In [1]:
from fluids.units import *
from thermo.units import Chemical
P1 = 2*u.bar # The full set of equations requires actual pressures not just the pressure difference, so an initial pressure of 2 bar is assumed.
P2 = P1 - 17*u.kPa
taps = 'D'
meter_type = 'ISO 5167 orifice'
Do = 50*u.mm
NPS, D, _, t_pipe = nearest_pipe(Do=80*u.mm, schedule=80)
D
Out[1]:
In [2]:
water = Chemical('water', T=15*u.degC, P=P1)
rho = water.rho
mu = water.mu
k = water.isentropic_exponent
rho, mu, k
Out[2]:
In [3]:
m = differential_pressure_meter_solver(D=D, D2=Do, P1=P1, P2=P2, rho=rho, mu=mu, k=k, meter_type=meter_type, taps=taps)
print(m)
Q = m/rho
print(Q.to('L/min'))
The answer given in TP410M is 478 gallons/minute; however the errata specifies this value is in units of liters/minute insteead.
This calculation matches their result well, given they did not include expansivity in their calculations and read a value of C from a graph.
A calculator at flow of fluids, which also does not include expansivity, gives an answer of 476.777 L/min, along with 7.93916 kg/s (http://www.flowoffluids.com/calculators/flow-through-orifices.aspx).