Question 1 of 8: Stirred Tank with Thermal Solid — Model, Transfer Function, and Large-$UA$ Limit

Problem 1: Stirred Tank with Thermal Solid — Model, Transfer Function, and Large-$UA$ Limit (20%)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A jacket-free CSTR whose liquid exchanges heat with an internal solid of large heat capacity. (The printed text calls the inlet temperature $T_0$ while the exam figure labels the feed $T_a$; they are the same variable, written $T_a$ below.)

Find. (a) the two coupled energy balances; (b) $\delta T(s)/\delta T_a(s)$; (c) the limiting behaviour as $UA\to\infty$.

Approach. Write an energy balance on the liquid and a separate one on the solid (two states $T,T_m$), Laplace-transform in deviation variables, eliminate $T_m$, and read the limit by letting $UA$ dominate every term it appears in.

1. (a) Liquid energy balance. With constant $\rho,C_p$ and negligible losses, accumulation $=$ convection in $-$ convection out $-$ heat to solid: $$\rho V C_p\frac{dT}{dt}=\rho F C_p\,(T_a-T)-UA\,(T-T_m).$$
2. (a) Solid energy balance. The solid only exchanges with the liquid: $$M C_m\frac{dT_m}{dt}=UA\,(T-T_m).$$ These two coupled ODEs are the fundamental model; both variables start at steady state.
3. Deviation variables and Laplace. Let $a=\rho V C_p$, $b=\rho F C_p$, $c=M C_m$, and use $\delta$ for deviations. Transforming (initial deviations zero): $$a\,s\,\delta T=b(\delta T_a-\delta T)-UA(\delta T-\delta T_m),\qquad c\,s\,\delta T_m=UA(\delta T-\delta T_m).$$
4. Eliminate the solid state. From the solid equation $\displaystyle \delta T_m=\frac{UA}{c\,s+UA}\,\delta T$. Substituting and collecting $\delta T$, the coupling term becomes $UA-\dfrac{UA^2}{cs+UA}=\dfrac{UA\,c\,s}{cs+UA}$, so $$\Big[a\,s+b+\frac{UA\,c\,s}{c\,s+UA}\Big]\delta T=b\,\delta T_a.$$
5. (b) Transfer function. Multiplying through by $(cs+UA)$: $$\boxed{\frac{\delta T(s)}{\delta T_a(s)}=\frac{b\,(c\,s+UA)}{a c\,s^{2}+\big(a\,UA+c(b+UA)\big)s+b\,UA}.}$$ Setting $s=0$ gives a steady-state gain $bUA/bUA=1$ — at steady state the outlet fully follows the inlet, as it must with no other heat source. The system is second order (two capacitances in series) with a numerator zero at $s=-UA/c$.
6. (c) Limit $UA\to\infty$. Perfect coupling forces $T_m\to T$: the solid and liquid become one lumped mass. Keeping only the terms carrying $UA$ in the numerator and denominator, $$\frac{\delta T}{\delta T_a}\xrightarrow{UA\to\infty}\frac{bUA}{UA\big[(a+c)s+b\big]}=\frac{1}{\tau s+1},\qquad \boxed{\tau=\frac{a+c}{b}=\frac{\rho V C_p+M C_m}{\rho F C_p}.}$$ The two model equations in (a) merge into a single balance $(\rho V C_p+M C_m)\,dT/dt=\rho F C_p(T_a-T)$, and the second-order transfer function of (b) collapses to first order: the solid’s heat capacity simply adds to the liquid’s, and the numerator zero cancels.