Thermocouple Response to a Triangular Pulse: 23-Chem-A6 Process Dynamics and Control, December 2016, Q3

Problem 3: Thermocouple Response to a Triangular 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 single-capacity sensor exchanging heat by convection:

Quantity	Symbol	Value
Mass	$m$	$0.25$ g
Heat capacity	$C$	$1\ \text{cal g}^{-1}\,{}^\circ\text{C}^{-1}$
Heat-transfer coefficient	$h$	$60\ \text{cal cm}^{-2}\,\text{h}^{-1}\,{}^\circ\text{C}^{-1}$
Surface area	$A$	$1\ \text{cm}^2$

Find. (1) $T(s)/T_L(s)$; (2) the reading $T(t)$ for the triangular bath input.

The bath (blue) ramps up then down as a symmetric triangle peaking at $300\,{}^\circ$C. The thermocouple reading (red) lags by the time constant: it reaches only $\approx285\,{}^\circ$C at the bath apex and its own maximum of $289.6\,{}^\circ$C occurs later, at $t=310.4$ s, then decays.

Approach. Write the lumped energy balance to get a first-order lag, evaluate its time constant, then decompose the triangular input into three ramps and superpose the known first-order ramp response.

Energy balance → transfer function. $mC\,\dfrac{dT}{dt}=hA\,(T_L-T)$. In deviation/Laplace form $$\frac{T(s)}{T_L(s)}=\frac{1}{\tau s+1},\qquad \tau=\frac{mC}{hA}.$$
Evaluate $\tau$. $\tau=\dfrac{(0.25)(1)}{(60)(1)}=4.167\times10^{-3}\ \text{h}\times3600\ \text{s/h}$, i.e. $$\boxed{\tau=15\ \text{s},\qquad \frac{T(s)}{T_L(s)}=\frac{1}{15s+1}.}$$
Decompose the input. The triangular pulse is three ramps of slope $1\,{}^\circ$C/s: $T_L(t)=r(t)-2\,r(t-300)+r(t-600)$, where $r(t)=t\,u(t)$.
Superpose the ramp response. A first-order lag driven by a unit-slope ramp gives $g(t)=t-\tau+\tau e^{-t/\tau}$ (for $t\ge0$). By linearity $$\boxed{T(t)=g(t)-2\,g(t-300)+g(t-600).}$$
Reading at the input apex ($t=300$ s). Only the first two ramps are active: $T(300)=\big(300-15+15e^{-20}\big)-2\big(0\big)=285\,{}^\circ$C. The reading trails the bath by $a\tau=15\,{}^\circ$C at this instant.
True peak. In $300

Quantity	Value
Transfer function	$1/(15s+1)$
Time constant	$\tau=15$ s
Reading at bath apex ($t=300$ s)	$285\,{}^\circ$C
True peak reading	$289.6\,{}^\circ$C at $t=310.4$ s

Concept & Theory

Topic. First-order thermal sensor; ramp response; superposition of shifted ramps.

Key relations. $\tau=mC/hA$; unit-ramp response $g(t)=t-\tau+\tau e^{-t/\tau}$; steady ramp error $=a\tau$; peak of a triangular response at $t=t_{\text{break}}+\tau\ln2$.

Why this works. The bead is a single thermal capacitance $mC$ charged through the film conductance $hA$, so it behaves as a pure first-order lag. A triangle is a superposition of ramps; because the system is linear, its response is the same superposition of ramp responses, and the reading peaks slightly after the input because the sensor must first stop chasing the still-rising input.

Common pitfall. Forgetting the hour→second conversion in $\tau$ (the coefficient is per hour), or reporting the peak at $t=300$ s; the true maximum is delayed by $\tau\ln2\approx10.4$ s.

Source. Coughanowr & LeBlanc, Process Systems Analysis and Control (3rd ed., McGraw-Hill) Ch. 5 (first-order dynamics, ramp forcing); Seborg, Edgar, Mellichamp & Doyle, Process Dynamics and Control (4th ed., Wiley) Ch. 5.

Question 3 of 8: Thermocouple Response to a Triangular Pulse

Problem 3: Thermocouple Response to a Triangular Pulse (20%)