Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams / EGBC — December 2013 — 04-Chem-A6 Process Dynamics & Control. Three-hour open-book examination; any non-communicating calculator is permitted. Eight problems are printed and any five constitute a complete paper (each worth 20%); all eight are solved below for completeness. The paper is almost entirely quantitative — linearisation, transfer functions, step/pulse/ramp responses, Routh and frequency-response (Bode/Nyquist) stability, and IMC design — and every boxed number.
Reference texts: D. E. Seborg, T. F. Edgar, D. A. Mellichamp & F. J. Doyle III, Process Dynamics and Control (4th ed., Wiley) — linearisation, transfer functions, frequency response, IMC design; G. Stephanopoulos, Chemical Process Control: An Introduction to Theory and Practice (Prentice Hall) — Routh array, Bode and Nyquist stability, dead-time systems; D. R. Coughanowr & S. E. LeBlanc, Process Systems Analysis and Control (3rd ed., McGraw-Hill) — first-order dynamics and step/pulse/ramp response. Standard control conventions are used (deviation variables about a steady state; unity valve/sensor gains unless stated).
Given. A radiatively-cooled heated rod obeying $mC\,\dfrac{dT}{dt}=Q-k(T^{4}-T_a^{4})$; the operating (steady) state $(T_s,\,T_{as},\,Q_s)$ satisfies $Q_s=k\,(T_s^{4}-T_{as}^{4})$. Deviation variables $\delta T=T-T_s$, $\delta Q=Q-Q_s$, $\delta T_a=T_a-T_{as}$.
Find. (a) $\delta T/\delta Q$ and $\delta T/\delta T_a$ in standard form (gain and time constant); (b) the stabilising sign of the proportional gain.
Problem 1: the Calrod element. Electrical power $Q$ enters the terminals and heats the U-shaped rod to temperature $T$; the rod loses heat to the surroundings at $T_a$ predominantly by radiation, which is why the loss term is proportional to $T^{4}-T_a^{4}$ rather than to $T-T_a$.
Approach. The only non-linearity is the $T^{4}$ radiation term; expand it in a first-order Taylor series about the steady state, collect the linear ODE into standard first-order form, and read off the gains and the (common) time constant. The controller sign then follows from the sign of the process gain.
Linearise the radiation term. Expanding $f(T,Q,T_a)=Q-k(T^{4}-T_a^{4})$ to first order about the steady state, $\left.\dfrac{\partial f}{\partial T}\right|_s=-4kT_s^{3}$ and $\left.\dfrac{\partial f}{\partial T_a}\right|_s=+4kT_{as}^{3}$, so the deviation model is $$mC\,\dfrac{d\,\delta T}{dt}=\delta Q-4kT_s^{3}\,\delta T+4kT_{as}^{3}\,\delta T_a.$$
Group into standard first-order form. Moving the $\delta T$ term left and dividing by $4kT_s^{3}$: $$\underbrace{\dfrac{mC}{4kT_s^{3}}}_{\tau}\dfrac{d\,\delta T}{dt}+\delta T=\underbrace{\dfrac{1}{4kT_s^{3}}}_{K_Q}\,\delta Q+\underbrace{\dfrac{T_{as}^{3}}{T_s^{3}}}_{K_a}\,\delta T_a,$$ giving a single time constant $\boxed{\tau=\dfrac{mC}{4kT_s^{3}}}$ shared by both inputs.
Transfer function $\delta T/\delta Q$. Setting $\delta T_a=0$ and Laplace-transforming, $$\dfrac{\delta T(s)}{\delta Q(s)}=\dfrac{K_Q}{\tau s+1},\qquad K_Q=\dfrac{1}{4kT_s^{3}}\;(>0).$$
Transfer function $\delta T/\delta T_a$. Setting $\delta Q=0$, $$\dfrac{\delta T(s)}{\delta T_a(s)}=\dfrac{K_a}{\tau s+1},\qquad K_a=\left(\dfrac{T_{as}}{T_s}\right)^{3}\;(0<K_a<1).$$ Both are first-order lags with the same $\tau$; the ambient gain $K_a$ is a pure ratio, always between 0 and 1 because $T_{as}<T_s$.
(b) Controller sign. The process gain $K_Q=1/(4kT_s^{3})$ is positive (more power ⇒ higher temperature). The closed-loop characteristic equation with a proportional controller $G_c=K_c$ is $\tau s+1+K_cK_Q=0$, i.e. a single pole at $s=-(1+K_cK_Q)/\tau$. Stability ($s<0$) needs $1+K_cK_Q>0$, so any $\boxed{K_c>0}$ (reverse-acting, negative feedback) keeps the loop stable — a first-order process cannot be destabilised by a correctly-signed proportional controller. A negative gain would be positive feedback and, for $K_c<-1/K_Q$, would drive the pole into the right half-plane.
