# Chapter 2: Principles of Momentum Transfer and Overall Balances¶

## Example 2.2-1, Page number 32¶

In :
#Units and Dimensions of Force
# Variable declaration
m = 3.                        #Mass of the body (lbm)
g_fps = 32.174                #Acceleration due to gravity in fps units (ft/s2)
confaclbmtolbf = 32.174       #Conversion factor from lbm to lbf(lbm.ft/lbf.s2)
confaclbmtog = 453.59         #Conversion factor from lbm to gram (g/lbm)
g_mks = 980.665               #Acceleration due to gravity in mks  units (cm/s2)
confaclbtokg = 2.2046         #Conversion factor from lbm to kg (kg/lbm)
g_SI = 9.80665                #Acceleration due to gravity in SI units (m/s2)
# Data SI Units
#Calculation
#(a)
#Fa = m*g_fps*ga
Fa = m*g_fps/confaclbmtolbf
#Fb = m*g_mks*mb
Fb = m*g_mks*confaclbmtog
#Fc = m*g_si*gc
Fc = m*g_SI/confaclbtokg
#Result
print "(a) The force in lbf is ",round(Fa,4),"lbf"
print '(b) The force in dynes is %6.4e'%(Fb),"dyne"
print "(c) The force in Newton is ",round(Fc,2),"N"
print 'The answers are correct because of book uses rounded numbers'

(a) The force in lbf is  3.0 lbf
(b) The force in dynes is 1.3345e+06 dyne
(c) The force in Newton is  13.34 N


## Example 2.2-2, Page number 34¶

In :
#Pressure in Storage Tank
# Variable declaration
Ht = 3.66             #Height of the tank (m)
P  = 1.               #Pressure exerted  (atm)
Hto = 3.05            #Height of tank filled with oil (m)
Htw = 0.61            #Height of tank filled with water (m)
Rho_oil = 917.        #Density of oil (kg/m3)
Rho_H2O = 1000.       #Density of H2O (kg/m3)
g = 9.8               #Gravitational acceleration (m/s2)
# Data
Patm_psia = 14.696        #Atmospheric pressure (psia)
P_pa   = 101325.          #Atmospheric pressure (pa)
Ht_ft  = 12.              #Height of the tank (ft)
Hto_ft  = 10.             #Height of tank filled with oil (ft)
Htw_ft = 2.               #Height of tank filled with water (ft)
confacsitofps = 0.06243   #conversion factor for density(kg/m3) to (lbm/ft3)
confacft2toin2 = 1./144.  #Conversion factor (ft2) to (in2)
# Calculation
#In fps units
#Pressure exerted by oil in fps units
#P1f = Hto_ft*Rho_oil*confacsitofps*confacft2toin2 + P_psia
P1f = (Hto_ft*Rho_oil*confacsitofps*confacft2toin2 + Patm_psia)
#P2f = Htw_ft*Rho_H2O*confacsitofps*confacft2toin2 + P1f
P2f=  Htw_ft*Rho_H2O*confacsitofps*confacft2toin2 + P1f
#Pressure exerted in SI units
#Pressure exerted by oil in SI units
P1si =  Hto*Rho_oil*g + P_pa
#Pressure exerted by water in SI
P2si = Htw*Rho_H2O*g + P1si
Pgage = P2f - Patm_psia
#Result
print "Pressure exerted by oil in fps ",round(P1f,2),"psia"
print 'Pressure exerted by oil in SI %6.3e Pa'%(P1si)
print "Pressure exerted by water in fps ",round(P2f,2),"psia"
print 'Pressure exerted by H2O in SI  %6.3e Pa'%(P2si)
print 'Gage pressure at the bottom of tank %5.2f psig'%(Pgage)
print 'The answers are correct because of book uses rounded numbers'

Pressure exerted by oil in fps  18.67 psia
Pressure exerted by oil in SI 1.287e+05 Pa
Pressure exerted by water in fps  19.54 psia
Pressure exerted by H2O in SI  1.347e+05 Pa
Gage pressure at the bottom of tank  4.84 psig


## Example 2.2-3, Page number 36¶

In :
#Conversion of a pressure to head of a fluid

# Variable declaration
P = 101325.  #Standard atmospheric pressure (kN/m2)
# Data
Rho_H2O = 1000.       #Density of H2O at 4 deg C (kg/m3)
Rho_Hg  = 13.5955     #Density of Hg at 0 deg C (g/cm3)
g = 9.80665           #Gravitational acceleration (m/s2)
Rho_H2O_cgs = 1.      #Density of H2O at 4 deg C (g/cm3)

# Calculation
#Pressure head in terms of H2O
# H_H2O = Pressure/(density of H2O*Gravitational acceleration)
H_H2O = P/(Rho_H2O*g)
#H_Hg = Pressure/(density of Hg*Gravitational acceleration)
H_Hg = H_H2O*Rho_H2O_cgs/Rho_Hg

#Result
print "Pressure head of H2O ",round(H_H2O,2),"m of water at 4°C"
print "Pressure head of Hg ",round(H_Hg,3),"mmHg"

Pressure head of H2O  10.33 m of water at 4°C
Pressure head of Hg  0.76 mmHg


## Example 2.2-4, Page number 37¶

In :
#Pressure Difference in a Manometer
# Variable declaration
R_cgs = 32.7    #Reading of manometer (cm)
# Data
Rho_H2O = 1000.         #Density of H2O  (g/cm3)
Rho_Hg  = 13600.        #Density of Hg  (g/cm3)
g = 9.80665             #Gravitational acceleration (m/s2)
R_SI = 0.327            #Reading of manometer (m)
# Calculation
#Pressure difference in SI units
#delP = R_SI*(Rho_Hg-Rho_H2O)*g
delP = R_SI*(Rho_Hg-Rho_H2O)*g
#Result
print ' The pressure difference in SI units  %10.3e'%(delP),"N/m2"
print 'The answer is correct because of book uses rounded numbers'

 The pressure difference in SI units   4.041e+04 N/m2


## Example 2.3-1, Page number 41¶

In :
#Molecular Transport of a Property at Steady State
# Variable declaration
C1 = 0.0137    #Concentration at point 1 (property/m3)
C2 = 0.0072    #Concentration at point 2 (property/m3)
z1 = 0         #Distance at point 1 (m)
z2 = 0.4       #Distance at point 2 (m)
z = 0.2        #Midpoint   (m)
delta = 0.013  #Diffusivty (m2/s)
# Calculation
#Calculation for part (a)
#Flux = delta*(C1-C2)/(z2-z1)
Flux = delta*(C1-C2)/(z2-z1)
#Calculation for part (c)
C = C1 +  Flux*(z1-z)/delta
#Result
print '(a) The flux is  %6.3e property/s.m2'%(Flux)
print '(c) The concentration at the midpoint is %6.3e property/m3'%(C)

(a) The flux is  2.113e-04 property/s.m2
(c) The concentration at the midpoint is 1.045e-02 property/m3


## Example 2.4-1, Page number 45¶

In :
#Calculation of Shear Stress in a Liquid
# Variable declaration
confaccgstofps = 0.0020875
confaccgstosi = 1/10.
dely = 0.5        #The distance between the two plates (cm)
delVz = 10.       #The velocity change (cm/s)
u = 0.0177        #The viscosity of liquid (g/cm.s)
#Calculation
#Calculation for part (a)
#Calculation of shear stress
#Tyz = u(delVz)/(dely)
Tyz_cgs = u*(delVz)/(dely)
#Calculation of Shear Rate
#Shear Rate = delVz/dely
Vyz_cgs = delVz/dely
#Calculation for part (b)
Tyz_fps = Tyz_cgs*confaccgstofps
#Calculation for part (c)
Tyz_SI = Tyz_cgs*confaccgstosi
#Result
print "(a)Shear stress in cgs is",round(Tyz_cgs,4),"dyne/cm2"
print "   Shear Rate in cgs is",round(Vyz_cgs,4),"1/s"
print '(b)Shear stress in fps %10.3e'%(Tyz_fps),"lbf/ft2"
print "   Shear Rate in fps is",round(Vyz_cgs,4),"1/s"
print "(c)Shear stress in SI is",round(Tyz_SI,4),"N/m2"
print "   Shear Rate in SI is",round(Vyz_cgs,4),"1/s"

(a)Shear stress in cgs is 0.354 dyne/cm2
Shear Rate in cgs is 20.0 1/s
(b)Shear stress in fps  7.390e-04 lbf/ft2
Shear Rate in fps is 20.0 1/s
(c)Shear stress in SI is 0.0354 N/m2
Shear Rate in SI is 20.0 1/s


## Example 2.5-1, Page number 49¶

In :
#Reynolds Number in a pipe
from math import pi

# Variable declaration
F = 10.                 #The flowrate of water (gal/min)
d = 2.067               #The inner diameter of pipe (in)
spg = 0.996             #specific gravity,M/M
d_SI = 0.0525           #Diameter in SI units (m)
mu = 0.8007             #Viscosity, cp

#Calculation English units
F_fps = F*(1./7.481)*(1./60)
d_fps = d/12
A = pi*d_fps**2/4
V = F_fps/A
mu_fps = mu*6.7197e-4    #convert in lbm/ft.s
rho = spg*62.43
Nre_fps = d_fps*V*rho/(mu_fps)

#Calculation SIh units

d_SI = d*2.54e-2
V_SI = V/3.2808
mu_SI = mu*1e-3         #convert in Pa.s
rho = spg*1000
Nre_SI = d_SI*V_SI*rho/(mu_SI)

#Result
print 'The reynolds number in fps units is %6.3e'%(Nre_fps)
print 'The reynolds number in SI units is  %6.3e'%(Nre_SI)
print 'The answers are correct because book uses rounded numbers'

The reynolds number in fps units is 1.903e+04
The reynolds number in SI units is  1.903e+04
The answers are correct because book uses rounded numbers


## Example 2.6-1, Page number 51¶

In :
#Flow of a Crude Oil and Mass Balance

#Variable declaration
rho = 892.              #Density of Crude Oil kh/m3
F = 1.388e-3            #Total Volumetric flow rate m3/s
d1 = 2.067              #Diamete of pipe 1 in inches
d3 = 1.610              #Diamete of pipe 3 in inches

#Calculations
intom = 0.0254                 #conversion factor for in to meter
A1 = pi*(d1*intom)**2/4        #Area of pipe 1
A3 = pi*(d3*intom)**2/4        #Area of pipe 3
m1 = F*rho                     #Mass flow rate of Crude Oil through pipe 1
m3 = m1/2                      #Mass flow rate of Crude Oil through pipe 3
v1 = m1/(rho*A1)               #velocity of Crude Oil through pipe 1
v3 = m3/(rho*A3)               #velocity of Crude Oil through pipe 3
G1 = m1/A1                     #Mass flux of cride oil through pipe 1

#Result
print "(a) Mass flow rate of Crude Oil through pipe 1:",round(m1,3), "kg/s"
print "    Mass flow rate through pipe 3:",round(m3,3), "kg/s"
print "(b) Velocity of Crude Oil through pipe 1:",round(v1,3), "m/s"
print "    Velocity of Crude Oil through pipe 3:",round(v3,3), "m/s"
print "(c) Mass velocity of Crude Oil through pipe 1:",round(G1), "m/s"

(a) Mass flow rate of Crude Oil through pipe 1: 1.238 kg/s
Mass flow rate through pipe 3: 0.619 kg/s
(b) Velocity of Crude Oil through pipe 1: 0.641 m/s
Velocity of Crude Oil through pipe 3: 0.528 m/s
(c) Mass velocity of Crude Oil through pipe 1: 572.0 m/s


## Example 2.7-1, Page number 61¶

In :
#Energy Balance on Steam boiler

#Variable Declaration
Twin = 18.33         #Inlet temperature of water to boiler, degC
Pwin = 137.9         #Inlet pressure of water to boiler, kPa
Vwin = 1.52          #Inlet temperature of water to boiler, m/s
Hwin = 0.            #Water Inlet height, m
Hsout = 15.2         #Height of steam outlet above water inlet, m
Psout = 137.9        #Outlet pressure of steam to boiler, kPa
Tsout = 148.9        #Inlet temperature of water to boiler, degC
Vsout = 9.14         #Inlet temperature of water to boiler, m/s
H1 = 76.97           #Entalpy of Incomming water to boiler, kJ/kg
H2 = 2771.4          #Entalpy of steam leaving boiler, kJ/kg
g = 9.80665          #Gravitational acceleration, m/s2
#Calcualtion

# Heat Added = Del PE + Del KE + Del Enthalpy
Q = (Hsout-Hwin)*g + (Vsout**2-Vwin**2)/2 + (H2-H1)*1000

#Result
print 'Heat Added per kg of water %5.4e'%(Q),"J/kg"
print 'The answers are correct because of book uses rounded of numbers'

Heat Added per kg of water 2.6946e+06 J/kg


## Example 2.7-2, Page number 62¶

In :
#Energy Balance on a flow system with a Pump

#Variable Declaration
F = 0.567            #Water pumped from the tank (m3/min.)
E1 = 7450.           #Energy supplied by the pump (W)
E2 = 1408000         #Energy lost in heat exchanger (W)
Rho = 968.5          #Density of water at 85 deg C (kg/m3)
Z1 = 0.              #Height of first second tank (m)
Z2 = 20.             #Height of second tank (m)
H1 = 355900          #Enthalpy of steam (kJ/kg)
delV = 0.            #The kintic energy changes
g = 9.80665          #Gravitational acceleration (m/s2)

#Calcualtion
m1 = F*Rho/60        #Water pumped from the tank (kg/s)
Ws = -E1/m1
Q = -E2/m1

# H2 - H1 + 1/2*delV + g(Z2 - Z1) = Q - Ws
# H2 = Q - Ws + H1 - 1/2*delV - g(Z2 - Z1)
H2 = Q - Ws + H1 - 0.5*delV - g*(Z2 - Z1)

#Result
print 'Heat Added per kg of water %5.3e'%(H2), "J/kg"
print "The value of heat added corresponds to  48.41 °C"

Heat Added per kg of water 2.027e+05 J/kg
The value of heat added corresponds to  48.41 °C


## Example 2.7-3, Page number 62¶

In :
#Energy Balance in flow Calorimeter
#Variable Declaration
F = 0.3964           #Flowrate of water entering the calorimeter  (kg/min)
E1 = 19.630          #Heat added to the water for complete vaporization  (kW)
Z1 = 0.              #Height of first second tank (m)
Z2 = 0.              #Height of second tank (m)
delV = 0.            #The kintic energy changes
g = 9.80665          #Gravitational acceleration  (m/s2)
Ws = 0.              #Shaft work
H1 = 0               #Initial Enthalpy
P1 = 150.            #Pressure at point 2 (kPa)
#Calcualtion
M1 = F/60            #Flowrate of water entering the calorimeter  (kg/s)
# Q = E1/M1
Q = E1/M1
# H2 - H1 + 0.5*delV + g(Z2 - Z1) = Q - Ws
# H2 = Q - Ws + H1 - 1/2*delV - g(Z2 - Z1)
H2 = Q - Ws + H1 - 0.5*delV - g*(Z2 - Z1)
#Result
print "Heat Added per kg of water",round(H2), "J/kg"
print "The corresponding value of enthalpy from steam table is 2972.7 kJ/kg"

Heat Added per kg of water 2971.0 J/kg
The corresponding value of enthalpy from steam table is 2972.7 kJ/kg


## Example 2.7-4, Page number 64¶

In :
#Mechanical-Energy Balance on Pumping System

#Variable Declaration
P1 = 68900.            #The entrance pressure of fluid (N/m2)
E = 155.4              #Energy supplied to the fluid (J/kg)
P2 = 137800.           #The exit pressure of fluid (N/m2)
Nre = 4000             #Reynolds number
Rho_H2O = 998.         #Density of water (kg/m3)
z1 = 0                 #Height of entrance pipe (m)
z2 = 3.05              #Height of exit pipe (m)
alpha = 1              #For turbulent flow value of alpha
Ws = -155.4            #Mechanical energy added to the fluid (J/kg)
g = 9.806              #Gravitational accleration (m/s2)
#Calcualtion
#(V1 - V2)/2*alpha = 0
#F = -Ws + 0 + g*(z1-z2) + (P1 - P2)/Rho_H2O
F = -Ws + 0 + g*(z1-z2) + (P1 - P2)/Rho_H2O
#Result
print "The frictional losses calculated are ",round(F,1), "J/kg"

The frictional losses calculated are  56.5 J/kg


## Example 2.7-5, Page number 65¶

In :
#Pump Horsepower in Flow System

#Variable Declaration
F = 69.1              #Volume of liquid drawn by the pump (gal/min)
F_fps = F/(60*7.481)  #Volume of liquid drawn bby the pump (ft3/s)
Rho = 114.8           #Density of liquid (lbm/ft3)
A1 = 0.05134          #Cross sectional area of suction  pipe (in2)
A2 = 0.0233           #Cross sectional area of discharge pipe (in2)
z1 = 0.               #Height of suction line (m)
z2 = 50.              #Height of discharge pipe (m)
f = 10.0              #Frictional losses (ft.lbf/lbm)
n = 0.65              #Efficiency of the pump
g = 32.174            #Gravitational accleration (m/s2)
gc = 32.174           #Conversion factor for mks to fps
v1 = 0.               #Velocity of liquid in storage feed tank (m/s)
alpha = 1.            #Since the flow is turbulent
fc = 550.             #Conversion factor from fps to horse power
z3 = 0.
z4 = 0.
ga = 144.             #Conversion factor for area from ft2 to in2

#Calcualtion
#v2 = flowrate/cross sectional area
v2 = F_fps/A2
#(P1-P2)/Rho = 0
#Ws = (z1-z2)*g/gc + (v1*v1 - v2*v2)/(2*gc) + 0 + f
Ws = (z1-z2)*g/gc + (v1*v1 - v2*v2)/(2*gc) + 0 - f
Wp = -Ws/n
m = F_fps*Rho
P = m*Wp/fc
v3 = F_fps/A1
v4 = F_fps/A2
#delP = P4 - P3
delP = Rho*((z4-z3)*g/gc +(v3*v3 - v4*v4)/(2*gc) - Ws - 0. )/ga
#Result
print "Power required by the pump ",round(P,0), "hp"
print "Pressure developed is ",round(delP,0),"lbf/in2"

Power required by the pump  3.0 hp
Pressure developed is  48.0 lbf/in2


## Example 2.8-2, Page number 73¶

In :
#Momentum Balance for Horizontal Nozzle
from math import pi
#Variable Declaration
F = 0.03154             #Flowrate of water (m3/s)
d1 = 0.0635             #Dia of pipe at section 1 (m)
d2 = 0.0286             #Dia of pipe at section 2 (m)
Rho_H2O = 1000.         #Density of water (kg/m3)
P2 = 0.                 #Pressure (N/m2)
#Calcualtion
m = F*Rho_H2O
A1 = pi*d1**2/4
A2 = pi*d2**2/4
v1 = F/A1
v2 = F/A2
P1 = Rho_H2O*((v2**2 - v1**2)/2 + P2/Rho_H2O)
Rx = m*(v2-v1) + P2*A2 - P1*A1
#Result
print "The resultant force on the nozzle is",round(Rx,1), "N"
print 'The answer is correct because of book uses rounded numbers'

The resultant force on the nozzle is -2425.3 N
The answer is correct because of book uses rounded numbers


## Example 2.8-5, Page number 76¶

In :
#Force of Free Jet on a Curved,Fixed Vane
from math import pi

#Variable Declaration
V = 30.5            #Velocity of jet stream of water (m/s)
Rho_H2O = 1000.     #Density of water (kg/m3)
d = 2.54e-2         #Diameter of pipe, m
alpha = 60          #Angle made by exit stream, deg

A = pi*d**2/4
m = V*A*Rho_H2O
#Calculation
#Rx is the x component of the force
#Rx = m*V*(cos60 - 1)
Rx = m*V*(cos(alpha*pi/180)-1)
#Ry is the y component of the force
#Ry = m*V*(sin60)
Ry = m*V*sin(alpha*pi/180)
#Result
print "The force in x direction is ",round(-Rx,1),"N"
print "The force in y direction is ",round(-Ry,1),"N"
print 'The answers are correct because of book uses rounded numbers'

The force in x direction is  235.7 N
The force in y direction is  -408.2 N


## Example 2.9-1, Page number 82¶

In :
#Falling Film Velocity and Thickness
#Variable Declaration
d = 0.0017          #Thickness of the film (m)
Rho_oil = 820       #Density of oil (kg/m3)
mu = 0.20            #Viscosity of oil (Pa.s)
g = 9.806           #Gravitational acceleration (m/s2)

#Calculation
Gama = Rho_oil*Rho_oil*d**3*g/(3*mu)
Nre = 4*Gama/mu
Vz = Rho_oil*g*d**2/(3*mu)

#Result
print "The mass flowrate per unit width of the wall",round(Gama,6),"kg/s.m"
print "Reynolds number is",round(Nre,3)
print "The average velocity is",round(Vz,6),"m/s"

The mass flowrate of the wall 0.05399 kg/s.m
Reynolds number is 1.08
The average velocity is 0.03873 m/s


## Example 2.10-1, Page number 85¶

In :
#Metering of Small Liquid flows
from math import pi
#Variable Declaration

d = 0.00222          #Diameter of capillary (m)
l = 0.317            #Length of capillary (m)
Rho = 875            #Density of liquid (kg/m3)
mu = 0.00113         #Viscosity of liquid (pa.s)
h = 0.0655           #Heightof water (m)
Rho_H2O = 996        #Density of water (kg/m3)
g = 9.80665          #Gravitational acceleration (m/s2)

#Calculation
delP = h*Rho_H2O*g
#v = delP*d*d/(32*mu*l)
#Flow assumed to be laminar
v = delP*d**2/(32*mu*l)
F = v*pi*d**2/(4)
#Checking whether the flow is laminar
Nre = d*v*Rho/(mu)
#Result
print 'The pressure drop over capillary is %3.0f N/m2'%(delP)
print 'The velocity through capillary is %4.3f m/s'%(v)
print 'The mass flowrate through the capillary is %5.3e m3/s'%(F)
print 'Reynolds number is %3.0f, Hence flow is laminar'%(Nre)
print 'The answers are correct because of book uses rounded numbers'

The pressure drop over capillary is 640 N/m2
The velocity through capillary is 0.275 m/s
The mass flowrate through the capillary is 1.065e-06 m3/s
Reynolds number is 473, Hence flow is laminar
The answers are correct because of book uses rounded numbers


## Example 2.10-2, Page number 87¶

In :
#Use of Friction Factor in Laminar Flow

#Variable Declaration
d = 0.00222          #Diameter of capillary (m)
l = 0.317            #Length of capillary (m)
V = 0.275            #Velocity of liquid (m/s)
Rho = 875            #Density of liquid (kg/m3)
mu = 0.00113         #Viscosity of liquid (pa.s)
h = 0.0655           #Heightof water (m)
Rho_H2O = 996        #Density of water (kg/m3)
g = 9.80665          #Gravitational acceleration (m/s2)
#Calculation
Nre = d*V*Rho/(mu)
#f = 16/Nre
f = 16/Nre
#delp = 4*f*Rho*l*V*V/(2*D)
delp = 4*f*Rho*l*V**2/(2*d)
#Result
print 'Reynolds number is %3.0f, Hence flow is laminar'%(Nre)
print 'friction factor is %0.4f'%(f)
print 'The pressure drop through capillary is %3.0f N/m2'%(delp)

Reynolds number is 473, Hence flow is laminar
friction factor is 0.0338
The pressure drop through capillary is 640 N/m2


## Example 2.10-3, Page number 89¶

In :
#Use of Friction Factor in Turbulent Flow

#Variable Declaration
d = 0.0525           #Diameter of capillary (m)
l = 36.6             #Length of capillary (m)
V = 4.57             #Velocity of liquid (m/s)
Rho = 801.           #Density of liquid (kg/m3)
mu = 4.46            #Viscosity of liquid (cp)
g = 9.806            #Gravitational acceleration (m/s2)
e = 0.000046         #Equvivalent Roughness (m)
f = 0.0060           #Friction factor
#Calculation
mu = mu*1e-3         #Convert Viscosity to  (kg/m.s)
Nre = d*V*Rho/(mu)
#delp = 4*f*Rho*l*V*V/(2*D)
delp = 4*f*Rho*l*V**2/(2*d)
#F = 4*f*l*V*V/(2*d)
F = 4*f*l*V**2/(2*d)
#Result
print 'Reynolds number is %3.0f, Hence flow is turbulent'%(Nre)
print 'friction factor from chart is %0.4f'%(f)
print "The friction loss is %3.1f J/kg" %(F)
print 'The answers are correct because of book uses rounded numbers, and varified using calculator'

Reynolds number is 43090, Hence flow is turbulent
friction factor from chart is 0.0060
The friction loss is 174.7 J/kg
The answers are correct because of book uses rounded numbers, and varified using calculator


## Example 2.10-4, Page number 90¶

In :
#Trial and Error Solution to Calculate Pipe Diameter
from scipy.optimize import root
from sympy import *
from math import pi, log

#Variable Declaration
l = 305.              #Length of steel pipe (m)
mu = 1.55e-3          #Viscosity of fluid (kg/m.s)
Q = 150.0             #Rate of flow of fluid (gal/min)
g = 9.80665           #Gravitational acceleration, m/s2
ep = 4.6e-5           #Roughness of commercial steel pipe, m
rho = 1000.           #Density of water, kg/m3
z = 6.1               #Head available, m

#Calculation
Qsi = Q*(1./7.481)*(1./60)*0.02831      #Multipliers in equation are conversion factors for conversion to SI Units
Ff = z*g
er = 12.
x = Symbol('x')
D = 0.089
while er >= 0.01:
epbyd = ep/D
A = pi*D**2/4.
u = Qsi/A
Re = D*u*rho/mu
sol = solve(1/sqrt(x)+4*(log(epbyd/3.7,10)+1.255/(Re*sqrt(x))),x)
f = sol
d = 4*f*l*u**2/(2*Ff)
er = abs((D-d)/D)
if er > 0.01:
print 'Error %6.5f and diameter calculated %6.5f'%(er, d)
D = float(raw_input("Not Converged...:Enter the value of guess diameter  "))
#Result
print '\nConverged...Diameter of the pipe is %4.3f m or %4.3f in'%(D,D*12./0.3048)
print 'The value calculated is accurate than answer in book\nbecause book reads the friction factor from chart whereas\nequations for friction factor are used in this'

Error 0.11457 and diameter calculated 0.09920
Not Converged...:Enter the value of guess diameter  0.092
Error 0.06269 and diameter calculated 0.08623
Not Converged...:Enter the value of guess diameter  0.093
Error 0.11416 and diameter calculated 0.08238
Not Converged...:Enter the value of guess diameter  0.094
Error 0.16230 and diameter calculated 0.07874
Not Converged...:Enter the value of guess diameter  0.094
Error 0.16230 and diameter calculated 0.07874
Not Converged...:Enter the value of guess diameter  0.098
Error 0.32616 and diameter calculated 0.06604
Not Converged...:Enter the value of guess diameter  0.099
Error 0.36096 and diameter calculated 0.06327
Not Converged...:Enter the value of guess diameter  0.1
Error 0.39363 and diameter calculated 0.06064
Not Converged...:Enter the value of guess diameter  0.14
Error 0.89524 and diameter calculated 0.01467
Not Converged...:Enter the value of guess diameter  0.15
Error 0.92689 and diameter calculated 0.01097
Not Converged...:Enter the value of guess diameter  0.16
Error 0.94777 and diameter calculated 0.00836
Not Converged...:Enter the value of guess diameter  0.091

Converged...Diameter of the pipe is 0.091 m or 3.583 in
The value calculated is accurate than answer in book
because book reads the friction factor from chart whereas
equations for friction factor are used in this


## Example 2.10-5, Page number 91¶

In :
#Flow of Gas in Line and Pressure Drop
#Variable Declaration
d = 0.010             #Inside diameter of tube (m)
G = 9.                #Rate of feed (kg/s.m2)
l = 200.              #Length of Tube (m)
mu = 0.0000177        #Viscosity of liquid (Pa.s)
R = 8314.3            #Gas constant (J/kg.mol.K)
P1 = 202650.          #Pressure at entrance (Pa)
f = 0.0090            #Friction factor
T1 = 25.              #Temperature at inlet (degC)
T2 = 298.15           #Temperature at outlet (degC)
M = 28.02             #Molecular weight of 1 kmol N2  (kmol)
#Calculation
Nre = d*G/mu
f = 0.079*Nre**(-0.25)   #for smooth pipes
P2 = sqrt(P1**(2) - 4*f*l*G**2*R*T2/(d*M))

#Result
print "for Reynolds Number ", round(Nre,2), "friction factor for smooth pipe is", round(f,6)
print 'The pressure at outlet %6.4e Pa, '%(P2)
print 'The answer is correct than the book because of book uses rounded numbers'

for Reynolds Number  5084.75 friction factor for smooth pipe is 0.009355
The pressure at outlet 1.8895e+05 Pa,
The answer is correct than the book because of book uses rounded numbers


## Example 2.10-6, Page number 94¶

In :
#Friction Losses and mechanical energy balance
#Variable Declaration
Q = 0.223                 #Desired flowrate of water (ft3/s)
Rho_fps = 60.52           #Density of water in fps units (lbm/m3)
u_fps = 0.000233          #Viscosity of water in fps units (lbm/ft.s)
D3 = 0.3353               #Diameter of the third pipe (ft)
D4 = 0.1722               #Diameter of the third pipe (ft)
A3 = 0.0884               #Area of third pipe (ft2)
A4 = 0.0233               #Area of third pipe (ft2)
g_fps = 32.174            #Gravitational acceleration in fps units (ft/s2)
e = 0.00015               #Roughness factor (ft)
f2 = 0.0047               #Fanning friction factor for 4-in pipe
f5 = 0.0048               #Fanning friction factor for 2-in pipe
delL_4 = 20.              #Length of 4-in pipe (ft)
delL_2 = 185.             #Length of 2-in pipe (ft)
delP = 0.                 #Since pressure at both end is atmspheric pressure
Kf = 0.75
alpha = 1.
Ws = 0.                   #Shaft work (j)
#Calculation
V1 = 0.
V3 = Q/A3
V4 = Q/A4
#
#(1) Contraction losses at tank exit
#Kc = 0.55*(1-A3/A1)
Kc1 = 0.55
hc1 = Kc1*V3**2/(2*g_fps)
#(2) Friction in 4-in pipe
Nre2 = D3*V3*Rho_fps/(u_fps)
F2 = 4*f2*delL_4*V3**2/(D3*2*g_fps)
#(3) Friction in 4-in elbow
hf = Kf*V3**2/(2*g_fps)
#(4) Contraction loss from 4-in to 2-in pipe
Kc4 = 0.55*(1-A4/A3)
hc4 = Kc4*V4**2/(2*g_fps)
#(5)Friction in 2-in pipe
Nre5 = D4*V4*Rho_fps/u_fps
F5 = 4*f5*delL_2*V4**2/(D4*2*g_fps)
#(6) Friction in 2-in elbows
hc6 = 2*Kf*V4**2/(2*g_fps)
#F = hc1 + F2 + hf + hc4 + F5 + hc6
F = hc1 + F2 + hf + hc4 + F5 + hc6
#delH*g_fps = delP/Rho_fps + (V4**2 - V1**2)/(2*alpha*g_fps) + Ws + F
#delH = (delP/Rho_fps + (V4**2 - V1**2)/(2*alpha*g_fps) + Ws + F)/g_fps
delH = (delP/Rho_fps + (V4**2 - V1**2)/(2*alpha*g_fps) + Ws + F)
#Result
print "The height of water level above the discharge outlet ",round(delH,2),"ft"
print 'The answer is correct because of book uses rounded numbers'

The height of water level above the discharge outlet  33.74 ft
The answer is correct because of book uses rounded numbers


## Example 2.10-7, Page number 97¶

In :
#Friction Losses with pump in mechanical energy balance
#Variable Declaration
Q = 0.005                 #Desired flowrate of water (m3/s)
Rho = 998.2               #Density of water in fps units (kg/m3)
u = 0.001005              #Viscosity of water in fps units (Pa/s)
D1 = 0.1023               #Diameter of the third pipe (m)
A1 = 0.008291             #Area of pipe (m2)
g = 9.806                 #Gravitational acceleration in fps units (ft/s2)
e = 0.000046              #Roughness factor (ft)
f2 = 0.0051               #Fanning friction factor for 4-in pipe
z1 = 0.                   #Height of storage tank (m)
z2 = 15.                  #Height of elevated tank (m)
delL = 170.               #Length of 4-in pipe (ft)
delL_2 = 185.             #Length of 2-in pipe (ft)
delP = 0.                 #Since pressure at both end is atmospheric pressure
delV2 = 0.
Kf = 0.75
Kex = 1.
alpha = 1.
n = .65                   #Efficiency of the pump
#Calculation
V1 = Q/A1
Nre = D1*V1*Rho/(u)

#(1) Contraction losses at tank exit
#Kc = 0.55*(1-A3/A1)
Kc1 = 0.55
hc1 = Kc1*V1**2/(2*alpha)
#(2) Friction in straight pipe

F2 = 4*f2*delL*V1**2/(D1*2)
#(3) Friction in the two elbows
hf = 2*V1**2/(2)
#(4) Expansion loss at tank entrance
hc4 = Kex*V1**2/(2)
#F = hc1 + F2 + hf + hc4
F = hc1 + F2 + hf + hc4
#1*(delV2)/(2*alpha) + g*(z2 - z1) +delP/Rho + F + Ws = 0.
Ws = -(1*(delV2)/(2*alpha) + g*(z2 - z1) +delP/Rho + F)
#m = Q*Rho
m = Q*Rho
#Ws = -n*Wp
Wp = Ws/-n
#P = m*Wp
P = m*Wp/1000
#Result
print "The power required by the pump is ",round(P,3),"kW"

The power required by the pump is  1.182 kW


## Example 2.10-8, Page number 100¶

In :
#Entry Length for a Fluid in a Pipe
#Variable Declaration
d = 0.010                #Diameter of tube (m)
V = 0.10                 #Velocity of water (m/s)
mu = 0.001005             #Viscosity of water (Pa.s)
Rho = 998.2              #Density of water (kg/m3)
#Calculation
Nre = d*V*Rho/mu
#For laminar flow
#Le = 0.0575*Nre

Le_l = 0.0575*Nre*d
Le_t = 50*d
#Result
print "Reynolds Number is",round(Nre,1)
print "For laminar flow the entry length is ",round(Le_l,3),"m"
print "For turbulent flow the entry length is ",round(Le_t,4),"m"

Reynolds Number is 993.2
For laminar flow the entry length is  0.571 m
For turbulent flow the entry length is  0.5 m


## Example 2.11-1, Page number 102¶

In :
#Compressible Flow of a gas in a pipe line
from math import log, pi
#Variable Declaration
d = 1.016                #Diameter of pipe (m)
G = 2.077                #The molar flow rate of the gas (kg mol/s)
mu = 1.04e-5             #Viscosity of methane at 288.8 k (Pa.s)
p2 = 170.3e3             #The pressure at the outlet (Pa)
L = 1.609e5              #The length of the pipe (m)
R = 8314.34              #Gas constant (N.m/kg.mol.K)
T = 288.8                #Temperature of methane gas (K)
M = 16.                  #Molecular weight of methane
epsilon = 4.6e-5         #Pipe roughness

#Calculation
A = pi*d**2/4
Gm = G*M/A
Nre = d*Gm/mu
f = 0.0027        #from friction factor chart
#P1**2 - P2**2 = (4*f*L*(G_mass**2)*R*T)/(D*M) + (2*(G**2)*R*T*log(p1/P2)/16)
#P1 = sqrt(P2**2 + (4*f*L*(G_mass**2)*R*T)/(D*M) + (2*(G**2)*R*T*log(p1/P2)/16))
er = 1.2
p1 = 500.e3
while er>0.001:
p1c = sqrt(p2**2 + 4*f*L*Gm**2*R*T/(d*M) + (2*Gm**2*R*T*log(p1/p2)/M))
er = abs(p1-p1c)/p1c
p1 = p1c
#Result
print 'The final pressure is %10.4e Pa.'%(p1)
print 'The answers are correct because of book uses rounded of numbers'

The final pressure is 6.7895e+05 Pa.
The answers are correct because of book uses rounded of numbers


## Example 2.11-2, Page number 103¶

In [ ]:
#Maximum Flow for Compressible Flow of a Gas
#Variable Declaration
R = 8314.              #Gas constant ()
T = 288.8              #Temperature of fluid (K)
M = 16.
P2 = 170300.           #Pressure at discharge tube (Pa)
G = 41.
#Calculation
Vmax = sqrt(R*T/M)
V2 = R*T*G/(P2*M)
#Result
print "The maximum velocity in the tube  is ",round(Vmax,1),"m/s"
print "The actual velocity is ",round(V2,2),"m/s"

The maximum velocity in the tube  is  387.4 m/s
The actual velocity is  36.13 m/s