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trying to perform a summation and its returning the value in separately in a matrice when i need a single answer.

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clayton holmes
clayton holmes 2021-12-6
评论： Walter Roberson 2021-12-6
syms x y z f g
%%%%%Variable Values%%%%%%%%%
theta=0;
n=0;
m=0;
k = (0.6569);
h = 11;
alpha = 2.57*10^(-7);
t=0;
L=(.051);
W=(.01);
H=(.005);
X=(.051/4);
Y=(.01/2);
Z=(.005/4);
beta = ((n+0.5)*pi);
eta = ((m/.01)*pi);
mew = (117);
lamda = (beta.*beta) +(eta.*eta)+ (mew.*mew);
%%%%%%%%DETERMINATION OF CONSTANT C%%%%%%%%%%%%%%%%
f(z,y,x) = cos(beta*z)*sin(mew*y)*cos(eta*x)*sin(eta*x);
C_nmp_numerator = int(int(int(f,z,[0 (H/2)]),y,[0 W]),x,[0 (L/2)]);
g(z,y,x) = (L/4)*(H/4)*((((-k*mew)/h)*(cos(mew*x))^2*sin(mew*x))+((sin(mew*x))^2*cos(mew*x)));
C_nmp_denominator = int(x,[0 (L/2)]);
C_nmp = C_nmp_numerator/C_nmp_denominator;
pretty(C_nmp);
%%%%%%%%DETERMINATION OF DEMENTIONLESS TEMP THETA%%%%%%%%%%%%%%%%
for n=0:10,m=0:10,t=0:10
theta = (C_nmp .* cos(beta.*z).*sin(eta.*y).*((-k.*mew)/h).*cos(mew.*x)+sin(mew.*x)).*exp(-(lamda.*lamda)*alpha.*t);
vpa(subs(theta, {x,y,z}, {X,Y,Z}));
end
theta;
vpa(subs(theta, {x,y,z}, {X,Y,Z}))
  1 个评论
Walter Roberson
Walter Roberson 2021-12-6
vpa(subs(theta, {x,y,z}, {X,Y,Z}));
That does the substitution, calculates a numeric approximation, and then throws the result away without displaying it because of the semicolon.

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