# 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.

    [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.

    [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