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Effacer les filtres

What is wrong with my code?

1 vue (au cours des 30 derniers jours)
Angelina Encinias
Angelina Encinias le 10 Mar 2022
Modifié(e) : Torsten le 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 commentaires
Torsten
Torsten le 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.
Angelina Encinias
Angelina Encinias le 10 Mar 2022
This is the question. I am pretty sure I need a for loop to display the graphs.

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Réponse acceptée

Torsten
Torsten le 10 Mar 2022
Modifié(e) : Torsten le 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])

Plus de réponses (1)

Benjamin Thompson
Benjamin Thompson le 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);
  2 commentaires
Angelina Encinias
Angelina Encinias le 10 Mar 2022
Yes I did. The graph is supposed to look like the graph on the left (image I attached). To make the image on the right I need to do yH(xH).
Angelina Encinias
Angelina Encinias le 10 Mar 2022
Image of graphs needed to display is attached.

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