Question

you can do by hand

Question 8: Compute the Laplace Transform Ts, then use the final value theorem to find the limiting temperature in each room after a long time. Use eye(3) for the 3x3 identity matrix I. Use inv() for the inverse matrix of the matrix. Declare s to be symbolic Name your result Ts. This is the solution for the model in the s-domain. l(T) = (sl-A)-1 T(0) +-. f >> Tinf vpa (limit (s*TS, 0), 4) = After you have computed Ts, you can obtain a nice form for the solution back in the time domain using: Tt = vpa (ílaplace (TS), 4) I found: 46.15 12.82*exp(-0.5*t)(cosh (0.3464*t) 1.593*sinh(0.3464*t) 13.33*exp(-0.5*t) 61.54 - 48.21xp(-0.5*t)(cosh(0.3464*t) 0.8845*sinh (0.3464*t)13.33 exp(-0.5*t) 13.33*exp(-0.5*61.03*exp(-0.5*t) (cosh(0.3464*t)1.033*sinh (0.3464t) 27.69 Does it agree with our previous solution? Ack! Those cosh and sinh terms are new! No problem. Let's convert them to exponentials using rewrite) Pt = vpa (simplify( rewrite (Pt, 'exp')), 4) T(t)[46.15 3.804 e-0.8464t13.33 e-0.5t -16.62 e-0.1536 t T(t)TM(t)61.54- 2.785 e-0.8464t13.33 e-05t 45.42 e-0.1536 t Ta(t) | L27.69+ 1.019 e-0.8464 t +13.33 e-ost-62.05 e-0.1536 t Now find the final values using the FVT. Use the limit command and take the limit at s = 0. Don't forget to multiply by s first! To earn your point for credit, record the final temperature in each room, using the final value theorem Question 8: The final temperature in each room is: TB() T46.15 The final temp in the basement is given for you. TA(o)

Answer 1

&Part Bmatrix form solution using eigenvalues and eigenvectors %Question 6-7 clear all close all clc Heating/Cooling constants k1-0.2 ; k2-0.1 ; k3-0.2 ; %cooling constant for the main room %cooling constants for attic and basement %Temperature and the heater TE-+40; TS--40 H20; TO- [ 20.0 ; 0.0 ; -20.01'; %Initial temperature in each room %Question 6.a. fprintf3X3 Matrix A-n') disp (vpa (A,4)) %Question 6.b. fprintf("Matrix f= disp(f) \n') syms s I-eye(3) Ts-(inv(s*I-A)) (TO+f/s) fprint f ("value for Ts=\n') disp (Ts) Tinf-vpa (limit (s Ts, 0),4); fprintf( 'Value for Tinf-In') disp (Tinf) Tt-vpa (ilaplace (Ts), 4) fprintf( 'Value for Tt n disp (Tt) Tt-vpa (simplify(rewrite(Tt, "exp,4); fprintf('value for Tt after simplyfication=\n') disp (Tt)

fprintf( 'Value for T final-\n') disp(Tinf(:,1)) 3X3 Matrix A- [-0.7, 0.2,01 [ 0.2,-0.5, 0.2) I0 0.2,-0.31 Matrix f- 20 16 Value for TS (4 (10*s 3)* (16/s + 20))/(200*s 3300 s2+ 126*s + 13) - (8 (4/s - 20))/(200*s 3300*s 2 126*s 13) (2* (20/s + 20)*(100*s 2 80*s 11))/(200 s3 300 s 2 + 126*s 13) (64 (10s3))/(s (200 s 3 300*s 2 126*s 13)32 (s (200 s 3 300 s 2 126*s 13)) (40 (100*s 2+80*s 11))/ (s (200 s 3+ 300*s 2 126*s 13)) s - 20))/(200*s 3 + 300*s 2 126 s 13) - (8*(4/s +20))/ (200 s3 +300 *s2 126s 13) (2* (20/s 20)*(100*s 2 80*s 11))/ (200*s 3 300*s2 126*s +13)1 ( (4*(10*s 3) *(20/s 20))/(200*s 3 300*s 2 + 126*s 13)- (4 (10*s7)*(4/s- 20))/(200*s 3300*s 2 126*s 13) + (2*(10*s + 3) (10 s 7) *(16/s 20))/(200*s 3 +300*s2 126*s + 13), (80 (10*s + 3))/(s (200*s 3 300*s 2+ 126*s+ 13)) - (16*(10*s + 7))/(s*(200*s 3 300s2126*s 13)) (32 (10 s 3)(10s + 7))/s*(200*s 3300*2 126*s + 13)), (4 (10*s 3) *(20/s - 20)) (200*s 3 300 s2 126*s 13) - (4*(10*s 7) *(4/s 20))/(200 s3 +300 *s 2 + 126*s13)+ (2* (10*s 3) (10 s7) (16/s - 20))/ (200*s 3 300 s2126*s 13) 1 (4* (10*s 3)(16 (8 (20/s +20))/(200 *s 3+300s2+ 126*s + 13) + (4* (10*s 7)*(16/s 20)/(200*s 3 300 s2 126*s 13) - (2* (4/ s - 20)* (100*s 2+120*s 31))/(200*s 3 +300 *s2 126 *s13), 160/(s* (200 s 3 300 s 2 126 s 13)) (64(10 s + 7))/s*(200*s 3+300*s2 126*s 13)) - (8 (100 s 2 120*s + 31)/(s (200 s 3 + 300*s 2 126s13)) (200*s 3 300s 2+126*s 13) (4 (10*s t 7) *(16/s - 20))/(200*s 3 +300 s 2+126*s 13) - (2* (4/s + 20) *(100 s 2 120*s 31))/ (200*s 3 300s2126*s 13)1 (8 (20/s - 20))/ Value for Tinf [ 46.15, 46.15, 46.15] [61.54, 61.54, 61.54 [ 27.69, 27.69, 27.69j Value for Tt- 46.15 -6.154 texp(-0.5*t) *(cosh (0.3464*t) + 1.443*sinh (0.3464*t)) - 20.0*exp(-0.5*t), 46.15 - 19.49*exp (-0.5*t) *(cosh (0.3464*t) +0.4558 sinh (0.3464*t)) - 26.67 exp(-0.5*t), 46.15 -

32.82 exp (-0.5*t) (cosh (0.3464*t) 0.2706*sinh (0.3464*t)) - 33.33*exp (-0.5*t) ] [ 61.54 - 21.54 *exp(-0.5*t)*(cosh(0.3464*t) 0.9073 *sinh(0. 3464*t)) -20·0+exp(-0 . 5 *t ) , 61 . 54-34. 8 7 *exp(-0 . 5 *t ) * (cosh( 0 . 3464 *t ) + 1.223 sinh 0.3464*t)) - 26.67 exp(-0.5*t), 61.54 - 48.21 *exp (-0.5*t)*(cosh (0.3464*t) + 1.364*sinh(0.3464*t)) 33.33 exp(-0.5*t)J [ 20.0 exp(-0.5*t- 27.69*exp(-0.5*t)* (cosh (0. 3464*t) + 1.026 sinh(0.3464*t)) 27.69, 26.67*exp(-0.5*t) - 54.36 *exp (-0.5*t) *(cosh (0.3464*t) 0.9477*sinh (0.3464*t))+ 27.69, 33.33*exp (-0.5*t) - 81.03 exp(-0.5*t)* (cosh(0.3464*t) + 0.9208*sinh(0.3464*t)) 27.691 Value for Tt after simplyfication- [ 1.364 *exp (-0.8464*t) - 7.518*exp(-0.1536*t) - 20.0 exp(-0.5*t) +46.15, 46.15 - 14.18*exp(-0.1536*t)-5.302*exp(-0.8464*t) - 26.67 *exp(-0.5*t), 46.15 - 20.85 *exp(-0.1536*t) - II.97*exp (-0.8464*t) - 33.33*exp(-0.5*t)) [ 61.54 - 20.54*exp(-0.1536*t) - 0.9987*exp(-0.8464*t) 20.0*exp(-0.5*t), 3.882 *exp (-0.8464*t) - 38.75*exp(-0.1536*t) - 26.67 *exp(-0.5*t) + 61.54, 8. 762*exp(-0.8464*t) - 56.97*exp(-0.1536*t)-33.33*exp(-0.5*t) 61.54] [ 20.0*exp (-0.5*t) - 28.06*exp (-0.1536*t) + 0.3655 exp (-0. 8464*t) +27.69, 26.67 exp(-0.5*t -52.94*exp(-0.1536*t) - 1.421 *exp(-0.8464*t) 27.69, 33.33*exp (-0.5*t) - 77.82*exp(-0.1536*t - 3.207*exp-0.8464*t) + 27.69] value for T final= 46.153846153873018920421600341797 61.538461538497358560562133789062 27.692307692312169820070266723633 Published with MATLAB R2018a

%%Part B : matrix form solution using eigenvalues and eigenvectors
%Question 6 - 7
clear all
close all
clc

%Heating/Cooling constants

k1 = 0.2; k2 = 0.1; k3 = 0.2; %Cooling constant for the main room
k0 = 0.5; k4 = 0.1;           %Cooling constants for attic and basement

%Temperature and the heater

TE = +40;
TS = -40;
H = 20;

T0 = [20.0 ; 0.0 ; -20.0]'; %Initial temperature in each room

%Question 6.a.

A = [-(k0+k1) k1 0; k1 -(k1+k2+k3) k3; 0 k3 -(k3+k4)];

fprintf('3X3 Matrix A= \n\n')
disp(vpa(A,4))

%Question 6.b.

f = [k0*TE; k2*TS+H; k4*TS];

fprintf('Matrix f= \n')
disp(f)

syms s
I=eye(3);

Ts=(inv(s*I-A))*(T0+f/s);
fprintf('Value for Ts=\n')
disp(Ts)

Tinf=vpa(limit(s*Ts,0),4);
fprintf('Value for Tinf=\n')
disp(Tinf)

Tt=vpa(ilaplace(Ts),4);
fprintf('Value for Tt=\n')
disp(Tt)

Tt=vpa(simplify(rewrite(Tt,'exp')),4);
fprintf('Value for Tt after simplyfication=\n')
disp(Tt)

fprintf('Value for T final=\n')
disp(Tinf(:,1))