Two Concentric Tanks — Second-Order Thermal Response: 23-Chem-A6 Process Dynamics and Control, May 2016, Q4

Problem 4: Two Concentric Tanks — Second-Order Thermal Response (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. Two well-mixed tanks coupled by a conducting wall:

Quantity	Symbol	Value
Thermal capacitance each tank	$\rho V c_p$	$50\cdot1\cdot2=100\ \mathrm{Btu/{}^\circ F}$
Wall conductance	$UA$	$10\cdot1=10\ \mathrm{Btu/(hr\,{}^\circ F)}$
Outer-flow capacity	$w c_p$	$10\cdot2=20\ \mathrm{Btu/(hr\,{}^\circ F)}$
Step heat input (inner)	$Q$	$0\to500\ \mathrm{Btu/hr}$
Feed & initial temperatures	$T_f,T_{1,0},T_{2,0}$	$100\ ^\circ$F

Find. (a) $T_1(s)$ (inner-tank deviation temperature); (b) $T_1$ at $t=0$ and $t=5$ hr.

Problem 4: the immersed heater adds $Q$ to the inner tank; heat conducts through the wall to the outer tank ($UA(T_1-T_2)$), which is swept by the through-flow $w$. Two capacitances in a feedback path give a second-order response.

Approach. Write an energy balance on each tank in deviation variables (feed constant, so its deviation is zero), Laplace-transform the pair, eliminate the outer-tank temperature to get $T_1(s)$, then invert by partial fractions and evaluate at the two times.

Energy balances (deviation form). With capacitance $100$, wall term $10$, flow term $20$ and $Q$ a step of $500$: $$100\frac{dT_1}{dt}=Q-10(T_1-T_2),\qquad 100\frac{dT_2}{dt}=-20T_2+10(T_1-T_2).$$ (Primes for deviation dropped; $T_f'=0$.)
Transform. $$(100s+10)T_1-10T_2=Q,\qquad -10T_1+(100s+30)T_2=0.$$ From the second equation $T_2=\dfrac{10\,T_1}{100s+30}$.
(a) Eliminate $T_2$. Substituting and clearing, $(100s+10)(100s+30)-100=10000s^2+4000s+200$, so with $Q=500/s$: $$\boxed{T_1(s)=\frac{500\,(100s+30)}{s\,(10000s^{2}+4000s+200)}=\frac{25\,(10s+3)}{s\,(50s^{2}+20s+1)}.}$$
Poles. $50s^2+20s+1=0$ gives $s=-0.0586,\,-0.3414\ \mathrm{hr^{-1}}$ (time constants $17.1$ and $2.93$ hr) — two real, over-damped roots, as expected for a two-capacitance thermal system.
(b) Partial-fraction inversion. Writing $T_1(s)=\dfrac{A}{s}+\dfrac{B}{s+0.0586}+\dfrac{C}{s+0.3414}$ gives $A=75$ (the steady offset), $B=-72.86$, $C=-2.14$, hence $$\boxed{T_1(t)=75-72.86\,e^{-0.0586t}-2.14\,e^{-0.3414t}\ \ (^\circ\text{F above }100).}$$
(b) Evaluate. At $t=0$: $75-72.86-2.14=0$, so $\boxed{T_1(0)=100,{}^ rc\text{F}}$ (matches the initial condition). At $t=5$ hr: $75-72.86e^{-0.293}-2.14e^{-1.707}=75-54.36-0.39=20.25$, so $\boxed{T_1(5)=120.3,{}^ rc\text{F}}$. The steady value is $175,{}^ rc$F, confirmed by the overall balance ($Q=wc_p\,\Delta T_2\Rightarrow T_2=125$; $Q=UA\,\Delta T\Rightarrow T_1=175$).

Problem 4(b): the inner tank climbs smoothly (over-damped, no overshoot) from $100,{}^ rc$F toward the steady $175,{}^ rc$F; at $t=5$ hr it has reached $120.3,{}^ rc$F. The slow $17.1$ hr mode dominates the long tail.

Quantity	Value
(a) $T_1(s)$	$\dfrac{25(10s+3)}{s(50s^2+20s+1)}$
Time constants	$17.1$ hr and $2.93$ hr
(b) $T_1(0)$	$100,{}^ rc$F
(b) $T_1(5\,\mathrm{hr})$	$120.3,{}^ rc$F
Steady-state $T_1$	$175,{}^ rc$F

Concept & Theory

Topic. Coupled energy balances; second-order (two-capacitance) thermal dynamics; Laplace inversion by partial fractions; final-value check.

Key relations. Lumped balance $\rho Vc_p\,dT/dt=\sum(\text{heat in}-\text{heat out})$; wall coupling $UA(T_1-T_2)$; final-value theorem $T_1(\infty)=\lim_{s\to0}sT_1(s)$.

Why this works. Each well-mixed tank is one first-order storage; the conducting wall links them, so the transfer function is second order. Because the two poles are real and negative the response is over-damped—a monotone rise—and the steady offset ($75,{}^ rc$F) is fixed by where the added power finally exits through the outer flow.

Common pitfall. Omitting the $-100$ cross-coupling term when eliminating $T_2$ (it comes from the $10T_2$ feedback), or forgetting that the inner tank has no flow term—only the wall conduction removes its heat.

Source. Coughanowr & LeBlanc, Ch. 7–8 (interacting/second-order thermal systems); Seborg et al., Ch. 2 & 6.

Question 4 of 8: Two Concentric Tanks — Second-Order Thermal Response

Problem 4: Two Concentric Tanks — Second-Order Thermal Response (20%)