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

Interstage Cooling

Elementary Exothermic Reversible Reaction:

\( A + B \rightarrow 2C \)

Process flow diagram of two plug flow reactors (PFRs) in series with interstage heat exchangers. Feed stream enters Reactor 1 at temperature T₀ and flow rate Fₐ₀, with initial conversion X₁. After Reactor 1, a heat exchanger adjusts the temperature back to T₀. The stream enters Reactor 2, exits with conversion X₂ and temperature T₂, and passes through another heat exchanger to return to temperature T₀.

\(-r_A' = k [C_A C_B - \frac{C_C^2}{K_C}]\)

At equilibrium:

\(-r_A' = 0 \Rightarrow K_C = \frac{C_Ce^2}{C_Ae C_Be}\)

We can stage the reactors using either the overall conversion X (easiest way) or the conversion per reactor X¢.

\( X = \frac{\text{total moles A reacted up to a point}}{\text{mole fed to first reactor}} \)

\([C_A = C_{A0}(1-X) \cdot y \cdot \frac{T_0}{T}], \, [C_B = C_{A0}(1-X) \cdot y \cdot \frac{T_0}{T}], \, [C_C = C_{A0}(0 + 2X) \cdot y \cdot \frac{T_0}{T}]\)

\( K_C = \frac{4X_e^2}{(1-X_e)^2} = K(T_2) \cdot \exp\left[ -\frac{\Delta H_{Rx}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \right] \)

Two-part diagram. Left: Equilibrium constant K_C decreases with increasing temperature T, indicating exothermic reaction behavior. Right: Corresponding plot of conversion X versus temperature T. Includes the equilibrium conversion line X_e = sqrt(K_C) / (2 + sqrt(K_C)) and a design equation line for X using heat capacities C̄_PA and C̄_PE. Arrows illustrate steps between two conversion points X₁ and X₂ across the temperature profile.

Adiabatic Energy Balance

\( X = \frac{\sum \theta_i \hat{C}_{p_i} (T - T_0)}{-\Delta H_{Rx}} = \frac{(C_{pA} + C_{pB}) (T - T_0)}{-\Delta H_{Rx}} \)

 

The following is an exercise for the more curious and is not necessary to study if you are pressed for time. What follows is in response to a question asked by Adarsh Radadia, a student in the undergraduate reaction engineering class of Winter 2001. Adarsh wanted to know why we didn’t include the reaction products in the balance to the second reactor. The answer is (as shown below) that there are two ways to work the problem. One method (shown above) is to use the overall conversion which is the easiest way to work the problem, and the other method (which follows) is to use the conversion for each reactor. In the case of using conversion for each reactor you include the reaction product from the first reactor as shown below in the answer to Adarsh’s question.

 

(2) Based on feed to second reactor

\( X' = \frac{\text{moles A reacted in second reactor}}{\text{mole A fed to second reactor}} \)

Flow diagram showing two sequential reactors. Reactant A and B enter Reactor 1 at equal molar flow rates (F_A0 and F_B0 = F_A0). The output from Reactor 1 includes species A1, B1, and C1. These enter Reactor 2 with adjusted inlet conditions: F_A01 = F_A1, F_B01 = F_A1, F_C01 = θ_C × F_A01. The final conversion is labeled X₁' and the flow goes through a separator or heat exchanger at the end.

\( C_A = C_{A0}(1 - X') \left( \frac{yT_0}{T} \right) \)

\( C_B = C_{A0}(1 - X') \left( \frac{yT_0}{T} \right) \)

\( C_{A01} = \frac{F_{A1}}{v_0} = \frac{F_{A0}(1 - X_1)}{v_0} = C_{A0}(1 - X_1) \)

\( C_C = C_{A0}( \theta_c + 2X' ) \left( \frac{T_0}{T} \right), \theta_c = \frac{E_{C01}}{F_{A01}} = \frac{2F_{A0}X_1}{F_{A0}(1 - X_1)} = \frac{2X_1}{(1 - X_1)} \)

\( K_C = \frac{(\theta_c + 2X')^2}{(1 - X_e)^2} \)

\( \sqrt{K_C} = \frac{\theta_c + 2X'}{1 - X_e} \)

\( \sqrt{K_C} - \sqrt{K_C} X_e - \theta_c = X_e \)

\( \frac{\sqrt{K_C} - \theta_c}{2 + \sqrt{K_C}} = X_e' \)

Graph showing conversion X versus temperature T for an exothermic reaction. Two equilibrium lines are shown: one for Xₑ = √K_C / (2 + √K_C) and another modified for θ_C. Tangent lines represent energy balance equations incorporating heat capacities (C_P) and the enthalpy change ΔH_RX. Horizontal dashed lines mark conversion values X₁ and X₂. This illustrates how introducing species C affects the equilibrium and conversion using multiple reactor stages.

\( X' = \frac{\sum \theta_i C_{p_i} (T - T_0)}{-\Delta H_{Rx}} \)

\( X' = \frac{\left( \tilde{C}_{pA} + \tilde{C}_{pB} + \theta_c \tilde{C}_{pc} \right) (T - T_0)}{-\Delta H_{Rx}} \)

We see which ever conversion we use, X or X¢, we obtain the same temperature and overall conversion exiting the second reactor.

\( X_2 = \frac{F_{A0} - F_{A2}}{F_{A0}} = \frac{F_{A0} - F_{A0} (1 - X_1)}{F_{A0}} = \frac{F_{A0} - F_{A0} (1 - X_1) (1 - X_1)}{F_{A0}} \)

\( X_2 = 1 - (1 - X_1)(1 - X_1) \)

 

 

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