How to multiply matrix by using for loop

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Luka Milat
Luka Milat am 11 Jan. 2020
Kommentiert: Luka Milat am 11 Jan. 2020
I am trying to calculate heat for 3 different stages and I have written my 3 cases in the form of a matrix. I want to get the final Q with all 3 stages but I am getting only first one calculated.
Qhc = 2000000*1000;%J , potrebna energija za grijanje u jednom satu
N = 200; %broj VK
Qg = Qhc/3600;
t = 3600; %vrijeme
Tz1 = 20; %°C
Tw1g = 50; %°C
Tw1h = 7; %°C
cpz = 1050; %J/kgK
cpw = 4187; %J/kgK
roz = 1.225; %kg/m3
row = 997; %kg/m3
Vw = 488 / ( 3600 * 1000 ); %m3/s protok vode
Cmax = row * Vw * cpw;
kA = [97.50664 156.195 234.9065]; %W/m2K
Vz = [ 340/3600 510/3600 680/3600 ]; %m3/s
Pvent = [ 11.99605 17.38172 29 ]; %W
Q = [ 0 0 0 ];
Qvk = 0;
P = 0;
for i = 1:length(Vz)
Cmin(i) = roz * Vz(i) * cpz; %W/K
NTU(i) = kA(i) / Cmin(i);
C(i) = Cmin(i) / Cmax;
eta(i) = NTU(i) .^ (-0.22);
epsilon(i) = 1 - exp ((1/(C(i)*eta(i))*(exp(-NTU(i)*C(i)*eta(i))-1)));
if Qhc > 0
Q(i) = Cmin(i) * epsilon(i) * (Tw1g - Tz1);
Q_J(i) = Q(i) * 3600; %energija koju preda u jednom satu
elseif Qhc < 0
Q(i) = Cmin(i) * epsilon(i) * (Tz1 - Tw1h);
Q_J(i) = Q(i) * 3600; %energija koju preda u jednom satu
end
end

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KALYAN ACHARJYA
KALYAN ACHARJYA am 11 Jan. 2020
Bearbeitet: KALYAN ACHARJYA am 11 Jan. 2020
clc;
clear;
Qhc=2000000*1000;%J , potrebna energija za grijanje u jednom satu
N=200; %broj VK
Qg=Qhc/3600;
t=3600; %vrijeme
Tz1=20; %°C
Tw1g=50; %°C
Tw1h=7; %°C
cpz=1050; %J/kgK
cpw=4187; %J/kgK
roz=1.225; %kg/m3
row=997; %kg/m3
Vw=488/(3600*1000 ); %m3/s protok vode
Cmax=row*Vw*cpw;
kA=[97.50664 156.195 234.9065]; %W/m2K
Vz=[340/3600 510/3600 680/3600 ]; %m3/s
Pvent=[11.99605 17.38172 29 ]; %W
Q=[0 0 0];
Qvk=0;
P=0;
for i=1:length(Vz)
Cmin(i)=roz*Vz(i)*cpz; %W/K
NTU(i)=kA(i)/Cmin(i);
C(i)=Cmin(i)/Cmax;
eta(i)=NTU(i).^(-0.22);
epsilon(i)=1-exp ((1/(C(i)*eta(i))*(exp(-NTU(i)*C(i)*eta(i))-1)));
if Qhc > 0
Q(i)=Cmin(i) * epsilon(i) * (Tw1g - Tz1);
Q_J(i)=Q(i) * 3600; %energija koju preda u jednom satu
else
Q(i) = Cmin(i) * epsilon(i) * (Tz1 - Tw1h);
Q_J(i) = Q(i) * 3600; %energija koju preda u jednom satu
end
end
Q
Q_J
Result:
Q =
1.0e+03 *
1.8956 2.8731 3.9813
Q_J =
1.0e+07 *
0.6824 1.0343 1.4333
  1 Kommentar
Luka Milat
Luka Milat am 11 Jan. 2020
Huh,
I would swear that something was wrong with the code. Thanks for your time anyway!

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