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


Sketch XEB versus T

For an adiabatic reaction with \( \dot{W}_s \) and \( \Delta C_p = 0 \), sketch the conversion, XEB as a function of temperature for:

a. an exothermic reaction

Blank plot template with axes labeled X_EB and T, and points marked T₀ and T along the x-axis.

Hint 1: What is the adiabatic energy balance?

Full Solution


b. an endothermic reaction

Graph template with Y-axis labeled X_EB and X-axis labeled T with T₀ marked; bold arrow at the end of the X-axis.

Full Solution


c. give XMB=f(T) using the combined mole balance, rate law, and stoichiometry for a first order irreversible reaction \( A \rightarrow B \)

Blank graph with Y-axis labeled X_MB and X-axis labeled T, with a bold arrow indicating increasing temperature.

i.e. show how to locate the steady state conversion and temperature

Full Solution

 


 

 

 

 

 

 

Hint 1

Energy Balance to Determine XEB=f(T)

\(\dot{Q} - W_s - F_{A0}X \left| \Delta H_{Rx}(T_R) + \hat{C}_p(T - T_R) \right| = F_{A0} \sum \theta_i C_p (T - T_0)\)

\(\dot{Q} = 0, W_s = 0, \Delta C_p = 0 \Rightarrow \Delta H_{Rx} = \Delta H_{Rx}^\circ\)

\(X[\Delta H_{Rx}(T_R)] = \left( \sum \theta_i C_p (T - T_0) \right)\)

\(X_{EB} = \left[ \sum \theta_i C_p \right] \left[ \frac{[T - T_0]}{-\Delta H_{Rx}} \right]\)

Back to Problem



 

 

 

 

 

 

 

 

 

 

 

 

For and adiabatic reaction with \( \dot{W}_s \) and \( \Delta C_p = 0 \), sketch conversion as a function of temperature.

Solution: Part A


Energy Balance to Determine XEB=f(T)

\(\dot{Q} - W_s - F_{A0}X \left| \Delta H_{Rx}(T_R) + \hat{C}_p(T - T_R) \right| = F_{A0} \sum \theta_i C_p (T - T_0)\)

\(\dot{Q} = 0, W_s = 0, \Delta C_p = 0 \Rightarrow \Delta H_{Rx} = \Delta H_{Rx}^\circ\)

\(X[\Delta H_{Rx}(T_R)] = \left( \sum \theta_i C_p (T - T_0) \right)\)

\(X_{EB} = \left[ \sum \theta_i C_p \right] \left[ \frac{[T - T_0]}{-\Delta H_{Rx}} \right]\)

For an exothermic reaction,ΔHRX is negative,[ e.g., ΔHRX= -100 kJ/mole A]. Therefore, [-ΔHRX] is a positive number.

\(X_{EB} = \frac{\left( \sum \theta_i C_p \right) (T - T_0)}{[ \text{a positive number} ]}\)

X increases with increasing T.

Graph of X versus T showing a diagonal line labeled X_EB, representing where the adiabatic energy balance is satisfied, starting at T₀.

Back to Problem

 

 

 

 

 

 

 

 

 

 

Solution: Part B

For an endothermic reaction, ΔHRX is positive (+), XEB increases with decreasing T. [e.g. ΔHRX=100 kJ/mol A] And therefore [-ΔHRX] is a negative number.

\(X_{EB} = \frac{\left( \sum \theta_i C_p \right) (T - T_0)}{[ \text{a positive number} ]}\)

Graph of X versus T showing a diagonal line labeled X_EB, representing where the adiabatic energy balance is satisfied, starting at T₀.

Back to Problem

 

 

 

 

 

 

 

 

 

 

 

 

Solution: Part C

Mole Balance to Determine XMB=f(T)

For a first order reaction

\( V = \frac{F_{A0} X}{-r_A} = \frac{v_0 C_{A0} X}{k C_{A0} (1 - X)}, \quad \tau = \frac{X}{k (1 - X)} \)

rearranging...

\( X_{MB} = \frac{\tau k}{1 + \tau k} = \frac{\tau A e^{-E/RT}}{1 + \tau A e^{-E/RT}} \) -> Graph of X_MB versus T showing a curved line with a note that the mole balance is satisfied all along the line.

At the intersection of XMB vs. T and XEB vs. T both the mole balance and the energy balances are satisfied. The steady-state conditions (XSS,TSS) occur here.

Graph showing X versus T with intersecting lines labeled X_EB and X_MB. Point X_SS is marked as the steady state where X_EB and X_MB intersect at T_SS.        Graph showing X versus T. The lines X_EB and X_MB intersect at steady state X_SS, occurring at T_SS. T_0 is marked before the intersection.

        Endothermic Reaction                                Exothermic Reaction

Both the mole and energy balances are satisfied when XMB=XEB. The steady state temperature and conversion are TSSand XSS respectively for an entering temperature TO.


Back to Chapter 11