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

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?




**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.


 0