Research is to see what everybody else sees, and to think what nobody else has thought.
—Albert SzentGyorgyi
In this section, we consider a tubular reactor in which heat is either added or removed through the cylindrical walls of the reactor (Figure 121). In modeling plugflow reactors, we shall assume that “there are no radial gradients” in the reactor and that the heat flux through the wall per unit volume of reactor is as shown in Figure 121.^{1}
^{1} Radial gradients are discussed in Section 12.7 and in Chapters 17 and 18.
We will carry out an energy balance on the volume ΔV. There is no work done, that is, ${\stackrel{\mathrm{.}}{W}}_{s}=0$, so Equation (1110) becomes
$\begin{array}{ccccccc}\begin{array}{c}\text{Heat}\\ \text{Added}\end{array}\text{}& +& \begin{array}{c}\begin{array}{c}\text{Energy}\end{array}\\ \text{In}\end{array}& & \begin{array}{c}\text{Energy}\\ \text{Out}\end{array}& =\text{0}& \text{}\\ \text{}\mathrm{\Delta}\dot{\mathit{\text{Q}}}& +& {\begin{array}{c}\text{}\mathrm{\Sigma}{\mathit{\text{F}}}_{i}{\mathit{\text{H}}}_{i}\end{array}}_{\mathit{\text{V}}}& & {\begin{array}{c}\text{}\mathrm{\Sigma}{\mathit{\text{F}}}_{i}{\mathit{\text{H}}}_{i}\end{array}}_{\mathit{\text{V}}+\mathrm{\Delta}\mathit{\text{V}}}& =\text{0}& \text{(121)}\end{array}$
The heat flow to the reactor, $\mathrm{\Delta}\stackrel{\mathrm{.}}{Q}$, is given in terms of the overall heattransfer coefficient, U, the heat exchange area, ΔA, and the difference between the ambient temperature, T_{a}, and the reactor temperature T
$\mathrm{\Delta}\dot{\mathit{\text{Q}}}=\mathit{\text{U}}\mathrm{\Delta}\mathit{\text{A}}({\mathit{\text{T}}}_{a}\mathit{\text{T}})=\mathit{\text{U}}a\mathrm{\Delta}\mathit{\text{V}}({\mathit{\text{T}}}_{a}\mathit{\text{T}})$
where a is the heat exchange area per unit volume of reactor. For a tubular reactor
$a=\frac{\mathit{\text{A}}}{\mathit{\text{V}}}=\frac{\pi \text{DL}}{\frac{\pi {\mathit{\text{D}}}^{2}\mathit{\text{L}}}{4}}=\frac{4}{\mathit{\text{D}}}$
where D is the reactor diameter. Substituting for $\mathrm{\Delta}\stackrel{\mathrm{.}}{Q}$ in Equation (121), dividing by ΔV, and then taking the limit as ΔV → 0, we get
$Ua({T}_{a}T)\frac{d\mathrm{\Sigma}\left({F}_{i}{H}_{i}\right)}{dV}=0$
Expanding
$\begin{array}{cc}Ua({T}_{a}T)\mathrm{\Sigma}\frac{{dF}_{i}}{dV}{H}_{i}\mathrm{\Sigma}{F}_{i}\frac{d{H}_{i}}{dV}=0& (122)\end{array}$
From a mole balance on species i, we have
$\begin{array}{cc}\frac{d{\mathit{\text{F}}}_{i}}{d\mathit{\text{V}}}={r}_{i}={v}_{i}({r}_{\mathit{\text{A}}})& \text{(123)}\end{array}$
Differentiating the enthalpy Equation (1119) with respect to V
$\begin{array}{cc}\frac{d{\mathit{\text{H}}}_{i}}{d\mathit{\text{V}}}={C}_{{\mathit{\text{P}}}_{i}}\frac{d\mathit{\text{T}}}{d\mathit{\text{V}}}& \text{(124)}\end{array}$
Substituting Equations (123) and (124) into Equation (122), we obtain
$\text{}\text{}\text{}\mathit{\text{U}}a({\mathit{\text{T}}}_{a}\mathit{\text{T}})\underset{\stackrel{\u23df}{\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}}}{\mathrm{\Sigma}{v}_{i}{\mathit{\text{H}}}_{i}}({\text{r}}_{\mathit{\text{A}}})\mathrm{\Sigma}{\mathit{\text{F}}}_{i}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{i}}\frac{d\mathit{\text{T}}}{d\mathit{\text{V}}}=0\text{}$
Rearranging, we arrive at
$\begin{array}{c}\hline \frac{d\mathit{\text{T}}}{d\mathit{\text{V}}}=\frac{\stackrel{\begin{array}{c}{\mathit{\text{Q}}}_{g}\\ \begin{array}{c}\begin{array}{c}\text{Heat}\end{array}\\ \u201c\text{Generated}\u201d\end{array}\end{array}}{\stackrel{\u23de}{{\text{r}}_{\mathit{\text{A}}}\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}}}\stackrel{\begin{array}{c}{\mathit{\text{Q}}}_{r}\\ \begin{array}{c}\begin{array}{c}\text{Heat}\end{array}\\ \u201c\mathrm{Removed}\u201d\end{array}\end{array}}{\stackrel{\u23de}{\mathit{\text{U}}a(\mathit{\text{T}}{\mathit{\text{T}}}_{a})}}}{\mathrm{\Sigma}{\mathit{\text{F}}}_{i}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{i}}}\\ \hline\end{array}\begin{array}{c}\text{(125)}\end{array}$
which is Equation (T111.G) in Table 111 on pages 547–549.
This form of the energy balance will also be applied to multiple reactions.
$\begin{array}{c}\hline \frac{d\mathit{\text{T}}}{d\mathit{\text{V}}}=\frac{{\mathit{\text{Q}}}_{g}{\mathit{\text{Q}}}_{r}}{\mathrm{\Sigma}{\mathit{\text{F}}}_{i}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{i}}}\text{}\\ \hline\end{array}\begin{array}{c}\text{(T111.G)}\end{array}$
where
$\begin{array}{cc}{Q}_{g}={r}_{\mathit{\text{A}}}\mathrm{\Delta}{H}_{\text{Rx}}\equiv ({r}_{\mathit{\text{A}}})(\mathrm{\Delta}{H}_{\text{Rx}})& (125\text{a})\end{array}$
$\begin{array}{cc}{Q}_{r}=Ua(T{T}_{a})& (125\text{b})\end{array}$
To help remember that Q_{r} is the “heat” removed from the reacting mixture, we note the driving force is from “T” to “T_{a}”, that is, Ua (T – T_{a}).
For exothermic reactions, Q_{g} will be a positive number. We note that when the heat “generated,” Q_{g}, is greater than the heat “removed,” Q_{r} (i.e., Q_{g} > Q_{r}), the temperature will increase down the reactor. When Q_{r} > Q_{g}, the temperature will decrease down the reactor.
For endothermic reactions ΔH_{Rx} will be a positive number and therefore the heat generated, Q_{g} in Equation (125a), will be a negative number. The heat removed, Q_{r} in Equation (125b), will also be a negative number because heat is added rather than removed T_{a} > T. See Chapter 12 Additional Material on the Web site for a sample calculation on endothermic reactions (http://www.umich.edu/~elements/6e/12chap/obj.html#/additionalmaterials/). This point is also illustrated in Example 122 for the case of constant ambient temperature.
We continue development of our algorithm by noting Equation (125) is coupled with the mole balances on each species, Equation (123). Next, we express r_{A} as a function of either the concentrations for liquid systems or molar flow rates for gas systems, as described in Chapter 6. We will use the molar flow rate form of the energy balance for membrane reactors and also extend this form to multiple reactions.
We could also write Equation (125) in terms of conversion by recalling F_{i} = F_{A0}(Θ_{i} + v_{i}X) and substituting this expression into the denominator of Equation (125).
PFR energy balance
$\begin{array}{c}\hline \frac{d\mathit{\text{T}}}{d\mathit{\text{V}}}=\frac{\stackrel{{\mathit{\text{Q}}}_{g}}{\text{}\stackrel{\u23de}{{\text{r}}_{\mathit{\text{A}}}\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}}}\stackrel{{\mathit{\text{Q}}}_{r}}{\stackrel{\u23de}{\mathit{\text{U}}a\left({\text{TT}}_{a}\right)}}}{{\mathit{\text{F}}}_{\text{A0}}(\mathrm{\Sigma}{\mathrm{\Theta}}_{i}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{i}}+\mathrm{\Delta}{\mathit{\text{C}}}_{\text{p}}\mathit{\text{X}})}=\frac{{\mathit{\text{Q}}}_{g}{\mathit{\text{Q}}}_{r}}{\mathrm{\Sigma}{\mathit{\text{F}}}_{i}{\mathit{\text{C}}}_{i}}\text{}\\ \hline\end{array}\begin{array}{c}\text{(126)}\end{array}$
For a packedbed reactor dW = ρ_{b} dV where ρ_{b} is the bulk density
PFR energy balance
$\begin{array}{c}\hline \text{}\frac{d\mathit{\text{T}}}{d\mathit{\text{W}}}=\frac{\stackrel{{\mathit{\text{Q}}}_{g}^{\prime}}{\stackrel{\u23de}{{\text{r}}_{\mathit{\text{A}}}^{\prime}\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}}}\frac{\stackrel{{\mathit{\text{Q}}}_{r}^{\prime}}{\stackrel{\u23de}{\mathit{\text{U}}a(\mathit{\text{T}}{\mathit{\text{T}}}_{a})}}}{\text{}{\rho}_{b}}}{\mathrm{\Sigma}{\mathit{\text{F}}}_{i}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{i}}}\\ \hline\end{array}\begin{array}{c}\text{(127)}\end{array}$
Equations (126) and (127) are also given in Table 111 as Equations (T111.D) and (T111.F). As noted earlier, having gone through the derivation to these equations, it will be easier to apply them accurately to CRE problems with heat effects.
We continue to use the algorithm described in the previous chapters and simply add a fifth building block, the energy balance.
Gas Phase
If the reaction is in gas phase and pressure drop is included, there are four differential equations that must be solved simultaneously. The differential equation describing the change in temperature with volume (i.e., distance) as we move down the reactor
Energy balance
$\begin{array}{cc}\frac{d\mathit{\text{T}}}{d\mathit{\text{V}}}=g\left({\text{X,T,T}}_{\text{a}}\right)& \text{(A)}\end{array}$
must be coupled with the mole balance
Mole balance
$\begin{array}{cc}\frac{d\mathit{\text{T}}}{d\mathit{\text{V}}}\text{=}\frac{{\text{r}}_{\mathit{\text{A}}}}{{\mathit{\text{F}}}_{\text{A0}}}\text{=}f\left(\text{X,T,}p\right)& \text{(B)}\end{array}$
and with the pressuredrop equation
Pressure drop
$\begin{array}{cc}\frac{d\text{p}}{d\mathit{\text{V}}}\text{=}\text{}\text{h}\left(\text{p,X,T}\right)& \text{(C)}\end{array}$
and solved simultaneously. If the temperature of the heatexchange fluid, T_{a}, varies down the reactor, we must add the energy balance on the heatexchange fluid. In the next section, we will derive the following equation for cocurrent heat transfer
Heat exchanger
$\begin{array}{cc}\text{}\frac{{d\mathit{\text{T}}}_{a}}{d\mathit{\text{V}}}\text{=}\frac{\mathit{\text{U}}a(\mathit{\text{T}}{\mathit{\text{T}}}_{a})}{{\dot{\text{m}}}_{\text{C0}}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\text{C0}}}}& \text{(D)}\end{array}$
Numerical integration of the coupled differential equations (A) to (D) is required.
along with the equation for countercurrent heat transfer. A variety of software packages (e.g., Polymath) can be used to solve these four coupled differential equations: (A), (B), (C), and (D).
Liquid Phase
For liquidphase reactions, the rate is not a function of total pressure, so our mole balance is
$\begin{array}{cc}\frac{d\mathit{\text{X}}}{d\mathit{\text{V}}}=\frac{{\text{r}}_{\mathit{\text{A}}}}{{\mathit{\text{F}}}_{\text{A0}}}=f\left(\text{X,T}\right)& \text{(E)}\end{array}$
Consequently, we need to only solve three equations (A), (D), and (E) simultaneously.
The heattransfer fluid will be a coolant for exothermic reactions and a heating medium for endothermic reactions. If the flow rate of the heattransfer fluid is sufficiently high with respect to the heat released (or absorbed) by the reacting mixture, then the heattransfer fluid temperature will be virtually constant along the length of the reactor. In the material that follows, we develop the basic equations for a coolant to remove heat from exothermic reactions; however, these same equations apply to endothermic reactions where a heating medium is used to supply heat.
We now carry out an energy balance on the coolant in the annulus between R_{1} and R_{2}, and axially between V and V + ΔV, as shown in Figure 122. The mass flow rate of the heatexchange fluid (e.g., coolant) is ${\stackrel{\mathrm{.}}{m}}_{c}$. We will consider the case when the reactor is cooled and the outer radius of the coolant channel R_{2} is insulated. Recall that by convention is the heat added to the system.
For cocurrent flow, the reactant and the coolant flow in the same direction.
The energy balance on the coolant in the volume between V and (V + ΔV) is
Coolant energy balance
$\text{}\begin{array}{cccccc}\left[\begin{array}{c}\begin{array}{c}\begin{array}{c}\text{Rate of engery}\end{array}\\ \text{in at V}\end{array}\end{array}\right]& & \left[\begin{array}{c}\begin{array}{c}\begin{array}{c}\text{Rate of energy}\end{array}\\ \text{out at V}+\text{}\mathrm{\Delta}\mathit{\text{V}}\end{array}\end{array}\right]& +& \left[\begin{array}{c}\begin{array}{c}\begin{array}{c}\begin{array}{c}\begin{array}{c}\text{Rate of heat added}\end{array}\\ \text{by conduction through}\end{array}\end{array}\\ \text{the inner wall}\end{array}\end{array}\right]& =\text{0}\\ {\dot{m}}_{c}{\mathit{\text{H}}}_{c}{}_{\mathit{\text{V}}}& & {\begin{array}{c}{\dot{m}}_{c}{\mathit{\text{H}}}_{c}\end{array}}_{\mathit{\text{V}}}& +& \mathit{\text{U}}a(\mathit{\text{T}}{\mathit{\text{T}}}_{a})\mathrm{\Delta}\mathit{\text{V}}& =0\end{array}$
where T_{a} is the temperature of the heat transfer fluid, that is, coolant, and T is the temperature of the reacting mixture in the inner tube.
Dividing by ΔV and taking limit as ΔV → 0
$\begin{array}{cc}\text{}{\dot{m}}_{c}\frac{d{\mathit{\text{H}}}_{c}}{d\mathit{\text{V}}}\text{+U}a(\mathit{\text{T}}{\mathit{\text{T}}}_{a})=0& \text{(128)}\end{array}$
Analogous to Equation (124), the change in enthalpy of the coolant can be written as
$\begin{array}{cc}\frac{{d\mathit{\text{H}}}_{c}}{d\mathit{\text{V}}}={\mathit{\text{C}}}_{{\mathit{\text{P}}}_{c}}\frac{{d\mathit{\text{T}}}_{a}}{d\mathit{\text{V}}}& \text{(129)}\end{array}$
The variation of coolant temperature T_{a} down the length of reactor is
$\begin{array}{c}\hline \frac{{d\mathit{\text{T}}}_{a}}{d\mathit{\text{V}}}=\frac{\mathit{\text{U}}a(\mathit{\text{T}}{\mathit{\text{T}}}_{a})}{{\dot{m}}_{c}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{c}}}\\ \hline\end{array}\begin{array}{c}\text{(1210)}\end{array}$
The equation is valid whether the heattransfer fluid is a coolant or a heating medium.
Typical heattransfer fluid temperature profiles for both exothermic and endothermic reactions when the heattransfer fluid enters at T_{a}_{0} are shown in Figure 123, parts (a) and (b) respectively.
In countercurrent heat exchange, the reacting mixture and the heattransfer fluid (e.g., coolant) flow in opposite directions. At the reactor entrance, V = 0, the reactants enter at temperature T_{0}, and the coolant exits at temperature T_{a}_{2}. At the end of the reactor, the reactants and products exit at temperature T, while the coolant enters at T_{a}_{0}.
Again, we write an energy balance on the coolant over a differential reactor volume to arrive at
$\begin{array}{c}\hline \frac{{d\mathit{\text{T}}}_{a}}{d\mathit{\text{V}}}=\frac{\mathit{\text{U}}a({\mathit{\text{T}}}_{a}\mathit{\text{T}})}{{\dot{m}}_{c}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{c}}}\\ \hline\end{array}\begin{array}{c}\text{(1211)}\end{array}$
At the entrance, V = 0 X = 0 and T_{a} = T_{a}_{2}.
At the exit, V = V_{f} T_{a} = T_{a}_{0}.
We note that the only difference between Equations (1210) and (1211) is a minus sign, that is, (T – T_{a}) versus (T_{a} – T).
The solution to a countercurrent flow problem to find the exit conversion and temperature requires a trialanderror procedure, as shown in Table 121.
Trialanderror procedure required
TABLE 121 PROCEDURE TO SOLVE FOR THE EXIT CONDITIONS FOR PFRS WITH COUNTERCURRENT HEAT EXCHANGE

We now have all the tools to solve reaction engineering problems involving heat effects in PFRs and PBRs for the cases of both constant and variable coolant temperatures.
Table 122 gives the algorithm for the design of PFRs and PBRs with heat exchange: In Case A Conversion is the reaction variable and in Case B MolarFlowRates are the reaction variables. The procedure in Case B must be used to analyze multiple reactions with heat effects.
The elementary gasphase reaction
$\text{A+B}\overrightarrow{\leftarrow}2\mathit{\text{C}}$
is to be carried out in a PBR with a cocurrent heat exchanger.
A. Conversion as the Reaction Variable
1. Mole Balance:
$\begin{array}{cc}\frac{d\mathit{\text{X}}}{d\mathit{\text{W}}}=\frac{{\text{r}}_{\mathit{\text{A}}}^{\prime}}{{\mathit{\text{F}}}_{0}}& \text{(T122.1)}\end{array}$
2. Rate Law:
$\begin{array}{cc}{\text{r}}_{\mathit{\text{A}}}^{\prime}={k}_{1}({\mathit{\text{C}}}_{\mathit{\text{A}}}{\mathit{\text{C}}}_{\mathit{\text{B}}}\frac{{\mathit{\text{C}}}_{\text{c}}^{2}}{{\mathit{\text{K}}}_{\mathit{\text{C}}}})& \text{(T122.2)}\end{array}$
$\begin{array}{cc}k={k}_{1}\left({\mathit{\text{T}}}_{1}\right)\mathrm{exp}\left[\frac{\mathit{\text{E}}}{\mathit{\text{R}}}(\frac{1}{{\mathit{\text{T}}}_{1}}\frac{1}{\mathit{\text{T}}})\right]& \text{(T122.3)}\end{array}$
$\begin{array}{cc}\begin{array}{cc}\text{for}\text{}\text{}\mathrm{\Delta}{\mathit{\text{C}}}_{\mathit{\text{P}}}\cong 0.& {\mathit{\text{K}}}_{\mathit{\text{C}}}={\mathit{\text{K}}}_{\text{C2}}\left({\mathit{\text{T}}}_{2}\right)\text{exp}\left[\frac{\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ}}{\mathit{\text{R}}}(\frac{1}{{\mathit{\text{T}}}_{2}}\frac{1}{\mathit{\text{T}}})\right]\end{array}& \text{(T122.4)}\end{array}$
Again we note that because δ ≡ 0, K_{C} ≡ K_{e}
3. Stoichiometry (gas phase):
$\begin{array}{cc}{\mathit{\text{C}}}_{\mathit{\text{A}}}=\frac{{\mathit{\text{C}}}_{\text{A0}\text{}(1\mathit{\text{X}})}}{\text{}\left(\text{1+}\epsilon \mathit{\text{X}}\right)}\frac{{\mathit{\text{T}}}_{0}}{\mathit{\text{T}}}\text{p}& \text{(S49)}\end{array}$
$\text{}\epsilon ={y}_{\text{A0}}\delta =\frac{1}{3}\left(\text{}\text{211}\right)=0$
$\begin{array}{cc}{\mathit{\text{C}}}_{\mathit{\text{A}}}={\mathit{\text{C}}}_{\text{A0}}(1\mathit{\text{X}})\frac{{\mathit{\text{T}}}_{0}}{\mathit{\text{T}}}\text{p}& \text{(T122.5)}\end{array}$
$\begin{array}{cc}{\mathit{\text{C}}}_{\mathit{\text{B}}}={\mathit{\text{C}}}_{\text{A0}}({\mathrm{\Theta}}_{\mathit{\text{B}}}\mathit{\text{X}})\frac{{\mathit{\text{T}}}_{0}}{\mathit{\text{T}}}\text{p}& \text{(T122.6)}\end{array}$
$\begin{array}{cc}{\mathit{\text{C}}}_{\mathit{\text{C}}}=2{\mathit{\text{C}}}_{\text{A0}}\mathit{\text{X}}\frac{{\mathit{\text{T}}}_{0}}{\mathit{\text{T}}}\text{p}& \text{(T122.7)}\end{array}$
$\begin{array}{cc}{\mathit{\text{C}}}_{\mathit{\text{I}}}={\mathit{\text{C}}}_{\text{I0}}\frac{{\mathit{\text{T}}}_{0}}{\mathit{\text{T}}}\text{p}& \text{(T122.8)}\end{array}$
$\begin{array}{cc}\text{}\frac{d\text{p}}{d\mathit{\text{W}}}=\text{}\frac{\alpha}{2\text{p}}\left(\text{1+}\epsilon \mathit{\text{X}}\right)\frac{\mathit{\text{T}}}{{\mathit{\text{T}}}_{0}}& \text{(530)}\end{array}$
δ = (2  1  1) = 0, ∴ ε = 0
$\begin{array}{cc}\text{}\frac{d\text{p}}{d\mathit{\text{W}}}=\text{}\frac{\alpha}{2\text{p}}\frac{\mathit{\text{T}}}{{\mathit{\text{T}}}_{0}}& \text{(T122.9)}\end{array}$
Mole Balance
Rate Law
Stoichiometry
Energy Balance
Parameters
Solution
Analysis
Combine Rate Law and Stoichiometry to find X_{e}
$\begin{array}{cc}{\text{r}}_{\mathit{\text{A}}}^{\prime}={k}_{1}{\mathit{\text{C}}}_{\text{A0}}^{2}\left[(1\mathit{\text{X}})({\mathrm{\Theta}}_{\mathit{\text{B}}}\mathit{\text{X}})\frac{4{\mathit{\text{X}}}^{2}}{{\mathit{\text{K}}}_{C}}\right]{\left(\frac{{\mathit{\text{T}}}_{0}}{\mathit{\text{T}}}\text{p}\right)}^{2}& \text{(T122.10)}\end{array}$
at equilibrium and ${r}_{\mathit{\text{A}}}^{\prime}=0$ and X = X_{e}
Solving for X_{e}
$\begin{array}{cc}\text{}{\mathit{\text{X}}}_{e}=\frac{({\mathrm{\Theta}}_{\mathit{\text{B}}}+1){\mathit{\text{K}}}_{\mathit{\text{C}}}{[{\left(({\mathrm{\Theta}}_{\mathit{\text{B}}}+1\text{}){\mathit{\text{K}}}_{\mathit{\text{C}}}\right)}^{2}4{\mathit{\text{K}}}_{\mathit{\text{C}}}{\mathrm{\Theta}}_{\mathit{\text{B}}}({\mathit{\text{K}}}_{\mathit{\text{C}}}4)]}^{1/2}}{\text{2}({\mathit{\text{K}}}_{\mathit{\text{C}}}4)}& \text{(T122.11)}\end{array}$
Because we are solving for the temperature as a function of catalyst weight down the reactor, we can use Equation (T122.4) to find the equilibrium constant, K_{C}, profile and then use Equation (T122.11) to find the equilibrium conversion profile, X_{e}.
4. Energy Balances:
$\begin{array}{cc}\text{Reactor:}\frac{d\mathit{\text{T}}}{d\mathit{\text{W}}}=\frac{{\mathit{\text{Q}}}_{g}^{\prime}{\mathit{\text{Q}}}_{r}^{\prime}}{\mathrm{\Sigma}{\mathit{\text{F}}}_{i}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{i}}}& \text{(T122.12)}\end{array}$
$\begin{array}{cc}{\mathit{\text{Q}}}_{g}^{\prime}=\text{}({\text{r}}_{\mathit{\text{A}}}^{\prime})(\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}})=\left({\text{r}}_{\mathit{\text{A}}}^{\prime}\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}\right)& \text{(T122.13)}\end{array}$
$\begin{array}{cc}{\mathit{\text{Q}}}_{r}^{\prime}=\left(\frac{\mathit{\text{U}}a}{{\rho}_{b}}\right)(\mathit{\text{T}}{\mathit{\text{T}}}_{a})& \text{(T122.14)}\end{array}$
$\begin{array}{cc}\mathrm{\Sigma}{F}_{i}{{C}_{P}}_{i}={F}_{\text{A0}}[{C}_{{P}_{\mathit{\text{A}}}+{\theta}_{\mathit{\text{B}}}}{C}_{{P}_{\mathit{\text{B}}}}+X\mathrm{\Delta}{C}_{p}]& \left(\text{T122.15}\right)\end{array}$
$\begin{array}{cc}\text{CoCurrent Coolant:}\frac{{d\mathit{\text{T}}}_{a}}{d\mathit{\text{W}}}=\frac{\left(\frac{\mathit{\text{U}}a}{{\rho}_{b}}\right)(\mathit{\text{T}}{\mathit{\text{T}}}_{a})}{{\dot{m}}_{c}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\text{cool}}}}& \text{(T122.16)}\end{array}$
B. Molar Flow Rates as the Reaction Variable
1. Mole Balance:
$\begin{array}{cc}\text{}\frac{{d\mathit{\text{F}}}_{\mathit{\text{A}}}}{d\mathit{\text{W}}}={\text{r}}_{\mathit{\text{A}}}^{\prime}& \text{(T122.17)}\end{array}$
$\begin{array}{cc}\text{}\frac{{d\mathit{\text{F}}}_{\mathit{\text{B}}}}{d\mathit{\text{W}}}={\text{r}}_{\mathit{\text{B}}}^{\prime}& \text{(T122.18)}\end{array}$
$\begin{array}{cc}\text{}\frac{{d\mathit{\text{F}}}_{\mathit{\text{C}}}}{d\mathit{\text{W}}}={\text{r}}_{\mathit{\text{C}}}^{\prime}& \text{(T122.19)}\end{array}$
$\begin{array}{cc}{F}_{\mathit{\text{I}}}={F}_{\mathit{\text{I}}0}& \left(\text{T122.20}\right)\end{array}$
2. Rate Law (elementary reaction):
$\begin{array}{cc}\text{}{\text{r}}_{\mathit{\text{A}}}^{\prime}={k}_{1}({\mathit{\text{C}}}_{\mathit{\text{A}}}{\mathit{\text{C}}}_{\mathit{\text{B}}}\frac{{\mathit{\text{C}}}_{\mathit{\text{C}}}^{2}}{{\mathit{\text{K}}}_{\mathit{\text{C}}}})& \text{(}\text{T122.2)}\end{array}$
$\begin{array}{cc}k={k}_{1}\left({\mathit{\text{T}}}_{1}\right)\text{}\text{exp}\left[\frac{\mathit{\text{E}}}{\mathit{\text{R}}}(\frac{1}{{\mathit{\text{T}}}_{1}}\frac{1}{1})\right]& \text{(T122.3)}\end{array}$
$\begin{array}{cc}{k}_{\mathit{\text{C}}}={k}_{\text{C2}}\left({\mathit{\text{T}}}_{2}\right)\text{}\text{exp}\left[\frac{\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ}}{\mathit{\text{R}}}(\frac{1}{{\mathit{\text{T}}}_{2}}\frac{1}{\mathit{\text{T}}})\right]& \text{(T122.4)}\end{array}$
3. Stoichiometry (gas phase):
$\begin{array}{cc}{r}_{\mathit{\text{B}}}^{\prime}={r}_{\mathit{\text{A}}}^{\prime}& \left(\text{T122.21}\right)\end{array}$
$\begin{array}{cc}{r}_{\mathit{\text{C}}}^{\prime}={2r}_{\mathit{\text{A}}}^{\prime}& \left(\text{T122.22}\right)\end{array}$
$\begin{array}{cc}{\mathit{\text{C}}}_{\mathit{\text{A}}}={\mathit{\text{C}}}_{\text{T0}}\frac{{\mathit{\text{T}}}_{\mathit{\text{A}}}}{{\mathit{\text{F}}}_{\mathit{\text{T}}}}\frac{{\mathit{\text{T}}}_{0}}{\mathit{\text{T}}}\text{p}& \text{(T122.23)}\end{array}$
$\begin{array}{cc}{\mathit{\text{C}}}_{\mathit{\text{B}}}={\mathit{\text{C}}}_{\text{T0}}\frac{{\mathit{\text{T}}}_{\mathit{\text{B}}}}{{\mathit{\text{F}}}_{\mathit{\text{T}}}}\frac{{\mathit{\text{T}}}_{0}}{\mathit{\text{T}}}\text{p}& \text{(T122.24)}\end{array}$
$\begin{array}{cc}{\mathit{\text{C}}}_{\mathit{\text{C}}}={\mathit{\text{C}}}_{\text{T0}}\frac{{\mathit{\text{T}}}_{\mathit{\text{C}}}}{{\mathit{\text{F}}}_{\mathit{\text{T}}}}\frac{{\mathit{\text{T}}}_{0}}{\mathit{\text{T}}}\text{p}& \text{(T122.25)}\end{array}$
$\begin{array}{cc}{F}_{\mathit{\text{T}}}={F}_{\mathit{\text{A}}}+{F}_{\mathit{\text{B}}}+{F}_{\mathit{\text{C}}}+{F}_{\mathit{\text{I}}}& \left(\text{T122.26}\right)\end{array}$
$\begin{array}{cc}\frac{d\text{p}}{d\text{w}}=\frac{\alpha}{2\text{p}}\frac{{\mathit{\text{F}}}_{\mathit{\text{T}}}}{{\mathit{\text{F}}}_{\text{T0}}}\frac{\mathit{\text{T}}}{{\mathit{\text{T}}}_{0}}& \text{(T122.27)}\end{array}$
4. Energy Balances:
Reactor:
$\begin{array}{cc}\begin{array}{c}\frac{d\mathit{\text{T}}}{d\mathit{\text{W}}}=\frac{{\mathit{\text{Q}}}_{g}^{\prime}{\mathit{\text{Q}}}_{r}^{\prime}}{\mathrm{\Sigma}{\mathit{\text{F}}}_{i}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{i}}}=\frac{\left({\text{r}}_{\mathit{\text{A}}}^{\prime}\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}\right)\left(\frac{\mathit{\text{U}}a}{{\rho}_{b}}\right)(\mathit{\text{T}}{\mathit{\text{T}}}_{a})}{{\mathit{\text{F}}}_{\mathit{\text{A}}}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{A}}}}+{\mathit{\text{F}}}_{\mathit{\text{B}}}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{B}}}}+{\mathit{\text{F}}}_{\mathit{\text{C}}}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{C}}}}+{\mathit{\text{F}}}_{\mathit{\text{I}}}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{I}}}}}\end{array}& \left(\text{T122.28}\right)\end{array}$
Heat Exchangers: Same as for A. Conversion as the Reaction Variable
Case A: Conversion as the Independent Variable Example Calculation
5. Parameter Evaluation:
Now we enter all the explicit equations with the appropriate parameter values.
k_{1}, E, R, C_{T0}, T_{a}, T_{0}, T_{1}, T_{2}, K_{C2}, Θ_{B}, Θ_{I}, $\mathrm{\Delta}{H}_{\text{Rx}}^{\xb0}$, C_{PA}, C_{PB}, C_{CA}, Ua, ρ_{b}
Base Case for Parameter Values
$\begin{array}{c}\begin{array}{cc}\begin{array}{c}\mathit{\text{E}}=25\text{kcal/mol}\\ \text{}\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}=\text{20 kcal/mol}\end{array}& {\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{I}}}}=40\text{cal}\text{/mol}\mathit{\text{K}}\\ {k}_{1}=0.004\frac{{\text{dm}}^{6}}{\text{mol}\cdot \text{kg}\cdot \text{s}}\text{@310}\mathit{\text{K}}& \text{}\left(\frac{\text{Ua}}{{\rho}_{b}}\right)=\text{Uarho}=0.5\text{}\frac{\text{cal}}{\text{kg}\cdot \text{s}\cdot \mathit{\text{K}}}\\ {\mathit{\text{K}}}_{\text{C2}}=1000@303\mathit{\text{K}}& {\dot{m}}_{c}=1000\text{}g/\text{s}\end{array}\\ \text{}\end{array}$
$\begin{array}{cc}\begin{array}{c}\begin{array}{c}\begin{array}{c}\begin{array}{c}\begin{array}{c}\alpha =0.0002/\text{kg}\end{array}\\ {F}_{\text{A0}}=5\text{mol/s}\end{array}\end{array}\end{array}\\ {C}_{{\mathit{\text{P}}}_{\text{Cool}}}=0.3{\text{mol/dm}}^{3}\\ {C}_{{\mathit{\text{P}}}_{\mathit{\text{A}}}}={C}_{{\mathit{\text{P}}}_{\mathit{\text{B}}}}={C}_{{\mathit{\text{P}}}_{\mathit{\text{C}}}}=20\text{cal/mol/K}\end{array}\hfill & \begin{array}{c}\begin{array}{c}\begin{array}{c}\begin{array}{c}{C}_{{\mathit{\text{P}}}_{\text{Cool}}}=18\text{cal/g/K}\end{array}\\ {\mathrm{\Theta}}_{\mathit{\text{B}}}=1\end{array}\end{array}\\ {\mathrm{\Theta}}_{1}=1\end{array}\hfill \end{array}$
with initial values T_{0} = 330 K, T_{a}_{0} = 320 K, p = 1, and X = 0 at W = 0 and final values: W_{f} = 5000 kg.
6. Solution:
Equations (T122.1)–(T122) are entered into the Polymath program along with the corresponding parameter values.
Differential equations
1 d(Ta)/d(W) = Uarho*(TTa)/(mc*Cpcool)
2 d(p)/d(W) = alpha/2*(T/To)/p
3 d(T)/d(W) = (QgQr)/(Fao*sumcp)
4 d(X)/d(W) = ra/Fao
Explicit equations
1 Hr = 20000
2 thetaB = 1
3 Uarho = 0.5
4 alpha = .0002
5 To = 330
6 Qr = Uarho*(TTa)
7 mc = 1000
8 Cpcool = 18
9 Kc = 1000*(exp(Hr/1.987*(1/3031/T)))
10 Fao = 5
11 thetaI = 1
12 CpI = 40
13 CpA = 20
14 yao = 1/(1+thetaB+thetaI)
15 CpB = 20
16 Cto = 0.3
17 Ea = 25000
18 Xe = ((thetaB+1)*Kc(((thetaB+1)*Kc)^24*(Kc4)*(Kc*thetaB))^0.5)/(2*(Kc4))
19 k = .004*exp(Ea/1.987*(1/3101/T))
20 Cao = yao*Cto
21 Cc = Cao*2*X*p*To/T
22 sumcp = (thetaI*CpI+CpA+thetaB*CpB)
23 Ca = Cao*(1X)*p*To/T
24 Cb = Cao*(thetaBX)*p*To/T
25 ra = k*(Ca*CbCc^2/Kc)
26 Qg = (ra)*(Hr)
http://www.umich.edu/~elements/6e/live/chapter12/T123/LEPT122.pol
Calculated values of DEQ variables
Variable  Initial value  Final value  

1  alpha  0.0002  0.0002 
2  Ca  0.1  0.0111092 
3  Cao  0.1  0.1 
4  Cb  0.1  0.0111092 
5  Cc  0  0.0255273 
6  CpA  20.  20. 
7  CpB  20.  20. 
8  Cpcool  18.  18. 
9  CpI  40.  40. 
10  Cto  0.3  0.3 
11  Ea  2.5E+04  2.5E+04 
12  Fao  5.  5. 
13  Hr  2.0E+04  2.0E+04 
14  k  0.046809  0.0303238 
15  Kc  66.01082  93.4225 
16  mc  1000.  1000. 
17  p  1.  0.2360408 
18  Qg  9.361802  0.0706176 
19  Qr  5.  1.615899 
20  ra  0.0004681  3.531E06 
21  sumcp  80.  80. 
22  T  330.  326.2846 
23  Ta  320.  323.0528 
24  thetaB  1.  1. 
25  thetaI  1.  1. 
26  To  330.  330. 
27  Uarho  0.5  0.5 
28  W  0  4500. 
29  X  0  0.5346512 
30  Xe  0.8024634  0.8285547 
31  yao  0.3333333  0.3333333 
Analysis:
The temperature profiles are shown in Figure T122.1. One notes that due to the exothermic heat of reaction, the temperature of the gas in the reactor increases sharply near the entrance then decreases as the reactants are consumed and the fluid is cooled as it moves down the reactor. Figure T122.2 shows the conversion X increases and approaches its equilibrium value X_{e} at approximately W = 1125 kg of catalyst where the reaction rate approaches zero. At this point the conversion cannot increase further unless X_{e} increases. The increase in X_{e} occurs because of the cooling of the reactor’s contents by the heat exchanger thus shifting the equilibrium to the right. By the end of the reactor, the equilibrium conversion has been increased to 82.9% and the conversion in the reactor to 53.4%. One notes from Figure T12.22, that owing to the cooling and consumption of reactants the conversion does not increase significantly above a catalyst weight of 2000 kg, even though X is much below X_{e} (see Problem P121_{A}).
The following figures show “representative” profiles that would result from solving the above equations for different parameter values. The reader is encouraged to download the Living Example Problem for Table 122 and use Wolfram to vary a number of parameters, as discussed in Problem P121_{A}. Be sure you can explain why these curves look the way they do.
Be sure you can explain why these curves look the way they do.
When plant engineer Maxwell Anthony looked up the vapor pressure at the exit to the adiabatic reactor in Example 113, where the temperature is 360 K, he learned the vapor pressure for isobutene was about 1.5 MPa, which is greater than the rupture pressure of the glass vessel the company had hoped to use. Fortunately, when Max looked in the storage shed, he found there was a bank of 10 tubular reactors, each of which was 5 m^{3}. The bank reactors were doublepipe heat exchangers with the reactants flowing in the inner pipe and with Ua = 5,000 kJ/m^{3}·h·K. Max also bought some thermodynamic data from one of the companies he found on the Internet that used Colorimeter experiments to find ΔH_{Rx} for various reactions. One of the companies had the value of ΔH_{Rx} for his reaction on sale this week for the low, low price of $25,000.00. For this value of ΔH_{Rx} the company said it is best to use an initial concentration of A of 1.86 mol/dm^{3}. The entering temperature of the reactants is 305 K and the entering coolant temperature is 315 K. The mass flow rate of the coolant, ${\stackrel{\mathrm{.}}{m}}_{c}$, is 500 kg/h and the heat capacity of the coolant, C_{P}_{C}, is 28 kJ/kg·K. The temperature in any one of the reactors cannot rise above 325 K. Carry out the following analyses with the newly purchased values from the Internet:
Cocurrent heat exchange: Plot X, X_{e}, T, T_{a}, and –r_{A}, down the length of the reactor.
Countercurrent heat exchange: Plot X, X_{e}, T, T_{a}, and –r_{A} down the length of the reactor.
Constant ambient temperature, T_{a}: Plot X, X_{e}, T, and –r_{A} down the length of the reactor.
Adiabatic operation: Plot X, X_{e}, T, T_{a}, and –r_{A}, down the length of the reactor.
Compare parts (a) through (d) above and write a paragraph describing what you find.
Additional information:
Recall from Example 113 that F_{T0} = 163 kmol/h and F_{A0} = 0.9 F_{T0}, C_{PA} = 141 kJ/kmol · K, C_{P0} = ΣΘ_{i}C_{P}_{i} = 159 kJ/kmol · K, and data from the company Maxwell got off the Internet are ΔH_{Rx} = −34500 kJ/kmol with ΔC_{P} = 0 and C_{A0} = 1.86 kmol/m^{3} and E/R = 7906 K with k = 31.1 h^{1} @ 360 K.
Solution
We shall first solve part (a), the cocurrent heat exchange case and then make small changes in the Polymath program for parts (b) through (d).
The molar flow rate of A to each of the 10 reactors in parallel
${\mathit{\text{F}}}_{\text{A0}}=\left(0.9\right)\left(\text{163 kmol/h}\right)\times \frac{1}{10}=\text{14.7}\frac{\text{kmol A}}{\text{h}}$
The mole balance, rate law, and stoichiometry are the same as in the adiabatic case previously discussed in Example 113; that is,
The Algorithm
Mole Balance:
$\begin{array}{c}\hline \frac{d\mathit{\text{X}}}{d\mathit{\text{V}}}=\frac{{\text{r}}_{\mathit{\text{A}}}}{{\mathit{\text{F}}}_{\text{A0}}}\\ \hline\end{array}\begin{array}{c}\text{(E113.1)}\end{array}$
Rate Law and Stoichiometry:
$\begin{array}{c}\hline {\text{r}}_{\mathit{\text{A}}}=k{\mathit{\text{C}}}_{\text{A0}}[1(1+\frac{1}{{\mathit{\text{K}}}_{\mathit{\text{C}}}})\mathit{\text{X}}]\\ \hline\end{array}\begin{array}{c}\text{(E113.7)}\end{array}$
with
$\begin{array}{c}\hline k=31.1\text{exp}\left[7906\left(\frac{\mathit{\text{T}}360}{360\mathit{\text{T}}}\right){\text{h}}^{1}\right]\\ \hline\end{array}\begin{array}{c}\text{(E113.10)}\end{array}$
$\begin{array}{c}\hline \text{}{k}_{\mathit{\text{C}}}=3.03\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{exp}\left[\frac{(\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}/\mathit{\text{R}})(\mathit{\text{T}}333)}{333\mathit{\text{T}}}\right]\\ \hline\end{array}\begin{array}{c}\text{(E113.11)}\end{array}$
Same as Example 113
The equilibrium conversion is
$\begin{array}{c}\hline {\mathit{\text{X}}}_{e}=\frac{{\mathit{\text{K}}}_{\text{c}}}{1+{\mathit{\text{K}}}_{\mathit{\text{C}}}}\\ \hline\end{array}\begin{array}{c}\text{(T113.12)}\end{array}$
Note: We could substitute for K_{C} using Equation (E113.12) to write the reaction rate as
$\begin{array}{cc}{\text{r}}_{\mathit{\text{A}}}=k{\mathit{\text{C}}}_{\text{A0}}(\text{1}\frac{\mathit{\text{X}}}{{\mathit{\text{X}}}_{e}})& \text{(E121.1)}\end{array}$
Energy Balance:
The energy balance on the reactor is
$\begin{array}{c}\hline \text{}\frac{d\mathit{\text{T}}}{d\mathit{\text{V}}}=\frac{{\mathit{\text{Q}}}_{g}{\mathit{\text{Q}}}_{r}}{\mathrm{\Sigma}{\mathit{\text{F}}}_{i}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{i}}}\\ \hline\end{array}\begin{array}{c}\text{(E121.2)}\end{array}$
$\begin{array}{cc}{Q}_{g}={r}_{\mathit{\text{A}}}\mathrm{\Delta}{H}_{\text{Rx}}& \left(\text{E121.3}\right)\end{array}$
Q_{g} = r_{A}ΔH_{Rx}
Q_{r} = Ua(T  T_{a})
$\frac{dT}{dV}=\frac{{Q}_{g}{Q}_{r}}{{F}_{\text{A0}}{C}_{\text{P0}}}$
$\begin{array}{cc}{Q}_{r}=Ua(T{T}_{a})& \left(\text{E121.4}\right)\end{array}$
$\begin{array}{cc}\mathrm{\Sigma}{\mathit{\text{F}}}_{i}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{i}}=\underset{\stackrel{\u23df}{{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{0}}}}{{\mathit{\text{F}}}_{\text{A0}}\mathrm{\Sigma}{\mathrm{\Theta}}_{i}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{i}}}={\mathit{\text{F}}}_{\text{A0}}{\mathit{\text{C}}}_{\text{P0}}& \text{(E121.5)}\end{array}$
Part (a) CoCurrent Heat Exchange
We are now going to solve the coupled, ordinary differential and explicit equations (E113.1), (E113.7), (E113.10), (E113.11), (E113.12), and (E121.2), and the appropriate heatexchange balance using Polymath. After entering these equations we will enter the parameter values. By using cocurrent heat exchange as our Polymath base case, we only need to change one line in the program for each of the other three cases and solve for the profiles of X, X_{e}, T, T_{a}, and –r_{A}, and not have to reenter the program.
For cocurrent flow, the balance on the heat transfer fluid is
$\begin{array}{c}\hline \frac{d{\mathit{\text{T}}}_{a}}{d\mathit{\text{V}}}=\frac{\mathit{\text{U}}a(\mathit{\text{T}}{\mathit{\text{T}}}_{a})}{{\dot{m}}_{\mathit{\text{C}}}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{C}}}}}\\ \hline\end{array}\begin{array}{c}\text{(E121.6)}\end{array}$
with T_{0} = 305 K and T_{a} = 315 K at V = 0. The Polymath program and solution are shown in Table E121.1.
Table E121.1 Part (a) COCURRENT HEAT EXCHANGE
Differential equations
1 d(Ta)/d(V) = Ua*(TTa)/m/Cpc
2 d(X)/d(V) = ra/Fa0
3 d(T)/d(V) = (QgQr)/(Cpo*Fa0)
Explicit equations
1 Cpc = 28
2 m = 500
3 Ua = 5000
4 deltaH = 34500
5 Qr = Ua*(TTa)
6 Ca0 = 1.86
7 Fa0 = 0.9*163*.1
8 Kc = 3.03*exp((deltaH/8.314)*((T333)/(T*333)))
9 k = 31.1*exp((7906)*(T360)/(T*360))
10 ra = k*Ca0*(1(1+1/Kc)*X)
11 Xe = Kc/(1+Kc)
12 Qg = ra*deltaH
13 Cpo = 159
14 rate = ra
Calculated values of DEQ variables
Variable  Initial value  Final value  

1  Ca0  1.86  1.86 
2  Cpc  28.  28. 
3  Cpo  159.  159. 
4  deltaH  3.45E+04  3.45E+04 
5  Fa0  14.67  14.67 
6  k  0.5927441  6.80861 
7  Kc  9.512006  2.641246 
8  m  500.  500. 
9  Qg  3.804E+04  4077.238 
10  Qr  5.0E+04  5076.445 
11  ra  1.102504  0.1181808 
12  rate  1.102504  0.1181808 
13  T  305.  336.7102 
14  Ta  315.  335.6949 
15  Ua  5000.  5000. 
16  V  0  5. 
17  X  0  0.7185996 
18  Xe  0.9048707  0.7253687 
Figure E121.1 shows the profiles for the reactor temperature, T, the coolant temperature, T_{a}, the conversion, X, the equilibrium conversion, X_{e}, and the reaction rate, –r_{A}. Note that at the entrance to the reactor the reactor temperature T is below the coolant temperature, but as reacting mixture moves into and through the reactor the mixture heats up and T becomes greater than T_{a}. When you load the Polymath, Python, or Wolfram Living Example Problem (LEP) code on your computer, one of the things you will want to explore is the value of the ambient temperature, T_{a}, or the entering temperature, T_{0}, below which the reaction might never take off or “ignite.”
Analysis: Part (a) CoCurrent Exchange: We note that reactor temperature goes through a maximum. Near the reactor entrance, the reactant concentrations are high and therefore the reaction rate is high (cf. Figure E121.1(a)) and Q_{g} > Q_{r}. Consequently, the temperature and conversion increase with increasing reactor volume while X_{e} decreases because of the increasing temperature. Eventually, X and X_{e} come close together (V = 0.95 m^{3}) and the rate becomes very small as the reaction approaches equilibrium. At this point, the reactant conversion X cannot increase unless X_{e} increases. We also note that when the ambient heatexchanger temperature, T_{a}, and the reactor temperature, T, are essentially equal, there is no longer a temperature driving force to cool the reactor. Consequently, the temperature does not change farther down the reactor, nor does the equilibrium conversion, which is only a function of temperature.
Part (b) Countercurrent Heat Exchange
For countercurrent flow, we only need to make two changes in the program. First, multiply the righthand side of Equation (E121.6) by minus one to obtain
$\begin{array}{cc}\frac{d{\mathit{\text{T}}}_{a}}{d\mathit{\text{V}}}=\frac{\mathit{\text{U}}a(\mathit{\text{T}}{\mathit{\text{T}}}_{a})}{{\dot{m}}_{\mathit{\text{C}}}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{C}}}}}& \text{(E121.7)}\end{array}$
Next, we guess T_{a} at V = 0 and see whether it matches T_{a}_{0} at V = 5 m^{3}. If it doesn’t, we guess again. In this example, we will guess T_{a} (V = 0) = 340.3 K and see whether T_{a} = T_{a}_{0} = 315 K at V = 5 m^{3}.
Good guess!
At V = 0, we guessed an entering coolant temperature of 340.3 K and found that at V = V_{f}, the calculation showed that it matched the emerging coolant temperature T_{a}_{0} = 315 K!! (Was that a lucky guess or what?!) The variable profiles are shown in Figure E121.2.
Figure E121.2 Profiles down the reactor for countercurrent heat exchange (a) temperature, (b) conversion, (c) reaction rate.
We observe that at V = 0.5 m^{3}, X = 0.36 and the rate, –r_{A}, drops to a low value as X approaches X_{e}. Because X can never be greater than X_{e} the reaction will not proceed further unless X_{e} is increased. In this case, the reactor is cooled, causing the temperature to decrease and resulting in an increase in the equilibrium conversion beyond a reactor volume of V = 0.5 m^{3}.
Analysis: Part (b) Countercurrent Exchange: We note that near the entrance to the reactor, the coolant temperature is above the reactant entrance temperature. However, as we move down the reactor, the reaction generates “heat” and the reactor temperature rises above the coolant temperature. We note that X_{e} reaches a minimum (corresponding to the reactor temperature maximum) near the entrance to the reactor. At this point (V = 0.5 m^{3}), X cannot increase above X_{e}. As we move down the reactor, the reactants are cooled and the reactor temperature decreases allowing X and X_{e} to increase. A higher exit conversion, X, and equilibrium conversion, X_{e}, are achieved in the countercurrent heatexchange system than for the cocurrent system.
Part (c) Constant T_{a}
For constant T_{a}, use the Polymath program in part (a), but multiply the right side of Equation (E121.6) by zero in the program, that is,
$\begin{array}{cc}\frac{d{\mathit{\text{T}}}_{a}}{d\mathit{\text{V}}}=\frac{\mathit{\text{U}}a(\mathit{\text{T}}{\mathit{\text{T}}}_{a})}{{\dot{m}}_{\mathit{\text{C}}}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{C}}}}}\text{*0}& \text{(E121.8)}\end{array}$
TABLE E121.3 PART (c) CONSTANT T_{a}
The initial and final values are shown in the Polymath report and the variable profiles are shown in Figure E121.3.
Figure E121.3 Profiles down the reactor for constant heatexchange fluid temperature T_{a}; (a) temperature, (b) conversion, (c) reaction rate.
We note that for the parameter values chosen, the profiles for T, X, X_{e}, and –r_{A} are similar to those for countercurrent flow. For example, as X and X_{e} become close to each other at V = 1.0 m^{3}, but beyond that point X_{e} increases because the reactor temperature decreases.
Analysis: Part (c) Constant T_{a}: When the coolant flow rate is sufficiently large, the coolant temperature, T_{a}, will be essentially constant. If the reactor volume is sufficiently large, the reactor temperature will eventually reach the coolant temperature, as is the case here. At this exit temperature, which is the lowest achieved in this example (i.e., Part (c)), the equilibrium conversion, X_{e}, is the largest of the four cases studied in this example.
Part (d) Adiabatic Operation
In Example 113, we solved for the temperature as a function of conversion and then used that relationship to calculate k and K_{C}. An easier way is to solve the general or base case of a heat exchanger for cocurrent flow and write the corresponding Polymath program. Next, use Polymath (part (a)), but multiply the parameter Ua by zero, that is,
Ua = 5,000*0
and run the simulation again.
$\frac{d\mathit{\text{T}}}{d\mathit{\text{V}}}=\frac{{\mathit{\text{Q}}}_{g}}{{\mathit{\text{F}}}_{\text{A0}}{{\mathit{\text{C}}}_{\text{}}}_{{\mathit{\text{P}}}_{0}}}$ Adiabatic
TABLE E121.4 PART (d) ADIABATIC OPERATION
Differential equations
1 d(Ta)/d(V) = Ua*(TTa)/(m*Cpc)
2 d(X)/d(V) = ra/Fa0
3 d(T)/d(V) = (QgQr)/(Cpo*Fa0)
Explicit equations
1 Cpc = 28
2 m = 500
3 Ua = 5000*0
4 deltaH = 34500
5 Qr = Ua*(TTa)
6 Ca0 = 1.86
7 Fa0 = 0.9*163*.1
8 Kc = 3.03*exp((deltaH/8.314)*((T333)/(T*333)))
9 k = 31.1*exp((7906)*(T360)/(T*360))
10 ra = k*Ca0*(1(1+1/Kc)*X)
11 Xe = Kc/(1+Kc)
12 Qg = ra*deltaH
13 Cpo = 159
14 rate = ra
Calculated values of DEQ variables
Variable  Initial value  Final value  

1  Ca0  1.86  1.86 
2  Cpc  28.  28. 
3  Cpo  159.  159. 
4  deltaH  3.45E+04  3.45E+04 
5  Fa0  14.67  14.67 
6  k  0.5927441  124.2488 
7  Kc  9.512006  0.5752071 
8  m  500.  500. 
9  Qg  3.804E+04  0.4109228 
10  Qr  0  0 
11  ra  1.102504  1.191E05 
12  rate  1.102504  1.191E05 
13  T  305.  384.2335 
14  Ta  315.  315. 
15  Ua  0  0 
16  V  0  5. 
17  X  0  0.3651629 
18  Xe  0.9048707  0.3651628 
The initial and exit conditions are shown in the Polymath report, while the profiles of T, X, X_{e}, and –r_{A} are shown in Figure E121.4.
Figure E121.4 Profiles down the reactor for adiabatic reactor; (a) temperature, (b) conversion, (c) reaction rate.
Analysis: Part (d) Adiabatic Operation: Because there is no cooling, the temperature of this exothermic reaction will continue to increase down the reactor until equilibrium is reached, X = X_{e} = 0.365 at T = 384 K, which is the adiabatic equilibrium temperature. The profiles for X and X_{e} are shown in Figure E121.4(b) where one observes that the adiabatic equilibrium conversion X_{e} decreases down the reactor because of the increasing temperature until it becomes equal to the reactor conversion (i.e., X ≡ X_{e}), which occurs at circa 0.9 m^{3}. There is no change in temperature, X or X_{e}, after this point because the reaction rate is virtually zero and thus the remaining reactor volume serves no purpose.
Finally, Figure E121.4(c) shows –r_{A} increases as we move down the reactor as the temperature increases, reaching a maximum and then decreasing until X and X_{e} approach each other and the rate becomes virtually zero.
Overall Analysis: This is an extremely important example, as we applied our CRE PFR algorithm with heat exchange to a reversible exothermic reaction. We analyzed four types of heatexchanger operations. We see that the countercurrent exchanger and the constant T_{a} cases give the highest conversion and adiabatic operation gives the lowest conversion.
In Example 121 we studied the four different types of heat exchanger on an exothermic reaction. In this section we carry out the same study on an endothermic reaction.
Jeffreys, in a treatment of the design of an acetic anhydride manufacturing facility, states that one of the key steps is the endothermic vaporphase cracking of acetone to ketene and methane is^{2}
^{2} G. V. Jeffreys, A Problem in Chemical Engineering Design: The Manufacture of Acetic Anhydride, 2nd ed. London: Institution of Chemical Engineers.
CH_{3}COCH_{3} → CH_{2}CO + CH_{4}
The article further states that this reaction is firstorder with respect to acetone and that the specific reaction rate is given by
$\begin{array}{cc}\text{ln}\text{}k=34.34\frac{34222}{\mathit{\text{T}}}& \text{(E122.1)}\end{array}$
where k is in reciprocal seconds and T is in Kelvin. It is desired to feed 7850 kg of acetone per hour to a tubular reactor. The reactor consists of a bank of 1000 oneinch, schedule40 tubes. We shall consider four cases of heatexchanger operation. The inlet temperature and pressure are the same for all cases at 1035 K and 162 kPa (1.6 atm) and the entering heatingfluid temperature available is 1250 K. The heatexchange fluid has a flow rate, ${\stackrel{\mathrm{.}}{m}}_{c}$, of 0.111 mol/s, with a heat capacity of 34.5 J/mol·K.
A bank of 1000 oneinch, schedule40 tubes, 1.79 m in length corresponds to 1.0 m^{3} (0.001 m^{3}/tube = 1.0 dm^{3}/tube) and gives 20% conversion. Ketene is unstable and tends to explode, which is a good reason to keep the conversion low. However, the pipe material and schedule size should be checked to learn whether they are suitable for these temperatures and pressures. In addition, the final design and operating conditions need to be cleared by the safety committee before operation can begin. Also check the Web site, Process Safety Across the Chemical Engineering Curriculum (http://umich.edu/~safeche/index.html).
Case 1 The reactor is operated adiabatically.
Case 2 Constant heatexchange fluid temperature T_{a} = 1250 K
Case 3 Cocurrent heat exchange with T_{a}_{0} = 1250 K
Case 4 Countercurrent heat exchange with T_{a}_{0} = 1250 K
Additional information:
CH_{3}COCH_{3} (A):H°_{A}(T_{R}) = –216.67 kJ/mol, C_{PA} = 163 J/mol · K
CH_{2}CO (B):H°_{B}(T_{R}) = –61.09 kJ/mol, C_{PB} = 83 J/mol · K
CH_{4} (C):H°_{C}(T_{R}) = –74.81 kJ/mol, C_{PC} = 71 J/mol · K
N_{2} (I):C_{PI} = —28.1 J/mol, · K
U = 110 J/S · m^{2} · K
The other parameters are given in Table E122.1.
Solution
Let A = CH_{3}COCH_{3}, B = CH_{2}CO, and C = CH_{4}. Rewriting the reaction symbolically gives us
A → B + C
Gasphase endothermic reaction examples:
Adiabatic
Heat exchange T_{a} is constant
Cocurrent heat exchange with variable T_{a}
Countercurrent exchange with variable T_{a}
Algorithm for a PFR with Heat Effects
1. Mole Balance:
$\begin{array}{cc}\frac{d\mathit{\text{X}}}{d\mathit{\text{V}}}=\frac{{\text{r}}_{\mathit{\text{A}}}}{{\mathit{\text{F}}}_{\text{A0}}}& \text{(E122.2)}\end{array}$
2. Rate Law:
${r}_{\text{A}}=k{C}_{\text{A}}\phantom{\rule[0.0ex]{6em}{0.0ex}}(\text{E122.3)}$
Rearranging (E122.1)
$\begin{array}{cc}k=8.2\times {10}^{14}\text{exp}\left[\text{}\frac{34222}{\mathit{\text{T}}}\right]\text{=3.58 exp}\left[34222\left[\frac{1}{1035}\text{}\frac{1}{\mathit{\text{T}}}\right]\right]& \text{(E122.4)}\end{array}$
3. Stoichiometry (gasphase reaction with no pressure drop):
$\begin{array}{cc}\begin{array}{c}{\mathit{\text{C}}}_{\mathit{\text{A}}}=\frac{{\mathit{\text{C}}}_{\text{A0}}(1\mathit{\text{X}}){\mathit{\text{T}}}_{0}}{(1+\u03f5\mathit{\text{X}})\text{}\mathit{\text{T}}}\\ \u03f5={y}_{\text{A0}}\delta =1(1+11)=1\end{array}& \text{(E122.5)}\end{array}$
4. Combining yields
$\begin{array}{cc}{\text{r}}_{\mathit{\text{A}}}=\frac{k{\mathit{\text{C}}}_{\text{A0}}(1\mathit{\text{X}})}{1+\mathit{\text{X}}}\frac{{\mathit{\text{T}}}_{0}}{\mathit{\text{T}}}& \text{(E122.5)}\end{array}$
Before combining Equations (E122.2) and (E122.6), it is first necessary to use the energy balance to determine T as a function of X.
5. Energy Balance:
Reactor balance
$\begin{array}{cc}\frac{d\mathit{\text{T}}}{d\mathit{\text{V}}}=\frac{\mathit{\text{U}}a({\mathit{\text{T}}}_{a}\mathit{\text{T}})+\left({\text{r}}_{\mathit{\text{A}}}\right)[\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ}+\mathrm{\Delta}{\mathit{\text{C}}}_{\mathit{\text{P}}}(\mathit{\text{T}}{\mathit{\text{T}}}_{\mathit{\text{R}}})]}{{\mathit{\text{F}}}_{\text{A0}}(\mathrm{\Sigma}{\mathrm{\Theta}}_{i}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{i}}+\mathit{\text{X}}\mathrm{\Delta}{\mathit{\text{C}}}_{\mathit{\text{P}}})}& \text{(E122.7)}\end{array}$
Heat Exchanger. We will use the heatexchange fluid balance for cocurrent flow as our base case. We will then show how we can very easily modify our ODE solver program (e.g., Polymath) to solve for the other cases by simply multiplying the appropriate line in the code by either zero or minus one.
For cocurrent flow:
$\begin{array}{cc}\frac{d{\mathit{\text{T}}}_{a}}{d\mathit{\text{V}}}=\frac{\mathit{\text{U}}a(\mathit{\text{T}}{\mathit{\text{T}}}_{a})}{\dot{m}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\text{c}}}}& \text{(E122.8)}\end{array}$
6. Calculation of Mole Balance Parameters on a Per Tube Basis:
$\begin{array}{c}\begin{array}{c}{\mathit{\text{F}}}_{\text{A0}}=\frac{7850\text{kg}/\text{h}}{58\text{kg}/\text{kmol}}\times \frac{1}{1000\text{Tubes}}=0.135\text{kmol/h}=0.0376\text{mol/s}\\ {\mathit{\text{C}}}_{\text{A0}}=\frac{{\mathit{\text{P}}}_{\text{A0}}}{\mathit{\text{R}}\mathit{\text{T}}}=\frac{162\text{b kPa}}{8.31\frac{\text{kPa}\text{}\cdot {\text{m}}^{3}}{\text{kmol}\cdot \mathit{\text{K}}}\left(1035\text{K}\right)}=0.0188\frac{\text{kmol}}{{\text{m}}^{3}}=18.8{\text{mol/m}}^{3}\end{array}\\ \begin{array}{cc}\begin{array}{c}\text{}{\upsilon}_{0}=\frac{{\mathit{\text{F}}}_{\text{A0}}}{{\mathit{\text{C}}}_{\text{A0}}}=\frac{0.0376}{0.0188}=2.0{\text{dm}}^{3}/\text{s,}\end{array}& \text{V=}\frac{1{\text{m}}^{3}}{1000\text{tubes}}=\frac{0.001{\text{m}}^{3}}{\text{tube}}=\frac{1.0{\text{dm}}^{3}}{\text{tube}}\end{array}\end{array}$
7. Calculation of Energy Balance Parameters:
Thermodynamics:
${\mathrm{\Delta}}_{\text{Rx}}^{\xb0}\left({T}_{\mathit{\text{R}}}\right):\text{At298K}$, using the standard heats of formation
$\begin{array}{ccc}\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ}\left({\mathit{\text{T}}}_{\mathit{\text{R}}}\right)& =& {\mathit{\text{H}}}_{\mathit{\text{B}}}^{\circ}\left({\mathit{\text{T}}}_{\mathit{\text{R}}}\right)+{\mathit{\text{H}}}_{\mathit{\text{C}}}^{\circ}\left({\mathit{\text{T}}}_{\mathit{\text{R}}}\right){\mathit{\text{H}}}_{\mathit{\text{A}}}^{\circ}\left({\mathit{\text{T}}}_{\mathit{\text{R}}}\right)\hfill \\ \text{}& =& \text{}\left(\text{}61.09\right)+\left(\text{}74.81\right)\left(\text{}216.67\right)\text{kJ}/\text{mol}\hfill \\ \text{}& =& 80.77\text{k}\mathit{\text{J}}/\text{mol}\hfill \end{array}$
ΔC_{p} : Using the mean heat capacities
ΔC_{p} = C_{PB} + C_{PC} – C_{PA} = (83 + 71 – 163) J/mol · K
ΔC_{p} = – 9 J/mol · K
$\text{Sum}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{i}}\text{}{\mathrm{\Theta}}_{i}:\text{}{\mathrm{\Theta}}_{i}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{i}}={\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{A}}}}+\text{}{\mathrm{\Theta}}_{\mathit{\text{I}}}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{I}}}}=163+28.1{\mathrm{\Theta}}_{\mathit{\text{I}}},{\left(\frac{\mathit{\text{J}}}{\text{mol}\cdot \text{K}}\right)}_{\text{}}$
Use Wolfram or Python to vary Θ_{I} in the LEP.
Heat Exchange:
Energy balance. The heattransfer area per unit volume of pipe is
$\begin{array}{c}\text{}a=\frac{\pi D\mathit{\text{L}}}{(\pi {\mathit{\text{D}}}^{2}/4)\mathit{\text{L}}}=\frac{4}{\mathit{\text{D}}}=\frac{4}{0.0266\text{}\text{m}}=150{\text{m}}^{1}\\ \mathit{\text{U}}=110{\text{J/m}}^{2}\text{}\cdot \text{s}\cdot \mathit{\text{K}}\end{array}$
Combining the overall heat transfer coefficient with the area yields
Ua = 16500 J/m^{3} · s · K
TABLE E122.1 SUMMARY OF PARAMETER VALUES
Parameter Values 

$\mathrm{\Delta}{H}_{\text{Rx}}^{\xb0}\left({T}_{R}\right)=80.77\text{kJ/mol}$  ${\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{I}}}}=28.1\frac{\mathit{\text{J}}}{\text{mol}\cdot \mathit{\text{K}}}$  T_{0} = 1035 K 
F_{A0} = 0.0376 mol/s  ΔC_{P} = =9 J/mol·K  T_{R} = 298 K 
C_{PA} = 163 J/mol A/K  C_{A0} = 18.8 mol/m^{3}  ${\dot{\text{m}}}_{\mathit{\text{C}}}=0.111\text{mol/s}$ 
C_{PCool} ≡ C_{Pc} = 34.5 J/mol/K  Ua = 16500 J/m^{3}·s·K  V_{f} = 0.001 m^{3} 
We will solve for all four cases of heatexchanger operation for this endothermic reaction example in the same way we did for the exothermic reaction in Example 121. That is, we will write the Polymath equations for the case of cocurrent heat exchange and use that as the base case. As noted in the LEP for this example one should use Wolfram or Python to explore the simulation. We will then manipulate the different terms in the heattransfer fluid balance (Equations 1210 and 1211) to solve for the other cases. In the four cases that follow we will take Θ_{I} = 0, but on the Wolfram and Python LEPs we allow the reader to vary Θ_{I}. We will start with the adiabatic case where we multiply the heat transfer coefficient in the base case by zero.
Case 1 Adiabatic
We are going to start with the adiabatic case first to show the dramatic effects of how the reaction dies out as the temperature drops. In fact, we are now going to extend the length of each tube to make the total reactor volume 5 dm^{3} in order to observe this effect of a reaction dying out as well as showing the necessity of adding a heat exchanger. For the adiabatic case, we simply multiply the value of Ua in our Polymath program by zero. No other changes are necessary. For the adiabatic case, the answer will be the same whether we use a bank of 1000 reactors, each a 1dm^{3} reactor, or one of 1 m^{3}. To illustrate how an endothermic reaction can virtually die out completely, let’s extend the singlepipe volume from 1 to 5 dm^{3}.
Ua = 16500*0
The Polymath program is shown in Table E122.2. Figure E122.1 shows the graphical output.
Adiabatic endothermic reaction in a PFR
TABLE E122.2 POLYMATH PROGRAM AND OUTPUT FOR ADIABATIC OPERATION
Differential equations
1 d(X)/d(V) = ra/Fao
2 d(T)/d(V) = (QgQr)/(Fao*(Cpa+X*delCp))
3 d(Ta)/d(V) = Ua*(TTa)/(mc*Cpc)
Explicit equations
1 To = 1035
2 Ua = 16500*0
3 Fao = .0376
4 Cpa = 163
5 delCp = 9
6 Cao = 18.8
7 ra = Cao*3.58*exp(34222*(1/To1/T))*(1X)*(To/T)/(1+X)
8 deltaH = 80770+delCp*(T298)
9 Qg = ra*deltaH
10 Qr = Ua*(TTa)
11 mc = .111
12 Cpc = 34.5
13 rate = ra
Calculated values of DEQ variables
Variable  Initial value  Final value  

1  Cao  1.88  1.88 
2  Cpa  163.  163. 
3  Cpc  34.5  34.4 
4  delCp  9.  9. 
5  deltaH  7.414E+04  7.414E+04 
6  Fao  0.0376  0.0376 
7  mc  0.111  0.111 
8  Qg  4.99E+06  2.79E+04 
9  Qr  0  0 
10  ra  67.304  0.3704982 
11  rate  67.304  0.3704982 
12  T  1035.  904.8156 
13  Ta  1250.  1250. 
14  To  1035.  1035. 
15  Ua  0  0 
16  V  0  0.005 
17  X  0  0.2817744 
Figure E122.1 Adiabatic conversion and temperature (a), and reaction rate (b) profiles.
Analysis: Case 1 Adiabatic Operation: As temperature drops, so does k and hence the rate, –r_{A}, drops to an insignificant value. Note that for this adiabatic endothermic reaction, the reaction virtually dies out after 3.5 dm^{3}, owing to the large drop in temperature, and we observe very little conversion is achieved beyond this point. One way to increase the conversion would be to add a diluent such as nitrogen, which could supply the sensible heat for this endothermic reaction. The Wolfram LEP allows the reader to vary the nitrogen in the feed. However, if too much diluent is added, the concentration, and hence the rate, will be quite low. On the other hand, if too little diluent is added, the temperature will drop and virtually extinguish the reaction. How much diluent to add is left as an exercise. Figures E122.1(a) and (b) give the reactor volume as 5 dm^{3} in order to show the reaction “dying out.” However, because the reaction is nearly complete near the entrance to the reactor, that is, –r_{A} ≅ 0, we are going to study and compare the heatexchange systems in a 1dm^{3} reactor (0.001 m^{3}) in each of the next three cases.
Case 2 Constant heatexchange fluid temperature, T_{a}
We make the following changes in our program on line 3 of the base case (a)
$\begin{array}{c}\begin{array}{c}\frac{{d\mathit{\text{T}}}_{a}}{d\mathit{\text{V}}}=\frac{{\mathit{\text{U}}}_{a}(\mathit{\text{T}}{\mathit{\text{T}}}_{a})}{\dot{m}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{c}}}\text{*}0\\ \mathit{\text{U}}a=16500{\text{J/m}}^{3}\text{/}\text{s/K}\end{array}\\ {\text{and V}}_{f}=0.001{\text{m}}^{3}\end{array}$
The Polymath Program is shown in Table E122.3 and the profiles for T, X, and –r_{A} are shown in Figure E122.2, parts (a), (b), and (c) respectively.
TABLE E122.3 POLYMATH PROGRAM AND OUTPUT FOR CONSTANT T_{a}
Differential equations
1 d(X)/d(V) = ra/Fao
2 d(T)/d(V) = (QgQr)/(Fao*(Cpa+X*delCp))
3 d(Ta)/d(V) = Ua*(TTa)/(mc*Cpc)*0
Explicit equations
1 To = 1035
2 Ua = 16500
3 Fao = .0376
4 Cpa = 163
5 delCp = 9
6 Cao = 18.8
7 ra = Cao*3.58*exp(34222*(1/To1/T))*(1X)*(To/T)/(1+X)
8 deltaH = 80770+delCp*(T298)
9 Qg = ra*deltaH
10 Qr = Ua*(TTa)
11 mc = .111
12 Cpc = 34.5
13 rate = ra
Variable  Initial value  Final value  

1  Cao  1.88  1.88 
2  Caa  163.  163. 
3  Cpc  34.5  34.5 
4  delCp  9.  9. 
5  deltaH  7.414E+04  7.343E+04 
6  Fao  0.0376  0.0376 
7  mc  0.111  0.111 
8  Qg  4.99E+06  1.211E+06 
9  Qr  3.548E+06  2.242E+06 
10  ra  67.304  16.48924 
11  rate  67.304  16.48924 
12  T  1035.  1114.093 
13  Ta  1250.  1250. 
14  To  1035.  1035. 
15  Ua  1.65E+04  1.65E+04 
16  V  0  0.001 
17  X  0  0.9508067 
Endothermic reaction profiles
Figure E122.2 Profiles for constant heat exchanger fluid temperature, T_{a}; (a) temperature, (b) conversion, (c) reaction rate.
Analysis: Case 2 Constant T_{a}: Just after the reactor entrance, the reaction temperature drops as the sensible heat from the reacting fluid supplies the energy for the endothermic reaction. This temperature decrease in the reactor also causes the rate of reaction to decrease. As we move farther down the reactor, the reaction rate drops further as the reactants are consumed. Beyond V = 0.08 dm^{3}, the heat supplied by the constant T_{a} heat exchanger becomes greater than that “consumed” by the endothermic reaction and the reactor temperature rises. In the range between V = 0.2 dm^{3} and V = 0.6 dm^{3}, the rate decreases slowly owing to the depletion of reactants, which is counteracted, to some extent, by the increase in temperature and hence the rate constant k. Consequently, we are eventually able to achieve an exit conversion of 95%.
Case 3 CoCurrent Heat Exchange
The energy balance on a cocurrent exchanger is
$\frac{{d\mathit{\text{T}}}_{a}}{d\mathit{\text{V}}}=\frac{\mathit{\text{U}}a\text{}(\mathit{\text{T}}{\mathit{\text{T}}}_{a})}{{\dot{m}}_{\mathit{\text{C}}}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{C}}}}}$
with T_{a}_{0} = 1250 K at V = 0.
The Polymath Program is shown in Table E122.4 and the variable profiles for T, T_{a}, X, and –r_{A} are shown in Figure E122.3, parts (a), (b), and (c) respectively. Because the reaction is endothermic, T_{a} needs to start off at a high temperature at V = 0.
TABLE E122.4 POLYMATH PROGRAM AND OUTPUT FOR COCURRENT EXCHANGE
Differential equations
1 d(X)/d(V) = ra/Fao
2 d(T)/d(V) = (QgQr)/(Fao*(Cpa+X*delCp))
3 d(Ta)/d(V) = Ua*(TTa)/(mc*Cpc)
Explicit equations
1 To = 1035
2 Ua = 16500
3 Fao = .0376
4 Cpa = 163
5 delCp = 9
6 Cao = 18.8
7 ra = Cao*3.58*exp(34222*(1/To1/T))*(1X)*(To/T)/(1+X)
8 deltaH = 80770+delCp*(T298)
9 Qg = ra*deltaH
10 Qr = Ua*(TTa)
11 mc = .111
12 Cpc = 34.5
13 rate = ra
Calculated values of DEQ variables
Variable  Initial value  Final value  

1  Cao  1.88  1.88 
2  Cpa  163.  163. 
3  Cpc  34.5  34.5 
4  delCp  9.  9. 
5  deltaH  7.414E+04  7.459E+04 
6  Fao  0.0376  0.0376 
7  mc  0.111  0.111 
8  Qg  4.99E+06  3.654E+05 
9  Qr  3.548E+06  1.881E+05 
10  ra  67.304  4.899078 
11  rate  67.304  4.899078 
12  T  1035.  984.8171 
13  Ta  1250.  996.215 
14  To  1035.  1035. 
15  Ua  1.65E+04  1.65E+04 
16  V  0  0.001 
17  X  0  0.456201 
Analysis: Case 3 CoCurrent Exchange: In cocurrent heat exchange, we see that the heatexchanger fluid temperature, T_{a}, decreases rapidly initially and then continues to decrease along the length of the reactor as it supplies the energy to the heat drawn by the endothermic reaction. Eventually T_{a} decreases to the point where it approaches T and the rate of heat exchange is small; as a result, the temperature of the reactor, T, continues to decrease, as does the rate, resulting in a small conversion. Because the reactor temperature for cocurrent exchange is lower than that for Case 2 constant T_{a}, the reaction rate will be lower. As a result, significantly less conversion will be achieved than in the case of constant heatexchange temperature T_{a}.
Endothermic reaction profiles
Figure E122.3 Profiles down the reactor for an endothermic reaction with cocurrent heat exchange; (a) temperature, (b) conversion, and (c) reaction rate.
Case 4 Countercurrent Heat Exchange
For countercurrent exchange, we first multiply the rhs of the cocurrent heatexchanger energy balance by –1, leaving the rest of the Polymath program in Table 122.5 the same.
$\frac{{d\mathit{\text{T}}}_{a}}{d\mathit{\text{V}}}=\frac{\mathit{\text{U}}a\text{}(\mathit{\text{T}}{\mathit{\text{T}}}_{a})}{{\dot{m}}_{\mathit{\text{C}}}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{C}}}}}$
Next, guess T_{a} (V = 0) = 995.15 K to obtain T_{a0} = 1250 K at V = 0.001m^{3}. (Don’t you believe for a moment 995.15 K was my first guess.) Once this match is obtained as shown in Table E122.5, we can report the profiles shown in Figure E122.4.
Good guess!
TABLE E122.5 POLYMATH PROGRAM AND OUTPUT FOR COUNTERCURRENT EXCHANGE
Differential equations
1. d(X)/d(V) = ra/Fao
2. d(T)/d(V) = (QgQr)/(Fao*(Cpa+X*delCp))
3. d(Ta)/d(V) = Ua*(TTa)/(mc*Cpc)
Explicit equations are the same as Case 3 CoCurrent Heat Exchange.
Figure E122.4 Profiles down the reactor for countercurrent heat exchange; (a) temperature, (b) conversion, (c) reaction rate.
Endothermic reaction profiles
Analysis: Case 4 Countercurrent Exchange: At the front of the reactor where the reactants enter, that is, V = 0, the reaction takes place very rapidly, drawing energy from the sensible heat of the gas and causing the gas temperature to drop because the heat exchanger cannot supply energy at an equal or greater rate to that being drawn by the endothermic reaction. Additional “heat” is lost at the entrance in the case of countercurrent exchange because the temperature of the exchange fluid, T_{a}, is below the entering reactor temperature, T. One notes there is a minimum in the reaction rate, –r_{A}, profile that is rather flat. In this flat region, the rate is “virtually” constant between V = 0.2 dm^{3} and V = 0.8 dm^{3}, because the increase in k caused by the increase in T is balanced by the decrease in rate brought about by the consumption of reactants. Just past the middle of the reactor, the rate begins to increase slowly as the reactants become depleted and the heat exchanger now supplies energy at a rate greater than the reaction draws energy and, as a result, the temperature eventually increases. This lower temperature coupled with the consumption of reactants causes the rate of reaction to be low in the plateau, resulting in a lower conversion than either the cocurrent or constant T_{a} heat exchange cases.
Analysis Summary of 4 Cases. The exit temperatures and conversion are shown in the Summary Table E122.6.
TABLE E122.6 SUMMARY TABLE
Heat Exclusion  X  T(K)  T_{a}(K) 

Adiabatic  0.28  905   
Constant T_{a}  0.95  1114  1250 
CoCurrent  0.456  985  996 
Countercurrent  0.35  1034  995 
For this endothermic reaction, the highest conversion is obtained for constant T_{a} and the lowest conversion is for an adiabatic reaction. These two extremes correspond to the greatest and least amount of heat transfer to the reacting fluid.
AspenTech: Example 122 has also been formulated in AspenTech and can be downloaded on your computer directly from the CRE Web site.
In this section we apply the general energy balance (Equation (1122)) to a CSTR at steady state. We then present example problems showing how the mole and energy balances are combined to design reactors operating adiabatically and nonadiabatically.
In Chapter 11 the steadystate energy balance was derived as
$\begin{array}{c}\hline \dot{\mathit{\text{Q}}}\text{}{\dot{\mathit{\text{W}}}}_{s}{\mathit{\text{F}}}_{\text{A0}}\mathrm{\Sigma}{\mathrm{\Theta}}_{i}{\mathit{\text{C}}}_{{\text{p}}_{i}}(\mathit{\text{T}}{\mathit{\text{T}}}_{i0})\left[\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ}\left({\mathit{\text{T}}}_{\mathit{\text{R}}}\right)+\mathrm{\Delta}{\mathit{\text{C}}}_{\mathit{\text{P}}}(\mathit{\text{T}}{\mathit{\text{T}}}_{\mathit{\text{R}}})\right]{\mathit{\text{F}}}_{\text{A0}}\mathit{\text{X}}=0\\ \hline\end{array}\begin{array}{c}\text{(1128)}\end{array}$
Recall that Ẇ_{s} is the shaft work, that is, the work done by a stirrer or mixer in the CSTR on the reacting fluid inside the CSTR. Consequently, because the convention that done by the system on the surroundings is positive, the CSTR stirrer work will be a negative number, for example, ${\stackrel{\mathrm{.}}{W}}_{s}=1,000\text{J/S}$. (See Problem P126_{B}, a California Professional Engineers’ Exam Problem.)
Note: In many calculations the CSTR mole balance derived in Chapter 2
(F_{A0}X = –r_{A}V)
These are the forms of the steadystate balance we will use.
will be used to replace the term following the brackets in Equation (1128), that is, (F_{A0}X) will be replaced by (–r_{A}V) to arrive at Equation (1212).
Rearranging yields the steadystate energy balance
$\begin{array}{c}\hline \dot{\mathit{\text{Q}}}\text{}{\dot{\mathit{\text{W}}}}_{s}{\mathit{\text{F}}}_{\text{A0}}\mathrm{\Sigma}{\mathrm{\Theta}}_{i}{\mathit{\text{C}}}_{{\text{p}}_{i}}(\mathit{\text{T}}{\mathit{\text{T}}}_{i0})+\left({r}_{\mathit{\text{A}}}\mathit{\text{V}}\right)\left(\text{}\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}\right)=0\\ \hline\end{array}\begin{array}{c}\text{(1212)}\end{array}$
Although the CSTR is well mixed and the temperature is uniform throughout the reaction vessel, these conditions do not mean that the reaction is carried out isothermally. Isothermal operation occurs when the feed temperature is identical to the temperature of the fluid inside the CSTR.
The $\dot{\mathit{\text{Q}}}$ Term in the CSTR
Figure 126 shows the schematics of a CSTR with a heat exchanger. The heattransfer fluid enters the exchanger at a mass flow rate (e.g., kg/s) at a temperature T_{a}_{1} and leaves at a temperature T_{a}_{2}. The rate of heat transfer from the exchanger to the reactor fluid at temperature T is^{3}
^{3} Information on the overall heat transfer coefficient may be found in J. R. Welty, G. L. Rorrer, and D. G. Foster, Fundamentals of Momentum Heat and Mass Transfer, 6th ed. New Jersey: Wiley, 2015, p. 370.
$\begin{array}{cc}\dot{\mathit{\text{Q}}}=\frac{\text{UA}({\mathit{\text{T}}}_{a1}{\mathit{\text{T}}}_{a2})}{\mathrm{l}\text{n}\left[\right(\mathit{\text{T}}{\mathit{\text{T}}}_{a1})/(\mathit{\text{T}}{\mathit{\text{T}}}_{a2})]}& \text{(1213)}\end{array}$
Figure 126 CSTR tank reactor with heat exchanger.
The following derivations, based on a coolant (exothermic reaction), apply also to heating mediums (endothermic reaction). As a first approximation, we assume a quasisteady state operation for the coolant flow and neglect the accumulation term (i.e., dT_{a}/dt = 0). An energy balance on the heatexchanger fluid entering and leaving the exchanger is
For exothermic reactions (T>T_{a}_{2}>T_{a}_{1})
For endothermic reactions (T_{1}_{a}>T_{2}_{a}>T)
Energy balance on heatexchanger fluid
$\begin{array}{cc}\begin{array}{cc}\begin{array}{cc}\begin{array}{cc}\begin{array}{cc}\begin{array}{cc}\left[\begin{array}{c}\begin{array}{c}\begin{array}{c}\text{Rate of}\\ \text{energy}\end{array}\\ \text{in}\end{array}\\ \text{by flow}\end{array}\right]& \end{array}& \left[\begin{array}{c}\begin{array}{c}\begin{array}{c}\begin{array}{c}\begin{array}{c}\begin{array}{c}\text{Rate of}\\ \text{energy}\end{array}\end{array}\\ \text{out}\end{array}\end{array}\\ \text{by flow}\end{array}\end{array}\right]\end{array}& \end{array}& \left[\begin{array}{c}\begin{array}{c}\begin{array}{c}\begin{array}{c}\begin{array}{c}\begin{array}{c}\begin{array}{c}\text{Rate of}\end{array}\\ \text{heat transfer}\end{array}\end{array}\\ from\text{exchanger}\end{array}\end{array}\\ \text{to reactor}\end{array}\end{array}\right]\end{array}& =0\end{array}& \text{(1214)}\end{array}$
$\begin{array}{cc}\begin{array}{cc}\begin{array}{cc}\begin{array}{cc}{\dot{m}}_{c}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{c}}({\mathit{\text{T}}}_{a1}\text{}{\mathit{\text{T}}}_{\mathit{\text{R}}})& \end{array}& {\dot{m}}_{c}\end{array}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{c}}({\mathit{\text{T}}}_{a2}{\mathit{\text{T}}}_{\mathit{\text{R}}})& \frac{\text{UA}({\mathit{\text{T}}}_{a1}{\mathit{\text{T}}}_{a2})}{1\text{n}[(\mathit{\text{T}}{\mathit{\text{T}}}_{a1})/(\mathit{\text{T}}{\mathit{\text{T}}}_{a2})]}=0\end{array}& \text{(1215)}\end{array}$
where C_{Pc} is the heat capacity of the heat exchanger fluid and T_{R} is the reference temperature. Simplifying gives us
$\begin{array}{cc}\begin{array}{cc}\text{}& \dot{\mathit{\text{Q}}}={\dot{m}}_{c}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{c}}({\mathit{\text{T}}}_{a1}{\mathit{\text{T}}}_{a2})=\frac{\text{UA}({\mathit{\text{T}}}_{a1}{\mathit{\text{T}}}_{a2})}{1\text{n}[(\mathit{\text{T}}{\mathit{\text{T}}}_{a1})/(\mathit{\text{T}}{\mathit{\text{T}}}_{a2})]}=0\end{array}& \text{(1216)}\end{array}$
Solving Equation (1216) for the exit temperature of the heatexchanger fluid yields
$\begin{array}{c}\hline {\mathit{\text{T}}}_{a2}=\text{}\mathit{\text{T}}(\mathit{\text{T}}{\mathit{\text{T}}}_{a1})\text{exp}\left(\frac{\text{UA}}{{\dot{m}}_{c}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{c}}}\right)\\ \hline\end{array}\begin{array}{c}\text{(1217)}\end{array}$
From Equation (1216)
$\begin{array}{cc}\dot{\mathit{\text{Q}}}={\dot{m}}_{c}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{c}}({\mathit{\text{T}}}_{a1}{\mathit{\text{T}}}_{a2})& \text{(1218)}\end{array}$
Substituting for T_{a}_{2} in Equation (1218), we obtain
$\begin{array}{c}\hline \dot{\mathit{\text{Q}}}={\dot{m}}_{c}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{c}}\text{}\left\{({\mathit{\text{T}}}_{a1}\mathit{\text{T}})[1\text{exp}\left(\frac{\text{UA}}{{\dot{m}}_{c}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{c}}}\right)]\right\}\\ \hline\end{array}\begin{array}{c}\text{(1219)}\end{array}$
Heat transfer to a CSTR
For large values of the heatexchanger fluid flow rate, ${\stackrel{\mathrm{.}}{m}}_{c}$, the exponent will be small and can be expanded in a Taylor series (e^{–x} = 1 – x + . . .) where secondorder terms are neglected in order to give
$\dot{\mathit{\text{Q}}}={\dot{m}}_{c}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{c}}({\mathit{\text{T}}}_{a1}\mathit{\text{T}})[1\left(\frac{\text{UA}}{{\dot{m}}_{c}{\mathit{\text{C}}}_{\mathit{\text{P}}}}\right)]$
Then
Valid only for large heattransfer fluid flow rates!!
$\begin{array}{c}\hline \dot{\mathit{\text{Q}}}=\text{UA}({\mathit{\text{T}}}_{a}\mathit{\text{T}})\\ \hline\end{array}\begin{array}{c}\text{(1220)}\end{array}$
where T_{a1} ≌ T_{a}_{2} = T_{a}.
With the exception of processes involving highly viscous materials such as in Problem P126_{B}, a California Professional Engineers’ Exam Problem, the work done by the stirrer can usually be neglected. Setting ${\stackrel{\mathrm{.}}{W}}_{s}$ equal to zero in Equation (1127), neglecting ΔC_{P}, in ΔH_{Rx} substituting for $\stackrel{\mathrm{.}}{Q}$, and rearranging, we have the following relationship between conversion and temperature in a CSTR:
$\begin{array}{cc}\frac{\text{UA}}{{\mathit{\text{F}}}_{\text{A0}}}({\mathit{\text{T}}}_{a}\mathit{\text{T}})\mathrm{\Sigma}{\mathrm{\Theta}}_{i}{\mathit{\text{C}}}_{{\text{p}}_{i}}(\mathit{\text{T}}{\mathit{\text{T}}}_{0})\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ}\mathit{\text{X}}=0& \text{(1221)}\end{array}$
Solving for X
$\begin{array}{c}\hline \mathit{\text{X}}=\frac{\frac{\text{UA}}{{\mathit{\text{F}}}_{\text{A0}}}(\mathit{\text{T}}{\mathit{\text{T}}}_{a})+\mathrm{\Sigma}{\mathrm{\Theta}}_{i}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{i}}(\mathit{\text{T}}{\mathit{\text{T}}}_{0})}{\text{}[\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ}\left({\mathit{\text{T}}}_{\mathit{\text{R}}}\right)]}\\ \hline\end{array}\begin{array}{c}\text{(1222)}\end{array}$
Equation (1222) is coupled with the mole balance equation
$\begin{array}{c}\hline \mathit{\text{V}}=\frac{{\mathit{\text{F}}}_{\text{A0}}\mathit{\text{X}}}{{\text{r}}_{\mathit{\text{A}}}\left(\text{X,T}\right)}\\ \hline\end{array}\begin{array}{c}\text{(1223)}\end{array}$
to design CSTRs.
To help us more easily see the effect of the operating parameters of T, T_{0}, and T_{a}, we collect terms and define three parameters: C_{P0}, κ, and T_{c}.
We now will further rearrange Equation (1221) after letting
$\begin{array}{c}\hline {C}_{{\mathit{\text{P}}}_{0}}={\mathrm{\Sigma}\mathrm{\Theta}}_{i}{C}_{{\mathit{\text{P}}}_{i}}\\ \hline\end{array}$
then
${\mathit{\text{C}}}_{{\mathit{\text{P}}}_{0}}\left(\frac{\text{UA}}{{\mathit{\text{F}}}_{\text{A0}}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{0}}}\right){\mathit{\text{T}}}_{a}+{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{0}}{\mathit{\text{T}}}_{0}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\text{0}}}(\frac{\text{UA}}{{\mathit{\text{F}}}_{\text{A0}}{\mathit{\text{C}}}_{{\text{p}}_{0}}}+1)\mathit{\text{T}}\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ}\mathit{\text{X}}=0$
Let κ and T_{C} be nonadiabatic parameters defined by
$\begin{array}{ccc}\begin{array}{c}\hline \kappa =\frac{UA}{{F}_{\text{A0}}{C}_{{\text{P}}_{0}}}\\ \hline\end{array}& \text{and}& \begin{array}{c}\hline {T}_{c}=\frac{\kappa {T}_{a}+{T}_{0}}{1+\kappa}\\ \hline\end{array}\end{array}$
Nonadiabatic CSTR heat exchange parameters: κ and T_{c}
Then
$\begin{array}{cc}\mathit{\text{X}}\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ}={\mathit{\text{C}}}_{{\mathit{\text{P}}}_{0}}(1+k)(\mathit{\text{T}}{\mathit{\text{T}}}_{c})& \text{(1224)}\end{array}$
The parameters κ and T_{c} are used to simplify the equations for nonadiabatic operation. Solving Equation (1224) for conversion X
$\begin{array}{c}\hline \mathit{\text{X}}=\frac{{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{0}}(1+k)(\mathit{\text{T}}{\mathit{\text{T}}}_{c})}{\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ}}\\ \hline\end{array}\begin{array}{c}\text{(1225)}\end{array}$
Solving Equation (1224) for the reactor temperature
$\begin{array}{c}\hline \mathit{\text{T}}={\mathit{\text{T}}}_{c}+\frac{(\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ})\left(\mathit{\text{X}}\right)}{{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{0}}(1+k)}\\ \hline\end{array}\begin{array}{c}\text{(1226)}\end{array}$
Table 123 shows three ways to specify the design of a CSTR. This procedure for nonisothermal CSTR design will be illustrated by considering a firstorder irreversible liquidphase reaction. To solve CSTR problems of this type, we have three variables X, T, and V and we specify one and then solve for the other two values. The algorithm for working through either cases A (X specified), B (T specified), or C (V specified) is shown in Table 123. Its application is illustrated in Example 123.
TABLE 123 WAYS TO SPECIFY THE SIZING OF A CSTR
Forms of the energy balance for a CSTR with heat exchange
Production, uses, and economics
Propylene glycol is produced by the hydrolysis of propylene oxide:
Over 900 million pounds of propylene glycol were produced in 2010 and the selling price was approximately $0.80 per pound. Propylene glycol makes up about 25% of the major derivatives of propylene oxide. The reaction takes place readily at room temperature when catalyzed by sulfuric acid.
You are the engineer in charge of an adiabatic CSTR producing propylene glycol by this method. Unfortunately, the reactor is beginning to leak, and you must replace it. (You told your boss several times that sulfuric acid was corrosive and that mild steel was a poor material for construction. He wouldn’t listen.) There is a nicelooking, bright, shiny overflow CSTR of 300gal capacity standing idle in Professor Köttlov’s storage shed at his mountain vacation home. It is glasslined, and you would like to use it.
We are going to work this problem in lb_{m}, s, ft^{3}, and lbmoles rather than g, mol, and m^{3} in order to give the reader more practice in working in both the English and metric systems. Why?? Many plants still use the English system of units.
You are feeding 2500 lb_{m}/h (43.04 lbmol/h) of propylene oxide (P.O.) to the reactor. The feed stream consists of (1) an equivolumetric mixture of propylene oxide (46.62 ft^{3}/h) and methanol (46.62 ft^{3}/h), and (2) water containing 0.1 wt % H_{2}SO_{4}. The volumetric flow rate of water is 233.1 ft^{3}/h, which is 2.5 times the methanol–P.O. volumetric flow rate. The corresponding molar feed rates of methanol and water are 71.87 and 802.8 lbmol/h, respectively. The water–propylene oxide–methanol mixture undergoes a slight decrease in volume upon mixing (approximately 3%), but you neglect this decrease in your calculations. The temperature of both feed streams is 58°F prior to mixing, but there is an immediate 17°F temperature rise upon mixing of the two feed streams caused by the heat of mixing. The entering temperature of all feed streams is thus taken to be 7°F (Figure E123.1).
Figure E123.1 Propylene glycol manufacture in a CSTR.
The Furusawa Engineering Team in Japan state that under conditions similar to those at which you are operating, the reaction is apparent firstorder in propylene oxide concentration and apparent zeroorder in excess of water with the specific reaction rate^{4}
^{4} T. Furusawa, H. Nishimura, and T. Miyauchi, J. Chem. Eng. Jpn., 2, 95.
k = Ae^{–E/RT} = 16.96 × 10^{12} (e ^{–32400/RT}) h^{–1}
The units of E are Btu/lbmol and T is in °R.
There is an important constraint on your operation. Propylene oxide is a rather lowboilingpoint substance. With the mixture you are using, the company’s safety team feels that you cannot exceed an operating temperature of 125°F, or you will lose too much propylene oxide by vaporization through the vent system.
Can you use the idle CSTR as a replacement for the leaking one if it will be operated adiabatically?
If so, what will be the expected conversion of propylene oxide to glycol?
Solution
(All data used in this problem were obtained from the CRC Handbook of Chemistry and Physics unless otherwise noted.) Let the reaction be represented by
A + B → C
where
A is propylene oxide (C_{PA} = 35 Btu/lbmol · °F)^{5}
^{5} C_{PA} and C_{PC} are estimated from the observation that the great majority of lowmolecularweight oxygencontaining organic liquids have a mass heat capacity of 0.6 cal/g · °C ± 15%.
B is water (C_{PB} = 18 Btu/lbmol· °F)
C is propylene glycol (C_{PC} = 46 Btu/lbmol · °F)
M is methanol (C_{PM} = 19.5 Btu/lbmol· °F)
In this problem, neither the exit conversion nor the temperature of the adiabatic reactor is given. By application of the mole and energy balances, we can solve two equations with two unknowns (X and T), as shown on the righthand pathway in Table 123. Solving these coupled equations, we determine the exit conversion and temperature for the glasslined reactor to see whether it can be used to replace the present reactor.
Mole Balance and Design Equation:
F_{A0} – F_{A} + r_{A}V = 0
The design equation in terms of X is
$\begin{array}{cc}\mathit{\text{V}}=\frac{{\mathit{\text{F}}}_{\text{A0}}\mathit{\text{X}}}{{\text{r}}_{\mathit{\text{A}}}}& \text{(E123.1)}\end{array}$
Rate Law:
$\begin{array}{cc}{r}_{\mathit{\text{A}}}={kC}_{\mathit{\text{A}}}& \left(\text{E123.2}\right)\end{array}$
k = 16.96 10^{12} exp[–32400/R/T] h^{–1}
Stoichiometry (liquid phase, (υ = υ_{0}):
$\begin{array}{cc}{C}_{\mathit{\text{A}}}={C}_{\text{A0}}(1X)& \left(\text{E123.3}\right)\end{array}$
Combining yields
$\begin{array}{cc}\mathit{\text{V}}=\frac{{\mathit{\text{F}}}_{\text{A0}}\mathit{\text{X}}}{k{\mathit{\text{C}}}_{\text{A0}}(1\mathit{\text{X}})}=\frac{{\upsilon}_{0}\mathit{\text{X}}}{k(1\mathit{\text{X}})}& \text{(E123.4)}\end{array}$
Solving for X as a function of T and recalling that τ = V/υ_{0} gives
$\begin{array}{c}\hline {\mathit{\text{X}}}_{\text{MB}}=\frac{\tau k}{1+\tau k}=\frac{{\tau {\mathit{\text{A}}}_{e}}^{\mathit{\text{E}}/\text{RT}}}{{1+\tau {\mathit{\text{A}}}_{e}}^{\mathit{\text{E}}/\text{RT}}}\\ \hline\end{array}\begin{array}{c}\text{(E123.5)}\end{array}$
This equation relates temperature and conversion through the mole balance.
Two equations, two unknowns
The energy balance for this adiabatic reaction in which there is negligible energy input provided by the stirrer is
$\begin{array}{c}\hline {\mathit{\text{X}}}_{\text{EB}}=\frac{\mathrm{\Sigma}{\mathrm{\Theta}}_{i}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{i}}(\mathit{\text{T}}{\mathit{\text{T}}}_{i0})}{\left[\text{}\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ}\left({\mathit{\text{T}}}_{\mathit{\text{R}}}\right)+\mathrm{\Delta}{\mathit{\text{C}}}_{\text{p}}(\mathit{\text{T}}{\mathit{\text{T}}}_{\mathit{\text{R}}})\right]}\\ \hline\end{array}\begin{array}{c}\text{(E123.6)}\end{array}$
This equation relates X and T through the energy balance. We see that two equations, Equations (E123.5) and (E123.6), and two unknowns, X and T, must be solved to find the conversion where X_{EB} = X_{MB} = X.
Calculations:
Rather than putting all those numbers in the mole and heat balance equations yourself, you can outsource this task, for a small fee, to Sven Köttlov Consulting Company, located on the third floor of the downtown Market Center Building (called the “MCB”) in Riça, Jofostan. The results of our outsourcing are given below.
Evaluate the mole balance terms (C_{A0}, Θ_{i}, τ): The total liquid volumetric flow rate entering the reactor is
$\begin{array}{cc}\begin{array}{cc}{\upsilon}_{0}& ={\upsilon}_{\text{A0}}+{\upsilon}_{\text{MD}}+{\upsilon}_{\text{B0}}\hfill \\ \text{}& =46.62+46.62+233.1=326.3\text{}{\text{ft}}^{3}/\text{h}\hfill \\ \mathit{\text{V}}& =300\text{gal}={\text{40.1ft}}^{3}\hfill \end{array}& \text{(E123.7)}\end{array}$
$\begin{array}{cc}\tau =\frac{\mathit{\text{V}}}{{\upsilon}_{0}}=\frac{40.1{\text{ft}}^{3}}{326.3{\text{ft}}^{3}/\text{h}}=0.123\text{h}& \text{(E123.8)}\end{array}$
$\begin{array}{cc}\begin{array}{c}{\mathit{\text{C}}}_{\text{A0}}=\frac{{\mathit{\text{F}}}_{\text{A0}}}{{\upsilon}_{0}}=\frac{43.0\text{1bmol/h}}{326.3{\text{ft}}^{3}\text{/h}}\\ =0.132{\text{1bmol/ft}}^{3}\end{array}& \text{(E123.9)}\end{array}$
I know these are tedious calculations, but someone’s gotta know how to do them.
For methanol: $\text{}{\mathrm{\Theta}}_{\mathit{\text{M}}}=\frac{{\mathit{\text{F}}}_{\text{m0}}}{{\mathit{\text{F}}}_{\text{A0}}}=\frac{71.87\text{1bmol/h}}{43.0\text{1bmol/h}}=1.67$
For water: $\text{}{\mathrm{\Theta}}_{\mathit{\text{M}}}=\frac{{\mathit{\text{F}}}_{\mathit{\text{B}}\text{0}}}{{\mathit{\text{F}}}_{\text{A0}}}=\frac{\text{802.8 1bmol/h}}{43.0\text{1bmol/h}}=18.65$
The conversion calculated from the mole balance, X_{MB}, is found from Equation (E123.5)
${\mathit{\text{X}}}_{\text{MB}}=\frac{(16.96\times {10}^{12}{\text{h}}^{1})\left(0.1229\text{h}\right)\text{exp}(32400/1.987\mathit{\text{T}})}{1+(16.96\times {10}^{\text{12}}{\text{h}}^{1})\left(0.1229\text{h}\right)\text{exp}\text{}\text{}(32400/1.987\mathit{\text{T}})}$
Plot X_{MB} as a function of temperature.
$\begin{array}{c}\hline {\mathit{\text{X}}}_{\text{MB}}=\frac{(2.084\times {10}^{12})\text{exp}(16306/\mathit{\text{T}})}{1+(2.084\times {10}^{12})\text{exp}(16306/\mathit{\text{T}})},{\text{T is in}}^{\circ}\mathit{\text{R}}\\ \hline\end{array}\begin{array}{c}\text{(E123.10)}\end{array}$
Evaluating the energy balance terms
(1) Heat of reaction at temperature T
$\begin{array}{cc}\begin{array}{c}\text{}\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}\left(\mathit{\text{T}}\right)=\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ}\left({\mathit{\text{T}}}_{\mathit{\text{R}}}\right)+\mathrm{\Delta}{\mathit{\text{C}}}_{\mathit{\text{P}}}(\mathit{\text{T}}{\mathit{\text{T}}}_{\mathit{\text{R}}})\\ \mathrm{\Delta}{\mathit{\text{C}}}_{\mathit{\text{P}}}={\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{C}}}}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{C}}}}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{A}}}}=461835=7{\text{Btu/lbmol/}}^{\circ}\mathit{\text{F}}\end{array}& \text{(1126)}\end{array}$
$\begin{array}{cc}{\mathrm{\Delta}H}_{\text{Rx}}=364007(T{T}_{\mathit{\text{R}}})& \left(\text{E123.11}\right)\end{array}$
(2) Heat capacity term
$\begin{array}{cc}\begin{array}{cc}\mathrm{\Sigma}{\mathrm{\Theta}}_{i}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{i}}& ={\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{A}}}}+{\mathrm{\Theta}}_{\mathit{\text{B}}}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{B}}}}+{\mathrm{\Theta}}_{\mathit{\text{M}}}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{M}}}}\hfill \\ \text{}& =35+\left(18.65\right)\left(18\right)+\left(1.67\right)\left(19.5\right)\hfill \\ \text{}& =403.3\text{Btu/lbmol}{\cdot}^{\circ}\mathit{\text{F}}\hfill \end{array}& \text{(E123.12)}\end{array}$
$\begin{array}{cc}\begin{array}{cc}\text{}\text{}{\mathit{\text{T}}}_{0}& ={\mathit{\text{T}}}_{00}+\mathrm{\Delta}{\mathit{\text{T}}}_{\text{mix}}={58}^{\circ}\mathit{\text{F}}+{17}^{\circ}\mathit{\text{F}}={75}^{\circ}\mathit{\text{F}}\hfill \\ \text{}& ={535}^{\circ}\mathit{\text{R}}\hfill \\ {\mathit{\text{T}}}_{\mathit{\text{R}}}& ={68}^{\circ}\mathit{\text{F}}={528}^{\circ}\mathit{\text{R}}\hfill \end{array}& \text{(E123.13)}\end{array}$
The conversion calculated from the energy balance, X_{EB}, for an adiabatic reaction is given by Equation (1129)
$\begin{array}{cc}{\mathit{\text{X}}}_{\text{EB}}=\frac{\mathrm{\Sigma}{\mathrm{\Theta}}_{i}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{i}}(\mathit{\text{T}}{\mathit{\text{T}}}_{i0)}}{\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ}\left({\mathit{\text{T}}}_{\mathit{\text{R}}}\right)+\mathrm{\Delta}{\mathit{\text{C}}}_{\mathit{\text{P}}}(\mathit{\text{T}}{\mathit{\text{T}}}_{\mathit{\text{R}}})}& \text{(1129)}\end{array}$
Substituting all the known quantities into the energy balance gives us
${\mathit{\text{X}}}_{\text{EB}}=\frac{\left(403.3\text{Btu/lbmol}{\cdot}^{\circ}\text{F}\right){(\mathit{\text{T}}535)}^{\circ}\text{F}}{[364007(\mathit{\text{T}}528\left)\right]\text{Btu/lbmol}}$
$\begin{array}{c}\hline {\mathit{\text{X}}}_{\text{EB}}=\frac{403.3{(\mathit{\text{T}}535)\text{}}^{\text{}}}{36400+7(\mathit{\text{T}}528)]}\\ \hline\end{array}\begin{array}{c}\text{(E123.14)}\end{array}$
Adiabatic CSTR
Solving: There are a number of different ways to solve these two simultaneous algebraic equations (E123.10) and (E123.14). The easiest way is to use the Polymath nonlinear equation solver. However, to give insight into the functional relationship between X and T for the mole and energy balances, we shall obtain a graphical solution. Here, X is plotted as a function of T for both the mole (X_{MB}) and energy (X_{EB}) balances, and the intersection of the two curves gives the solution where both the mole and energy balance solutions are satisfied, that is, X_{EB} = X_{MB}. In addition, by plotting these two curves we can learn whether there is more than one intersection (i.e., multiple steady states) for which both the energy balance and mole balance are satisfied. If numerical rootfinding techniques were used to solve for X and T, it would be quite possible that you would find only one root when there is actually more than one. If Polymath were used, you could learn whether multiple roots exist by changing your initial guesses in the nonlinear equation solver. We shall discuss multiple steady states further in Section 125. We now choose T and then calculate X_{MB} and X_{EB} (Table E123.1). The calculations for X_{MB} and X_{EB} are plotted in Figure E123.2. The virtually straight line corresponds to the energy balance, Equation (E123.14), and the curved line corresponds to the mole balance, Equation (E123.10). We observe from this plot that the only intersection
TABLE E123.1 CALCULATIONS OF X_{EB} AND X_{MB} AS A FUNCTION OF T
T (°R)  X_{MB} (Eq. (E123.10))  X_{EB} (Eq. (E123.14)) 

535  0.108  0.000 
550  0.217  0.166 
565  0.379  0.330 
575  0.500  0.440 
585  0.620  0.550 
595  0.723  0.656 
605  0.800  0.764 
615  0.860  0.872 
625  0.900  0.980 
The reactor cannot be used because it will exceed the specified maximum temperature of 585°R.
Figure E123.2 The conversions X_{EB} and X_{MB} as a function of temperature.
point is at 83% conversion and 613=R. At this point, both the energy balance and mole balance are satisfied. Because the temperature must remain below 125°F (585°R), we cannot use the 300gal reactor as it is now.
Analysis: After using Equations (E123.10) and (E123.14) to make a plot of conversion as a function of temperature, we see that there is only one intersection of X_{EB}(T) and X_{MB}(T), and consequently only one steady state. The exit conversion is 83% and the exit temperature (i.e., the reactor temperature) is 613°R (153°F), which is above the acceptable limit of 585°R (125°F) and we thus cannot use the CSTR operating at these conditions. Be sure to go to the LEP 123 and use Wolfram to see the X_{MB} and X_{EB} versus T lines change along with their intersection change as you change the parameters in the slider box.
Ouch! Looks like our plant will not be able to be completed and our multimilliondollar profit has flown the coop. But wait, don’t give up, let’s ask reaction engineer Maxwell Anthony to fly to our company’s plant in the country of Jofostan to look for a cooling coil heat exchanger that we could place in the reactor. See what Max found in Example 124.
Fantastic! Max has located a cooling coil in an equipment storage shed in the small, mountainous village of Ölofasis, Jofostan, for use in the hydrolysis of propylene oxide discussed in Example 123. The cooling coil has 40 ft^{2} of cooling surface and the coolingwater flow rate inside the coil is sufficiently large that a constant coolant temperature of 85°F can be maintained. A typical overall heattransfer coefficient for such a coil is 100 Btu/h·ft^{2}·°F. Will the reactor satisfy the previous constraint of 125°F maximum temperature if the cooling coil is used?
Solution
If we assume that the cooling coil takes up negligible reactor volume, the conversion calculated as a function of temperature from the mole balance is the same as that in Example 123, Equation (E123.10).
Combining the mole balance, stoichiometry, and rate law, we have, from Example 123
$\begin{array}{c}\hline {\mathit{\text{X}}}_{\text{EB}}=\frac{\text{}\tau k}{1+\tau k}=\frac{(2.084\times {10}^{\text{12}})\text{exp (}16306\text{/T)}}{1+(2.084\times {10}^{\text{12}})\text{exp (}16306\text{/T)}}\\ \hline\end{array}\begin{array}{c}\text{(E123.10)}\end{array}$
T is in °R.
Energy balance: Neglecting the work by the stirrer, we combine Equations (1127) and (1220) to write
$\begin{array}{cc}\frac{\text{UA(}{\mathit{\text{T}}}_{\text{}a}\mathit{\text{T}})}{{\mathit{\text{F}}}_{\text{A0}}}\mathit{\text{X}}\left[\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ}\left({\mathit{\text{T}}}_{\mathit{\text{R}}}\right)+\mathrm{\Delta}{\mathit{\text{C}}}_{\mathit{\text{P}}}(\mathit{\text{T}}{\mathit{\text{T}}}_{\mathit{\text{R}}}\right]=\mathrm{\Sigma}{\mathrm{\Theta}}_{i}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{i}}(\mathit{\text{T}}{\mathit{\text{T}}}_{0})& \text{(E124.1)}\end{array}$
Solving the energy balance for X_{EB} yields
$\begin{array}{c}\hline {\mathit{\text{X}}}_{\text{EB}}=\frac{\mathrm{\Sigma}{\mathrm{\Theta}}_{i}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{i}}(\mathit{\text{T}}{\mathit{\text{T}}}_{0})+\left[\text{UA}(\mathit{\text{T}}{\mathit{\text{T}}}_{a})/{\mathit{\text{F}}}_{\text{A0}}\right]}{\left[\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ}\left({\mathit{\text{T}}}_{\mathit{\text{R}}}\right)+\mathrm{\Delta}{\mathit{\text{C}}}_{\mathit{\text{P}}}(\mathit{\text{T}}{\mathit{\text{T}}}_{\mathit{\text{R}}})\right]}\\ \hline\end{array}\begin{array}{c}\text{(E124.2)}\end{array}$
The coolingcoil term in Equation (E124.2) is
$\begin{array}{cc}\text{}\text{}\frac{\text{UA}}{{\mathit{\text{F}}}_{\text{A0}}}=\left(100\frac{\text{Btu}}{\text{h}\cdot {\text{ft}}^{\text{2}}\text{}{\cdot}^{\circ}\mathit{\text{F}}}\right)\frac{\left(40\text{}{\text{ft}}^{2}\right)}{\left(43.04\text{lbmol/h}\right)}=\frac{92.9\text{Btu}}{\text{1bmol}{\cdot}^{\circ}\mathit{\text{F}}}& \text{(E124.3)}\end{array}$
Recall that the cooling temperature is
T_{a} = 85°F = 545°R
The numerical values of all other terms of Equation (E124.2) are identical to those given in Equation (E123.13), but with the addition of the heat exchange term, X_{EB} becomes
$\begin{array}{c}\hline {\mathit{\text{X}}}_{\text{EB}}=\frac{403.3(\mathit{\text{T}}535)+92.9(\mathit{\text{T}}545)}{36400+7(\mathit{\text{T}}528)}\\ \hline\end{array}\begin{array}{c}\text{(E124.4)}\end{array}$
We now have two equations, Equations (E123.10) and (E124.4), and two unknowns, X and T, which we can solve with Polymath. Recall Examples 45 and 86 to review how to solve nonlinear, simultaneous equations of this type with Polymath. (See Problem P121_{A}(f) on pages 661–662 to plot X versus T on Figure E123.2.) We could generate Figure E123.2 by “fooling” Polymath to plot X and T, as explained in the tutorial on the Web site (http://www.umich.edu/~elements/6e/software/Polymath_fooling_tutorial.pdf ).
TABLE E124.1 POLYMATH: CSTR WITH HEAT EXCHANGE
The Polymath program and solution to these two Equations (E123.10) for X_{MB}, and (E124.4) for X_{EB}, are given in Table E124.1. The exiting temperature and conversion are 103.7°F (563.7°R) and 36.4%, respectively, that is,
$\begin{array}{c}\hline T=564\xb0\mathit{\text{R}}\text{and}\text{}\mathit{\text{X}}\mathit{\text{=0.36}}\mathit{\text{}}\\ \hline\end{array}$
Be sure to go to the LEP 123 and use Wolfram or Python to see how X_{MB} and X_{EB} vary as you change the parameters in the slider box.
Analysis: We are grateful to the people of the village of Ölofasis in Jofostan for their help in finding this shiny new heat exchanger. By adding heat exchange to the CSTR, the X_{MB}(T) curve is unchanged but the slope of the X_{EB}(T) line in Figure E123.2 increases and intersects the X_{MB} curve at X = 0.36 and T = 564°R. This conversion is low! We could try to reduce the cooling by increasing T_{a} or T_{0} to raise the reactor temperature closer to 585°R, but not above this temperature. The higher the temperature in this irreversible reaction, the greater the conversion.
We will see in the next section that there may be multiple exit values of conversion and temperature (multiple steady states, MSS) that satisfy the parameter values and entrance conditions.
In this section, we consider the steadystate operation of a CSTR in which a firstorder reaction is taking place. An excellent experimental investigation that demonstrates the multiplicity of steady states was carried out by Vejtasa and Schmitz.^{6} They studied the reaction between sodium thiosulfate and hydrogen peroxide
^{6} S. A. Vejtasa and R. A. Schmitz, AIChE J., 16 (3), 415.
2Na_{2}S_{2}O_{3} + 4H_{2}O_{2} → Na_{2}S_{3}O_{6} + Na_{2}SO_{4} + 4H_{2}O
in a CSTR operated adiabatically. The multiple steadystate temperatures were examined by varying the flow rate over a range of space times, τ.
To illustrate the concept of MSS, reconsider the X_{MB}(T) curve, Equation (E123.10), shown in Figure E123.2, which has been redrawn and shown as dashed lines in Figure E123.2A. Now consider what would happen if the volumetric flow rate ν_{0} is increased (τ decreased) just a little. The energy balance line, X_{EB}(T), remains unchanged, but the mole balance line, X_{MB}, moves to the right, as shown by the curved, solid line in Figure E123.2A. This shift of X_{MB}(T) to the right results in the X_{EB}(T) and X_{MB}(T) intersecting three times, X_{MB} = X_{EB}, indicating three possible steadystate conditions at which the reactor could operate.
When more than one intersection occurs, there is more than one set of conditions that satisfy both the energy balance and mole balance (i.e., X_{EB} = X_{MB}); consequently, there will be multiple steady states at which the reactor may operate. These three steady states are easily determined from a graphical solution, but only one could show up in the Polymath equation solver solution. Thus, when using the Polymath nonlinear equation solver, we need to either choose different initial guesses to find whether there are other solutions that exist for which X_{MB} = X_{EB}. Or We could also fool Polymath to obtain the
Figure E123.2A Plots of X_{EB}(T) and X_{MB}(T) for different spaces times τ.
plot in Figure E123.2A by using the ODE solver and setting $\frac{d\mathit{\text{T}}}{d\text{t}}=0.1$ and then plot X_{MB} and X_{EB} versus T, as in Example 123. Tutorials on “How to Fool Polymath” and generate G(T) and R(T) plots can be found at http://www.umich.edu/~elements/6e/software/Polymath_fooling_tutorial.pdf.
We begin by recalling Equation (1224), which applies when one neglects shaft work and ΔC_{P} (i.e., ΔC_{P} = 0 and therefore $\mathrm{\Delta}{H}_{\text{Rx}}=\mathrm{\Delta}{H}_{\text{Rx}}^{\xb0}$)
$\begin{array}{cc}\mathit{\text{X}}\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ}={\mathit{\text{C}}}_{\text{P0}}(1+\kappa )(\mathit{\text{T}}{\mathit{\text{T}}}_{c})& \text{(1224)}\end{array}$
where
$\begin{array}{c}\hline {C}_{\text{P0}}=\mathrm{\Sigma}{\mathrm{\Theta}}_{i}{C}_{{\mathit{\text{P}}}_{i}}\\ \hline\end{array}\begin{array}{c}(1226\text{a})\end{array}$
$\begin{array}{c}\begin{array}{c}\hline \kappa =\frac{\text{UA}}{{\mathit{\text{C}}}_{\text{P0}}{\mathit{\text{F}}}_{\text{A0}}}\\ \hline\end{array}\end{array}\begin{array}{c}(1226\text{b})\end{array}$
“Fooling” Polymath
and
$\begin{array}{c}\hline {\mathit{\text{T}}}_{c}=\frac{{\mathit{\text{T}}}_{0}{\mathit{\text{F}}}_{\text{A0}}{\mathit{\text{C}}}_{\text{P0}}+{\text{UAT}}_{a}}{\text{UA}+{\mathit{\text{C}}}_{\text{P0}}{\mathit{\text{F}}}_{\text{A0}}}=\frac{\kappa {\mathit{\text{T}}}_{a}+{\mathit{\text{T}}}_{0}}{1+\kappa}\\ \hline\end{array}\begin{array}{c}\text{(1227)}\end{array}$
Using the CSTR mole balance $\mathit{\text{X}}=\frac{{\text{r}}_{\mathit{\text{A}}}\mathit{\text{V}}}{{\mathit{\text{F}}}_{\text{A0}}}$, Equation (1224) may be rewritten as
$\begin{array}{c}\hline ({\text{r}}_{\mathit{\text{A}}}\text{V/}{\mathit{\text{F}}}_{\text{A0}})(\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ})={\mathit{\text{C}}}_{\text{P0}}(1+\kappa )(\mathit{\text{T}}{\mathit{\text{T}}}_{c})\\ \hline\end{array}\begin{array}{c}\text{(1228)}\end{array}$
The lefthand side is referred to as the heatgenerated term
G(T) = Heatgenerated term
$\begin{array}{c}\hline \mathit{\text{G}}\left(\mathit{\text{T}}\right)=(\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ})({\text{r}}_{\mathit{\text{A}}}\text{V/}{\mathit{\text{F}}}_{\text{A0}})\\ \hline\end{array}\begin{array}{c}\text{(1229)}\end{array}$
The righthand side of Equation (1228) is referred to as the heatremoved term (by flow and heat exchange) R(T)
R(T) = Heatremoved term
$\begin{array}{c}\hline R\left(T\right)={C}_{\text{P0}}(1\kappa )(T{T}_{c})\\ \hline\end{array}\begin{array}{c}(1230)\end{array}$
To study the multiplicity of steady states, we shall plot both R(T) and G(T) as a function of temperature on the same graph and analyze the circumstances under which we will obtain multiple intersections of R(T) and G(T).
Vary Entering Temperature. From Equation (1230), we see that R(T) increases linearly with temperature, with slope and intercept T_{c}. As the entering temperature T_{0} is increased, the line retains the same slope but shifts to the right as the intercept T_{c} increases, as shown in Figure 127.
Heatremoved curve R(T)
Figure 127 Variation of heatremoved line with inlet temperature.
Vary Nonadiabatic Parameter κ. If one increases κ by either decreasing the molar flow rate, F_{A0}, or increasing the heatexchange area, A, the slope increases and for the case of T_{a} < T_{0} the ordinate intercept moves to the left, as shown in Figure 128.
Figure 128 Variation of heatremoved line with κ (κ = UA/C_{P0}F_{A0}).
$\begin{array}{c}\hline \text{}\kappa =\frac{\text{UA}}{{\mathit{\text{C}}}_{\text{P0}}{\mathit{\text{F}}}_{\text{A0}}}\\ {\mathit{\text{T}}}_{c}=\frac{{\mathit{\text{T}}}_{0}+\kappa {\mathit{\text{T}}}_{a}}{1+\kappa}\\ \hline\end{array}$
On the other hand, if T_{a} < T_{0}, the intercept will move to the right as κ increases.
The heatgenerated term, Equation (1229), can be written in terms of conversion. (Recall that X = –r_{A}V/F_{A0}.)
$\begin{array}{cc}\mathit{\text{G}}\left(\mathit{\text{T}}\right)=(\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ})\mathit{\text{X}}& \text{(1231)}\end{array}$
To obtain a plot of heat generated, G(T), as a function of temperature, we must solve for X as a function of T using the CSTR mole balance, the rate law, and stoichiometry. For example, for a firstorder liquidphase reaction, the CSTR mole balance becomes
$\mathit{\text{V}}=\frac{{\mathit{\text{F}}}_{\text{A0}}\mathit{\text{X}}}{k{\mathit{\text{C}}}_{\mathit{\text{A}}}}=\frac{{\upsilon}_{0}{\mathit{\text{C}}}_{\text{A0}}\mathit{\text{X}}}{k{\mathit{\text{C}}}_{\text{A0}}(1\mathit{\text{X}})}$
Solving for X yields
$\begin{array}{cc}\mathit{\text{X}}=\frac{\tau k}{1+\text{}\tau k}& \text{(58)}\end{array}$
Firstorder reaction
Substituting for X in Equation (1231), we obtain
$\begin{array}{cc}\mathit{\text{G}}\left(\mathit{\text{T}}\right)=\frac{\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ}\tau k}{1+\tau k}& \text{(1232)}\end{array}$
Finally, substituting for k in terms of the Arrhenius equation, we obtain
$\begin{array}{c}\hline \mathit{\text{G}}\left(\mathit{\text{T}}\right)=\frac{\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ}{\tau \mathit{\text{A}}e}^{\text{E/RT}}}{1+{\tau \mathit{\text{A}}e}^{\text{E/RT}}}\\ \hline\end{array}\begin{array}{c}\text{(1233)}\end{array}$
Note that equations analogous to Equation (1233) for G(T) can be derived for other reaction orders and for reversible reactions simply by solving the CSTR mole balance for X. For example, for the secondorder liquidphase reaction
Secondorder reaction
$\mathit{\text{X}}=\frac{(\text{2}\tau k{\mathit{\text{C}}}_{\text{A0}}+1)\sqrt{4\tau k{\mathit{\text{C}}}_{\text{A0}}+1}}{2\tau k{\mathit{\text{C}}}_{\text{A0}}}$
the corresponding heatgenerated term is
$\begin{array}{c}\hline \mathit{\text{G}}\left(\mathit{\text{T}}\right)=\frac{\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ}\left[(2\text{}\tau {\mathit{\text{C}}}_{\text{A0}}{\mathit{\text{A}}e}^{\text{E/RT}}+1)\right]\sqrt{4\tau {\mathit{\text{C}}}_{\text{A0}}{\mathit{\text{A}}e}^{\text{E/RT}}+1}}{2\tau {\mathit{\text{C}}}_{\text{A0}}t{\mathit{\text{A}}e}^{\text{E/RT}}}\\ \hline\end{array}\begin{array}{c}\text{(1234)}\end{array}$
Let’s now return to our firstorder reaction, Equation (1233), and examine the behavior of the G(T) curve. At very low temperatures, the second term in the denominator of Equation (1233) for the firstorder reaction can be neglected, so that G(T) varies as
Low T
$\mathit{\text{G}}\left(\mathit{\text{T}}\right)=\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ}{\tau \mathit{\text{A}}e}^{\text{E/RT}}$
(Recall that ${\mathrm{\Delta}H}_{\text{Rx}}^{\xb0}$ means that the standard heat of reaction is evaluated at T_{R}.)
At very high temperatures, the second term on the right side in the denominator is large and dominates in Equation (1233) so that the τk in the numerator and denominator cancel, and G(T) is reduced to
High T
$\mathit{\text{G}}\left(\mathit{\text{T}}\right)=\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ}$
G(T) is shown as a function of T for two different activation energies, E, in Figure 129. If the flow rate is decreased or the reactor volume increased so as to increase τ, the heatgenerated term, G(T), changes, as shown in Figure 1210.
Figure 129 Variation of G(T) curve with activation energy.
Figure 1210 Variation of G(T) curve with space time.
Heatgenerated curves, G(T)
Can you combine Figures 1210 and 128 to explain why a Bunsen burner flame goes out when you turn up the gas flow to a very high rate? As an exercise, plot R(T) verses G(T) for different values of T_{0} and κ.
The points of intersection of R(T) and G(T) give us the temperature at which the reactor can operate at steady state. This intersection is the point at which both the energy balance, R(T), and the mole balance, G(T), are satisfied. Suppose that we begin to operate our reactor at some relatively low temperature, T_{01}. If we construct our G(T) and R(T) curves, illustrated by curve y = G(T) and line a = R(T), respectively, in Figure 1211, we see that there will be only one point of intersection, point 1. From this point of intersection, one can find the steadystate temperature in the reactor, T_{s}_{1}, by following a vertical line down to the Taxis and reading off the temperature, T_{s}_{1}, as shown in Figure 1211.
Ignition: The point of no return
If one were now to increase the entering temperature to T_{02}, the G(T) curve, y, would remain unchanged, but the R(T) curve would move to the right, as shown by line b in Figure 1211, and will now intersect the G(T) at point 2 and also be tangent at point 3. Consequently, we see from Figure 1211 that there are two steadystate temperatures, T_{s}_{2} and T_{s}_{3}, that can be realized in the CSTR for an entering temperature T_{02}. If the entering temperature is further increased to T_{03}, the R(T) curve, line c (Figure 1212), intersects the G(T) curve three times and there are three steadystate temperatures, T_{s}_{4}, T_{s}_{5}, and T_{s}_{6}. As we continue to increase T_{0}, we finally reach line e, in which there are only two steadystate temperatures, a point of tangency at T_{s}_{10} and an intersection at T_{s}_{11}. If the entering temperature is slightly increased beyond T_{05}, say to T_{06}, then the point of tangency disappears and there is a jump in reactor temperature, from T_{s}_{10} up to T_{s}_{12}. This jump is called the ignition point for the reaction. By further increasing T_{0}, we reach line f, corresponding to T_{06}, in which we have only one reactor temperature that will satisfy both the mole and energy balances, T_{s}_{12}. For the six entering temperatures, we can form Table 124, relating the entering temperature to the possible reactor operating temperatures.
By plotting steadystate reactor temperature T_{s} as a function of entering temperature T_{0}, we obtain the wellknown ignition–extinction curve shown in
Both the mole and energy balances are satisfied at the points of intersection or tangency.
Figure 1211 Finding multiple steady states with T_{0} varied.
Figure 1212 Finding multiple steady states with T_{0} varied.
TABLE 124 MULTIPLE STEADYSTATE TEMPERATURES
Entering Temperature  Reactor Temperatures  

T_{01} 
T_{s1} 

T_{02} 
T_{s}_{2} 
T_{s}_{3} 

T_{03} 
T_{s}_{4} 
T_{s}_{5} 
T_{s}_{6} 

T_{04} 
T_{s}_{7} 
T_{s}_{8} 
T_{s}_{9} 

T_{05} 
T_{s}_{10} 
T_{s}_{11} 

T_{06} 
T_{s}_{12} 
Figure 1213. From this figure, we see that as the entering temperature T_{0}_{} is increased, the steadystate temperature T_{s} increases along the bottom line until T_{05} is reached. Any fractionofadegree increase in temperature beyond T_{05} the point of tangency disappears and the steadystate reactor temperature T_{s} will jump up from T_{s10} to T_{s11}, as shown in Figure 1213. The temperature at which this jump occurs is called the ignition temperature. That is, we must exceed a certain feed temperature, T_{05}, to operate at the upper steady state where the temperature and conversion are higher. The ignition point temperature is sometimes also called the onset temperature for a runaway reaction or the point of no return, especially when the upper steady state is at a very high temperature.
Runaway reaction?
If a reactor were operating at T_{s12} and we began to cool the entering temperature down from T_{06}, the steadystate reactor temperature, T_{s3}, would eventually be reached, corresponding to an entering temperature, T_{02}. Any slight decrease below T_{02} the point of tangency on line b in Figure 1212 would disappear and the temperature would drop the steadystate reactor temperature to immediately the lower steadystate value T_{s}_{2}. Consequently, T_{02} is called the extinction temperature.
The middle points 5 and 8 in Figures 1212 and 1213 represent unstable steadystate temperatures. Consider the heatremoved line d in Figure 1212, along with the heatgenerated curve y, which are replotted in Figure 1214. If we were operating at the middle steadystate temperature T_{s}_{8}, for example, and a
Figure 1213 Temperature ignition–extinction curve.
Figure 1214 Stability of multiple steadystate temperatures.
Point of No Return
Beyond the onset point/temperature, if no action is taken the reaction will proceed to its maximum temperature.
pulse increase in reactor temperature occurred, we would find ourselves at the temperature shown by vertical line , between points 8 and 9. We see that along this vertical line , the heatgenerated curve, y ≡ G(T), is greater than the heatremoved line d ≡ R(T), that is, (G > R). Consequently, the temperature in the reactor would continue to increase until point 9 is reached at the upper steady state. On the other hand, if we had a pulse decrease in temperature from point 8, we would find ourselves on a vertical line between points 7 and 8. Here, we see that the heatremoved curve d is greater than the heatgenerated curve y (R > G), so the temperature will continue to decrease until the lower steady state is reached. That is, a small change in temperature either above or below the middle steadystate temperature, T_{s8}, will cause the reactor temperature to move away from this middle steady state. Steady states that behave in this manner are said to be unstable.
In contrast to these unstable operating points, there are stable operating points. Consider what would happen if a reactor operating at T_{s9} were subjected to a pulse increase in reactor temperature indicated by line in Figure 1214. We see that the heatremoved line d is greater than the heatgenerated curve y (R > G), so that the reactor temperature will decrease and return to T_{s9}. On the other hand, if there is a sudden drop in temperature below T_{s9}, as indicated by line , we see the heatgenerated curve y is greater than the heatremoved line d (G > R), and the reactor temperature will increase and return to the upper steady state at T_{s9}. Consequently, T_{s9} is a stable steady state.
Next, let’s look at what happens when the lower steadystate temperature at T_{s7} is subjected to pulse increase to the temperature shown as line in Figure 1214. Here, we again see that the heat removed, R, is greater than the heat generated, G, so that the reactor temperature will drop and return to T_{s7}. If there is a sudden decrease in temperature below T_{s7} to the temperature indicated by line , we see that the heat generated is greater than the heat removed (G > R), and that the reactor temperature will increase until it returns to T_{s7}. Consequently, T_{s7} is a stable steady state. A similar analysis could be carried out for temperatures T_{s1}, T_{s2}, T_{s4}, T_{s6}, T_{s11}, and T_{s12}, and one would find that reactor temperatures would always return to locally stable steadystate values when subjected to both positive and negative fluctuations.
While these points are locally stable, they are not necessarily globally stable. That is, a large perturbation in temperature or concentration may be sufficient to cause the reactor to fall from the upper steady state (corresponding to high conversion and temperature, such as point 9 in Figure 1214), to the lower steady state (corresponding to low temperature and conversion, point 7).
Most reacting systems involve more than one reaction and do not operate isothermally. This section is one of the most important, if not the most important, sections of the book. It ties together all the previous chapters to analyze multiple reactions that do not take place isothermally.
In this section we give the energy balance for multiple reactions. We begin by recalling the energy balance for a single reaction taking place in a PFR, which is given by Equation (125)
$\begin{array}{cc}\frac{d\mathit{\text{T}}}{d\mathit{\text{V}}}=\frac{({\text{r}}_{\mathit{\text{A}}})[\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}\left(\mathit{\text{T}}\right)]\mathit{\text{U}}a(\mathit{\text{T}}{\mathit{\text{T}}}_{a})}{\underset{j=1}{\overset{m}{\mathrm{\Sigma}}}{\mathit{\text{F}}}_{j}{\mathit{\text{C}}}_{{\text{p}}_{j}}}& \text{(125)}\end{array}$
When we have multiple reactions occurring, we have to account for, and sum up, all the heats of reaction in the reactor for each and every reaction. For q multiple reactions taking place in a PFR where there are m species, it is easily shown that Equation (125) can be generalized to (cf. Problem P121(j))
Energy balance for multiple reactions
$\begin{array}{c}\hline \text{}\frac{d\mathit{\text{T}}}{d\mathit{\text{V}}}\text{=}\frac{\underset{\text{i=1}}{\overset{q}{\mathrm{\Sigma}}}({\text{r}}_{\text{ij}})[{\mathrm{\Delta}\mathit{\text{H}}}_{\text{Rxij}}\left(\mathit{\text{T}}\right)]\mathit{\text{U}}a(\mathit{\text{T}}{\mathit{\text{T}}}_{a})}{\underset{\text{j=1}}{\overset{\text{m}}{\mathrm{\Sigma}}}{\mathit{\text{F}}}_{\text{j}}{\mathit{\text{C}}}_{{\text{p}}_{\text{j}}}}\\ \hline\end{array}\begin{array}{c}\text{(1235)}\end{array}$
i = Reaction number
j = Species
Note that we now have two subscripts on the heat of reaction, the reaction number “i” and species “j”. The heat of reaction for reaction i must be referenced to the same species in the rate, r_{ij}, by which ΔH_{Rxij} is multiplied, which is
$\begin{array}{cc}[{\text{r}}_{\text{ij}}][\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rxij}}]=\left[\frac{\text{Moles of j reaction in reaction i}}{\text{Volume}\cdot \text{time}}\right]\times \left[\frac{\text{Joulesed \u201creleased\u201d in reaction i}}{\text{Moles of j reacted in reaction i}}\right]& \text{}\\ =\left[\frac{\text{Joules \u201creleased\u201d in reaction i}}{\text{Volume}\cdot \text{time}}\right]& \text{(1236)}\end{array}$
where again we note the subscript j refers to the species, the subscript i refers to the particular reaction, q is the number of independent reactions, and m is the number of species. We are going to let
$\begin{array}{c}\hline {\mathit{\text{Q}}}_{\mathit{\text{R}}}=\text{}\underset{\text{i=l}}{\overset{\text{q}}{\text{}\mathrm{\Sigma}}}({\text{r}}_{\text{ij}})[\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rxij}}\left(\mathit{\text{T}}\right)]\\ \hline\end{array}$
and
$\begin{array}{c}\hline {Q}_{r}=Ua(T{T}_{a})\\ \hline\end{array}$
Then Equation (1235) becomes
$\begin{array}{c}\hline \frac{d\mathit{\text{T}}}{d\mathit{\text{V}}}=\frac{{\mathit{\text{Q}}}_{\text{g}}{\mathit{\text{Q}}}_{\text{r}}}{\underset{\text{j=l}}{\overset{\text{m}}{\text{}\mathrm{\Sigma}}}{\mathit{\text{F}}}_{\text{j}}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\text{j}}}}\\ \hline\end{array}\begin{array}{c}\text{(1237)}\end{array}$
Equation (1237) represents a nice compact form of the energy balance for multiple reactions.
Consider the following series reaction sequence carried out in a PFR:
$\begin{array}{c}\begin{array}{cc}\mathrm{Re}\text{action}\text{1:}& \mathit{\text{A}}\stackrel{{k}_{1}}{\to}\mathit{\text{B}}\end{array}\\ \begin{array}{cc}\mathrm{Re}\text{action}\text{1:}& \mathit{\text{B}}\stackrel{{k}_{2}}{\to}\mathit{\text{C}}\end{array}\end{array}$
The PFR energy balance becomes
One of the major goals of this text is that the reader will be able to solve multiple reactions with heat effects, and this section shows how!
$\begin{array}{c}{Q}_{g}=(\mathrm{\Delta}{H}_{\text{Rx1A}})({r}_{1\mathit{\text{A}}})+({r}_{2\mathit{\text{B}}})(\mathrm{\Delta}{H}_{\text{Rx2}\mathit{\text{B}}})\\ {Q}_{r}=Ua(T{T}_{a})\hfill \end{array}$
Substituting for Q_{g} and –Q_{r} in Equation (1237) we obtain
$\begin{array}{cc}\text{}\frac{d\mathit{\text{T}}}{d\mathit{\text{V}}}=\frac{\text{Ua}(\mathit{\text{T}}{\mathit{\text{T}}}_{a})+({\text{r}}_{1\mathit{\text{A}}})(\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx1A}})+({\text{r}}_{\text{2B}})(\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx2B}})}{{\mathit{\text{F}}}_{\mathit{\text{A}}}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{A}}}}+{\mathit{\text{F}}}_{\mathit{\text{B}}}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{B}}}}+{\mathit{\text{F}}}_{\mathit{\text{C}}}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\text{c}}}}& \text{(1238)}\end{array}$
where ΔH_{Rx1A} = [J/mol of A reacted in reaction 1] and
ΔH_{Rx2B} = [J/mol of B reacted in reaction 2].
We will now give three examples of multiple reactions with heat effects: Example 125 discusses parallel reactions, Example 126 discusses series reactions, and Example 127 discusses complex reactions.
The following gasphase reactions occur in a PFR:
$\begin{array}{cc}\begin{array}{ccc}\text{Reaction: 1}& \mathit{\text{A}}\text{}\stackrel{{k}_{1}}{\to}\mathit{\text{B}}& {\text{r}}_{1\mathit{\text{A}}}={k}_{1\mathit{\text{A}}}{\mathit{\text{C}}}_{\mathit{\text{A}}}\end{array}& \text{(E125.1)}\end{array}$
$\begin{array}{cc}\begin{array}{ccc}\text{Reaction: 2}& 2\mathit{\text{A}}\stackrel{{k}_{2}}{\to}\mathit{\text{C}}& {\text{r}}_{2\mathit{\text{A}}}={k}_{2\mathit{\text{A}}}{\mathit{\text{C}}}_{\mathit{\text{A}}}^{2}\end{array}& \text{(E125.2)}\end{array}$
Pure A is fed at a rate of 100 mol/s, at a temperature of 150=C and at a concentration of 0.1 mol/dm^{3}. Neglect pressure drop and determine the temperature and molar flow rate profiles down the reactor.
Additional information:
Δ_{H}_{Rx1A} = – 20000 J/(mol = of A reacted in reaction 1)
Δ_{H}_{Rx2A} = – 60000 J/(mol = of A reacted in reaction 2)
$\begin{array}{cc}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{A}}}}=\text{90J/mol}{\cdot}^{\circ}\mathit{\text{C}}& {k}_{1\mathit{\text{A}}}=10\text{exp}\left[\frac{{\mathit{\text{E}}}_{1}}{\mathit{\text{R}}}(\frac{1}{300}\frac{1}{\mathit{\text{T}}})\right]{\text{s}}^{1}\hfill \\ {\mathit{\text{C}}}_{{\mathit{\text{P}}}_{B}}=90\text{J/mol}{\cdot}^{\circ}\mathit{\text{C}}& {\mathit{\text{E}}}_{1}\text{/R}=4000\mathit{\text{K}}\hfill \\ {\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{C}}}}=180\text{J/mol}{\cdot}^{\circ}\mathit{\text{C}}& {k}_{2\mathit{\text{A}}}=0.09\text{exp}\left[\frac{{\mathit{\text{E}}}_{2}}{\mathit{\text{R}}}(\frac{1}{300}\frac{1}{\mathit{\text{T}}})\right]\frac{{\text{dm}}^{3}}{\text{mol}\cdot \mathit{\text{S}}}\hfill \\ \mathit{\text{U}}a=4000{\text{J/m}}^{3}\text{}\cdot \text{s}{\cdot}^{\circ}\mathit{\text{C}}& {\mathit{\text{E}}}_{0}\text{/}\text{R=9000K}\hfill \\ {\mathit{\text{T}}}_{a}={100}^{\circ}\text{C}\left(\text{Constant}\right)& \text{}\end{array}$
Solution
0. Number Each Reaction: This step was given in the problem statement.
$\begin{array}{cc}\left(1\right)& \mathit{\text{A}}\stackrel{{k}_{1\mathit{\text{A}}}}{\to}\mathit{\text{B}}\\ \left(2\right)& 2\mathit{\text{A}}\stackrel{{k}_{2\mathit{\text{A}}}}{\to}\mathit{\text{C}}\end{array}$
Mole Balances:
$\begin{array}{cc}\text{}\frac{d{\mathit{\text{F}}}_{\mathit{\text{A}}}}{d\mathit{\text{V}}}={\text{r}}_{\mathit{\text{A}}}\text{}& \text{(E125.3)}\end{array}$
$\begin{array}{cc}\text{}\frac{d{\mathit{\text{F}}}_{\mathit{\text{B}}}}{d\mathit{\text{V}}}={\text{r}}_{\mathit{\text{B}}}\text{}& \text{(E125.4)}\end{array}$
$\begin{array}{cc}\text{}\frac{d{\mathit{\text{F}}}_{\mathit{\text{C}}}}{d\mathit{\text{V}}}={\text{r}}_{\mathit{\text{C}}}\text{}& \text{(E125.5)}\end{array}$
Rates:
Rate laws
$\begin{array}{cc}{r}_{1\mathit{\text{A}}}={k}_{1\mathit{\text{A}}}{C}_{\mathit{\text{A}}}& \left(\text{E125.1}\right)\end{array}$
$\begin{array}{cc}{r}_{\text{2A}}={k}_{\text{1A}}{C}_{\mathit{\text{A}}}^{2}& \left(\text{E125.2}\right)\end{array}$
Relative rates
$\begin{array}{cc}\begin{array}{cc}\text{Reaction 1:}& \frac{{\text{r}}_{1\mathit{\text{A}}}}{1}=\frac{{\text{r}}_{1\mathit{\text{B}}}}{1};\end{array}& {\text{r}}_{1\mathit{\text{B}}}\end{array}={\text{r}}_{1\mathit{\text{A}}}={k}_{1\mathit{\text{A}}}{\mathit{\text{C}}}_{\mathit{\text{A}}}$
$\begin{array}{cc}\begin{array}{cc}\text{Reaction 2:}& \frac{{\text{r}}_{\text{2A}}}{2}=\frac{{\text{r}}_{\text{2B}}}{1};\end{array}& {\text{r}}_{2\mathit{\text{C}}}\end{array}=\text{}\frac{1}{2}{\text{r}}_{2\mathit{\text{A}}}=\frac{{k}_{2\mathit{\text{A}}}}{2}{\mathit{\text{C}}}_{\mathit{\text{A}}}^{2}$
Net rates
$\begin{array}{cc}{\text{r}}_{\mathit{\text{A}}}={\text{r}}_{1\mathit{\text{A}}}+{\text{r}}_{2\mathit{\text{A}}}={k}_{1\mathit{\text{A}}}{\mathit{\text{C}}}_{\mathit{\text{A}}}{k}_{2\mathit{\text{A}}}{\mathit{\text{C}}}_{\mathit{\text{A}}}^{2}& \text{}\left(\text{E125.6}\right)\end{array}$
$\begin{array}{cc}{r}_{\mathit{\text{B}}}={r}_{1\mathit{\text{B}}}={k}_{1\mathit{\text{A}}}{C}_{\mathit{\text{A}}}& \left(\text{E125.7}\right)\end{array}$
$\begin{array}{cc}{\text{r}}_{\mathit{\text{C}}}={\text{r}}_{2\mathit{\text{C}}}=\frac{1}{2}{k}_{2\mathit{\text{A}}}{\mathit{\text{C}}}_{\mathit{\text{A}}}^{2}& \left(\text{E125.8}\right)\end{array}$
Stoichiometry (gasphase but ΔP = 0, i.e., p = 1):
$\begin{array}{cc}{\mathit{\text{C}}}_{\mathit{\text{A}}}={\mathit{\text{C}}}_{\text{T0}}\left(\frac{{\mathit{\text{F}}}_{\mathit{\text{A}}}}{{\mathit{\text{F}}}_{\mathit{\text{T}}}}\right)\left(\frac{{\mathit{\text{T}}}_{0}}{\mathit{\text{T}}}\right)& \text{(E125.9)}\end{array}$
$\begin{array}{cc}{\mathit{\text{C}}}_{\mathit{\text{B}}}={\mathit{\text{C}}}_{\text{T0}}\left(\frac{{\mathit{\text{F}}}_{\mathit{\text{B}}}}{{\mathit{\text{F}}}_{\mathit{\text{T}}}}\right)\left(\frac{{\mathit{\text{T}}}_{0}}{\mathit{\text{T}}}\right)& \text{(E125.10)}\end{array}$
$\begin{array}{cc}{\mathit{\text{C}}}_{\mathit{\text{C}}}={\mathit{\text{C}}}_{\text{T0}}\left(\frac{{\mathit{\text{F}}}_{\mathit{\text{C}}}}{{\mathit{\text{F}}}_{\mathit{\text{T}}}}\right)\left(\frac{{\mathit{\text{T}}}_{0}}{\mathit{\text{T}}}\right)& \text{(E125.11)}\end{array}$
$\begin{array}{cc}{F}_{\mathit{\text{T}}}={F}_{\mathit{\text{A}}}+{F}_{\mathit{\text{B}}}+{F}_{\mathit{\text{C}}}& \left(\text{E125.12}\right)\end{array}$
$\begin{array}{cc}{k}_{1\mathit{\text{A}}}=10\text{exp}\left[4000(\frac{1}{300}\frac{1}{\mathit{\text{T}}}){\text{s}}^{1}\right]& \text{(E125.13)}\end{array}$
(T in K)
$\begin{array}{cc}{k}_{\text{2A}}=0.09\text{exp}\left[9000(\frac{1}{300}\frac{1}{\mathit{\text{T}}})\right]\frac{{\text{dm}}^{3}}{\text{mol}\cdot \text{s}}& \text{(E125.14)}\end{array}$
Selectivity:
$\begin{array}{cc}{\tilde{\mathit{\text{S}}}}_{\text{B/C}}=\frac{{\mathit{\text{F}}}_{\mathit{\text{B}}}}{{\mathit{\text{F}}}_{\mathit{\text{C}}}}& \text{(E125.15)}\end{array}$
Major goal of CRE: Analyze Multiple Reactions with Heat Effects
Energy Balance:
The PFR energy balance (cf. Equation (1235))
$\begin{array}{cc}\frac{d\mathit{\text{T}}}{d\mathit{\text{V}}}=\frac{\mathit{\text{U}}a({\mathit{\text{T}}}_{a}\mathit{\text{T}})+({\text{r}}_{1\mathit{\text{A}}})(\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx1A}})+({\text{r}}_{2\mathit{\text{A}}})(\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx2A}})}{{\mathit{\text{F}}}_{\mathit{\text{A}}}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{A}}}}+{\mathit{\text{F}}}_{\mathit{\text{B}}}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{B}}}}+{\mathit{\text{F}}}_{\mathit{\text{C}}}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{C}}}}}& \text{(E125.16)}\end{array}$
becomes
$\begin{array}{cc}\frac{d\mathit{\text{T}}}{d\mathit{\text{V}}}=\frac{4000(373\mathit{\text{T}})+({\text{r}}_{1\mathit{\text{A}}})\left(2000\right)+({\text{r}}_{2\mathit{\text{A}}})\left(60000\right)}{{90\mathit{\text{F}}}_{\mathit{\text{A}}}+90{\mathit{\text{F}}}_{\mathit{\text{B}}}+180{\mathit{\text{F}}}_{\mathit{\text{C}}}}& \text{(E125.17)}\end{array}$
Evaluation:
The Polymath program and its graphical outputs are shown in Table E125.1 and Figures E125.1 and E125.2.
TABLE E125.1 POLYMATH PROGRAM
Differential equations
1 d(Fa)/d(V) = r1a+r2a
2 d(Fb)/d(V) = r1a
3 d(Fc)/d(V) = r2a/2
4 d(T)/d(V) = (QgQr)/(90*Fa+90*Fb+180*Fc)
Explicit equations
1 Qr = 4000*(T373)
2 To = 423
3 k1a = 10*exp(4000*(1/3001/T))
4 k2a = 0.09*exp(9000*(1/3001/T))
5 Cto = 0.1
6 deltaH1 = 20000
7 deltaH2 = 60000
8 Ft = Fa+Fb+Fc
9 Ca = Cto*(Fa/Ft)*(To/T)
10 r2a = k2a*Ca^2
11 Cb = Cto*(Fb/Ft)*(To/T)
12 Cc = Cto*(Fc/Ft)*(To/T
13 r1a = k1a*Ca
14 Qg = (r1a)*(deltaH1)+(r2a)*(deltaH2)
Calculated values of DEQ variables
Variable  Initial value  Final value  

1  Ca  0.1  2.069E09 
2  Cb  0  0.0415941 
3  Cc  0  0.016986 
4  Cto  0.1  0.1 
5  Fa  100.  2.738E06 
6  Fb  0  55.04326 
7  Fc  0  22.47837 
8  Ft  100.  77.52163 
9  k1a  482.8247  2.426E+04 
10  k2a  553.0557  3.716E+06 
11  Qg  1.297E+06  1.003708 
12  Qr  2.0E+05  1.396E+06 
13  r1a  48.28247  5.019E05 
14  r2a  5.530557  1.591E11 
15  T  423.  722.0882 
16  To  423.  423. 
17  V  0  1. 
Why does the temperature go through a maximum value?
Figure E125.1 Temperature profile.
Figure E125.2 Profile of molar flow rates F_{A}, F_{B} and F_{C}
Analysis: From Figure E125.1 we see that the temperature increases slowly up to a reactor volume of 0.4 dm^{3} then suddenly jumps up (ignites) to a temperature of 850 K. Correspondingly, from Figure E125.2 the reactant molar flow rate decreases sharply at this point. The reactant is virtually consumed by the time it reaches a reactor volume V = 0.45 dm^{3}; beyond this point,Q_{r} > Q_{g}, and the reactor temperature begins to drop. In addition, the selectivity ${\tilde{\mathit{\text{S}}}}_{\text{B/}\mathit{\text{C}}}={\mathit{\text{F}}}_{\mathit{\text{B}}}/{\mathit{\text{F}}}_{\mathit{\text{C}}}=\text{55/22.5}=\text{2.44}$ remains constant after this point. If a high selectivity is required, then the reactor should be shortened to V = 0.3 dm^{3}, at which point the selectivity is ${\tilde{\mathit{\text{S}}}}_{\text{B/}\mathit{\text{C}}}=21.9\text{/2.5}=\mathrm{8.9.}$.
Recall that for the steadystate mole balance in a CSTR with a single reaction [–F_{A0}X = r_{A}V], and that $\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}\left(\mathit{\text{T}}\right)=\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ}+\mathrm{\Delta}{\mathit{\text{C}}}_{\mathit{\text{P}}}(\mathit{\text{T}}{\mathit{\text{T}}}_{\mathit{\text{R}}})$, so that for T_{0} = T_{i}_{0} Equation (1127) may be rewritten as
$\begin{array}{cc}\dot{\mathit{\text{Q}}}{\dot{\mathit{\text{W}}}}_{\text{s}}{\mathit{\text{F}}}_{\text{A0}}\text{}\mathrm{\Sigma}{\mathrm{\Theta}}_{j}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{j}}(\mathit{\text{T}}{\mathit{\text{T}}}_{0})+\left[\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}\left(\mathit{\text{T}}\right)\right]\left[{\text{r}}_{\mathit{\text{A}}}\mathit{\text{V}}\right]=0& \text{(1127A)}\end{array}$
Again, we must account for the “heat generated” by all the reactions in the reactor. For q multiple reactions and m species, the CSTR energy balance becomes
$\begin{array}{c}\hline \dot{\mathit{\text{Q}}}{\dot{\mathit{\text{W}}}}_{\text{s}}{\mathit{\text{F}}}_{\text{A0}}\underset{\text{j=1}}{\overset{\text{m}}{\mathrm{\Sigma}}}{\mathrm{\Theta}}_{\text{j}}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\text{j}}}(\mathit{\text{T}}{\mathit{\text{T}}}_{0})+\mathit{\text{V}}\underset{\text{i=l}}{\overset{\text{q}}{\mathrm{\Sigma}}}{\text{r}}_{\text{ij}}\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rxij}}\left(\mathit{\text{T}}\right)=0\\ \hline\end{array}\begin{array}{c}\text{(1239)}\end{array}$
Energy balance for multiple reactions in a CSTR
Substituting Equation (1220) for $\stackrel{\mathrm{.}}{Q}$, neglecting the work term, and assuming constant heat capacities and large coolant flow rates ${\stackrel{\mathrm{.}}{m}}_{c}$, Equation (1239) becomes
$\begin{array}{c}\hline \mathit{\text{U}}\mathit{\text{A}}({\mathit{\text{T}}}_{a}\mathit{\text{T}}){\mathit{\text{F}}}_{\text{A0}}\underset{\text{j=1}}{\overset{\text{m}}{\mathrm{\Sigma}}}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\text{j}}}{\mathrm{\Theta}}_{\text{j}}(\mathit{\text{T}}{\mathit{\text{T}}}_{0})+\mathit{\text{V}}\underset{\text{i=l}}{\overset{\text{q}}{\mathrm{\Sigma}}}{\text{r}}_{\text{ij}}\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rxij}}\left(\mathit{\text{T}}\right)=0\\ \hline\end{array}\begin{array}{c}\text{(1240)}\end{array}$
For the two parallel reactions described in Example 125, the CSTR energy balance is
$\begin{array}{cc}\mathit{\text{U}}\mathit{\text{A}}({\mathit{\text{T}}}_{a}\mathit{\text{T}}){\mathit{\text{F}}}_{\text{A0}}\underset{\text{j=1}}{\overset{\text{m}}{\mathrm{\Sigma}}}{\mathrm{\Theta}}_{\text{j}}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\text{j}}}(\mathit{\text{T}}{\mathit{\text{T}}}_{0})+{\text{Vr}}_{1\mathit{\text{A}}}\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx1}\mathit{\text{A}}}\left(\mathit{\text{T}}\right)+{\text{Vr}}_{2\mathit{\text{A}}}\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx2}\mathit{\text{A}}}\left(\mathit{\text{T}}\right)=0& \text{(1241)}\end{array}$
Major goal of CRE
As stated previously, one of the major goals of this text is to have the reader solve problems involving multiple reactions with heat effects (cf. Problems P1223_{B}, P1224_{B}, P1225_{C}, and P1226_{C}). That’s exactly what we are doing in the next two examples!
The elementary liquidphase reactions
$\mathit{\text{A}}\stackrel{{k}_{1}}{\text{}\to}\mathit{\text{B}}\stackrel{{k}_{2}}{\text{}\to}\mathit{\text{C}}$
take place in a 10dm^{3} CSTR. What are the effluent concentrations for a volumetric feed rate of 1000 dm^{3}/min at a concentration of A of 0.3 mol/dm^{3}? The inlet temperature is 283 K.
Additional information:
$\begin{array}{c}\begin{array}{c}\begin{array}{c}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{A}}}}={\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{B}}}}={\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{C}}}}=200\text{J/mol}\cdot \mathit{\text{K}}\hfill \\ {k}_{1}=3.3\text{}{\text{min}}^{1}\text{at300}{\text{K,withE}}_{1}=\text{9900cal/mol}\hfill \end{array}\hfill \\ {k}_{\text{2}}=4.58\text{}{\text{min}}^{1}\text{at500}{\text{K,withE}}_{\text{2}}=27000\text{cal/mol}\hfill \\ \begin{array}{cc}\begin{array}{c}\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx1}\mathit{\text{A}}}=55000\text{J/molA}\end{array}& \text{UA}=40000\text{J/min}\cdot {\text{KwithT}}_{a}={57}^{\circ}\end{array}\mathit{\text{C}}\end{array}\\ \mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx2B}}=\text{71500J/molB}\hfill \end{array}$
Solution
The Algorithm:
0. Number Each Reaction:
$\begin{array}{c}\begin{array}{cc}\mathrm{Re}\text{action}\left(1\right)& \mathit{\text{A}}\stackrel{{k}_{1}}{\to}\mathit{\text{B}}\end{array}\\ \begin{array}{cc}\mathrm{Re}\text{action}\left(2\right)& \mathit{\text{B}}\stackrel{{k}_{1}}{\to}\mathit{\text{C}}\end{array}\\ \end{array}$
1. Mole Balance on Every Species:
Species A: Combined mole balance and rate law for A
$\begin{array}{cc}\mathit{\text{V}}=\frac{{\mathit{\text{F}}}_{\text{A0}}{\mathit{\text{F}}}_{\mathit{\text{A}}}}{{\text{r}}_{\mathit{\text{A}}}}=\frac{{\upsilon}_{0}[{\mathit{\text{C}}}_{\text{A0}}{\mathit{\text{C}}}_{\mathit{\text{A}}}]}{{\text{r}}_{1\mathit{\text{A}}}}=\frac{{\upsilon}_{0}[{\mathit{\text{C}}}_{\text{A0}}{\mathit{\text{C}}}_{\mathit{\text{A}}}]}{{k}_{1}{\mathit{\text{C}}}_{\mathit{\text{A}}}}& \text{(E126.1)}\end{array}$
Solving for C_{A} gives us
$\begin{array}{cc}{\mathit{\text{C}}}_{\mathit{\text{A}}}=\frac{{\mathit{\text{C}}}_{\text{A0}}}{1+{\tau k}_{1}}& \text{(E126.2)}\end{array}$
Species B: Combined mole balance and rate law for B
$\begin{array}{cc}\mathit{\text{V}}=\frac{0{\mathit{\text{C}}}_{\mathit{\text{B}}}{\upsilon}_{0}}{{\text{r}}_{\mathit{\text{B}}}}=\frac{{\mathit{\text{C}}}_{\mathit{\text{B}}}{\upsilon}_{0}}{{\text{r}}_{\mathit{\text{B}}}}& \text{(E126.3)}\end{array}$
2. Rates:
The reactions follow elementary rate laws
$\begin{array}{ccc}\left(\text{a}\right)\text{Laws}& \left(\text{b}\right)\text{Relative Rates}& \left(\begin{array}{c}\mathit{\text{C}}\end{array}\right)\text{Net Rates}\\ {\text{r}}_{1\mathit{\text{A}}={k}_{1\mathit{\text{A}}}{\mathit{\text{C}}}_{\mathit{\text{A}}}\equiv {k}_{1}{\mathit{\text{C}}}_{\mathit{\text{A}}}}& {\text{r}}_{1\mathit{\text{B}}}={\text{r}}_{1\mathit{\text{A}}}& {\text{r}}_{\mathit{\text{A}}}={\text{r}}_{1\mathit{\text{A}}}\\ {\text{r}}_{2\mathit{\text{B}}={k}_{2\mathit{\text{B}}}{\mathit{\text{C}}}_{\mathit{\text{B}}}\equiv {k}_{2}{\mathit{\text{C}}}_{\mathit{\text{B}}}}& {\text{r}}_{2\mathit{\text{C}}}={\text{r}}_{\text{2}\mathit{\text{B}}}& {\text{r}}_{\mathit{\text{B}}}={\text{r}}_{1\mathit{\text{B}}}+{\text{r}}_{2\mathit{\text{B}}}\end{array}$
3. Combine:
Substituting for r_{1B} and r_{2B} in Equation (E126.3) gives
$\begin{array}{cc}\mathit{\text{V}}=\frac{{\mathit{\text{C}}}_{\mathit{\text{B}}}{\upsilon}_{0}}{{k}_{1}{\mathit{\text{C}}}_{\mathit{\text{A}}}{k}_{2}{\mathit{\text{C}}}_{\mathit{\text{B}}}}& \text{(E126.4)}\end{array}$
Solving for C_{B} yields
$\begin{array}{cc}{\mathit{\text{C}}}_{\mathit{\text{B}}}=\frac{\tau {k}_{1}{\mathit{\text{C}}}_{\mathit{\text{A}}}}{1+\tau {k}_{2}}=\frac{\tau {k}_{1}{\mathit{\text{C}}}_{\text{A0}}}{(1+\tau {k}_{1})(1+\tau {k}_{2})}& \text{(E126.5)}\end{array}$
$\begin{array}{c}\hline {\text{r}}_{1\mathit{\text{A}}}={k}_{1}{\mathit{\text{C}}}_{\mathit{\text{A}}}=\frac{{k}_{1}{\mathit{\text{C}}}_{\text{A0}}}{1+\tau {k}_{1}}\\ \hline\end{array}\begin{array}{c}\text{(E126.6)}\end{array}$
$\begin{array}{c}\hline {\text{r}}_{2\mathit{\text{B}}}={k}_{2}{\mathit{\text{C}}}_{\mathit{\text{B}}}=\frac{{k}_{2}\tau {k}_{\text{1}}{\mathit{\text{C}}}_{\text{A0}}}{(1+\tau {k}_{1})+\left(\right(1+\tau {k}_{\text{2}}\left)\right)}\\ \hline\end{array}\begin{array}{c}\text{(E126.7)}\end{array}$
4. Energy Balances:
Applying Equation (1241) to this system gives
$\begin{array}{cc}[{\text{r}}_{\text{1A}}\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx1A}}+{\text{r}}_{\text{2B}}\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx2B}}]\mathit{\text{V}}\text{UA}(\mathit{\text{T}}{\mathit{\text{T}}}_{\text{}a}){\mathit{\text{F}}}_{\text{A0}}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{A}}}}(\mathit{\text{T}}{\mathit{\text{T}}}_{0})=0& \text{(E126.8)}\end{array}$
Substituting for F_{A0} = υ_{0}C_{A0}, r_{1A}, and r_{2B} and rearranging, we have
$\begin{array}{cc}\stackrel{\text{G(T)}}{\stackrel{\u23de}{[\frac{\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx1A}}\tau {k}_{1}}{1+\tau {k}_{1}}\frac{\tau {k}_{1}\tau {k}_{2}\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx2B}}}{(1+\tau {k}_{1})(1+\tau {k}_{2})}]}}=\frac{\mathit{\text{R}}\left(\mathit{\text{T}}\right)}{{\text{c}}_{{\mathit{\text{P}}}_{\mathit{\text{A}}}}(1+\kappa )[\mathit{\text{T}}{\mathit{\text{T}}}_{e}]}& \text{(E126.9)}\end{array}$
$\begin{array}{c}\hline \mathit{\text{G}}\left(\mathit{\text{T}}\right)=[\frac{\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx1A}}\tau {k}_{1}}{1+\tau {k}_{1}}\frac{\tau {k}_{1}\tau {k}_{2}\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx2B}}}{(1+\tau {k}_{1})(1+\tau {k}_{2})}]\\ \hline\end{array}\begin{array}{c}\text{}\left(\text{E126.10}\right)\end{array}$
$\begin{array}{c}\hline \mathit{\text{R}}\left(\mathit{\text{T}}\right)={\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{A}}}}(1+\kappa )[\mathit{\text{T}}{\mathit{\text{T}}}_{c}]\\ \hline\end{array}\begin{array}{c}\text{(E126.11)}\end{array}$
5. Parameter Evaluation
$\begin{array}{cc}\kappa =\frac{\text{UA}}{{\mathit{\text{F}}}_{\text{A0}}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{A}}}}}=\frac{40000\text{J/min}\cdot \mathit{\text{K}}}{\left(0.3{\text{mol/dm}}^{3}\right)\left(1000{\text{dm}}^{3}\text{/min}\right)200\text{J/mol}\cdot \mathit{\text{K}}}=0.667& \text{(E126.12}\end{array}$
$\begin{array}{cc}{\mathit{\text{T}}}_{e}=\frac{{\mathit{\text{T}}}_{0}+\kappa {\mathit{\text{T}}}_{a}}{1+\kappa}=\frac{283+\left(0.666\right)\left(330\right)}{1+0.667}=301.8\text{K}& \text{(E126.13)}\end{array}$
We are going to generate G(T) and R(T) by "fooling" Polymath to first generate T as a function of a dummy variable, t. We then use our plotting options to convert T(t) to G(T) and R(T). The Polymath program to plot R(T) and G(T) vs. T is shown in Table E126.1, and the resulting graph is shown in Figure E126.1 (http://www.umich.edu/~elements/5e/tutorials/Polymath_fooling_tutorial_Example126.pdf).
Fooling Polymath
Incrementing temperature in this manner is an easy way to generate R(T) and G(T) plots.
TABLE E126.1 POLYMATH: PROGRAM AND OUTPUT
When F = 0, then G(T) = R(T) and the steady states can be found.
Wow!
Five (5) multiple steady states!
Figure E126.1 Heatremoved and heatgenerated curves.
TABLE E126.2 EFFLUENT CONCENTRATIONS AND TEMPERATURES
SS 
T(K) 
C_{A} (mol/dm^{3}) 
C_{B} (mol/dm^{3}) 
C_{C} (mol/dm^{3}) 

1  310  0.284  0.016  0 
2  363  0.189  0.111  0 
3  449  0.033  0.267  0.00056 
4  558  0.0041  0.167  0.129 
5  677  0.00087  0.0053  0.294 
Analysis: Wow! We see that five steady states (SS) exist!! The exit concentrations and temperatures listed in Table E126.2 were determined from the tabular output of the Polymath program. Steady states 1, 3, and 5 are stable steady states, while 2 and 4 are unstable. The selectivity at steady state 3 is $\text{}{\tilde{\mathit{\text{S}}}}_{\text{B/C}}=\frac{0.0053}{0.294}=477$, while at steady state 5 the selectivity is $\text{}{\tilde{\mathit{\text{S}}}}_{\text{B/C}}=\frac{0.0053}{0.294}=0.018$ and is far too small. Consequently, to operate at a large selectivity we either have to operate at steady state 3 or find a different set of operating conditions. What do you think of the value of tau, that is, τ = 0.01 min? Is it a realistic number? (See Problem P121_{A} (h).)
We will use the reactions that we discussed in Chapter 8 that were coded with dummy names, A, B, C, and D for reasons of national security.
The following complex gasphase reactions follow elementary rate laws
$\begin{array}{c}\begin{array}{cc}\begin{array}{cc}\begin{array}{cc}\left(1\right)& \mathit{\text{A}}+\text{2}\mathit{\text{B}}\to \mathit{\text{C}}\end{array}& {\text{r}}_{1\mathit{\text{A}}}={k}_{1\mathit{\text{A}}}{\mathit{\text{C}}}_{\mathit{\text{A}}}{\mathit{\text{C}}}_{\mathit{\text{B}}}^{2}\end{array}& \mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx1}\mathit{\text{B}}}=15000\text{cal/molB}\end{array}\\ \begin{array}{cc}\begin{array}{cc}\begin{array}{cc}\left(2\right)& 2\mathit{\text{A}}+\text{3C}\to \mathit{\text{D}}\end{array}& {\text{r}}_{\text{2C}}={k}_{\text{2C}}{\mathit{\text{C}}}_{\mathit{\text{A}}}^{2}{\mathit{\text{C}}}_{\mathit{\text{C}}}^{2}\end{array}& \mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx2}\mathit{\text{A}}}=10000\text{cal/molA}\end{array}\end{array}$
and take place in a PFR. The feed is stoichiometric for reaction (1) in A and B with F_{A0} = 5 mol/min. The reactor volume is 10 dm^{3} and the total entering concentration is C_{T0} = 0.2 mol/dm^{3}. The entering pressure is 100 atm and the entering temperature is 300 K. The coolant flow rate is 50 mol/min and the entering coolant fluid has a heat capacity of C_{PC0} = 10 cal/mol · K and enters at a temperature of 325 K.
Parameters
${k}_{1\mathit{\text{A}}}=40{\left(\frac{{\text{dm}}^{3}}{\text{mol}}\right)}^{2}{\text{/min at 300 K with E}}_{1}=8000\text{cal/mol}$
${k}_{\text{2}\mathit{\text{C}}}=\text{2}{\left(\frac{{\text{dm}}^{3}}{\text{mol}}\right)}^{4}{\text{/min at 300 K with E}}_{\text{2}}=12000\text{cal/mol}$
$\begin{array}{cc}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{A}}}}=\text{10 cal/mol/K}& \mathit{\text{U}}a=\text{80}\frac{\text{cal}}{{\text{m}}^{3}\text{}\cdot \text{min}\cdot \mathit{\text{K}}}\\ {\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{B}}}}=\text{12 cal/mol/K}& {\mathit{\text{T}}}_{\text{a0}}=325\text{K}\\ {\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{C}}}}=\text{14 cal/mol/K}& {\dot{\text{m}}}_{c}=50\text{mol/min}\\ {\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{C}}}}=\text{16 cal/mol/K}& {\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\text{C0}}}=10\text{cal/mol/K}\end{array}$
Plot F_{A}, F_{B}, F_{C}, F_{D}, p, T, and T_{a} as a function of V for
(a) Cocurrent heat exchange
(b) Countercurrent heat exchange
(c) Constant T_{a}
(d) Adiabatic operation
Solution
Gas Phase PFR No Pressure Drop (p = 1)
1. Mole Balances:
$\begin{array}{cc}\begin{array}{ccc}\left(\text{1}\right)& \frac{d{\mathit{\text{F}}}_{\mathit{\text{A}}}}{d\mathit{\text{V}}}={\text{r}}_{\mathit{\text{A}}}& ({\mathit{\text{F}}}_{\text{A0}}=\text{5 mol/min})\end{array}& \text{(E127.1)}\end{array}$
$\begin{array}{cc}\begin{array}{ccc}\left(\text{2}\right)& \frac{d{\mathit{\text{F}}}_{\mathit{\text{B}}}}{d\mathit{\text{V}}}={\text{r}}_{\mathit{\text{B}}}& ({\mathit{\text{F}}}_{\text{B0}}=\text{10}\text{mol/min})\end{array}& \text{(E127.2)}\end{array}$
$\begin{array}{cc}\begin{array}{ccc}\left(\text{3}\right)& \frac{d{\mathit{\text{F}}}_{\mathit{\text{C}}}}{d\mathit{\text{V}}}={\text{r}}_{\mathit{\text{C}}}& {\mathit{\text{V}}}_{f}={\text{10dm}}^{3}\end{array}& \text{(E127.3)}\end{array}$
$\begin{array}{cc}\begin{array}{ccc}\left(\text{4}\right)& \frac{d{\mathit{\text{F}}}_{\mathit{\text{D}}}}{d\mathit{\text{V}}}={\text{r}}_{\mathit{\text{D}}}& \text{}\end{array}& \text{(E127.4)}\end{array}$
2. Rates:
2a. Rate Laws
$\begin{array}{cc}\begin{array}{cc}\text{}\left(\text{5}\right)& {\text{r}}_{1\mathit{\text{A}}}={k}_{1\mathit{\text{A}}}{\mathit{\text{C}}}_{\mathit{\text{A}}}{\mathit{\text{C}}}_{\mathit{\text{B}}}^{\text{2}}\end{array}& \text{(E127.5)}\end{array}$
$\begin{array}{cc}\begin{array}{cc}\text{}\left(\text{6}\right)& {\text{r}}_{\text{2C}}={k}_{\text{2C}}{\mathit{\text{C}}}_{\mathit{\text{A}}}^{\text{2}}{\mathit{\text{C}}}_{\mathit{\text{C}}}^{\text{2}}\end{array}& \text{(E127.6)}\end{array}$
2b. Relative Rates
$\begin{array}{cc}\begin{array}{cc}\left(\text{7}\right)& {\text{r}}_{\text{1B}}\end{array}={\text{2r}}_{1\mathit{\text{A}}}& \text{(}\text{E127.7)}\end{array}$
$\begin{array}{cc}\begin{array}{cc}\left(\text{8}\right)& {\text{r}}_{1\mathit{\text{C}}}\end{array}={\text{r}}_{\text{1A}}& \text{(E127.8)}\end{array}$
$\begin{array}{cc}\begin{array}{cc}\left(\text{9}\right)& {\text{r}}_{\text{2A}}\end{array}=\frac{\text{2}}{\text{3}}{\text{r}}_{\text{2C}}=\frac{\text{2}}{\text{3}}{k}_{\text{2}\mathit{\text{C}}}{\mathit{\text{C}}}_{\mathit{\text{A}}}^{2}{\mathit{\text{C}}}_{\mathit{\text{C}}}^{\text{3}}& \text{(E127.9)}\end{array}$
$\begin{array}{cc}\begin{array}{cc}\left(\text{10}\right)& {\text{r}}_{\text{2D}}\end{array}\text{=}\frac{1}{\text{3}}{\text{r}}_{\text{2}\mathit{\text{C}}}\text{=}\frac{1}{3}{k}_{\text{2C}}{\mathit{\text{C}}}_{\mathit{\text{A}}}^{\text{2}}{\mathit{\text{C}}}_{\mathit{\text{C}}}^{\text{3}}& \text{(E127.10)}\end{array}$
2c. Net Rates of reaction for species A, B, C, and D are
$\begin{array}{c}\begin{array}{c}\begin{array}{c}\begin{array}{cc}\begin{array}{cc}\left(\text{11}\right)& {\text{r}}_{\mathit{\text{A}}}\end{array}={\text{r}}_{1\mathit{\text{A}}}+{\text{r}}_{\text{2A}}={k}_{1\mathit{\text{A}}}{\mathit{\text{C}}}_{\mathit{\text{A}}}{\mathit{\text{C}}}_{\mathit{\text{B}}}^{2}\frac{2}{\text{3}}{k}_{\text{2C}}{\mathit{\text{C}}}_{\mathit{\text{A}}}^{\text{2}}{\mathit{\text{C}}}_{\mathit{\text{C}}}^{\text{3}}\text{}& \text{(E127.11)}\end{array}\hfill \\ \begin{array}{cc}\begin{array}{cc}\left(\text{12}\right)\hfill & {\text{r}}_{\mathit{\text{B}}}\end{array}={\text{r}}_{1\mathit{\text{B}}}=\text{2}{k}_{1\mathit{\text{A}}}{\mathit{\text{C}}}_{\mathit{\text{A}}}{\mathit{\text{C}}}_{\mathit{\text{B}}}^{2}\text{}& \text{(E127.12)}\end{array}\hfill \end{array}\hfill \\ \begin{array}{cc}\begin{array}{cc}\text{}\left(\text{13}\right)& {\text{r}}_{\mathit{\text{C}}}={\text{r}}_{\text{1C}}+{\text{r}}_{\text{2C}}={k}_{\text{1A}}{\mathit{\text{C}}}_{\mathit{\text{A}}}{\mathit{\text{C}}}_{\mathit{\text{A}}}^{\text{2}}{\mathit{\text{C}}}_{\mathit{\text{B}}}^{\text{2}}{k}_{\text{2C}}{\mathit{\text{C}}}_{\mathit{\text{A}}}^{\text{2}}{\mathit{\text{C}}}_{\mathit{\text{C}}}^{\text{3}}\end{array}& \text{(E127.13)}\end{array}\hfill \end{array}\\ \begin{array}{cc}\begin{array}{cc}\text{}\left(\text{14}\right)& {\text{r}}_{\mathit{\text{D}}}\end{array}={\text{r}}_{\text{2D}}=\frac{\text{1}}{\text{3}}{k}_{\text{2C}}{\mathit{\text{C}}}_{\mathit{\text{A}}}^{\text{2}}{\mathit{\text{C}}}_{\mathit{\text{C}}}^{\text{2}}& \text{(E127.14)}\end{array}\hfill \end{array}$
3. Selectivity: ${\tilde{\mathit{\text{S}}}}_{\text{C/D}}={\mathit{\text{F}}}_{\mathit{\text{C}}}{\text{/F}}_{\mathit{\text{D}}}$
At V = 0, F_{D} = 0 causing S_{C/D} to go to infinity. Therefore, we set S_{C/D} = 0 between V = 0 and a very small number, say, V = 0.0001 dm^{3} to prevent the ODE solver from crashing.
$\begin{array}{cc}\begin{array}{cc}\left(\text{15}\right)& {\mathit{\text{S}}}_{\text{C/D}}=\text{if}\left(\text{V>0.0001}\right)\text{then}\left(\frac{{\mathit{\text{F}}}_{\mathit{\text{C}}}}{{\mathit{\text{F}}}_{\mathit{\text{D}}}}\right)\text{else}\left(\text{0}\right)\end{array}& \text{(E127.15)}\end{array}$
Major goal of CRE
4. Stoichiometry:
$\begin{array}{cc}\begin{array}{cc}\text{}\left(\text{16}\right)& {\mathit{\text{C}}}_{\mathit{\text{A}}}={\mathit{\text{C}}}_{\text{T0}}\text{}\left(\frac{{\mathit{\text{F}}}_{\mathit{\text{A}}}}{{\mathit{\text{F}}}_{\mathit{\text{T}}}}\right)\mathit{\text{P}}\left(\frac{{\mathit{\text{T}}}_{0}}{\mathit{\text{T}}}\right)\end{array}& \text{(E127.16)}\end{array}$
$\begin{array}{cc}\begin{array}{cc}\text{}\left(\text{17}\right)& {\mathit{\text{C}}}_{\mathit{\text{B}}}={\mathit{\text{C}}}_{\text{T0}}\text{}\left(\frac{{\mathit{\text{F}}}_{\mathit{\text{B}}}}{{\mathit{\text{F}}}_{\mathit{\text{T}}}}\right)\mathit{\text{P}}\left(\frac{{\mathit{\text{T}}}_{0}}{\mathit{\text{T}}}\right)\end{array}& \text{(E127.17)}\end{array}$
$\begin{array}{cc}\begin{array}{cc}\text{}\left(\text{18}\right)& {\mathit{\text{C}}}_{\mathit{\text{C}}}={\mathit{\text{C}}}_{\text{T0}}\text{}\left(\frac{{\mathit{\text{F}}}_{\mathit{\text{C}}}}{{\mathit{\text{F}}}_{\mathit{\text{T}}}}\right)\mathit{\text{P}}\left(\frac{{\mathit{\text{T}}}_{0}}{\mathit{\text{T}}}\right)\end{array}& \text{(E127.18)}\end{array}$
$\begin{array}{cc}\begin{array}{cc}\text{}\left(\text{19}\right)& {\mathit{\text{C}}}_{\mathit{\text{D}}}={\mathit{\text{C}}}_{\text{T0}}\text{}\left(\frac{{\mathit{\text{F}}}_{\mathit{\text{D}}}}{{\mathit{\text{F}}}_{\mathit{\text{T}}}}\right)\mathit{\text{P}}\left(\frac{{\mathit{\text{T}}}_{0}}{\mathit{\text{T}}}\right)\end{array}& \text{(E127.19)}\end{array}$
Neglect pressure drop
$\begin{array}{cc}\begin{array}{cc}\left(\text{20}\right)& \mathit{\text{P}}\end{array}=1& \text{(E127.20)}\end{array}$
$\begin{array}{cc}\begin{array}{cc}\left(\text{21}\right)& {\mathit{\text{F}}}_{\mathit{\text{T}}}\end{array}={\mathit{\text{F}}}_{\mathit{\text{A}}}+{\mathit{\text{F}}}_{\mathit{\text{B}}}+{\mathit{\text{F}}}_{\mathit{\text{C}}}+{\mathit{\text{F}}}_{\mathit{\text{D}}}& \text{(E127.21)}\end{array}$
5. Parameters:
$\text{}\begin{array}{cc}\begin{array}{cc}\left(\text{22}\right)& {k}_{\text{1}\mathit{\text{A}}}\end{array}=40\text{exp}\left[\frac{{\mathit{\text{E}}}_{\text{1}}}{\mathit{\text{R}}}(\frac{1}{300}\frac{1}{\mathit{\text{T}}})\right]{\left({\text{dm}}^{3}\text{/}\text{mol}\right)}^{\text{2}}/\text{min}& \text{(E127.22)}\end{array}$
$\begin{array}{cc}\begin{array}{cc}\left(\text{23}\right)& {k}_{\text{2C}}\end{array}=2\text{}\text{exp}\left[\frac{{\mathit{\text{E}}}_{\text{2}}}{\mathit{\text{R}}}(\frac{1}{300}\frac{1}{\mathit{\text{T}}})\right]{\left({\text{dm}}^{3}\text{/}\text{}\text{mol}\right)}^{4}/\text{min}& \text{(E127.23)}\end{array}$
(24) C_{A0} = 0.2 mol/dm^{3}
(25) R= 1.987 cal/mol/K
(26) E_{1} = 8,000 cal/mol
(27) E_{2} = 12,000 cal/mol
Other parameters are given in the problem statement, that is, Equations (28) to (34).
(28) Cp_{A}, (29) Cp_{B}, (30) Cp_{c}, (31) ${\dot{\text{m}}}_{c}$, (32) $\text{}\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx1B}}^{\circ}$,
(33) $\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx2A}}^{\circ}$, (34) Cp_{C0}
6. Energy Balance:
Recalling Equation (1237)
$\begin{array}{cc}\begin{array}{cc}\left(\begin{array}{c}35\end{array}\right)& \frac{d\mathit{\text{T}}\text{}}{d\mathit{\text{V}}}=\frac{{\mathit{\text{Q}}}_{g}{\mathit{\text{Q}}}_{r}}{\text{}\mathrm{\Sigma}{\mathit{\text{F}}}_{j}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{j}}}\end{array}& \text{(E127.35)}\end{array}$
The denominator of Equation (E127.35) is
$\begin{array}{cc}\begin{array}{cc}\left(36\right)& \mathrm{\Sigma}{F}_{j}\end{array}{C}_{{p}_{j}}={F}_{\mathit{\text{A}}}{C}_{{p}_{\mathit{\text{A}}}}+{F}_{\mathit{\text{B}}}{C}_{{p}_{\mathit{\text{B}}}}+{F}_{\mathit{\text{C}}}{C}_{{p}_{\mathit{\text{C}}}}+{F}_{D}{C}_{{p}_{\mathit{\text{D}}}}& \left(\text{E127.36}\right)\end{array}$
The “heat removed” term is
$\begin{array}{cc}\begin{array}{cc}\left(37\right)& {Q}_{r}=Ua(T{T}_{a})\end{array}& \left(\text{E127.37}\right)\end{array}$
The “heat generated” is
$\begin{array}{cc}\begin{array}{cc}\left(38\right)& {Q}_{r}=\mathrm{\Sigma}{r}_{ij}{\mathrm{\Delta}H}_{\text{Rxij}}={r}_{1\mathit{\text{B}}}{\mathrm{\Delta}H}_{\text{Rx1B}}+{r}_{\text{2B}}{\mathrm{\Delta}H}_{\text{Rx2B}}\end{array}& \left(\text{E127.38}\right)\end{array}$
(a) Cocurrent heat exchange
The heat exchange balance for cocurrent exchange is
$\begin{array}{cc}\begin{array}{cc}\left(\text{39}\right)& \frac{{d\mathit{\text{T}}}_{a}}{d\mathit{\text{V}}}=\frac{\mathit{\text{U}}a(\mathit{\text{T}}{\mathit{\text{T}}}_{a})}{{\dot{\text{m}}}_{\mathit{\text{C}}}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\text{C0}}}}\end{array}& \text{(E127.39)}\end{array}$
Part (a) Cocurrent flow: Plot and analyze the molar flow rates, and the reactor and coolant temperatures as a function of reactor volume. The Polymath code and output for cocurrent flow are shown in Table E127.1.
Cocurrent heat exchange
Solution
TABLE E127.1 POLYMATH PROGRAM AND OUTPUT FOR COCURRENT EXCHANGE
Figure E127.1 Profiles for cocurrent heat exchange; (a) temperature (b) molar flow rates. Note: The molar flow rate FD is very small and is essentially the same as the bottom axis.
The temperature and molar flow rate profiles are shown in Figure E127.1.
Analysis: Part (a): We also note that the reactor temperature, T, increases when Q_{g} > Q_{r} and reaches a maximum, T = 886 K, at approximately V = 6 dm^{3}. After that, Q_{r} > Q_{g} the reactor temperature decreases and approaches T_{a} at the end of the reactor. For cocurrent heat exchange, the selectivity ${\tilde{\mathit{\text{S}}}}_{\text{C/D}}=\frac{4.59}{0.0035}=1258$ is really quite good.
Part (b) Countercurrent heat exchange: Solution: We will use the same program as part (a), but will change the sign of the heatexchange balance and guess T_{a} at V = 0 to be 507 K and see whether the value gives us a T_{a0} of 325 K at V = V_{f}.
$\frac{{d\mathit{\text{T}}}_{a}}{d\mathit{\text{V}}}=\frac{\mathit{\text{U}}a(\mathit{\text{T}}{\mathit{\text{T}}}_{a})}{{\dot{\text{m}}}_{\mathit{\text{C}}}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\text{Cool}}}}$
We find our guess of 507 K matches T_{a0} = 325 K. Are we lucky or what?!
The Polymath Program is shown in Table E127.2.
Countercurrent heat exchange
TABLE E127.2 POLYMATH PROGRAM AND OUTPUT FOR COUNTERCURRENT EXCHANGE
Figure E127.2 Profiles for countercurrent heat exchange; (a) temperature (b) molar flow rates.
Analysis: Part (b): For countercurrent exchange, the coolant temperature reaches a maximum at V = 1.3 dm^{3} while the reactor temperature reaches a maximum at V = 2.1 dm^{3}. The reactor with a countercurrent exchanger reaches a maximum reactor temperature of 1100 K, which is greater than that for the cocurrent exchanger, (i.e., 930 K). Consequently, if there is a concern about additional side reactions occurring at this maximum temperature of 1100 K, one should use a cocurrent exchanger or if possible use a large coolant flow rate to maintain constant T_{a} in the exchanger. In Figure 127.2(a) we see that the reactor temperature approaches the coolant entrance temperature at the end of the reactor. The selectivity for the countercurrent systems, ${\tilde{\mathit{\text{S}}}}_{\text{C/}\mathit{\text{D}}}=1181$, is slightly lower than that for the cocurrent exchange.
Part (c) Constant T_{a}: Solution: To solve the case of constant heatingfluid temperature, we simply multiply the righthand side of the heatexchanger balance by zero, that is,
$\text{}\frac{{d\mathit{\text{T}}}_{a}}{d\mathit{\text{V}}}=\frac{\mathit{\text{U}}a(\mathit{\text{T}}{\mathit{\text{T}}}_{a})}{{\dot{\text{m}}}_{\mathit{\text{C}}}{\mathit{\text{C}}}_{\mathit{\text{P}}}}\text{*}0$
and use Equations (E127.1)–(E127.39). The Polymath Progam is shown in Table E127.3 and the temperature and molar flow rate profiles are shown in Figures E127.3(a) and E127.3(b), respectively.
Constant T_{a}
TABLE E127.3 POLYMATH PROGRAM AND OUTPUT FOR CONSTANT T_{a}
Figure E127.3 Profiles for constant T_{a}; (a) temperature (b) molar flow rates.
Analysis: Part (c): For constant T_{a}, the maximum reactor temperature, 837 K, is less than either cocurrent or countercurrent exchange while the selectivity, ${\tilde{\mathit{\text{S}}}}_{\text{C/D}}=1861$ is greater than either cocurrent or countercurrent exchange. Consequently, one should investigate how to achieve sufficiently high mass flow of the coolant in order to maintain constant T_{a}.
Part (d) Adiabatic: To solve for the adiabatic case, we simply multiply the overall heat transfer coefficient by zero.
Ua = 80*0
The Polymath Program is shown in Table E127.4 while the temperature and molar flow rate profiles are shown in Figures E127.4(a) and E127.4(b), respectively.
Adiabatic operation
TABLE E127.4 POLYMATH PROGRAM AND OUTPUT FOR ADIABATIC OPERATION
Figure E127.4 Profiles for adiabatic operation.
Analysis: Part (d): For the adiabatic case, the maximum temperature, which is the exit temperature, is higher than the other three exchange systems, and the selectivity is the lowest with ${\tilde{\mathit{\text{S}}}}_{\text{C/D}}=\text{700}$. At this high temperature, the occurrence of unwanted side reactions certainly is a concern.
Overall Analysis Parts (a) to (d): Suppose the maximum temperature in each of these cases is outside the safety limit of 750 K for this system. Problem P121_{A} (i) asks how you can keep the maximum temperature below 750 K.
In the previous sections, we have assumed that there were no radial variations in velocity, concentration, temperature, or reaction rate in the tubular and packedbed reactors so that the axial profiles could be determined using an ordinary differential equation (ODE) solver. However, we consider both radial and axial variations in our mole and energy balance, we obtain partial differential equations (PDEs), which cannot be solved with Polymath and Wolfram. Consequently, it was decided to move this discussion on radial variation to Chapter 18 where we use the PDE solver COMSOL to explore twodimensional problems of this type.
^{†} Also see http://umich.edu/~safeche.
The Process Safety Across the Chemical Engineering Curriculum Web site (http://umich.edu/~safeche/index.html) provides a safety module for every core course in chemical engineering. A module consists of viewing a Chemical Safety Board (CSB) video, carrying out a safety analysis of the incident and doing a calculation related to a specific course. The Web site includes tutorials on a number of safety algorithms such as the BowTie Diagram (Section 6.7), the NFPA Diamond (Section 2.7), and the process Safety Pyramid (Section 9.5). This Web site was developed by the author of this text and the tutorials are also given at the end of the various chapters in a section titled “A Word From Our Sponsor—Safety.” The CRE safety modules include the explosions at Monsanto, Synthron, and T2 Laboratories. The CRE and safety Web sites are open to all as they have no passwords or other to be accessed.
Scaling up exothermic chemical reactions can be very tricky. Tables 125 and 126 give reactions that have resulted in accidents and their causes, respectively.^{7} The reader should review the case histories of these reactions to learn how to avoid similar accidents.
^{7} Courtesy of J. Singh, Chemical Engineering, 92 (1997) and B. Venugopal, Chemical Engineering, 54 (2002).
TABLE 125 INCIDENCE OF BATCH PROCESS ACCIDENTS
Process Type  Number of Incidents in United Kingdom, 1962–1987 

Polymerization 
64 
Nitration 
15 
Sulfurization 
13 
Hydrolysis 
10 
Salt formation 
8 
Halogenation 
8 
Alkylation (FriedelCrafts) 
5 
Amination 
4 
Diazolization 
4 
Oxidation 
2 
Esterification 
1 
Total: 
134 
Source: Courtesy of J. Singh, Chemical Engineering, 92 (1997).
TABLE 126 CAUSES OF BATCH REACTOR ACCIDENTS IN TABLE 125
Cause  Contribution, % 

Lack of knowledge of reaction chemistry 
20 
Problems with material quality 
9 
Temperaturecontrol problems 
19 
Agitation problems 
10 
Mischarging of reactants or catalyst 
21 
Poor maintenance 
15 
Operator error 
5 
Source: B. Venugopal, Chemical Engineering, 54 (2002).
Runaway reactions are the most dangerous in reactor operation, and a thorough understanding of how and when they could occur is part of the chemical reaction engineer’s responsibility. The reaction in Example 127 could be thought of as running away. Recall that as we moved down the length of the reactor, none of the cooling arrangements could keep the reactor from reaching an extremely high temperature (e.g., 800 K). In Chapter 13, we study case histories of two runaway reactions. One is the nitroaniline explosion discussed in Example E132 and the other is Example E136, concerning the explosion at T2 Laboratories. See Example 137 and videos on T2 Laboratories’ explosion and what caused it to happen (https://www.csb.gov/t2laboratoriesincreactivechemicalexplosion/ and https://www.youtube.com/watch?v=C561PCq5E1g).
There are many resources available for additional information on reactor safety and the management of chemical reactivity hazards. Guidelines for managing chemical reactivity hazards and other fire, explosion, and toxic release hazards are developed and published by the Center for Chemical Process Safety (CCPS) of the American Institute of Chemical Engineers. CCPS books and other resources are available at www.aiche.org/ccps. For example, the book Essential Practices for Managing Chemical Reactivity Hazards, written by a team of industry experts, is also provided free of charge by CCPS on the site at https://app.knovel.com/web/toc.v/cid:kpEPMCRH02/viewerType:toc/. A concise and easytouse software program that can be used to determine the reactivity of substances or mixtures of substances, the Chemical Reactivity Worksheet, is provided by the National Oceanic and Atmospheric Administration (NOAA) for free on its Web site, www.noaa.gov.
The Safety and Chemical Engineering Education (SAChE) program was formed in 1992 as a cooperative effort between the AIChE, CCPS, and engineering schools to provide teaching materials and programs that bring elements of process safety into the education of undergraduate and graduate students studying chemical and biochemical products and processes. The SAChE Web site (www.sache.org) has a great discussion of reactor safety with examples as well as information on reactive materials. These materials are also suitable for training purposes in an industrial setting.
The following instruction modules are available on the SAChE Web site (www.sache.org).
Chemical Reactivity Hazards: This Webbased instructional module contains about 100 Web pages with extensive links, graphics, videos, and supplemental slides. It can be used either for classroom presentation or as a selfpaced tutorial. The module is designed to supplement a junior or senior chemical engineering course by showing how uncontrolled chemical reactions in industry can lead to serious harm, and by introducing key concepts for avoiding unintended reactions and controlling intended reactions.
Runaway Reactions: Experimental Characterization and Vent Sizing: This instruction module describes the ARSST and its operation, and illustrates how this instrument can easily be used to experimentally determine the transient characteristics of runaway reactions, and how the resulting data can be analyzed and used to size the relief vent for such systems. The theory along with an example behind the ARSST is given on the CRE Web site under Additional Material for Chapter 13 and the LEPs for Chapter 13, that is, PRS Example R131.
Rupture of a Nitroaniline Reactor: This case study demonstrates the concept of runaway reactions and how they are characterized and controlled to prevent major losses.
Seveso Accidental Release Case History: This presentation describes a widely discussed case history that illustrates how minor engineering errors can cause significant problems; problems that should not be repeated. The accident was in Seveso, Italy, in 1976. It was a small release of a dioxin that caused many serious injuries.
Note: Access is restricted to SAChE members and universities.
Membership in SAChE is required to view these materials. Virtually all U.S. universities and many nonU.S. universities are members of SAChE—contact your university SAChE representative, listed on the SAChE Web site, or your instructor or department chair to learn your university’s username and password. Companies can also become members—see the SAChE Web site for details.
Certificate Program
SAChE also offers several certificate programs that are available to all chemical engineering students. Students can study the material, take an online test, and receive a certificate of completion. The following two certificate programs are of value for reaction engineering:
Runaway Reactions: This certificate focuses on managing chemical reaction hazards, particularly runaway reactions.
Chemical Reactivity Hazards: This is a Webbased certificate that provides an overview of the basic understanding of chemical reactivity hazards.
Many students are taking the certificate test online and put the fact that they successfully obtained the certificate on their résumés.
More information on safety is given in the Summary Notes and Professional Reference Shelf on the Web. Particularly study the use of the ARSST to detect potential problems. These will be discussed in Chapter 13 Professional Reference Shelf R13.1 on the CRE Web site.
1. For single reactions, the energy balance on a PFR/PBR in terms of molar flow rate is
$\begin{array}{c}\hline \frac{d\mathit{\text{T}}}{d\mathit{\text{V}}}=\frac{\left({\text{r}}_{\mathit{\text{A}}}\right)\left[\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}\left(\mathit{\text{T}}\right)\right]\mathit{\text{U}}a(\mathit{\text{T}}{\mathit{\text{T}}}_{a})}{\mathrm{\Sigma}{\mathit{\text{F}}}_{i}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{i}}}=\frac{{\mathit{\text{Q}}}_{g}{\mathit{\text{Q}}}_{r}}{\text{}\mathrm{\Sigma}{\mathit{\text{F}}}_{i}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{i}}}\\ \hline\end{array}\begin{array}{c}\text{(S121)}\end{array}$
In terms of conversion
$\begin{array}{c}\hline \frac{d\mathit{\text{T}}}{d\mathit{\text{V}}}=\frac{\left({\text{r}}_{\mathit{\text{A}}}\right)\left[\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}\left(\mathit{\text{T}}\right)\right]\mathit{\text{U}}a(\mathit{\text{T}}{\mathit{\text{T}}}_{a})}{{\mathit{\text{F}}}_{\text{A0}}(\mathrm{\Sigma}{\mathrm{\Theta}}_{j}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{i}}+\mathit{\text{X}}\mathrm{\Delta}{\mathit{\text{C}}}_{\mathit{\text{P}}})}=\frac{{\mathit{\text{Q}}}_{g}{\mathit{\text{Q}}}_{r}}{{\mathit{\text{F}}}_{\text{A0}}(\mathrm{\Sigma}{\mathrm{\Theta}}_{j}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{j}}+\mathit{\text{X}}\mathrm{\Delta}{\mathit{\text{C}}}_{\text{p}})}\\ \hline\end{array}\begin{array}{c}\text{(S122)}\end{array}$
2. The temperature dependence of the specific reaction rate is given in the form
$\begin{array}{c}\hline k\left(\mathit{\text{T}}\right)=k\left({\mathit{\text{T}}}_{1}\right)\text{exp}\left[\frac{\mathit{\text{E}}}{\mathit{\text{R}}}(\frac{1}{{\mathit{\text{T}}}_{1}}\frac{1}{\mathit{\text{T}}})\right]=k{\text{\hspace{0.17em}}\text{(}\mathit{\text{T}}}_{1}\text{)}\text{}\text{exp}\left[\text{}\frac{\mathit{\text{E}}}{\mathit{\text{R}}}\left(\frac{\mathit{\text{T}}{\mathit{\text{T}}}_{1}}{{\text{TT}}_{1}}\right)\right]\\ \hline\end{array}\begin{array}{c}\text{(S123)}\end{array}$
3. The temperature dependence of the equilibrium constant is given by van’t Hoff’s equation for ΔC_{p} = 0
$\begin{array}{c}\hline {\mathit{\text{K}}}_{\mathit{\text{P}}}\left(\mathit{\text{T}}\right)={\mathit{\text{K}}}_{\mathit{\text{P}}}\left({\mathit{\text{T}}}_{\text{2}}\right)\text{exp}\left[\frac{\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ}}{\mathit{\text{R}}}(\frac{1}{{\mathit{\text{T}}}_{2}}\frac{1}{\mathit{\text{T}}})\right]\\ \hline\end{array}\begin{array}{c}\text{(S124)}\end{array}$
4. Neglecting changes in potential energy, kinetic energy, and viscous dissipation, and for the case of no work done on or by the system, large coolant flow rates (m_{c}), and all species entering at the same temperature, the steadystate CSTR energy balance for single reactions is
$\begin{array}{c}\hline \frac{\text{UA}}{{\mathit{\text{F}}}_{\text{A0}}}({\mathit{\text{T}}}_{a}\mathit{\text{T}})\mathit{\text{X}}\left[\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ}\left({\mathit{\text{T}}}_{\mathit{\text{R}}}\right)+\mathrm{\Delta}{\mathit{\text{C}}}_{\mathit{\text{P}}}(\mathit{\text{T}}{\mathit{\text{T}}}_{\mathit{\text{R}}})\right]=\mathrm{\Sigma}{\mathrm{\Theta}}_{j}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{j}}(\mathit{\text{T}}{\mathit{\text{T}}}_{i0})\\ \hline\end{array}\begin{array}{c}\text{(S125)}\end{array}$
5. Multiple steady states
$\begin{array}{c}\hline \mathit{\text{G}}\left(\mathit{\text{T}}\right)=(\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ})\left(\frac{{\text{r}}_{\mathit{\text{A}}}\mathit{\text{V}}}{{\mathit{\text{F}}}_{\text{A0}}}\right)=(\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ})\left(\mathit{\text{X}}\right)\\ \hline\end{array}\begin{array}{c}\text{(S126)}\end{array}$
$\begin{array}{c}\hline \mathit{\text{R}}\left(\mathit{\text{T}}\right)={\mathit{\text{C}}}_{\text{P0}}(1+\kappa )(\mathit{\text{T}}{\mathit{\text{T}}}_{c})\\ \hline\end{array}\begin{array}{c}\text{(S127)}\end{array}$
where $\text{}\kappa =\frac{\text{UA}}{{\mathit{\text{C}}}_{{\text{p}}_{0}}{\mathit{\text{F}}}_{\text{A0}}}{\text{and T}}_{c}=\frac{\kappa {\mathit{\text{T}}}_{a}+{\mathit{\text{T}}}_{0}\text{}}{1+\kappa}$
6. When q multiple reactions are taking place and there are m species
$\begin{array}{c}\hline \frac{d\mathit{\text{T}}}{d\mathit{\text{V}}}=\frac{\underset{i=1}{\overset{\text{q}}{\mathrm{\Sigma}}}\left({\text{r}}_{\mathit{\text{ij}}}\right)\left[\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}ij}\left(\mathit{\text{T}}\right)\right]\mathit{\text{U}}a(\mathit{\text{T}}{\mathit{\text{T}}}_{a})}{\underset{j=1}{\overset{\text{m}}{\mathrm{\Sigma}}}{\mathit{\text{F}}}_{j}{\mathit{\text{C}}}_{\mathit{\text{P}}j}}=\frac{{\mathit{\text{Q}}}_{g}{\mathit{\text{Q}}}_{r}}{\underset{j=1}{\overset{\text{m}}{\mathrm{\Sigma}}}\text{}{\mathit{\text{F}}}_{j}{\mathit{\text{C}}}_{\mathit{\text{P}}j}}\\ \hline\end{array}\begin{array}{c}\text{(S128)}\end{array}$
Interactive Computer Games that Talk to Each Other
Heat Effects I (http://www.umich.edu/~elements/6e/icm/heatfx1.html)
Heat Effects II (http://www.umich.edu/~elements/6e/icm/heatfx2.html)
The derivation of the userfriendly energy balances are color coded and derived in the tutorials in Heat Effects I and Heat Effects II. Here you see animation of the various terms as they move around the screen and talk to each other to arrive at the final form of the balance equation.
A stepbystep AspenTech tutorial is given on the CRE Web site.
See Example 122 (http://www.umich.edu/~elements/6e/software/aspenexample122.html). Formulated in AspenTech: Download AspenTech directly from the CRE Web site.
The subscript to each of the problem numbers indicates the level of difficulty: A, least difficult; D, most difficult.
In each of the questions and problems, rather than just drawing a box around your answer, write a sentence or two describing how you solved the problem, the assumptions you made, the reasonableness of your answer, what you learned, and any other facts that you want to include. See Preface Section G.2 for additional generic parts (x), (y), and (z) to the home problems.
Before solving the problems, state or sketch qualitatively the expected results or trends.
Q121_{A} QBR (Question Before Reading). Before reading this chapter, identify the differences in applying the energy balance for cocurrent heat exchange and counter current heat exchange.
Q122_{A} i>clicker. Go to the Web site (http://www.umich.edu/~elements/6e/12chap/iclicker_ch12_q1.html) and view five i>clicker questions. Choose one that could be used as is, or a variation thereof, to be included on the next exam. You also could consider the opposite case: explaining why the question should not be on the next exam. In either case, explain your reasoning.
Q123_{A} Review Figure 1213. Use this figure to write a few sentences (or at least draw an analogy) explaining why, when you strike the head of a safety match slowly on its pumice with little pressure, it may heat up a little, but does not ignite, yet when you put pressure on it and strike it rapidly, it does ignite. Thanks to Oscar Piedrahita, Medellín, Colombia.
Q124_{A} Read over the problems at the end of this chapter. Make up an original problem that uses the concepts presented in this chapter. To obtain a solution:
(a) Make up your data and reaction.
(b) Use a real reaction and real data. See Problem P51_{A} for guidelines.
(c) Prepare a list of safety considerations for designing and operating chemical reactors. (See www.sache.org.) The August 1985 issue of Chemical Engineering Progress may be useful for part (c).
Q125_{A} Go to the LearnChemE screencast link for Chapter 12 (http://www.umich.edu/~elements/6e/12chap/learnchemevideos.html). View one or more of the screencast 5 to 6minute videos and write a twosentence evaluation. What would you suggest changing in the video to increase the comment from Recommended to Highly Recommended?
Q126_{A} What did the section And Now… A Word From Our Sponsor point out or emphasize that was not done in the other AWFOS sections? What were the top two takeaway points in Chapter 12’s AWFOS?
We will use the Living Example on the CRE Web site extensively to carry out simulations. Why carry out simulations to vary the parameters in the Living Example Problems? We do it in order to
Get a more intuitive feel reactor system.
Gain insight about the most sensitive parameters (e.g., E, K_{C}) and how they affect outlet conditions.
Learn how reactors are affected by different operating conditions.
Simulate dangerous situations such as potential runaway reactions.
Compare the model and parameters with experimental data.
Optimize the reaction system.
P121_{B} (a) LEP Table 122: Exothermic Reaction with Heat Exchange
Download the Polymath, MATLAB, Python, or Wolfram codes for the algorithm and data given in Table T122 for the exothermic gas phase reversible reaction
A + B ⇄ C
given on the Web in the Living Example Problems (LEPs).
Vary the following parameters in the ranges shown in parts (i) through (xi). Write a paragraph describing the trends you find for each parameter variation and why they look the way they do. Use the base case for parameters not varied. The feedback from students on this problem is that one should use Wolfram on the Web in the LEP T122 as much as possible to carry out the parameter variations. For each part, write two or more sentences describing the trends.
Wolfram and Python
(i) This is a Stop and Smell the Roses Simulation. View the base case X, X_{e}, and T profiles and explain why the conversion (X and X_{e}) and temperature profiles look the way they do.
(ii) The slider parameters Ua/rho, Θ_{I}, F_{A0}, and ΔH_{Rx} all can be moved to make the X and X_{e} profiles to come together and to separate. What is the reason these sliders affect the profiles similarly?
(iii) Starting with the base case, determine which parameters when changed only a small amount most dramatically affects the conversion and temperature profiles.
(iv) What parameters separate X and X_{e} the most?
(v) Finally, write at least three conclusions about what you found in your experiments (i) through (iv).
Polymath
(vi) Vary T_{0}: 310 K ≤ T_{0} ≤ 350 K and write a conclusion.
(vii) Vary T_{a}: 300 K ≤ T_{a} ≤ 340 K and write a conclusion.
Repeat (i) for countercurrent coolant flow.
Hint: In analyzing the trends it might be helpful to plot X, X_{e}, and p as a function of W on the same graph and T and T_{a} as a function of W on the same graph.
(viii) Repeat this problem for the case of constant T_{a} and adiabatic operation, and describe the most dramatic affect you find.
(b) Example 121: Butane Isomerization
Wolfram and Python CoCurrent
This is another Stop and Smell the Roses Simulation.
(i) Describe the differences of the X, X_{e}, and T profiles between the four cases of heat exchange, cocurrent, counter current, constant T_{a}, and adiabatic.
(ii) Explain why the temperature (T and T_{a}) and conversion (X and X_{e}) look the way they do, and explain what happens to the X, X_{e}, T, and T_{a} profiles as you move slider y_{A0}.
(iii) What parameter value brings T and T_{a} profiles close together?
(iv) What parameter when varied separates X and X_{e} the most?
(v) What parameter keeps the X and X_{e} profiles the closest together?
(vi) Write at least three conclusions about what you found in your experiments (i) through (v).
Polymath CoCurrent
(vii) What is the entering value of the temperature T_{a}_{0} of the heat exchanger fluid below which the reaction will never “ignite”?
(viii) Vary some of the other parameters and see whether you can find unsafe operating conditions.
(ix) Plot Q_{r} and T_{a} as a function of V necessary to maintain isothermal operation.
Wolfram and Python Countercurrent
(x) Explain why T, T_{a}, X, and X_{e} profiles look the way they do for the base case.
(xi) Vary Ua and T_{a}_{0} and describe what you find.
(xii) Vary y_{A0} and then one of the other parameters and describe what you find.
(xiii) Describe how Q_{r} and Q_{g} and their intersection change when you vary the molar flow rates of coolant, inert and y_{A0}.
(xiv) Write at least three conclusions about what you found in your experiments (x) through (xiii).
Polymath Countercurrent
(xv) Describe and explain what happens to the X and X_{e} profiles as you vary ΔH_{Rx}.
(xvi) Describe what happens to the temperature profile T and T_{a} as you vary F_{A0}.
(xvii) Compare the variations in the profiles for X, X_{e}, T, and T_{a} when you change the parameter values for all four cases: adiabatic, countercurrent exchange, cocurrent exchange, and constant T_{a}. What parameters when changed only a small amount dramatically change the profiles?
(xviii) The heat exchanger is designed for a maximum temperature of 370 K. Which parameter will you vary so that at least 75% conversion is still achieved while maintaining temperature under a safe limit?
Wolfram and Python Constant T_{a}
(xix) Vary y_{A0} and one other parameter of your choice, and describe how the X, X_{e}, and T profiles change.
(xx) Vary Ua and T_{a} and describe what you find.
(xxi) Vary ${\mathrm{\Delta}H}_{\text{Rx}}^{\xb0}$ and then one of the other parameters (e.g., y_{A0}) and describe what you find.
(xxii) Write at least three conclusions about what you found in your experiments (xix)–(xxi).
Wolfram and Python Adiabatic Operation
(xxiii) Why do the shape of the profiles change in the way they do as you change y_{A0}?
(xxiv) Write a set of three conclusions, one of which compares adiabatic operation with the other three modes of heat transfer (e.g., cocurrent).
(c) Example 122: Production of Acetic AnhydrideEndothermic Reaction Wolfram and Python
CoCurrent
(i) Explain why the temperatures (T and T_{a}) and conversions (X and X_{e}) look the way they do for the base case slider values.
(ii) What parameter value when increased or decreased causes the reaction to die out the most quickly near the reactor entrance?
(iii) Which parameter, when varied most, drastically changes the profiles?
Adiabatic
(iv) How does the conversion change as heat capacity of A is increased?
Constant T_{a}
(v) You find that conversion has decreased after 6 months of operation. You checked the flow rate and material properties and found that these values have not changed. Which parameter would have changed?
Countercurrent
(vi) Explain why the temperatures (T and T_{a}) and conversions (X and X_{e}) look the way they do for the base case slider values.
(vii) Discuss the most profound differences between the four heat exchange modes (e.g., cocurrent, adiabatic) for this endothermic reaction.
(viii) Write two conclusions of what you learned from your experiments (i) through (vii).
Polymath
(ix) Let Q_{g} = r_{A}ΔH_{Rx} and Q_{r} = Ua (T – T_{a}), and then plot Q_{g} and Q_{r} on the same figure as a function of V.
(x) Repeat (vi) for V = 5 m^{3}.
(xi) Plot Q_{g}, Q_{r}, and –r_{A} versus V for all four cases on the same figure and describe what you find.
(xii) For each of the four heat exchanger cases, investigate the addition of an inert I with a heat capacity of 50 J/mol · K, keeping F_{A0} constant and letting the other inlet conditions adjust accordingly (e.g., ε).
(xiii) Vary the inert molar flow rate (i.e., Θ_{I}, 0.0 < Θ_{I} < 3.0 mol/s). Plot X and analyze versus Θ_{I}.
(xiv) Finally, vary the heatexchange fluid temperature T_{a}_{0} (1000 K < T_{a}_{0} < 1350 K). Write a paragraph describing what you find, noting interesting profiles or results.
(d) Example 122: AspenTech Formulation. Go to the Aspen LEP and repeat Example 122 using AspenTech.
(e) Example 123: Production of Propylene Glycol in a CSTR
Wolfram and Python
(i) Vary activation energy (E) and find the values of E for which there are at least two solutions.
(ii) Vary the flow rate of F_{B0} to find the temperature at which the conversion is 0.8.
(iii) What is the operating range of inlet temperatures such that at least one steady state solution exists while maintaining the reactor temperature below 640°R? Describe how your answers would change if the molar flow rate of methanol were increased by a factor of 4.
(iv) Vary V and find the points of (1) tangency and (2) intersection of X_{EB} and X_{MB}.
(v) Write a set of conclusions based on your experiments (i) through (iv).
(f) Example 124: CSTR with a Cooling Coil
Wolfram and Python
(i) Explore how variations in the activation energy (E) and in the heat of reaction, at reference temperature ($\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ}$), affect the conversions X_{EB} and X_{MB}.
(ii) What variables most affect the intersection of and the points of tangency of X_{EB} and X_{MB}?
(iii) Find a value of E and of $\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ}$ that increase conversion to more than 80% while maintaining the constraint of 125°F maximum temperature.
(iv) Write two conclusions about what you found in your experiments (i) through (iii).
Polymath
(v) The space time was calculated to be 0.01 minutes. Is this realistic? Decrease ν_{0} and both reaction rate constants by a factor of 100 and describe what you find.
(vi) Use Figure E123.2 and Equation (E124.4) to plot X versus T to find the new exit conversion and temperature.
(vii) Other data on the Jofostan chemical engineering Web site show $\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ}=\text{38700}$ Btu/lbmol and ΔC_{P} = 29 Btu/lbmol/°F. How would these values change your results?
(viii) Make a plot of conversion as a function of heat exchanger area. [0 < A < 200 ft^{2}] and write a conclusion.
(g) Example 125: Parallel Reactions in a PFR with Heat Effects
Wolfram and Python
(i) Vary the sliders for the rate constants k_{1A0} and k_{2A0} and describe what you find.
(ii) What should be the initial concentration of A so that the company can produce equal flow rates of B and C?
(iii) Describe how the profiles for T, F_{A}, F_{B}, and F_{C} change as you move the sliders. Comment specifically on the drastic changes that occur when you change the overall heat transfer coefficient.
(iv) What set of conditions give you the greatest selectivity, ${\tilde{\mathit{\text{S}}}}_{\text{B/C}}$ at the exit and what is the corresponding conversion.
(v) Write at least three conclusions about what you found in your experiments (i) through (iv).
Polymath
(vi) Why is there a maximum in temperature? How would your results change if there is a pressure drop with α = 1.05 dm^{–3}?
(vii) What if the reaction is reversible with K_{C}= 10 at 450 K?
(viii) How would the selectivity change if Ua is increased? Decreased?
(h) Example 126: Multiple Reactions in a CSTR (Use the LEP)
Wolfram and Python
(i) What is the minimum value of T_{a} that causes the reactor temperature to jump to the single, upper steadystate value?
(ii) What is the operating range for the entering temperature T_{0} for which there are only three steady states?
(iii) Vary UA between 10,000 and 80,000, and describe how the number of steadystate solutions changes.
(iv) Write a set of conclusions from carrying out experiments (i) through (iii).
(i) Example 127: Complex Reactions with Heat Effects in a PFR—Safety
Wolfram and Python
(i) Cocurrent: Explore the effect of the coolant flow rate ${\dot{\text{m}}}_{c}$ on the reactant temperature and coolant temperature profiles, and describe what you find.
(ii) Cocurrent: Explore the effect of Ua on the selectivity and describe what you find.
(iii) Constant T_{a}: How does the overall heat transfer coefficient affect the product flow rate distribution? Should it be kept minimum or maximum? Explain.
(iv) Constant T_{a}: Which are the variables that have virtually no effect on the profiles? Explain the reason.
(v) Adiabatic: You want to save capital cost by using a smaller volume reactor, that is, 5 dm^{3} instead of 10 dm^{3}. Which parameters will you vary to achieve exit reactant and product flow rates same as base case?
(vi) Describe how the selectivity, ${\tilde{\mathit{\text{S}}}}_{\text{C/D}}$ changes as you change the parameters and operating conditions. What two variable slider(s) affect ${\tilde{\mathit{\text{S}}}}_{\text{C/D}}$ the most?
(vii) Write a set of conclusions of what you learned in experiments (i) through (vi).
Polymath
(viii) Plot Q_{g} and Q_{r} as a function of V. How can you keep the maximum temperature below 700 K? Would adding inerts help, and if so what should the flow rate be if C_{PI} = 10 cal/mol/K?
(ix) Look at the figures. What happened to species D? What conditions would you suggest to make more species D?
(x) Make a table of the temperature (e.g., maximum T, T_{a}) and molar flow rates at two or three volumes, comparing the different heatexchanger operations.
(xi) Why do you think the molar flow rate of C does not go through a maximum? Vary some of the parameters to learn whether there are conditions where it goes through a maximum. Start by increasing F_{A0} by a factor of 5.
(xii) Include pressure in this problem. Vary the pressuredrop parameter (0 < αρ_{b} < 0.0999 dm^{–3}) and describe what you find and write a conclusion.
(j) Review the steps and procedure by which we derived Equation (125) and then, by analogy, derive Equation (1235).
(k) CRE Web site SO_{2} Example PRSR12.41. Download the SO_{2} oxidation LEP R121. How would your results change if (1) the catalyst particle diameter were cut in half? (2) The pressure were doubled? At what particle size does pressure drop become important for the same catalyst weight, assuming the porosity doesn’t change? (3) You vary the initial temperature and the coolant temperature? Write a paragraph with at least two conclusions describing what you find.
(l) SAChE. Go to the SAChE Web site, www.sache.org. Your instructor or department chair should have the username and password to enter the SAChE Web site in order to obtain the modules with the problems. On the lefthand menu, select “SAChE Products.” Select the “All” tab and go to the module titled, “Safety, Health and the Environment” (S, H & E). The problems are for KINETICS (i.e., CRE). There are some example problems marked “K” and explanations in each of the above S, H & E selections. Solutions to the problems are in a different section of the site. Specifically look at: Loss of Cooling Water (K1), Runaway Reactions (HT1), Design of Relief Values (D2), Temperature Control and Runaway (K4) and (K5), and Runaway and the Critical Temperature Region (K7). Go through the K problems and write a paragraph on what you have learned.
P122_{B} (a) Example 122 Production of Acetic Anhydride Endothermic Reaction
Case 1 Adiabatic
(i) Vary inlet flow rate of inert and observe the conversion and temperature profiles. Describe what you find.
(ii) Find inlet temperature, T_{0}, for which the reaction rate at V = 0.001 m^{3} is 0.1 times the rate at inlet. Note the conversion at the reactor exit for this temperature.
(iii) After varying two other parameters, write a set of conclusions from your experiments in (i) and (ii).
Case 2 Constant T_{a}
(iv) To reduce operating cost, the inlet temperature of heating fluid can be reduced. Can you find a minimum temperature of heating fluid at which exit conversion is virtually 100% for constant T_{a}.
(v) Find maximum value of the inlet concentration for which the conversion at exit is virtually 100% constant T_{a}.
(vi) Write a set of conclusions from your experiments in (iv) and (v).
P123_{B} OEQ (Old Exam Question). Computer Experiments on a PFR for Reaction in Table 122.2. In order to develop a greater understanding of the temperature effects in PBRs, download the Living Example Problem Table 122, LEP T122, from the Web site to learn how changing the different parameters changes the conversion and temperature profiles.
This is a Sherlock Holmes Problem. Starting with the base case, one, and only one, parameter was varied between its maximum and minimum values to give the following figures. You need to decide what operating variable (e.g., T_{0}) property value (e.g., C_{PA}), or engineering variable (e.g., $\frac{{\mathit{\text{U}}}_{a}}{{\rho}_{b}}$) was varied. Choose from the following:
${k}_{1},{\text{E, R, C}}_{\text{T0}},{\mathit{\text{T}}}_{a},{\mathit{\text{T}}}_{0},{\mathit{\text{T}}}_{1},{\mathit{\text{T}}}_{2},{\mathit{\text{K}}}_{\text{C2}},{\mathit{\Theta}}_{\mathit{\text{B}}},{\mathrm{\Theta}}_{1},\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ},{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{A}}}},{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{B}}}},{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{C}}}},\mathit{\text{U}}a,{\rho}_{a}\text{}$
P124_{A} Download the Interactive Computer Games (ICG) from the CRE Web site. Play the game, and then record your performance number for the module, which indicates your mastery of the material. Note: For simulation (b), only do the first three reactors, as reactors 4 and above do not work because of the technician tinkering with them.
(a) ICG Heat Effects Basketball 1 Performance # ________________.
(b) ICG Heat Effects Simulation 2 Performance # ________________.
Before attempting to play the ICG/Interactive Computer Games simulations, be sure to first go through the reviews to see the equations talking to each other.
P125_{C} OEQ (Old Exam Question). Safety Problem. The following is an excerpt from The Morning News, Wilmington, Delaware (August 3, 1977): “Investigators sift through the debris from blast in quest for the cause [that destroyed the new nitrous oxide plant]. A company spokesman said it appears more likely that the [fatal] blast was caused by another gas—ammonium nitrate—used to produce nitrous oxide.” An 83% (wt) ammonium nitrate and 17% water solution is fed at 200°F to the CSTR operated at a temperature of about 510°F. Molten ammonium nitrate decomposes directly to produce gaseous nitrous oxide and steam. It is believed that pressure fluctuations were observed in the system and, as a result, the molten ammonium nitrate feed to the reactor may have been shut off approximately 4 min prior to the explosion.
Assume that at the time the feed to the CSTR stopped, there was 500 lb_{m} of ammonium nitrate in the reactor. The conversion in the reactor is believed to be virtually complete at about 99.99%.
Additional information (approximate but close to the real case):
$\begin{array}{ccc}\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ}& =& 336{\text{Btu/lb}}_{\text{m}}{\text{ammoniumnitrateat500}}^{\circ}\text{F}\text{\hspace{0.17em}}\text{(}\text{constant)}\hfill \\ {\mathit{\text{C}}}_{\text{P}}& =& 0.38{\text{Btu/lb}}_{\text{m}}\text{ammoniumnitrate}{\cdot}^{\circ}\text{F}\hfill \\ {\mathit{\text{C}}}_{\text{P}}& =& 0.47{\text{Btu/lb}}_{\text{m}}\text{of}\text{steam}{\cdot}^{\circ}\text{F}\hfill \\ {\text{r}}_{\mathit{\text{A}}}\mathit{\text{V}}& =& k{\mathit{\text{C}}}_{\mathit{\text{A}}}\mathit{\text{V}}=k\frac{\mathit{\text{M}}}{\mathit{\text{V}}}\mathit{\text{V}}=k\mathit{\text{M}}({\text{lb}}_{\text{m}}/\text{h})\hfill \end{array}$
where M is the mass of ammonium nitrate in the CSTR (lb_{m}) and k is given by the relationship below.
T (°F) 
510 
560 
k (h^{1}) 
0.307 
2.912 
The enthalpies of water and steam are
$\begin{array}{c}{\mathit{\text{H}}}_{w}\left({200}^{\circ}\mathit{\text{F}}\right)=168{\text{Btu/lb}}_{\text{m}}\\ {\mathit{\text{H}}}_{g}\left({500}^{\circ}\mathit{\text{F}}\right)=1202{\text{Btu/lb}}_{\text{m}}\end{array}$
(a) Can you explain the cause of the blast? Hint: See Problem P133_{B}.
(b) If the feed rate to the reactor just before shutoff was 310 lb_{m} of solution per hour, what was the exact temperature in the reactor just prior to shutdown? Hint: Plot Q_{r} and Q_{g} as a function of temperature on the same plot.
(c) How would you start up or shut down and control such a reaction? Hint: See Problem P132_{B}.
(d) Explore this problem and describe what you find. For example, add a heat exchanger UA (T – T_{a}), choose values of UA and T_{a}, and then plot R(T) versus G(T)?
(e) Discuss what you believe to be the point of the problem. The idea for this problem originated from an article by Ben Horowitz.
P126_{B} OEQ (Old Exam Question)—Taken from California Professional Engineer’s Exam. The endothermic liquidphase elementary reaction
A + B → 2C
proceeds, substantially, to completion in a single steamjacketed, continuousstirred reactor (Table P126_{B}). From the following data, calculate the steadystate reactor temperature:
Reactor volume: 125 gal
Steam jacket area: 10 ft^{2}
Jacket steam: 150 psig (365.9°F saturation temperature)
Overall heattransfer coefficient of jacket, U: 150 Btu/h · ft^{2} · °F
Agitator shaft horsepower: 25 hp
Heat of reaction, $\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ}=+20000$ Btu/lbmol of A (independent of temperature)
TABLE P126_{B} FEED CONDITIONS AND PROPERTIES
Component 

A 
B 
C 

Feed (lbmol/hr) 
10.0 
10.0 
0 
Feed temperature (°F) 
80 
80 
— 
Specific heat (Btu/lbmol·°F)^{*} 
51.0 
44.0 
47.5 
Molecular weight 
128 
94 
111 
Density (lb_{m}/ft^{3}) 
63.0 
67.2 
65.0 
^{*} Independent of temperature. (Ans: T = 199°F)
(Courtesy of the California Board of Registration for Professional & land surveyors.)
P127_{B} OEQ (Old Exam Question). Use the data in Problem P114_{A} for the following reaction. The elementary, irreversible, organic liquidphase reaction
A + B → C
is carried out in a flow reactor. An equal molar feed in A and B enters at 27°C, and the volumetric flow rate is 2 dm^{3}/s and C_{A0} = 0.1 kmol/m^{3}.
Additional information:
$\begin{array}{ccc}{\mathit{\text{H}}}_{\mathit{\text{A}}}^{\circ}\left(273\mathit{\text{K}}\right)=20\text{kcal/mol,}& {\mathit{\text{H}}}_{\mathit{\text{B}}}^{\circ}\left(273\mathit{\text{K}}\right)=15\text{kcal/mol,}& {\mathit{\text{H}}}_{\mathit{\text{C}}}^{\circ}\left(273\mathit{\text{K}}\right)=41\text{kcal/mol}\\ {\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{A}}}}={\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{B}}}}=15\text{cal/mol}\cdot \mathit{\text{K}}& {\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{C}}}}=30\text{cal/mol}\cdot \mathit{\text{K}}& \text{}\\ k=0.01\frac{{\text{dm}}^{3}}{\text{mol}\cdot \text{s}}\text{at 300 K}& \mathit{\text{E}}=10000\text{cal/mol}& \text{}\\ \mathit{\text{U}}a=20{\text{cal/m}}^{3}/\text{s/}\mathit{\text{K}}& {\dot{\text{m}}}_{\mathit{\text{C}}}={50}_{g\text{/s}}& \text{}\\ {\mathit{\text{T}}}_{a0}=450\mathit{\text{K}}& {\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\text{Cool}}}=1\text{cal/g/K}& \text{}\end{array}$
(a) Calculate the conversion when the reaction is carried out adiabatically in one 500dm^{3} CSTR and then compare the results with the two adiabatic 250dm^{3} CSTRs in series.
The reversible reaction (part (d) of Problem P114_{A}) is now carried out in a PFR with a heat exchanger. Plot and then analyze X, X_{e}, T, T_{a}, Q_{r}, Q_{g}, and the rate, –r_{A}, for the following cases:
(b) Constant heatexchanger temperature T_{a}
(c) Cocurrent heat exchanger T_{a} (Ans: At V = 10 m^{3} then X = 0.36 and T = 442 K)
(d) Countercurrent heat exchanger T_{a} (Ans: At V = 10 m^{3} then X = 0.364 and T = 450 K)
(e) Adiabatic operation
(f) Make a table comparing all your results (e.g., X, X_{e}, T, T_{a}). Write a paragraph describing what you find.
(g) Plot Q_{r} and T_{a} as a function of V necessary to maintain isothermal operation.
P128_{A} The gasphase reversible reaction as discussed in Problem P117_{B}
A ⇄ B
is now carried out under high pressure in a packedbed reactor with pressure drop. The feed consists of both inerts I and species A with the ratio of inerts to the species A being 2 to 1. The entering molar flow rate of A is 5 mol/min at a temperature of 300 K and a concentration of 2 mol/dm^{3}. Work this problem in terms of volume. Hint: $\mathit{\text{V}}=\text{W/}{\rho}_{\mathit{\text{B}}},{r}_{\mathit{\text{A}}}={\rho}_{\mathit{\text{B}}}{r}_{\mathit{\text{A}}}^{\prime}$
Additional information:
$\begin{array}{cccc}{\mathit{\text{F}}}_{\text{A0}}=5.0\text{mol/min}\hfill & {\mathit{\text{T}}}_{0}=300\mathit{\text{K}}\hfill & \mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}=20000\text{cal/mol}\hfill & {\begin{array}{c}\alpha \rho \end{array}}_{\text{b}}=0.02{\text{dm}}^{3}\hfill \\ {\mathit{\text{C}}}_{\text{A0}}=2{\text{mol/dm}}^{3}\hfill & {\mathit{\text{T}}}_{1}=300\mathit{\text{K}}\hfill & {\mathit{\text{K}}}_{\text{C}}=1000\text{at300K}\hfill & \text{Coolant}\hfill \\ {\mathit{\text{C}}}_{1}=2{\mathit{\text{C}}}_{\text{A0}}\hfill & {k}_{1}=0.1{\text{min}}^{2}\text{at300K}\hfill & {\mathit{\text{C}}}_{{{\text{P}}_{\text{B}}}_{\text{}}}=160\text{cal/mol/K}\hfill & {\dot{\text{m}}}_{\mathit{\text{C}}}=50\text{mol/min}\hfill \\ {\mathit{\text{C}}}_{{\text{P}}_{\text{I}}}=18\text{cal/mol/K}\hfill & \mathit{\text{U}}a=150{\text{cal/dm}}^{3}/\text{min/K}\hfill & {p}_{\text{B}}=1.2{\text{kg/dm}}^{3}\hfill & {\mathit{\text{C}}}_{{\text{P}}_{\text{Cool}}}=20\text{cal/mol/K}\hfill \\ {\mathit{\text{C}}}_{{\text{P}}_{\text{A}}}=160\text{cal/mol/}\mathit{\text{K}}\hfill & {\mathit{\text{T}}}_{\text{}a0}=300\mathit{\text{K}}\hfill & \text{}& \\ E\text{\hspace{0.17em}}\text{= 10000cal/mol}\hfill & \mathit{\text{V}}=40{\text{dm}}^{3}\hfill & \text{}& \end{array}$
Plot and then analyze X, X_{e}, T, T_{a}, and the rate (–r_{A}) profiles in a PFR for the following cases. In each case, explain why the curves look the way they do.
(a) Cocurrent heat exchange
(b) Countercurrent heat exchange (Ans: When V = 20 dm^{3} then X = 0.86 and X_{e} = 0.94)
(c) Constant heatexchanger temperature T_{a}
(d) Compare and contrast each of the above results and the results for adiabatic operation (e.g., make a plot or a table of X and X_{e} obtained in each case).
(e) Vary some of the parameters, for example, (0 < Θ_{I} < 10) and describe what you find.
(f) Plot Q_{r} and T_{a} as a function of V necessary to maintain isothermal operation.
P129_{A} Algorithm for reaction in a PBR with heat effects and pressure drop
The elementary gasphase reaction
A + B ⇄ 2C
in Problem P118_{B} is now continued and carried out in packedbed reactor. The entering molar flow rates are F_{A0} = 5 mol/s, F_{B0} = 2F_{A0}, and F_{I} = 2F_{A0} with C_{A0} = 0.2 mol/dm^{3}. The entering temperature is 325 K and a coolant fluid is available at 300 K.
Additional information:
$\begin{array}{ccc}{\mathit{\text{C}}}_{{\text{P}}_{\text{A}}}={\mathit{\text{C}}}_{{\text{P}}_{\text{B}}}={\mathit{\text{C}}}_{{\text{P}}_{\text{C}}}=20\text{cal/mol/}\mathit{\text{K}}& k=0.0002\frac{{\text{dm}}^{6}}{\text{kg}\cdot \text{mol}\cdot \text{s}}\text{@300K}& \text{}\\ {\mathit{\text{C}}}_{{\text{P}}_{\text{A}}}=18\text{cal/mol/K}\hfill & \alpha =0.00015{\text{kg}}^{1}\hfill & \mathit{\text{U}}a=320\frac{\text{Cal}}{\text{s}\cdot {\text{m}}^{3}\text{}\text{}\cdot \mathit{\text{K}}}\hfill \\ \mathit{\text{E}}=25\frac{\text{kcal}}{\text{mol}}\hfill & {\dot{\text{m}}}_{\mathit{\text{C}}}=18\text{mol/s}\hfill & {\rho}_{b}=1400\frac{\text{kg}}{{\text{m}}^{3}}\hfill \\ \mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}=20\frac{\text{kcal}}{\text{mol}}@298\text{\hspace{0.17em}}\text{K}\hfill & {\mathit{\text{C}}}_{{\text{P}}_{\text{Cool}}}=18\text{cal/mol}\left(\text{coolant}\right)\hfill & \text{}\\ \text{}& {\mathit{\text{K}}}_{\text{C}}=1000@305\text{K}\hfill & \text{}\end{array}$
Plot X, X_{e}, T, T_{a}, and –r_{A} down the length of the PFR for the following cases:
(a) Cocurrent heat exchange
(b) Countercurrent heat exchange
(c) Constant heatexchanger temperature T_{a}
(d) Compare and contrast your results for (a), (b), and (c) along with those for adiabatic operation and write a paragraph describing what you find.
P1210_{B} Use the data and reaction in Problems P114_{A} and P127_{B} for the following reaction:
A+B → C+D
(a) Plot and then analyze the conversion, Q_{r}, Q_{g}, and temperature profiles up to a PFR reactor volume of 10 dm^{3} for the case when the reaction is reversible with K_{C} = 10 m^{3}/kmol at 450 K. Plot and then analyze the equilibrium conversion profile.
(b) Repeat (a) when a heat exchanger is added, Ua = 20 cal/m^{3}/s/K, and the coolant temperature is constant at T_{a} = 450 K.
(c) Repeat (b) for both a cocurrent and a countercurrent heat exchanger. The coolant flow rate is 50 g/s, C_{P}_{c} = 1 cal/g · K, and the inlet coolant temperature is T_{a}_{0} = 450 K. Vary the coolant rate $(10<{\dot{\text{m}}}_{c}<{1000}_{g/\text{s}}\text{})$.
(d) Plot Q_{r} and T_{a} as a function of V necessary to maintain isothermal operation.
(e) Compare your answers to (a) through (d) and describe what you find. What generalizations can you make?
(f) Repeat (c) and (d) when the reaction is irreversible but endothermic with ${\mathrm{\Delta}H}_{\text{Rx}}^{\xb0}=6000\text{cal/mol}$. Choose T_{a}_{0} = 450 K.
P1211_{B} OEQ (Old Exam Question). Use the reaction data in Problems P114_{A} and P127_{B} for the case when heat is removed by a heat exchanger jacketing the reactor. The flow rate of coolant through the jacket is sufficiently high that the ambient exchanger temperature is constant at T_{a} = 50°C.
A+B → C
(a) (1) Plot and then analyze the temperature conversion, Q_{r}, and Q_{g} profiles for a PBR with where
$\frac{\mathit{\text{U}}a}{{\rho}_{b}}=0.08\frac{\mathit{\text{J}}}{\text{s}\cdot \text{kgcat}\cdot \mathit{\text{K}}}$
where
ρ_{b} = bulk density of the catalyst (kg/m^{3})
a = heatexchange area per unit volume of reactor (m^{2}/m^{3})
U = overall heat transfer coefficient (J/s · m^{2} · K)
(2) How would the profiles change if Ua/=_{b} were increased by a factor of 3000?
(3) If there is a pressure drop with α = 0.019 kg^{–1}?
(b) Repeat part (a) for cocurrent and countercurrent flow and adiabatic operation with ${\dot{\text{m}}}_{c}=0.2\text{kg/s}$, C_{Pc} = 5000 J/kg K and an entering coolant temperature of 50°C.
(c) Find X and T for a “fluidized” CSTR with 80 kg of catalyst.
$\begin{array}{cc}UA=500\frac{\mathit{\text{J}}}{\text{s}\cdot \mathit{\text{K}}},& {\rho}_{b}\end{array}=1{\text{kg/m}}^{3}$
(d) Repeat parts (a) and (b) for W = 80.0 kg, assuming a reversible reaction with a reverse specific reaction rate of
$\begin{array}{cc}{k}_{r}=0.2\text{exp}\left[\frac{{\mathit{\text{E}}}_{r}}{\mathit{\text{R}}}(\frac{1}{450}\frac{1}{\mathit{\text{T}}})\right]\left(\frac{{\text{dm}}^{6}}{\text{kgcat}\cdot \text{mol}\cdot \text{s}}\right);& {\mathit{\text{E}}}_{r}=51.4\text{kJ/mol}\end{array}$
Vary the entering temperature, T_{0}, and describe what you find.
(e) Use or modify the data in this problem to suggest another question or calculation. Explain why your question requires either critical thinking or creative thinking. See Preface Section G and http://www.umich.edu/~scps.
P1212_{C} Derive the energy balance for a packedbed membrane reactor. Apply the balance to the reaction in Problem P115_{A}
A ⇄ B + C
for the case when it is reversible with K_{C} = 1.0 mol/dm^{3} at 300 K. Species C diffuses out of the membrane with k_{C} = 1.5 s^{–1}.
(a) Plot and then analyze the concentration profiles for different values of K_{C} when the reaction is carried out adiabatically.
(b) Repeat part (a) when the heat transfer coefficient is Ua = 30 J/s·kgcat·K with T_{a} = 50°C.
P1213_{B} OEQ (Old Exam Question). Circle the correct answer.
(a) The elementary reversible isomerization of A to B was carried out in a packedbed reactor. The following profiles were obtained:
If the total entering volumetric flow rate remains constant, the addition of inerts to the feed stream will most likely
A) Increase conversion.
B) Decrease conversion.
C) Have no effect.
D) Insufficient information to tell
(b)
Which of the following statements are true?
A) The above reaction could be adiabatic.
B) The above reaction could be exothermic with constant cooling temperature.
C) The above reaction could be endothermic with constant heating temperature.
D) The above reaction could be second order.
(c) The conversion is shown below as a function of catalyst weight down a PBR.
Which of the following statements are false?
A) The reaction could be firstorder endothermic and carried out adiabatically.
B) The reaction could be firstorder endothermic and reactor is heated along the length with T_{a} being constant.
C) The reaction could be secondorder exothermic and cooled along the length of the reactor with T_{a} being constant.
D) The reaction could be secondorder exothermic and carried out adiabatically.
E) The reaction could be irreversible.
To view more conceptual problems similar to (a)–(c) above go to http://www.umich.edu/~elements/6e/12chap/iclicker_ch12_q1.html.
P1214_{A} OEQ (Old Exam Question). The irreversible reaction
A+B → C+D
is carried out adiabatically in a CSTR. The “heat generated” G(T) and the “heat removed” R(T) curves are shown in Figure P1214_{A}.
Figure P1214_{A} Heat removed R(T) and heat “generated” G(T) curves.
(a) What is the ΔH_{Rx} of the reaction?
(b) What are the inlet ignition and extinction temperatures?
(c) What are all the temperatures in the reactor corresponding to the inlet ignition and extinction temperatures?
(d) What are the conversions at the ignition and extinction temperatures?
P1215_{B} The firstorder, irreversible, exothermic liquidphase reaction
A → B
is to be carried out in a jacketed CSTR. Species A and an inert I are fed to the reactor in equimolar amounts. The molar feed rate of A is 80 mol/min.
Additional information:
$\begin{array}{cc}\text{Heatcapacityoftheinert:30cal/mol}{\cdot}^{\circ}\mathit{\text{C}}\hfill & \tau =100\text{min}\hfill \\ \text{HeatcapacityofAandB:20cal/mol}{\cdot}^{\circ}\mathit{\text{C}}\hfill & \mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ}=7500\text{cal/mol}\hfill \\ \text{UA: 8000cal/min}{\cdot}^{\circ}\mathit{\text{C}}\hfill & k=6.6\times {10}^{3}{\text{min}}^{1}\text{at350K}\hfill \\ \text{Ambienttemperature,}{\mathit{\text{T}}}_{a}:300\text{K}\hfill & \mathit{\text{E}}=40000\text{cal/mol}\cdot \mathit{\text{K}}\hfill \end{array}$
(a) What is the reactor temperature for a feed temperature of 450 K?
(b) Plot and then analyze the reactor temperature as a function of the feed temperature.
(c) To what inlet temperature must the fluid be preheated for the reactor to operate at a high conversion? What are the corresponding temperature and conversion of the fluid in the CSTR at this inlet temperature?
(d) Suppose that the fluid inlet temperature is now heated 5°C above the reactor temperature in part (c) and then cooled 20°C, where it remains. What will be the conversion?
(e) What is the inlet extinction temperature for this reaction system? (Ans: T_{0} = 87°C)
P1216_{B} The elementary reversible liquidphase reaction
takes place in a CSTR with a heat exchanger. Pure A enters the reactor.
(a) Derive an expression (or set of expressions) to calculate G(T) as a function of the heat of reaction, equilibrium constant, temperature, and so on. Show a sample calculation for G(T) at T = 400 K.
(b) What are the steadystate temperatures? (Ans: 310, 377, 418 K)
(c) Which steady states are locally stable?
(d) What is the conversion corresponding to the upper steady state?
(e) Vary the ambient temperature T_{a} and make a plot of the reactor temperature as a function of T_{a}, identifying the ignition and extinction temperatures.
(f) If the heat exchanger in the reactor suddenly fails (i.e., UA = 0), what would be the conversion and the reactor temperature when the new upper steady state is reached? (Ans: 431 K)
(g) What heat exchanger product, UA, will give the maximum conversion?
(h) Write a question that requires critical thinking and then explain why your question requires critical thinking. Hint: See Preface Section G.
(i) What is the adiabatic blowout flow rate, υ_{0}?
(j) Suppose that you want to operate at the lower steady state. What parameter values would you suggest to prevent runaway, for example, the upper SS?
Additional information:
$\begin{array}{cc}UA=3600\text{cal/min}\cdot \mathit{\text{K}}\hfill & \text{E/R}=20000\mathit{\text{K}}\hfill \\ {\mathit{\text{C}}}_{{\text{P}}_{\text{A}}}={\mathit{\text{C}}}_{{\text{P}}_{\text{B}}}=40\text{cal/mol}\cdot \mathit{\text{K}}\hfill & \mathit{\text{V}}=10{\text{dm}}^{3}\hfill \\ \mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}^{\circ}=80000\text{cal/molA}\hfill & {\upsilon}_{0}=1{\text{dm}}^{3}/\mathrm{min}\hfill \\ {\mathit{\text{K}}}_{\text{C}}=100\text{at400}\text{k}\hfill & {\mathit{\text{F}}}_{\text{A0}}=10\text{mol/min}\hfill \\ k=1{\text{min}}^{1}\text{at400K}\hfill & \\ \text{Ambienttemperature,}{\mathit{\text{T}}}_{a}={37}^{\circ}\hfill & \text{Feedtemperature,}{\mathit{\text{T}}}_{0}={\text{37}}^{\circ}\mathit{\text{C}}\hfill \end{array}$
P1217_{B} OEQ (Old Exam Question). The reversible liquidphase reaction
A ⇄ B
is carried out in a 12dm^{3} CSTR with heat exchange. Both the entering temperature, T_{0}, and the heat exchange fluid, T_{a}, are at 330 K. An equal molar mixture of inerts and A enter the reactor.
(a) Choose a temperature, T, and carry out a calculation to find G(T) to show that your calculation agrees with the corresponding G(T) value on the curve shown in Figure P1217_{B} at the temperature you choose.
(b) Find the exit conversion and temperature from the CSTR. X = _____ T = _____ Answers
(c) What entering temperature T_{0} would give you the maximum conversion? T_{0} = _____ X = _____
(d) What would the exit conversion and temperature be if the heatexchange system failed (i.e., U = 0)?
(e) Can you find the inlet ignition and extinction temperatures? If yes, what are they? If not, go on to the next problem.
(f) Use Preface Section G to ask another question.
Additional information:
The G(T) curve for this reaction is shown in Figure P1217_{B}.
$\begin{array}{cc}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{A}}}}={\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{B}}}}=100{\text{cal/mol/K,C}}_{{\mathit{\text{P}}}_{\mathit{\text{I}}}}=150\text{cal/mol/K}\hfill & k=0.001{\text{h}}^{1}\text{at}\text{300KwithE=30000cal/mol}\hfill \\ {\mathit{\text{F}}}_{\text{A0}}={\text{10mol/h,C}}_{\text{A0}}={\text{1mol/dm}}^{3},\text{}{\upsilon}_{0}=10{\text{dm}}^{3}\text{/h}\hfill & {\mathit{\text{K}}}_{\mathit{\text{C}}}=5,000,000\text{at300K}\hfill \\ \mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx}}=42000\text{cal/mol}\hfill & \text{UA}=5000\text{cal/h/K}\hfill \end{array}$
Figure P1217_{B} Heat removed curve, G(T), for a reversible reaction.
P1218_{C} The elementary gasphase reaction
2A ⇄ C
is carried out in a packedbed reactor. Pure A enters the reactor at a 450K flow rate of 10 mol/s, and a concentration of 0.25 mol/dm^{3}. The PBR contains 90 kg of catalyst and is surrounded by a heat exchanger for which cooling fluid is available at 500 K. Compare the conversion achieved for the four types of heat exchanger operation: adiabatic, constant T_{a}, cocurrent flow, and countercurrent flow.
Additional information:
$\begin{array}{cc}\alpha =0.019\text{/kgcat}\hfill & {\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\text{C}}}=20\text{J/mol/K}\\ \mathit{\text{U}}a/{\rho}_{b}=0.8\text{J/kgcat}\cdot \text{s}\cdot \mathit{\text{K}}\hfill & {\mathit{\text{F}}}_{\text{A}\text{0}}=10\text{mol/h}\\ \mathrm{\Delta}{H}_{\text{Rx}}=20000\text{J/mol}\hfill & {\mathit{\text{C}}}_{\text{A0}}={\text{1mol/dm}}^{3}\\ {\mathit{\text{C}}}_{{\text{P}}_{\text{A}}}=10\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{J/mol}\cdot \mathit{\text{K}}\hfill & \text{}{\upsilon}_{0}=10{\text{dm}}^{3}/\text{h}\end{array}$
P1219_{C} A reaction is to be carried out in the packedbed reactor shown in Figure P1219_{C}.
Figure P1219_{C} PFR with heat exchange.
The reactants enter the annular space between an outer insulated tube and an inner tube containing the catalyst. No reaction takes place in the annular region. Heat transfer between the gas in this packedbed reactor and the gas flowing countercurrently in the annular space occurs along the length of the reactor. The overall heattransfer coefficient is 5 W/m^{2} · K. Plot the conversion and temperature as a function of reactor length for the data given in Problem P127_{B}.
P1220_{B} OEQ (Old Exam Question). The reaction
$\mathit{\text{A}}+\mathit{\text{B}}\overrightarrow{\leftarrow}2\mathit{\text{C}}$
is carried out in a packedbed reactor. Match the following temperature and conversion profiles for the four different heatexchange cases: adiabatic, constant T_{a}, cocurrent exchange, and countercurrent exchange.
(a) Figure 1 matches Figure ___
(b) Figure 2 matches Figure ___
(c) Figure 3 matches Figure ___
(d) Figure 4 matches Figure ___
P1221_{B} OEQ (Old Exam Question). Also Hall of Fame Problem. The irreversible liquidphase reactions
$\begin{array}{ccc}\text{Reaction(1)}& \mathit{\text{A}}+\mathit{\text{B}}\to \text{2}\mathit{\text{C}}& {r}_{1\mathit{\text{C}}}={k}_{1C}{\mathit{\text{C}}}_{\mathit{\text{A}}}{\mathit{\text{C}}}_{\mathit{\text{B}}}\\ \text{Reaction(2)}& 2\mathit{\text{B}}+\mathit{\text{C}}\to \mathit{\text{D}}& {r}_{2\mathit{\text{D}}}={k}_{2\mathit{\text{D}}}{\mathit{\text{C}}}_{\mathit{\text{B}}}{\mathit{\text{C}}}_{\mathit{\text{C}}}\end{array}$
are carried out in a PFR with heat exchange. The temperature profiles shown in Figure P1221_{B} were obtained for the reactor and the coolant stream:
Figure P1221_{B} Reactant temperature T and coolant temperature T_{a} profiles.
The concentrations of A, B, C, and D were measured at the point down the reactor where the liquid temperature, T, reached a maximum, and they were found to be C_{A} = 0.1, C_{B} = 0.2, C_{C} = 0.5, and C_{D} = 1.5, all in mol/dm^{3}. The product of the overall heat transfer coefficient and the heatexchanger area per unit volume, Ua, is 10 cal/s · dm^{3} · K. The entering molar flow rate of A is 10 mol/s.
Additional information:
$\begin{array}{cc}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\text{A}}}={\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\text{B}}}={\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\text{C}}}=30\text{cal/mol/K}\hfill & {\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\text{D}}}=90\text{cal/mol/K,}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\text{I}}}=100\text{cal/mol/}\mathit{\text{K}}\hfill \\ \mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx1}\text{A}}^{\circ}=50000\text{cal/molA}\hfill & {k}_{1\text{C}}=0.043\frac{{\text{dm}}^{3}}{\text{mol}\cdot \text{S}}\text{at400K}\hfill \\ \mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx2}\text{B}}^{\circ}=+5000\text{cal/molB}\hfill & {k}_{\text{2D}}=0.4\frac{{\text{dm}}^{3}}{\text{mol}\cdot \text{S}}\text{at500K,with}\frac{\text{E}}{\text{R}}=5000\text{K}\hfill \end{array}$
(a) What is the activation energy for Reaction (1)?
P1222_{B} The following elementary reactions are to be carried out in a PFR with a heat exchange with constant T_{a}:
$\begin{array}{ccc}\text{2A}+\mathit{\text{B}}\to \mathit{\text{C}}& \mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx1}\mathit{\text{B}}}=10\frac{\text{kJ}}{\text{mol}\text{B}}& \text{}\\ \mathit{\text{A}}\to \mathit{\text{D}}& \mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx2A}}=+10\frac{\text{kJ}}{\text{mol A}}& \text{}\\ \mathit{\text{B}}+2\mathit{\text{C}}\to \mathit{\text{E}}& \mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx3C}}=20\frac{\text{kJ}}{\text{mol C}}& \text{}\end{array}$
The reactants all enter at 400 K. Only A and B enter the reactor. The entering concentration of A and B are 3 molar and 1 molar at a volumetric flow rate of 10 dm^{3}/s.
Additional information:
$\begin{array}{cc}\mathit{\text{U}}a={\text{100J/dm}}^{3}/\text{S/K}\hfill & {\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\text{A}}}=10\text{J/mol/K}\hfill \\ {k}_{1\text{A}}\left(400\text{K}\right)=1.0{\left(\frac{{\text{dm}}^{3}}{\text{mol}}\right)}^{2}/\text{S}\hfill & {\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\text{B}}}=20\text{J/mol/K}\hfill \\ {k}_{2\text{A}}\left(400\text{K}\right)={1.333}_{{\text{S}}^{1}}\hfill & {\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\text{C}}}=40\text{J/mol/K}\hfill \\ {k}_{3\text{B}}\left(400\text{K}\right)=2{\left(\frac{{\text{dm}}^{3}}{\text{mol}}\right)}^{2}/\text{S}\hfill & {\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\text{D}}}=20\text{J/mol/K}\hfill \\ \text{}& {\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\text{E}}}=100\text{J/mol/K}\hfill \end{array}$
What constant coolant temperature, T_{a}, is necessary such that at the reactor entrance, that is, V = 0, $\frac{d\mathit{\text{T}}}{d\mathit{\text{V}}}=0$?
P1223_{B} The complex gasphase reactions are elementary
$\begin{array}{cccc}\left(\text{1}\right)\text{}& 2\mathit{\text{A}}\overrightarrow{\leftarrow}\text{2}\mathit{\text{B}}& {r}_{\text{1A}}={k}_{1\mathit{\text{A}}}[{\mathit{\text{C}}}_{\mathit{\text{A}}}^{2}\frac{{\mathit{\text{C}}}_{\mathit{\text{B}}}}{{\mathit{\text{K}}}_{{\mathit{\text{C}}}_{\mathit{\text{A}}}}}]& \mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx1A}}=20\text{kJ/mol A}\\ \left(\text{2}\right)& 2\mathit{\text{B}}+\mathit{\text{A}}\to \mathit{\text{C}}& {r}_{2\mathit{\text{C}}}={k}_{2\mathit{\text{C}}}\left[{\mathit{\text{C}}}_{\mathit{\text{B}}}^{\text{2}}{\mathit{\text{C}}}_{\mathit{\text{A}}}\right]& \text{}\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx2B}}=+30\text{kJ/mol B}\end{array}$
and carried out in a PFR with a heat exchanger. Pure A enters at a rate of 5 mol/min, a concentration of 0.2 mol/dm^{3}, and temperature 300 K. The entering temperature of an available coolant is 320 K.
Additional information:
k_{1A1} = 50 dm^{3}/mol · [email protected] K with E_{1} = 8000 J/mol
k_{2C2} = 4000 dm^{9}/mol^{3} · [email protected] K with E_{2} = 4000 J/mol
K_{CA} = 10 dm^{3}/[email protected] K
Note: This is the equilibrium constant with respect to A in reaction 1 when using van’t Hoff’s equation, that is, (S124),
$\begin{array}{cccc}\mathit{\text{U}}a=200\frac{\mathit{\text{J}}}{{\text{dm}}^{3}\text{}\cdot \text{min K}}& {\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\text{Cool}}}=10\text{/mol/K}& {\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{B}}}}=80\text{J/mol/K}& {\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{C}}}}=100\text{J/mol/K}\\ {\dot{\text{m}}}_{\mathit{\text{C}}}=50g\text{/min}& {\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\mathit{\text{A}}}}=20\text{J/mol/K}& \mathit{\text{R}}=8.31\text{J/mol/K}& \text{}\end{array}$
The reactor volume is 10 dm^{3}.
(a) Plot (F_{A}, F_{B}, F_{C}) on one graph and (T and T_{a}) on another along the length of the reactor for adiabatic operation, heat exchange with constant T_{a}, and cocurrent and countercurrent heat exchange with variable T_{a}. Only turn in a copy of your code and output for cocurrent exchange.
Adiabatic operation
(b) What is the maximum temperature and at what reactor volume is it reached?
(c) At what reactor volume is the flow rate of B a maximum, and what is F_{Bmax} at this value?
Constant Ta
(d) What is the maximum temperature and at what reactor volume is it reached?
(e) At which reactor volume is the flow rate of B a maximum, and what is F_{Bmax} at this volume?
Cocurrent exchange
(f) At what reactor volume does T_{a} become greater than T? Why does it become greater?
Countercurrent exchange
(g) At what reactor volume does T_{a} become greater than T?
Hint: Guess T_{a} at entrance around 350 K.
P1224_{B} The elementary liquidphase reactions
(1) A+ 2B → 2C
(2) A+C→ 2D
are carried out adiabatically in a 10 dm^{3} PFR. After streams A and B mix, species A enters the reactor at a concentration of C_{A0} = 2 mol/dm^{3} and species B at a concentration of 4 mol/dm^{3}. The entering volumetric flow rate is 10 dm^{3}/s.
Assuming you could vary the entering temperature between 300 K and 600 K, what entering temperture would you recommend to maximize the concentration of species C exiting the reactor? (±25°K). Assume all species have the same density.
Additional information:
$\begin{array}{c}{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\text{A}}}={\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\text{B}}}=20{\text{cal/mol/K,C}}_{{\mathit{\text{P}}}_{\text{C}}}=60{\text{cal/mol/K,C}}_{{\mathit{\text{P}}}_{\text{D}}}=80\text{cal/mol/K}\\ \mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx1A}}=20000\text{cal/molA,}\mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx2A}}=10000\text{cal/molA}\hfill \\ {k}_{\text{1A}}=0.001\frac{{\text{dm}}^{\text{6}}}{{\text{mol}}^{2}\text{}\cdot \text{S}}at\text{300KwithE = 5000cal/mol}\hfill \\ {k}_{2\text{A}}=0.001\frac{{\text{dm}}^{3}}{\text{mol}\cdot \text{S}}at\text{300KwithE = 7500cal/mol}\hfill \\ \text{}\\ \text{}\end{array}$
P1225_{C} OTHEQ (Old Take Home Exam Question). Multiple reactions with heat effects. Xylene has three major isomers, mxylene (A), oxylene (B), and pxylene (C). When mxylene (A) is passed over a Cryotite catalyst, the following elementary reactions are observed. The reaction to form pxylene is irreversible:
The feed to the reactor is pure mxylene (A). For a total feed rate of 2 mol/min and the reaction conditions below, plot the temperature and the molar flow rates of each species as a function of catalyst weight up to a weight of 100 kg.
(a) Plot the concentrations of each of xylenes down the length (i.e., V) of a PBR.
(b) Find the lowest concentration of oxylene achieved in the reactor.
(c) Find the maximum concentration of oxylene in the reactor.
(d) Repeat part (a) for a pure feed of oxylene (B). What is the maximum concentration of metaxylene and where does it occur in the reactor?
(e) Vary some of the system parameters and describe what you learn.
(f) What do you believe to be the point of this problem?
Additional information^{8}
^{8} Obtained from inviscid pericosity measurements.
All heat capacities are virtually the same at 100 J/mol · K.
$\begin{array}{cc}{\mathit{\text{C}}}_{\text{T0}}={\text{2mol/dm}}^{3}\hfill & {\mathit{\text{K}}}_{\text{C}}=10\text{exp}\left[\text{4.8}(430\text{/}\mathit{\text{T}}1.5)\right]\hfill \\ \mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx10}}^{\circ}=1800\text{J/mol}\text{}\circ \text{xylene}\hfill & {\mathit{\text{T}}}_{0}=330\text{K}{\mathit{\text{T}}}_{a}=500\text{}\mathit{\text{K}}\hfill \\ \mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx30}}^{\circ}=1100\text{J/mol}\circ \text{xylene}\hfill & {k}_{2}={k}_{\text{1}}\text{/}{K}_{\text{C}}\hfill \\ {k}_{1}=0.5\text{exp}\left[\text{2}(\text{1}320\text{/T})\right]{\text{dm}}^{3}\text{/kgcat}\cdot \text{min,}\left(\text{TisinK}\right)\hfill & \mathit{\text{U}}a/{\rho}_{b}=16\text{J/kgcat}\cdot \mathrm{min}\text{}{\cdot}^{\circ}\text{C}\hfill \\ {k}_{3}=0.005\text{exp}\left\{\left[4.6\left(\text{1}\left(460\text{/T}\right)\right)\right]\right\}{\text{dm}}^{3}\text{/kgcat}\cdot \text{min}\hfill & \mathit{\text{W}}=100\text{kg}\hfill \end{array}$
P1226_{C} OTHEQ (Old Take Home Exam Question). Comprehensive problem on multiple reactions with heat effects. Styrene can be produced from ethylbenzene by the following reaction:
$\begin{array}{cc}\text{ethylbenzene}\leftrightarrow \text{styrene}+{\text{H}}_{2}& \text{(1)}\end{array}$
However, several irreversible side reactions also occur:
$\begin{array}{c}\begin{array}{cc}\text{ethylbenzene}\to \text{benzene}+\text{ethylene}& \text{(2)}\end{array}\\ \begin{array}{cc}\begin{array}{c}\text{ethylbenzene}+{\text{H}}_{2}\to \text{toluene}+\text{methane}\end{array}& \text{(3)}\end{array}\end{array}$
(J. Snyder and B. Subramaniam, Chem. Eng. Sci., 49, 5585 (1994)). Ethylbenzene is fed at a rate of 0.00344 kmol/s to a 10.0m^{3} PFR (PBR), along with inert steam at a total pressure of 2.4 atm. The steam/ethylbenzene molar ratio is initially, that is, parts (a) to (c), 14.5:1 but can be varied.
Given the following data, find the exiting molar flow rates of styrene, benzene, and toluene along with ${\tilde{\mathit{\text{S}}}}_{\text{St/BT}}$ for the following inlet temperatures when the reactor is operated adiabatically:
(a) T_{0} = 800 K
(b) T_{0} = 930 K
(c) T_{0} = 1100 K
(d) Find the ideal inlet temperature for the production of styrene for a steam/ethylbenzene ratio of 58:1. Hint: Plot the molar flow rate of styrene versus T_{0}. Explain why your curve looks the way it does.
(e) Find the ideal steam/ethylbenzene ratio for the production of styrene at 900 K. Hint: See part (d).
(f) It is proposed to add a countercurrent heat exchanger with Ua = 100 kJ/m^{3}/min/K, where T_{a} is virtually constant at 1000 K. For an entering stream to ethylbenzene ratio of 20, what would you suggest as an entering temperature? Plot the molar flow rates and ${\tilde{\mathit{\text{S}}}}_{\text{St/BT}}$.
(g) What do you believe to be the major points of this problem?
(h) Ask another question or suggest another calculation that can be made for this problem.
Note: Whenever I teach Chemical Reaction Engineering, I always assign this problem.
Additional information
Heat capacities 

Methane 
68 J/mol · K 
Ethylene 
90 J/mol·K 
Benzene 
201 J/mol · K 
Toluene 
249 J/mol · K 
Styrene 
273 J/mol · K 
Ethylbenzene 
299 J/mol·K 
Hydrogen 
30 J/mol · K 
Steam 
40 J/mol · K 
$\begin{array}{c}\begin{array}{c}\begin{array}{c}\begin{array}{c}\rho =2137{\text{kg/m}}^{3}\text{ofpellet}\\ \mathrm{\varphi}=0.4\hfill \end{array}\hfill \\ \mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx1EB}}^{\circ}=118000\text{kJ/kmolethylbenzene}\end{array}\\ \mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx2EB}}^{\circ}=105200\text{}\text{kJ/kmolethylbenzene}\end{array}\\ \mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx3EB}}^{\circ}=53900\text{}\text{kJ/kmolethylbenzene}\end{array}$
$\begin{array}{cc}{\mathit{\text{K}}}_{\text{P1}}=\text{exp}\{{b}_{1}+\frac{{b}_{2}}{\mathit{\text{T}}}+{b}_{3}\text{}\mathrm{ln}\left(\mathit{\text{T}}\right)+\left[({b}_{4}\mathit{\text{T}}+{b}_{5})\mathit{\text{T}}+{b}_{\text{6}}\right]\mathit{\text{T}}\}\text{atmwith}& \text{}\\ {b}_{1}=17.34\hfill & {b}_{4}=2.314\times {10}^{10}{\mathit{\text{K}}}^{3}\hfill \\ {b}_{2}=1.302\times {10}^{\text{4}}\mathit{\text{K}}\hfill & {b}_{5}=1.302\times {10}^{6}{\mathit{\text{K}}}^{2}\hfill \\ {b}_{3}=5.051\hfill & {b}_{6}=4.931\times {10}^{3}{\mathit{\text{K}}}^{1}\hfill \end{array}$
The kinetic rate laws for the formation of styrene (St), benzene (B), and toluene (T), respectively, are as follows (EB = ethylbenzene).
$\begin{array}{c}\begin{array}{c}\begin{array}{cc}\begin{array}{c}{r}_{1\text{St}}=\rho (1\varphi )\mathrm{exp}(0.08539\frac{10925\text{}\mathit{\text{K}}}{\mathit{\text{T}}})({P}_{EB}\frac{{P}_{St}{P}_{{\mathit{\text{H}}}_{2}}}{{K}_{p1}})\end{array}& ({\text{kmol/m}}^{3}\cdot \text{s})\end{array}\\ \begin{array}{cc}\begin{array}{c}{r}_{\text{2}\text{B}}=\rho (1\varphi )\mathrm{exp}(13.2392\frac{25000\text{}\mathit{\text{K}}}{\mathit{\text{T}}})\left({P}_{EB}\right)\end{array}& ({\text{kmol/m}}^{3}\cdot \text{s})\end{array}\hfill \end{array}\\ \begin{array}{cc}\begin{array}{c}{r}_{3\text{T}}=\rho (1\varphi )\mathrm{exp}(0.2961\frac{11000\text{}\mathit{\text{K}}}{\mathit{\text{T}}})\left({P}_{EB}{P}_{{\text{H}}_{2}}\right)\end{array}& ({\text{kmol/m}}^{3}\cdot \text{s})\end{array}\hfill \end{array}$
The temperature T is in Kelvin and P_{i} is in atm.
P1227_{B} OTHEQ (Old Take Home Exam Question). The liquidphase, dimerquadmer series addition reaction
4 A → 2A_{2} → A_{4}
can be written as
$\begin{array}{ccc}\text{2A}\to {\text{A}}_{2}& {r}_{1\text{A}}={k}_{1\text{A}}{\mathit{\text{C}}}_{\text{A}}^{2}& \mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx1A}}=32.5\frac{\text{kcal}}{\text{molA}}\\ {\text{2A}}_{2}\to {\text{A}}_{\text{4}}& {r}_{2{\text{A}}_{2}}={k}_{2{\text{A}}_{2}}{\mathit{\text{C}}}_{{\text{A}}_{2}}^{2}& \mathrm{\Delta}{\mathit{\text{H}}}_{\text{Rx2}{\text{A}}_{2}}=27.5\text{}\frac{\text{kcal}}{{\text{molA}}_{2}}\end{array}$
and is carried out in a 10dm^{3} PFR. The mass flow rate through the heat exchanger surrounding the reactor is sufficiently large so that the ambient temperature of the exchanger is constant at T_{a} = 315 K. The reactants enter at a temperature T_{0}, of 300 K. Pure A is fed to the rector at a volumetric flow rate of 50 dm^{3}/s and a concentration of 2 mol/dm^{3}.
(a) Plot, compare, and analyze the profiles F_{A}, F_{A2}, and F_{A4} down the length of the reactor up to 10 dm^{3}.
(b) The desired product is A_{2} and it has been suggested that the current reactor may be too large. What reactor volume would you recommend to maximize F_{A2}?
(c) What operating variables (e.g., T_{0}, T_{a}) would you change and how would you change them to make the reactor volume as small as possible and to still maximize F_{A2}? Note any opposing factors in maximum production of A_{2}. The ambient temperature and the inlet temperature must be kept between 0°C and 177°C.
Additional information:
$\begin{array}{c}\begin{array}{c}\begin{array}{c}{k}_{1\text{A}}=0.6\frac{{\text{dm}}^{3}}{\text{mol}\cdot \mathit{\text{S}}}{\text{at300KwithE}}_{1}=\text{4000}\frac{\text{cal}}{\text{mol}}\hfill \\ {k}_{2{\text{A}}_{2}}=0.35\frac{{\text{dm}}^{3}}{\text{mol}\cdot \mathit{\text{S}}}{\text{at320KwithE}}_{\text{2}}=5\text{000}\frac{\text{cal}}{\text{mol}}\hfill \\ \text{}\end{array}\hfill \\ {\mathit{\text{C}}}_{{\mathit{\text{P}}}_{\text{A}}}=\text{25}\frac{\text{cal}}{\text{molA}\cdot \mathit{\text{K}}},{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{{\text{A}}_{2}}}=50\frac{\text{cal}}{{\text{molA}}_{2}\text{}\cdot \mathit{\text{K}}},{\mathit{\text{C}}}_{{\mathit{\text{P}}}_{{\text{A}}_{4}}}=100\frac{\text{cal}}{\text{mol}{\text{A}}_{4}\cdot \mathit{\text{K}}}\end{array}\\ \mathit{\text{U}}a=1000\frac{\text{cal}}{{\text{dm}}^{3}\cdot \mathit{\text{S}}\cdot \mathit{\text{K}}}\hfill \end{array}$
Turn in your recommendation of reactor volume to maximize and the molar flow rate at this maximum.
1. An excellent development of the energy balance is presented in
R. ARIS, Elementary Chemical Reactor Analysis. Upper Saddle River, NJ: Prentice Hall, 1969, Chaps. 3 and 6. JOSEPH FOGLER, A Reaction Engineer’s Handbook of Thermochemical Data. Riça, Jofostan: Jofostan Press, (2025).
2. Safety
CENTER FOR CHEMICAL PROCESS SAFETY (CCPS), Guidelines for Chemical Reactivity Evaluation and Application to Process Design. New York: American Institute of Chemical Engineers (AIChE), 1995.
DANIEL A. CROWL and JOSEPH F. LOUVAR, Chemical Process Safety: Fundamentals with Applications, 3rd ed. Upper Saddle River, NJ: Prentice Hall, 2011.
G. A. MELHEM and H. G. FISHER, International Symposium on Runaway Reactions and Pressure Relief Design. New York: Center for Chemical Process Safety (CCPS) of the American Institute of Chemical Engineers (AIChE) and The Institution of Chemical Engineers, 1995.
See the Center for Chemical Process Safety (CCPS) Web site, www.aiche.org/ccps.
3. A number of example problems dealing with nonisothermal reactors may or may not be found in
THORNTON W. BURGESS, The Adventures of Jerry Muskrat. New York: Dover Publications, Inc., 1914.
G. F. FROMENT and K. B. BISCHOFF, Chemical Reactor Analysis and Design, 3rd ed. New York: Wiley, 2010.
S. M. WALAS, Chemical Reaction Engineering Handbook of Solved Problems. Amsterdam: Gordon and Breach, 1995. See the following solved problems: Problem 4.10.1, page 444; Problem 4.10.08, page 450; Problem 4.10.09, page 451; Problem 4.10.13, page 454; Problem 4.11.02, page 456; Problem 4.11.09, page 462; Problem 4.11.03, page 459; Problem 4.10.11, page 463.
4. A review of the multiplicity of the steady state and reactor stability is discussed by
D. D. PERLMUTTER, Stability of Chemical Reactors. Upper Saddle River, NJ: Prentice Hall, 1972.
5. The heats of formation, (H_{i} T), Gibbs free energies, (G_{i} T_{R}), and the heat capacities of various compounds can be found in
DON W. GREEN and ROBERT H. PERRY, Perry’s Chemical Engineers’ Handbook, 8th ed. (Chemical Engineers Handbook). New York: McGrawHill, 2008.
DAVID R. LIDE, CRC Handbook of Chemistry and Physics, 90th ed. Boca Raton, FL: CRC Press, 2009.