This is example problem 8.4 in “Elements of CRE” by H.S. Fogler

Problem description: Propylene oxide reacts with water to form propylene glycol. This is done in a

CSTR. Propylene oxide is dissolved in methanol (inert material). Water is supplied in excess, and

mixed just before entering the CSTR. The water has a bit of sulfuric acid (0.1 wt%) which is a catalyst.

However, the properties of the water can be used in the calculations directly, without accounting for

the presence of sulfuric acid. There is a small ‘heat of mixing’, but we can account for this and say that

the feed is effectively at a slightly higher temperature. The reaction is proposed to be conducted under

adiabatic conditions.

Q1. Given the feed data, enthalpy and Cp values, for a desired conversion (x = 0.85), find the volume

of the reactor needed. (Also find the operating temperature).

Q2. If the volume of the reactor is given, determine the conversion. (Also find the operating

temperature).

Q3. There is an issue with adiabatic operation. The operating temperature is too high. (e.g. the product

or reactant may evaporate, the tank may get corroded easily at higher temperature, other unwanted

reactions may occur to a significant extent, …). So, we are also given an upper limit for operating

temperature. The question then is to cool the reactor, perhaps with a cooling coil. In this case, we are

given U, A and Tc , i.e. overall heat transfer coefficient, cooling coil area and cooling fluid temperature.

We are also told that Tc does not vary much because the cooling fluid flow rate is very high and it has a

high specific heat capacity.

Q4. The cooling coil doesn’t work for some reason. We want to know if we can adjust the flow rate of

propylene oxide (while keeping other flow rates the same) and maintain the temperature to be within

limits. So we want to calculate the temperature and conversion as a function of the FA-in.

The reaction is given by A + B � C. A is propylene oxide, B is water and C is propylene glycol.

Methanol (M) is solvent for A.

Data: FA-in = 19.54 kmol/h, FB-in = 364.47 kmol/h, FM-in = 32.63 kmol/h, CpA = 146.4 J/mol/K, CpB =

75.3 J/mol/K, CpM = 81.6 J/mol/K, CpC = 192.4 J/mol/K, ∆HRxn at 20 °C = -84,591 J/mol of A reacted.

Feed temperature = 24 °C (after accounting for heat of mixing), 9084

12 -116.96 10 hTk e−

= × where T is in

K.

The volumetric flow rates of methanol and propylene oxide were 1.365 kL/h each and that of water is

6.51 kL/h. Qin = 9.24 kL/h

Solution:

Q1.

For a CSTR, the design equation under steady state conditions is A in A A outoutF V r F− −+ = , and this can

be rewritten as follows.

Since ( )1A A in AF F x−= −

A

A in A out

xV

F r−

=−

From the units of rate constant, we know that this is a first order reaction. So, this can be written as

( )1

V x

Q k x=

−. Note that we have dropped the subscript ‘A’ of the conversion for the sake of simplicity.

Also note that the value of k depends on the operating temperature and unless that is known, the above

equation can not be solved.

Heat balance:

Heat released during the reaction = heat needed to raise feed temperature to ‘T’ which is the operating

temperature

To calculate the heat released during the reaction:

∆HRxn at temperature T = ∆HRxn at 293K + (Cp-products – Cp-reactants ) (T-293)

∆HRxn at temperature T = ∆HRxn at 293K + ∆Cp (T-293)

Here, ∆Cp = Cp-C – Cp-A – Cp-B = -29.3 J/mol/K

Therefore ∆HRxn at temperature T = -(84,591 +29.3 (T-293) )

This is the heat absorbed per unit mole of A reacted. The negative sign indicates that heat is actually

released.

Given FA-in = 19.54 kmol/h and the conversion desired is 0.85, the total heat released

= FA-in × x × -∆HRxn at T

= 19.54 × 0.85 × (84,591 + 29.3 (T-293) )

= 14,04, 967 + 486.6 (T-293)

Heat added to raise the feed to ‘T’ from 24 °C is

FA-in Cp-A (T-297) + FB-in Cp-B (T-297)+ FM-in Cp-M (T-297) = 32955 (T-297)

Equating the above two equations, we get

T = 340 K = 67 °C.

(Note: Had we assumed that ∆HRxn at temperature T ≃ ∆HRxn at 293K, we would have arrived at

more or less the same result, i.e. T = 66.6 °C).

When T = 340 K,

the value of k = 42.23 h-1

, and V = 1241 lit (i.e. 1.25 m3)

Q2. If the volume of the reactor is 1.25 m3, determine the conversion. Here, we know V, but we do

not know T or x. The basic equations remain the same. i.e. ( )1

MB

MB

xV

Q k x=

− and

FA-in × xEB × -∆HRxn at T = FA-in Cp-A (T-297) + FB-in Cp-B (T-297)+ FC-in Cp-C (T-297) = 32955 (T-

297)

The first equation can be written as

( )

1

9084

12

1.250.135

9.24116.96 10 1

MB

TMB

xh

e x

τ −

−= = =

× −

which can be re-arranged as

9084

12

9084

12

2.2896 10

11 2.2896 10

T

MB

T

k ex

ke

τ

τ

−

−

×= =

++ ×

The second equation can be written as

i.e. [1,652902 + 572.5 (T-293)] × x = 32966 × (T-297)

In the energy balance equation, the relationship between x and T are linear. In the mass balance

(design) equation, it is highly nonlinear. If we solve them together, we will get the value of T and x.

Another way is to choose values of T (e.g. from 300 to 350) and then plot the value of x given by mass

balance equation (xMB) and by the energy balance equation (xEB). When they intersect, we will note

that value as the solution for both T and x. Or, we can tabulate xMB and xEB and find where they

intersect.

Rewrite the energy balance equation as

( )( )

32966 297

1,652,902 572.5 293EB

Tx

T

−=

+ −

T (K) XMB XEB

300 0.139 0.06

310 0.301 0.258

320 0.518 0.454

330 0.718 0.650

340 0.851 0.844

350 0.924 1.037

Note that for any positive order reaction, the mass balance equation ensures that the conversion is

within 0 to 1. The energy balance equation does not have such constraints, and one can get values

which are not physically meaningful. (e.g. if T = 295, XEB = -0.04).

Q3. We are informed that the temperature should not exceed 52 °C because there will be a loss of the

product due to vaporization. The CSTR is 1.25 kL. It comes with a cooling system which has U = 2.

044 MJ m-2

h-1

K-1

and A = 3.716 m2. The cooling water can be maintained at 29.4 °C (302.4 K).

Determine the operating temperature and conversion for this system.

The mass balance (design) equation remains the same, but the heat balance equation would change

now.

[1,652,902 + 572.5 (T-293)] × xEB = U × A × (T-Tc) + 32966 × (T-297).

i.e. heat generated = heat removed by the cooling system + heat needed to raise the feed to ‘T’

But, U × A = 7,595.504 KJ/h/K, Tc= 302.4 K.

∴ ( ) ( )

( )

7,595.504 302.4 32966 297

1,652,902 572.5 293EB

T Tx

T

× − + × −=

+ × −

Using this and the mass balance equation, we can find that T is between 310 and 315 K, and the value

is approximately 40 °C. The corresponding conversion is about 0.36. Thus the temperature will be

below the limit.

Q4. It turns out that our cooling system is dysfunctional. So, instead of using cooling coils, can we

modify the flow rate of propylene oxide, and maintain a temperature of 52 °C or less under adiabatic

operating conditions?

Assume that the flow rate of other streams and all other operating parameters are the same, and only

flow rate (molar as well as volumetric) of propylene oxide is changed. Note that the density of

propylene oxide is 0.83 g/ml and the molecular weight is 58 g/mol.

In the original problem, the volumetric flow rates of methanol and propylene oxide were 1.365 kL/h

each and that of water was 6.51 kL/h.

Let the new molar flow rate be FA-in-new kmol/h.

The new volumetric flow rate of propylene oxide = FA-in-new × MWA / ρA.

= FA-in-new 58/0.83 = ( 69.87962 × FA-in-new ) lit/h

Hence the total volumetric flow rate is (7,875 + 69.87962 × FA-in-new) lit/h

The mass balance equation becomes

( )1

V x

Q k x=

−. i.e.

( )( )

9084

12A in new7,875 69.87962 F 16.96 10 1T

V x

e x

−

− −

=+ ×

× −

The volume of the reactor is 1.25 m3. Therefore the equation becomes

( )( )

9084

12A in new

1250

7,875 69.87962 F 16.96 10 1T

x

e x

−

− −

=+ ×

× −

After re-arranging, this becomes

( )( )

9084

12

A in new

1250 16.96 10 17,875 69.87962 F

Te x

x

−

− −

× × −= + ×

( )( )

9084

16

A in new

2.12 10 17,875 69.87962 F

Te x

x

−

− −

× × −= + ×

This can be rearranged as

A in new

9084

16

1

7,875 69.87962 F1

2.12 10 T

x

e

− −

−

=

+ × +

× ×

The energy balance equation: Heat released by the reaction is

= A in newF − − × x × -∆HRxn at 325

= A in newF − − × x × (84,591 + 29.3 × (T-293) ) kJ/h

Heat added to raise the feed to T K from 24 °C is

FA-in-new Cp-A (T-297) + FB-in Cp-B (T-297) + FM-in Cp-M (T-297)

= FA-in-new × 146.4 × (T-297) + 364.47 × 75.3 × (T-297)+ 32.63 × 81.6 × (T-297)

= (146.4 × FA-in-new + 30,107) (T-297) kJ/h

Therefore, under adiabatic operation,

A in newF − − × x × (84,591 + 29.3 × (T-293) ) = (146.4 × FA-in-new + 30,107) (T-297)

Re arranging we get,

( )( )

( )( )A in new

A in new

146.4F 30107 297

F 84591 29.3 293

Tx

T

− −

− −

+ −=

+ −

Now, given a A in newF − − , we should be able to find the T and x, by simultaneously solving the mass and

energy balance equations. Now, we can plot the temperature (and conversion) as a function of A in newF − − .

The matlab code is given in the example. The optimization algorithm is not very robust, and at times

we get incorrect answers. For solving them simultaneously, we can use ‘fmincon’ and a reasonable

temperature / conversion + random number as seed values.

Note that temperature and conversion increase rapidly near FA-in = 16 kmol/h. After that, the

temperature keeps increasing and conversion remains at 100%, as we increase FA-in. This is contrary to

what one would expect under isothermal conditions.

Q5: If we conduct the reaction in a PFR, what will be the volume required for 85% conversion?

Data: FA-in = 19.54 kmol/h, FB-in = 364.47 kmol/h, FM-in = 32.63 kmol/h, CpA = 146.4 J/mol/K, CpB =

75.3 J/mol/K, CpM = 81.6 J/mol/K, CpC = 192.4 J/mol/K, ∆HRxn at 20 °C = -84,591 J/mol of A reacted.

Feed temperature = 24 °C (after accounting for heat of mixing), 9084

12 -116.96 10 hTk e−

= × where T is in

K.

The volumetric flow rates of methanol and propylene oxide were 1.365 kL/h each and that of water is

6.51 kL/h. Qin = 9.24 kL/h

Soln: The design equation for a PFR is

( )

( )( )

( )

0.85

0

. 1

1 .1

1

AA A in A in

dF dxr F kC x

dV dV

dx dxQ k x dV Q

dV k x

dxV Q

k x

− −= = −

= − =−

=−∫

Along with this, the heat balance tells

( )in

T

A in i in p iTiT

F x H F C dT− − −∆ = ∑∫

A in TF x H− ∆ = 19.54 × x × (84,591 + 29.3 (T-293) )=(1652902

+ 572.5 (T-293)) × x

FA-in Cp-A (T-297) + FB-in Cp-B (T-297)+ FM-in Cp-M (T-297) = 32966 (T-297)

∴ ( )

( )

1652902 32966 297 572.5 293

32966 572.5

x xT

x

× + × − × ×=

− ×

We can create a table and then use calculator-based integration or do the integration by numerical

methods implemented in software such as Matlab®.

x T k K(1-x) Q/(k(1-x))

0 297 0.8834 0.8834 10.4596

0.2 307 2.41 1.928 4.792

0.4 317 6.21 3.73 2.48

0.6 327 15.17 6.07 1.522

0.8 338 35.33 7.07 1.30

0.85 340 43.3 6.5 1.4212

A very rough integration gives that the PFR volume needed to get a conversion of 80% (not 85%) itself

is 2.4 m3. In comparison, a CSTR of 1.25 m

3 was needed to get 85% conversion.