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How to do optimization?

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WASIM ASHRAF
WASIM ASHRAF on 2 Jul 2021
Answered: Alan Weiss on 2 Jul 2021
optimization of efficiency is required for variation of following parameters,
d = 5: 1: 15; I_t = 400: 100: 1200; U = 2: 0.1: 3; Fr = 0.5: 0.1: 0.9;
Ti = 300; To = 350; x =0.9; m = 0.5; Cp = 4200; F = 0.8;
eta = Q/I_t*A; %efficiency
A = pi*(d^2)/4;
Q = A*Fr*(S - U (Ti - To));
Fr = m*Cp*(1-exp((-F*U*A)/(m*Cp)));
S = x*I_t;

Answers (1)

Alan Weiss
Alan Weiss on 2 Jul 2021
I am not 100% sure that I understand you, because you have two different definitions of Fr, namely
Fr = 0.5: 0.1: 0.9;
Fr = m*Cp*(1-exp((-F*U*A)/(m*Cp)));
In any case, it seems that your problem is straightforward.
eta = Q/I_t*A; %efficiency
I suppose that you are trying to maximize eta. In that case, you should maximize Q and A and minimize I_t.
I_t = 400: 100: 1200;
So minimizing I_t means taking I_t = 400.
A = pi*(d^2)/4;
d = 5: 1: 15;
Maximizing A means taking d = 15.
Q = A*Fr*(S - U (Ti - To));
Fr = 0.5: 0.1: 0.9;
Ti = 300; To = 350; % So Ti - To = -50
U = 2: 0.1: 3; % So -U*(Ti - To) = 50*3 at a maximum
S = x*I_t; x =0.9; % So Q/I_t = A*Fr*x + A*Fr*50*3/I_t
% So indeed you want to minimize I_t to maximize Q
Maybe I misunderstand something, but it doesn't seem to me that there are any non-monotone things here.
Alan Weiss
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