# 7.9 Laminar Flow in Valves, Fittings, and Pipe

SAE 30 oil at 100 deg F flows through a 5" schedule 40 pipe at 400 gal/min.

There is a sketch in the Crane document of the system.

Find the velocity in feet/second and the pressure drop.

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In [1]:

from fluids.units import *
from math import pi
from fluids.constants import g
g = g*u.m/u.s**2

dH = 50*u.foot
L = (175+50+75)*u.foot
mu = 130*u.cP
rho = (62.364*0.87)*u.lb/u.ft**3
NPS, Di, Do, t = nearest_pipe(Do=5*u.inch, schedule='40')

Q = 400*u.gallon/u.min
A = 0.25*pi*Di**2

v = Q/A
Re = rho*v*Di/mu
fd = friction_factor(Re=Re, eD=0.0018*u.inch/Di)
ft = ft_Crane(Di)

K_gate = K_gate_valve_Crane(D1=Di, D2=Di, fd=ft, angle=0.0*u.degrees)
K_angle = K_angle_valve_Crane(D1=Di, D2=Di, fd=ft, style=1)
K_elbow = bend_rounded(Di=Di, bend_diameters=1, angle=90*u.degrees, Re=Re, method='Crane')

K_tot = K_elbow + K_gate + K_angle
K_tot += K_from_f(fd=fd, L=L, D=Di)

dP = dP_from_K(K=K_tot, rho=rho, V=v) + rho*g*dH
dP.to(u.psi), v.to(u.foot/u.s)

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Out[1]:

(26.063989664869815 <Unit('pound_force_per_square_inch')>,
6.414177533606955 <Unit('foot / second')>)

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## Intermediate calculations

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In [2]:

K_gate/ft, K_angle/ft, K_elbow/ft

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Out[2]:

(8.0 <Unit('dimensionless')>,
150.0 <Unit('dimensionless')>,
20.00000000000006 <Unit('dimensionless')>)

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In [3]:

K_tot

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Out[3]:

\[30.280117239939614\ dimensionless\]

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In [4]:

v.to(u.ft/u.s)

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Out[4]:

\[6.414177533606955\ foot/second\]

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In [5]:

fd, ft

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Out[5]:

(0.038600064113563516 <Unit('dimensionless')>,
0.01543941031464228 <Unit('dimensionless')>)

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## Proof system is laminar

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In [6]:

Re.to(u.dimensionless)

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Out[6]:

\[1675.616699190696\ dimensionless\]

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