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Please, I would appreciate any assistance on debugging the attached code.

2 Ansichten (letzte 30 Tage)
Isaac
Isaac am 1 Apr. 2016
Bearbeitet: Walter Roberson am 2 Apr. 2016
The first section of the code is okay, but I am having it difficult solving for uu and ud in the attached code. I want to continue solving for the values of uu and ud as long the conditions in the while loops are valid. Pls, see the attached file.
This is the pseudo code:
for j=1:N
k=1;
uu(k,j)=0;
Lou(k,j)=2;
p(k,j)=6000;
pQ(k,j)=asin(1+Lou(k,j));
Lh2(j)=200;
k22(k,j)=Lh2(j)./Lou(k,j);
if p(k,j)>pQ(k,j)
while Lou(k,j)<Lh2
k22f(k,j)=Lh2(j)./(Lou(k,j)+uus);
funLu(k,j)=asin(1+k22f(k,j));
testfunLu=@(x)eval(subs(funLu(k,j),uus,x))-p(k,j);
uuf(k,j)=fsolve(testfunLu,55);
uu(k+1,j)=uuf(k,j);
Lou(k+1,j)=Lou(k,j)+uu(k+1,j);
k22(k+1,j)=Lh2(j)./(Lou(k+1,j)+uu(k+1,j));
p(k+1,j)=p(k,j);
k=k+1
end
end
end
  9 Kommentare
Isaac
Isaac am 1 Apr. 2016
Bearbeitet: per isakson am 2 Apr. 2016
I have tried using the fsolve this way also
testfunLu=@(x)eval(subs(funLu(k,j),uus,x))-p(k,j);
uuf(k,j)=fsolve(testfunLu,55);
Isaac
Isaac am 1 Apr. 2016
Bearbeitet: per isakson am 2 Apr. 2016
This is the pseudo code:
for j=1:N
k=1;
uu(k,j)=0;
Lou(k,j)=2;
p(k,j)=6000;
pQ(k,j)=asin(1+Lou(k,j));
Lh2(j)=200;
k22(k,j)=Lh2(j)./Lou(k,j)
if p(k,j)>pQ(k,j)
while Lou(k,j)<Lh2
k22f(k,j)=Lh2(j)./(Lou(k,j)+uus);
funLu(k,j)=asin(1+k22f(k,j));
testfunLu=@(x)eval(subs(funLu(k,j),uus,x))-p(k,j);
uuf(k,j)=fsolve(testfunLu,55);
uu(k+1,j)=uuf(k,j);
Lou(k+1,j)=Lou(k,j)+uu(k+1,j);
k22(k+1,j)=Lh2(j)./(Lou(k+1,j)+uu(k+1,j));
p(k+1,j)=p(k,j);
k=k+1
end
end
end

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