Question 5 of 8: Thermocouple Dynamics — Response to a Triangular Bath Pulse

Problem 5: Thermocouple Dynamics — Response to a Triangular Bath 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. A lumped-capacitance thermocouple:

Find. (1) the transfer function $T_{tc}/T_L$; (2) the registered temperature $T_{tc}(t)$.

Approach. Write a lumped energy balance on the thermocouple bead to get a first-order lag and its time constant; then decompose the triangular bath pulse into three ramps and superpose the first-order ramp responses.

1. (1) Energy balance and transfer function. With no internal gradients, $m c_p\dfrac{dT_{tc}}{dt}=hA\,(T_L-T_{tc})$. Rearranging into standard form, $\tau\dfrac{dT_{tc}}{dt}+T_{tc}=T_L$ with $$\tau=\dfrac{m c_p}{hA}=\dfrac{(0.25)(1)}{(60)(1)}=\dfrac{1}{240}\ \mathrm{h}=\boxed{15\ \mathrm{s}},$$ so $$\dfrac{T_{tc}(s)}{T_L(s)}=\dfrac{1}{\tau s+1}=\dfrac{1}{15s+1}$$ (unity gain — the thermocouple eventually reads the true bath temperature).
2. Decompose the triangular pulse. The bath temperature has slope $+1\,{}^{\circ}\mathrm{C/s}$ for $0$–$300\ \mathrm{s}$ then $-1\,{}^{\circ}\mathrm{C/s}$ for $300$–$600\ \mathrm{s}$. As shifted ramps, $$T_L(t)=t\,\mathcal{U}(t)-2(t-300)\,\mathcal{U}(t-300)+(t-600)\,\mathcal{U}(t-600),$$ i.e. $T_L(s)=\dfrac{1}{s^2}\big[1-2e^{-300s}+e^{-600s}\big]$.
3. First-order ramp response. For a unit-slope ramp into $1/(\tau s+1)$, $\mathcal{L}^{-1}\!\big\{\tfrac{1}{s^2(\tau s+1)}\big\}=g(t)\equiv t-\tau+\tau e^{-t/\tau}$. The thermocouple reading is the same superposition: $$\boxed{T_{tc}(t)=g(t)-2\,g(t-300)+g(t-600)},\quad g(t)=t-15+15e^{-t/15}.$$
4. Evaluate key points. During the rise ($t\gg\tau$) $T_{tc}\approx t-15$, a constant $15\,{}^{\circ}\mathrm{C}$ lag. At the bath peak, $T_{tc}(300)=300-15\approx\boxed{285\,{}^{\circ}\mathrm{C}}$. The reading keeps rising briefly, peaking at $t\approx300+\tau\ln2=310\ \mathrm{s}$ with $T_{tc}\approx\boxed{289.6\,{}^{\circ}\mathrm{C}}$ — it never reaches the true $300\,{}^{\circ}\mathrm{C}$. When the bath returns to $0$ at $t=600\ \mathrm{s}$, the thermocouple still reads $T_{tc}(600)\approx15\,{}^{\circ}\mathrm{C}$.
5. Tail ($t>600\ \mathrm{s}$). With the bath at zero the linear parts cancel and only a decaying exponential remains: $$T_{tc}(t)\approx 15\,e^{-(t-600)/15}\,{}^{\circ}\mathrm{C},$$ so the reading relaxes to zero with the same $15\ \mathrm{s}$ time constant. The thermocouple thus rounds and lags the sharp triangular pulse, under-reading the peak by $\approx10\,{}^{\circ}\mathrm{C}$ and trailing it by roughly one time constant.