# Chapter 10 : Boiling and Condensation¶

### Example 10.1¶

In :
import math
# Nucleate Boiling of Water in a Pan
Ts = 108.			#Temp of surface of bottom of pan[degree Celcius]
Tsat = 100.			#Saturation temp of water[degree Celcius]
D = 0.3 			#Diameter[m]
#Properties of water at the saturation temp
rho_l = 957.9			#Density of liquid[kg/m**3]
rho_v = 0.6			#Density of vapour[kg/m**3]
Pr_l = 1.75			#Prandtl no of liquid
mu_l = 0.282*10**(-3)			#Vismath.cosity of liquid[kg/m.s]
Cp_l = 4217			#Specific Heat of liquid[J/kg.degree Celcius]
h_fg = 2257*10**3			#[J/kg]
sigma = 0.0589			#[N/m]
g = 9.81			#Acc due to gravity[m/s**2]
Csf = 0.0130
n = 1.0;

# Calculations and Results
q_nuc = mu_l*h_fg*((g*(rho_l-rho_v)/sigma)**(1./2))*((Cp_l*(Ts-Tsat)/(Csf*h_fg*(Pr_l**n)))**3)			#[W/m**2]
A = math.pi*(D**2)/4			#Surface Area of bottom of the pan[m**2]
Q_boiling = A*q_nuc 			#[W]
print "(a) The rate of heat transfer during nucleate boiling becomes ",Q_boiling,"W"
#Solution(b):-
m = Q_boiling/h_fg			#[kg/s]
print "The rate of Evaporation of water is",m,"kg/s"

(a) The rate of heat transfer during nucleate boiling becomes  5095.13655169 W
The rate of Evaporation of water is 0.00225748185719 kg/s


### Example 10.2¶

In :
import math
# Peak Heat Flux in Nucleate Boiling

# Variables
D = 0.01			#[m]
Tsat = 100			#Saturation Temperature[degree Celcius]
sigma = 0.0589			#[N/m]
#Properties of water at saturation temperature
rho_l = 957.9			#[kg/m**3]
rho_v = 0.6			#[kg/m**3]
h_fg = 2257*10**3			#[J/kg]
mu_l = 0.282*10**(-3)			#[kg/m.s]
Pr_l = 1.75			#Prandtl number
Cp_l = 4217			#[J/kg.degree Celcius]
Csf = 0.0130
n = 1.0;
g = 9.81			#[m/s**2]

# Calculations and Results
L_ = (D/2)*((g*(rho_l-rho_v)/sigma)**(1./2))			#dimensionless Parameter
#For this value of L_ we have
C_cr = 0.12			#Constant
q_max = C_cr*h_fg*((sigma*g*(rho_v**2)*(rho_l-rho_v))**(1./4))			#[W/m**2]
print "The maximum or critical heat flux is",q_max,"W/m**2"

Ts = (((q_max/(mu_l*h_fg*((g*(rho_l-rho_v)/sigma)**(1./2))))**(1./3))*(Csf*h_fg*Pr_l**n)/Cp_l)+Tsat			#[degree Celcius]
print "The surface temperature is",round(Ts),"degree Celcius"

The maximum or critical heat flux is 1017411.21501 W/m**2
The surface temperature is 119.0 degree Celcius


### Example 10.3¶

In :
import math
# Film Boiling of Water on a Heating Element

# Variables
D = 0.005			#[m]
e = 0.05			#Emissivity
Ts = 350			#Surface temperature[degree Celcius]
Tsat = 100			#[degree Celcius]
Tf = (Ts+Tsat)/2			#[degree Celcius]
g = 9.81			#[m/s**2]
#Properties of water at Tsat
rho_l = 957.9			#[kg/m**3]
h_fg = 2257*10**3			#[J/kg]
#Properties of vapor at film temp
rho_v = 0.444			#[kg/m**3]
Cp_v = 1951			#[J/kg.degree Celcius]
mu_v = 1.75*10**(-5)			#[kg/m.s]
k_v = 0.0388			#[W/m.degree Celcius]

# Calculations and Results
q_film = 0.62*(((g*(k_v**3)*rho_v*(rho_l-rho_v)*(h_fg+(0.4*Cp_v*(Ts-Tsat))))/(mu_v*D*(Ts-Tsat)))**(1./4))*(Ts-Tsat)			#[W/m**2]
print "The film boiling heat flux is",q_film,"W/m**2)"

print "The total heat flux is",q_total,"W/m**2"

Q_total = (math.pi*D*1)*q_total			#[W]
print "The rate of heat transfer from the heating element to the water is",Q_total,"W"

The film boiling heat flux is 63008.0039829 W/m**2)
The radiation heat flux is 372.19933107 W/m**2
The total heat flux is 63287.1534812 W/m**2
The rate of heat transfer from the heating element to the water is 994.112282215 W


### Example 10.4¶

In :
import math
# Condensation of steam on a Vertical Plate

# Variables
Tsat = 100
Ts = 80			#[degree Celcius]
Tf = (Ts+Tsat)/2			#[degree Celcius]
L = 2
w = 3			#Dimensions of Plate[m]
g = 9.81			#[m/s**2]
#Properties of water at Tsat
h_fg = 2257*10**3			#[J/kg]
rho_v = 0.60			#[kg/m**3]
#Properties of liquid water at Tf
rho_l = 965.3			#[kg/m**3]
mu_l = 0.315*10**(-3)			#[kg/m.s
Cp_l = 4206			#[J/kg.degree Celcius]
k_l = 0.675			#[W/m.degree Celcius]
nu_l = 0.326*10**(-6)			#[m**2/s]

# Calculations and Results
h_fg_m = h_fg+0.68*Cp_l*(Tsat-Ts)			#[J/kg]
print "The modified latent heat of vapourization is",h_fg_m,"J/kg"

Re = ((4.81+((3.70*L*k_l*(Tsat-Ts)*((g/nu_l**2)**(1./3)))/(mu_l*h_fg_m)))**(0.820));
print "For wavy laminar flow Reynolds number is",Re

h = (Re*k_l*((g/nu_l**2)**(1./3)))/((1.08*(Re**(1.22)))-5.2)			#[W/m**2.degree Celcius]
print "The conensation heat transfer coefficient is",h,"W/m**2.degree Celcius"

As = w*L			        #[m**2]
Q = h*As*(Tsat-Ts)			#[W]
print "The rate of heat transfer during condensation process is",Q,"W"

#Solution (b)
m = Q/h_fg_m			#[kg/s]
print "The rate of condensation of steam is",m,"kg/s"

The modified latent heat of vapourization is 2314201.6 J/kg
For wavy laminar flow Reynolds number is 1287.24272174
The conensation heat transfer coefficient is 5850.17651749 W/m**2.degree Celcius
The rate of heat transfer during condensation process is 702021.182099 W
The rate of condensation of steam is 0.303353511682 kg/s


### Example 10.5¶

In :
import math
# Condensation of steam on a Vertical Tilted Plate

# Variables
Tsat = 100
Ts = 80			#[degree Celcius]
Tf = (Ts+Tsat)/2			#[degree Celcius]
L = 2
w = 3			#Dimensions of Plate[m]
g = 9.81			#[m/s**2]
#Properties of water at Tsat
h_fg = 2257*10**3			#[J/kg]
rho_v = 0.60			#[kg/m**3]
#Properties of liquid water at Tf
rho_l = 965.3			#[kg/m**3]
mu_l = 0.315*10**(-3)			#[kg/m.s
Cp_l = 4206			#[J/kg.degree Celcius]
k_l = 0.675			#[W/m.degree Celcius]
nu_l = 0.326*10**(-6)			#[m**2/s]
theta = (math.pi/6)			#Angle at which plate is tilted[radians]

# Calculations and Results
h_fg_m = h_fg+0.68*Cp_l*(Tsat-Ts)			#[J/kg]
print "The modified latent heat of vapourization is",h_fg_m,"J/kg"

Re = ((4.81+((3.70*L*k_l*(Tsat-Ts)*((g/nu_l**2)**(1./3)))/(mu_l*h_fg_m)))**(0.820));
print "For wavy laminar flow Reynolds number is",Re

h = ((Re*k_l*((g/nu_l**2)**(1./3)))/((1.08*(Re**(1.22)))-5.2))*((math.cos(theta))**(1./4))			#[W/m**2.degree Celcius]
print "The conensation heat transfer coefficient is",h,"W/m**2.degree Celcius"

As = w*L			        #[m**2]
Q = h*As*(Tsat-Ts)			#[W]
print "The rate of heat transfer during condensation process is",Q,"W"

#Solution (b)
m = Q/h_fg_m			#[kg/s]
print "The rate of condensation of steam is",m,"kg/s"

The modified latent heat of vapourization is 2314201.6 J/kg
For wavy laminar flow Reynolds number is 1287.24272174
The conensation heat transfer coefficient is 5643.54026792 W/m**2.degree Celcius
The rate of heat transfer during condensation process is 677224.83215 W
The rate of condensation of steam is 0.292638650043 kg/s


### Example 10.6¶

In :
import math
# Condensation of Steam on horizontal Tubes

# Variables
Tsat = 40   		    	#[degree Celcius]
D = 0.03	    		    #[m]
Ts = 30 		        	#Outer Surface temperature of tube[degree Celcius]
Tf = (Ts+Tsat)/2			#Film Temperature[degree Celcius]
g = 9.81			        #[m/s**2]
#Properties of water at the saturation temp
h_fg = 2407*10**3			#[J/kg]
rho_v = 0.05	    		#[kg/m**3]
#Properties of liquid water at the film temperature
rho_l = 994 	    		#[kg/m**3]
Cp_l = 4178	    	    	#[J/kg.degree Celcius]
mu_l = 0.720*10**(-3)		#[kg/m.s]
k_l = 0.623			        #[W/m.degree Celcius]

# Calculations and Results
h_fg_m = h_fg+0.68*Cp_l*(Tsat-Ts)			#[J/kg]
print "(a) The modified latent heat of vapourisation is",h_fg_m,"J/kg"

h_hori = 0.729*(((g*(rho_l**2)*h_fg_m*(k_l**3))/(mu_l*D*(Tsat-Ts)))**(1./4))			#[W/m**2.degree Celcius]
print "The heat transfer coefficient for condensation on a single horizontal tube is",h_hori,"W/m**2.degree Celcius"

As = math.pi*D*1			#[m**2]
Q = h_hori*As*(Tsat-Ts)			#[W]
print "The rate of heat transfer during condensation Process is",Q,"W"

#Solution (b)
m = Q/h_fg_m			#[kg/s]
print "(b) The rate of condensation of steam is",m,"kg/s"

(a) The modified latent heat of vapourisation is 2435410.4 J/kg
The heat transfer coefficient for condensation on a single horizontal tube is 9294.6621572 W/m**2.degree Celcius
The rate of heat transfer during condensation Process is 8760.0127052 W
(b) The rate of condensation of steam is 0.00359693491709 kg/s


### Example 10.7¶

In :
import math
# Condensation of Steam on horizontal Tube Banks

# Variables
Tsat = 40	    		#[degree Celcius]
D = 0.03		    	#[m]
Ts = 30			        #Outer Surface temperature of tube[degree Celcius]
Tf = (Ts+Tsat)/2			#Film Temperature[degree Celcius]
g = 9.81		    	#[m/s**2]
N = 3			        #No of tubes in a vertical tier
N_total = 12			#Total number of tubes
#Properties of water at the saturation temp
h_fg = 2407*10**3			#[J/kg]
rho_v = 0.05			#[kg/m**3]
#Properties of liquid water at the film temperature
rho_l = 994			#[kg/m**3]
Cp_l = 4178			#[J/kg.degree Celcius]
mu_l = 0.720*10**(-3)			#[kg/m.s]
k_l = 0.623			#[W/m.degree Celcius]

# Calculations and Results
h_fg_m = h_fg+0.68*Cp_l*(Tsat-Ts)			#[J/kg]
print "(a) The modified latent heat of vapourisation is",h_fg_m,"J/kg"

h_hori_N = (0.729*(((g*(rho_l**2)*h_fg_m*(k_l**3))/(mu_l*D*(Tsat-Ts)))**(1./4)))*(1/(N**(1./4)))			#[W/m**2.degree Celcius]
print "The heat transfer coefficient for condensation 12 horizontal tube is",h_hori_N,"W/m**2.degree Celcius"

As = math.pi*D*1*N_total			#[m**2]
Q = h_hori_N*As*(Tsat-Ts)			#[W]
print "The rate of heat transfer during condensation Process is",Q,"W"

#Solution (b)
m = Q/h_fg_m			#[kg/s]
print "(b) The rate of condensation of steam is",m,"kg/s"

(a) The modified latent heat of vapourisation is 2435410.4 J/kg
The heat transfer coefficient for condensation 12 horizontal tube is 7062.41599312 W/m**2.degree Celcius
The rate of heat transfer during condensation Process is 79874.0431221 W
(b) The rate of condensation of steam is 0.0327969541076 kg/s


### Example 10.8¶

In :
import math
# Replacing a Heat Pipe by a Copper Rod

# Variables
L = 0.3			#[m]
D = 0.006			#[m]
Q = 180.			#[W]
del_T = 3.			#Temperature Difference [degree Celcius]
#Properties of copper at room temperature
rho = 8933.			#[kg/m**3]
k = 401.			#[W/m.degree Celcius]

# Calculations and Results
A = Q*L/(k*del_T)			#[m**2]
d = math.sqrt(4*A/math.pi)			#[m]
print "The diameter of the copper pipe is",100*d,"cm"
m = rho*A*L			#[kg]
print "Mass of the copper rod is",round(m),"kg"

The diameter of the copper pipe is 23.9066721377 cm
Mass of the copper rod is 120.0 kg