for loop to evaluate every minute

4 visualizaciones (últimos 30 días)
Aaron LaBrash el 29 de Abr. de 2022
Respondida: Riccardo Scorretti el 30 de Abr. de 2022
I have a for loop written that evaluates the volume of water in a tank given a new fill rate every hour, one pump that turns on/off based on percentage fill, and one pump that it given on/off per hour. I need to make it run a new iteration every minute, but I am unsure how to do so while still breaking it into hours. Any help is greatly appreciated.
data =
20 1
20 1
0 0
375 0
200 0
20 1
50 0
150 0
50 1
60 1
100 0
150 0
180 1
170 0
100 1
250 0
100 1
210 0
170 1
150 0
50 1
25 1
25 1
50 0
s = 1000;
pC = 60;
v = s*pC*.01;
a1 = s.*0.85; % tank 85% full
a2 = s.*0.20; % tank 20% full
kt = 1:length(data);
for n = 1:length(kt)
if v>=(a1);
a = 1;
elseif v<=(a2);
a = 0;
else a = 0;
end
if data(n,2) == 1;
b = 1;
else
b = 0;
end
v = v+(data(n,1))-150.*a-100.*b
end
11 comentariosMostrar 9 comentarios más antiguosOcultar 9 comentarios más antiguos
Aaron LaBrash el 29 de Abr. de 2022
yes, that worked great. I did have another issue, where I was instructed to pump a (first if loop) to turn on at 85% and run until it reaches 20%, but I don't know if that's even possible.
Riccardo Scorretti el 30 de Abr. de 2022
For this, I'm afraid I cannot help. Do you mean "if it's even possible" physically?
By the way, if you are satisfied with the code, I'll post it as answer so the question can be closed.

Iniciar sesión para comentar.

Respuestas (1)

Riccardo Scorretti el 30 de Abr. de 2022
% Setup variables
data = [
20 1
20 1
0 0
375 0
200 0
20 1
50 0
150 0
50 1
60 1
100 0
150 0
180 1
170 0
100 1
250 0
100 1
210 0
170 1
150 0
50 1
25 1
25 1
50 0
];
s = 1000;
pC = 60;
v = s*pC*.01;
a_ = [0]; % Just to record a and b
b_ = [0];
% Run the program
a1 = s.*0.85; % tank 85% full
a2 = s.*0.20; % tank 20% full
% kt = 1:length(data);
kt = 1:size(data,1);
for nh = 1:length(kt)
for nm = 1 : 60
if v(end)>=(a1);
a = 1;
elseif v(end)<=(a2);
a = 0;
else
a = 0;
end
if data(nh,2) == 1;
b = 1;
else
b = 0;
end
% v = v+data(nh,1)/60-150/60.*a-100/60.*b
v(end+1) = v(end) + data(nh,1)/60-150/60.*a-100/60.*b;
a_(end+1) = a;
b_(end+1) = b;
end
end
figure
subplot(2, 1, 1);
plot((1:numel(v))/60, v) ; hold on
plot([1 numel(v)]/60, [a1 a1], 'r--');
plot([1 numel(v)]/60, [a2 a2], 'r--');
grid on
xlabel('time (hour)') ; ylabel('volume');
subplot(2, 1, 2);
plot((1:numel(v))/60, a_, (1:numel(v))/60, b_) ; grid on ; legend('a', 'b')
xlabel('time (hour)') ; ylabel('a, b');
0 comentariosMostrar -2 comentarios más antiguosOcultar -2 comentarios más antiguos

Iniciar sesión para comentar.

Categorías

Más información sobre Loops and Conditional Statements en Help Center y File Exchange.

R2021b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by