Chemical Engineering Chapter 9 Solutions For Unsteady State Nonisothermal Reactor Design Individualized Solution Example The

9-1

Solutions for Chapter 9 – Unsteady State

Nonisothermal Reactor Design

P9-1 Individualized solution

P9-2 (a) Example 9-1

The new T0 of 20 ˚F (497 ˚R) gives a new ΔHRn and T. With T=497+89.8X the polymath

P9-2 (b) Example 9-2

To show that no explosion occurred without cooling failure.

9-2

P9-2 (c) Example 9-3

Decreasing the electric heating rate (Tedot in polymath program from example 9-3) by a factor of

10 gives a conversion of 9.72 % at the onset temperature. For a decrease by a factor 10 the

P9-2 (d) Example 9-4

Decreasing the coolant rate to 10 kg/s gives a weak cooling effect and the maximum temperature

in the reactor becomes 315 K. An increase of the coolant rate to 1000 kg/s gives a Tmax of 312 K.

P9-2 (e) Example 9-5

9-3

P9-2 (f) Example 9-6

Using the code from Example 9-5, we could produce the following graphs either by changing TO

9-4

9-5

P9-2 (g) Example 9-7

The temperature trajectory changes significantly. Instead of a maximum in temperature, there is

now a minimum.

P9-2 (h) Example RE9-1

P9-2 (i) Example RE9-2

9-6

P9-2 (j) No solution will be given

P9-3

See Polymath program P9-3.pol

POLYMATH Results

Calculated values of the DEQ variables

Variable initial value minimal value maximal value final value

t 0 0 2.9 2.9

T 970 970 1.119E+23 1.119E+23

dH -336 -336 –336 -336

ODE Report (RKF45)

Differential equations as entered by the user

[1] d(T)/d(t) = -dH*k/Cpa

9-7

Explicit equations as entered by the user

[1] dH = -336

P9-4

9-8

See Polymath program P9-4.pol

POLYMATH Results

No Title 08-11-2005, Rev5.1.233

Calculated values of the DEQ variables

Variable initial value minimal value maximal value final value

t 0 0 100 100

Na 1.0E-09 1.0E-09 326.10947 312.51175

T 400 400 801.376 800

Cpa 20 20 20 20

Vra -1.9E-10 -101.76832 –1.9E-10 -100

ODE Report (STIFF)

Differential equations as entered by the user

[1] d(Na)/d(t) = Fao+Vra

[2] d(T)/d(t) = ((Vra*dHr)-Fao*Cpa*(T-To))/(Na*Cpa)

Explicit equations as entered by the user

[1] Fao = 100

[2] dHr = -8000

[3] Cpa = 20

9-9

P9-5 (a)

9-10

POLYMATH Results

Calculated values of the DEQ variables

Variable initial value minimal value maximal value final value

t 0 0 120 120

vo 1 0 1 0

V 50 50 100 100

ra 0 -0.3365379 0 -1.969E-05

ODE Report (RKF45)

Differential equations as entered by the user

[1] d(Na)/d(t) = ra*V

[2] d(Nb)/d(t) = ra*V+Fbo

[3] d(Nc)/d(t) = -ra*V

Explicit equations as entered by the user

[1] Fbo = if(t<50)then(10)else(0)

[2] Nao = 500

[3] Cbo = 10

[4] k = .01*exp((10000/1.987)*(1/300-1/T))

P9-5 (b)

This is the same as part (a) except the energy balance.

9-11

See Polymath program P9-5-b.pol

P9-5 (c)

This is the same as part (b) except the reaction is now reversible.

9-12

P9-6 (a)

P9-6 (b)

9-13

P9-7

9-14

See Polymath program P9-7.pol

POLYMATH Results

Calculated values of the DEQ variables

Variable initial value minimal value maximal value final value

t 0 0 10 10

X 0 0 0.2504829 0.2504829

T 373 373 562.91803 562.91802

Ca 0.1 0.0749517 0.1 0.0749517

k2 3.0E-05 3.0E-05 366.75159 366.75083

ODE Report (RKF45)

Differential equations as entered by the user

[1] d(X)/d(t) = –ra/.1

[2] d(T)/d(t) = ((40000+(10*(T-298)))*(-ra)*(1/.1))/(56.25-(10*X))

Explicit equations as entered by the user

[1] k1 = .002*exp((100000/8.314)*(1/373-1/T))

[2] Ca = .1*(1-X)

[3] Cb = .1*(1.25-X)

[4] k2 = .00003*exp((150000/8.314)*(1/373-1/T))

9-15

P9-8 (a)

Use Polymath to solve the differential equations developed from the unsteady state heat and mass

balances.

See Polymath program P9-8-a.pol

POLYMATH Results

No Title 08-11-2005, Rev5.1.233

Calculated values of the DEQ variables

Variable initial value minimal value maximal value final value

t 0 0 300 300

Cc 0.1 0.1 205.63558 205.63558

Cs 300 43.080524 300 43.080524

T 278 278 333.55016 333.55016

mu 0.0426159 2.912E-04 0.1725264 2.912E-04

Q 0 0 0 0

Cps 74 74 74 74

Cpc 74 74 74 74

rs -0.005327 -30.908876 –0.005327 -0.074831

9-16

ODE Report (STIFF)

Differential equations as entered by the user

[1] d(Cc)/d(t) = rg

[2] d(Cs)/d(t) = -rg/Ycs

Explicit equations as entered by the user

[1] Iprime = (0.0038*T*exp(21.6-6700/T))/(1+exp(153-48000/T))

[2] Km = 5.0

[3] mu1max = 0.5

[4] Ycs = 0.8

[5] mu = mu1max*Iprime*(Cs/(Km+Cs))

[6] Q = 0

[7] Cps = 74

[8] Hrxn = -20000

9-17

P9-8 (b)

When we change the initial temperature we find that the outlet concentration of species C has a

P9-8 (c)

Cc can be maximized with respect to T0 (inlet temp), Ta (coolant/heating temperature), and heat

exchanger area. Therefore, if we are to find the optimal heat exchanger area the inlet temperature

P9-9

First order liquid phase, CSTR

First solve the steady state problem for the heat exchange area A for normal operaiton T = 358K.

9-18

See Polymath program P9-9.pol

POLYMATH Results

Calculated values of the DEQ variables

Variable initial value minimal value maximal value final value

t 0 0 10 10

Ca 180 4.1007565 180 4.1007565

tau 0.4 0.4 0.4 0.4

k 1.1 1.1 107.23721 107.23721

Na 0.9 0.0205038 0.9 0.0205038

9-19

ODE Report (RKF45)

Differential equations as entered by the user

[1] d(Ca)/d(t) = ((180-Ca)/tau)+ra

[2] d(Cb)/d(t) = -ra-(Cb/tau)

Explicit equations as entered by the user

[1] tau = .4

[2] A = 22.696

[3] k = 1.1*exp(11409*(1/313-1/T))

[7] Nb = Cb/V

P9-10 (a)

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Remember that CB0 is not constant here. Similar to CA, use equations (1) and (4)

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