7.13 Flow given in English Units - Oil

Fuel oil at a specific gravity of 0.815 (kinematic viscosity of 2.7 centistokes) flows at 2 inch, schedule 40 steel pipe 100 foot long at a rate of 2 US gallons/second.

Calculate the pressure drop in bars and psi.



In [1]:

    
from math import *
from fluids.units import *
from thermo import *

SG = 0.815
rho = SG*999.1*u.kg/u.m**3
nu = 2.7*u.centistokes
mu = nu_mu_converter(rho,  nu=nu)
Q = 2*u.gal/u.s
L = 100*u.foot

NPS, D_pipe, Do_pipe, t = nearest_pipe(Di=2*u.inch)
v = Q/(pi/4*D_pipe**2)
Re = Reynolds(rho=rho, mu=mu, D=D_pipe, V=v)
fd = friction_factor(Re=Re, eD=0.0018*u.inch/D_pipe)
print('Darcy friction factor = %s' %fd)
K_friction = K_from_f(fd=fd, L=L, D=D_pipe)
dP = dP_from_K(K=K_friction, rho=rho, V=v)
print('Pressure drop = %s' %dP.to(u.Pa))
print('In imperial, pressure drop = %s' %dP.to(u.psi))









    



Darcy friction factor = 0.0609076579823 dimensionless
Pressure drop = 176425.270238 pascal
In imperial, pressure drop = 25.5883220737 pound_force_per_square_inch

The pressure drop calculated in the example is 66500 Pa (9.65 psi). The discrepancy is from their friction factor; they use 0.0230. The result is matched exactly if their friction factor is used.



In [2]:

    
fd = 0.023
print('Darcy friction factor = %s' %fd)
K_friction = K_from_f(fd=fd, L=L, D=D_pipe)
dP = dP_from_K(K=K_friction, rho=rho, V=v)
print('Pressure drop = %s' %dP.to(u.Pa))
print('In imperial, pressure drop = %s' %dP.to(u.psi))









    



Darcy friction factor = 0.023
Pressure drop = 66621.8559355 pascal
In imperial, pressure drop = 9.66268326827 pound_force_per_square_inch