0 个投票

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 个评论
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AndresVar
AndresVar 2022-3-10
Ran in:
Ran in:
  • xH and yH are not vectors
  • this T should be in celcius or kelvin? it returns -1 -1 -1 -1...
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)
xH = 0.8441
xT=1-xH;
fBPH=@(T2) (((xH*fPvapH1(T2))/P)+((xT*fPvapT1(T2))/P))-1;
BPH=fzero(fBPH,340);
yH=xH*fPvapH1(BPH)/P
yH = 0.9144
yT=xT*fPvapT1(BPH)/P;
T=linspace(55,80,100);
fBPH(T)
ans = 1×100
-1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1
Angelina Encinias
Angelina Encinias 2022-3-10
I attached the graphs that the code should display. I am unsure if I need for loops.
Angelina Encinias
Angelina Encinias 2022-3-10
T should be Celsius
Torsten
Torsten 2022-3-10
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 2022-3-10
This is the question. I am pretty sure I need a for loop to display the graphs.

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 采纳的回答

Torsten
Torsten 2022-3-10
编辑：Torsten 2022-3-11

0 个投票

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])

更多回答（1 个）

Benjamin Thompson
Benjamin Thompson 2022-3-10

0 个投票

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 个评论
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Angelina Encinias
Angelina Encinias 2022-3-10
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 2022-3-10
Image of graphs needed to display is attached.

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