Control Systems · State-space, controllability, and observability
Two lumped thermal masses exchange heat by conduction and lose heat to a constant ambient
Problem
Two lumped thermal masses exchange heat by conduction and lose heat to a constant ambient. Mass 1 has capacitance \(C_1=2.00\times 10^3\,\mathrm{J\,K^{-1}}\) and temperature \(T_1\) (kelvin). Mass 2 has capacitance \(C_2=8.00\times 10^2\,\mathrm{J\,K^{-1}}\) and temperature \(T_2\). The conduction path between the masses is a linear thermal resistance \(R_{12}=0.0500\,\mathrm{K\,W^{-1}}\), with heat flow \((T_1-T_2)/R_{12}\) positive from 1 to 2. Mass 2 loses heat \((T_2-T_a)/R_{2a}\) to a large ambient at constant \(T_a\), where \(R_{2a}=0.100\,\mathrm{K\,W^{-1}}\). An electrical heater supplies a heat rate \(u(t)\) (watts) into mass 1 only. Mass 1 has no direct ambient path. Energy balances are \(C_i\dot T_i=\)net heat into mass \(i\). Radiation, spatial gradients, and temperature dependence of the resistances are neglected. (a) Write the energy balances in the absolute temperatures \(T_1,T_2\). Identify every constant-equilibrium pair \((T_{1*},T_{2*})\) corresponding to a constant heater power \(u_*\) and the given constant \(T_a\). (b) Define deviation variables \(\delta T_1=T_1-T_a\), \(\delta T_2=T_2-T_a\), and keep \(u\) as the input (the heater bias is not subtracted). Show that the deviation system is linear and unforced by \(T_a\). Report the state-space matrices for \(\mathbf{x}=(\delta T_1,\delta T_2)^\mathrm{T}\), input \(u\), and output \(y=\delta T_2\), with units. (c) Compute the monic characteristic polynomial and the eigenvalues. Compute the zero-state transfer function \(H(s)=Y(s)/U(s)\) as a ratio of coprime real polynomials with monic denominator, and state its units. (d) For \(u_*=40.0\,\mathrm{W}\) and \(T_a=293.15\,\mathrm{K}\), compute the unique equilibrium \((T_{1*},T_{2*})\) in kelvin. Confirm that \(\delta T_{2*}=H(0)\,u_*\).
Hint
** Energy in each mass changes by net heat; the ambient appears only in mass 2's loss term. Measuring temperatures above ambient cancels \(T_a\) identically because the conduction laws are linear.
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