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- Mark M. Clark

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Exercise 1.1

1

Solutions Chapter 1

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

2

Solutions Chapter 1

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Solutions Chapter 1

Exercise 1.2 Assuming an incompressible fluid, Eq.1.5 can be used: N N dV = ∑ Qin,i − ∑ Qout , j dt i =1 j =1

In this problem, Eq 1.5 reduces to:

dV = Qin − Qout dt Where Qin = 0.2 m3/s, Qout = 0.001[V − 200] m3/s, with the initial condition: V=200m3 @ t=0. So

dV = 0.2 − 0.001(V − 200) dt dV = 0.4 − 0.001V dt Rearranging and separating variables, dV

∫ 0.4 − 0.001V = ∫ dt

So

Integrating (Calculus refresher:

∫

dx = ln x , redefine x=0.4-0.001V), x

So 1 d (0.4 − 0.001V ) = ∫ dt ∫ 0.001 0.4 − 0.001V ln(0.4 − 0.001V ) = −0.001t + c1 −

0.4 − 0.001V = e −0.001t +c1 0.4 − 0.001V = e −0.001t ⋅ c 2 V = 400 − e −0.001t ⋅ c3

(Note that c1, c2 and c3 are constants to be determined. They are not the same. ) Now applying initial condition that V=200m3 @ t=0, 200 = 400 − c3 ⇒ c 3 = 200 So

V = 400 − 200e −0.001t Note: at t → ∞, e −0.001t → 0 , so V → 400 m3/s (i.e. at steady state)

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Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Exercise 1.3

4

Solutions Chapter 1

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

5

Solutions Chapter 1

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

6

Solutions Chapter 1

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Exercise 1.4

7

Solutions Chapter 1

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Exercise 1.5

8

Solutions Chapter 1

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Exercise 1.6

9

Solutions Chapter 1

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Solutions Chapter 1

Exercise 1.7 DIAGRAM: Q=1 m3/s C=100 mg/L

q= 0.001 m2/s

Q=2 m3/s Q=1 m3/s 0

1000

2000 3000 X, length (m)

4000

5000

(1) We will divide the whole stretch of river into the following segments: 0-1000 m, 1000-2000 m, 2000-3000 m, and 3000-5000m. The contaminant concentration will be calculated at each of the 4 segments. (2) For 0 ≤ x < 1000 m, no contaminant is added into the river, so the concentration will be 0. (3) For 1000 ≤ x ≤ 2000 m: Systems Approach: choose the control volume as indicated in the shaded area below. Qin,2 =1 m3/s Cin,2 =100 mg/L

q= 0.001 m2/s Q

Qin,1 =2 m3/s x

Qout,1 =1 m3/s 0

1000

2000

3000

X, length (m) Assuming constant volume and steady state, Eq. 1.1 becomes V

dC sys dt

N

N

i =1

j =1

= ∑ Qin,i C in,i − ∑ Qout, j C out , j = 0

10

4000

5000

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Solutions Chapter 1

So N

N

i =1

j =1

∑ Qin,i Cin,i − ∑ Qout , j Cout , j = 0

(Eq. 1.3)

Now apply the assumption of fluid incompressibility, Eq 1.5 can be used. N N dV = ∑ Qin,i − ∑ Qout , j dt i =1 j =1

Because of the constant volume,

dV =0 dt

N

N

i =1

j =1

∑ Qin,i − ∑ Qout , j = 0

(Eq. 1.6)

Apply Eq 1.6 and Eq 1.3 to the system control volume

Qin,1 + Qin, 2 = Q x or, (2 m3/s) + (1 m3/s) = Q x = 3 m3/s Qin,1C in,1 + Qin, 2 C in, 2 = Q x C x

Cx =

Qin,1C in,1 + Qin, 2 C in, 2

=

Qx

(2

m3 mg m3 mg )(0 ) + (1 )(100 ) s L s L m3 (3 ) s

C x = 33.3

m3 s

(4) For 2000 ≤ x ≤ 3000 m: Systems Approach: choose the control volume as indicated in the shaded area below. Qin,2 =1 m3/s Cin,2 =100 mg/L

q= 0.001 m2/s Qx, Cx

Qin,1 =2 m3/s

Q2000 m

C2000m

0

1000

2000

x Qout,1 =1 m3/s 3000

X, length (m)

11

4000

5000

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Solutions Chapter 1

Apply Eq 1.6 and Eq 1.3 to the system control volume Q2000m + q( x − 2000) = Q x

(3 m3/s) + (0.001 m2/s) (x-2000)m = Q x = (0.001x+1) m3/s

Q2000m C 2000m + 0 = Q x C x

Cx =

Q2000m C 2000m Qx

Cx = (

m3 mg )(33.3 ) s L = m3 (0.001x + 1) s (3

100 m3 ) 0.001x + 1 s

For example, @2500m, C 2500m = (

100 m3 m3 ) = 28.6 0.001× 2500 + 1 s s

@3000m, C 3000m = (

100 m3 m3 = 25 ) 0.001× 3000 + 1 s s

(5) For 3000 ≤ x ≤ 5000 m, no contaminant or flow is added, so the concentration will no change, i.e. C x = C 3000m = 25

m3 s

ANSWER: Plot concentration (mg/L) vs. length (m) 35

Note slight curvature

concentration (mg/L)

30 25 20 15 10 5 0 0

1000

2000

3000 length (m)

12

4000

5000

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Exercise 1.8

13

Solutions Chapter 1

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Solutions Chapter 1

Exercise 1.9 (a) Assume No prior contamination, perfect mixing, i.e. Csys = Cout, and constant volume

106 m3 Lake

Qout = 1 m3/s

Qin = 1 m3/s Cin = 25 mg/L

a) System approach assuming constant volume

dC sys dt

=

1 1 Qin C in − Qout C out V V

Qin = Qout = Q and C sys = C out

So V dC out = C in − C out Q dt

Separating variables and integrating

V dC out = dt Q C in − C out

−

V ln(C in − C out ) = t + const Q

Cin − Cout = Ke

−

Qt V

Applying initial condition: Cout = 0 @ t = 0, C in − 0 = K ⋅ 1 , so K = C in Cin − Cout = Cin ⋅ e Cout = Cin (1 − e

−

−

Qt V

Qt V

)

In this problem t

Cout = 25

b) As t → ∞, e −t / 10 → 0 C out → 25 6

− 6 mg (1 − e 10 ) L

mg L

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Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Solutions Chapter 1

Exercise 1.10 a.

Here we can use Eq. 1.5,

dV = ∑ Qin,i − ∑ Qout , j dt i j However, since there is no outflow and only one inflow, the equation reduces to dV = Qin dt Separating and integrating with initial condition that V=200 at t=0, yields, V=Qin t + 200

Starting with Eq. 1.1, and (1) applying the chain rule to the first term, and (2) letting Qout=0,

b.

Csys

dCsys dV +V = QinCin dt dt

However, from the answer to part a, V=Qint + 200 and dV/dt = Qin, and substituting also some of the known flows and concentrations, we get (0.2 t + 200) c.

dCsys dt

+ 0.2Csys = 2

Fortunately this equation separates nicely into, dCsys

∫ 2 − 0.2 C

=∫ sys

dt 200 + 0.2t

Here we apply to both sides of the equation the simple formula from integrals tables: dx ∫ x = ln x + const. Using the initial condition that Csys = 2 at t = 0 yields Csys = 10 − d.

320 40 + 0.04 t

From the governing differential equation in Part b, dCsys/dt must approach zero as time goes to infinity; hence, Csys=10. In addition, from Part c, as time goes to infinity the term on the right hand side of the negative sign goes to zero, hence we find the same answer, Csys=10.

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Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Solutions Chapter 1

Exercise 1.11 (a) Applying the systems approach,

V

dCsys dt

= QCin − QCout

However if we restrict to times great than zero and assume mixing is good enough to ensure homogeneity of soap concentration (the perfect mixing assumption, Chapter 10, i.e., Csys=Cout) dCout Q = − dt Cout V

All we need is an initial condition for Cout which with our assumptions can be Cout=C0 at t=0; hence, we find

C t = exp − C0 t where t = V/Q (b) As t → ∞ Cout → 0 from the above solution. You reach the same conclusion from the governing mass balance V

dCsys dt

= QCin − QCout

if you let the left hand side go to zero as t → ∞.

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Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Solutions Chapter 1

New exercise 1.12 In the first method, a mass balance yields the following soap concentration, Cf, in the glass after it is filled with clean water the first time:

Cf =

0.01 V0 C 0 = 0.01C0 V0

This is also the soap concentration in the film after the glass is emptied. Similarly, a mass balance for subsequent rinses yields: Number rinses 0 1 2 3 etc.

Film soap concentration C0 0.01 C0 0.0001 C0 0.000001 C0 etc.

Clean water used -V0 2 V0 3 V0 etc.

In the second method we start adding clean water and when the glass is just filled the soap concentration in the glass is 0.01 C0. At this point the glass starts over flowing at exactly the same rate that clean water is entering; we can use the result of Exercise 1.11 at this point, t = exp − 0.01C0 t Cf

or

t C f = 0.01C0 exp − t

Where t = V0/Q. There are a number of ways to compare the two methods, but the most direct is to just make a table and compare the two predictions for different amounts of Clean water: Clean water used 0 V0 2 V0

Cf first method C0 0.01 C0 0.0001 C0

3 V0

0.000001 C0

etc.

etc.

Time for second method 0 0 t

Cf second method C0 0.01 C0 (initial condition) 0.00368 C0 0.00135 C0

2t etc.

etc.

So by the time two glasses of rinse water are used, the film in the second method is 37 times as dirty as the first method; by the time two glasses of rinse water are used, the film in the second method is 1350 times as dirty as the first method. Therefore the first method is more efficient.

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Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Solutions Chapter 1

New Exercise 1.13

V

dCsys

= QinCin − Qout Cout

dt

∫C

dCsys

in

− Csys

=

− ln ( Cin − Csys ) =

Q dt V ∫

Q t + const. V

10 Cin (mg/L) 5 2.5

0

1

2 3 Time, hours

4

5

At t=2 (call it t=0), Cin = 10 and Csys = 2.5 then

const. = − ln 7.5

Q Csys = 10 − 7.5exp − t V Here “t” only pertains to time between 2 and 3 hours. At t=3 hours (t=1 for above solution), Csys=4.77 mg/L.

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Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Exercise 1.14

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Solutions Chapter 1

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

20

Solutions Chapter 1

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

21

Solutions Chapter 1

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Exercise 1.15

22

Solutions Chapter 1

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Solutions Chapter 1

Exercise 1.16 In this problem we need to use the control volume approach with increasing control volume size:

n

u

Cbulk

Cbl

δ

C=0 ∂

∫∫ C n ⋅ udA + ∂t ∫∫∫ CdV = 0

C ,S

C ,V

But C values are constant, and there is mass inflow only at top surface, so dV − uCbulk A + C bl =0 dt C dV = uA bulk dt C bl

or But V = δA , so

C dδ = u bulk dt C bl

Now separating and integrating, C bulk udt C bl ∫

∫ dδ = δ=

However, V p = ∫ uAdt = A∫ udt , so

C bulk udt C bl ∫

Vp A

δ=

= ∫ udt and finally,

C bulk V p C bl A

23

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Solutions Chapter 1

Exercise 1.17

I n θ I

a) Examining the figure, the power production of the panel is

I • n dA When the sun is at either horizon (θ = 90 or 270o), cos θ = 0, so the dot product is zero. When the sun is at its zenith, θ = 180o, and cos θ = -1; this is the maximum power, -I dA. (The negative sign here is correct and simply means that the energy is being absorbed rather than radiated by the panel.) So maximum power production occurs when the sun is directly overhead. b) It is useful at this point to recall that power has the units of energy per time; therefore, energy is power times time. (Your energy bill may therefore be quoted in units of kilowatt-hours for instance.) The angle of the sun is a function of time, however, so we need to connect θ and time before going much further. For the horizontal panel, the sun starts at θ = π and ends at θ = 3π/2, and travels π radians during the effective daylight period, T. So a simple expression for at θ as a function of time is π π θ= + t 2 T Then we can set up a simple integral for the energy T

energy = A ∫ I • n dt 0

where A is the area of the panel. Substituting in for the dot product and θ,

π π energy = A I ∫ cos + 2 T 0 T

AIT π π t dt = sin + π 2 T

T

2A I T t =− π 0

Where the negative simply means energy is being absorbed.

24

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Solutions Chapter 1

For the vertically placed panel, consider the following sketch:

I

Energy collected on this side only

n θ

I

θ travels from π to π/2 during the period T/2, hence

θ =π −

π

T

t

Substituting as before: T /2

energy = A I

∫ 0

T /2

π AIT π cos π − t dt = − sin π − t T π T 0

=−

AIT

π

So the vertically placed panel produces only half the energy of the horizontally placed panel. If the vertical panel absorbed energy on both sides, the two scenarios would be identical. But that would probably cost twice as much.

25

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Solutions Chapter 1

Exercise 1.18 Redraw figure as shown below: Equation 1.14 can recast in terms of whatever causes the mass transfer; here, that would be the sedimentation velocity of the particles, Vp. An appropriate control volume is shown by the dotted lines.

I II

3 cm

30 o

At steady state, Equation 1.14 can be adapted to this problem: - ∫∫ c (n • Vp) dA = ∫∫ c (n • Vp) dA I

II

Where Section I is the sloped control surface and Section II is the horizontal surface. Noting the geometry of the control volume we find,

3 -C (V p cos 150 ) = CVp cos (0) (3) cos30 Which simply gives the identity: 3 C Vp = 3 C Vp

So the steady state (unit depth into page) mass transfer through the opening is mg cm mg mass transfer = 3 C Vp = 50 0.5 3 cm = 0.075 L s cm − s

26

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Solutions Chapter 1

Exercise 1.19 Eq. 1.25 is ∂C + C ∇ • v + v • ∇C = 0 ∂t

Note from appendix II that ∇ •v =

∂vx ∂v y ∂vz + + ∂x ∂y ∂z

and ∇C =

∂C ∂C ∂C i+ j+ k ∂x ∂y ∂z

and of course

v = vx i + v y j + vz k So Eq. 1.25 becomes

0=

∂v ∂v ∂v ∂C dC ∂C ∂C + C x + y + z + ( vx i + v y j + vz k ) • i+ j+ k dt ∂y ∂z ∂x ∂x ∂y ∂z

Focusing on the dot product, recall that terms like vx i • like vx i •

∂C ∂C , and terms i = vx ∂x ∂x

∂C j = 0 , (i.e., dot product of collinear unit vectors is one, and dot product of ∂y

perpendicular unit vectors is zero – Appendix II) leaving

0=

∂v ∂v y ∂vz ∂C dC ∂C ∂C +C x + + + vy + vz + vx dt ∂y ∂z ∂x ∂y ∂z ∂x

27

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Solutions Chapter 1

Exercise 1.20 We would like to know that if the interior of the microbe is approximated as water, is 1 t p , the given function is identical to Eq. 10.68, where time is simply offset by the delay time in the plug flow element, t p . Note also that the relevant mean detention time for t > t p must then be t s = Vs / Q . This must be the correct mean detention time ∞

since only that time will result in ∫ f (t ) dt = 1 . 0

Exercise 10.6 (a) As explained in Sections 10.2 and10.6, t10 corresponds to W=0.9; so 100 m3 t10 − ln 0.9* t = − ln 0.9* = W = 0.9 = exp - ⇒ t10 = 0.105 hr. 100 m3 / hr t (b)

100 m3 t W = 0.9 = exp - 10 ⇒ t10 = − ln 0.9* t = − ln 0.9* = 0.070 hr. 150 m3 / hr t

Exercise 10.7 If Eq. 4.52 is correct, then it is a proper F function for this system, i.e., t F (t ) = 1 − exp − tR t t 1 t d d 1 Then, f (t ) = F (t ) = 1 − exp − 0 + exp − =+ = + exp − dt dt tR t R tR tR tR

6

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Solutions Chapter 10

Exercise 10.8 (a) From problem 10.1, fit the N tanks-in-series model to the discrete residence-time data using the time domain fitting procedure. Using Eq 10.69 f (t ) =

N N t N −1 Nt exp(− ) N (t ) ( N − 1)! t

t =10.35 min from Problem 10.1

Using a spreadsheet, f(ti) (column 3) is calculated using equation 10.66 SSEi = [fi(model) – fi(data)]2 in column 4 and SSE =

∑ SSE

i

i

By trying different N values, the SSE is a minimum for N=4(4.02 E-5) tRTD (min)=

N=

SSE=

2nd moment=

10.35

4

4.02021E-05

25.66436

data

Model

Time (min)

f(i)

f(i)

SSE(i)

Sigma(i)

0

0

0

0

0

1

0

0.002525

6.37343E-06

0.499335

2

0.012745

0.013724

9.57293E-07

1.299555

3

0.029412

0.031472

4.24553E-06

1.933209

4

0.052941

0.050691

5.06255E-06

2.048012

5

0.066667

0.067275

3.69526E-07

1.707914

6

0.078431

0.078992

3.14296E-07

1.2001

7

0.083333

0.085234

3.61185E-06

0.697742

8

0.088235

0.086452

3.17924E-06

0.297582

9

0.085294

0.083642

2.73078E-06

0.060131

10

0.080392

0.077962

5.90531E-06

0.001665

11

0.071569

0.07051

1.12096E-06

0.088512

12

0.062745

0.062202

2.94988E-07

0.271416

13

0.054902

0.053738

1.35553E-06

0.495508

14

0.045098

0.045606

2.57977E-07

0.708492

15

0.037255

0.038115

7.40342E-07

0.922384

16

0.032353

0.031432

8.47814E-07

1.093189

17

0.02549

0.025618

1.6402E-08

1.15213

7

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Solutions Chapter 10

18

0.019608

0.020664

1.11493E-06

1.204142

19

0.016667

0.016514

2.34406E-08

1.271708

20

0.013725

0.013088

4.06962E-07

1.262043

21

0.010784

0.010295

2.39693E-07

1.157498

22

0.007843

0.008043

3.99119E-08

1.012769

23

0.005882

0.006245

1.31358E-07

0.932147

24

0.004902

0.004821

6.52025E-09

0.88309

25

0.003922

0.003703

4.78565E-08

0.787373

26

0.002941

0.00283

1.23121E-08

0.639116

27

0.001961

0.002154

3.72053E-08

0.576579

28

0.001961

0.001632

1.08024E-07

0.48434

29

0.00098

0.001232

6.33756E-08

0.359458

30

0.00098

0.000927

2.86494E-09

0.397984

31

0.00098

0.000695

8.15013E-08

0.219235

32

0

0.000519

2.6975E-07

0

33

0

0.000387

1.49804E-07

0

34

0

0.000288

8.27349E-08

0

SUM=

SUM=

4.02021E-05

25.6643589

(b) Column 5 is σ i2 calculated using t + t 2

f ( ti ) + f ( ti +1 ) (ti +1 − ti ) 2 2

i i +1 σ i2 = − t

σ2 =

σ ∑= 2 i

25.66

Equation 10.65 and 10.70 ---

σ= τ 2

N =

σ2

25.66 = = 0.2395 2 (10.35) 2 t 4 (must be an integer) 1 4.17 = 2

στ

Get the same number, N=4

8

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Solutions Chapter 10

Comparison of the data and the model (N=4) 0.1 0.09 0.08 0.07 f(i)

0.06

fi-- data

0.05

fi--model

0.04 0.03 0.02 0.01 0 0

10

20 time (min)

9

30

40

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Solutions Chapter 10

Exercise 10.9 By minimizing the SSE for the proposed model, we arrive at

= t s 7min = and t p 3.35 min . However, as shown in the plot (next page), this model is inferior compared with the N tanks-in-series fit of Exercise 10.8.

10

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

11

Solutions Chapter 10

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Exercise 10.10 In Example 10.4 In Example 10.4

12

Solutions Chapter 10

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Exercise 10.11

13

Solutions Chapter 10

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Solutions Chapter 10

Exercise 10.12 We can use Figure 10.20 directly, where the product of the mean detention time and the first-order kinetic constant is

60 min =3 hr Going into Figure 10.20, we find the highest conversion (plug flow reactor) is about 0.95, and the lowest conversion (perfect mixer) is about 0.75. From the definition of conversion, k * t = 0.05 min -1 *1 hr *

C = Cin (1 − conversion) out Therefore the minimum concentration leaving the facility is 0.05 Cin and the maximum concentration is 0.25 Cin.

Exercise 10.13 We know for the N tanks-in series model,

N N t N −1 Nt exp − N (t ) ( N − 1)! t

f (t ) =

(10.68)

and for irreversible first-order reaction,

Aout ] ∞ [= exp ( −kt ) f (t ) dt [ Ain ] ∫0

(10.109)

Substituting 10.68 into 10.109,

[ Aout ] =∞ exp −kt N N t N −1 exp − Nt dt = ( ) N (t ) ( N − 1)! [ Ain ] ∫0 t ∞

NN Nt exp ( −kt ) t N −1 exp − dt N ∫ (t ) ( N − 1)! 0 t Using Mathematica or an integrals table, the solution is

[ Aout ] = [ Ain ]

Γ( N )

NN N

N t ( N − 1)! k + t

N

Then since Γ( N ) =( N − 1)!

[ Aout ] = [ Ain ]

N

N

( N − 1)! NN t NN t = = = N N N N N N N t ( N − 1)! t t + + kt N kt N k + t NN

(

14

)

(

NN

) ( kt + N )

N

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

15

Solutions Chapter 10

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Solutions Chapter 10

Exercise 10.14 (a) In Figure 2.9, y = R = Fractional Removal, given by Eq. 2.36. In Figure 10.20, y = conv. = Conversion for [A], given by Eq. 10.113 And x is vt/v0 and kt for R and conv. (respectively):

1 1− y= R= v 1+ t V0

vt x= V0

1 1− y= conv. = 1 + kt

x= kt

The two curves have the same pattern as y = 1 − so if the two curves overlay each other, = x

but we know that

replacing

so

1 , 1+ x

vt = kt V0

V= 0

Q V /t D = = A A t

ks =

vt D

vt = kt , V0

vt = ks t , D/t

(b) [ A]out 1 N Cout = (= ) [ A]in Cin kt 1+ N Now let k =

vt V and t = Q D

Cout 1 1 1 (= )N ( = )N ( )N = v V v v Cin t 1+ t 1+ 1+ t DNQ NQ / A NV0 where V0 = Q/A and A=V/D

16

(Eq. 10. 116)

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Solutions Chapter 10

Extra: Plot the derived equation for N-1, 2, 5, 10 and 20 Equations: Quiescent Sedimentation: = R

Tanks in Series: R = 1 − (

vt V0

for vt ≤ V0

1 )N vt 1+ NV0

(Exercise 2.12)

Plug Flow – Vertical Mixing: R = 1 − exp(−

or as N → ∞, R= {1 − (

vt ) V0

v 1 ) N } → {1 − exp(− t )} vt V0 1+ NV0

1

fractional removal, R

0.8

0.6

quiescent perfect mix(N=1) N=2 N=5 N=10 N=20 plug flow - vertical mixing ( N=infinite)

0.4

0.2

0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Vt/Vo

These curves make sense because the bigger N is, the more plug-flow characteristics the system has, and the higher fractional removal the system achieves.

17

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Solutions Chapter 10

Exercise 10.15 Perfect mixer performance equation (Eq. 10121):

[A]in − [A]out

t=

− rA

Now, for second order irreversible reactions: k

A + B→R

If A and B are in stoichiometric equivalence,

[A] = [B] and therefore:

d [A] 2 = −k [A][B ] = −k [A] dt

Thus, in a perfect mixer t=

[A]in − [A]out 2 k [A]out

⇒ kt =

[A]in − [A]out 2 [A]out

Multiplying the equation by [A]in, we get the quadratic equation: [A]in [A]out

[A]in − k t[A] = 0 − in [A]out 2

Solving the quadratic equation gives:

[A]in [A]out

=

1 + 1 + 4k t [A]in ⇒ 2

18

[A]out [A]in

=

2 1 + 1 + 4k t [A]in

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Solutions Chapter 10

Exercise 10.16 (a) At equilibrium, ra = rd, therefore, ka M sC = kd C p . Then defining

= R

Cp ka and K p = Ms kd

and substituting into the above, R = K pC . (b) We can use the following schemes: Particle-associated phase

Dissolved phase

Cin

Cp,in

C

adsorption desorption

Cp

sedimentation

Writing Eq. 10.120 for every phase: Dissolved phase:

Cin − C + kd C p − ka M sC = −t ( kd C p − ka M sC ) 0 → Cin − C = t

(1)

Particle-associated phase:

C p ,in − C p t

+ kd C p − ka M sC − k sC p = −t ( −kd C p + ka M sC − k sC p ) 0 → C p ,in − C p =

Adding Eqs. 1 and 2:

(C (C

p ,in

+ Cin ) − ( C + C p ) = −t ( kd C p − ka M sC − kd C p + ka M sC − k sC p )

p ,in

+ Cin ) − ( C + C p ) = t ksC p

Then define CT= Cin + CP ,in (3) and CT= C + CP (4) ,in Substituting Eqs. 3 and 4 into the above,

CT ,in − CT = t ksC p

or CT ,in = CT + t ksC p

(5)

And using the relationships from the top of the page,

= C p RM = K pCM s s

(6)

Then substituting Eq. 4 into Eq. 6,

C (7) CT = C + K pCM s = C (1 + K p M s ) or C =T 1+ Kp Ms 19

(2)

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Solutions Chapter 10

Combining Eqs. 6 and 7,

Cp =

K p M sCT 1+ Kp Ms

(8)

Then combining Eqs. 5 and 8,

t ks ( K p M sCT ) CT ,in = CT + 1+ Kp Ms

tkK M CT ,in or CT ,in = CT 1 + s p s or CT = 1+ Kp Ms tkK M 1+ s p s 1+ Kp Ms

Exercise 10.17 (a) Making the suggested substitution,

c A,out

∞ =c A,in − ( c A,in − c A,∞ ) ∫ f (t ) dt − 0

c A,out

∞ − c A,in = − ( c A,in − c A,∞ ) 1 − ∫ 0

c A,out − c A,in c A,in − c A,∞

∞ = − 1 − ∫ 0

qn2 t 6 α ( α +1)exp − ∞ ∞ τ d f (t ) dt ∫0 ∑ 9 + 9α + qn2α 2 n =1

qn2 t α α 6 ( +1)exp − ∞ τ d f (t )dt ∑ 9 + 9α + qn2α 2 n =1

qn2 t 6 α ( α +1)exp − ∞ τ d f (t )dt or ∑ 2 2 9 + 9α + qnα n =1

qn2 t 6 α ( α +1)exp − ∞ ∞ c A,in − c A,out τ d f (t ) dt = 1− ∫ ∑ c A,in − c A,∞ 9 + 9α + qn2α 2 0 n =1 (b) Substituting as suggested qn2 t 6 ( +1)exp α α − ∞ ∞ c A,in − c A,out τ d N N t N −1 Nt 1− ∫ ∑ exp − = dt N 2 2 α α c A,in − c A,∞ q 9 9 + + t 1 = n ( 1)! t N − n 0 This solution is similar to Exercise 10.13:

20

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

c A,in − c A,out 6α (α +1) N N = 1− N c A,in − c A,∞ t ( N − 1)!

∞

n =1

Γ( N )

∑ 9 + 9α + q α 2 n

Solutions Chapter 10

1 N 2 2 qn N τ + t

c A,in − c A,out NN Γ( N ) 6α (α +1) ∞ 1 = 1− ∑ 2 2 N c A,in − c A,∞ ( N − 1)! n=1 9 + 9α + qnα t q 2 N N n τ + t c A,in − c A,out Γ( N ) 6α (α +1) ∞ 1 = 1− ∑ N 2 2 N c A,in − c A,∞ ( N − 1)! n=1 9 + 9α + qnα t q 2 N n NN τ + t

c A,in − c A,out Γ( N ) 6α (α +1) ∞ 1 = 1− ∑ N 2 2 c A,in − c A,∞ ( N − 1)! n=1 9 + 9α + qnα t q 2 N n N τ + t

or

c A,in − c A,out Γ( N ) 6α (α +1) ∞ 1 = 1− ∑ N 2 2 c A,in − c A,∞ ( N − 1)! n=1 9 + 9α + qnα t q 2 n Nτ + 1 To plot the equation we will need values for the qn introduced and defined in Section 6.7.4: tan qn =

3 qn . 3 + α qn2

To do this, first use Excel to plot the function= f ( x)

21

3x − tan x (where α = 1.0). 3 + x2

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Solutions Chapter 10

From the plot, it looks like there are non-zero roots at the following approximate values, x= 3.75, 6.75, 9.75, etc. Then we can go to Excel Solver and see what the exact roots are (with the above values as first guesses). Guess root 3.75 6.75 9.75 12.75 15.75 18.75 21.75 24.75 28.3 31.3

qn q1 q2 q3 q4 q5 q6 q7 q8 q9 q10

Exact root (Solver) 3.726 6.681 9.716 12.792 15.892 19.005 22.125 25.250 28.379 31.511

The following three spreadsheets (N=1, 2 and 5) provide values (column “1-sum”) for

c A,in − c A,out Γ( N ) 6α (α +1) ∞ 1 = 1− ∑ N 2 2 c A,in − c A,∞ ( N − 1)! n=1 9 + 9α + qnα t q 2 n 1 + Nτ a series of dimensionless time values, “tbar/tau” and α = 1. N=1 qsubn:

3.726

6.681

9.716

12.792

15.892

19.005

22.125

25.25

28.379

31.511

1

2

3

4

5

6

7

8

9

10

0

0.3763752

0.1915838

0.1067609

0.0660665

0.0443532

0.0316464

0.0236446

0.0183049

0.0145743

0.0118701

0.8852 0.1148

0.025

0.2074131

0.0427928

0.0094565

0.0025491

0.0008291

0.0003146

0.0001349

0.0183049

3.263E-05

1.78E-05

0.2818 0.7182

0.05

0.131134

0.0183431

0.003263

0.0007837

0.0002388

8.712E-05

3.643E-05

1.693E-05

8.558E-06

4.627E-06

0.1539 0.8461

0.1

0.0659843

0.0064181

0.0009795

0.0002191

6.434E-05

2.297E-05

9.476E-06

4.365E-06

2.192E-06

1.18E-06

0.0737 0.9263

0.2

0.0263886

0.0019441

0.0002701

5.808E-05

1.672E-05

5.9E-06

2.417E-06

1.108E-06

5.548E-07

2.98E-07

0.0287 0.9713

0.4

0.0087642

0.0005389

7.106E-05

1.496E-05

4.261E-06

1.495E-06

6.105E-07

2.793E-07

1.396E-07

7.487E-08

0.0094 0.9906

0.6

0.0043239

0.0002482

3.213E-05

6.716E-06

1.906E-06

6.677E-07

2.722E-07

1.244E-07

6.216E-08

3.333E-08

0.0046 0.9954

0.8

0.002568

0.0001422

1.823E-05

3.797E-06

1.076E-06

3.764E-07

1.534E-07

7.009E-08

3.5E-08

1.876E-08

0.0027 0.9973

1

0.0016992

9.199E-05

1.173E-05

2.437E-06

6.899E-07

2.412E-07

9.827E-08

4.489E-08

2.241E-08

1.202E-08

1.2

0.0012069

6.435E-05

8.175E-06

1.696E-06

4.797E-07

1.677E-07

6.829E-08

3.119E-08

1.557E-08

8.347E-09

1.4

0.0009012

4.753E-05

6.021E-06

1.248E-06

3.528E-07

1.233E-07

5.02E-08

2.292E-08

1.144E-08

6.134E-09

1.6

0.0006985

3.653E-05

4.618E-06

9.565E-07

2.703E-07

9.443E-08

3.845E-08

1.756E-08

8.764E-09

4.697E-09

0.0007 0.9993

1.8

0.0005572

2.895E-05

3.654E-06

7.564E-07

2.137E-07

7.464E-08

3.039E-08

1.387E-08

6.926E-09

3.712E-09

0.0006 0.9994

2

0.0004548

2.351E-05

2.964E-06

6.131E-07

1.732E-07

6.048E-08

2.462E-08

1.124E-08

5.61E-09

3.007E-09

0.0005 0.9995

n:

sum

1-sum

tbar/tau

22

0.0018

0.9982

0.0013 0.9987 0.001

0.999

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Solutions Chapter 10

N=2 qsubn:

3.726

6.681

9.716

12.792

15.892

19.005

22.125

25.25

28.379

31.511

1

2

3

4

5

6

7

8

9

10

0

0.3763752

0.1915838

0.1067609

0.0660665

0.0443532

0.0316464

0.0236446

0.0183049

0.0145743

0.0118701

0.8852 0.1148

0.025

0.2732917

0.0789321

0.0224645

0.0071233

0.0025667

0.0316464

0.0004666

0.0002275

0.000119

6.599E-05

0.4169 0.5831

0.05

0.2074131

0.0427928

0.0094565

0.0025491

0.0008291

0.0003146

0.0001349

6.38E-05

3.263E-05

1.78E-05

0.2636 0.7364

0.1

0.131134

0.0183431

0.003263

0.0007837

0.0002388

8.712E-05

3.643E-05

1.693E-05

8.558E-06

4.627E-06

0.1539 0.8461

0.2

0.0659843

0.0064181

0.0009795

0.0002191

6.434E-05

2.297E-05

9.476E-06

4.365E-06

2.192E-06

1.18E-06

0.0737 0.9263

0.4

0.0263886

0.0019441

0.0002701

5.808E-05

1.672E-05

5.9E-06

2.417E-06

1.108E-06

5.548E-07

2.98E-07

0.0287 0.9713

0.6

0.0141089

0.0009251

0.0001242

2.633E-05

7.526E-06

2.646E-06

1.082E-06

4.952E-07

2.476E-07

1.329E-07

0.0152 0.9848

0.8

0.0087642

0.0005389

7.106E-05

1.496E-05

4.261E-06

1.495E-06

6.105E-07

2.793E-07

1.396E-07

7.487E-08

0.0094 0.9906

1

0.0059678

0.0003524

4.595E-05

9.632E-06

2.738E-06

9.597E-07

3.915E-07

1.79E-07

8.943E-08

4.796E-08

1.2

0.0043239

0.0002482

3.213E-05

6.716E-06

1.906E-06

6.677E-07

2.722E-07

1.244E-07

6.216E-08

3.333E-08

0.0046 0.9954

1.4

0.0032763

0.0001843

2.373E-05

4.949E-06

1.403E-06

4.912E-07

2.002E-07

9.149E-08

4.569E-08

2.45E-08

0.0035 0.9965

1.6

0.002568

0.0001422

1.823E-05

3.797E-06

1.076E-06

3.764E-07

1.534E-07

7.009E-08

3.5E-08

1.876E-08

0.0027 0.9973

1.8

0.0020668

0.000113

1.445E-05

3.005E-06

8.51E-07

2.976E-07

1.213E-07

5.54E-08

2.766E-08

1.483E-08

0.0022 0.9978

2

0.0016992

9.199E-05

1.173E-05

2.437E-06

6.899E-07

2.412E-07

9.827E-08

4.489E-08

2.241E-08

1.202E-08

0.0018 0.9982

3.726

6.681

9.716

12.792

15.892

19.005

22.125

25.25

28.379

31.511

1

2

3

4

5

6

7

8

9

10

0

0.3763752

0.1915838

0.1067609

0.0660665

0.0443532

0.0316464

0.0236446

0.0183049

0.0145743

0.0118701

0.025

0.3291002

0.1280498

0.0492714

0.0199852

0.0086624

0.0040194

0.0019893

0.0010437

0.0005768

0.0003336

0.05

0.2902035

0.0915815

0.0282499

0.0095055

0.0035684

0.0014879

0.0006804

0.0003365

0.0001778

9.937E-05

0.4259 0.5741

0.1

0.2305629

0.0534796

0.0128001

0.0036189

0.0012113

0.0004679

0.0002031

9.68E-05

4.98E-05

2.728E-05

0.3025 0.6975

0.2

0.1555893

0.024693

0.0046804

0.0011604

0.0003598

0.0001326

5.582E-05

2.606E-05

1.321E-05

7.16E-06

0.1867 0.8133

0.4

0.0844871

0.0091699

0.0014597

0.0003327

9.864E-05

3.541E-05

1.466E-05

6.768E-06

3.404E-06

1.835E-06

0.0956 0.9044

0.6

0.0529555

0.0047419

0.0007025

0.0001551

4.525E-05

1.609E-05

6.625E-06

3.047E-06

1.529E-06

8.222E-07

0.0586 0.9414

0.8

0.0362711

0.0028902

0.0004117

8.942E-05

2.587E-05

9.156E-06

3.758E-06

1.725E-06

8.643E-07

4.644E-07

0.0397 0.9603

1

0.0263886

0.0019441

0.0002701

5.808E-05

1.672E-05

5.9E-06

2.417E-06

1.108E-06

5.548E-07

2.98E-07

0.0287

0.9713

1.2

0.0200566

0.0013965

0.0001908

4.073E-05

1.168E-05

4.116E-06

1.684E-06

7.717E-07

3.861E-07

2.073E-07

0.0217

0.9783

1.4

0.0157576

0.0010515

0.0001419

3.014E-05

8.624E-06

3.034E-06

1.24E-06

5.68E-07

2.841E-07

1.525E-07

1.6

0.0127061

0.0008202

0.0001096

2.32E-05

6.626E-06

2.328E-06

9.514E-07

4.355E-07

2.177E-07

1.168E-07

0.0137 0.9863

1.8

0.0104622

0.0006576

8.723E-05

1.841E-05

5.249E-06

1.843E-06

7.528E-07

3.445E-07

1.722E-07

9.238E-08

0.0112 0.9888

2

0.0087642

0.0005389

7.106E-05

1.496E-05

4.261E-06

1.495E-06

6.105E-07

2.793E-07

1.396E-07

7.487E-08

0.0094 0.9906

n:

sum

1-sum

tbar/tau

0.0064

0.9936

N=5 qsubn: n:

sum

1-sum

tbar/tau

The plot is shown below, along with the solution of the Adham et al. model (part c of problem).

23

0.8852 0.1148 0.543

0.017

0.457

0.983

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Solutions Chapter 10

The spreadsheet for the Adham et al. model follows: Adham Model n→ tbar/tau 0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

sum

1-sum

0.60796 0.15199 0.06755 0.038 0.0243 0.0169 0.0124 0.009499 0.0075 0.00608 0.005 0.0042 0.0036 0.0031 0.0027 0.96085 0.03915

0.025 0.48765 0.0765 0.02098 0.00768 0.0034 0.0017 0.0009 0.000566 0.0004 0.00024 0.0002 0.0001 8E-05

6E-05

5E-05 0.60048 0.39952

0.05

0.40709 0.05111 0.01242 0.00427 0.0018 0.0009 0.0005 0.000292 0.0002 0.00012 8E-05

3E-05

2E-05 0.47893 0.52107

0.1

0.30599 0.03072 0.00684 0.00226 0.0009 0.0005 0.0003 0.000148 9E-05 6.1E-05 4E-05

3E-05

2E-05

2E-05

1E-05 0.34789 0.65211

0.2

0.20444 0.01709 0.0036 0.00117 0.0005 0.0002 0.0001 7.46E-05 5E-05 3.1E-05 2E-05

1E-05

1E-05

8E-06

6E-06 0.22735 0.77265

0.4

0.12288 0.00905 0.00185 0.00059 0.0002 0.0001 6E-05 3.75E-05 2E-05 1.5E-05 1E-05

7E-06

5E-06

4E-06

3E-06 0.13491 0.86509

0.6

0.08784 0.00616 0.00124 0.0004 0.0002 8E-05

4E-05

7E-06

5E-06

4E-06

3E-06

2E-06 0.09599 0.90401

0.8

0.06835 0.00467 0.00094 0.0003 0.0001 6E-05

3E-05 1.88E-05 1E-05 7.7E-06 5E-06

4E-06

3E-06

2E-06

2E-06 0.07452 0.92548

1

0.05594 0.00376 0.00075 0.00024 1E-04 5E-05

3E-05

1.5E-05 9E-06 6.2E-06 4E-06

3E-06

2E-06

2E-06

1E-06

1.2

0.04734 0.00314 0.00063 0.0002 8E-05 4E-05

2E-05 1.25E-05 8E-06 5.1E-06 4E-06

2E-06

2E-06

1E-06

1E-06 0.05149 0.94851

1.4

0.04103 0.0027 0.00054 0.00017 7E-05 3E-05

2E-05 1.07E-05 7E-06 4.4E-06 3E-06

2E-06

2E-06

1E-06

9E-07

1.6

0.03621 0.00237 0.00047 0.00015 6E-05 3E-05

2E-05 9.39E-06 6E-06 3.8E-06 3E-06

2E-06

1E-06

1E-06

8E-07 0.03933 0.96067

1.8

0.0324 0.00211 0.00042 0.00013 5E-05 3E-05

1E-05 8.35E-06 5E-06 3.4E-06 2E-06

2E-06

1E-06

9E-07

7E-07 0.03518 0.96482

2

0.02932 0.0019 0.00038 0.00012 5E-05 2E-05

1E-05 7.51E-06 5E-06 3.1E-06 2E-06

1E-06

1E-06

8E-07

6E-07 0.03182 0.96818

2.5E-05 2E-05

1E-05

6E-05

4E-05

I am guessing that none of the models reach zero conversion and tbar/tau =0 because not enough terms were included in the summation. (Conversion for the remaining times converges nicely with 10-15 terms in the summation.) Summary of ploting values

tbar/tau 0

series 1

series 2

series 3

Adham

micro N=1

micro N=2

0.0391508 0.114820071 0.114820071

series 4 micro N=5 0.114820071

0.025

0.3995176 0.718154425

0.583096334 0.456967994

0.05

0.5210657 0.846083728

0.73639552 0.574109404

0.1

0.6521124 0.926294482

0.846083728 0.697482285

0.2

0.7726506 0.971312184

0.926294482 0.813282158

0.4

0.8650938 0.990604023

0.971312184 0.904389886

0.6

0.9040081 0.995385992

0.984803335 0.941371676

0.8

0.9254848 0.997266115

0.990604023

1

0.9391045 0.998193567

0.993619888 0.971312184

1.2

0.9485134 0.998718152

0.995385992 0.978296536

1.4

0.9554034 0.999043447

0.996508528 0.983004968

1.6

0.9606666 0.999258967

0.997266115

1.8

0.9648185 0.999409076

0.997801399 0.988766163

2

0.9681775 0.999517796

0.998193567 0.990604023

0.96029579

0.98633026

24

0.0609 0.0446

0.9391 0.9554

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Solutions Chapter 10

Comments: First regarding part (b) models, the highest conversion is seen for N = 1 (Series 2 above) , the exponential distribution of residence time. This appears to say that for the adsorption “reaction,” when there is complete segregation, conversion decreases as the (segregated) flow residence time approaches plug flow (e.g., Series 3 and 4 above). An interesting result. Regarding the question in part (c), the completely segregated reactor with exponential residence time (Series 2 above) has higher conversion than the perfectly mixed reactor (Series 1 above). This is a little reminiscent of Figure 10.21 where we noted that the completely segregated reactor with exponential residence time outperfromed the perfectly mixed reactor (for a second order reaction).

Exercise 10.18 The simplified equation for Brownian motion coagulation (Eq. 3.55) is identical mathematically to Eq. 9.19. Therefore the analysis of Section 10.11.2 applies to coagulation here. So the “conversion” of the coagulation reaction is given by Eqs. 10.123 and 10.132, which are plotted in Figure 10.21. The coagulation “constant” in Figure 10.21 is simply, 4kT 3µ

25

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Solutions Chapter 10

Exercise 10.19 Regardless of order, consider the first reactor in series “reactor 1” and the second

[ A]in and [ A]out refer to the overall reactor.

“reactor 2”, while

(a) Plug flow reactor followed by perfect mixer. We are given k t [ A]in = 1 for the overall reaction and t p= t = 0.5 t where p stand for the plug flow element and s for s the perfect mix unit. For plug flow reactor (Eq. 10.131),

[ A]out ,1 [ A]out ,1 = = [ A]in,1 [ A]in

1

1 1 = = = 0.667 1 + kt p [ A]in 1 + 0.5 kt [ A]in 1.5

(∗)

since [ A]in ,1 = [ A]in . For the perfect mix reactor (Eq. 10.122), [ A]out ,2 2 = [ A]in,2 1 + 1 + 4kt s [ A]in,2 However, since [ A]out ,1 = [ A]in ,2 ,

[ A]out ,2 [ A]in,2

2 2 = = 1 + 1 + 4k (0.5) t (0.667) [ A]in ,1 1 + 1 + (0.5) 4k t [ A]in ,1 (0.667) 2

2 = = 0.791 1 1 (0.5)(1) (0.667) + + 1 + 1 + (0.5) 4k t [ A]in (0.667)

(∗∗)

Then multiplying Eqs.* and **, and recognizing that [ A]in ,1 = [ A]in and

[ A]out ,1 [ A]out ,2 [ A]out ,1 [ A]out ,2 [ A]out ,2 [ A]out = = = = [ A]in,1 [ A]in,2 [ A]in [ A]out ,1 [ A]in [ A]in

[ A]out ,1 = [ A]in,2

= 0.528 0.6670.791

Now, for the reversed configuration (perfect mixer followed by plug flow reactor), the conversion for perfect mixer is

[ A]out ,1 [ A]out ,1 = = [ A]in,1 [ A]in

2

2 = = 1 + 1 + 4kt s [ A]in 1 + 1 + 4k (0.5) t [ A]in

2

2 = = 0.732 1 + 1 + 4(0.5) kt [ A]in 1 + 1 + 4(0.5)(1)

Then for the plug flow component,

26

(∗∗∗)

Transport Modeling for Environmental Engineers and Scientists (2nd Edition)

Solutions Chapter 10

[ A]out ,2 1 1 1 1 = = = = = [ A]in,2 1 + kt p [ A]in,2 1 + 0.5 kt [ A]in,2 1 + 0.5 kt [ A]out ,1 1 + 0.5 kt (0.732) [ A]in 1 1 = = 0.732 1 + 0.5(0.732) kt [ A]in 1 + 0.5(0.732) (1)

(∗∗∗∗)

Then multiplying Eqs.*** and ****, and recognizing that [ A]in ,1 = [ A]in and

[ A]out ,1 = [ A]in,2 [ A]out ,1 [ A]out ,2 [ A]out ,1 [ A]out ,2 [ A]out = = = [ A]in,1 [ A]in,2 [ A]in [ A]out ,1 [ A]in

0.732 0.732 = 0.536

(b) Consider the higher value k t [ A]in = 4 . Similar calculations for the plug flow reactor before the perfect mix reactor result in

[ A]out ,1 [ A]out ,2 [ A]out ,1 [ A]out ,2 [ A]out ,2 [ A]out = = = = [ A]in,1 [ A]in,2 [ A]in [ A]out ,1 [ A]in [ A]in

= 0.228 0.333 0.686

while for the perfect mix reactor followed by the plug flow reactor (maximum mixedness),

[ A]out ,1 [ A]out ,2 [ A]out ,1 [ A]out ,2 [ A]out = = = [ A]in,1 [ A]in,2 [ A]in [ A]out ,1 [ A]in

0.50 = 0.50 0.25

(c) We appears that the maximum mixedness reactor has slightly lower conversion than the late molecular mixing system. It also appears that the difference between the two configuration may grow as k t [ A]in for the overall reactor increases.

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