# Chapter 22 : Industrial Application¶

## Example 22.4 Page no 298¶

In [1]:
from __future__ import division
from math import sqrt,pi
print "Example 22.4 page no 298\n\n"
#a centrifugal pump operating at 1800 rpm ,we have to find the impeller diameter needed to develop a head of 200 ft
h=200#height,ft
g=32.2#gravitational acc. ft/s**2
v=sqrt(2*g*h)#velocity needed to develop a head of 200 ft
print "\n velocity v=%.2f ft/s"%(v)#
N=1800#pump operating at this rotational speed,in rpm
c=v*60/N#the number of feet that the impeller travels in one rotations
#this c represents the circumference of the impeller  since it is equal to one rotation
print "\n circumference c=%.2f ft/rotation"%(c)#
D=c/pi#diameter of the impeller
print "\n diameter D=%.2f ft"%(D)#
Example 22.4 page no 298

velocity v=113.49 ft/s

circumference c=3.78 ft/rotation

diameter D=1.20 ft

## Example 22.5 Page no 299¶

In [2]:
from __future__ import division
from math import sqrt,pi
print "Example 22.5 page no 299\n\n"
#water for a processing plant is required to be stored in a reservoir
#assume the properties of water at 20 deg C are
rho=998#density,kg/m**3
meu=0.001#viscosity,N.s/m**2
L=120#length of pipe,m
D=0.15#diameter  of pipe,m
S=(pi/4)*D**2#cross sectional area of pipe
#given:
q=1.2/60#volumetric flow rate,m**3/s
v=q/S#flow velocity,m/s
R_e=D*v*rho/meu#reynolds no
print "\n reynolds no R_e=%.2f "%(R_e)#
#from value of R_e ,flow is clearly turbulent
k=0.0005#roughness factor for galvanized iron
K_r=k/D#relative roughness
f=0.0053#fricion factor from fig. 14.2
h_f=4*f*(L/D)*(v**2/2)#friction loss of energy
print "\n h_f frictional  loss=%.2f J "%(h_f)#
#for right elbows (from table 18.1),the estimated value of resistance coff. K for one regular 90 deg elbows is 0.5
K=4#resstance coeff.
e_l=K*V_h#the total loss from the elbows
print "\n e_l total elbow loss=%.2f J/kg"%(e_l)#
#the energy to move 1 kg of water against a head of 22m of water is
z=22#height,m
g=9.81#grav. acc,m/s**2
PE=z*g
print "\n potential energy PE=%.2f J/kg"%(PE)#
TE = h_f + e_l  + PE#total requirement per kg
print "\n total energy TE=%.2f J/kg"%(TE)#
W_dot_s= TE*q*rho#theoretical power requirement
print "\n theoritical power W_dot_s=%.2f J/s"%(W_dot_s)#
h=TE/g#head equivalent to the energy requirement
print "\n equivalent head h=%.2f m "%(h)#
Example 22.5 page no 299

reynolds no R_e=169425.74

h_f frictional  loss=10.86 J

e_l total elbow loss=2.56 J/kg

potential energy PE=215.82 J/kg

total energy TE=229.24 J/kg

theoritical power W_dot_s=4575.71 J/s

## Example 22.6 Page no 301¶

In [3]:
from __future__ import division
print "Example 22.6 page no 301\n\n"
#oil is flowing through a standard 3/2 inch steel pipe containing a 1 inch square edged orifice
v_gal=400#orifice velocity of oil in gal/hr
v_o=400*144/(0.785*3600*7.48)#orifice velocity in ft/hr
D_o=1/12#diameter of orifice
rho=0.87*62.4#density of oil
meu=20.6*0.000672#viscosity of oil
R_e=D_o*v_o*rho/meu
print "\n reynolds no =%.2e "%(R_e)#
D_r=0.62#ratio of orifice plate to pipe diametersD_o/D1 = 1/1.61
C_d=0.76#discharge coeff. fro fig 19.8
g=32.2#grav. acc. ft/s**2
h=(v_o**2/(2*g*(C_d)**2))*(1-D_r**4)#height of oil in gauge reading
print "\n gauge reading h=%.2f ft "%(h)#
Example 22.6 page no 301

reynolds no =8.91e+02

## Example 22.7 Page no 302¶

In [4]:
from __future__ import division
from math import sqrt,pi
print "Example 22.7 page no 302\n\n "
#natural gas consisting of essentially pure methane flows through a long straight standard 10 inch steel pipe into which is inserted a square edged orifice 2.50 inches in diameter ,with pressure taps ,each 5 inch from the orifice plate
#manometer is attached across the orifice reads 1.60 in H20
D_o=2.50#diameter of orifice
D_1=10.15#diameter of plate
D_r=D_o/D_1#ratio of diameters
#assuming the reynolds no R_e in the orifice to be over 30,000
C_o=0.61#coeff. of discharge from R_e value
g=32.2#garv. acc ft/s**2
rho_m=62.4#density of medium (water)
rho=0.054#density of methane gas,lb/ft**3
meu=12*0.011*0.000672#viscosity
v_o= C_o*sqrt((2*g*h*rho_m)/(12*rho))# orifice velocity
print "\n orifice veloctiy v_o=%.2f ft/s"%(v_o)#
R_e_o=D_o*v_o*rho/meu#reynolds no in the orifice
print "\n R_e_o reynolds no =%.2f "%(R_e_o)#
#from R_e_o value C_o=0.61 is permissible
m_dot=round(v_o*(pi/4)*(D_o**2)*rho*(3600/144))#mass flow rate
print "\n mass flow rate m_dot=%.2f lb/hr"%(m_dot)#
Example 22.7 page no 302

orifice veloctiy v_o=60.76 ft/s

R_e_o reynolds no =92475.80

mass flow rate m_dot=403.00 lb/hr

## Example 22.8 Page no 303¶

In [5]:
from __future__ import division
from math import sqrt,pi
print "Example 22.8 page no 303\n\n"
#refer to fig 22.1
D1=.1#upstream diameter(at station 1),m
D2=.06#downstream diameter(station 2),m
S2=(pi/4)*D2**2#cross sectional area at point 2
rho=1.22#density of air  from ideal gas law
rho_m=827#density of medium,kg.m**3
g=9.8#gravitational acc.
#from bernoulli equation
v2=sqrt(2*g*h*((rho_m/rho)-1))#velocity at point 2
v1=v2*(D2/D1)**2#velocity at point 1
q=v2*S2#volumatric flow rate
print "\n vol.flow rate q=%.2f m**3/s"%(q)#
#calculation of mach number from equation 15.1
T=293#temperature in k
c=20*sqrt(T)#speed of light at this temperature,m/s
M_a=v2/c#mach no.
print "\n mach no. M_a =%.2f "%(M_a)#
#noting that M_a=0.095 < 0.3 , we can conclude that flow is incompressible#given
P1=130000 #absolute pressure at point 1,pa
#by using bernoulli eq for P2
P2=P1-rho*v2**2*(1-(D2/D1)**4)/2#pressure at point 2
print "\n pressure at point 2=%.2f Pa"%(P2)#
Example 22.8 page no 303

vol.flow rate q=0.09 m**3/s

mach no. M_a =0.10

pressure at point 2=129436.49 Pa

## Example 22.9 Page no 305¶

In [6]:
from __future__ import division

print "\n Example 22.9 page no 305\n\n"
#water is flowing from an elevated reservoir through a conduit to a turbine at a lower level and out of the turbine through a similar conduit
#refer to fig 22.2
#since the diameter of the conduit is the same at location 1 and 2 ,kinetic energy effects can be  neglected and bernoulli eq. takes the form
#P/rho  + z(g/g_c)  -h_s + h_f  = 0
P1=30#/pressure at point 1,psia
z1=300#height of point 1,ft
P2=18#pressure at point 2,psia
z2=-10#height of point 2,ft
rho=62.4#density
m_dot=3600#mass flow rate,tons/hr
W_dot =1000#output at the shaft of turbine,hp
neta=0.9#efficiency of turbine
h_s=W_dot*550*3600/(neta*m_dot*2000)#
print "\n h_s =%.2f ft.lbf/lb"%(h_s)#
#put this value in bernoulli eq.
h_f=(P2-P1)*144/rho + (z2-z1) -h_s#frictional loss
print "\n frictional loss h_f=%.2f ft.lbf/lb"%(h_f)
Example 22.9 page no 305

h_s =305.56 ft.lbf/lb

frictional loss h_f=-643.25 ft.lbf/lb

## Example 22.10 Page no 306¶

In [7]:
from __future__ import division
from math import sqrt,pi
print "\n Example 22.10 page no 306\n\n"
#benzene is pumped from a large tank to a delivery station
#refer fig 22.3
q=0.003#vol. flow rate,m**3/s
#tank is at atmosphric pressure
D=0.03#diameter of suction and discharge line,m
v_2=q/((pi/4)*D**2)#discharge velocity,m/s
#since all diameters are same likewise velocities are same
v_3=v_2
g=9.807#grav. acc.
print "\n dynamic head D_h=%.2f m"%(D_h)#
z1=0#height at point 1,tank level
z2=1.8#height at point 3
#applying bernoulli's eq. between the top of the tank(open to theatomsphere)and the inlet to the pump(station3)
rho=865#density of benzene,kg/m**3
P3=101325-(z2+D_h)*(rho*g)#ptressure at point 3
print "\n pressure at point 3 P3=%.2f Pa"%(P3)#
P_v=26200#vapor pressure of benzene,Pa
NPSH =  (P3 - P_v)/(rho*g) +  D_h
print "\n  NPSH=%.2f m"%(NPSH)
#the manufacturer NPSH is 8 m, which is greater than the calculated NPSH  of 7.06m,therfore, the suction point of pump must be lowered
#calculation of new pressure
NPSH_m=8#NPSH by manufacturer
P3_n_ab=8*(rho*g)-D_h*(rho*g) + P_v
print "\n new pressure at point 3 P3_n_ab=%.2f Pa absolute"%(P3_n_ab)#
P3_n_bz=-1.77#pressure in terms of benzene height,m
z3=-P3_n_bz -D_h#desired height of point 3
print "\n height z3=%.2f m"%(z3)#
Example 22.10 page no 306

pressure at point 3 P3=78265.03 Pa

NPSH=7.06 m

new pressure at point 3 P3_n_ab=86273.97 Pa absolute

height z3=0.85 m

## Example 22.11 Page no 308¶

In [8]:
from __future__ import division
from math import sqrt,pi
print "Example 22.11 page no 308\n\n"
#a storage tank on top of a building pumps 60 deg F water through an open pipe to it from a reservoir
q=1.36#vol. flow rate ,ft**3/s
D=0.333#diameter of pipe,ft
S=pi/4*D**2#cross sectional area,ft**2
v2=q/S#flow velocity,ft/s
rho=62.37#density of water,lb/ft**3
meu=1.129*6.72e-4#viscosity of water
R_e=D*v2*rho/meu#reynolds no.
print "\n reynolds no. R_e=%.2f"%(R_e )#
#from R_e we can conclude that flow is turbulent
k=0.0018#roughness factor
K_r=k/D#relative roughness
f=0.0046#friction factor
L=525#length of pipe,ft
g_c=32.174#grav. acc
h_fp=(4*f*L*v2**2)/(D*2*g_c)#frictional loss due to the length of pipe
print "\n frictional loss h_fp=%.2f ft.lbf/lb"%(h_fp)#
#friction due to the fitings from table 18.1
K_ff_gate=2*0.11#loss coeff. due to gates
K_ff_elbows=5*0.64#loss coeff. due to elbows
#friction due to the sudden contraction is obtained from eq. 18.10 .
#note that D1/D2=0,since the upstream diameter is singnificantly larger than the downward diameter
K_c=0.42#coeff. of sudden contraction
K_e=1#coeff. of sudden expansion
K_s=K_ff_gate +K_ff_elbows +K_e +K_c#sum of loss coeff.
h_f=K_s*v2**2/(2*g_c)#friction losses due to fitting,expansion,contraction
h_f_total=h_fp + h_f#total frictional losses
print "\n total frictional loss h_f_total=%.2f ft.lbf/lb"%(h_f_total)#
v1=0
P_drop=0#pressure drop
z1=0#reservoir water level
z2=200#height of reservoir
W_s=(v2**2-v1**2)/(2*g_c)  + (z2-z1)  + h_f_total#power requirement
m_dot=q*rho#mass flow rate,lb/s
neta=0.6#efficiency of pump
W_dot_s=m_dot*W_s/(550*neta)#actual horsepower requirement
print "\n W_dot_s=%.2f hp"%(W_dot_s)#
Example 22.11 page no 308

reynolds no. R_e=427481.50

frictional loss h_fp=109.93 ft.lbf/lb

total frictional loss h_f_total=128.27 ft.lbf/lb

W_dot_s=85.35 hp

## Example 22.12 Page no 311¶

In [9]:
from __future__ import division
from math import sqrt,pi
print "Example 22.12 page no 311\n\n"
#turpentine is being moved  from a large storage tank to a blender through a 700 ft pipeline
rho=62.4#density
SG=0.872#specific gravity of terpentine
rho_t=SG*rho#density of turpentine
v=12.67#av. velocity of the turpentine in the line,ft/s
z1=20#height of top surface in the storage tank above floor level,ft
z2=90#height of discharge end of pipe,ft
neta=0.74#efficiency of pump
W_s=401.9#average energy delivered by pump,ft/lbf/lb
g_c=32.174#grav.acc
L=700#length of pipeline
#from bernoulli eq.
h_f= neta*W_s - v**2/(2*g_c) - (z2-z1)#frictional loss  if there is no pressure drop
print "\n frictional loss h_f =%.2f ft.lbf/lb"%(h_f)#
k_c=0.4#coeff. of contraction
k_e=0.9#coeff. of expansion
k_f=0.2#coeff. of bends and valve
#making equation(1) from the friction coeff. due to fittings between f and D,f=0.0293*D
#making another equation(2) from Reynolds number in terms D ,R_e=582250*D
#from trial and error method we get D
D=0.184#diameter
S=pi*D**2/4#cross sectional area
S=0.0266
q=v*S#volumetric flow rate
print "\n q=%.2f ft**3/s "%(q)#
m_dot=rho_t*q#mass flow rate
bhp =m_dot*W_s/(550*neta)#brake horse power
print "\n brake horse power bhp=%.2f hp"%(bhp)#
Example 22.12 page no 311

frictional loss h_f =224.91 ft.lbf/lb

q=0.34 ft**3/s

brake horse power bhp=18.11 hp

## Example 22.13 Page no 313¶

In [10]:
from __future__ import division
from math import sqrt,pi
print "Example 22.13 page no 313\n\n"
#hydrogen flows through a horizontal pipe
#properties of hydrogen at 20 deg C from table A.3 in the appendix
rho=0.0838#density of hydrogen,kg/m**3
meu=9.05e-6#viscosity,kg/m.s
D=0.08#diameter of pipe,m
L=1#unit length of pipe,m
q=0.0004#vol. flow rate ,m**3/s
S=.000503#cross sectional area
v=q/S#flow velocity,m/s
m_dot=rho*q#mass flow rate,kg/s
R_e=(D*v*rho/meu)#reynolds no.
print "\n R_e reynolds no=%.2f "%(R_e)#
#since R_e is 593<2100, flow is laminar
#since the tube is horizontal z1=z2,calculation of pressure gradient(P/L)
v_max=2*v#m/s
#calculation of fanning friction factor
#since the flow is laminar
f=16/R_e#fanning friction factor
print "\n fanning friction factor f=%.2f "%(f)#
f_d=4*f#darcy friction factor
print "\n darcy friction factor f_d=%.2f "%(f_d)#
g=9.807#grav. acc.
h_f=f_d*(L/D)*(v**2/(2*g))#friction loss
print "\n friction loss h_f=%.2f m"%(h_f)#
W_f = m_dot*g*h_f#friction power loss
print "\n friction power loss W_f=%.2f W"%(W_f)#
Example 22.13 page no 313

R_e reynolds no=589.08

fanning friction factor f=0.03

darcy friction factor f_d=0.11

friction loss h_f=0.04 m

friction power loss W_f=0.00 W

## Example 22.14 Page no 315¶

In [11]:
from __future__ import division
from math import sqrt,pi
print "Example 22.14 page no 315\n\n"
#gasoline is pump through a horizontal cast iron pipe
L=30#length of pipe
D=0.2#diameter of pipe,m
S=(pi/4)*D**2#cross sectional area
q=0.3#vol. flow rate ,m**3/s
v=q/S#flow velocity,m/s
rho=680#density of gasoline,kg/m**3
meu=2.92e-4#viscosity of gasoline,kg/m.s
R_e=D*v*rho/meu#reynolds no.
print "\n reynolds no R_e=%.2f "%(R_e)#
#since R_e is >4000 flow is turbulent
k=0.00026#roughness factor from table 14.1 for cast iron,m
K_r=k/D#relative roughness
f=0.00525#fanning friction factor from fig 14.2
#Note that the flow corresponds to complete turbulence in the rough pipe
g=9.807#gravitational acceleration
h_f=14.647
#applying bernoulli equation to the fluid in the pipe
#in this case the pipe is horizontal (z1=z2) with constant diameter (v1=v2) and no shaft head (h_s=0)
#first convert the friction head to a pressure difference
P_diff=rho*g*h_f#pressure difference
P_diff= 97.68*10**3#after round off
W_s_id=q*P_diff#ideal shaft work
print "\n ideal shaft work W_s_id=%.2f W "%(W_s_id)#
neta=0.8#efficiency of pump
W_s_ac=W_s_id/neta#actual shaft work
print "\n actual shaft work W_s_ac=%.2f W"%(W_s_ac)#
f_s=0.009#friction factor smooth
f_r=0.021#friction factor roughnes
k=f_r/f_s
f_inc=100*(k-1)#percentage increment in f due to roughness
print "\n f_inc=%.2f "%(f_inc)#
Example 22.14 page no 315

reynolds no R_e=4447617.59

ideal shaft work W_s_id=29304.00 W

actual shaft work W_s_ac=36630.00 W

f_inc=133.33

## Example 22.15 Page no 316¶

In [12]:
from __future__ import division
from math import sqrt,pi
print "\n Example 22.15 page no 316\n\n"
#liquid benzene flows through a smooth horizontal iron pipe
D=2.3#diameter of pipe,m
L=146.304#length of pipe,m
S=(pi/4)*D**2#cross sectional area,m**2
q=4000#vol. flow rate,gal/min
v=q/(S*264.17*60)#flow velocity
print "\n flow velocity v=%.2f m/s"%(v)#
rho=899#density of benzene
meu=0.0008#viscosity of benzene,kg/m.s
R_e = D*v*rho/meu#reynolds no
print "\n reynolds no R_e=%.2f "%(R_e)#
#since the reynolds number falls in the turbulent regime,determine the fanning friction factor from fig. 14.2
f=0.0032#fanning friction factor
# calculation of pressure drop with the assumption of no height and velocity change , and no pump work
#since only frictional losses are to be considered
#applying eq. 14.3
P_drop = 4*f*(L/D)*(v**2/2)*rho#pressure drop
print "\n pressure drop P_drop=%.2f Pa"%(P_drop)#
W_dot_f=q*P_drop/(264.17*60)#friction power loss
print "\n friction power loss W_dot_f=%.2f W"%(W_dot_f)#
Example 22.15 page no 316

flow velocity v=0.06 m/s

reynolds no R_e=156991.90

pressure drop P_drop=1.35 Pa

friction power loss W_dot_f=0.34 W

## Example 22.16 Page no 317¶

In [13]:
from __future__ import division

print "\n Example 22.16 page no 317\n\n"
#a power plant employs steam to generate power
z1=0#steam vertical position at inlet,ft
z2=-20#steam vertical position at outlet,ft
v1=120#steam velocity at inlet,ft/s
v2=330#steam velocity at outlet,ft/s
H1=1505.4#steam enthalpy at inlet
H2=940#steam enthalpy at outlet
g_c=32.174#grav .acc
#applying energy equation
W_s=-(z2/778) - v2**2/(2*g_c*778) - H2  +z1 + v1**2/(2*g_c*778) + H1#work extracted from system
print "\n work extracted from the system W_s=%.2f Btu/lb "%(W_s)#
m_dot=450000#mass flow rate ,lb/h
W_dot_s=m_dot*W_s#total power generated by the turbine
print "\n W_dot_s =%.2f Btu/h"%(W_dot_s)##approx calculation in book
W_hp=W_dot_s*3.927e-4#power generated in horsepower hp
print "\n power generated W_hp=%.2f hp"%(W_hp)##approx calculation in book
Example 22.16 page no 317

work extracted from the system W_s=563.54 Btu/lb

W_dot_s =253592134.06 Btu/h

power generated W_hp=99585.63 hp