isentropic table, we get M1 = 2.76 For M1 = 2.76, from normal shock table, we have p02 = 0.40283 p01 Using this in Eq. (1), we get A∗1 = 0.40283 A∗2

110

Normal Shock Waves Ae A∗2

=

Ae Ai A∗1 Ai A∗1 A∗2

=

4 × 1.6875 × 0.40283 = 2.7191

For this area ratio, from isentropic table (subsonic solution), we get Me = 0.22 The exit pressure pe may be expressed as pe

=

pe p02 p01 pi p02 p01 pi

=

pe p02 p0i pi , p02 p01 pi

=

0.96684 × 0.40283 ×

=

since p01 = p0i 1 × 80 0.1278

243.8 kPa

Thus, the back pressure required is 243.8 kPa. 5.28 When a pitot tube is placed in a supersonic stream, there will be a detached shock standing at its nose. At the nose where the pressure tap is located, the shock may be treated as a normal shock and hence what the pitot tube measures is the pitot pressure p02 downstream of the shock. The wall pressure measured by a pressure tap may be treated as the actual static pressure of the stream. Thus we may take the static pressure upstream of the shock as p1 = 112 kPa. Thus, p02 2895 = 25.848 = p1 112 p02 = 25.848, we get p1 T2 M1 = 4.44 , = 4.7706 T1 Now from isentropic table, for M1 = 4.44, we get

From normal shock table, for

T1 T01

=

0.2023

T1

=

0.2023 × 500 = 101.15 K

a1

=

/

γ R T1 = 201.6 m/s

111 Thus, V1

=

M1 a1 = 895.1 m/s

5.29 At 10,000 m altitude, from standard atmospheric table, we have p = 26.452 kPa, T = 223.15 K During steady–state operation, mass flow through the test–section is given by m ˙

/ p A M γRT RT

=

ρ AV =

=

√ π(0.25)2 26.452 × 103 × × 2.4 1.4 × 287 × 223.15 287 × 223.15 4

=

14.57 kg/s

The stagnation temperature during steady–state operation is T0 =

T0 T T

For, M = 2.4,

T0 1 = = 2.152 T 0.46468

Thus, T0 = 2.152 × 223.15 = 480.22 K For the present geometry of fixed angle diffuser, the optimum condition for steady state operation is a normal shock at the diffuser throat. The diffuser throat area is A∗ A A∗ = A From normal shock table, for M1 = 2.4, M2 = 0.52. From isentropic table, for M2 = 0.52, we have A2 = 1.3 A∗2 Here A∗2 is the area of the second throat and A is the area of test-section. A2 Therefore, looking in to the corresponding supersonic Mach number for ∗ = A2 1.3, we get M = 1.66

112

Normal Shock Waves

This is the Mach number just upstream of the second throat with a shock. From normal shock table, for M1 = 1.66, we have p02 = 0.872 p01 This pressure loss must be compensated by the compressor. (a) The power required for the compressor is given by Power = h0 − hi = Cp (T0 − Ti ) where the subscripts 0 and i refer to outlet and inlet conditions. For an isentropic compression, ! " γ−1 p0 γ T0 = Ti pi * '! " γ−1 p0 γ T0 − Ti = Ti −1 pi '! * "0.286 1 = 480.22 − 1 = 19.18 K 0.872 Thus, the power input required for mbox mass of air becomes Power = 1004.5 × 19.18 = 19.266 kJ/kg The total power required for the compressor is m ˙ 19266 Power = 746 =

376.3 hp

This is the running power required for the compressor. (b) During start-up, M1 = 2.4 in the test-section. The corresponding total pressure ratio across the normal shock is p02 = 0.5401 p01 The isentropic work required for the compressor during start-up to drive the shock out of the test-section per mbox mass of air is '! * "0.286 1 Power = Cp T0i −1 0.5401 '! * "0.286 1 = 1004.5 × 480.22 −1 0.5401 =

92929.87 J/kg

113 The power required to start the tunnel is Power

=

m ˙ × 92929.87 14.57 × 92929.87 = 746 746

=

1815 hp

5.30 Let subscripts 1 and 2 refer to conditions just upstream and downstream of the normal shock. The Mach number just upstream of the shock is 1.85. From isentropic table, for M1 = 1.85, we get p1 A1 = 0.1612, ∗ = 1.495 p01 A1 p1 = 0.1612 × 400 = 64.48 kPa A1 = 1.495 × 10 = 14.95 cm2

Now from normal shock table, for M1 = 1.85, we have p02 = 0.79023 p01

M2 = 0.60,

Again from isentropic table, for M1 = 0.6, we get A2 = 1.1882 A∗2 But A2 = A1 = 14.95 cm2 . Therefore, A∗2 =

14.95 = 12.58 cm2 1.1882

We want the Mach number and pressure at a state 3 where A3 = 2 A1 = 2 × 14.95 = 29.9 cm2 . Downstream of the shock, A∗2 = A∗3 . Thus, A3 A∗3

=

29.9 = 2.377 ≈ 2.4 12.58

From isentropic table, for this area ratio (from subsonic solution), M3 = 0.25

and

p3 = 0.95745 p03

But p03 = p02 = 316.1 kPa. Therefore, p3 = 0.95745 × 316.1 = 302.65 Note: This problem has been solved using isentropic table and hence the results obtained are only approximate. For exact results, we have to use the actual relations.

114

Normal Shock Waves

5.31 (a) Given, m ˙ = 0.9 kg/s, p0 = 4×101325 = 405300 Pa, T0 = 30+273 = 303 ∆p = 2. K and p1 The pressure ratio across the shock is p2 − p 1 = p1 p2 p1

2

=

2+1

=

3

p2 For = 3, from normal shock table, p1 M1 = 1.65, M2 = 0.654,

T2 = 1.4228 T1

From isentropic table, for M1 = 1.65, T1 A = 0.6475, ∗ = 1.292 T0 A Therefore, T1

T2

=

0.6475T0

=

0.6475 × 303

=

196.19 K

=

1.4228T1

=

1.4228 × 196.19

=

279.14 K

The mass flow rate is 0.9 kg/s, therefore, m ˙

=

Ath

= = =

0.9 =

0.6847 p0 Ath √ RT0

√ 0.9 RT0 0.6847 p0 √ 0.9 287 × 303 0.6847 × 405300 9.56 cm2

115 Therefore, the exit area becomes 1.292 Ath = 1.292 × 9.56

=

Ae

12.35 cm2

= (b) The exit velocity is V2

=

M2 a 2 = M2

=

0.654 ×

=

√

/

γRT2

1.4 × 287 × 279.14

219 m/s

5.32 a) Given, (p2 −p1 )/p1 = 3.5, where p1 and p2 are the pressures ahead of and behind the shock. Therefore, the pressure ratio across the shock is p2 /p1 = 4.5. For p2 /p1 = 4.5, from normal shock table, M2 = 2,

ρ2 = 2.67 ρ1

Therefore, the shock speed becomes Cs

=

M1 a 1 = 2 ×

=

2×

=

694.4 m/s

√

/

γRT

1.4 × 287 × 300

The piston speed is given by Vp

= = = = =

V1 − V2 = V1 !

!

V2 1− V1

" ρ1 Cs 1 − ρ2 ! " 1 694.4 × 1 − 2.67 694.4 × (1 − 0.375) 434 m/s

"

116

Normal Shock Waves

(b) Given (p2 − p1 )/p1 = 7. Therefore, p2 /p1 = 8. For p2 /p1 = 8, from normal shock table, ρ2 M2 = 2.65, = 3.5047 ρ1 Therefore, the shock speed becomes √ Cs = 2.65 × 1.4 × 287 × 300 =

920.05 m/s

=

! 920.05 × 1 −

=

920.05 × 0.7147

Thus, Vp

=

1 3.5047

"

657.56 m/s

5.33 (a) Let subscripts 1 and 2 refer to states ahead of and behind the shock, respectively and subscript e refer to the nozzle exit. Given, p01 − p02 × 100 = 12.4 p01 1−

p02 p01

=

0.124

p02 p01

=

0.876

From normal shock table, for p02 /p01 = 0.876, M1 = 1.65 , M2 = 0.654,

T2 = 1.4228 T1

(b) For M1 = 1.65 from isentropic table, we have T1 T01

=

0.6475

T1

=

0.6475 × T01

=

0.6475 × 330

=

213.7 K

117 Therefore, T2 = 1.4228 × 213.7 = 304 K By energy equation, we have he +

Ve2 2

=

h0e

c p Te +

Ve2 2

=

cp T0e

But the flow process across the shock is adiabatic, therefore, T01 = T02 = T0e . Hence the flow speed at the nozzle exit becomes

Ve

= = =

0 /

2 × cp (T0e − Te ) 2 × 1004.5 × (330 − 300)

245.5 m/s

The flow speed behind the shock is V2

=

M2 a 2 = M2 ×

=

0.654 ×

=

√

/

γ R T2

1.4 × 287 × 304

228.57 m/s

(c) The mass flow rate is m ˙

=

0.6847 p01 Ath √ RT01

=

0.6847 × (5 × 101325) × (5 × 10−4 ) √ 287 × 330

=

0.5636 kg/s

118

Normal Shock Waves

Chapter 6

Oblique Shock and Expansion Waves 6.1 Given, M1 = 2 and β = 40◦ , therefore, M1n = 2.0 sin 40 = 1.29. From normal shock tables, for M1n = 1.29, we have p2 p1

=

1.775

T2 p1

=

1.185 and

M2n

=

0.7911

Therefore, p2

T2

=

0.5 × 1.775 × 105

=

0.8875 × 105 Pa

= =

1.185 × 273 323.5 K

For the adiabatic process, Tt1 = Tt2 . From isentropic tables, for Mach 2, we have T1 /Tt = 0.556. Thus, T1t = T2t =

273 = 491 K 0.556

From isentropic table, for T2 /T2t = 0.659, we have M2 = 1.61 119

120

Oblique Shock and Expansion Waves

Also, u2 w2

=

0.791 0.791 = 0.4913 = M2 1.61

β−θ

=

29.43

=⇒ θ

=

10.57 deg

sin (β − θ) =

Thus,

the wedge angle = 2θ = 21.14◦

6.2 For M1 = 2.0, from isentropic table, we have ν1 = 26.38◦ . Prandtl-Meyer function after expansion is ν2 = ν1 + θ = 26.38 + 5 = 31.38 deg Therefore, for ν2 = 31.38◦ the corresponding Mach number from isentropic table is M2 ≈ 2.18 . From isentropic tables for M2 = 2.18, we have p2 p02 T2 T02 ρ2 ρ02

=

0.0965;

=

0.5127;

=

0.1882;

p2

= p1 × =

T2

ρ2

6.3 (a)

p1 p01 T1 T01 ρ1 ρ01 0.0965 0.0965 = 98 × 0.1278 0.1278

74 kPa

= 300 ×

0.5127 0.5556

=

276.8 K

=

74 × 103 287 × 276.8

=

0.9315 kg/m3

=

0.1278

=

0.5556

=

0.2300

121 pe = 1 atm Me = 3

p0 = 70 × 105 kPa

Figure S6.3a

M1 = 3.0

=⇒

p1 = 0.02722 p01

With p1 = pe = patm , p01 =

1.01325 × 105 = 37.2 × 105 Pa. 0.02722

Therefore, the supply pressure for which oblique shock wave will first appear in the exhaust jet is

≤ 37.2 × 105 Pa (b)

0.45 D p0 = 70 × 105 kPa

0.1 D 0.45 D D

97 D 1.0

0.45 D D

Figure S6.3b

sin β

=

0.45 = 0.41 1.097

M1n

=

M1 sin β = 3 × 0.41 = 1.23

p2 p1

=

1.5984; p2 = pe

122

Oblique Shock and Expansion Waves

Therefore, p1

M1

=

atm pe = 1.5984 1.5984

=

0.634 × 105 Pa

=

3.0 =⇒

p1 = 0.02722 p01

Therefore, p01

=

0.634 × 105 0.02722

=

23.3 × 105 Pa

Therefore, minimum supply pressure for obtaining the desired test region is 23.3 × 105 Pa. (c) pe = 1 atm M1 = 3

p0 = 70 × 105 kPa 1

2

Figure S6.3c

M1 p2 p1 p1 p01

=

3.0

p1

=

=

10.33

p2

=

=

0.02722

p01

=

patm = 0.098105 Pa 10.333 pe = patm 0.098 × 105 = 3.6 × 105 Pa 0.02722

Therefore, the minimum supply pressure for which a normal shock will appear at the nozzle exit is 3.6 × 105 Pa Note: The static pressure after the shock has to be equal to the back pressure, namely the atmospheric pressure. This is because, subsonic jets are always correctly expanded. Thus, the total pressure of this subsonic flow is higher than

123 its static pressure. Hence, the flow will move some distance downstream of the nozzle exit before coming to rest. 6.4 The relation between θ and β is 1 2 2 cot β M12 sin2 β − 1 tan θ = M12 (γ + cos 2β) + 2 For θ = 15◦ , M1 = 2.0, solution by iteration yields β = 79.8◦ . This is the strong shock solution. (a) β = 79.8 deg (b) p2 p1

=

2γ γ−1 M 2 sin2 β − γ+1 1 γ+1

=

4.354

(c) T2 T1

=

p2 ρ2 / p1 ρ1

=

1.662

(d) β−θ

=

64.8◦

ρ2 ρ1

=

tan β tan (β − θ)

=

2.615

(e) M12 sin2 (β − θ)

=

γ−1 2 2 2 M1 sin β γ−1 M12 sin 2 β − 2

1+ γ

=

1 + 0.2 × 4 × (sin 79.8◦ × sin 79.8◦ ) 1.4 × 4 × (sin 79.8◦ )2 − 0.2

=

1.775 = 0.34 5.224

124

Oblique Shock and Expansion Waves M22

=

0.34 0.34 = 0.415 = (sin 64.8)2 0.819

M2

=

0.644

Weak shock (a) Solving θ − β − M relation we can obtain βweak = 45.3◦ (b) p2 p1

=

2γ γ−1 M 2 sin2 β − γ+1 γ+1

=

2.191

(c) T2 T1

=

1.267

(d) β−θ

=

45.3 − 15 = 30.3 deg

ρ2 ρ1

=

tan 45.3◦ tan 30.3◦

=

1.729

(e) M2 = 1.446

Aliter: Using Normal shock tables: M1n = M1 sin β , 1.97 Strong shock solution lution

Weak shock so-

125

γ = 1.4

M1n = 1.97

M1n = 2sin 45.3 deg = 1.42

γ = 1.4

p2 p1

=

4.361

M2n

=

0.7314

T2 T1

=

1.663

p2 p1

=

2.186

M2n

=

0.583

ρ2 ρ1

=

1.724

ρ2 ρ1

=

2.622

T2 T1

=

1.268

M2n

=

M2 sin (β − θ)

M2

=

M2

=

M2 n sin (β − θ)

M2n sin (β − θ)

=

0.7314 = 1.45 sin 30.3

=

0.583 = 0.644 sin(79.8 − 15)

Note: The solution obtained with oblique shock relations may also be obtained using oblique shock tables, which will result in considerable time saving. 6.5 2 1

Figure S6.5

From isentropic tables, for M1 = 2.0, we have p1 = 0.1278. Since, pt1 0.75 = 0.0599. 0.1278 × 1.60

pt1 = pt2

for this isentropic expansion,

For this pressure ratio, from isentropic table, we get

M2 = 2.48 .

p2 = pt2

126

Oblique Shock and Expansion Waves

From isentropic table, the Prandtl-Meyer function for Mach 2.0 and 2.48, respectively are ν1

=

26.38◦

ν2

=

38.655◦

Therefore, the flow turning angle becomes ν12 = ν2 − ν1 = 12.275◦ 6.6 (a) Angles for which the oblique shock remains attached to the wedge (from oblique shock table) are At

At

M1 θmax

= =

2.0 22◦

M1 θmax

= 3.0 =

34◦

θd 15◦ 25◦

M1min 1.65 2.11

40◦

4.45

(b)

6.7 M1 = 3.5

θ

M2 45◦ θ

Figure S6.7

M1

=

3.5

β

=

45 deg

127 28.158◦

=⇒ θ

=

M1n

=

M1 sin β = 3.5 sin 45 = 2.47

M2n

=

0.51592

M2

=

M2n sin (β − θ)

=

1.78

6.8 M2

=

4.0

ν2

=

65.785◦

ν2

=

ν1 + |∆θ|

ν1

= ν2 − |∆θ|

Therefore,

(a) |∆θ|

=

60 deg −15 deg = 45 deg

Therefore, ν1

=

=⇒ M1

=

65.785 deg −45 deg = 20.785 deg 1.8022

(b) |∆θ|

=

60 deg −30 deg = 30 deg

Therefore, ν1

=

=⇒ M1

=

65.785 deg −30 deg = 35.785 deg 2.360

128

Oblique Shock and Expansion Waves

(c) |∆θ|

=

0 deg

Therefore, nu1

=

ν2 = 65.785 deg

For this value of Prandtl Meyer function, from isentropic table, we get M1 = 4.0 . (d) |∆θ|

=

60 deg +15 deg = 75 deg

ν1

=

65.785 deg −75 deg = − 9.215 deg

A negative ν is not possible. The flow downstream can exist only upto |∆θ| = 65.785 deg for which ν1 = 0 =⇒ M1 = 1.0. 6.9 p1 , M 1

p2 ,

M2

θ

Figure S6.9

For M1 = 2.0, from isentropic table, we have p1 /p0 = 0.1278. Therefore, p2 p1 1 p2 = × = × 0.1278 = 0.0639 p0 p1 p0 2 For p2 /p0 = 0.0639, from isentropic table, we have M2 = 2.444 . M1

=

2.0 =⇒ ν1 = 26.38 deg

M2

=

2.444 =⇒ ν2 = 37.81 deg

ν2

=

ν1 + |∆θ| = ν1 + θ2

θ2

=

ν2 − ν1 = 37.81 − 26.38

=

11.43 deg

129 6.10

M 1 , p1

p0

θ Me

Figure S6.10

(a)

=⇒ and

M1

=

3.0

p1 p0

=

0.02722

pe p0

=

1.01325 × 105 70 × 105

=

0.0145
0 for supersonic flow, it can be visualized that equation (2) is also of the form of the classical wave equation. Hence, a solution to equation (2) can be expressed as, / / 2 − 1y) + g(x + 2 − 1y) M∞ (3) φ(x, y) = f (x − M∞

For the problem under consideration, only left running waves are present and therefore, / 2 − 1y) = 0 g(x + M∞

Thus,

φ(x, y) = f (x − and

/ 2 − 1y) M∞

< ∂φ ; # = f (x − βy) (−β) ∂y

By boundary condition, at the wall

∂φ ∂y

w = U∞ dy dx

x x x dyw = k(1 − ) − k = k − 2k dx l l l 165

(4) (5)

166

Two Dimensional Compressible Flows

Therefore,

% ∂φ x& = U∞ k − 2k ∂y l

Substituting this into Eq.(5), we get, % # x& = −βf (x − βy) U∞ k − 2k l # U∞ % x& f (x) = − k − 2k β l

(6) (7)

Integrating Eq.(7) with respect to its argument, [Note that the argument is (x − βy), but with y = 0], we have ! " U∞ x2 f (x) = − kx − k + constant (8) β l At x = 0, f = 0, which gives constant = 0. Therefore, ! " U∞ x2 f (x) = − kx − k β l

(9)

Since f (x) is defined throughout the flow, not just at the wall, and because it has the form of Eq.(8a), where x represents the argument of f , Eq.(4) can be written as & % x−βy k φ(x, y) = f (x − βy) = − U∞ β (x − βy) 1 − l 9.2 λu

= =

λL

=

CL

= = =

!

dz dx

"

= u

1 −α 7

0 ≤ x ≤ 0.7c

1 − −α 0.7c ≤ x ≤ c 3 ! " dz = −α dx l $ c 2 − (λu + λL ) dx βc 0 " " , +$ 0.7c ! $ c ! $ c 1 2 1 − − α dx + −αdx − − α dx + βc 0 7 3 0.7c 0 " ! " , +! 1 2 1 − α 0.7c − + α 0.3c − αc − βc 7 3

167 = β

=

α

=

4α β / M 2 − 1 = 2.29 2 deg =

2π = 0.0349 180

Thus, CL

CD

=

4 × 0.0349 2.29

=

0.06096

=

2 βc

$

c 0

'!

1

2 λ2u + λ2L dx

1 −α 7

"2

!

1 +α 3

"2

=

2 βc

=

2 [0.0081583 + 0.0406787 + 0.001218] 2.29

=

2 × 0.050055 2.29

=

0.7c +

2

0.3c + α c

*

0.04372

9.3 0 ≤ x ≤ 0.3c λu

= =

λl

0.3c ≤ x ≤ c

" 0.1 −α 0.3 0.333 − α " ! 0.03 −α 0.3 0.1 − α !

= = α

=

2 deg = 2

β

=

√

λu

= =

λl

= =

π ≈ 0.035 radian 180

8 = 2.83

" 0.1 −α 0.7 −0.143 − α " ! 0.03 −α − 0.7 −0.043 − α !

−

168

Two Dimensional Compressible Flows φ(x, y)

=

f (x − βy)

∂φ | ∂y y=0

=

−βf & (x) = U∞ λ

Therefore, U∞ λ β f & (x − βy)

f & (x)

=

φx (x, y)

=

φx (x, 0)

=

f & (x) = −

Cp

=

−2

−

U∞ λ β

φx λ =2 U∞ β

For 0 ≤ x ≤ 0.3c Cpu

=

2

= C pl

=

0.298 × 2 λu = β 2.83 0.211

2

=

0.065 × 2 λl = β 2.83 0.046

For 0.3 ≤ x ≤ c C pu

=

2

− 0.1258

= C pl

= =

0.178 × 2 λu =− β 2.83

2

0.078 × 2 λl =− β 2.83 −0.0551

Chapter 10

Prandtl-Meyer Flow 10.1 Let subscripts 1 and 2 refer to conditions upstream and downstream of the Prandtl-Meyer fan. p0

=

33.5 × 105 Pa

M1

=

2.0

From isentropic table, for M1 = 2.0, ν1

=

26.38 deg

p1 p0

=

0.1278

p2 p0

=

1.0133 = 0.03025 33.5

where subscript 2 refer to conditions downstream of the expansion fan. From p2 isentropic table, for = 0.03025, p0 M2

=

2.93

ν2

=

48.388 deg

But ν2

=

ν1 + ∆θ

Therefore, ∆θ

= =

48.388 − 26.38 22 deg

169

170

Prandtl-Meyer Flow

That is, after initial expansion, the flow direction is 22 deg with respect to nozzle axis. 10.2 For M1 = 2.3 and θ = 12 deg, from oblique shock chart, β = 36.5 deg. M1n

=

M1 sin β = 2.3 × sin 36.5 deg

=

1.368

For M1n = 1.368, normal shock table gives: M2n = 0.7527, and Thus, M2

T2

=

M2n sin (β − θ)

=

1.815

=

1.235 × 500

=

617.5K

T2 T1

= 1.235.

For M2 = 1.815, from isentropic table, ν2 = 21 deg. Also, |θ| = 24 deg. Therefore, ν3

=

ν2 + |θ| = 21 + 24

=

45 deg

For ν3 = 45 deg, isentropic table gives, M3 = 2.764 , and fore,

T3 = 0.396. ThereT0

T3 T3 T02 = T2 T02 T2

But T02 = T03 , thus, we have T3 T2

=

0.396 = 0.657 0.603

T3

=

0.657 × 617.5

=

405.7 K

For M3 = 2.764 and θ = 12 deg, oblique shock chart gives, β = 32 deg, and M3n = 2.764 × sin 32 = 1.465. Now, for M3n = 1.465, the normal shock table

171 gives M4n = 0.7157, and

T4 = 1.2938. Thus, T3 M4

=

T4

=

0.7157 = 2.09 sin 20 deg 524.9 K

Note: In this problem, oblique shock table may be used in places where oblique shock chart is used. 10.3 Let the subscripts 1 and 2 refer to conditions upstream and downstream of expansion. M1

=

2.0

p1 p01

=

0.1278

For isentropic compression p02 = p01 . Therefore, ! " 760 − 240 p2 = 0.1278 × = 0.0642 p02 760 + 275 =⇒ M2

=

2.44

=⇒ ν1

=

26.38 deg

=⇒ ν2

=

37.71 deg

or, the flow has been turned through ν2 − ν1 = 11.33 deg . Note: Compare this problem with 6.8.

172

Prandtl-Meyer Flow

Chapter 11

Flow with Friction and Heat Transfer 11.1

M∗ = 1

M 1 , p1 p∗ p01

p02

Figure S11.1

This problem has to be solved by trial and error. The pressure ratio written as

Given,

p1 can be p∗

p1 p01 p02 p1 = p∗ p01 p02 p∗

p01 p02 = 10. We want ∗ = 1.893 (critical pressure ratio). Thus, p02 p p1 p∗

=

18.93

p1 p01

or p1 /p∗ p1 /p01

=

18.93

p1 p1 and for different M1 and check for the ratio to be 18.93. For ∗ p p01 that, use the relations

Calculate

p1 p∗

=

1 M1

!

1.2 1 + 0.2M12

173

" 12

174

Flow with Friction and Heat Transfer

p1 p01

=

Trial

!

1+

M1

1 2 3 4 5

0.1 0.06 0.055 0.059 0.058

γ "− γ−1 γ−1 2 M1 2 % & p1 p1 p∗ / p01

10.944/0.99 18.25/0.977 19.90/0.998 18.56/0.998 18.88/0.998

= = = = =

11.05 18.29 19.95 18.60 18.92

It is seen that M1 = 0.058 is the required Mach number. Now use the relation ' * γ+1 2 M 4f¯L∗ 1 − M12 γ+1 1 2 = ln + 2 D γM12 2γ 1 + γ−1 2 M1 to calculate L∗ , 4f¯L∗ D

L∗

!

0.0040368 1.0006728

=

211.6 + 0.857 ln

=

211.6 − 4.72 = 206.88

=

206.88 × 0.1 4 × 0.005

=

1034.4 m

p1

=

T1 p2 m ˙

= = =

"

11.2

L = f¯ =

3.5 × 105 mboxP a 300 K 1.4 × 105 mboxP a 0.09 kg/s 600 m 0.004 m ˙

=

ρ1 V1 A

π p1 / γRT1 M1 D2 RT1 4

=

0.090

√ 3.5 × 105 π × 20.04 300 × × M1 D2 287 × 300 4

=

0.090

M1 D 2

=

8.12 × 10−5

175 For an isothermal flow with friction, it can be shown that ' ! "2 * ! "2 p2 p2 4f L 1 = 1− + ln 2 D γM1 p1 p1 ' * ! ! "2 "2 1.4 1.4 4 × 0.004 × 600 1 = 1 − + ln D γM12 3.5 3.5 9.6 D

=

0.6

1 − 1.832 M12

From Eqs. (1) and (2), we obtain 9.6 D

=

9.6 D

= 9.10 × 107 D4 − 1.832

0.6 D4 − 1.832 (8.12)2 × 10−10

Solving equation (3), We obtain D = 0.0402 m . 11.3 Energy equation gives, h+

V2 = h0 = constant 2

Also, for a perfect gas, V2 a2 + γ−1 2 ! " γ−1 2 2 M a 1+ 2 ! " γ−1 2 M T1 1 + 2

=

a20 γ−1

=

a20

=

T0

From continuity equation for a perfect gas, G = = =

m ˙ = ρ1 V1 = ρ1 a1 M1 A p1 / γRT1 M1 RT1 3 γ 1 √ p1 M1 R T1

From Eq. (2), we have T1

=

γp21 M12 RG2

176

Flow with Friction and Heat Transfer

Using this in Eq. (1), we get ! " γ p21 M12 γ−1 2 M1 1+ R G2 2 ! " γ−1 2 M1 M12 1 + 2 M1

!

γ−1 2 M1 1+ 2

= =

" 12

=

T0 R G2 T0 2 γ p1 4 R G T0 γ p1

Now, G

=

m ˙ m ˙∗ = A A

=

Γ A∗ √ p0 γRT0 A

where Γ

=

γ

!

2 γ+1

γ+1 " 2(γ−1)

Using Eq. (4) in Eq. (3), we get, M1

!

γ−1 2 M1 1+ 2

" 12

=

4

=

Γ p0 A∗ γ p1 A1

R Γ A∗ 1 T0 p0 γ γRT0 A1 p1

i.e. ! "1 γ γ−1 2 2 M1 1 + M1 Γ 2 p1 p0

=

A A∗

=

=

p0 A∗ p1 A1

2.38 × 105 = 0.0354 67.3 × 105 ! "2 ! "2 D 0.0127 = = 2.082 = 4.33 D∗ 0.0061

Therefore, p1 A p0 A∗

=

0.0354 × 4.33 = 0.1533

177 =⇒ M1 !

=⇒

4f L∗ D

"

=

2.51

=

0.4341

=

4.85 × 105 = 0.072 67.3 × 105

1

p2 p0 Therefore,

=⇒

p2 A × ∗ p0 A

=

0.072 × 4.33 = 0.312

=⇒ M2

=

1.53

=

0.147

!

4f L∗ D

"

2

Therefore, !

"

=

0.4341 − 0.147

∆L D

=

29.60 − 1.75 = 27.85

4f L∗ D

Hence, the average friction coefficient is f¯ = =

0.2871 4(27.85) 0.002577

11.4 (a)

Ae /A∗ = 2 1

p0 T0

Figure S11.4a

pb

Shock M =1 A∗

L

178

Flow with Friction and Heat Transfer

Normal shocks can exist only for M1 > 1. Therefore, the interpretation that a normal shock stands in the throat implies M = 1 at the throat and downstream is subsonic. Therefore, At = A∗ . A∗ Ae

0.50

M1 0.306

ρ1 ρ0

T1 T0

a1 a0

p1 p0

0.9547

0.9816

0.9907

0.9371

p0 7 × 105 = 8.13 kg/m3 = RT0 287 × 300 / √ γRT0 = 1.4 × 287 × 300 = 347 m/s

ρ0

=

a0

=

ρ1

=

(0.9547)ρ0 = (0.9547) × (8.13) = 7.7617 kg/m3

a1

=

(0.9907)a0 = (0.9907) × (347) = 343.8 m/s

p1

=

(0.9371)p0 = (0.9371) × (7 × 105 ) = 6.5597 × 105 Pa % ¯ ∗& 4f L p1 M1 D p∗ M1

0.306

5.0776

3.547

p1 = 1.849 × 105 Pa 3.547 Therefore, p∗ < pB =⇒ duct length L < L∗ and p2 = pB at L. p∗

p2 p∗

=

pB 2.8 × 105 = = 1.514 p∗ 1.849 × 105 % ¯ ∗&

=

p2 p∗

1.514 Therefore, 4f¯L D

4f L D

M2

0.220

=

! ¯ " ! ¯ " 4f L 4f L − D M1 D M2

=

5.0776 − 0.220

= G = = =

4.8576 ρ1 V1 = ρ1 a1 M1 (7.7617)(343.8)(0.306) 816.5 kg/m2 s

179 (b)

1 Shock

M =1 A∗

L Figure S11.4b

Upstream of normal shock wave A∗ Ae

p p0

ρ ρ0

T T0

a a0

M

0.5

0.09396

0.1847

0.5088

0.7133

2.197

p

=

ρ = = T a

= =

(0.09396)(7 × 105 ) = 0.6577 × 105 Pa (0.1847)(8.13) 1.5 kg/m3

(0.5088)(300) = 152.6 K (0.7132)(347) = 247.5 m/s

Downstream of normal shock wave This zone is referred by subscript 1.

M 2.197

p1 p 5.465

ρ1 ρ 2.947

T1 T 1.854

a1 a 1.362

M1 0.5475

Therefore, p1 ρ1

= =

(5.465)(0.6577 × 105 ) = 3.5943 × 105 Pa (2.947)(1.5) = 4.4205 kg/m3

T1 a1

= =

(1.854)(152.6) = 282.9 K (1.362)(247.5) = 337.1 m/s % ¯ ∗& 4f L p1 M1 D p∗ M1

0.5475

0.7451

1.9434

180

Flow with Friction and Heat Transfer

Thus,

3.5943 × 105 p1 = = 1.85 × 105 Pa 1.9434 1.9434 Since p∗ < pB =⇒ duct length L < L∗ and p2 = pB at L. p∗ =

p2 pB 2.8 × 105 = = = 1.514 p∗ p∗ 1.85 × 105 % ¯ ∗& p2 p∗

4f L D

1.514

Therefore, 4f¯L D

= = =

G

M2

0.220

! ¯ " ! ¯ " 4f L 4f L − D M1 D M2 0.7451 − 0.220 0.5251

=

ρ1 V1 = ρ1 a1 M1

=

(4.4205)(337.1)(0.5475)

=

815.9 kg/m2 s

(c)

1

Shock M =1 A∗

Figure S11.4c

Properties at station 1 are the same as properties in Part-b. Upstream of normal shock wave Therefore, p1 ρ1

= =

0.6577 × 105 Pa 1.5 kg/m3

T1 a1 M1

= = =

152.6 K 247.5 m/s 2.197

181 M1 2.197 Thus, p∗ =

%

4f¯L∗ D

&

p1 p∗

M1

0.36011

0.35566

p1 = 1.849 × 105 Pa. 0.35566

Therefore, p∗ < pB =⇒ duct length L < L∗ and p2 = pa at L downstream of normal shock wave. From the diagram in Fig. s11.4d, it is clear that the Mach number Mx at point x jumps.

h

pb

=

0 28

a kP

*

1

s

Figure S11.4d

The normal shock pressure jump and the Fanno line equation are given by =

2γ γ−1 Mx2 − γ+1 γ+1

px p∗

=

1 Mx

2.8 × 105 1.849 × 105

=

Let

=

p3 px

f (Mx )

-

1+

γ+1 2 γ−1 2 2 Mx

.1/2

2 1 1 1.1667Mx2 − 0.1667 Mx 2 1 1 1.1667Mx2 − 0.1667 Mx

! !

1.2 1 + 0.2Mx2 1.2 1 + 0.2Mx2

Use Secant method to find Mx as follows. ! " xj − xj−1 xj+1 = xj − f (xj ) f (xj ) − f (xj−1 ) Trial 1 x1

=

1.6

"1/2 "1/2

− 1.5143

182

Flow with Friction and Heat Transfer =

x0

1.5

x2

=

1.6 − 0.0559

x2

=

1.53

f (x2 )

=

0.000752

!

1.6 − 1.5 0.0559 − (−0.0236)

x3

=

1.53 − 0.000752

x3

=

1.53,

!

1.53 − 1.6 0.000752 − 0.0553

Therefore, Mx = 1.53. M2

! ¯ ∗" 4f L D M2

1.53

0.14699

Therefore, 4f¯L D

=

! ¯ " ! ¯ " 4f L 4f L − D M1 D M2

=

0.36011 − 0.14699

=

G

0.21312

=

ρ1 V1 = ρ1 a1 M1

=

(2.197)(247.5)(1.5)

=

815.6 kg/(m2 s)

11.5 The speed of sound a1 is given by, a1

= =

M1

= =

/

"

√ γRT1 = 20.04 333.3 366 m/s V1 73.15 = a1 366 0.2

"

183 From tables Rayleigh flow tables, for M1 = 0.2, we have T01 T0∗ T1 T∗ p01 p∗0 p1 p∗ V1 V∗

=

0.17355

=

0.20611

=

1.2346

=

2.2727

=

0.09091

From isentropic tables, for M1 = 0.2, we have p1 = 0.97250 p01 Therefore, p01

= =

T1 T01

=

0.5516 × 105 0.9725 5.672 × 104 Pa 0.99206

Therefore, T01

=

333.3 = 336 K 0.99206

(a) T02 ∆h

= =

stagnation temperature after combustion cP (T02 − T01 )

Therefore, =

T02

= =

∆h cP 1395.5 × 103 336 + 1004.5 1725.2 K

T01 +

(b) T02 T01

=

1725.2 = 5.1345 336.0

184

Flow with Friction and Heat Transfer

T02 T0∗

=⇒ M2 T2 T∗ p02 p∗0 p2 p∗ V2 V∗

=

T01 T02 T0∗ T01

=

(0.17355)(5.1345) = 0.89109

=

0.68

=

0.98144

=

1.0489

=

1.4569

=

0.67366

(c) T2

=

T2 T ∗ T1 T ∗ T1

=

0.98144 × 333.3 0.20661

=

1583.2 K

(d) p02

∆p0

=

p02 p∗ p01 p∗ p01

=

1.0489 × 5.672 × 104 = 4.819 × 104 Pa 1.2346

=

p02 − p01 = (4.819 − 5.672) × 104 −8530 Pa

= (e) s2 − s1

= = =

cp ln

!

T02 T01

"

− R ln

!

p02 p01

"

1004.5 ln (5.1345) − 287 ln 1690.1 J/kg.K

!

4.819 5.672

"

185 (f ) V2

=

V2 V ∗ V1 V ∗ V1

=

0.67366 × 73.15 0.09091

=

542.1 m/s

(g) The initial conditions can be maintained till the flow is choked at the duct T01 exit after combustion. That is, M2 = 1 and TT02∗ = 1 and since ∗ = 0.17355, 0 T0 and T01 = 336K, we have T02

=

T02 T0∗ T01 T0∗ T01

=

336.0 0.17355

= 1936 K Maximum heat of reaction q is given by q

= =

cP ∆T0 = 1004.5(1936 − 336) 1607.2 kJ/kg

11.6 The flow process is adiabatic and therefore, it can be treated as a Fanno flow. The velocity at station 1 is / V 1 = M1 a 1 = M1 γ R T 1 =

√ 0.2 1.4 × 287 × 300 = 69.44 m/s

From Fanno flow table, for M1 = 0.2, we have p1 = 5.4554, p∗

T1 = 1.1905, T∗

V1 = 0.21822, V∗

p01 = 2.9635 p∗0

Again, from Fanno flow table, for M2 = 0.5, we have p2 = 2.1381, p∗

T2 = 1.1429, T∗

V2 = 0.53452, V∗

p02 = 1.3398 p∗0

186

Flow with Friction and Heat Transfer

Thus, p2

T2

V2

p02

=

p2 p∗ 2.1381 × 5 × 101325 p1 = p∗ p1 5.4554

=

198.558 kPa

=

T2 T ∗ 1.1429 × 300 T1 = ∗ T T1 1.1905

=

288 K

=

V2 V ∗ 0.53452 × 69.44 V1 = V ∗ V1 0.21832

=

170.09 m/s

=

p02 p∗0 1.3398 × 520.951 p01 = p∗0 p01 2.9635

=

235.52 kPa

11.7 For, M1 = 0.2, from isentropic table, we have, T1 = 0.99206 T01 Therefore,

72 + 273.15 T1 = = 347.9 K 0.99206 0.99206 Since the tube is perfectly insulated, T01 = T02 , thus, T01 =

T02 = 347.9 K The initial density is ρ1

=

p1 2 × 101325 = R T1 287 × 345.15

=

2.046 kg/m3

since, 1 atm = 101325 Pa. Thus, the mass flow rate through the tube is m ˙

=

ρ1 A1 V1 = ρ1 A1 M1 a1

=

2.046 × (0.1 × 0.1) × 0.2 ×

/

γ R T1

187

=

√ 0.004092 1.4 × 287 × 345.15

=

1.524 kg/s

Now, assuming a control volume between the sections 1 and 2, we can write the force balance equation as (p1 − p2 ) A + F = ρ1 A1 V1 (V2 − V1 ) where F is the frictional drag and A is the cross-sectional area of the tube. For M2 = 0.76, from isentropic table, we have T2 = 0.89644 T02 Therefore, T2

=

0.89644 T02 = 0.89644 × 347.9

= 311.87 K a2

=

√

1.4 × 287 × 311.87

= 354 m/s V2

= M2 a2 = 0.76 × 354 = 269.04 m/s

For obtaining p2 , let us use the Fanno flow table. For M1 = 0.2 and M2 = 0.76, respectively, from Fanno flow table, we have p1 = 5.4554, p∗

p2 = 1.3647 p∗

Therefore, p2

=

p2 p∗ p1 p∗ p1

=

1.3647 × 2 = 0.5 atm 5.4554

Thus, F

=

ρ1 A1 V1 (V2 − V1 ) − (p1 − p2 ) A

188

Flow with Friction and Heat Transfer

=

1.524(269.04 − 74.52) − (2 − 0.5) × 101325 × (0.1 × 0.1) = 296.45 − 1519.88

=

− 1223.43 N

Hence, Drag = 1223.43 N 11.8 Given, L/D = 50, V1 = 195 √ m/s, T1 = 310 K and Me = 1. The speed of sound at the entrance is a1 = γRT1 . For carbon dioxide, γ = 1.3 and molecular weight is 44, thus, R

8314 Ru = M 44 188.95 J/(kg K)

= =

Therefore, a1

= =

M1

= =

√ 1.3 × 188.95 × 310

275.95 m/s V1 195 = a1 275.95 0.71

From Fanno flow table for γ = 1.3, for M1 = 0.71, we have 4 f Lmax D

=

0.20993

Thus, f

= =

0.20993 50 × 4 0.00105

11.9 From Fanno flow table, for M2 = 0.8, we have 4 f L∗2 D p2 p∗ T2 T∗ 4 f L∗1 D

=

0.07229

=

1.2893

=

1.0638

= =

4 f L∗2 4 fL + D D 4 × 0.005 × 51 0.07229 + = 40.87229 0.025

189 From Fanno flow table, for

4 f L∗ 1 D

= 40.87229, we have

M1

=

0.13

p1 p∗

=

8.457

T1 T∗

=

1.1959

p1

=

p1 p∗ 8.457 ×1 p2 = ∗ p p2 1.2893

=

6.56 atm

=

T1 T ∗ 1.1959 × 270 T2 = T ∗ T2 1.0638

=

303.52 K

T1

11.10 By energy equation, we have V12 2 Treating air as a perfect gas, we can express h = cp T and hence h0 = h1 +

c p T0 = c P T1 +

V12 2

For air cp = 1004.5 J/(kg K). Therefore, T1

a1

=

T0 −

=

!

= =

V12 2cp

" 1352 K = 349.9 K 2 × 1004.5 / √ γRT1 = 1.4 × 287 × 349.9 359 −

374.95 m/s

Therefore, M1 =

V1 135 = 0.36 = a1 374.95

(a) For M1 = 0.36, from Fanno flow table, we have 4f Lmax = 3.1801 D

190

Flow with Friction and Heat Transfer

Thus, Lmax

=

3.1801 × 5 × 10−2 4 × 0.02

=

1.99 m

This is the minimum length of the tube for the flow to choke. (b) For L2 = 0.6 m, !

4f L D

"

=

2

!

4f L D

"

1

−

!

4f L D

"

12

where 1 and 2 stands for the inlet and exit of the tube ! " 4f L 4 × 0.02 × 0.6 = 3.1801 − D 2 5 × 10−2 3.1801 − 0.96 = 2.22

=

The corresponding Mach number, from Fanno flow table, is M2 ≈ 0.41

and

For M1 = 0.36, from Fanno flow table, p02

p02 = 1.5587 p∗0

p02 = 1.7358. Therefore, p∗0

=

p02 p∗0 p01 p∗0 p01

=

1.5587 × 135 1.7358

=

121.23 kPa

For M1 = 0.36, from isentropic table, p1 p01

=

0.91433

p1

=

0.91433 × 135 = 123.43 kPa

ρ1

=

p1 123.43 × 103 = RT1 287 × 349.9

=

1.229 kg/m3

191 Thus, the mass flow rate through the tube m ˙ is m ˙

ρ1 A1 V1 = 1.229 × 135 ×

= =

π × 52 × 10−4 4

0.326 kg/s

11.11 For hydrogen, M = 2.016 and γ = 1.4 R=

8314 = 4124 J/(kg K) 2.016

At the tube inlet, the Mach number is M1

=

V1 V1 200 =√ =√ a1 γRT1 1.4 × 4124 × 303

=

0.15

From Fanno flow table, for M1 = 0.15, we get 4f¯Lmax D p1 p∗

=

27.932

=

7.2866

Thus, the tube length required for the flow to choke is Lmax

pexit

=

27.932D 4f¯

=

27.932 × 25 × 10−3 4 × 0.03

=

5.82 m p1 7.286

=

p∗ =

=

250 7.2866

=

34.31 kPa

11.12 At the pipe entrance, Mach number M1 = V1 = 200 m/s

V1 a1 ,

where

192

Flow with Friction and Heat Transfer

and a1

=

/

=

349 m/s

γRT1 =

√

1.4 × 287 × 303.15

Thus,

200 = 0.57 349 From Fanno flow table, for M1 = 0.57, we have M1 =

4f Lmax = 0.62288 D Therefore, Lmax =

0.62288 × 20 × 10−3 = 0.156 m 4 × 0.02

Thus, the length of the pipe at which the flow would be sonic is 15.6 cm 11.13 For methane, γ = 1.3, R

M1

=

8314 Ru = M 16.04

=

518 J/(kg K)

=

V1 V1 =√ a1 γRT1

=

√

=

0.0538

25 25 = 464.2 1.3 × 518 × 320

From Fanno flow equations, we have 4f¯L∗1 D

= = =

1 − M12 (γ + 1) M12 γ+1 2 ln 1 + 2 2 γM1 2γ 2 1 + γ−1 2 M1 ! " 0.00666 0.997 + 0.8846 ln = 265.16 − 5.046 0.00376 2 260.114

The pipe length at which the flow chokes is L∗

=

260.114 × 25 × 10−3 4 × 0.004

=

406.4 m

193 Again by Fanno flow relations, we have ' * 12 p1 1 γ+1 2 2 1 = p∗ M1 2 1 + γ−1 M1 2 =

1 0.0538

=

19.93

!

2.3 2

" 12

Thus, p∗

T1 T∗

=

1 × 106 p1 = 19.93 19.93

=

50.18 kPa

=

γ+1 2.3 2= 1 2 2 2 1 + γ−1 M 1 2

= 1.15 Thus, T∗

V∗

=

320 T1 = 1.15 1.15

=

278.16 K

=

a∗ =

=

/

γRT ∗ =

√

1.3 × 518 × 278.26

432.87 m/s

11.14 For argon, γ = 1.67. The given flow is a Fanno flow. From Fanno flow table, for M1 = 0.6, we have " ! 4f Lmax = 0.4908 D 1 p1 p∗

=

1.76336

T1 T∗

=

1.1194

For the given duct, 4f L 4 × 0.02 × 1.1194 = = 0.2984 D 0.3

194

Flow with Friction and Heat Transfer

Also, 4f L D

= =

!

4f Lmax D

"

1

−

!

4f Lmax D

"

2

0.2984

where subscript 2 refers to duct exit. From the above equation, we have ! " 4f Lmax = 0.4908 − 0.2984 D 2

For

!

4f Lmax D

"

=

0.1924

= 0.1924, from Fanno flow table, we get 2

M2

=

0.73

p2 p∗

=

1.426

T2 T∗

=

1.084

Thus, p2

T2

=

p2 p∗ 1.426 × 90 p1 = ∗ p p1 1.76336

=

72.81 kPa

=

T2 T ∗ 1.084 × 300 T1 = T ∗ T1 1.1194

=

290.62 K

11.15 %L

=

D 4f

-!

4f l D

"

M1

−

!

=

D (14.533 − 0) 4f

=

50 × 10−3 × 14.533 4 × 0.006

=

30.277 m

4f l D

"

M2

.

195 11.16 A =

π × 0.12 = 78.539 × 10−4 m2 4

ρ =

p = 1.94 kg/m3 RT

Therefore, the inlet velocity becomes V

=

m ˙ = 177.2 m/s ρA

M

=

0.49

From Fanno flow table, for M = 0.49, we have 4f l D

=

1.1539

p p∗

=

2.1838

T T∗

=

1.145

Thus, L

p∗

T∗

=

1.1539 × 40.1 4 × 0.006

=

4.8 m

=

1.8 × 105 2.1838

=

82.4 kPa

=

323.15 = 282.23 K 1.145

=

9.08◦ C

At half way before the chocking location, L = 2.4 m. Thus, 4f L 4 × 0.006 × 2.4 = = 0.576 D 0.1 From Fanno flow table, for

4f L D

p p∗

= 0.576, we get =

1.8282

196

Flow with Friction and Heat Transfer

T T∗

=

p

=

150.6 kPa

T

=

44.19◦ C

1.1244

11.17 (a) From Fanno flow table, for M = 0.2, we have 4f Lmax = 14.533 D where Lmax the distance from the Mach 0.2 location at which the Mach number becomes unity. Therefore, Lmax

=

14.533 × 0.05 4 × 0.00375

=

48.44 m

(b) This problem has to be solved by finding the chocking location for initial Mach numbers 0.2 and 0.6. LM =0.6 = (Lmax )M =0.2 − (Lmax )M =0.6 From Fanno flow table, for M = 0.6, we have 4f Lmax D

=

0.49082

Lmax

=

0.49082 × 0.05 = 1.64 m 4 × 0.00375

Thus, LM =0.6 = 48.44 − 1.64 = 46.8 m 11.18 (a) Given, T0 = 380 K. By energy equation, we have h0 = h1 +

V12 2

where h0 is the stagnation enthalpy and h1 and V1 are the static enthalpy and velocity, respectively, at the duct entrance. Treating air to be a perfect gas, we

197 have, h0 = cp T0 and h1 = cp T1 . Therefore, the energy equation becomes, T0

=

T1 +

V12 2 cp

T1

=

T0 −

V12 302 = 380 − 2 cp 2 × 1004.5

=

379.6 K

since cp = 1004.5 J/(kg K) for air. The speed of sound is / a1 = γ R T1 = 390.54 im/s M1 =

V1 30 ≈ 0.08 = a1 390.54

From Fanno flow table, for M1 ≈ 0.08, we have ! " 4f L = 106.72 D 1 Thus, D

=

4 × 0.02 × 55 m 106.72

=

4.12 cm

(b) The inlet velocity V1 = 90 m/s. Therefore, T1

=

380 −

a1

=

/

M1

=

902 = 376 K 2 × 1004.5

γ R T1 = 388.7 m/s

90 = 0.23 388.7

From Fanno flow table, for M1 = 0.23, we have 4f L D

=

10.416

D

=

4 × 0.02 × 55 1046

=

0.422 m

198

Flow with Friction and Heat Transfer

(c) The inlet velocity V1 = 425 m/s. Therefore, T1

=

380 −

a1

=

/

4252 = 290.1 K 2 × 1004.5

γ R T1 = 341.4 m/s

425 = 1.24 341.4 From Fanno flow table, for M1 = 1.24, we have M1

=

4f L D

=

0.04547

D

=

4 × 0.02 × 55 0.04547

=

96.77 m

Note: For the supersonic flow at the duct entrance, the diameter comes out to be more than the length, for the present data. 11.19 The hydraulic diameter of the duct is Dh

=

4 × 0.03 × 0.03 4 × cross-sectional area = perimeter 4 × 0.03

=

0.03 m

At the duct entrance the flow velocity is V1 = 1000 m/s. The local speed of sound is / √ a1 = γRT1 = 1.4 × 287 × 350 = 375 m/s Thus, the Mach number at duct entrance is M1

=

V1 1000 = 2.67 == a1 375

For M1 = 2.67, from Fanno flow table, we have ! " 4f Lmax = 0.46619 D 1 Therefore, the duct length required to decelerate the flow to Mach 1.0 is Lmax

=

0.46619 × 0.03 4 × 0.0025

=

1.40 m

199 11.20 For mass flow rate m ˙ to be maximum, the exit Mach number M2 should be unity, i.e. the flow is choked. For the given duct, 4f L D

=

4 × 0.023 × 0.25 0.025

=

0.92

4f L D ,

For this value of if the flow at the exit has to choke, from Fanno flow table, we have M1 = 0.52. From isentropic table, for M1 = 0.52, we have p1 = 0.83165 p01 T1 T01

=

0.94869

p1

=

0.83165 × 50 × 101325 = 4.21 MPa

T1

=

0.94869 × 320 = 303.58 K

Therefore, m ˙ max

=

ρ1 A1 V1 = ρ1 A1 M1 a1

=

48.32 ×

=

/ 22 π1 25 × 10−3 × 0.52 × γRT1 4 √ 0.012334 1.4 × 285 × 303.58

=

4.3 kg/s

Further, p∗ p1

=

1 = 0.487 2.0519

Thus, p∗

=

0.487 × 4.21 = 2.05 MPa

Therefore, the mass flow rate will remain maximum for the back pressure range 0 < pb < 2.05 MPa 11.21 (a) Let subscripts 1 and 2 refer to duct entry and exit conditions. The duct length ∆L required to accelerate the from Mach 0.2 to 0.5 can be determined from ! ∗" ! ∗" fL fL f ∆L = − D D M =0.2 D M =0.5

200

Flow with Friction and Heat Transfer

where f is friction factor, D is duct diameter and M is Mach number. From Fanno flow table, we have ! ∗" fL = 14.533 D M =0.2 ! ∗" fL = 1.0691 D M =0.5 Thus, 0.025∆L 30 × 10−3 ∆L

= =

14.533 − 1.0691 16.157 m

(b) From Fanno flow table, we have ! ∗" fL D M =1

≈

0

Thus, f ∆L D

=

14.533

∆L

=

14.533 × 30 × 10−3 0.025

=

17.44 m

11.22 For the given pipe, we have 4f Le De

=

4 × 0.02 × 18 5 × 10−2

=

28.8

For 4fDLe e = 28.8, from Fanno flow table, Me = 0.15. The speed of sound at pipe exit is √ ae = 1.4 × 287 × 468 = 433.74 m/s The exit velocity and density are Ve

=

0.15 × 433.74 = 65.062 m/s

ρe

=

pe 101325 = 0.754 kg/m3 = RTe 287 × 468

201 Therefore, m ˙

= =

ρe Ae Ve = 0.754 ×

π × 0.052 × 65.062 4

0.096 kg/s

11.23 Let subscripts 1 and 2 refer to conditions at inlet and exit of the tube, respectively. Given, pe = 0 Pa, T01 = 300 K and p01 = 6 × 101325 Pa, since 1 atm = 101325 Pa. 2 is much lower than the choking limit of 0.48667(for γ = (a) L = 0 m and pp01 1.67), therefore, the flow is choked. Thus, the mass flow rate is 22 1 0.7266 × 6 × 101325 π 30 × 10−2 √ m ˙ = 4 RT01

For argon, molecular weight is 39.944, gas constant R is 208 J/(kg K) and γ = 1.67. Therefore, the mass flow rate becomes m ˙ = 125 kg/s (b) L = 2.22 m. Therefore, 4f Lmax D

=

4 × 0.005 × 2.22 30 × 10−2

=

0.148

From Fanno flow table, for γ = 1.67 and

4f Lmax D

= 0.148, we have M1 = 0.71.

Now, from isentropic table, for M1 = 0.71(γ = 1.67), we have p1 = 0.67778 p01 T1 T01

=

0.85552

p1

=

6 × 101325 × 0.67778 = 412056.35 Pa

T1

=

300 × 0.85552 = 256.66 K

The density at the inlet is ρ1

=

p1 412056.35 = RT1 208 × 256.66

=

7.72 kg/m3

202

Flow with Friction and Heat Transfer

Thus, the mass flow rate is m ˙

=

ρ1 A1 V1 = ρ1 A1 M1 a1

=

7.72 ×

=

√ π0.32 × 0.71 1.67 × 208 × 256.66 4

115.68 kg/s

11.24 Given, p1 /p2 = 15 and p1 = 150 atm. Therefore, p2 =

150 = 10 atm 15

where p1 is the storage tank static pressure and p2 is the settling chamber static pressure. Let p2 be the pressure at which the flow chokes, i.e. p2 = p∗ . Thus, p1 = 15 p∗ From Fanno flow table, for p1 /p∗ = 15, we have 4f L∗ D

=

140.66

Thus, L∗

=

140.66 × 0.1 4 × 0.005

=

703.3 m

That is, the pipe will choke at a length of 703.3 m. 11.25 For the given piping system, p01 600 = 13.33 = p4 45 This pressure ratio is much more than the pressure ratio required for the flow to choke. Therefore, at the exit, the Mach number M4 can be taken as unity. Also, ! " 4f L 4 × 0.013 × 1.1 = D B 3.10 × 10−2 =

1.845

203 For this value, from Fanno Table, we get M3 ≈ 0.43. From isentropic table, for M3 = 0.43, we have A3 A∗

=

1.5

Also, A2 A3

=

!

A2 A2 ∗

=

A2 A3 , A3 A3 ∗

"2

6 3.1

= 3.75 since A2 ∗ = A3 ∗

Thus, A2 A2 ∗

3.75 × 1.5 = 5.625

=

From isentropic table, for AA22∗ = 5.625, we have M2 ≈ 0.1 and from from Fanno flow table, for M2 = 0.1, we have 4f L = 66.922 D For tube A, !

4f L D

"

=

4 × 0.015 × 0.9 6 × 10−2

=

0.9

A

Also, !

4f L D

"

=

1−2

!

4f L D

"

1

−

!

4f L D

"

2

Therefore, !

4f L D

"

= 1

!

4f L D

"

+ 1−2

!

4f L D

=

"

= 0.9 + 66.922 2

67.822 % & From Fanno flow table, for this 4fDL = 67.822, we have M1 ≈ 0.1. From isentropic table, for M1 = 0.1, we get ρ1 ρ01

=

0.99502

T1 T01

=

0.998

204

Flow with Friction and Heat Transfer

The stagnation density is ρ0

=

p0 600 × 103 = RT0 287 × 390

=

5.36 kg/m3

Thus, ρ1

=

0.99502 × 5.36 = 5.33 kg/m3

T1

=

0.998 × 390 = 389.22 K

a1

=

/

V1

=

M1 a1 = 39.55 m/s

A1

=

π × 0.062 = 28.27 × 10−4 m2 4

γRT = 395.5 m/s

Thus, the mass flow rate is m ˙

= =

ρ1 A1 V1 = 5.33 × 28.27 × 10−4 × 39.55 0.596 kg/s

Chapter 12

MOC

205

206

MOC

Chapter 13

Measurements in Compressible Flow 13.1 (a) p

= (760 − 500) = 260 mm of Hg

p0

= (760 − 350) = 410 mm of Hg

p p0

=

260 = 0.634 410

From isentropic table the corresponding Mach number is M = 0.835 (b) p0

=

760 + 275 = 1035 mm of Hg

p p0

=

260 = 0.251 1035

From isentropic table the corresponding Mach number is M = 1.56 13.2 pgauge

=

3.6 × 104 Pa

patm

=

0.756 × 9.81 × 13.6 × 103

=

1.0086 × 105 Pa 207

208

Measurements in Compressible Flow

=

50 cm of Hg = 0.5 × 9.81 × 13.6 × 103

=

6.67 × 104 Pa

T0

=

300 K

p0,gauge

=

1.027 × 105 Pa

p0abs

=

(1.027 + 1.0086) × 105 Pa

p0 − p

Therefore, ρ0 =

p0 RT0

=

2.0356 × 105 287 × 300

=

2.36 kg/m3

(a) By compressible Bernoulli equation, we have γ p u2 + 2 γ−1ρ

=

γ p0 γ − 1 ρ0

=

!

p p0

=

!

1.3686 × 105 2.0356 × 105

ρ

=

0.753 × 2.36 = 1.78 kg/m3

u2 2

=

3.5

u

=

ρ ρ0

" γ1

!

1 " 1.4

= 0.753

2.0356 1.3686 − 2.36 1.78

"

× 105

256.06 m/s

(b) For incompressible flow, ρ = ρ0 . Therefore, u = =

4

2 (p0 − p) = ρ

237.55 m/s

3

2 × 6.67 × 104 2.364

209 Note: The error committed in assuming the flow to be incompressible in this problem is 7.23%. 13.3 From isentropic table, for M1 = 0.9, we have p1 = 0.5913 p0 For M2 = 0.2, again from isentropic table, we have p2 = 0.9725 p0 Therefore, p2 p1

=

0.9725 = 1.6448 0.5913

p2

=

1.6447 × 4.15 × 105 = 6.826 × 105 Pa

Therefore, p2 − p1 = (6.826 − 4.15) × 105 = 2.676 × 105 Pa 13.4 (a) T = 500◦ C = 773 K. Therefore, √ a = 1.4 × 287 × 773 = 557.3 m/s M

=

400 V = = 0.718 a 557.3

From isentropic table, for M = 0.718, we have p0

= =

p0 p

= 1.4098. Therefore,

1.4098 × 1.01325 × 105 1.428 × 105 Pa

(b) T

=

−50◦ C = 273 − 50 = 223 K

a

=

√

M

=

p0

=

p0

=

1.4 × 287 × 223 = 299.33 m/s

400 = 1.336 299.33