Question 4 of 8: Thermocouple Dynamics — Response to a Triangular Input

Problem 4: Thermocouple Dynamics — Response to a Triangular Input (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 lumped thermocouple bead exchanging heat with a stirred bath:

Find. (a) $T_{tc}/T_\ell$; (b) the registered temperature $T_{tc}(t)$ for the triangular liquid profile.

Approach. Write the bead energy balance to get a unity-gain first-order lag, then decompose the triangular input into three ramps (slopes $+1,-2,+1$) and superpose the standard first-order ramp response.

1. (a) Transfer function. Energy balance on the bead: $mC\dfrac{dT_{tc}}{dt}=hA\left(T_\ell-T_{tc}\right)$. Dividing by $hA$: $$\frac{mC}{hA}\frac{dT_{tc}}{dt}+T_{tc}=T_\ell\ \Rightarrow\ \boxed{\frac{T_{tc}(s)}{T_\ell(s)}=\frac{1}{\tau s+1}},\ \ \tau=\frac{mC}{hA}.$$
2. Time constant. Convert $h$ to per-second: $hA=60\ \tfrac{\mathrm{cal}}{\mathrm{cm^2\,h\,{}^\circ C}}\times1\ \mathrm{cm^2}=60\ \tfrac{\mathrm{cal}}{\mathrm{h\,{}^\circ C}}=\dfrac{60}{3600}=0.01667\ \tfrac{\mathrm{cal}}{\mathrm{s\,{}^\circ C}}$; $mC=0.25\ \mathrm{cal/^\circ C}$. Hence $$\tau=\frac{0.25}{0.01667}=\boxed{15\ \mathrm s}\quad(\text{unity gain}).$$
3. (b) Decompose the input. The triangular liquid temperature is the superposition of three ramps: slope $+1\,\mathrm{^\circ C/s}$ starting at $t=0$; slope $-2\,\mathrm{^\circ C/s}$ starting at $t=300$ (turning the rise into a fall); slope $+1\,\mathrm{^\circ C/s}$ starting at $t=600$ (levelling off at $0$).
4. First-order ramp response. For a ramp of slope $a$ applied at $t_0$, the first-order lag output is $r_a(t)=a\big[(t-t_0)-\tau(1-e^{-(t-t_0)/\tau})\big]$ for $t\ge t_0$. Superposing the three ramps: $$\boxed{T_{tc}(t)=r_{+1}(t;0)+r_{-2}(t;300)+r_{+1}(t;600)},\ \tau=15\ \mathrm s.$$
5. Interpret — lag and peak. During the first ramp the reading trails the input by the ramp offset $a\tau=1\times15=15\,{}^\circ\mathrm C$, so at the input apex $t=300\ \mathrm s$ the thermocouple reads $\approx300-15=285\,{}^\circ\mathrm C$ — it has not caught up. Its own peak occurs slightly later, at $t=300+\tau\ln2=310.4\ \mathrm s$, where $T_{tc}\approx289.6\,{}^\circ\mathrm C$; thereafter it lags the falling input and relaxes toward $0$. The bead never registers the true $300\,{}^\circ\mathrm C$ because $\tau$ is comparable to the ramp duration.