23-Chem-A6 Process Dynamics and Control · December 2017
Question 2 of 8: Linearising a Radiative Heater
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Reference texts: D. E. Seborg, T. F. Edgar, D. A. Mellichamp & F. J. Doyle III, Process Dynamics and Control (4th ed., Wiley) — Laplace-domain modelling, linearisation, transfer functions, feedback stability, frequency response and IMC design; G. Stephanopoulos, Chemical Process Control: An Introduction to Theory and Practice (Prentice Hall) — Routh test, Nyquist criterion and dead-time systems; D. R. Coughanowr & S. E. LeBlanc, Process Systems Analysis and Control (3rd ed., McGraw-Hill) — first-order thermal/level dynamics, step/ramp/pulse response and block-diagram algebra. Standard control conventions (deviation variables; unity valve/sensor gains unless stated) are used throughout. This December 2017 sitting is an open-book paper of eight problems of which any five constitute a complete exam; all eight are solved here.
Given. Nonlinear energy balance $mC\,\dot T=Q-k(T^4-T_a^4)$; steady operating point $(\,\overline T,\overline{T_a},\overline Q\,)$ with $\overline Q=k(\overline T^{\,4}-\overline{T_a}^{\,4})$.
Find. First-order transfer functions $\delta T/\delta Q$ and $\delta T/\delta T_a$ (gain + time constant), then the stabilising sign of $K_c$.
The rod stores energy ($mC\,\dot T$), gains it from the electrical input $Q$, and loses it to the surroundings as radiation $k(T^4-T_a^4)$. Only the radiation term is nonlinear, so it is what we linearise.
Approach. Expand the radiation term in a first-order Taylor series about the steady state, subtract the steady-state balance to obtain deviation variables, and Laplace-transform to read off gain and time constant.
Taylor-expand the nonlinearity. $k(T^4-T_a^4)\approx k(\overline T^{\,4}-\overline{T_a}^{\,4})+4k\overline T^{\,3}\,\delta T-4k\overline{T_a}^{\,3}\,\delta T_a$, where $\delta T=T-\overline T$, $\delta T_a=T_a-\overline{T_a}$.
Form the deviation model. Subtracting the steady balance leaves $$mC\,\frac{d\,\delta T}{dt}=\delta Q-4k\overline T^{\,3}\,\delta T+4k\overline{T_a}^{\,3}\,\delta T_a.$$
Laplace transform & group. $\big(mC\,s+4k\overline T^{\,3}\big)\delta T(s)=\delta Q(s)+4k\overline{T_a}^{\,3}\,\delta T_a(s)$. Dividing by $4k\overline T^{\,3}$ puts it in standard form with $$\boxed{\tau=\frac{mC}{4k\overline T^{\,3}}.}$$
(a) Transfer function to $Q$. With $\delta T_a=0$: $$\boxed{\frac{\delta T}{\delta Q}=\frac{K_Q}{\tau s+1},\qquad K_Q=\frac{1}{4k\overline T^{\,3}}>0.}$$
(a) Transfer function to $T_a$. With $\delta Q=0$: $$\boxed{\frac{\delta T}{\delta T_a}=\frac{K_a}{\tau s+1},\qquad K_a=\frac{4k\overline{T_a}^{\,3}}{4k\overline T^{\,3}}=\Big(\frac{\overline{T_a}}{\overline T}\Big)^{3}.}$$ Because $\overline{T_a}<\overline T$, this ambient gain is a positive fraction less than one.
(b) Controller sign. Raising $Q$ raises $T$ ($K_Q>0$), so the process is direct-acting; a stabilising negative feedback loop needs a controller of matching sign, i.e. $\boxed{K_c>0}$. The closed-loop characteristic equation $\tau s+1+K_cK_Q=0$ gives the single pole $s=-(1+K_cK_Q)/\tau$, which is negative for every $K_c>0$ — a first-order process is unconditionally stable under proportional control of the correct sign.
