SAE oil is discharged at 15 degrees Celsius from a tank with 7 m of head to atmosphere through:
This is the same problem as 7.6, except the properties of the fluid are sufficiently viscous to put it into the laminar regime although not by much.
In [1]:
from fluids.units import *
from math import pi
rho = 875.*u.kg/u.m**3
mu = 78*u.cP
H = 7*u.m
L = 60*u.m
NPS, D_pipe, Do_pipe, t = nearest_pipe(Do=80*u.mm)
fd = 0.017
Re = 1E5
for i in range(20):
K = K_from_f(fd=fd, L=L, D=D_pipe)
K += entrance_sharp()
K += exit_normal()
K += 6*bend_rounded(D_pipe, angle=90*u.degrees, fd=fd, bend_diameters=0.65)
ball_valve_angle = 0.5*(15+30)*u.degrees # use the average angle
K += K_ball_valve_Crane(D1=D_pipe, D2=60*u.mm, angle=ball_valve_angle, fd=fd)
v = (2*u.gravity*H/K)**0.5
Q = v*pi/4*D_pipe**2
print('Flow rate = %s, Reynolds number = %s' %(Q.to(u.L/u.min), Re))
Re = Reynolds(D=D_pipe, rho=rho, mu=mu, V=v)
fd = friction_factor(Re=Re, eD=0.0018*u.inch/D_pipe)
No solution is actually presented in the example; but the result of their initial guess of a velocity of 1.5 m/s gives 511.2 L/min.