The elementary irreversible organic liquid-phase reaction A+B --> C is carried out adiabatically in a flow reactor. An equal molar feed of A and B enters at 27oC the reactor. The volumetric flow rate is 2 dm3 /s and CA0 = 0.1kmol/m3

a.) Calculate the PFR and CSTR volumes necessary to achieve 85% conversion.

b.) What is the maximum inlet temperature on could have so that the boiling point of the liquid (550K) would not be exceeded even for complete conversion?

c.) Plot the conversion and temperature as a function of PFR volume (i. e., distance down the reactor)

d.) Calculate the conversion that can be achieved in one 500dm3 CSTR and in two 250dm3 CSTRs in series.

Additional information:

A B C
H (273K) (kcal/mol) -20 -15 -41
Cp (cal/mol*K) 15 15 30
k=0.01 dm3 /mol. s at 300K, E=10,000 cal/mol

a.) and

b.)

c.) See attached picture.

d.)

500 dm -> X=0.977

250dm-> X_1=0.967 and X_2=0.992

Explanation:

Hello,

a.) At first, it is possible to find the PFR volume as shown below, considering its design equation:

Thus, since this is a nonisothermal reactor, it is defined as:

Thus, by leaving the formula in terms of conversion and temperature, we obtain (equal molar feed):

Nonetheless, by applying the energy balance for ΔCp=0 under adiabatic conditions:

Hence, integrating the dX/dV, we obtain:

Now, for the CSTR, just the design equation is changed by:

Thus:

b.) In this case, we assume that the outlet temperature is 550K or 276.85°C, thus, one modifies the energy balance in terms of the inlet temperature as:

c.) See attached picture.

d.) In this case, for 500 dm³, we solve the following nonlinear equation (design equation):

Considering that:

Whereas T is defined in the enegy balance:

Thus, solving by using a solver:

X=0.977.

Finally, for two series reactors, for the first one, the outlet conversion (using the same nonlinear equation and solver is:

So, now, the initial flow, concentrations and temperature entering to the new reactor are:

So the equation changes by:

and:

Thus, we obtain for the same k(T):

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