Changing the value of only one parameter causes a problem (possibly with while loop)

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Mahmut Cenk
Mahmut Cenk 2024-9-26
评论： Mahmut Cenk 2024-9-26
Hello everyone. Firstly, as I do not have access to my other computer right now, I am using the online compiler, if that makes any difference (but I do not think so). In the attaced code, in the first section, when I set the d(1:10) value to 0.04, it just can't proceed further after the while loop and keeps calculating forever, as seen in screenshot (1). At least I assume so; the thing is, I haven't used matlab much and when I open the debugger (screenshot(2)) and step in and continue, it keeps doing the while loop over and over again. Overall, I do not get any crucial error that seemingly affect the working mechanism of the code: Only "consider preallocating for speed suggestions", so I cannot understand what is wrong with it. By the way, this happens as the value gets smaller: When d is smaller than 0.07. When I set it to something like 150, it still works. What might be the reasoon?
Thanks in advance!
%% Constants and Assumptions
g=9.81;
hgelgit=0.5;
he=0.3;
Vmin=0.6;
Vmax=0.9;
f=0.03;
s=0.25;
Cq(1)=0.61;
rho0=1008.301;
rhoa=998.83;
H=20;
Lana=1500;
d(1:10)=0.04;
hf(1)=0;
%% Initial Hydraulic Head
h=(rhoa*(H+hgelgit+he))/(rho0);
E(1)=h-H;
%% First Port
a(1)=pi*(d(1))^2./4;
q(1)=Cq(1)*a(1)*sqrt(2*g*E(1));
Q(1)=q(1);
D(1)=floor(20*(sqrt(4*Q(1)/(pi*Vmin))))/20;
A(1)=pi*D(1)^2./4;
V(1)=Q(1)/(pi*(D(1)^2./4)); %[m/s]
hizyuk(1)=0;
U(1)=q(1)/(pi*d(1)^2./4);
Dmax(1)=(sqrt(4*Q(1)/(pi*Vmin)));
Dmin(1)=(sqrt(4*Q(1)/(pi*Vmax)));
%% Other Ports
for i=2:10
hf(i)=f*(s/D(i-1))*(V(i-1)^2./(2*g));
E(i)=E(i-1)+hf(i);
Cq(i)=0.63-0.58*(((V(i-1))^2)/(2*g*E(i)));
a(i)=pi*(d(i))^2./4;
x=cumsum(a);
q(i)=Cq(i)*a(i)*sqrt(2*g*E(i));
Q(i)=Q(i-1)+q(i);
Dmax(i)=(sqrt(4*Q(i)/(pi*Vmin)));
Dmin(i)=(sqrt(4*Q(i)/(pi*Vmax)));
D(i)=D(i-1);
A(i)=pi*D(i)^2./4;
V(i)=Q(i)/(pi*(D(i)^2./4));
hizyuk(i)=V(i-1)^2./(2*g*E(i));
while (V(i) >= Vmax)
D(i)=floor(20*(sqrt(4*Q(i)/(pi*Vmin))))/20;
V(i)=Q(i)/(pi*(D(i)^2./4));
end
U(i)=q(i)/(pi*d(i)^2./4);
end
y=x./(pi*D.^2./4);
%% Results
results=zeros(10,16);
results(:,1)=transpose(hf);
results(:,2)=transpose(E);
results(:,3)=transpose(hizyuk);
results(:,4)=transpose(Cq);
results(:,5)=transpose(a);
results(:,6)=transpose(x);
results(:,7)=transpose(q);
results(:,8)=transpose(U);
results(:,9)=transpose(Q);
results(:,10)=transpose(Dmin);
results(:,11)=transpose(Dmax);
results(:,12)=transpose(D);
results(:,13)=transpose(A);
results(:,14)=transpose(y);
results(:,15)=transpose(V);
results(:,16)=transpose(Q)*1000;
max(q)/min(q)
results
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Cris LaPierre
Cris LaPierre 2024-9-26
The value of d is used to calculate a, which is used to calculate q, which is turned into Q, which is used to compute D.
It would appear, then, when d is small, V(i) is less the Vmax, and when it is large, V(i) is greater than or equal to Vmax. There is no difference is how the values of D or V are calculated inside or outside your loop, so I suspect all that is happening is your loop runs infinitely, or not at all.

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采纳的回答

Steven Lord
Steven Lord 2024-9-26
Let's look at the while loop.
while (V(i) >= Vmax)
D(i)=floor(20*(sqrt(4*Q(i)/(pi*Vmin))))/20;
V(i)=Q(i)/(pi*(D(i)^2./4));
end
In order for the truth of the condition to change, either V(i) and/or Vmax must change. Nothing inside the loop assigns a value to Vmax so that means V(i) must change. Since we assign a value to V(i) in the loop that means this can't be an infinite loop, right?
Well, in order for V(i) to change one of Q(i) or D(i) must change. [pi is not going to change :)] You assign a value to D(i) in the loop but don't assign to Q(i) so if V(i) changes it must be because D(i) did.
D(i) depends on Q(i), pi, and Vmin. None of those change in the loop. So you're always going to get the same value of D(i). That means that the next line will always give you the same value for V(i). That means the while condition's truth cannot change unless the value of D(i) and/or the value of V(i) changes at the first iteration through the loop.
The code in the loop matches the code that was used to define V(i) two lines before the loop, so it's not going to change in that first iteration. That means If you enter the while loop, unless the value of D(i) changes at that first iteration, you're never going to exit the loop.
From looking at your code I suspect that the value of D(i) doesn't change either, but you have enough variables involved in the calculations that it would take a little while to prove that it doesn't and I'll leave that for you to check when you determine how to modify your code.

更多回答（1 个）

Alan Stevens
Alan Stevens 2024-9-26
Ran in:
You can see from the following that when i = 3, 4 and 7, V(i) stays at a constant value greater than Vmax
%% Constants and Assumptions
g=9.81;
hgelgit=0.5;
he=0.3;
Vmin=0.6;
Vmax=0.9;
f=0.03;
s=0.25;
Cq(1)=0.61;
rho0=1008.301;
rhoa=998.83;
H=20;
Lana=1500;
d=0.04;
hf(1)=0;
%% Initial Hydraulic Head
h=(rhoa*(H+hgelgit+he))/(rho0);
E(1)=h-H;
%% First Port
a(1)=pi*(d)^2./4;
q(1)=Cq(1)*a(1)*sqrt(2*g*E(1));
Q(1)=q(1);
D(1)=floor(20*(sqrt(4*Q(1)/(pi*Vmin))))/20;
A(1)=pi*D(1)^2./4;
V(1)=Q(1)/(pi*(D(1)^2./4)); %[m/s]
hizyuk(1)=0;
U(1)=q(1)/(pi*d^2./4);
Dmax(1)=(sqrt(4*Q(1)/(pi*Vmin)));
Dmin(1)=(sqrt(4*Q(1)/(pi*Vmax)));
%% Other Ports
for i=2:10
hf(i)=f*(s/D(i-1))*(V(i-1)^2./(2*g));
E(i)=E(i-1)+hf(i);
Cq(i)=0.63-0.58*(((V(i-1))^2)/(2*g*E(i)));
a(i)=pi*(d)^2./4;
x=cumsum(a);
q(i)=Cq(i)*a(i)*sqrt(2*g*E(i));
Q(i)=Q(i-1)+q(i);
Dmax(i)=(sqrt(4*Q(i)/(pi*Vmin)));
Dmin(i)=(sqrt(4*Q(i)/(pi*Vmax)));
D(i)=D(i-1);
A(i)=pi*D(i)^2./4;
V(i)=Q(i)/(pi*(D(i)^2./4));
hizyuk(i)=V(i-1)^2./(2*g*E(i));
steps = 0;
while (V(i) >= Vmax) & (steps<4)
steps= steps+1;
D(i)=floor(20*(sqrt(4*Q(i)/(pi*Vmin))))/20;
disp(['i = ', int2str(i),' step = ', int2str(steps)])
disp(['V(i)in= ',num2str(V(i))])
V(i)=Q(i)/(pi*(D(i)^2./4));
disp(['V(i)out= ',num2str(V(i))])
end
U(i)=q(i)/(pi*d^2./4);
end
i = 2 step = 1
V(i)in= 2.5565
V(i)out= 0.63911
i = 3 step = 1
V(i)in= 0.97981
V(i)out= 0.97981
i = 3 step = 2
V(i)in= 0.97981
V(i)out= 0.97981
i = 3 step = 3
V(i)in= 0.97981
V(i)out= 0.97981
i = 3 step = 4
V(i)in= 0.97981
V(i)out= 0.97981
i = 4 step = 1
V(i)in= 1.307
V(i)out= 1.307
i = 4 step = 2
V(i)in= 1.307
V(i)out= 1.307
i = 4 step = 3
V(i)in= 1.307
V(i)out= 1.307
i = 4 step = 4
V(i)in= 1.307
V(i)out= 1.307
i = 5 step = 1
V(i)in= 1.6163
V(i)out= 0.71836
i = 7 step = 1
V(i)in= 1.0191
V(i)out= 1.0191
i = 7 step = 2
V(i)in= 1.0191
V(i)out= 1.0191
i = 7 step = 3
V(i)in= 1.0191
V(i)out= 1.0191
i = 7 step = 4
V(i)in= 1.0191
V(i)out= 1.0191
i = 8 step = 1
V(i)in= 1.1652
V(i)out= 0.65545

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