- 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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Angelina Encinias
il 10 Mar 2022
Modificato: Torsten
il 11 Mar 2022
Code needs to display:
Code is displaying a blank graph.
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 Commenti
Torsten
il 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.
Risposta accettata
Torsten
il 10 Mar 2022
Modificato: Torsten
il 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 Commenti
Più risposte (1)
Benjamin Thompson
il 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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