# Chapter 13 : Laminar Flow in Pipe¶

## Example 13.1 Page no 136¶

In [1]:
from __future__ import division
print "Example 13.1 page no 136\n\n"
#calculate average velocities for which th flow will be viscous,laminar
#(a) water at 60 deg F in a 2-inch standard pipe
R_e=2100#reynolds number <2100, for laminar flow
meu_w=6.72e-4#viscosity of water,lb/ft.s
rho_w=62.4#density of water,lb/ft**3
D_w=2.067#diameter  of pipe,ft
v_w=(R_e*meu_w)/((D_w/12)*rho_w)#velocity of water
print "\n velocity v_w=%0.2f ft/s"%(v_w)#
#(b) air at 60 deg F and 5 psig in a 2 inch standard pipe
meu_a=12.1e-6#viscosity of air ,lb/ft.s
rho_a=.1024# density of air,lb/ft**3
D_a=0.17225#diameter of pipe ,ft
v_a=(R_e*meu_a)/(D_a*rho_a)#velocity of air
print "\n velocity of air v_a=%0.2f ft/s"%(v_a)#
#(c) oil of a viscosity of 300 cP and SG of .92 in a 4 inch standard pipe
meu_o=300*6.72e-4#viscosity of oil ,lb/ft.s
rho_o=0.92*62.4#density of oil, lb/ft**3
D_o=.3355#diameter of pipe,ft
v_o=round((R_e*meu_o)/(D_o*rho_o))#velocity of oil
print "\n velocity of oil v_o=%0.2f ft/s"%(v_o)#

Example 13.1 page no 136

velocity v_w=0.13 ft/s

velocity of air v_a=1.44 ft/s

velocity of oil v_o=22.00 ft/s


## Example 13.2 Page no 137¶

In [2]:
from __future__ import division

print " Example 13.2 page no 137\n\n"
#refer to part a of example 1
#appplying Hagen-Poiseuille equation
meu=6.72e-4#viscosity of water
v=0.13#velocity of water
D=2.067/12#diameter of pipe
P_l=32*meu*v/(D**2)
print "\n pressure drop per unit length P_l=%0.2f psf/ft"%(P_l)#

 Example 13.2 page no 137

pressure drop per unit length P_l=0.09 psf/ft


## Example 13.4 Page no 138¶

In [3]:
from __future__ import division

print " Example 13.4 page no 138\n\n "
#an air conducting duct has a rectangular cross section
w=1#width of rectangular section
h=0.25#height of rectangular section
D=2*w*h/(w+h)#equivalent or hydraulic diameter
print "\n hydraulic diameter D=%0.2f m"%(D)
R_e=2300#critical reynolds no
neu=1e-5#kinematic viscosity of air
v=R_e*neu/D#velocity
print "\n velocity of air v=%0.2f m/s"%(v)#

 Example 13.4 page no 138

hydraulic diameter D=0.40 m

velocity of air v=0.06 m/s


## Example 13.5 Page no 139¶

In [4]:
from __future__ import division

from math import pi
print " Example 13.5 page no 139\n\n"
#a circulsr horizontal tube cntains asphalt
D=0.1667#diameter of tube,ft
s=pi*D**2/4#surface area of tube,ft**2
q=0.486#volumatric flow rate,ft**3/s
v=q/s#flow velocity
print "flow velocity v=%0.2f ft/s"%(v)#
g=32.174
print "\n dynamic viscosity meu=%0.2f lb/ft.s"%(meu)#
#check on the laminar flow
rho=70#density,lb/ft**3
R_e=D*v*rho/meu#reynlods number
print "\n reynolds no R_e=%0.2f "%(R_e)#
f=16/R_e#fanning friction factor
print "\n friction factor f=%0.2f "%(f)#
#the pipe must be longer than the entrance length to have fully developed flow
L_e=0.05*D*R_e#entrance length
print "\n entance length L_e=%0.2f ft"%(L_e)#

 Example 13.5 page no 139

flow velocity v=22.27 ft/s

dynamic viscosity meu=0.18 lb/ft.s

reynolds no R_e=1438.12

friction factor f=0.01

entance length L_e=11.99 ft


## Example 13.6 Page no 140¶

In [5]:
from __future__ import division
from math import pi
print " Example 13.6 page no 140\n\n"
#liquid glycerin flows in a tube
#to obtain the properties of glycerine  use table A.2 in the appendix
rho=1260#density,kg/m**3
meu=1.49#viscosity,kg/ms
neu=meu/rho#kinematic viscosity,m**2/s
R=0.02#by no slip condition radius of tube,m
q=32*pi*quad(lambda r:r-2500*r**3,[0,R]) #volumatric flow rate from the given parabolic velocity distribution
print "vol. flow rate q=%0.2f m**3/s"%(q)#
r=0#for average velocity for laminar flow
v_av=16*(1-2500*r**2)/2#average velocity
q=0.010#approximation
m_dot=q*rho#mass flow rate
G=rho*v_av#mass flux
M_dot=m_dot*v_av#inear momentum flux
print "\n av. velocity v_av=%0.2f m/s\n mass flow rate m_dot=%0.2f kg/s\n mass flux G=%0.2f kg/m**2.s\n linear mometum flux M_dot=%0.2f N "%(v_av,m_dot,G,M_dot)#

 Example 13.6 page no 140

vol. flow rate q=0.01 m**3/s

av. velocity v_av=8.00 m/s
mass flow rate m_dot=12.60 kg/s
mass flux G=10080.00 kg/m**2.s
linear mometum flux M_dot=100.80 N


## Example 13.7 Page no 142¶

In [6]:
from __future__ import division

print "Example 13.7 page no 142\n\n"
#refer to example 13.6
rho=1260#density,kg/m**3
v=8#flow velocity,m**2/s
D=0.02#diameter,m
meu=1.49#viscosity
R_e=rho*v*D/meu#reynolds no
print "\n reynolds no R_e=%0.2f "%(R_e)#
V=14000#volume in gallons of glycerine pass through a cross section of tube
q=159.6#flow rate
t=V/q#time
print "\n time t=%0.2f min"%(t)#

Example 13.7 page no 142

reynolds no R_e=135.30

time t=87.72 min