# Chapter 10: Free Convection¶

### ILLUSTRATIVE EXAMPLE 10.1, Page number: 163¶

In :
#Variable declaration:
Gr = 100.0                  #Grashof number
Re = 50.0                   #Reynolds number

#Calculation:
LT = Gr/Re**2               #Measure of influence of convection effect

#Result:
if (LT<1.0):
print "The free convection effects can be neglected."
elif (LT>1.0):
print "The free convection effects can not be neglected."

The free convection effects can be neglected.


### ILLUSTRATIVE EXAMPLE 10.2, Page number: 166¶

In :
#Variable declaration:
Ts = 110.0+273.0                #Surface temperature of plate (K)
Too = 30.0+273.0                #Ambient air temperature (K)
L = 3.5                         #Height of plate (m)
g = 9.807                       #Gravitational acceleration (m^2/s)

#Calculation:
Tf = (Ts+Too)/2                 #Film temperature (K)
DT = Ts - Too                   #Temperature difference between surface and air (K)
#From appendix:
v = 2.0*10**-5                  #Kinematic viscosity for air (m^2/s)
k = 0.029                       #Thermal conductivity for air (W/m.K)
Pr = 0.7                        #Prandtl number
B = 1.0/Tf                      #Coefficient of expansion (K^-1)
Gr = g*B*DT*L**3/v**2           #Grashof number
Ra = Gr*Pr                      #Rayleigh number

#Result:
print "The Grashof number is :",round(Gr/10**11,2)," x 10^11 ."
print "The Rayleigh number is :",round(Ra/10**11,2)," x 10^11 ."

The Grashof number is : 2.45  x 10^11 .
The Rayleigh number is : 1.72  x 10^11 .


### ILLUSTRATIVE EXAMPLE 10.3, Page number: 166¶

In :
#Variable declaration:
#From example 10.2:
Ra = 1.71*10**11                #Rayleigh number

#Result:
if (Ra>10**9):
print "The convection flow category is turbulent."
elif(Ra<10**9):
print "The convection flow category is laminar."

The convection flow category is turbulent.


### ILLUSTRATIVE EXAMPLE 10.4, Page number: 167¶

In :
#Variable declaration:
#From Table 10.1:
c = 0.1                             #Constant c
m = 1.0/3.0                         #Constant for turbulent free conection
#From example 10.2:
Ra = 1.71*10**11                    #Rayleigh number
k = 0.029                           #Thermal conductivity (W/m.K)
L = 3.5                             #Thickness of plate (m)

#Calculation:
Nu = c*Ra**m                        #Average Nusselt number
h = Nu*k/L                          #Average heat transfer coefficient (W/m^2.K)

#Result:
print "The average heat transfer coefficient is :",round(h,1)," W/m^2.K ."

The average heat transfer coefficient is : 4.6  W/m^2.K .


### ILLUSTRATIVE EXAMPLE 10.6, Page number: 167¶

In :
from math import log,e

#Variable declaration:
Ts = 200.0+460.0                #Surface temperature of pipe (°R)
Too = 70.0+460.0                #Air temperature (°R)
D = 0.5                         #Diameter of pipe (ft)
R = 0.73                        #Universal gas constant (ft^3.atm.R^−1.lb.mol^−1)
P = 1.0                         #Atmospheric pressure (Pa)
MW = 29.0                       #Molecular weight of fluid (mol)
#From Appendix:
mu = 1.28*10**-5                #Absolute viscosity (lb/ft.s)
k = 0.016/3600.0                #Thermal conductivity (Btu/s.ft.°F)
g = 32.174                      #Gravitational acceleration (ft/s^2)

#Calculation:
Tav = (Ts+Too)/2                #Average temperature (°R)
v = R*Tav/P                     #kinematic viscosity (ft^3/lbmol)
p = MW/v                        #Air density (lb/ft^3)
B = 1.0/Tav                     #Coefficient of expansion (°R^-1)
DT = Ts-Too                     #Temperature difference (°R)
Gr = D**3*p**2*g*B*DT/mu**2     #Grashof number
#From equation 10.5:
Cp = 0.25                       #Air heat capacity (Btu/lb.°F)
Pr = Cp*mu/k                    #Prandtl number
GrPr = 10**8.24                 #Rayleigh number
#From Holman^(3):
Nu = 10**(1.5)                  #Nusselt number
h = Nu*(k/D)*3600.0             #Air heat transfer film coefficient (Btu/h.ft.°F)

#Result:
print "The required air heat transfer film coefficient is :",round(h,2)," Btu/h.ft.°F ."

The required air heat transfer film coefficient is : 1.01  Btu/h.ft.°F .


### ILLUSTRATIVE EXAMPLE 10.7, Page number: 168¶

In :
#Variable declaration:
Ts = 120.0+460                  #Surface temperature of plate (°R)
Too = 60.0+460                  #Ambient temperature of nitrogen (°R)
L = 6                           #Height of plate (ft)
#From Appendix:
p = 0.0713                      #Air density (lb/ft^3)
k = 0.01514                     #Thermal conductivity (Btu/h.ft.°F)
v = 16.82*10**-5                #Kinematic viscosity (ft^2/s)
Pr = 0.713                      #Prandtl number
g = 32.2                        #Gravitational acceleration (ft/s^2)

#Calculation:
Tf = (Ts+Too)/2                 #Mean film temperature (°R)
B = 1.0/Tf                      #Coefficient of expansion (°R^-1)
Gr = g*B*(Ts-Too)*L**3/v**2     #Grashof number
Ra = Gr*Pr                      #Rayleigh number
#From equation 10.13(Table 10.2) and costants from Table 10.1:
h = 0.10*(k/L)*Ra**(1.0/3.0)    #Free convection heat transfer coefficient (Btu/h.ft^2.°F)

#Result:
print "The free convection heat transfer coefficient is :",round(h,3)," Btu/h.ft^2.°F ."
print "There is a calculation mistake in the book for calculating 'Gr', so, value of 'h' alters from that given."

The free convection heat transfer coefficient is : 0.675  Btu/h.ft^2.°F .
There is a calculation mistake in the book for calculating 'Gr', so, value of 'h' alters from that given.


### ILLUSTRATIVE EXAMPLE 10.8, Page number: 169¶

In :
#Variable declaration:
#From example:
h = 0.675                       #Free convection heat transfer coefficient (Btu/h.ft^2.°F)
A = 6.0*8.0                     #Area of plate (ft^2)
Ts = 120.0                      #Surface temperature of plate (°F)
Too = 60.0                      #Ambient temperature of nitrogen (°F)

#Calculation:
Q = h*A*(Ts-Too)                #Heat loss (Btu/h)

#Result:
print "The heat loss is :",round(Q,-1)," Btu/h ."
print " The 'h' obtained in the previous example differs, therefore, 'Q' obtained here also fiffers from that given in book."

The heat loss is : 1940.0  Btu/h .
The 'h' obtained in the previous example differs, therefore, 'Q' obtained here also fiffers from that given in book.


### ILLUSTRATIVE EXAMPLE 10.9, Page number: 169¶

In :
#Variable declaration:
Ts = 113.0+273.0                    #Surface temperature of bulb (K)
Too = 31.0+273.0                    #Ambient air temperature (K)
D = 0.06                            #Diameter of sphere (m)
g = 9.8                             #Gravitational acceleration (m/s^2)

#Calculation:
Tf = (Ts+Too)/2                     #Mean temperature (K)
#From Appendix:
v = (22.38*10**-5)*0.0929           #Kinematic viscosity (m^2/s)
Pr = 0.70                           #Prandtl number
k = 0.01735*1.729                   #Thermal conductivity (W/m.K)
B = 1.0/(Tf)                        #Coefficient of expansion (K^-1)
Gr = g*B*(Ts-Too)*D**3/v**2         #Grashof number
Ra = Gr*Pr                          #Rayleigh number

#From equation 10.13:
h = (k/D)*0.6*Ra**(1.0/4.0)         #Heat transferred from bulb (W/m^2.K)

#Result:
print "The heat transferred from bulb to air is :",round(h,2)," W/m^2.K ."

The heat transferred from bulb to air is : 9.01  W/m^2.K .


### ILLUSTRATIVE EXAMPLE 10.10, Page number: 170¶

In :
from math import pi

#Variable declaration:
#From example 10.9:
h = 9.01                    #Heat transferred from bulb (W/m^2.K)
D = 0.06                    #Diameter of sphere (m)
Ts = 113.0+273.0            #Surface temperature of bulb (K)
Too = 31.0+273.0            #Ambient air temperature (K)

#Calculation:
A = pi*D**2                 #Surface area of bulb (m^2)
Q = h*A*(Ts-Too)            #Heat transfer lost by free convection from light bulb (W)

#Result:
print "The heat transfer lost by free convection from light bulb is :",round(Q,2)," W ."

The heat transfer lost by free convection from light bulb is : 8.36  W .


### ILLUSTRATIVE EXAMPLE 10.11, Page number: 170¶

In :
#Variable declaration:
#From example 10.9-10.10:
Q = 8.36                        #Heat transfer lost by free convection from light bulb (W)

#Calculation:
E = Q/100.0*(100.0)             #Percent energy lost by free convection (%)

#Result:
print "The percentage of the energy lost by free convection is :",round(E,2)," % ."
print "The energy lost fraction is :",round(E/100.0,4)," ."

The percentage of the energy lost by free convection is : 8.36  % .
The energy lost fraction is : 0.0836  .


### ILLUSTRATIVE EXAMPLE 10.13, Page number: 175¶

In :
#Variable declaration:
F = 50.0                        #Buoyancy flux of gas (m^4/s^3)
u = 4.0                         #wind speed (m/s)

#Calculation:
xc = 14*F**(5.0/8.0)            #Downward distance (m)
xf = 3.5*xc                     #distance of transition from first stage of rise to the second stage of rise (m)
Dh = 1.6*F**(1.0/3.0)*u**-1*xf**(2.0/3.0)   #Plume rise (m)

#Result:
print "The plume rise is :",round(Dh)," m ."

The plume rise is : 101.0  m .