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

Topics

  1. Why Use the Energy Balance?
  2. Overview of User Friendly Energy Balance Equations
  3. Manipulating the Energy Balance, DHRx
  4. Reversible Reactions
  5. Adiabatic Reactions
  6. Interstage Cooling/Heating

Energy Balances, Rationale and Overview

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Link Reaction YouTube Video: Nitrogen Triiodide

Let's calculate the volume necessary to achieve a conversion, X, in a PFR for a first-order, exothermic reaction carried out adiabatically. For an adiabatic, exothermic reaction the temperature profile might look something like this:

Graph of T versus V showing a sigmoidal curve: temperature T increases slowly with V at first, then rapidly, and finally levels off.   Graph of k versus V showing a sigmoidal curve: rate constant k increases slowly with V at first, then rapidly, and finally levels off.   Graph of conversion X versus volume V showing a sigmoidal curve: conversion increases slowly with V, then more steeply, and finally plateaus.

The combined mole balance, rate law, and stoichiometry yield:

\( \frac{dX}{dV} = \frac{-r_A}{F_{A0}} \)
\( r_A = -k_i \exp\left[ \frac{E}{R} \left( \frac{1}{T_1} - \frac{1}{T} \right) \right] C_A \)
\( \frac{dX}{dV} = \frac{k_i \exp\left[ \frac{E}{R} \left( \frac{1}{T_1} - \frac{1}{T} \right) \right]}{F_{A0}} C_A (1 - X) \)

To solve this equation we need to relate X and T.

We will use the Energy Balance to relate X and T. For example, for an adiabatic reaction, e.g.,\( A -> B\), in which no inerts the energy balance yields

\( T = T_0 + \frac{(-\Delta H_{rx}) X}{C_{P_A}} \)

We can now form a table like we did in Chapter 2,

choose \(X\) \(\xrightarrow{\text{calculate}}\) \(T\) \(\xrightarrow{\text{calculate}}\) \(k\) \(\xrightarrow{\text{calculate}}\) \(K_c\) \(\xrightarrow{\text{calculate}}\) \(-r_A\) \(\xrightarrow{\text{calculate}}\) \(\left( \frac{F_{A0}}{-r_A} \right)\) increment \(X\) and repeat

Plot of F_A0 divided by negative r_A versus conversion X. The curve is U-shaped, indicating the rate first decreases with conversion, reaches a minimum, and then increases again.

User Friendly Energy Balance Equations

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The user friendly forms of the energy balance we will focus on are outlined in the following table.


User friendly equations relating X and T, and Fi and T

1. Adiabatic CSTR, PFR, Batch, PBR achieve this:

\( \dot{W}_s = \Delta \hat{C}_P = 0 \)

\( \chi_{EB} = \frac{\sum \theta_i \hat{C}_{P_i} (T - T_0)}{-\Delta H_{Rx}} \)

\( \chi = \frac{\hat{C}_P A (T - T_0)}{-\Delta H_{Rx}} \)

\( T = T_0 + \frac{-\Delta H_{Rx} \chi}{\sum \theta_i C_{P_i}} \)

(1.A)

 

(1.B)

2. CSTR with heat exchanger, UA(Ta-T) and large coolant flow rate.

\( \chi_{EB} = \frac{\left( \frac{UA}{F_{A0}} (T - T_a) \right) + \sum \theta_i \hat{C}_{P_i} (T - T_0)}{-\Delta H_{Rx}} \)

(2)


3 . PFR/PBR with heat exchange

Schematic of a tubular reactor with counter-current coolant flow. Reactant F_A0 enters at temperature T_0 from the left. The reactor shows catalyst pellets inside. Coolant enters from the right, creating a temperature gradient along the reactor. The label T_a indicates axial temperature in the reactor.

3A. In terms of conversion, X

\( \frac{dT}{dW} = \frac{ \frac{U_a}{\rho_B} (T_a - T) + r_A \Delta H_{Rx} (T)}{ F_{A0} \left( \sum \theta_i \hat{C}_{P_i} + \Delta C_p X \right)} \)

(3.A)

3B. In terms of molar flow rates, Fi 

\( \frac{dT}{dW} = \frac{ \frac{U_a}{\rho_B} (T_a - T) + r_A \Delta H_{Rx}(T)}{ \sum F_i C_{P_i} } \)

(3.B)

4. For Multiple Reactions

\( \frac{dT}{dV} = \frac{ \frac{U_a}{\rho_B} (T_a - T) + \sum_{ij} \Delta H_{Rx_{ij}} }{ \sum F_i C_{P_i} } \)

(4)

5. Coolant Balance

Co-Current Flow

\( \frac{dT_A}{dV} = \frac{ U_a (T - T_a) }{ m_c C_{P_c} } \)

(5)

These equations are derived in the text.  These are the equations that we will use to solve reaction engineering problems with heat effects.

Energy Balance

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In the material that follows, we will derive the above equations.

Energy Balance:
Energy balance on an open system. The equation shows that the rate of accumulation of energy within the system equals the rate of heat flow into the system minus the rate of work done by the system, plus the rate of energy added by mass flow into the system, minus the rate of energy leaving by mass flow. Mathematically: dÊ_sys/dt = Q̇ - Ẇ + F_in E_in - F_out E_out, all in units of J/s.

Typical units for each term are J/s; i.e. Watts

Biography James Prescott Joule (1818-1889)
Biography James Watt (1736-1819)
Figure 11-1. Schematic of energy balance on a well-mixed open system. The diagram shows a control volume with input and output mass flows (Fi and Hi) on the left and right, a heat input Q̇ from the top, and shaft work Ẇs into the system from the bottom left. Example inputs include F_A0 and H_A0; example outputs include F_A and H_A.

\( \dot{Q} - \dot{W}_s + \sum F_{i,0} E_{i0} |_{in} - \sum F_i E_i |_{out} = \frac{dE_{system}}{dt} \)    (1)

OK folks, here is what we are going to do to put this above equation into a usable form.

  1. Replace Ei by Ei=Hi-PVi

  2. Express Hi in terms of enthalpies of formation and heat capacities

  3. Express Fi in terms of either conversion or rates of reaction 

  4. Define DHRX

  5. Define DCP

  6. Manipulate so that the overall energy balance is either in terms of the User Friendly Equations (yellow box) 1.A, 1.B, 2, 3A, 3B, or 4 depending on the application



Step 1:

Substitute

\( E_i = U_i, \)

\( \dot{W} = \dot{W}_s + \sum F_{i,0} P_{i0} \dot{V}_i0 - \sum F_i P_i \dot{V}_i, \)

\( H_i = U_i + P_i \dot{V}_i \)

into equation (1) to obtain the General Energy Balance Equation.

General Energy Balance:

\( \dot{Q} - \dot{W}_s + \sum F_{i0} H_{i0} - \sum F_i H_i = \frac{dE_{system}}{dt} \)

For steady state operation:

\( \dot{Q} - \dot{W}_s + \sum F_{i0} H_{i0} - \sum F_i H_i = 0 \)

We need to put the above equation into a form that we can easily use to relate X and T in order to size reactors. To achieve this goal, we write the molar flow rates in terms of conversion and the enthalpies as a function of temperature. We now will "dissect" both Fi and Hi. [Note: For an animated derivation of the following equations, see the Interactive Computer Modules (ICMs) Heat Effects 1 and Heat Effects 2.]

Flow Rates, Fi

For the generalized reaction:

\( A + \frac{b}{a} B \rightarrow \frac{c}{a} C + \frac{d}{a} D \)

\( F_A = F_{A0} (1 - X), \quad F_B = F_{A0} \left( \frac{b}{a} - \frac{b}{a} X \right) \)

In general,

\( F_i = F_{A0} \left( \nu_i + \nu_i X \right) \)

\( \nu_A = -1, \quad \nu_B = -\frac{b}{a}, \quad \nu_C = \frac{c}{a}, \quad \nu_D = \frac{d}{a} \)


Enthalpies, Hi

Assuming no phase change:

\( H_i = H_i^\circ (T_R) + \int_{T_R}^{T} C_{p,i} dT \)
\( \sum_i H_i = \Delta H_R (T) = \Delta H_R^\circ (T_R) + \int_{T_R}^{T} \Delta C_p dT \)
\( \Delta H_{RX} = \frac{d}{a} H_D + \frac{c}{a} H_C - \frac{b}{a} H_B - H_A \)
\( \Delta C_p = \frac{d}{a} C_{pD} + \frac{c}{a} C_{pC} - \frac{b}{a} C_{pB} - C_{pA} \)
\( C_{p,i} = \alpha_i + \beta_i T + \chi_i T^2 \)  


Mean heat capacities:

\( \hat{C}_{p,i} = \int_{T_R}^{T} C_{p,i} \, dT \) \( \tilde{C}_{p,i} = \frac{T_{io}}{(T - T_{io})} \)
\( H_i = H_i^\circ(T_R) + \hat{C}_{p,i}(T - T_R) \) \( \Delta H_R(T) = \Delta H_R^\circ(T_R) + \hat{C}_p(T - T_R) \)
\( \sum_u \hat{C}_{p,i} = \Delta C_p = \frac{d}{a} \hat{C}_{p,FD} + \frac{c}{a} \hat{C}_{p,FC} - \frac{b}{a} \hat{C}_{p,FB} - \hat{C}_{p,FA} \)


Energy Balance with "dissected" enthalpies:

\( \dot{Q} - \dot{W}_s - F_{A0} \int_{T_R}^{T} \sum_u \hat{C}_{p,i} dT - F_{A0} X \left[ \Delta H_R^\circ(T_R) + \int_{T_R}^{T} \Delta C_p dT \right] = 0 \)


For constant or mean heat capacities:

\( \dot{Q} - \dot{W}_s - F_{A0} X \left[ \Delta H_R^\circ(T_R) + \Delta C_p (T - T_R) \right] = F_{A0} \sum_u \hat{C}_{p,i} (T - T_0) \)

Adiabatic Energy Balance:

\( T = T_0 - \frac{X \left[ \Delta H_R^\circ(T_R) + \Delta C_p (T_0 - T_R) \right]}{\sum_u \hat{C}_{p,i} + X \Delta C_p} = T_0 - \frac{X \left[ \Delta H_R(T) \right]}{\sum_u \hat{C}_{p,i} + X \Delta C_p} \)
Plot showing temperature T increasing linearly with conversion X, starting from initial temperature T0.

Adiabatic Energy Balance for variable heat capacities:

\( X = \frac{\sum_u \alpha_i (T_0 - T) + \frac{1}{2} \sum_u \beta_i (T_0^2 - T^2) + \frac{1}{3} \sum_u \gamma_i (T_0^3 - T^3)}{\Delta H_{RX} (T_R) + \Delta \alpha (T - T_R) + \frac{1}{2} \Delta \beta (T^2 - T_R^2) + \frac{1}{3} \Delta \gamma (T^3 - T_R^3)} \)

For constant heat capacities:

\( \beta = \gamma = 0, \ \alpha = C_{pi} \)

\( X = \frac{\sum_u \theta_i C_{pi} (T - T_0)}{-\Delta H_{RX}} \)

We will only be considering constant heat capacities for now.

Reversible Reactions

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Consider the reversible gas phase elementary reaction.

\( A \rightleftharpoons B \)

The rate law for this gas phase reaction will follow an elementary rate law.

\( -r_A = k \left[ C_A - \frac{C_B}{K_c} \right] \)

Where Kc is the concentration equilibrium constant. We know from Le Chaltlier's Law that if the reaction is exothermic, Kc will decrease as the temperature is increased and the reaction will be shifted back to the left. If the reaction is endothermic and the temperature is increased, Kcwill increase and the reaction will shift to the right.

at equilibrium

\( -r_A = 0 \)

\( K_c = \frac{C_{B\beta}}{C_{A\beta}} = \frac{C_{A0} X_\beta \frac{T_0}{T} p}{C_{A0} (1 - X_\beta) \frac{T_0}{T} p} = \frac{X_\beta}{1 - X_\beta} \)

\( X_\beta = \frac{K_c}{1 + K_c} \)

\( K_c = \frac{K_P}{(RT)^\delta} \)

Van't Hoff Equation

\( \frac{d \ln K_P}{dT} = \frac{\Delta H_R(T)}{RT^2} = \frac{\Delta H_R^\circ(T_R) + \Delta \hat{C}_P (T - T_R)}{RT^2} \)

For the special case of \( \Delta \hat{C}_P = 0 \)

Integrating the Van't Hoff Equation gives:

\( K_P(T_2) = K_P(T_1) \exp \left[ \frac{\Delta H_R^\circ(T_R)}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \right] \)

Graph of K_P versus temperature (T). The red curve shows K_P decreasing with increasing T for an exothermic reaction. The green curve shows K_P increasing with T for an endothermic reaction. Labels 'endothermic reaction' and 'exothermic reaction' correspond to the respective curves. Graph of equilibrium conversion (Xe) versus temperature (T). The red curve for the exothermic reaction shows Xe decreasing as T increases. The green curve for the endothermic reaction shows Xe increasing with temperature.

\( \delta = 0 \)

\( K_c(T) = K_c(T_1) \exp \left[ \frac{\Delta H_{RX}}{R} \left( \frac{1}{T_1} - \frac{1}{T} \right) \right] \)

Adiabatic Equilibrium

Conversion on Temperature

Exothermic ΔH is negative

Adiabatic Equilibrium temperature (Tadia) and conversion (Xeadia

Graph of conversion X versus temperature T for an exothermic reaction. Shows adiabatic line intersecting the equilibrium curve at point (Tadia, Xeadia). Includes equations: T = T0 + (−ΔH_RX X) / C_PA and Xe = Kc / (1 + Kc).

Endothermic ΔH is positive

Graph of conversion X versus temperature T for an endothermic reaction. The equilibrium curve rises with temperature. The adiabatic line intersects the equilibrium curve at the point (Tadia, Xeadia), indicating the maximum conversion under adiabatic conditions. Also shows the equations: Xe = Kc / (1 + Kc) and T = T0 + (−ΔHRX X) / CPA.


Adiabatic Reactions

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Biography Van't Hoff

Algorithm Adiabatic Reactions:

Suppose we have the Gas Phase Reaction

\( A \rightleftharpoons 2B \)

that follows an elementary rate law. To generate a Levenspiel plot to size CSTRs and PFRs we use the following steps or as we will see later use POLYMATH.

1.   Choose X 
     Calculate T \( T = T_0 - \frac{\Delta H_{rx} X}{\sum \theta_i C_{pi}} \)
     Calculate k \( k = k_1 \exp \left( \frac{E}{R} \left( \frac{1}{T_1} - \frac{1}{T} \right) \right) \)
     Calculate KC \( K_c = K_{c2} \exp \left( \frac{\Delta H_{rx}}{R} \left( \frac{1}{T_2} - \frac{1}{T} \right) \right) \)
     Calculate To/T
     Calculate CA \( C_A = \frac{C_{A0}(1 - X)}{1 + \epsilon X} \left( \frac{T_0}{T} \right) \)
     Calculate CB \( C_B = \frac{C_{A0} X}{1 + \epsilon X} \left( \frac{T_0}{T} \right) \)
     Calculate -rA \( -r_A = k \left( C_A - \frac{C_B^2}{K_c} \right) \)
2. Increment X and then repeat calculations.
3. When finished, plot \( \frac{F_{A0}}{-r_A} \) vs. X or use some numerical technique to find V.
Graph of FA0 divided by -rA versus X showing a U-shaped curve, indicating that the value decreases with X initially, reaches a minimum, and then increases.
Levenspiel Plot for an
exothermic, adiabatic reaction.

Reactor Sizing

We can now use the techniques developed in Chapter 2 to size reactors and reactors in series to compare and size CSTRs and PFRs.  

Consider:

CSTR: \( V_1 = \frac{F_{A0}}{-r_A} X_1 \)

\( V_2 = \frac{F_{A0}}{-r_{A2}} (X_2 - X_1) \)

PFR: \( V_1 = \int_0^{X_1} \frac{F_{A0}}{-r_A} dX \)

\( V_2 = \int_{X_1}^{X_2} \frac{F_{A0}}{-r_A} dX \)


Diagram of a Plug Flow Reactor (PFR) showing flow entering from the left and exiting from the right. PFR Shaded area is the volume.

For an exit conversion of 40%

For an exit conversion of 70%
Plot of FA0 over negative rA versus X. A shaded region under the curve from X = 0 to X = 0.4 represents the reactor volume required for conversion. Graph of FA0 over negative rA versus X, showing a shaded region under the curve from X = 0 to X = 0.7, representing the reactor volume required for a CSTR achieving 70% conversion.

CSTR Diagram of a Continuous Stirred-Tank Reactor (CSTR) with an inlet stream entering from the top and an outlet stream exiting from the side.Shaded area is the reactor volume.

For an exit conversion of 40%

For an exit conversion of 70%
Levenspiel plot showing 1 over negative rA versus conversion X. A gray rectangle highlights a CSTR design volume at X = 0.4, under a concave-up curve. Levenspiel plot showing FA0 over negative rA versus conversion X. The entire area under a wide rectangle up to a high conversion is shaded, representing the total volume required for multiple CSTRs or an inefficient CSTR design.                     

We see for 40% conversion very little volume is required.

CSTR+PFR

Reactor system schematic showing a CSTR followed by a PFR. The CSTR achieves 0.4 conversion before entering the plug flow reactor. Reactor system schematic showing a PFR followed by a CSTR. The conversion at the end of the plug flow reactor is 0.4 before entering the continuous stirred-tank reactor.
            (a)             (b)

For an intermediate conversion of 40% and exit conversion of 70%
Graph showing F_A0 divided by -r_A versus conversion X. The graph has a U-shaped curve with shaded rectangular and hatched areas under it, representing a CSTR operating to X = 0.4 followed by a PFR taking conversion to X = 0.7. Graph of F_A0 divided by -r_A versus X. The graph shows a U-shaped curve. The area under the curve is shaded from 0 to X = 0.4 (dark gray rectangle for PFR), and from X = 0.4 to X = 0.7 the area is hatched (CSTR represented as a rectangle intersecting the curve at X = 0.4).
            (a)             (b)

Looks like the best arrangement is a CSTR with a 40% conversion followed by a PFR up to 70% conversion.

Interstage Cooling/Heating

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Curve A: Reaction rate slow, conversion dictated by rate of reaction and reactor volume. As temperature increases rate increases and therefore conversion increases.
Curve B: Reaction rate very rapid. Virtual equilibrium reached in reaction conversion dictated by equilibrium conversion.

Optimum Inlet Temperature:

Fixed Volume Exothermic Reactor

Graph showing conversion X vs. temperature T. The purple line labeled Xe represents equilibrium conversion, decreasing with T. Region A (orange) forms a bell-shaped curve peaking at Topt. Region B (purple) drops sharply after Topt. Topt is marked as the optimal temperature for maximum conversion.

Interstage Cooling:

Series of three reactor stages, each followed by a heat exchanger represented by coils with arrows labeled Q1 and Q2 indicating heat addition. Initial feed F_A0 at T0 goes through the first reactor to yield F_A1 and conversion X1, then through Q1, repeated for the second and third reactors, ending with F_A3 and conversion X3.
Graph showing equilibrium conversion curve Xe versus temperature T, with multiple entry points labeled X1, X2, X3 and corresponding exit conversions X_EB along red lines. Includes energy balance equation relating X to temperature and heat capacities.

Example Interstage Cooling Calculations in terms of x and x'

CSTR Algorithm

1.)

Given X
Find T and V


Solution:
Linear progression of calc T -> cal k -> calc KC -> calc -rA -> calc V

Example Adiabatic Liquid Phase in A CSTR

2.)

Given T
Find X and V


Solution:
Linear progression: calc k -> cal KC -> calc X -> calc -rA -> calc V

Example Adiabatic Liquid Phase in A CSTR

3.)

Given V
Find X at T
Solution: plot XEB vs. T and XMB vs. T on the same graph:

Graph of conversion X versus temperature T showing two curves: one labeled X_MB in red with a gradual rise and plateau, and a straight line labeled X_EB representing an energy balance.

Self Test Heats of Reaction
Self Test Effects of inerts on T and X
Self Test Reactor staging with cool inerts
Self Test Sketch XEB versus T
Self Test Second Order Reaction in a CSTR
Self Assess Object Assessment of Chapter 11

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