- xH and yH are not vectors
- this T should be in celcius or kelvin? it returns -1 -1 -1 -1...
What is wrong with my code?
2 ビュー (過去 30 日間)
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clear,clc
R=8.314; T1=60+273.15; PvapH=.7583;PvapT=.3843;HvapH=29000;HvapT=31000;P=.7;
fPvapH1=@(T2) PvapH*exp(-1*HvapH/R*((1./T2)-(1/T1)));
fPvapT1=@(T2) PvapT*exp(-1*HvapT/R*((1./T2)-(1/T1)));
fxH=@(x) (x*PvapH/P)+((1-x)*PvapT/P)-1;
xH=fzero(fxH,.7);
xT=1-xH;
fBPH=@(T2) (((xH*fPvapH1(T2))/P)+((xT*fPvapT1(T2))/P))-1;
BPH=fzero(fBPH,340);
yH=xH*fPvapH1(BPH)/P;
yT=xT*fPvapT1(BPH)/P;
T=linspace(55,80,100);
figure(1)
plot(xH,fBPH(T))
hold on
plot(yH,fBPH(T))
hold off
axis([0 1 55 80])
5 件のコメント
Torsten
2022 年 3 月 10 日
In the equation
fxH=@(x) (x*PvapH/P)+((1-x)*PvapT/P)-1;
all parameters are given constants except x.
But PvapH and PvapT should depend on T, I guess.
採用された回答
Torsten
2022 年 3 月 10 日
編集済み: Torsten
2022 年 3 月 11 日
R = 8.314;
T1 = 60 + 273.15;
PvapH = .7583;
PvapT = .3843;
HvapH = 29000;
HvapT = 31000;
P = .7;
fPvapH = @(T) PvapH*exp(-HvapH/R*(1./T - 1/T1));
fPvapT = @(T) PvapT*exp(-HvapT/R*(1./T - 1/T1));
XH = 0:0.01:1;
Tstart = 340;
for i=1:numel(XH)
xH = XH(i);
fun = @(T) xH*fPvapH(T)/P + (1-xH)*fPvapT(T)/P - 1;
T(i) = fzero(fun,Tstart);
Tstart = T(i)
end
YH = XH.*fPvapH(T)/P;
XT = 1-XH;
YT = XT.*fPvapT(T)/P;
T = T - 273.15;
figure(1)
plot(XH,T)
hold on
plot(YH,T)
hold off
axis([0 1 55 80])
figure(2)
plot(XT,T)
hold on
plot(YT,T)
hold off
axis([0 1 55 80])
0 件のコメント
その他の回答 (1 件)
Benjamin Thompson
2022 年 3 月 10 日
xH, yH, and BPH are all scalar values. They must be the same length as fBPH(T). In this line to you mean to pass a vector argument to the fxH function handle?
xH=fzero(fxH,.7);
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