plotting a for loop returns Empty Graph

I cannot seem to get the output graph to work. I need to show at what pressure ratios (Rp and Pr) the thermal efficiency is at its heighest (Nth_IC).

%% Variables

Rp = 50; %% pressure ratio

Pr = 50; %% P2/P1

%% for loop

for i = 0:0.5:Rp

for j = 0:0.5:Pr

%% Low Pressure Compressor

T2 = T1*Pr^(0.4/1.4);

W2s = 1005*(T1-T2); %% specific work from 1 to 2s

W2 = W2s * Nc;

%% Intercooler

T3 = T1; %% assume ideal intercooler

q3 = 1005*(T3 - T2);

%% High Pressure Compressor

T4 = T3*(Rp/Pr)^(0.4/1.4);

W4s = 1005*(T3 - T4);

W4 = W4s * Nc;

%% Combustion Chamber

q5 = 1005*(T5 - T4);

%% Turbine

T6 = T5*(1/Rp)^(0.4/1.4);

W6s = 1005*(T5 - T6);

W6 = W6s * Nt;

%% Calculations for Intercooled

NWO_IC = W6 + W2; %% Net Work Output for Intercooled

Nth_IC = NWO_IC/q5; %% Thermal Efficiency for Intercooled

Wout_IC = m*(W4 + W2 + W6); %% Output Power for Intercooled

figure(1); plot(j,Nth_IC);

end

figure(2); plot(i,Nth_IC);

end

Answer 1

[John D'Errico]

You are not plotting ANYTHING in a for loop. You may think you are. Where are the plot statements? AFTER THE LOOP TERMINATES. After the end statement for the loop.

So we have two plots made:

figure(1); plot(i,Nth_IC);
figure(2); plot(j,Nth_IC);

What are i, j, and Nth_IC, after the loop terminates?

>> i
i =
50

>> Nth_IC
Nth_IC =
0.293033227395743

>> j
j =
50

What does plot do when you tell it to plot, but with no third argument?

It connects the points with a simple line. Not even a symbol.

How many points are in that plot? ONE point. So what did plot do? It connected that single point with a line. Only one point, so all you see is a blank figure.

Note that all of ths could have been seen by you, had you simply looked at what are the arguments to those plot calls, and what values do they take on.

What should you have done? It is often a good idea to simply collect all of those values in the loop in a vector, and then plot them all after the loop has terminated in one call to plot. If you want to connect them with a line, then you cannot do that when you plot only one point at a time, even if you do have the plots inside your loop.

Answer 2

[Star Strider]

I’m not certain what you want, however it is necessary to subscript the varialbes to be plotted.

This runs without error and produces some plots, however they may not be the desired results.

%% INTERCOOLED BRAYTON CYCLE
%% Variables
Rp = 50; %% pressure ratio
m = 1; %% mass flow rate
Pr = 50; %% P2/P1
T5 = 1496; %% max value for T5
%% Given Conditions
T1 = 300;
P1 = 97E+3;
Nc = 0.9; %% compressor efficiency
Nt = 0.9; %% turbine efficiency
Power = 36e+9; %% power Generation
Ne = 0.96; %% electric generator efficiency
Net_Power = Power*Ne;
%% for loop
iv = 0:0.5:Rp;
jv = 0:0.5:Pr;
for i = 1:numel(iv)
    for j = 1:numel(jv) 
        Rp - iv(i);
        Pr = jv(j);
        %% Low Pressure Compressor
        T2 = T1*Pr^(0.4/1.4);
        W2s = 1005*(T1-T2); %% specific work from 1 to 2s
        W2 = W2s * Nc;
        %% Intercooler
        T3 = T1; %% assume ideal intercooler
        q3 = 1005*(T3 - T2);
        %% High Pressure Compressor
        T4 = T3*(Rp/Pr)^(0.4/1.4);
        W4s = 1005*(T3 - T4);
        W4 = W4s * Nc;
        %% Combustion Chamber
        q5 = 1005*(T5 - T4);
        %% Turbine 
        T6 = T5*(1/Rp)^(0.4/1.4);
        W6s = 1005*(T5 - T6);
        W6 = W6s * Nt;
        %% Calculations for Intercooled
        NWO_IC = W6 + W2; %% Net Work Output for Intercooled
        Nth_IC(i,j) = NWO_IC/q5; %% Thermal Efficiency for Intercooled
        Wout_IC = m*(W4 + W2 + W6); %% Output Power for Intercooled
    end
end
figure(1); plot(iv,Nth_IC);

figure(2); plot(jv,Nth_IC.');

figure(3)
surf(iv,jv,Nth_IC, 'EdgeColor','none')
view(30,30)
xlabel('Rp')
ylabel('Pr')
zlabel('Nth_{IC}')

Experimment to get the desired result.

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