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I am trying to calculate the power of a radiator and the final temperature. the equation is Planck's law I do not know what is wrong with it. could you please help me

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Zahra Kamali Khanghah
回答： Neelanshu 2024-2-6
h=6.626176e-34; %J/s
c=2.99792458e8; %m/s
k=1.380649e-23; %J/K
A=20*1e-4; %m^2
er=0.95; %radation emissivity according to measurement Fig.1 e & f
T=300; %Kelvin,because of the cover and vacuum condition of cooling.
tc=1; % tc=transmittance of the cover
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%calculations:
%% radiator
%direction
syms theta
f = (1/2)*sin(2*theta);
intf = int(f,0,pi/2);
%wavelength, Planck's law, Blackbody Values For Vector Lam
Lam=(0:inf).*1e-6; % meters
dLam = Lam(2) - Lam(1); % Delta Lambda
I1 =(2*h*c^2)./((Lam.^5).*(exp((h.*c)./(k.*T.*Lam))-1));
% plotI1 = I1*1e-12;
% plotLam = Lam/1e-6;
% plot(plotLam,plotI1,'linewidth',4)
Ibb=trapz(Lam,I1);
%the power of radiator
Pr = 2*A*pi*intf*Ibb*er; % W. m-2
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Zahra Kamali Khanghah
Zahra Kamali Khanghah 2022-3-1
thanks. As the final answer should be somthing around 330, could you tell me what is the reseason I cannot get it like a usual number?

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回答（1 个）

Neelanshu
Neelanshu 2024-2-6
Hi Zahra,
I understand from your query that you are interested in finding the reason why you're not getting a numeric value for the power of the radiator.
You can follow these steps to ensure you get a numeric value for the power :
  • As Torsten pointed out you need to change the "Lam" variable range to a vector of finite range and convert "intf" variable to double datatype.
intf = double(intf)
  • Additionally you have to choose a very small starting value for "Lam" variable instead of 0. In MATLAB, division by zero leads to a NaN value for I1, which propagates when it undergoes numerical integration.
% Define wavelength range, starting from a small positive number to avoid division by zero
Lam = linspace(1e-10, 1e-6, 10000); % Meters
By making these adjustments, you should be able to obtain a numeric value for the power.
Hope this helps.

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