# Chapter 12: Condensation and Boiling¶

### ILLUSTRATIVE EXAMPLE 12.2, Page number: 206¶

In [2]:
#Variable declaration:
C = 1                       #Number of constituents
P = 1                       #Number of phases

#Calculation:
F = C-P+2                   #Number of degrees of freedom

#Result:
print "The number of degrees of freedom is :",F," ."

The number of degrees of freedom is : 2  .


### ILLUSTRATIVE EXAMPLE 12.4, Page number: 209¶

In [4]:
#Variable declaration:

U1 = 1237.1                     #Internnal energy of gas (Btu/lb)
U2_g = 1112.2                   #Internal energy of gas (Btu/lb)
U2_l = 343.15                   #Internal energy of liquid (Btu/lb)

#Calculation:
Q = 0.5*(U2_g+U2_l)-1*U1        #Heat removed (Btu/lb)

#Result:
print "Heat removed from the system during the process is :",round(Q,1)," Btu/lb ."

Heat removed from the system during the process is : -509.4  Btu/lb .


### ILLUSTRATIVE EXAMPLE 12.5, Page number: 212¶

In [14]:
from __future__ import division
from sympy import symbols,solve

#Variable declaration:
T1 = 99.0                           #Mean film temperature (°C)
T2 = 98.0                           #Plate surface temperature (°C)
g = 9.807                           #Gravitational acceleration (m/s^2)
#From Appendix:
T3 = 100.0                          #Saturation temperatre (°C)
h_vap1 = 970.3                      #Latent heat of steam in Btu/lb (Btu/lb)
h_vap2 = 2.255*10**6                #Latent heat of steam in J/kg (J/kg)
p_v = 0.577                         #Density of steam (kg/m^3)
p_l = 960.0                         #Density of liquid water condensate (kg/m^3)
mu_l = 2.82*10**-4                  #Absolute viscosity of liquid water condensate (kg/m.s)
k = 0.68                            #Thermal conductivity of water (W/m.K)
#From table 12.2
Z = 0.4                             #Height of rectangular plate (m)
Pw = 0.2                            #Wetted perimeter of rectangular plate (m)
h = symbols('h')                    #Average heat transfer coefficient (W/m^2.K)

#Calculation:
A = Z*Pw                            #Heat transfer area of plate (m^2)
R = A/Pw                            #Ratio A/Pw (m)
v_l = mu_l/p_l                      #Kinematic viscosity of liquid water condensate (m^2/s)
Co1 = (h/k)*(v_l**2/g/(1-p_v/p_l))**(1/3)    #Condensation number (in terms of the average heat transfer coefficient)
Re = 4*h*Z*(T3-T2)/(mu_l*h_vap2)    #Reynolds number in terms of the average heat transfer coefficient
#From equation 12.14:
CO1 = 0.0077*Re**Z                  #Co in terms of Reynolds number for flow type 1
x1 = solve(Co1-CO1,h)               #Solving heat transfer coefficient (W/m^2.K)
h1 =x1[1];                          #Average heat transfer coefficient for flow type 1 (W/m^2.K)
Re1 = Re.subs(h,h1);                #Reynolds number for flow type 1
CO2 = 1.874*Re**(-1/3)              #Co in terms of Reynolds number for flow tupe 2
x2 = solve(Co1-CO2,h)               #Solving average heat transfer coefficient for flow type 2 (W/m^2.K)
h2 = x2[0];                         #Average heat transfer coefficient for flow type 2 (W/m^2.K)
Re2 = Re.subs(h,h2)                 #Reynolds number for flow type 2

#Result:
print "The type of condensation flow type 2 is laminar."
print "And the condensation heat transfer coefficient is :",round(h2,-1)," W/m^2.K ."

The type of condensation flow type 2 is laminar.
And the condensation heat transfer coefficient is : 14700.0  W/m^2.K .


### ILLUSTRATIVE EXAMPLE 12.6, Page number: 214¶

In [6]:
#Variable declaration:
#From example 12.5:
Re = 73.9                       #Reynolds number
mu_l = 2.82*10**-4              #Absolute viscosity of liquid water condensate (kg/m.s)
Pw = 0.2                        #Wetted perimeter of rectangular plate (m)
h = 14700.0                     #Heat transfer coefficient (W/m^2.K)
T_sat = 100.0                   #Saturation temperature (°C)
Ts = 98.0                       #Surface temperature (°C)
A = 0.2*0.4                     #Heat transfer area of plate (m^2)

#Calculation:
m1 = Re*mu_l/4.0                #Mass flow rate of condensate (kg/m.s)
m = Pw*m1                       #Mass flow rate of condensate (kg/s)
Co = (3.038*10**-5)*h           #Condensation number
Q = h*A*(T_sat-Ts)              #Heat transfer rate (W)

#Result:
print "1. The mass flow rate of condensate is :",round(m1,4)," kg/m.s . "
print "2. The heat transfer rate is :",round(Q/10**3,2)," kW . "

1. The mass flow rate of condensate is : 0.0052  kg/m.s .
2. The heat transfer rate is : 2.35  kW .


### ILLUSTRATIVE EXAMPLE 12.7, Page number: 215¶

In [7]:
#Variable declaration:
T_sat = 126.0                   #Saturation temperature (°F)
T = 64.0                        #Surface temperature of tube (°F)
g = 32.2                        #Gravitational acceleration (ft^2/s)
D = 4.0/12.0                    #Outside diameter of tube (ft)

#Calculation:
Tf = (T_sat+T)/2.0               #Mean film temperature (°F)
#From approximate values of key properties:
h_vap = 1022.0                  #Latent heat of steam (Btu/lb)
p_v = 0.00576                   #Density of steam (lb/ft^3)
p_l = 62.03                     #Density of liquid (lb/ft^3)
k_l = 0.364                     #Thermal conductivity of liquid (Btu/h.ft.°F)
mu_l = 4.26*10**-4              #Absolute viscosity of liquid water condensate (lb/ft.s)
h = 0.725*((p_l*(p_l-p_v)*g*h_vap*k_l**3)/(mu_l*D*(T_sat-T)/3600.0))**(1.0/4.0)   #Average heat transfer coefficient (Btu/h.ft^2.°F)

#Result:
print "The average heat transfer coefficient is :",round(h,1)," Btu/h.ft^2.°F ."

The average heat transfer coefficient is : 911.4  Btu/h.ft^2.°F .


### ILLUSTRATIVE EXAMPLE 12.9, Page number: 222¶

In [9]:
#Variable declaration:
Qs1 = 9800.0                        #Heat flux (W/m^2)
Ts1 = 102.0                         #Original surface temperature (°C)
Ts2 = 103.0                         #New surface temperature (°C)
Tsat = 100.0                        #Saturation temperature (°C)

#Calculation:
h1 = Qs1/(Ts1-Tsat)                 #Original heat transfer coefficient (W/m^2.K)
DT1 = (Ts1 - Tsat)                  #Original excess temperature (°C)
DT2 = (Ts2 - Tsat)                  #New excess temperature (°C)
n = 0.25                            #Value of n for laminar flow
h2 = h1*(DT2/DT1)**(n)              #New heat transfer coefficient (W/m^2.K)
Qs2 = h2*(Ts2-Tsat)                 #New heat flux (W/m^2)

#Result:
print "The new heat flux is :",round(Qs2)," W/m^2.K . "

The new heat flux is : 16268.0  W/m^2.K .


### ILLUSTRATIVE EXAMPLE 12.10, Page number: 223¶

In [10]:
#Variable declaration:
#From example 12.9:
Ts1 = 102.0                         #Original surface temperature (°C)
Ts2 = 103.0                         #New surface temperature (°C)
Tsat = 100.0                        #Saturation temperature (°C)

#Calculation:
DTe1 = (Ts1 - Tsat)                 #Original excess temperature (°C)
DTe2 = (Ts2 - Tsat)                 #New excess temperature (°C)

#Result:
print "The original excess temperature is: DTe = ",DTe1," °C ."
print "The new excess temperature is: DTe = ",DTe2," °C ."
if ((DTe1 < 5) and (DTe2 < 5)):
print "The assumption of the free convection mechanism is valid since DTe < 5°C."

The original excess temperature is: DTe =  2.0  °C .
The new excess temperature is: DTe =  3.0  °C .
The assumption of the free convection mechanism is valid since DTe < 5°C.


### ILLUSTRATIVE EXAMPLE 12.11, Page number: 223¶

In [11]:
#Variable declaration:
#From example 12.9:
Cp = 4127.0                         #heat capacity (J/kg . K)
DTe = 3.0                           #New excess temperature (°C)
h_vap = 2.26*10**6                  #latent heat of vaporization (J/kg)

#Calculation:
Ja_L = Cp*DTe/h_vap                 #Liquid Jakob number

#Result:
print "The liquid Jakob number is :",round(Ja_L,5)," ."

The liquid Jakob number is : 0.00548  .


### ILLUSTRATIVE EXAMPLE 12.12, Page number: 223¶

In [12]:
#Variable declaration:
Ts = 106.0                      #Surface temperature (°C)
Tsat = 100.0                    #Saturation temperature (°C)

#Calculation:
DTe = Ts-Tsat                   #Excess temperature (°C)
#From table 12.5:
C1 = 5.56                       #Constant C1
n1 = 3.0                        #Constant n1
C2 = 1040.0                     #Constant C2
n2 = 1.0/3.0                    #Constant n2
P = 1.0                         #Absolute pressure (atm)
Pa = 1.0                        #Ambient absolute pressure (atm)

#Calculation:
h1 = C1*DTe**n1*(P/Pa)**0.4     #Boiling water heat transfer coefficient (W/m^2)
Qs1 = h1*DTe                    #Surface flux (W/m^2)
h2 = C2*DTe**n2*(P/Pa)**0.4     #Second Boiling water heat transfer coefficient (W/m^2)
Qs2 = h2*DTe                   #Second Surface flux (W/m^2)

#Result:

if (Qs1/10**3 > 15.8 and Qs1/10**3 < 236):
print "The boiling regime is :",round(Qs1/10**3,1)," kW/m^2 ."
print "The heat transfer coefficient is :",round(h1), " W/m^2 ."
elif (Qs1/10**3 < 15.8):
print "The boiling regime is :",round(Qs2/10**3,2)," kW/m^2 ."
print "The heat transfer coefficient is :",round(h2), " W/m^2 ."

The boiling regime is : 11.34  kW/m^2 .
The heat transfer coefficient is : 1890.0  W/m^2 .


### ILLUSTRATIVE EXAMPLE 12.13, Page number: 224¶

In [13]:
from math import pi

#Variable declaration:
#From example 12.12:
Qs1 = 11340.0                   #Surface flux (W/m^2)
D = 0.3                         #Diameter of electric heater (m)

#Calculation:
A = pi*(D/2.0)**2               #Surface area of heater (m^2)
Qs = Qs1*A                      #Heat transfer rate (W)

#Result:
print "The rate of heat transfer is :",round(Qs)," W ."

The rate of heat transfer is : 802.0  W .