# Chapter 21 : Academic Application¶

## Example 21.7 Page no 284¶

In [1]:
from __future__ import division
print "Example 21.7 page no 284\n\n"
# water is flowing through a 3/8 in schedule 40 brass pipe
D=0.0411#diameter of pipe,ft
S=0.00133#cross section area of pipe,ft**2
meu=6.598e-4#viscosity of water from table A.4 in the appendix,lb/ft.s
rho=62.4#density,lb/ft**3
q_gpm=2#vol.flow rate
q=q_gpm*0.00228#volumatric flow rate in ft**3s
v=q/S#velocity of fluid
print "\n veloctiy of fluid v=%.2f ft/s"%(v)#
R_e=D*v*rho/meu#reynolds no.
print "\n reynolds no R_e=%.2f "%(R_e)#

Example 21.7 page no 284

veloctiy of fluid v=3.43 ft/s

reynolds no R_e=13326.84


## Example 21.8 Page no 285¶

In [2]:
from __future__ import division
print "Example 21.8 page no 285\n\n"
#water flowing through a pipe
rho=62.4#density of water,lb/ft**3
meu=6.72e-4#viscosity of water,lb/ft.s
q_1gpm=1.5#vol. flow rate in gpm
q_2gpm=6#vol. flow rate in gpm
D_1=0.493#internal diameter of 3/8 in schdule pipe
v11=(0.409*q_1gpm)/(D_1**2)#flow velocity for an 3/8 in pipe with 1.5 gpm flow rate
v12=(0.409*q_2gpm)/(D_1**2)#flow velocity for an 3/8 pipe with 6 gpm flow
R_e11=D_1*v11*rho/meu#reynolds no for case 11
R_e12=D_1*v12*rho/meu#reynolds no for case 12
print "\n reynolds no R_e11=%.2f\n reynolds no R_e12=%.2f  "%(R_e11,R_e12)##printing mistake in book
D_2=0.622#internal diameter of  1/2 in schdule pipe
v21=(0.409*q_1gpm)/D_2**2#flow velocity for 1/2 pipe with 1.5 gpm
v22=(0.409*q_2gpm)/D_2**2#flow velocity for 1/2 pipe with 6 gpm
R_e21=D_2*v21*rho/meu#reynolds no for case 21
R_e22=D_2*v22*rho/meu#reynolds no foe case 22
print "\n reynolds no R_e21=%.2f\n reynolds no R_e22=%.2f"%(R_e21,R_e22)#
#printing mistake in value of R_e

Example 21.8 page no 285

reynolds no R_e11=115553.46
reynolds no R_e12=462213.85

reynolds no R_e21=91588.19
reynolds no R_e22=366352.78


## Example 21.9 page no 286¶

In [3]:
from __future__ import division
print "Example no 21.9 page no 286\n\n"
#water is flowing in a vertical pipe
#assume constant velocity
P_drop=-4.5#pressure drop from bottom to top
rho=62.4 #density of water
z2=15#height of pipe
z1=0#bottom level
#applying bernoulli equation
h_f=(P_drop/rho)+(z2-z1)#frictional loss
print "\n frictional loss h_f=%.2f ft.lbf/lb "%(h_f)

Example no 21.9 page no 286

frictional loss h_f=14.93 ft.lbf/lb


## Example 21.10 Page no 286¶

In [4]:
from __future__ import division
print "Example 21.10 page no 286\n\n"
#a centrifugal pump is needed to transport water from sea level to 10000 feet above sea level
#using bernoulli equation
#neglectiing kinetic energy effects and frictional losses
P1=14.7#atmospheric pressure at sea level,psi
P2=10.2#atmospheric pressure at 10000 feet,psi
z1=0#at sea level,ft
z2=10000#height above sea level,ft
rho=62.4#density of water
g=32.2#gravitational acc.
g_c=32.2#gravitational constant
h_s=((P2-P1)*144/(rho)  + (z2-z1)*(g/g_c))#work deliverd by the pump to the water,in ft.lbf/lb
h_s=9990#ft.lbf/lb
h_sf=h_s*50#in ft.lbf
print "\n work h_sf=%.2f ft.lbf/s"%(h_sf)#
#actual pump work is calculated by dividing the above terms by the frictional afficiency
neta=0.65#frictional efficiency
W_p=round((h_sf/550)/neta)#actual work
print "\n actual work W_p=%.2f hp"%(W_p)#

Example 21.10 page no 286

work h_sf=499500.00 ft.lbf/s

actual work W_p=1397.00 hp


## Example 21.12 Page no 288¶

In [5]:
from __future__ import division
print "Example 21.12 page no 288\n\n"
#refer to illustrative Example 21.4
# if the pipe contains two globe valves and one straight through tee,what is the friction loss
K_f_globe=6
K_f_tee=0.4
v=2.53# flow velocity
g_c=32.2
f=5/4#friction factor
L=144#lenth of pipe
D=62.4#diameter
h_f=4*f*(L/D) + (2*K_f_globe + K_f_tee)*(v**2/(2*g_c))
print "\n frictional loss h_f=%.2f ft.lbf/lb"%(h_f)#


Example 21.12 page no 288

frictional loss h_f=12.77 ft.lbf/lb


## Example 21.13 Page no 289¶

In [6]:
from __future__ import division
from math import sqrt, pi
print "Example 21.13 page no 289 fig 21.1  \n\n\n"
#a pitot tube is inserted in acircular pipe  to measure the flow velocity
# the tube is inserted so that it points upstream into the flow and the pressure sensed by thre probeis the stagnation pressure
#the change in elevation between the tip of the pitot and the wall pressure tap is negligible
#the flowing fluid is soyabean oil at 20 deg C and the fluid in manometer tube is mercury
#point 2 is a stagnation point ,P2>P1 and the manometer fluid should be higher on th eleft side(h<0)
rho_m=13600#density of mercury,kg/m**3
h=0.04#height of mercury,
rho=919#density of oil kg/m**3
g=9.804
D=0.055#diameter of pipe,m
meu=0.04#viscosity of oil,kg.m.s
v=sqrt(2*g*h*((rho_m/rho)-1))#flow velocity
print "\n flow velocity v=%.2f m/s"%(v)#
#assuming uniform velocity
S=(pi/4)*D**2
m_dot=rho*v*S#mass flow rate
R_e=(D*v*rho)/meu#reynolds no
print "\n reynolds no R_e=%.2f "%(R_e)#
print "\n mass flow rate m_dot=%.2f kg/s"%(m_dot)#

Example 21.13 page no 289 fig 21.1

flow velocity v=3.29 m/s

reynolds no R_e=4157.04

mass flow rate m_dot=7.18 kg/s


## Example 21.14 Page no 290¶

In [7]:
from __future__ import division
print "Example 21.14 page no 290\n\n"
#given: a 50 ft pipe with flowing water ,we have to determine the flow rate if there is an expansion from 3/8 inch to 1/8 inch and immediatly back to 3/8n inch with an overall pressure loss no greater than 2lbf/ft**2
#from table A.5 in the appendix
S1=0.00133#cross sectional area of 3/8 inch pipe,ft**2
S2=0.00211#cross sectional area of 1/2 inch pipe,ft**2
K_e=(1-S1/S2)**2#expansion constant
K_c=0.4*(1-S2/S1)**2#contraction constant
L=50#length of pipe
D=0.03125#diameter of pipe
v=1.93#velocity ,ft/s
f=0.01124#friction factor from table 21.3,for velocity estimated to be 1.93 ft/s
g_c=32.2
h_f=(4*f*L/D  + K_e + K_c)*(v**2*g_c)#frictional loss
print "\n frictional loss h_f=%.2f ft.lbf/lb "%(h_f)#

Example 21.14 page no 290

frictional loss h_f=8661.02 ft.lbf/lb


## Example 21.16 Page no 291¶

In [8]:
from __future__ import division
print "Example 21.16 page no 291\n\n"
#water flows in a concrete pipe
v_p=0.02# flow velocity,m/s
D_p=1.5#diameter of pipe
L_p=20#length of pipe,m
rho_p=1000#density of water,kg/m**3
meu_p=0.001#viscosity of water,kg/m.s
K_p=0.003#roughnes factor,m
#this prototype is to be modeled in a lab using a 1/3o th scale pipe
D_m=D_p/30#D_m is diameter of modeled pipe
L_m=L_p*(D_m/D_p)#length of modeled pipe
K_m=K_p*(D_m/D_p)#roughness factor for modeled pipe
#the fluid in the model is caster oil
rho_m=961.3#densiy of oil, kg/m**3
meu_m=0.0721#viscosity of oil,kg/m.s
#since R_e = (rho_m*v_m*D_m)/meu_m   =  (rho_p*v_p*D_p)/meu_p
v_m = (rho_p*v_p*D_p*meu_m)/(rho_m*D_m*meu_p)# flow velcity in molded pipe
print "\n flow velocity v_m=%.2f m/s"%(v_m)#
#pressure drop in prototype
P_drop_m=1e+5#pressure drop in model
P_drop_p=(P_drop_m*rho_p*(v_p)**2)/(rho_m*(v_m)**2)#pressure drop in prototype
print "\n pressure drop in prototype P_drop_p=%.3f Pa"%(P_drop_p)#

Example 21.16 page no 291

flow velocity v_m=45.00 m/s

pressure drop in prototype P_drop_p=0.021 Pa