# Chapter 12:Momentum,Heat and MAss Transfer¶

## Example no:12.1,Page no:714¶

In [3]:
#Variable declaration
d=250e-3  #internal diameter of pipe
u=15  #Velocity of air through the pipe
y1=50e-3  #First point where velocity is to be found out
y2=5e-3  #Second point where velocity is to be found out
rho_air=1.10  #Density of air
Meu_air=20e-6  #Viscosity of air

#Calculation
import math
Re=d*u*rho_air/Meu_air
#Hence, from Figure 3.7: X=R/(rho*u**2)=0.0018
X=0.0018
u_s=u/0.817
u_star=u*X**0.5
#At 50 mm from the wall:
y1_r=2*y1/d # y/r
#Hence, from equation 12.34:
u_x1=u_s+2.5*u_star*math.log(y1_r)
#At 5 mm from the wall:
y2_r=2*y2/d # y/r
#Hence, from equation 12.34:
u_x2=u_s+2.5*u_star*math.log(y2_r)
#The thickness of the laminar sub-layer is given by equation 12.54:
del_b=5*d/(Re*X**0.5)

#Result
print"\n The fluid velocity at 50 mm from the wall =",round(u_x1,1),"m/s"
print"\n The fluid velocity at 5 mm from the wall =",round(u_x2,1),"m/s"
print"\n The thickness of the laminar sub-layer =",round(del_b*1e3,3),"mm"

 The fluid velocity at 50 mm from the wall = 16.9 m/s

The fluid velocity at 5 mm from the wall = 13.2 m/s

The thickness of the laminar sub-layer = 0.143 mm


## Example no:12.2,Page no:722¶

In [6]:
#Variable declaration
u=10  #Velocity of air
T=330  #Temperature of air
d=25e-3  #Inner diameter of pipe
T_p=415  #Temperature at which the pipe is maintained
DP_l=80  #Drop of static pressure along the pipe per unit length

import math
#From equations 12.98 and 3.18:
#we get h=0.05*Cp
#The heat taken up per unit time by the air dQ=0.0052*Cp*dT......(i)
#The heat transferred through the pipe wall is also given by: = 0.039*Cp*(415-T)......(ii)
#Equating (i) & (ii)
#On integrating we get
T_0=415-(85/math.exp(0.45))
print"\n The required air Temperature =%d"%T_0,"K"

 The required air Temperature =360 K


## Example no:12.2,Page no:732¶

In [5]:
#Variable declaration
u=0.5     #m/s
l=20.0/1000.0        #m lemngth of tube
mu=10**-3
rho=1000.0     #kg/m**3
Sc=2330.0
#Calculation
Re=l*u*rho/mu
Hd=u*0.032*Re**(-1/4.0)*(1+2.0*(Sc-1)*Re**(-1/8.0))**-1

#result
print"Mass Transfer coefficient is: %.3e"%Hd,"kmol/m**2 s= %.3e"%Hd,"m/s"

Mass Transfer coefficient is: 1.085e-06 kmol/m**2 s= 1.085e-06 m/s


## Example no:12.3,Page no:733¶

In [6]:
#Variable declaration
u=3.5  #Velocity of water
d=25e-3  #Diameter of the pipe
l=6  #Length of the pipe
T1=300  #Temperature at enterance
T2=330  #Temperature at exit
rho=1000  #density of water at 310 K
Meu=0.7e-3  #Viscosity of water at 310 K
#Taking the fluid properties at 310 K and assuming that fully developed flow exists
Cp=4.18e3  #heat capapcity
k=0.65  #Thermal conductivity

#Calculation
Re=d*u*rho/Meu
Pr=Cp*Meu/k
import math
h1=0.032*(Re**-0.25)*Cp*rho*u #....Equation 12.139
# on solving we get final equation as
theta_dash1=330-10**(math.log10(30)-(0.0654*h1*1e-3/2.303))
h2=0.032*(Re**-0.25)*(1+2*Re**(-1.0/8.0)*(Pr-1))**-1*Cp*rho*u
# on solving we get final equation as
theta_dash2=330-10**(math.log10(30)-(0.0654*h2*1e-3/2.303)) #....Equation 12.140
h3=0.032*(Re**-0.25)*(1+0.82*Re**(-1.0/8.0)*((Pr-1)+math.log(0.83*Pr+0.17)))**-1*Cp*rho*u #...equation 12.141
# on solving we get final equation as
theta_dash3=330-10**(math.log10(30)-(0.0654*h3*1e-3/2.303))
h4=k/d*0.023*Re**0.8*Pr**0.33
# on solving we get final equation as
theta_dash4=330-10**(math.log10(30)-(0.0654*h4*1e-3/2.303))

#Result
print"\n (a) Reynolds analogy"
print"\n The outlet temperature =",round(theta_dash1,1),"K"
print"\n\n (b) Taylor Prandtl Equation"
print"\n The outlet temperature =",round(theta_dash2,1),"K"
print"\n\n (c) Universal velocity profile equation"
print"\n The outlet temperature =",round(theta_dash3,1),"K"
print"\n\n (d) Nu=0.023*Re**0.8*Pr**0.33"
print"\n The outlet temperature = %d"%theta_dash4,"K"

 (a) Reynolds analogy

The outlet temperature = 324.1 K

(b) Taylor Prandtl Equation

The outlet temperature = 313.9 K

(c) Universal velocity profile equation

The outlet temperature = 317.2 K

(d) Nu=0.023*Re**0.8*Pr**0.33

The outlet temperature = 316 K


## Example no:12.4,Page no:734¶

In [2]:
#Variable declaration
u=3.5  #Velocity of air
d=25e-3  #Diameter of the pipe
l=6  #Length of the pipe
T1=290  #Temperature at enterance
T2=350  #Temperature at exit
rho=29/22.4*273/310  #density of air at 310 K
Meu=0.018e-3  #Viscosity of air at 310 K
#Taking the physical properties at 310 K and assuming that fully developed flow exists
Cp=1.003e3  #heat capapcity
k=0.024  #Thermal conductivity

#Calculation
import math
Re=d*u*rho/Meu
Pr=Cp*Meu/k
h1=0.032*(Re**-0.25)*Cp*rho*u #....Equation 12.139
# on solving we get final equation as
theta_dash1=350-10**(math.log10(60)-(239.88*h1*1e-3/2.303))
h2=0.032*(Re**-0.25)*(1+2*Re**(-1.0/8.0)*(Pr-1))**-1*Cp*rho*u
# on solving we get final equation as
theta_dash2=350-10**(math.log10(60)-(239.88*h2*1e-3/2.303)) #....Equation 12.140
h3=0.032*(Re**-0.25)*(1+0.82*Re**(-1.0/8.0)*((Pr-1)+math.log(0.83*Pr+0.17)))**-1*Cp*rho*u #...equation 12.141
# on solving we get final equation as
theta_dash3=350-10**(math.log10(60.0)-(239.88*h3*1e-3/2.303))
h4=k/d*0.023*Re**0.8*Pr**0.33
# on solving we get final equation as
theta_dash4=350-10**(math.log10(60)-(239.88*h4*1e-3/2.303))
print"\n (a) Reynolds analogy"
print"\n The outlet temperature =",round(theta_dash1,1),"K"
print"\n\n (b) Taylor Prandtl Equation"
print"\n The outlet temperature =",round(theta_dash2),"K"
print"\n\n (c) Universal velocity profile equation"
print"\n The outlet temperature = ",round(theta_dash3,1),"K"
print"\n\n (d) Nu=0.023*Re**0.8*Pr**0.33"
print"\n The outlet temperature =",round(theta_dash4,1),"K"

 (a) Reynolds analogy

The outlet temperature = 348.3 K

(b) Taylor Prandtl Equation

The outlet temperature = 349.0 K

(c) Universal velocity profile equation

The outlet temperature =  349.0 K

(d) Nu=0.023*Re**0.8*Pr**0.33

The outlet temperature = 349.5 K