How can the reaction heat be ignored in an energy balance?

Question

I have received from my professor the following (summarized) problem:

A first-order exothermic reaction $\ce{A->B}$ took place in a typical tubular plug flow reactor (PFR). The tube was cooled to a constant wall temperature $T_\mathrm{w}$ and the total amount of heat removed from the reactor $Q$ was given in Watts. All needed constants and geometrical parameters and $T_\text{inlet}$ and $T_\text{outlet}$ were provided.

The goal was to find the (constant) value of the wall temperature $T_\mathrm{w}$ that ensures heat $Q$ is being removed from the reactor.

My question is not about the final value but rather the logical aspects. In my professor's solution, they simply used the following formula from a heat exchanger:

$$ Q = U \times A \times LMTD $$

where $LMTD$ is the logarithmic mean temperature difference based on $T_\text{inlet}$, $T_\text{outlet}$ and $T_w$ (to be determined). $Q$, $U$ and $A$ are given.

This approach makes sense for a tubular heat exchanger but as soon as a reaction is happening inside the inner duct, the energy balance doesn't result in the $LMTD$ anymore. The PFR heat balance in this scenario is:

$$\frac{\mathrm{d}T}{\mathrm{d}z} = \frac{k_{W}}{u\rho c_P}\frac{4}{D}\left(T_\mathrm{w}-T\right)+\frac{r(-\Delta_R H)}{u\rho c_P}$$

If the reaction were non-existing ($r=0$), one could derive the $LMTD$ approach to solving this from the balance above. But that's not the case.

What do you think? Given $Q$ (along the entire reactor), $T_\text{inlet}$, $T_\text{outlet}$ and all other constants/parameter, can you calculate $T_\mathrm{w}$? Is my professor's approach correct? If so, how can one derive this formula from the energy balance as one does for a simple heat exchanger?

Answer 1

Update: Comments reveal I missed a major point below, which invalidates this answer, which is that the reaction rate $r$ depends at least on $z$ and probably also on $T$. For now I'm keeping the answer up in case it's useful to anyone else; I (or someone else) might delete it soon.

WRONG ANSWER BELOW

The energy balance you have is:

$$\frac{dT}{dz} = \frac{k_W}{u \rho c_P} \frac{4}{D}(T_W - T) + \frac{r\left(-\Delta_RH \right)}{u \rho c_P}$$

This is a non-homogenous differential equation in $T$. "Non-homogenous" means the equation is like $\frac{dT}{dz} = f(T) + K$, where $K$ is a term that does not contain any dependence on $T$. (In the general case, $K$ may be a function of the dependent variable $z$, but you enjoy a simpler case where $K$ is just a constant.)

To be specific, for your energy balance, the term not dependent on $T$ is:

$$K = \frac{k_W}{u \rho c_P} \frac{4}{D}T_W + \frac{r\left(-\Delta_RH \right)}{u \rho c_P}$$

It's instructive to compare to the case with no reaction. The non-homogeneous part of the equation gets simpler:

$$K_{\text{no reaction}} = \frac{k_W}{u \rho c_P} \frac{4}{D}T_W$$

However, in neither case does this non-homogeneous (some call it hetereogenous) part depend on $T$, or on $z$.

But, the "homogeneous" part of the equation is the same, that is, in either case,

$$f(T) = - \frac{k_W}{u \rho c_P} \frac{4}{D} T$$.

What I would recommend doing is to just substitute the symbol $K$ for the non-homogeneous part and solve the differential equation

$$\frac{dT}{dz} = -\frac{k_W}{u \rho c_P} \frac{4}{D} T + K$$

After you solve it, and generate the log-mean temperature difference equation, you can then substitute whatever symbols you'd like for $K$ into your solution; that is, you can generate solutions for both the no-reaction heat exchanger case as well as the case with the reaction.

Hopefully this will convince you that your professor was right. If you know the inlet and outlet temperatures, you can just use the simple equation. It's obviously true, however, that if you compared two PFRs, one acting as a simple heat exchanger with no reaction, and the other with a reaction, and both PFRs had the same flow rate, heat capacities, and inlet temperatures, the outlet temperature would be very different in the reacting PFR vs. the inert heat exchanger one.