HELP- MATLAB FREEZING/ NOT OUTPUTTING CODE

Hi, matlab seems to be struggling to run my code, i cant see any obvious issues and it is not outputting anything just kind of freezing? can anyone spot what the issue may be? I have attached my code

clc 
clear 
close all

%%question C 
Y=1.4; %heat capacity ratio
r=8:0.01:24;%compression ratio array V1/V2
re=4:0.01:20;%Expanison ratio array V4/V3
ro=10; %compression ratio of otto engine
etaotto=1-ro^(1-Y); %efficency of otto engine
eta=zeros(length(r),length(re)); %empty matrix
etaotto=zeros(length(r),length(re));%empty matrix
for i=1:length(r) %loop of all r values 
    for j=1:length(re)%loop of all re values 
        if r(i)>re(j)%checks if the r>re
        eta(j,i)=1-((1/Y).*((re(j).^(-Y))-(r(i).^(-Y)))./((1./re(j))-(1./r(i))));
        %Equation was derived in part A
        end 
        if etaotto>eta(j,i)
            %checks to see whether the diesel engine is less efficent than
            %the otto engine with compression ratio
            eta(j,i)=0; %sets diesel engine efficency to 0 
        end
        
    end
    
end

imagesc(r,re,eta)%specifies image location
set(gca,'YDir','normal'); %Flips the y axis
title('Otto Engines V Diesel Engines')%title of the graph
xlabel('Compression ratios (V1/V2)')%x-axis label
ylabel('Expanison Ratios (V4/V3)')%y-axis label 
x=colorbar;%shows the colour scale of the vertical colorbar
x.Label.String=('Efficency, \eta'); %label for colorbar
colormap winter %changes graph colour

[Luna]

You have 2 loops and doing 1601*1601 = 2.563.201 iterations, so it would obviously takes long time.

What is the operation you need to do in a mathematical way? Can't you just do it by matrix operations instead of using for loops?

Answer 1

[Jan]

It will increase your code, if you clean it up: Look at the orange marks in the editor, which show you, that etaotto have been defined twice. If you want the scalar value 1-ro^(1-Y) omit the redefinition as a matrix. Currently the comparison etaotto>eta(j,i) wastes a lot of time, because etaotto is a large matrix. With the scalar the code runs much faster.

By the way, there is a typo:

eta=zeros(length(r),length(re));
for i=1:length(r) %loop of all r values 
    for j=1:length(re)%loop of all re values 
        ...
        eta(j,i) = ...

You have defined eta with [length(r) x length(re)], but in the loop the indices are swapped. I guess, you want:

    eta(i, j) = ...

A vectorization will help also:

Y  = 1.4; %heat capacity ratio
r  = 8:0.01:24;%compression ratio array V1/V2
re = 4:0.01:20;%Expanison ratio array V4/V3
ro = 10; %compression ratio of otto engine
etaotto = 1-ro^(1-Y); %efficency of otto engine
eta = zeros(length(re),length(r));  % Swapped dimensions
for i = 1:length(r) %loop of all r values 
    index         = (r(i) > re);
    eta(index, i) = 1-((1/Y).*((re(index).^(-Y))- ...
                     (r(i).^(-Y))) ./ ((1./re(index))-(1./r(i))));
end

This needs 0.2 seconds on my machine.

By the way, I'd reduce the number of parenthesis:

eta2(index, i) = 1 - (1 / Y .* (re(index).^(-Y) - r(i).^(-Y)) / Y ./ ...
                (1 ./ re(index) - 1 ./ r(i)));

The comment is misleading:

eta=zeros(length(r),length(re)); %empty matrix

No, zeros are not "empty".