A system is designed with a flow rate of 225 gpm of water at 60 F flowing through 6" schedule 40 pipe. A long-radius nozzle flow has been requested. A measured head loss of 4 feet is the design measurement, with 1D upstream and 1/2D downstream taps.
Find the required diameter of the nozzle.
In [1]:
from fluids.units import *
from math import pi
mu = 1.1*u.cP
rho = 62.364*u.lb/u.ft**3
NPS, Di, Do, t = nearest_pipe(NPS=6, schedule='40')
A = 0.25*pi*Di*Di
dP = 4*u.feet_H2O
P1 = 10*u.bar # assumed, not very important
P2 = P1 - dP
k = 1.3 # not important
Q = 225*u.gal/u.min
m = Q*rho
In [2]:
Do = differential_pressure_meter_solver(D=Di, rho=rho, mu=mu, k=k, P1=P1, P2=P2,
m=m, meter_type='long radius nozzle',
taps='D')
print('Nozzle diameter found to be %s.' %(Do.to(u.inch)))
The solution given in Crane is 2.40 inches after two iterations and has an error of 0.4 gpm, whereas the answer above has almost zero theoretical error.
The Non-recoverable pressure drop (NRPD) is the permanent pressure drop associated with the flow through the measurement device. Find the NPRD for example 7.30.
In [3]:
C, epsilon = differential_pressure_meter_C_epsilon(D=Di, D2=Do, m=m, P1=P1, P2=P2, rho=rho, mu=mu, k=k,
meter_type='long radius nozzle')
C
Out[3]:
In [4]:
dP = differential_pressure_meter_dP(D=Di, D2=Do, P1=P1, P2=P2, C=C,
meter_type='long radius nozzle')
dP.to(u.psi)
Out[4]:
The solution given in Crane is 1.272 psi.