Linearising a Radiative (Calrod) Heater: 23-Chem-A6 Process Dynamics and Control, December 2015, Q1

Problem 1: Linearising a Radiative (Calrod) Heater (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, radiation-dominated heater:

Quantity	Symbol	Role
Energy balance	—	$mC\,dT/dt=Q-k(T^4-T_a^4)$
Heater mass × specific heat	$mC$	thermal capacitance
Radiation coefficient	$k$	$\dot q_{rad}=k(T^4-T_a^4)$
Manipulated input	$Q$	electrical power
Disturbance	$T_a$	ambient temperature

Find. (a) $\dfrac{\delta T}{\delta Q}$ and $\dfrac{\delta T}{\delta T_a}$ in the form $\dfrac{K}{\tau s+1}$; (b) the controller sign for a stable loop.

Problem 1: electrical power $Q$ heats a rod of capacitance $mC$; the rod loses heat to the surroundings by radiation, $\dot q_{rad}=k(T^4-T_a^4)$. The strong non-linearity ($T^4$) is linearised about the operating point $(T_s,T_{as})$.

Approach. Expand the non-linear radiation term in a first-order Taylor series about the steady operating point, cast the result as a first-order lag in deviation variables, then read off gain and time constant for each input.

Steady state and deviation variables. Define deviations $\delta T=T-T_s$, $\delta Q=Q-Q_s$, $\delta T_a=T_a-T_{as}$. At steady state $Q_s=k\!\left(T_s^4-T_{as}^4\right)$, so subtracting removes the constant terms and leaves an equation purely in deviations.
Linearise the $T^4$ term. Taylor-expand each quartic about its steady value: $T^4\approx T_s^4+4T_s^3\,\delta T$ and $T_a^4\approx T_{as}^4+4T_{as}^3\,\delta T_a$. Substituting into the balance gives $$mC\,\frac{d(\delta T)}{dt}=\delta Q-4kT_s^3\,\delta T+4kT_{as}^3\,\delta T_a.$$
Standard first-order form. Collect the $\delta T$ terms and divide by the radiative conductance $4kT_s^3$: $$\boxed{\tau\,\frac{d(\delta T)}{dt}+\delta T=K_Q\,\delta Q+K_a\,\delta T_a},\qquad \tau=\frac{mC}{4kT_s^{3}}.$$ Both the manipulated input and the disturbance pass through the same lag $\tau$.
Transfer function to $\delta Q$. Laplace-transforming with $\delta T_a=0$: $$\boxed{\frac{\delta T(s)}{\delta Q(s)}=\frac{K_Q}{\tau s+1}},\qquad K_Q=\frac{1}{4kT_s^{3}}\ \left[\mathrm{K/W}\right].$$ The gain is the inverse radiative conductance — a hotter set point ($T_s\!\uparrow$) makes the heater both faster ($\tau\!\downarrow$) and less sensitive ($K_Q\!\downarrow$) because radiative loss stiffens as $T_s^3$.
Transfer function to $\delta T_a$. With $\delta Q=0$: $$\boxed{\frac{\delta T(s)}{\delta T_a(s)}=\frac{K_a}{\tau s+1}},\qquad K_a=\frac{4kT_{as}^{3}}{4kT_s^{3}}=\left(\frac{T_{as}}{T_s}\right)^{3}.$$ The disturbance gain is the dimensionless cube of the temperature ratio — always $<1$ for a heater run above ambient, so ambient swings are strongly attenuated.
(b) Controller sign. With proportional control $\delta Q=K_c\,\delta e=K_c(\delta T_{sp}-\delta T)$, the closed-loop characteristic equation is $\tau s+1+K_cK_Q=0$, giving the pole $$s=-\frac{1+K_cK_Q}{\tau}.$$ Since $K_Q>0$ and $\tau>0$, choosing $\boxed{K_c>0\ \text{(positive / direct-acting on error)}}$ keeps $1+K_cK_Q>0$, so the pole stays in the left half-plane for any positive gain — the loop is unconditionally stable. Physically: the process gain is positive (more $Q\Rightarrow$ higher $T$), so negative feedback requires a positive controller gain; a negative $K_c$ would only threaten stability once $K_c<-1/K_Q$.

Result	Expression
Time constant	$\tau=mC/(4kT_s^3)$
$\delta T/\delta Q$	$K_Q/(\tau s+1)$, $K_Q=1/(4kT_s^3)$
$\delta T/\delta T_a$	$K_a/(\tau s+1)$, $K_a=(T_{as}/T_s)^3$
(b) Stabilising controller sign	$K_c>0$ (positive)

Check — sanity check of the gains

With illustrative numbers $mC=25\ \mathrm{J/K}$, $k=2\times10^{-8}\ \mathrm{W/K^4}$, $T_s=800\ \mathrm K$, $T_{as}=300\ \mathrm K$: $4kT_s^3=40.96\ \mathrm{W/K}$, so $\tau=0.61\ \mathrm s$, $K_Q=0.0244\ \mathrm{K/W}$, $K_a=(300/800)^3=0.0527$. The pole $-(1+K_cK_Q)/\tau$ is negative for all $K_c>0$. (Exact symbolic forms are what the exam requires; numbers only illustrate the signs.)

Concept & Theory

Topic. Linearisation of a non-linear energy balance; first-order lag; process gain versus disturbance gain; controller-sign selection.

Key relations. Taylor linearisation $f(x)\approx f(x_s)+f'(x_s)\delta x$; standard lag $\tau\,dy/dt+y=Ku$ with $\tau=(\text{capacitance})/(\text{conductance})$; radiative conductance $\partial\dot q/\partial T=4kT_s^3$; closed-loop pole $-(1+K_cK)/\tau$.

Why this works. Near an operating point the quartic loss behaves like a linear resistance of slope $4kT_s^3$, so the heater is a first-order thermal capacitance–resistance system. Because the same conductance sets both $\tau$ and the denominator of each gain, $Q$ and $T_a$ share the lag but scale differently (Watt input vs. Kelvin disturbance).

Common pitfall. Forgetting the sign on the $T_a^4$ term (it enters with $+$ because losses decrease when ambient rises), or reporting $K_a$ with units — it is dimensionless. In (b), quoting "negative gain" by analogy with a cooling loop; here the process gain is positive so $K_c>0$.

Source. Seborg et al., Process Dynamics and Control, Ch. 4 (linearisation, deviation variables) and Ch. 11 (proportional feedback, sign/stability); Coughanowr & LeBlanc, Ch. 4–5 (thermal first-order systems).

Question 1 of 8: Linearising a Radiative (Calrod) Heater

Problem 1: Linearising a Radiative (Calrod) Heater (20%)