2nd Order Reaction in CSTR

Question :

PROBLEM STATEMENT A second order reaction where A → B takes place in an isothermal, jacketed CSTR. The exothermic reaction occurs at 200°C. Heat transfer fluid enters the jacket at 35°C. Component A is fed into the reactor at 500 mol/m and a temperature of 50°C. The reaction rate constant is 0.001 m3/(mol.s) and conversion is 82%.

AH, = -12.5 J/mol

CPA = 10 J/(mol-K)

a. Based on mole balance and a volume of 5 cubic meters, what is the initial flow rate of A? b. What is the heat transfer area if the heat transfer coefficient is 1.85 kJ/(m2K.h)?

Answer :

Given Data:

Reaction Temperature = T = 200^oC,       Component A Concentration = C_A0 = 500 mol/m³

Heat Transfer |Fluid Entering Temperature = T₁ = 35^oC

Component A entering temperature = T₀ = 50^oC

Reaction Rate Constant = k = 0.001m³/(mol.s),                     % Conversion = 82%                       

Reactor volume = 5m³

C_PA = 10 J/(mol.K)                  Heat Transfer |Coefficient = 1.85 kJ/(m².K.h)

Solution part a:

For second order reaction, the rate law/rate of reaction can be expressed as follows,

r_A = -k*[C_A]²   (1)

Applying mass balance for component A

Accumulation = Molar flow of A entering – Molar flow of A leaving + generation

Considering steady state situation (Accumulation = 0). CA is the concentration of A at the outlet of reactor, because the reaction is 82% complete so there will some unreacted component A at the exit of reactor.

0 = F_A0 – F_A + r_A*V    (V is the volume of reactor)

0 = v_o*C_Ao – v_o*C_A – k*[C_A]²*V (2)

(vo is the volumetric flow rate and since accumulation is zero so it’s constant at the inlet and exit of reactor). 

The conversion in the CSTR reactor can be defined as follows:

X = (C_Ao – C_A)/C_Ao                  CA = C_Ao(1– X)                     C_A = 90 mol/m³

Now substitute CA = 90 mol/m3 in equation 2, we get

40.5 = 410 v_o      v_o = 0.0988 m³/s or 355.61 m³/hr

Solution part b:

Applying energy balance taking reactor temperature T0 as reference temperature

Assumptions:

No phase change, constant heat capacity 

Heat Accumulation = Heat Flow in – Heat Flow Out + Q

 (Where Q is the heat transfer across the wall of the rector due to the flow of coolant in the reactor jacket)

The final simplified equation of energy balance can be written as 

          0 = -F_Ao*C_PA*(T-T_o) + ΔHrx*r_A*V + Q

Q = v_o*C_Ao*C_PA*(T-T₀) + ΔHrx*k*C_A²*V

Putting numerical values in above equation, we get

Q = 73593.75 kJ/hr

The heat transfer equation for equipment can be written as:

Q = U*A*dT (dT is the temperature difference between the reactor and coolant temperatures) 

Q is the heat flow; A is the area and U is the heat transfer coefficient.

Solving above equation for Area (A), we get 

Area = A = 241m²