Question 7.9.2: Consider the wall cross section shown in Figure 7.9.3. The t......
Consider the wall cross section shown in Figure 7.9.3. The temperature model was developed in Example 7.8.6. Use the following values and plot the temperatures versus time for the case where the inside temperature is constant at {{T}}_{i} = 20°C and the outside temperature {T}_{o} decreases linearly from 5°C to −10°C in 1 h. The initial wall temperatures are 10°C.
The resistance values in °C/W are
\begin{array}{l l}{{R_{a}=0.018~~~~~~R_{b}=2.023~~~~~~R_{c}=2.204~~~~~}}\\ {{R_{d}=0.223~~~~~~R_{e}=0.019~~~~~~~}}\end{array}The capacitance values in J/°C are
\begin{array}{l l}{{C_{1}=8720~~~~~~C_{2}=6210}}\\ {{C_{3}=6637~~~~~~C_{4}=2.08\times10^{4}}}\end{array}
The model was developed in Example 7.8.6.
The given information shows that the outside temperature is described by
T_{o}(t)=5-15t\quad0\leq t\leq3600\,\mathrm{s}The following MATLAB program creates the required plots. Recall that the total response is the sum of the forced and the free responses.
Script File
% htwall.m Heat transfer thru a multilayer wall.
% Resistance and capacitance data.
Ra = 0.018; Rb = 2.023; Rc = 2.204; Rd = 0.223; Re = 0.019;
C1 = 8720; C2 = 6210; C3 = 6637; C4 = 20800;
% Compute the matrix coefficients.
a11 = -(Ra+Rb)/(C1*Ra*Rb); a12 = 1/(C1*Rb);
a21 = 1/(C2*Rb); a22 = -(Rb+Rc)/(C2*Rb*Rc); a23 = 1/(C2*Rc);
a32 = 1/(C3*Rc); a33 = -(Rc+Rd)/(C3*Rc*Rd); a34 = 1/(C3*Rd);
a43 = 1/(C4*Rd); a44 = -(Rd+Re)/(C4*Rd*Re);
b11 = 1/(C1*Ra); b42 = 1/(C4*Re);
% Define the A and B matrices.
A = [a11,a12,0,0; a21,a22,a23,0; 0,a32,a33,a34; 0,0,a43,a44];
B = [b11,0; 0,0; 0,0; 0,b42];
% Define the C and D matrices.
% The outputs are the four wall temperatures.
C = eye(4);
D = zeros(size(B));
% Create the LTI model.
sys = ss(A,B,C,D);
% Create the time vector for 1 hour (3600 seconds).
t = (0:1:3600);
% Create the input vector.
u = [20*ones(size(t));(5-15*ones(size(t)).*t/3600)];
% Compute the forced response.
[yforced,t] = lsim(sys,u,t);
% Compute the free response.
[yfree,t] = initial(sys,[10,10,10,10],t);
% Plot the response along with the outside temperature.
plot(t,yforced+yfree,t,u(2,:))
% Compute the time constants.
tau =(-1./real(eig(A)))/60
The plot is shown in Figure7.9.4. Note how {T}_{1} follows the inside temperature, while {T}_{4} follows the outside temperature, as expected. The time constants are 2.6, 6, 24, and 117 min.
From the plot note that {T}_{1} reaches steady state in less than 1000 s, but the dominant time constant is 117 min, or 7020 s. The discrepancy is explained by examining the transfer functions for numerator dynamics. After running htwall, which puts the matrices A and B in the workspace, the following MATLAB session can be used to obtain the transfer functions for {T}_{1} and {T}_{2}. The C and D matrices need to be replaced to obtain only {T}_{1} and {T}_{2} as the outputs.
C1 = [1,0,0,0]; D1=[0,0];
sys1 = ss(A,B,C1,D1);
C2 = [0,1,0,0]; D2 = [0,0];
sys2 = ss(A,B,C2,D2);
tf1 = tf(sys1)
tf2 = tf(sys2)
This session produces four transfer functions. Two of them are
{\frac{T_{1}(s)}{T_{i}(s)}}={\frac{6.371\times10^{-3}s^{3}+2.321\times10^{-5}s^{2}+1.545\times10^{-8}s+1.758\times10^{-12}}{P(s)}}
{\frac{T_{2}(s)}{T_{i}(s)}}={\frac{5.071\times10^{-7}s^{2}+1.77\times10^{-9}s+9.622\times10^{-13}}{P(s)}}
where the denominator is
P(s)=s^{4}+0.01007s^{3}+2.583\times10^{-5}s^{2}+1.585\times10^{-8}s+1.765\times10^{-12}
The transfer function T_{1}(s)/T_{i}(s) has higher order numerator dynamics than T_{2}(s)/T_{i}(s). This accounts for the faster response of T_{1}.
