Linearization of a Radiative Calrod Heating Element: 20-Bio-A2 Process Dynamics and Control, May 2018, Q5

Problem 5: Linearization of a Radiative Calrod Heating Element (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. $mC\,dT/dt=Q-k(T^4-T_a^4)$; steady state $(T_s,T_{as},Q_s)$ with $Q_s=k(T_s^4-T_{as}^4)$; $m$, $C$, $k$ all positive physical constants.

Find. (a) $\delta T(s)/\delta Q(s)$ and $\delta T(s)/\delta T_a(s)$ in standard gain/time-constant form; (b) the required sign of the proportional controller gain $K_c$.

Approach. Perturb about the steady state, linearize the $T^4$ and $T_a^4$ terms by a first-order Taylor expansion ($T^4\approx T_s^4+4T_s^3\delta T$), Laplace-transform the resulting linear ODE, and solve for each transfer function in turn (holding the other input at its steady value). For (b), examine the sign of the process gain and its effect on the closed-loop pole.

Linearize. $T^4\approx T_s^4+4T_s^3\,\delta T$ and $T_a^4\approx T_{as}^4+4T_{as}^3\,\delta T_a$. Substituting and subtracting the steady-state balance leaves $$mC\frac{d(\delta T)}{dt}+4kT_s^3\,\delta T=\delta Q+4kT_{as}^3\,\delta T_a.$$
Laplace transform. $(mCs+4kT_s^3)\,\delta T(s)=\delta Q(s)+4kT_{as}^3\,\delta T_a(s)$.
(a) $\delta T/\delta Q$ (with $\delta T_a=0$). $$\boxed{\frac{\delta T(s)}{\delta Q(s)}=\frac{1}{mCs+4kT_s^3}=\frac{K_1}{\tau s+1},\qquad K_1=\frac{1}{4kT_s^3},\quad \tau=\frac{mC}{4kT_s^3}.}$$
(a) $\delta T/\delta T_a$ (with $\delta Q=0$). $$\boxed{\frac{\delta T(s)}{\delta T_a(s)}=\frac{4kT_{as}^3}{mCs+4kT_s^3}=\frac{K_2}{\tau s+1},\qquad K_2=\left(\frac{T_{as}}{T_s}\right)^3,}$$ using the same time constant $\tau$ as above (both inputs enter the same first-order lag).
(b) Sign of $K_c$. The process gain $K_1=1/(4kT_s^3)$ is strictly positive (all physical quantities are positive): more heat input $Q$ always raises $T$. With proportional feedback, the closed-loop characteristic equation is $\tau s+1+K_cK_1=0\Rightarrow s=-(1+K_cK_1)/\tau$. If $T$ falls below setpoint (positive error), the controller must increase $Q$ to correct it — requiring $K_c\gt0$ for the correct (negative-feedback) control action; a negative $K_c$ would decrease $Q$ on a low-temperature error, driving $T$ further away and, for sufficiently negative $K_c$, pushing $1+K_cK_1<0$ and the pole into the right half-plane. $$\boxed{K_c\gt0\ \text{is required (direct-acting process, correct-sign reverse control action).}}$$

Result	Value
Time constant	$\tau=mC/(4kT_s^3)$
Gain $\delta T/\delta Q$	$K_1=1/(4kT_s^3)$
Gain $\delta T/\delta T_a$	$K_2=(T_{as}/T_s)^3$
Controller sign	$K_c\gt0$

Concept & Theory

Topic. Taylor-series linearization of a nonlinear radiative-heat-transfer energy balance.

Key relations. $x^4\approx x_s^4+4x_s^3\,\delta x$; first-order lag standard form $K/(\tau s+1)$; closed-loop pole $-(1+K_cK)/\tau$ for P-control of a first-order process.

Why this works. A first-order Taylor expansion of the fourth-power radiative terms about the operating point converts the strongly nonlinear balance into a linear ODE valid for small deviations, and the resulting model is a classic first-order-lag with two inputs sharing the same time constant (because both enter the same energy-storage term $mC\,dT/dt$).

Common pitfall. Linearizing $T^4$ but forgetting to also linearize $T_a^4$ (dropping the $\delta T_a$ path entirely), or mis-assigning the sign of $K_c$ by not checking the sign of the process gain first.

Source. Seborg et al., Process Dynamics and Control, Ch. 4 (linearization of nonlinear models) and Ch. 8 (proportional-control sign/action selection).

Question 5 of 8: Linearization of a Radiative Calrod Heating Element

Problem 5: Linearization of a Radiative Calrod Heating Element (20%)