MATLAB: Hey guys can anyone help me to see where I am wrong in the code? I have to get the unknowns within the ranges of 02300. I have attached the code so far.

MATLABmechanical engineering

Original Equations to solve for Unknowns are:

1. V=0.35/((pi*D^2)/4)

2. Re=(VD)/0.00001265

3. 20=f*(150/D)*((V^2)/(2*9.81))

4. 1/sqrt(f)=-log_10(2.51/(Re*sqrt(f)))

My code is as follows:

syms V D Re f

V=[0 50]

D=[0 10]

f=[0 1]

Re>2300

%declaring equations

eqn1 = (V*pi*D.^2)/1.4 == 0;

eqn2 = Re * 0.00001265 – V*D == 0

eqn3 = 20*D*2*9.81 – 150*f*V.^2 == 0

eqn4 = 1/sqrt(f) + log10(2.51/Re*sqrt(f)) == 0

sol = solve([eqn1, eqn2, eqn3, eqn4], [V,D,Re, f]);

vSol = vpa(sol.V, 3)

dSol = vpa(sol.D, 3)

ReSol = vpa(sol.Re, 3)

fSol = vpa(sol.f, 3)

%displaying results

disp(vSol)

disp(dSol)

disp(ReSol)

disp(fSol)

Best Answer

Use the vpa funciton:

... CODE ...
vSol = vpa(sol.V, 3)
dSol = vpa(sol.D, 3)
ReSol = vpa(sol.Re, 3)
fSol = vpa(sol.f, 3)
vSol =
0.00169
dSol =
16.2
ReSol =
2177.0
fSol =
7.45e5
0.00169
16.2
2177.0
7.45e5

EDIT —

The Symbolic Math Toolbox is likely not the best option if you want to impose constraints. Creating an anonymous function (see the appropriate section in Function Basics) from your equations and using the fmincon function would probably do what you want.

Related Solutions

MATLAB: I still have errors in the code and would like help in fixing these errors. I have to get the unknowns within the ranges of 02300. I have attached the code so far along with the equations given to be solved

Re>2300

is just going to display the same thing. Your Re is symbolic so the result of that calculation is going to be a symbolic inequality that will then be displayed (because the line does not end in semi-colon) and then discarded (because you do not assign to a variable.) If you are trying to create a constraint, then you should be using assume()

You have

V=[0 50]
D=[0 10]

and then

eqn1 = (V*pi*D.^2)/1.4 == 0;

With V being a 1 x 2 vector, V*Pi is going to be a 1 x 2 vector. D is a 1 x 2 vector so D.^2 is a 1 x 2 vector. Now you have a 1 x 2 vector "*" a 1 x 2 vector. But the "*" operator is linear algebra matrix multiplication, so the "inner dimensions" must agree -- you can multiply a 1 x 2 by a 2 x 1 (getting a 1 x 1 result) or you can multiply a 2 x 1 by a 1 x 2 (getting a 2 x 2 result), but you cannot multiply a 1 x 2 by a 1 x 2.

Perhaps you want the .* operator instead of the * operator, to do element-by-element multiplication.

eqn1 = (V.*pi.*D.^2)./1.4 == 0;

However, you used constants for V and D, so everything in the (V.*pi.*D.^2)./1.4 expression is going to have a fixed numeric result, which you then compare to 0. The result is going to be a 1 x 2 logical vector, not an equation.

MATLAB: Hello, I have written the following code (as seen below). The answer i obtain is what i am after, however the answer is displayed to too many decimal places. Is there a way of displaying the answers to 3 decimal places

vpa(FSN1, 3)

Best Answer

Related Solutions

MATLAB: I still have errors in the code and would like help in fixing these errors. I have to get the unknowns within the ranges of 02300. I have attached the code so far along with the equations given to be solved

MATLAB: Hello, I have written the following code (as seen below). The answer i obtain is what i am after, however the answer is displayed to too many decimal places. Is there a way of displaying the answers to 3 decimal places

Related Question