- 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?
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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 Comments
Torsten
on 10 Mar 2022
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.
Accepted Answer
Torsten
on 10 Mar 2022
Edited: Torsten
on 11 Mar 2022
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 Comments
More Answers (1)
Benjamin Thompson
on 10 Mar 2022
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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