MATLAB Answers

How can I solve this set of complex equations?

조회 수: 2(최근 30일)
Stefanos Nistor
Stefanos Nistor 30 Jul 2020
Hey guys,
I am fairly new to matlab and I am currently using it to process some data. I have some power and thrust coefficient values (20 Cp and 20 Ct values). I have to use them in order to solve a set of four equations (See attachment Equations_1 and Equations_2).
Cp, Ct and beta are known values. From these 4 equations I need to extract values for and and plug them into the next set of equations (see attachment Equations_3) in order to get new values for Cp, Ct and lamda.
I forgot to mention that lamda values are also known.
My main issue is how to solve the first set of 4 equations.
Can someone help?
Thanks in advance,
Stef

  댓글 수: 2

John D'Errico
John D'Errico 30 Jul 2020
Sorry, but this makes relatively little sense.
In the first page, we see three equations. They involve the variables, or possibly functions. I can't be positive if they are unknown constants or what, but I see:
u_T, u_1, beta, u_2, V0, C_T
You tell us that C_T and C_P are knowns, with 20 values for each. C_P seems irrelevant at this point, so it must enter into the last page. but beta is also a known constant.
So that leaves me with 4 unknowns (u_T, u_1, u_2, V0), and 3 equations in the first page, but 20 different values for C_T. Got it. That some of those equations could be simplified by quite a bit seems interesting, but that is just algebra. I do see that we could immediately eliminate V0 from the first 3 equations.
Then we see the second page, which also seems to beg for some serious simplification. It also has the variable V0' in it. Is that the derivative of V0? With respect to what? Is it just a new variable? Should we somehow know? And while this 4th equation seems to be truly begging for simplification, though this is again just algebra.
Regardless, the 4th equation does nothing for you, since it just provides a formula to compute V0', which we see in no other place. If that is a differential equation, where V0' is a derivative, that could make sense, but we are not told that V0 is an unknown function.
We do see t appear in various places. Is there a difference between little t and big T? My guess it is just random capitalization, since the figures show T, but your question uses t. Do these indicate time? Temperature?
What do the 20 values of C_T mean?
And then on page 3, we see primes attached to the variables. Are they derivatives too? Or does the prime just indicate they are somehow new values of those variables? And since this is the only place where C_P ever appears, it is irrelevant. The same thing applies to lambda, which also seems to have a known list of values.
Highly confusing.
Stefanos Nistor
Stefanos Nistor 30 Jul 2020
Sorry for not making this clear enough. Cp (Power coefficient) and Ct (thrust coefficient) are extracted together with lamda values (tip speed ratio) from some graphs using a plot digitiser. I should also mention that these data are from a wind turbine experiment in a wind tunnel. The initial 4 equations are blockage correction equations. Basically, you use the Cp values together with beta (blockage ratio) in the first set of 4 equations. You somehow have to solve for and . The third picture is showing the equations I have to use in order to get the corrected values.
As you explained above the unknows are quite a few and that's where I got confused too. I don't know how to solve this kind of system.
Is there anything else that didn't make sense?

댓글을 달려면 로그인하십시오.

답변(1개)

David Goodmanson
David Goodmanson 31 Jul 2020
편집: David Goodmanson 31 Jul 2020
Hi Stefanos,
As has already been mentioned, the 4th equation is just the definition of V0' and is independent of the first three equaitions. The same is true for the last three equations, so it's all down to the first three equations. Three of the variables appear as u2/u1, uT/u1 and V0//u1, so they are all scaled by u1. Letting u2 stand for u2/u1 and so forth (or just pretending that u1 = 1) we have
uT = (-1 + sqrt(1 + beta*(u2^2-1)))./(beta*(u2-1));
V0 = u2 - beta*uT.*(u2-1);
V0 = sqrt((u2.^2-1)/cT)
so despite apprearances it's really three equations, three unknowns. Leaving the rhs of the third eqn as is, and doing some substitution on the lhs results in
u2 + 1 - sqrt(1 + beta*(u2^2-1)) = sqrt((u2.^2-1)/cT)
and this can be solved with fzero. Here I just took a guess at some values.
beta = .1;
cT = .7;
f1 = @(x) ( x + 1 - sqrt(1+beta*(x.^2-1)));
f2 = @(x) sqrt((x.^2-1)/cT);
f12 = @(x) f1(x) - f2(x);
u2 = fzero(@(x) f12(x),[1 100])
% back substitute
uT = (-1 + sqrt(1 + beta*(u2^2-1)))./(beta*(u2-1));
V0 = u2 - beta*uT.*(u2-1);
V0 - sqrt((u2root.^2-1)/cT) % consistency check, should be small
% plot for visual purposes
u2vec = 1:0.001:5;
plot(u2vec,f1(u2vec),u2vec,f2(u2vec))
Here u2 really means u2/u1 etc. Fortunately the last four equations (with eqn 4 in the form V0'/V0 = (...) ) are all in terms of ratios, so the particular value of u1 cancels out and is not needed. The procedure can be enclosed by a for loop for different values of cT.

  댓글 수: 0

댓글을 달려면 로그인하십시오.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by