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Based on your contour plot please answer the following question: what is the minimum possible square area (and its width in miles) that gives us the average electric output of the array greater than or equal to the current USA electric consumption of 5´1011 W=500 Gigawatts at the solar panel efficiency of 15%


Matlab Code:

clear all;
close all;
%   Vector of available area widths (a 1D array of numbers)
Wmiles  = [1:1:150];    %   width of the square area in miles  (vector or 1D array)
Wmeters = Wmiles*1609;  %   width of the square area in meters (vector or 1D array)
A  =  Wmeters.*Wmeters; % A.*  area in m^2 (vector or 1D array); note element-by-element 
                        %   vector multiplication!
%   Vector of available array efficiencies (a 1D array of numbers)
E   = [1:1:20];         %   efficiency percentage (vector or 1D array)         
%   Matrix of all available total output powers of the array (a 2D array of numbers)
p   = 270.9;                 %   average daily solar radiation in W per m^2 (scalar)
for i = 1:length(A)          %   index i corresponds to different areas A
    for j = 1:length(E)      %   index j corresponds to different efficiencies E 
        P(i,j) = p*A(i)*E(j)*0.01;  %total average daily power (2D matrix)  
%   "Plot" the matrix - a contour plot of all available total output powers
v     = [2 10 50 100 300 500 1000]; %   create contour plot levels in GWatts
C = contourf(P*1e-9, v);       %   create contour plot of output powers in GWatts 
clabel(C, 'FontSize', 11, 'Color', 'r'); %   be fancy (do not have to)
grid on; xlabel('Solar panel efficiency percentage, E'); ylabel('Width of the square area in miles, W');
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Look at the line labeled 500 on your graph. Trace that line until you meet 20 on the x axis and that is the minimum possible width. We square this to get the final answer.


The minimum possible width would be about 70 miles, and the minimum possible area would be 70^2 = 4900 square miles.

Edited by AresScripts
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