Stirred Tank with a Solid Thermal Mass: 23-Chem-A6 Process Dynamics and Control, December 2014, Q1

Problem 1: Stirred Tank with a Solid Thermal Mass (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 stirred tank (liquid inventory $V$, density $\rho$, specific heat $C_p$, throughput $F$) with an immersed solid block (mass $M$, specific heat $C_s$, temperature $T_m$), coupled by convective heat transfer $UA(T-T_m)$:

Quantity	Symbol	Role
Liquid volume / density / specific heat	$V,\ \rho,\ C_p$	liquid energy store
Volumetric throughflow	$F$	convective in/out
Solid mass / specific heat	$M,\ C_s$	second energy store
Liquid→solid heat transfer	$UA(T-T_m)$	internal coupling
Input / output	$T_o(t)\ /\ T(t)$	inlet vs tank temperature

Find. (a) the coupled energy-balance ODEs; (b) the transfer function $\delta T/\delta T_o$; (c) its limiting form as $UA\to\infty$.

Problem 1: a well-mixed liquid (temperature $T$, one energy store) exchanging heat $UA(T-T_m)$ with an immersed solid block (temperature $T_m$, a second energy store). Two independent stores → second-order dynamics.

Approach. Write one energy balance on each independent store (liquid and solid), move to deviation variables, Laplace-transform and eliminate $T_m$ to obtain $\delta T/\delta T_o$; then take $UA\to\infty$ to see the two stores lump into one.

(a) Energy balance on the liquid. Accumulation = convective in−out − heat lost to the solid: $$\rho V C_p\,\frac{dT}{dt}=\rho F C_p\,(T_o-T)-UA\,(T-T_m).$$ The $UA$ term leaves the liquid and enters the solid, coupling the two states.
(a) Energy balance on the solid. The block has no flow term — it only receives the convective heat: $$\boxed{M C_s\,\frac{dT_m}{dt}=UA\,(T-T_m).}$$ These two coupled ODEs are the fundamental model. At steady state $dT_m/dt=0\Rightarrow T_s=T_{ms}$, and the liquid balance then gives $T_{os}=T_s$ (no driving force at rest).
Deviation variables. With $T'=T-T_s$, $T_m'=T_m-T_{ms}$, $T_o'=T_o-T_{os}$ the (already linear) equations carry over unchanged: $\rho V C_p\,\dot T'=\rho F C_p(T_o'-T')-UA(T'-T_m')$ and $M C_s\,\dot T_m'=UA(T'-T_m')$.
(b) Eliminate the solid state. Transforming the solid balance, $(M C_s s+UA)T_m'=UA\,T'$, so $T_m'=\dfrac{UA}{M C_s s+UA}T'$. Substituting into the transformed liquid balance $(\rho V C_p s+\rho F C_p+UA)T'=\rho F C_p T_o'+UA\,T_m'$ and clearing the denominator gives the second-order transfer function $$\boxed{\dfrac{\delta T}{\delta T_o}=\dfrac{\rho F C_p\,(M C_s s+UA)}{\rho V C_p M C_s\,s^{2}+\big(\rho V C_p\,UA+\rho F C_p M C_s+UA\,M C_s\big)s+\rho F C_p\,UA}.}$$ Two energy stores → two poles. Setting $s=0$ gives a steady-state gain of exactly 1 ($\delta T=\delta T_o$ at rest).
(c) Limit $UA\to\infty$. An infinite heat-transfer coefficient forces $T_m\to T$: the solid tracks the liquid perfectly, so the two capacitances merge into one lumped store $\rho V C_p+M C_s$. Adding the two balances (the $UA$ terms cancel) leaves a single first-order model $$\boxed{\dfrac{\delta T}{\delta T_o}=\dfrac{1}{\tau s+1},\qquad \tau=\dfrac{\rho V C_p+M C_s}{\rho F C_p}.}$$ The response degenerates from two lags to a single first-order lag whose time constant is inflated by the solid's heat capacity (the same limit falls out of the second-order form when $UA\to\infty$).

Result	Expression
(a) Liquid balance	$\rho V C_p\,\dot T=\rho F C_p(T_o-T)-UA(T-T_m)$
(a) Solid balance	$M C_s\,\dot T_m=UA(T-T_m)$
(b) Transfer function	2nd-order, DC gain $=1$ (boxed above)
(c) $UA\to\infty$	$\dfrac{1}{\tau s+1},\ \tau=\dfrac{\rho V C_p+M C_s}{\rho F C_p}$

Concept & Theory

Topic. Multi-capacitance thermal dynamics; model order set by the number of independent energy stores; lumping under strong coupling.

Key relations. Energy balance per store $C\,dT/dt=\sum \dot Q$; each independent store contributes one pole; convective coupling $UA(T-T_m)$; series capacitances lump when the coupling resistance $\to 0$.

Why this works. A stirred tank alone is first-order; adding a solid that exchanges heat with the liquid introduces a second state $T_m$ and a second pole. The coupling $UA$ sets how the poles split — finite $UA$ keeps them distinct, while $UA\to\infty$ locks $T_m=T$ so only the summed capacitance survives and the model collapses to first order. Recognising how many stores are truly independent is the core modelling judgement.

Common pitfall. Forgetting the solid's own balance (giving a spurious first-order model), or dropping the $UA$ coupling term from one of the two equations so the DC gain is no longer 1.

Source. Coughanowr & LeBlanc, Ch. 4–6 (multi-capacity systems); Seborg et al., Ch. 2, 4–6 (energy balances, second-order dynamics).

Question 1 of 8: Stirred Tank with a Solid Thermal Mass

Problem 1: Stirred Tank with a Solid Thermal Mass (20%)