# Chapter 3: Principles of Momentum Transfer and Applications¶

## Example 3.1-1, Page number 117¶

In :
#Force on submerged sphere
from math import pi

# Variable declaration
Rho_air = 1.137             #Density of air at 37.8 degC (kg/m3)
u = 0.000019                #Viscosity of air (Pa.s)
Dp = 0.042                  #Diameter of the sphere (m)
V = 23.0                    #velocity of the sphere (m/s)

# Data SI Units
#Calculation
Nre = Dp*Rho_air*V/(u)
Cd = 0.47                   #Drag coefficient from fig 3.1-2
Ap = pi*Dp**2/4
Fd = Cd*V**2*Rho_air*Ap/2
#Result
print 'Reynolds number %4.3e'%Nre
print "The drag coefficient of the sphere is",Cd
print "The force on the cylinder is", round(Fd,4),"N"

Reynolds number 5.781e+04
The drag coefficient of the sphere is 0.47
The force on the cylinder is 0.1958 N


## Example 3.1-2, Page number 117¶

In :
#Force on a cylinder in a tunnel
# Variable declaration
V = 1.                            #Velocity of water in the tunnel (m/s)
d = 0.09                          #Diameter of cylinder (m)
Rho_H2O = 997.2                   #Density of water (kg/m3)
L = 1.                            #Length of the tunnel (m)
mu = 0.0009142                     #Viscosity of water (Pa.s)
T = 24.                           #Temerature of water (°C)

# Data SI Units
#Calculation
Nre = d*Rho_H2O*V/(mu)
Cd = 1.4                          #Drag Coefficient for cylinder from fig 3.1-2
Ap = L*d
Fd = Cd*V**2*Rho_H2O*Ap/2
#Result
print 'Reynolds number %4.3e'%Nre
print "The drag coefficient of the cylinder is",Cd
print "The force on the sphere is ",round(Fd,2),"N"

Reynolds number 9.817e+04
The drag coefficient of the cylinder is 1.4
The force on the sphere is  62.82 N


## Example 3.1-3, Page number 119¶

In :
#Surface area in Packed Bed of Cylinders
# Variable declaration
D = 0.02                   #Diamter of the cylinder (m)
h = 0.02                   #Length of the cylinder (m)
d_pb = 962.                #Density of packed bed (kg/m3)
d_sc = 1600.               #Density of solid cylinders (kg/m3)
m_pb = 962.                #Mass of the packed bed (kg)
v_pb = 1.                  #Volume of packed bed (m3)
v_t = 1.                   #Total volume of the bed (m3)
# Calculation
v_sc = m_pb/d_sc
#(a) Calculate void fraction
#e = Volume of voids in bed/total volume of bed
e = (v_pb - v_sc)/v_t
#(b)Calculate the effective Diameter of the particle
#Dp = 6/(6/D)
Dp = 6/(6/D)
#(c) Calculate the value of "a"
# a = 6*(1-e)/Dp
a = 6*(1-e)/Dp
#Result

print "(a) The void fraction of bed is",round(e,3)
print "(b) The effective diameter of the particle is",round(Dp,4),"m"
print "(c) The calculated value of 'a' is ",round(a,2), "1/m"
print 'The answers are correct because of book uses rounded numbers'

(a) The void fraction of bed is 0.399
(b) The effective diameter of the particle is 0.02 m
(c) The calculated value of 'a' is  180.38 1/m
The answers are correct because of book uses rounded numbers


## Example 3.1-4, Page number 121¶

In :
#Pressure drop and flow of gases through packed beds
from math import pi
from scipy.optimize import root
# Variable declaration
d = 0.0127            #Diameter of the pipe (m)
e = 0.38              #Void fraction
d_bed = 0.61          #Bed diameter (m)
h = 2.44              #Height of bed (m)
p1 = 1.1              #Pressure P1 (atm)
T = 311.              #Temperature of water (K)
R = 8314.34           #Gas constant
M = 28.97             #Molecular weight of air
L = 2.44              #Length of the bed (m)
Mdot = 0.358          #Mass rate of air, kg/s

# Data
mu = 0.000019          #Viscosity of water (Pa.s)
# Calculation
A = pi*d_bed**2/4
G = Mdot/A
Nre = d*G/((1-e)*mu)
#Rho_av = M*Pav/(R*T)
p1 = p1*101325.       #convert to pascal
er = 0.1

f = lambda dp:dp*(M*(p1+0.5*dp)/(R*T))*d*e**3/(G**2*L*(1-e))-(150./Nre + 1.75)
sol = root(f,0.049e5)
dp = sol.x

#Result
print "The calclated Reynolds number is",round(Nre,0)
print 'The calculated pressure drop is %6.5e Pa'%(dp)
print 'The answer is difference than book because of method of solution, and is more correct.\nBook assumes a pressure drop and calculates density at average pressure'

The calclated Reynolds number is 1321.0
The calculated pressure drop is 4.75988e+03 Pa
The answer is differencethan book because of method of solution and is more correct.
Book assumes a pressure drop and calculates density at average pressure


## Example 3.1-5, Page number 122¶

In :
#Mean Diameter for a Particle
# Variable declaration
# Data
x1 = 0.25
x2 = 0.40
x3 = 0.35
Dp1 = 25.
Dp2 = 50.
Dp3 = 75.
fi = 0.68

# Calculation
#Dpm = 1/(x1/(fi*Dp1) + x2/(fi*Dp2) + x3/(x3/(fi*Dp3))
Dpm = 1/((x1/(fi*Dp1)) + (x2/(fi*Dp2)) + (x3/(fi*Dp3)))
#Result
print "The calculated effective mean diameter is ",round(Dpm,4),"mm"

The calculated effective mean diameter is  30.0 mm


## Example 3.1-6, Page number 125¶

In :
#Minimum Velocity for Fluidization
from scipy.optimize import root

# Variable declaration
d = 0.12                      #Size of the particle (mm)
fi = 0.88                     #Shape of the particle
P = 2.                        #Pressure of the fluidized bed (atm)
v = 0.42                      #Voidage
A = 0.30                      #Cross section of the bed (m2)
M = 300.                      #Weight of the solid (kg)
ep1 = 0.
ep_mf = 0.42                  #voidage at minimum fluidisation
#Data
mua = 1.85e-5                 #Viscosity of air (Pa.s)
rhoa = 2.374                  #Density of air at 2 atm, kg/m3
p = 2.0165e5                  #Pressure, Pa
rhop = 1000                   #Density of particle, kg/m3
Dp = 0.00012                  #Particle diameter, m
g = 9.80665                   #m/s2
# Calculation

#Part A
V = M/rhop
L1 = V/A
Lmf = L1*(1. - ep1)/(1 - ep_mf)

#Part B
delp = Lmf*(1.-ep_mf)*(rhop-rhoa)*g

#Part C
#Nremf = d*umf*rhoa/mua
Nrea = Dp*rhoa/mua

f = lambda u: 1.75*(Nrea*u)**2/(fi*ep_mf**3)+150*(1.-ep_mf)*(Nrea*u)/(fi**2*ep_mf**3)-Dp**3*rhoa*(rhop-rhoa)*g/mua**2
sol = root(f,0.001)
umf = sol.x
Nremf = Dp*rhoa*umf/mua

f1 = lambda u: Dp*rhoa*u/mua - sqrt(33.7**2 + 0.0408*Dp**3*rhoa*(rhop-rhoa)*g/mua**2) + 33.7
sol = root(f1,umf)
umfn = sol.x
#Result
print "(a) The minimum height of the fluidized bed is",round(Lmf,3),"m"
print '(b) Pressure drop under conditions of minimum fluidization is %5.4e'%(delp),"Pa"
print "(c) Minimum fluidozation velocity",round(umf,6), "m/s"
print "    Reynold number at Minimum fluidization is",round(Nremf,5)
print '(d) Minimum fluidozation velocity using simplified relation %7.6f'%umfn
print 'The answer is correct because of book uses rounded numbers'

(a) The minimum height of the fluidized bed is 1.724 m
(b) Pressure drop under conditions of minimum fluidization is 9.7834e+03 Pa
(c) Minimum fluidozation velocity 0.005015 m/s
Reynold number at Minimum fluidization is 0.07723
(d) Minimum fluidozation velocity using simplified relation 0.004605


## Example 3.1-7, Page number 126¶

In :
#Expansion of Fluidized Bed
from scipy.optimize import root

# Variable declaration
Nre_mf = 0.07723            #From previous problem
vmf = 0.005015
epmf = 0.42

#Calculation
f = lambda K:vmf-K*epmf**3/(1-epmf)
sol = root(f,0.01)
K = sol.x
vop = 3*vmf

f1 = lambda ep:vop-K*ep**3/(1-ep)
sol = root(f1,0.1)
ep1 = sol.x
#Result
print 'Operating velocity %6.5f m/s'%vop
print 'Voidage of bed at operating velocity %4.3f m/s'%(ep1)

Operating velocity 0.01505 m/s
Voidage of bed at operating velocity 0.555 m/s


## Example 3.2-1, Page number 128¶

In :
#Flow Measurement Using a Pitot Tube
from math import pi

# Variable declaration
d = 600.          #Diameter of circular duct (m)
dh1 = 10.7        #Pitot tube reading (mm)
dh2 = 205.        #Pitot tube reading (mm)
Cp = 0.98         #Pitot tube coefficient
mu = 0.0000203    #Viscosity of air (Pa.s)
rhoa = 1.043      #Density of air (kg/m3)
delh = 0.205      #Head loss in height of water (m)
rhow = 1000.      #Density of water (kg/m3)
g = 9.80665       #Gravitational accleration (m/s2)

#Calculation
d = 600./1e3
dp1 = (dh2/1e3)*(rhow - rhoa)*g
p1abs = 101325 + dp1
rhoac = rhoa*p1abs/101325
dp = (dh1/1e3)*g*(rhow - rhoa)
v = Cp*sqrt(2*dp/rhoa)
Nre = d*v*rhoa/mu
A = 3.14*d**2/4
vav = 0.85*v
Q = A*vav

#Result
print 'Total absolute pressure %7.1f Pa'%p1abs
print 'Corrected density of air %5.4f kg/m3'%rhoac
print 'Velocity at centre %5.3f m/s'%v
print 'Average velocity through pipe %3.2f m/s'%vav
print 'Reynolds number %5.3f'%Nre
print "The Volumetric flow rate is",round(Q,4),"m/s"
print 'The answers are correct because of book uses rounded numbers'

Total absolute pressure 103333.3 Pa
Corrected density of air 1.0637 kg/m3
Velocity at centre 13.894 m/s
Average velocity through pipe 11.81 m/s
Reynolds number 428315.769
The Volumetric flow rate is 3.3375 m/s


## Example 3.2-2, Page number 132¶

In :
#Metering Oil Flow by an Orifice
from math import pi

#Variable declaration
d0 = 0.0566                     #Diameter of orifice (m)
d1 = 0.1541                     #Diameter of pipe (m)
Rhooil = 878.                   #Density of oil (kg/m3)
mu = 0.0041                     #Viscoity of oil (cp)
dP = 93200.                     #Pressure diffrence across orifice (kN/m2)
Co = 0.61                       #Coeficient of discharge of orifice

#Calculations
beta = d0/d1
v = Co*sqrt(2*dP/Rhooil)/sqrt(1 - beta**4)
Q = v*pi*d0**2/4
Nre = d0*v*Rhooil/mu

#Result
print 'Velocity of oil %5.3f m/s'%v
print "The volumetric flowrate",round(Q,5), "m3/s"
print 'Reynolds number %5.4e'%Nre

Velocity of oil 8.970 m/s
The volumetric flowrate 0.02257 m3/s
Reynolds number 1.0872e+05


## Example 3.3-1, Page number 135¶

In :
#Calculation of Brake Horse Power of a Pump

#Variable declaration
m = 40.                 #Flowrate for the Pump (gal/min)
mfps = 5.56             #Flowrate of the Pump in fps units (lbm/s)
D = 62.4                #Density of water in fps units (lbm/ft3)
n = .60                 #Efficiency of the pump
H = 38.5                #Developed Head (ft)

#Calculations
mfps = m*(1./60)*(1./7.481)*62.4
Ws = -H
Bhp = -Ws*mfps/(n*550)

#Result
print "The Calculated Brake Horse Power is ",round(Bhp,2), "hp"

The Calculated Brake Horse Power is  0.65 hp


## Example 3.3-2, Page number 137¶

In :
#Brake-kQ Power of a Centrifugal Fan
#Variable declaration
m = 28.32               #Flowrate of the air (m3/min)
Ma = 28.97              #Molecular weight of air
V = 22.414              #Volume of 1 kg air at 101.3 atm pressure and 273.2 K
T1 = 273.2              #Temperature of air at atmospheric pressure (K)
T2 = 366.3              #Temperature of air at suction (K)
T3 = 294.1
P1 = 760.               #Atmospheric pressure of air (mm Hg)
P2 = 741.7              #Suction Pressure of air (mm Hg)
P3 = 769.6              #Discharge Pressure of air (mm Hg)
n = 60.                 #Efficiency of the pump
confac = 101325.        #Conversion factor pressure from (mm Hg) to (N/m2) (N/m2.atm)
rhow = 0.958            #Density of water (kg/m3)
V1 = 0.                 #Velocity of air at suction (m/s2)
V2= 45.7                #Velocity of air at discharge (m/s2)
z1 = 0.                 #Height at suction (m)
z2 = 0.                 #Height at discharge (m)
g = 9.8                 #Gravitational accleration (m/s)
F = 0.
#Calculations
n = n/100
Rho1 = Ma*T1*P2/(V*T2*P1)
Rho2 = Rho1*P3/P2
Rhoavg = (Rho1 + Rho2)/2
mmks = m*(1./60)*(1./22.414)*(T1/T3)*(Ma)
P = (P3 - P2)*confac/(P1*rhow)

#z1*g + V1**2/2 + P1/Rho_avg + Ws = z2*g + V2**2/2 + P2/Rho_avg + F
#-Ws = -z1*g - V1**2/2 - P1/Rho_avg +z2*g + V2**2/2 + P2/Rho_avg + F
Ws = -z1*g - V1**2/2 + P +z2*g + V2**2/2 + F
BkW = Ws*mmks/(n*1000)

#Result
print "The Calculated Brake Horse Power is",round(BkW,2), "kW"

The Calculated Brake Horse Power is 4.65 kW


## Example 3.3-3, Page number 140¶

In :
#Compression of Methane
from math import log
#Variable Declaration
Q = 0.00756                #Flowrate of methane (kg.mol/s)
P1 = 137.9                 #Initial pressure (kPa)
P2 = 551.6                 #Final pressure (kPa)
T1 = 26.7                  #Temperature of methane (degC)
M_CH4 = 16.                #Molecular weight of Methane
gama = 1.31
R = 8314.3                 #Gas constant
n = .8                     #Efficiency of compressor
#Calculation
#(a) Calculation for part (a)
T1_K = T1 + 273.2
Q_mks = Q*M_CH4
Ws = gama*R*T1_K*((P2/P1)**((gama - 1)/gama) - 1)/((gama - 1)*M_CH4)
B_kW = Ws*Q_mks/(n*1000)
#(b) Calculation for part (b)
# Ws_b = 2.3026*R*T1_K*math.log10(P2/P1)/M_CH4
Ws_b = (2.3026/M_CH4)*R*T1_K*log(P2/P1,10)
B_kW_b = Ws_b*Q_mks/(n*1000)
#Result
print "The power required for adiabatic compression",round(B_kW,2),"kW"
print "The power required for isothermal compression",round(B_kW_b,2),"kW"
print 'The answers are correct because of book uses rounded numbers'

The power required for adiabatic compression 38.66 kW
The power required for isothermal compression 32.67 kW
The answers are correct because of book uses rounded numbers


## Example 3.4-1, Page number 145¶

In :
#Power Consumption by an Agitator

#Variable Declaration
Dt = 1.83            #The tank diameter (m)
W = 0.122            #Width of the tank (m)
Da = 0.61            #The turbine diameter (m)
J = 0.15             #Width of baffle (m)
Vrpm = 90.           #Speed of the turbine (rpm)
Rho = 929.           #Density of liquid (kg/m3)
mu = 10.             #Viscosity of liquid (cp)
mub = 100000.        #Viscosity of the in part (b)liquid (cp)

#Calcualtion
mu = mu*1e-3
mub = mub*1e-3
#Calculation for (a)
Vrps = Vrpm/60.
Nre = Da**2*Vrps*Rho/mu
NpT = 5.
P = NpT*Rho*Vrps**3*Da**5/1000.
#Calculation for (b)
Nreb = Da*Vrps*Rho/mub
NpL = 14.
Pb = NpL*Rho*Vrps**3*Da**5/1000.

#Result
print "(a)The power required by the mixer is",round(P,3),"kW"
print "(b)The power requird by the mixer is for higher viscosity",round(Pb,2),"kW"

(a)The power required by the mixer is 1.324 kW
(b)The power requird by the mixer is for higher viscosity 3.71 kW


## Example 3.4-3, Page number 149¶

In :
#Scale-up of Turbine Agitation System
from math import  pi
#Variable Declaration
Dt1 = 1.83                 #Tank diameter (m)
Da1 = 0.61                 #Diameter of turbine (m)
W1 = 0.122                 #Width of turbine (m)
J1 = 0.15                  #Width of baffle (m)
N1 = 1.5                   #Speed of the turbine (rev/s)
Rho = 929.                 #Density of fluid (kg/m3)
mu = 0.01                  #Viscosity of the fluid (Pa.s)
H1 = Dt1
Np = 5.0

#Calcualtion
V1 = pi*Dt1**2*H1/4.
V2 = 3.*V1
R = (V2/V1)**(1./3)
Dt2 = R*Dt1
Da2 = R*Da1
W2 = R*W1
J2 = R*J1
P1 = Np*Rho*N1**3*Da1**5
P1_kW = P1/1000.
#(a)Calculation for equal rate of mass transfer
n = 2./3
N2 = N1*(1/R)**n
Nre = Da2**2*N2*Rho/mu

P2 = Np*Rho*N2**3*Da2**5
P2_kW = P2/1000.
#The power per unit volume
P_v1 = P1_kW/V1
P_v2 = P2_kW/V2
#(b)Calculation for equal liquid motion
n= 1.
N2_b = N1*(1./R)**n
P2_b = Np*Rho*N2_b**3*Da2**5
P2_vb = P2_b/V2

#Result
print "(a) For equal mass transfer"
print '    Total power for smaller tank %5.3f W and \n    power required per unit volume is %5.4f W' %(P1,P_v1)
print '    Total power for larger tank %5.3f W and \n    power required per unit volume is %5.4f W' %(P2,P_v2)
print "(b) The power per unit volume for equal liquid motion is",round(P2_vb/1000,4),"kW"
print 'The answers are different than book because book uses rounded numbers'

(a) For equal mass transfer
Total power for smaller tank 1324.063 W and
power required per unit volume is 0.2751 W
Total power for larger tank 3972.189 W and
power required per unit volume is 0.2751 W
(b) The power per unit volume for equal liquid motion is 0.1907 kW
The answers are correct because of book uses rounded numbers


## Example 3.5-1, Page number 158¶

In :
#Pressure Drop of Power-Law Fluid in Laminar Flow
#Variable Declaration
Rho = 1041.              #Density of fluid (kg/m3)
L = 14.9                 #Length of the tube (m)
D = 0.0524               #Inside diameter of the tube (m)
V = 0.0728               #Average Velocity of the flowing fluid (m/s)
K = 15.23                #Rheological properties of fluid
n = 0.4

#Calcualtion
#Calculation for part (a)
delP = (K*4*L/D)*(8*V/D)**n
Ff = delP/Rho
Nre = ((D**n)*(V**(2-n))*Rho)/(K*8**(n-1))
print 'Nre = %5.4f is in Laminar range' %Nre
#Calculation for part (b)
f = 16./Nre
delPb = 4*f*Rho*L*V**2/(D*2)

#Result
print "(a) The pressure drop calculated is",round(delP,0),"N/m2"
print "    The friction loss calculated is",round(Ff,2),"J/kg"
print "(b) The pressure drop calculated by using the friction factor method is",round(delP/1000,2),"kN/m2"
print 'The answers are different than book because book uses rounded numbers'

Nre = 1.1060 is in Laminar range
(a) The pressure drop calculated is 45391.0 N/m2
The friction loss calculated is 43.6 J/kg
(b) The pressure drop calculated by using the friction factor method is 45.39 kN/m2


## Example 3.5-2, Page number 160¶

In :
#Turbulent Flow of Power-Law Fluid
#Variable Declaration
Rho = 961.              #Density of fluid (kg/m3)
Di = 0.0508             #Inner diameter of circular tube (m)
V_av = 6.1              #Average velocity of fluid (m/s)
K = 2.744               #Consisitency index
n = 0.30                #Flow behaiviour index
L = 30.5                #Length of tubing, m
f = 0.0032              #Friction factor for turbulent flow
#Calcualtion
Nre = Di**n*V_av**(2-n)*Rho/K*8**(n-1)
delP = 4*f*Rho*L*V_av**2/(2*Di)
delP_kN = delP/1000
#Result
print "The frictional pressure drop for turbine is",round(delP_kN,1), "kN/m2"

The frictional pressure drop for turbine is 137.4 kN/m2