How could I make a script for this ecuation where i can solve for f?

[Walter Roberson]

What are the inputs? What is the output?

Does stand for the real part of the square root of f, or does it stand for a variable named Re times the square root of f, or does it stand for a variable named R times exp(sqrt(f)) ?

[Torsten]

@Walter Roberson

Re is Reynolds number in fluid flow.

Looks like the Colebrook-White equation to compute the pressure loss coefficient f from the above implicit equation in f.

@Jorge

Use "fzero" to solve

x + 2*log10(epsilon/D/3.7+2.51*x/Re) = 0

for x.

[Jorge]

Yes thank you

Would it show the value of x in the workspace?

[Walter Roberson]

rng(655321)
epsilon = rand

epsilon = 0.3140

D = rand

D = 0.3864

Re = rand

Re = 0.1855

eqn = @(x) x + 2*log10(epsilon/D/3.7+2.51*x/Re);
X = fzero(eqn, [0 1e5])

X = 0.0533

[Torsten]

...  
f = 1/X^2

Answer 1

[Walter Roberson]

syms f epsilon D Re
Q = @(v) sym(v);
eqn = 1/sqrt(f) == -Q(2.0) * log(epsilon/D/Q(3.7) + Q(2.51)/(Re*sqrt(f)))
F = solve(eqn, f)

Warning: Solutions are only valid under certain conditions. To include parameters and conditions in the solution, specify the 'ReturnConditions' value as 'true'.

disp(char(F))

(86248369*D^2)/(1000*Re*epsilon - 18574*wrightOmega((500*Re*epsilon)/(9287*D) - log(251/(50*Re)))*D)^2

wrightOmega is a function in the Symbolic Toolbox

rng(655321)
epsilon = rand

epsilon = 0.3140

D = rand

D = 0.3864

Re = rand

Re = 0.1855

f = subs(F)
disp(char(vpa(f)))

322.8497917132456879511724468922

subs(eqn)
disp(char(vpa(ans)))

0.055654430712142653567619972363646 == 0.055654430712142653567619972363646