davissmch09

7/21/2019 davissmch09

http://slidepdf.com/reader/full/davissmch09 1/14

Chapter 9

Exercise 1:

As in example 9.3.2, the material balance on the batch reactor is:

)1](exp[10)1( )(104 4

AT R A A f f k

dt

df g

−=−= − (1)

where T is given by the energy balance:

∑∆

+=

i

pi A

A

o

AT r o

i

o

C n

f n H T T

)( ν

(2)

For this problem, ∆ H r = -40,000 cal/mol. Therefore the energy balance becomes:

)192862944288(1

)2381(40000300

A

A

f

f T

−−

−+= (3)

After substituting equation (3) into equation (1) and solving the differential equation

(initial condition is f A = 0 at t = 0) the time required to achieve 50% conversion is 417 s.

The final temperature is 347oC. This is a very short time for a large scale system, since

loading and unloading must be considered.

Exercise 2:

The same energy balance and material balance are used here as in exercise 1. The time to

reach 90% conversion is 531 s while the final temperature is 392oC.

Exercise 3:

The material and energy balances for the adiabatic system are shown below, respectively:

)150)(100()65)(2.1)(100(65)1(100

)100)(15000(

)2.1)(1()5.0()](exp[105 13001200003

A A A

A

A AT R A

f f f

dt

df

dt

dT

f f dt

df g

+−+−=

−−⋅⋅−×= −

(1)

9.1



The only difference between example 9.3.3 and this exercise is that here UA H = 0.

Solution to this set of differential equations is shown below (@ t = 0, T = 300 K and f A =0). The dashed curves are solutions to example 9.3.3.

100 200 300 400 500

0.2

0.4

0.6

0.8

f A

Time (s)

100 200 300 400 500

320

340

360

380

T ( K )

Time (s)

The amount of energy that is removed from the reactor can be calculated with thefollowing formula from the energy balance:

T C f nC f nC f n f n H C B A p A

o

A p A

o

A p A

o

A A

o

Ar ∆+−+−+∆= ])2.1()1([removedenergy

For example, with ∆T = 31.6 at f A = 0.8 the amount of energy removed is 698,000 J.

9.2



Exercise 4:

After substituting the following relationship for the wall temperature into the energy

balance in example 9.3.3 and solving the set of differential equations, the following

conversion and temperature profiles were found. The dashed lines correspond to the

constant wall temperature case.

100 200 300 400 500

0.2

0.4

0.6

0.8

A

Time (s)

100 200 300 400 500

310

320

330

340

350

T ( K )

Time (s)

Exercise 5:

The exit temperature is calculated from the energy balance. The energy balance is not a

function of ε B. Therefore, the equation for the exit temperature does not change from thatin example 9.4.2. The exit temperature is calculated with the following formula:

9.3



B

B

f

f T

5.257

30000723

++= (1)

However the material balance changes slightly. The new material balance is thefollowing:

[ ]∫ −=

1.0

027232

)(1T B

o

B

B

f kC

df τ (2)

where C and the rate constant is the same as in example 9.4.2. After

substituting the rate constant and equation (1) into equation (2) and solving, the space

time is 49.2 s and at 10% conversion the temperature is 775 K. If molar expansion is

taken into account, as in example 9.4.2, the space time is 47.1 s. Therefore, about 4%error is introduced by not taking into account molar expansion.

mol/L10x42.8 3−=o

B

Exercise 6:

If exit temperature effects are ignored in the reactor described in example 9.4.2, the

material balance reduces to:

[ ]∫−−

=1.0

0

2

5.01

1

B

B

f

f o

B

B

kC

df τ (1)

where C . The energy balance is the same as that in example 9.4.2.

After substituting the expression for the rate constant and the energy balance into

equation (1) and solving, the space time is 44.5 s. If exit temperature effects wereincluded, then the space time is 47.1 s. Therefore, 6% error was introduced by excluding

exit temperature effects (due to low conversion).

mol/L10x42.8 3−=o

B

9.4



Exercise 7:

The following plots show the solution of the set of equations that described the adiabatic

PFR:

0.2 0.4 0.6 0.8 1

0.2

0.4

0.6

0.8

1

y

x

0.2 0.4 0.6 0.8 1

1.02

1.04

1.06

1.08

1.1

θ

x

The dashed lines are for the case when the heat of reaction is doubled. The energy balance changes when the heat of reaction is doubled. The new energy balance is the

following:

([θ

θ 1118exp4.0 −= y

dx

d )] (1)

9.5



Exercise 8:

First write a material balance on A in the CSTR:

r )( A A

o

A

V

f F υ −= (1)

The rate expression for this reversible reaction is the following:

(2)2

2

2

1r P A C k C k −=

Since the conversion of A in the CSTR is 0.4, the concentrations of P and A at the outlet

of the CSTR can be determined.

M C C C C

M f C C

o

A A

o

P P

A

o

A A

6.0)5.19.0(1))((

9.0)4.01(5.1)1(

22

=−−=−+=

=−=−=

−

Now substitute equation (2) into equation (1) along with C A and C P and solve for V .

V = 0.824 L

Use the energy balance on the CSTR to determine the rate of heat removed.

The energy balance on the CSTR is the following:

)()()(

o

P P

o

P A A

Ar

o

A

T T C F T T C F

f H F

Q P A

−+−+−

∆

= υ &

(3)

Note that T-T o = 60 K, (-ν A) = 2, F A = 60 mol min

-1and F P = 40 mol min

-1. Therefore, the

heat transfer rate can be determined:

mincal58000=Q&

Now write a mass balance on the PFR:

r A

R

Ao

AdV df F υ −= (4)

Write r in terms of f A:

9.6



(5)

22

2222

2

2

1

225.0)1(5.67r

))1(5.15.1(1.0))1(5.1(30r

A A

A A P A

f f

f f C k C k

−−=

−−−−=−=

Substitute the rate equation into equation (4) and simplify:

∫ =−−

9.0

4.0

22 1007.2)1(5.67

R

A A

A V

f f

df

V required is 6.90 L

The heat removed can be calculated from the energy balance on an isothermal PFR:

A

o

A

A

r df F H

Qd ν

∆−=& (1)

Therefore the change in heat removed in the PFR is 250,000 cal min-1

.

It is not reasonable to remove this much heat from vessels so small in any conventional

way.

Exercise 9:

Part a)

First mix the stream of ester (E) and the stream of base (B). A material balance leads to

the following concentrations at the inlet of the reactor:

M33.1 M,67.0 == o

B

o

E C C

Note that E is the limiting reactant and the total volumetric flow rate is 30 L s-1

.

A material balance on E in the CSTR is:

(1)V C C E E

o

E r vv0 υ −−=

where B E B E C C C kC )exp(10r 3031100014 −==

Write equation (1) in terms of fractional conversion of E:

9.7



V f C C f C f C C E

o

E

o

B E

o

E E

o

E

o

E ))(1()exp(10)1(vv0 3031100014 −−−−−−= (2)

After substituting the known initial concentrations, the volume of the reactor, and the

volumetric flow rate, the fractional conversion of E is 0.755.

Part b)

The energy balance for the CSTR is the following:

)(Cv)r ( p

o

or T T V H Q −+∆= ρ & (3)

However, the reactor operates isothermally. Therefore, T = T o.

Since the rate can be expressed as

B E B E C C C )exp(10 3031100014 −=kC r = (4)

and the fractional conversion of E is 0.755 (from part a),

r = 0.00232 mol L-1

s-1

Therefore, the heat removed is 670 kcal s-1

.

Sincethe rate of heat removed is known, the following equation can be used to determine

the area required.

) (5)( C H T T UAQ −=− &

Therefore, the area required is 3.77 m

2.

part c)

If the coolant supply fails, 0=Q&

Therefore the energy balance reduces to the following equation:

)(Cv)r (0 po

or T T V H −+∆= ρ (6)

From the material balance:

(7))(vr E

o

E C C V −=

substitute equation (7) into equation (6)

9.8



)(Cv)v(0 p oo E

o

E r T T C C H −+−∆= ρ (8)

From part (a) (at 30oC) C E = 0.163 M. However, if the temperature goes up it will cause

C E to go even lower.

Therefore C . Equation (8) is simplified to the following: E

o

E C >>

)(Cv)v(0 p oo

o

E r T T C H −+∆= ρ (9)

Now substitute the known values,

)30)()(130()67.0)(30)(45(0CL

kcalsL

Lmol

sL

molkcal

o −+−= T (10)

Therefore, the temperature is 60oC.

Exercise 10:

First write a material balance on A in the reactor:

0 (1)r vv =−− V C C A

o

A

The first-order rate expression is the following:

A A C T

kC −

== 12500exp10r 13 (2)

After substituting equation (2) into equation (1), notice that there are two unknowns, T

and C A. Therefore the energy balance is needed. The energy balance is:

)(Cv)r (0 p

o

or T T V H Q −+∆== ρ & (3)

Equation (3) becomes:

))(1)()(1)(10)(1000(L)10000()exp(10)70(0kcal

4.18kJ

1000L

m

K kg

kcal

s

L

m

kg1250013

mol

kJ 3

3

o

AT

T T C −+−= ⋅

−

where T o is the feed temperature.

After simplification, the material and energy balances reduce to the following:

0)exp(101050 1250017 =−−− AT A C C (4)

9.9



0)(8.41)exp(107 1250018 =−+−×− o

AT T T C (5)

Solve equation (4) for C A and substitute it into equation (5).

0)(8.41)exp(1010

)50)(exp(107

1250017

1250018

=−++

×−−

−o

T

T

T T (6)

g Q&− r Q&

where the subscript g denotes the heat generated and the subscript r denotes heatremoved.

Plot the heat generated and heat removed curves vs. T for each condition.

-1000

0

1000

2000

3000

4000

5000

250 270 290 310 330 350 370 390 410 430

Exit Temperature (K)

Q r o r Q g

Qr at

To = 300K Qg

Qr at

To

= 280K Qr at

To = 320K

9.10



If To = 280 K, then the reactor temperature at the outlet is approximately 280 K.

If To = 300 K, then the reactor temperature at the outlet can be 301 K or 382 K (it cannot

be 337 K-this state is unstable). The temperature of the reactor will depend on the startup

of the reactor.

If To = 320 K, then the reactor temperature at the outlet is 403 K.

Now use equation (4) to find the exit concentrations:

If To = 280 K, T = 280 K and CA = 4.99 mol L

-1


-1


-1

If T

o

= 300 K, T = 337 K and CA = 2.81 mol L

-1

however this state is unstable.If To = 300 K, T = 382 K and CA = 0.080 mol L

-1

part b)

The energy balance on the reactor is the following:

)(Cv)r ()( p

o

or C H T T V H T T UAQ −+∆=−−= ρ & (7)

where the coolant is at 290 K (T C ).

Use the material balance to find T.

Since the fractional conversion of A is 0.9, the concentration of A out of the reactor is0.5M.

The material balance is:

(8)V C C A

o

A r vv0 −−=

Substitute the known values

)L10)(5.0)(exp(10)5.0)(10()5)(10(0 4L

mol1250013L

molsL

Lmol

sL

T −−−= (9)

Therefore, the exit temperature is 361 K. This is below 373 K, so it is possible to get90% conversion.

In order to evaluate start up problems, use the energy balance from equation (7) and solve

for UA H since the exit temperature is now known.

9.11



K s

kJ UA H

⋅= 04.21

Plot the energy balance to see if there are multiple steady states. Note that the material

balance yields the following expression for the concentration of A as a function oftemperature:

)exp(1010

50

1250017T

AC −+

= (10)

Substitute the concentration of A into the energy balance:

)exp(1010

)exp(105.3)320(8.41)290(21

1250017

1250020

T

T T T −

−

+

×=−+− (11)

Q r g Q−

-1000

-500

0

500

1000

1500

2000

2500

3000

3500

4000

250 270 290 310 330 350 370 390 410

Exit Temperature

Q r o r Q g

Qg

Qr

9.12



Since there are 3 steady states, in order to reach 90% conversion, you must start up the

reactor hot.

Exercise 11:

Write the material balance on A in the CSTR

0r =−− τ A

o

A C C (1)

where B A C k C k 21r −=

Rewrite the concentrations in terms of the fractional conversion of A:

(2))1( A

o

A A f C C −=

(3) A

o

A B f C C =

The equilibrium constant, K , is the ratio of the forward rate constant over the reverse rate

constant (k 1 at 300 K is 0.240 s-1

). Therefore, k 2 = 0.03 s-1

.

Substitute the rate expression and equations (2) and (3) into equation (1).

Therefore, the fractional conversion of A is 0.88.

In order to find the conversion at 500 K, the rate constants at 500 K must be determined.

1

50025003

1 74.6)exp(10 −−

== sk

Since K

k k 1

2 = , the equilibrium constant at 500 K must be determined in order to find k 2.

The equilibrium constant is related to the Gibbs free energy in the following way:

(4) K RT G ln−=∆ In addition the Gibbs free energy is related to the heat of reaction:

(5)S T H G ∆−∆=∆ After substitution of equation (5) into equation (4) and assuming that the entropy and heat

of reaction are not functions of temperature, the equilibrium constant is the following:

∆−=

T R

H K K

g

r o exp (6)

9.13



where K o is a constant.

Therefore if K is known at one temperature, K o can be determined. Then, K can be

determined at a second temperature.

K at 500 K is 0.00975 and k 2 is 691.3 s-1

.

Substitute the new rate constants into the material balance and solve for f A.

The fractional conversion of A at 500 K is 0.0097.

part b)

The energy balance on the reactor is

τ

ρ )(Cr

p oo

r

T T H

V

Q −+∆=

&

(7)

where r is the following:

(8) A

o

A A

o

A B A f C f C C k C k 03.0)1(240.0r 21 −−=−=

From part (a) the fractional conversion is 0.88, thereforeLs

mol012.0r ⋅= .

Substitute all known quantities into the energy balance and solve forV Q&

Therefore,sL

kcal17.0 ⋅−=V

Q& or

sLkcal17. ⋅0 of heat is removed.

9.14

davissmch09

Documents

Transcript of davissmch09