# 7.30 Nozzle Sizing Calculations

## Objective: Size a long-radius venturi nozzle flow meter

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.

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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

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In [2]:

Do = differential_pressure_meter_solver(D=Di, rho=rho, mu=mu, k=k, P1=P1, P2=P2,
taps='D')
print('Nozzle diameter found to be %s.' %(Do.to(u.inch)))

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Nozzle diameter found to be 2.405956560384632 inch.

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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.

# 7.31 NPRD Calculations

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.

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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,
C

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Out[3]:

\[0.9839001432570909\ dimensionless\]

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In [4]:

dP = differential_pressure_meter_dP(D=Di, D2=Do, P1=P1, P2=P2, C=C,

dP.to(u.psi)

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Out[4]:

\[1.2691031737597311\ pound\_force\_per\_square\_inch\]

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The solution given in Crane is 1.272 psi.