Question

using matlab code to solve the Colebrook Equation: 1/sqrtf= −2.0 ??? ((?/?)/3.7)+2.51/??√?) The following situation should be considered: 3. Given Q, L, hf, ρ, µ, and ?, compute the diameter d of the pipe (sizing problem). 4. Given Q, d, hf, ρ, µ, and ?, compute the pipe length L.

Answer 1

Validation: Example problem: kg Q=0.342m/sec E = 0.06 L=100m p=950- h; Ns 8m u = 0.0199 d = 0.3m m m

function d=pipedia(Q,L,hf,rho,mu,e)
d=0.1;
A=(pi/4)*d.^2;
V=Q/A;
Re=rho*V*d/mu;
f=hf*(2*9.81*d)/(L*V.^2);
LHS=1/sqrt(f);
RHS=-2*log(((e/d)/3.7)+(2.51/(Re*sqrt(f))));
while round(LHS,1)~=round(RHS,1)
d=d+0.01;
A=(pi/4)*d.^2;
V=Q/A;
Re=rho*V*d/mu;
f=hf*(2*9.81*d)/(L*V.^2);
LHS=1/sqrt(f);
RHS=-2*log(((e/d)/3.7)+(2.51/(Re*sqrt(f))));
end
end
  

Matlab Result:

>> D=pipedia(0.342,100,8,950,0.019,0.06)

D =

0.3200

function L=pipelength(Q,d,hf,rho,mu,e)
L=1;
A=(pi/4)*d.^2;
V=Q/A;
Re=rho*V*d/mu;
f=hf*(2*9.81*d)/(L*V.^2);
LHS=1/sqrt(f);
RHS=-2*log(((e/d)/3.7)+(2.51/(Re*sqrt(f))));
while round(LHS,2)~=round(RHS,2)
L=L+0.1;
A=(pi/4)*d.^2;
V=Q/A;
Re=rho*V*d/mu;
f=hf*(2*9.81*d)/(L*V.^2);
LHS=1/sqrt(f);
RHS=-2*log(((e/d)/3.7)+(2.51/(Re*sqrt(f))));
end
end
  

Matlab Result:

>> L=pipelength(0.342,0.32,8,950,0.019,0.06)

L =

98.5000