Chapter 11: Nonisothermal Reactor Design: The Steady State Energy Balance and Adiabatic PFR Applications

Second Order Reaction in a CSTR. 

The acid catalyzed irreversible liquid phase reaction

\( A \rightarrow B \)

is carried out adiabatically in a CSTR. The reaction is second order in A. The feed, which is equal molar in water (which contains the catalyst) and A, enters the reactor at a temperature of 52˚C and a total volumetric flow rate of 10 dm3/min. The concentration of A entering the reactor is 4 molar.

a) What is the reactor volume to achieve 80% conversion

b) What conversion can be achieved in a 1000 dm3 CSTR? What is the exit temperature?

Process flow diagram of a reactor. Feed stream A enters the reactor at a volumetric flow rate of 10 dm³/min, a concentration of 4 mol/dm³, and a temperature of 325K. The stream splits from line W into the reactor. The reactor is shown as a vertical cylindrical tank with an outlet arrow to the right indicating flow of the product stream. 

Additional Information:

\( \Delta H_{\text{Rx}} = -3000 \, \text{cal/mol} \)

\( C_{P_A} = 15 \, \text{cal/mol} \cdot \text{°C} \)

\( C_{P_B} = 15 \, \text{cal/mol} \cdot \text{°C} \)

\( C_{P_W} = 18 \, \text{cal/mol} \cdot \text{°C} \)

\( k = 0.0005 \, \text{at} \, 25 \, \text{°C} \)

\( E = 15000 \, \text{cal/mol} \)

Part (a)

Hint 1: What is the combined mole balance, rate law and stoichiometry that gives the reactor value as a function of temperature and conversion?

Hint 2: What is the equation that gives X, solely as a function of t, k, and CA0?

Hint 3: What is the equation that one obtains from the energy balance that gives X as a function of T?

Hint 4: What is the specific reaction rate 3?

Hint 5: What CSTR reactor volume is necessary to achieve 80% conversion?

Part (b)

Hint 1: Sketch the conversion obtained from the mole balance and energy balance as a function of T.

Hint 2: Write a Polymath program to obtain the exit conversion and temperature. Use the output to plot XEB and EMB.

  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Hint 1

Mole Balance

\( V = \frac{F_{A0} X}{-r_A} \)

\( -r_A = k C_A^2 \)

\( C_A = C_{A0} (1 - X) , F_{A0} = C_{A0} \nu_0 , \tau = \frac{V}{\nu_0} \)

\( V = \frac{F_{A0} X}{k C_{A0}^2 (1 - X)^2} , \tau = \frac{1}{k C_{A0}} \frac{X}{(1 - X)^2} \)

Back to Part a

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Hint 2

\( \tau k C_{A0} = \frac{X}{(1 - X)^2} \)

\( \tau k C_{A0} - 2 \tau k C_{A0} X + \tau k C_{A0} X^2 = X \)

\( \tau k C_{A0} X^2 - (2 \tau k C_{A0} + 1) X + \tau k C_{A0} = 0 \)

\( X = \frac{(2 \tau k C_{A0} + 1) \pm \sqrt{(2 \tau k C_{A0} + 1)^2 + 4 \tau k C_{A0}}}{2 \tau k C_{A0}} \)

\( X = \frac{2 \tau k C_{A0} + 1 - \sqrt{4 (\tau k C_{A0})^2 + 4 \tau k C_{A0} + 1 - 4 (\tau k C_{A0})^2}}{2 \tau k C_{A0}} \)

Let

\( D\alpha = \tau k C_{A0} \)

\( X_{MB} = \frac{(2 \tau k C_{A0} + 1) - \sqrt{4 \tau k C_{A0} + 1}}{2 \tau k C_{A0}} = \frac{(2 D \alpha + 1) - \sqrt{4 D \alpha + 1}}{2 D \alpha} \)

Back to Part a

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Hint 3

Energy Balance

\( X_{EB} = \sum \Theta_i C_{A} \left( \frac{T - T_0}{\Delta H_{Rx}} \right) \)

\( X = \frac{(C_{PA} + C_{PW}) (T - T_0)}{\Delta H_{Rx}} \)

\( T = T_0 + \left( \frac{-\Delta H_{Rx}}{C_{PA} + C_{PW}} \right) X \)

\( T = 325 \, \text{K} + \left( \frac{-3000 \, \text{cal}}{\text{mol}} \right) \left( \frac{1}{15 + 18 \, \text{cal/mol · K}} \right) X \)

\( T = 325 + 90.1X \)

For 80% conversion, \( T = 325 + 72.7 = 397.7 \)

Back to Part a

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Hint 4

\( @397.7 \, k = 0.005 \exp \left[ \frac{15,000 \, \text{cal/mol}}{1.987 \, \text{cal/mol · K}} \left( \frac{1}{298} - \frac{1}{397.7} \right) \right] = (0.005)(572.8) \, \frac{\text{dm}^3}{\text{mol · s}} \)

\( = 0.29 \, \frac{\text{dm}^3}{\text{mol · s}} \)

Back to Part a

 

 

 

 

 

 

 

 

 

 

 

Hint 5

\( V = \frac{1}{k C_{A0}} \left( \frac{X}{(4 - X)^2} \right) = \frac{1}{(0.29)(4)(1 - 0.8)^2} = 8.62 \, \text{dm}^3 \)

Back to Part a

  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Part (b)  What conversion can be achieved in a 1000 dm3 CSTR?

Hint 1

\( X_{MB} = \frac{(2 D_\alpha + 1) - \sqrt{4 D_\alpha + 1}}{2 D_\alpha} \)

where \( D_\alpha = \tau k C_{A0} = C_{A0} k_1 e^{\frac{E}{R} \left( \frac{1}{T_1} - \frac{1}{T} \right)} \)

\( X_{EB} = T_0 + \frac{-\Delta H_{Rx} X}{C_{pA} + C_{pW}} \)

  Graph showing conversion X versus temperature T. Two curves are plotted: one nonlinear curve labeled X_MB that increases then plateaus, and a linear diagonal line labeled X_EB that intersects the plateau of X_MB. Horizontal and vertical dashed lines indicate where the curves intersect and the corresponding T and X values.

The exit temperature and conversion are determined from the intersection of XEB and XMB

Back to Part b

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Hint 2

We can generate XEB and X vs. T curves by incrementing T and calculating the conversion.

Equations and parameters for simulating an adiabatic CSTR, including rate, temperature, conversion, and Damköhler number expressions.

Table of initial, maximum, minimum, and final values for variables in an adiabatic CSTR model, including T, τ, Cₐ₀, ΔHₓ, E, R, k, Xeb, Da, and X.

  Table of temperature vs conversion (X) and Xeb values for an adiabatic CSTR, ranging from 300K to 400K.

Plot of temperature vs conversion (X) and Xeb for an adiabatic CSTR, showing two curves increasing with temperature and intersecting near 390K.

Equation for Figure

\( v_0 = 10 \, \text{dm}^3/\text{min} \)

\( C_{A0} = 4 \, \text{mol}/\text{dm}^3 \)

\( T_0 = 325 \, \text{K} \)

Back to Part b

 

Back to Chapter 11