Question 3 of 8: Thermocouple Response to a Triangular Bath-Temperature Pulse

Problem 3: Thermocouple Response to a Triangular Bath-Temperature Pulse (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. $m=0.25\ \text{g}$, $C=1\ \text{cal/g}\,{}^{\circ}\text{C}$, $h=60\ \text{cal/cm}^2\,\text{h}\,{}^{\circ}\text{C}$, $A=1\ \text{cm}^2$; $T_{liquid}$ rises linearly $0\to300^{\circ}\text{C}$ over $0\le t\le300\ \text{s}$, falls linearly $300\to0^{\circ}\text{C}$ over $300\le t\le600\ \text{s}$, constant thereafter.

Find. (a) $T'(s)/T'_{liquid}(s)$; (b) $T'(t)$, the thermocouple-registered temperature (deviation variable), for the full triangular pulse.

Approach. Write the lumped energy balance on the bead ($mC\,dT/dt=hA(T_{liquid}-T)$), Laplace-transform to a standard first-order lag, decompose the triangular pulse into three superposed ramp functions (starting at $t=0,300,600\ \text{s}$), and sum the known first-order ramp response for each.

1. (a) Energy balance and transfer function. $mC\dfrac{dT}{dt}=hA(T_{liquid}-T)\Rightarrow \tau\dfrac{dT'}{dt}+T'=T'_{liquid}$, with $$\tau=\frac{mC}{hA}=\frac{0.25\ \text{cal/}{}^{\circ}\text{C}}{60\ \text{cal/(cm}^2\text{h}\,{}^{\circ}\text{C})\times1\ \text{cm}^2}=\frac{0.25}{60}\ \text{h}=15\ \text{s}.$$ So $$\boxed{\frac{T'(s)}{T'_{liquid}(s)}=\frac{1}{15s+1}\quad(\tau=15\ \text{s},\ K=1).}$$
2. (b) Decompose the triangular pulse into ramps. With unit-slope ramps $r(t-t_0)=(t-t_0)u(t-t_0)$: $$T'_{liquid}(t)=r(t)-2r(t-300)+r(t-600),$$ since slope $+1^{\circ}\text{C/s}$ on $[0,300]$, then $-1-(+1)=-2^{\circ}\text{C/s}$ change of slope at $t=300$ (net slope $-1$), then $+1$ change back to $0$ slope at $t=600$. (Check: at $t=300$, value $=300$; at $t=600$, value $=300-2(300)+0=-300+300=0$; for $t>600$, value $=t-2(t-300)+(t-600)=0$. All match the figure.)
3. Ramp response of a first-order lag. $\mathcal L^{-1}\!\left[\dfrac{1}{s^2(\tau s+1)}\right]=t-\tau(1-e^{-t/\tau})$ for $t\ge0$ (zero before the ramp starts).
4. (b) Superpose the three ramp responses. $$\boxed{T'(t)=\big[t-\tau(1-e^{-t/\tau})\big]u(t)-2\big[(t-300)-\tau(1-e^{-(t-300)/\tau})\big]u(t-300)+\big[(t-600)-\tau(1-e^{-(t-600)/\tau})\big]u(t-600),\ \ \tau=15\ \text{s}.}$$
5. Evaluate at representative times. $T'(150)=135.0^{\circ}\text{C}$ (rising, lagging the $150^{\circ}\text{C}$ input by the first-order dynamics), $T'(300)=285.0^{\circ}\text{C}$ (thermocouple has not yet caught up to the $300^{\circ}\text{C}$ input peak), the thermocouple's own peak occurs slightly after the input peak, at $t\approx310.4\ \text{s}$, $T'_{\max}\approx289.6^{\circ}\text{C}$ (attenuated and delayed relative to the true $300^{\circ}\text{C}$ peak — classic first-order lag behind a ramp-down/up input), $T'(600)=15.0^{\circ}\text{C}$ (still cooling toward the return-to-baseline), and $T'(t)\to0$ as $t\to\infty$.