# Chapter 8: Condensation and boiling¶

## Example 8.1 , Page no:318¶

In [1]:
import math
from __future__ import division

#Variable declaration
Ts = 80 ; #C
Tw = 70 ; #C
L = 1 ; #m
g = 9.8 ; #m/s^2
#From table A.1
rho = 978.8 ; #kg/m^3
k = 0.672 ; #W/m K
hfg = 2309 ; #At 80 C,kJ/kg

#calculations
Tm = (Ts + Tw)/2 ; #Assuming condensate film is laminar and Re < 30
u = 381 *10**-6 ; #kg/m s
v = u/rho ;
#Substituting in eqn 8.3.9, we get
h = 0.943*(( hfg *1000*( rho**2)*g*(k**3)) /(( Ts -Tw)*u*L) )**(1/4) ; #W/m^2 K
rate = h*L*(Ts -Tw)/( hfg *1000) ; #kg/m s
Re = 4* rate /u;
#Substituting h = Re*(lambda*1000)*u/(4*L*(Ts-Tw)), in eqn 8.3.12
Re_1 = (((4* L*(Ts -Tw)*k/( hfg *1000* u)*(g/(v**2) )**(1/3) )+5.2)/1.08)**(1/1.22) ; #Substituting h = Re*(hfg*1000)*u/(4*L*(Ts-Tw))
#From eqn 8.3.12
h_1 = ((Re /(1.08*( Re**1.22) -5.2) )*k *(( g/v**2)**(1/3) )); #W/m^2 K
m = h_1*L *10/( hfg *1000) ; #rate of condensation,kg/m s

#result
print"Assuming condensate film is laminar and Re < 30";
print"h =",round(h,4),"W/m^2 K";
print"ReL =",round(Re,4);
print"Initial assumption was wrong, Now considering the effect of ripples, we get";
print"Re =",round(Re_1,4);
print"Heat Transfer Cofficient =",round(h_1,4),"W/m^2 K";
print"Rate of condensation =",round(m,6),"kg/m s";
Assuming condensate film is laminar and Re < 30
h = 6078.7864 W/m^2 K
ReL = 276.3936
Initial assumption was wrong, Now considering the effect of ripples, we get
Re = 320.4829
Heat Transfer Cofficient = 7287.8478 W/m^2 K
Rate of condensation = 0.031563 kg/m s

## Example 8.2 , Page no:321¶

In [2]:
import math
from __future__ import division

#Variable declaration
Ts = 262 ; #K
D = 0.022 ; #m
Tw = 258 ; #K
#Properties at Tm
rho = 1324 ; #kg/m^3
k = 0.1008 ; #W/m K
g = 9.81 ; #m/s^2

#calculations
Tm = (Ts+Tw) /2;
v = 1.90*10**-7 ; #m^2/s
hfg = 215.1*10**3 ; #J/kg
u = v*rho ; #Viscosity
#From eqn 8.4.1
h = 0.725*( hfg *( rho**2) *g*(k**3) /(( Ts -Tw)*u*D))**(1/4) ;
rate = h*3.14*D*(Ts -Tw) / hfg ; #kg/s m
Re = 4* rate /u ;

#result
print"Heat transfer coefficient =",round(h,4),"W/m^2 K";
print"Condensation rate per unit length =",round(rate,6),"kg/s m";
print"Film Reynolds number =",round(Re,4);
Heat transfer coefficient = 2622.2475 W/m^2 K
Condensation rate per unit length = 0.003369 kg/s m
Film Reynolds number = 53.5629

## Example 8.3 , Page no:322¶

In [3]:
import math
from __future__ import division

#Variable declaration
m = 25/60 ; #kg/sec
ID = 0.025 ; #m
OD = 0.029 ; #m
Tci = 30 ; #C
Tce = 70 ; #C
g = 9.8 ; #m/s^2
Ts = 100 ; #C
#Assuming 5.3.2 is valid, properties at 50 C
#Properties at Tm
rho = 988.1 ; #kg/m^3
k = 0.648 ; #W/m K
Pr = 3.54 ;
#From eqn 4.6.4a
f = 0.005635;
#From eqn 5.3.2
Nu = 198.39 ;
Tw = 90 ; #Assuming average wall temperature = 90 C
#Properties at Tm
#Properties at Tm
rho = 961.9 ; #kg/m^3
k = 0.682 ; #W/m K
l = 0; #initial guess, assumed value for fsolve function

#calculations
v = 0.556*10**-6; #m^2/s
Re = 4*m/(3.14*ID*rho *v);
h = Nu*k/ID ;
u = 298.6*10**-6 ; #kg/m s
hfg = 2257*10**3 ; #J/kg
#Equating the heat flow from the condensing steam to the tube wall, to the heat flow from the tube wall to the flowing water.
#Solving the simplified equation
h = 0.725*(hfg *( rho**2) *g*(k**3) /(( Ts -Tw)*u*OD))**(1/4) ;
#By solving trial and error method, the temperature value we get
T=86.964984;# in oC
#Therefore
hc = 21338.77/(100 - T)**(1/4) ; #W/m^2 K
#Now, equating the heat flowing from the condensing steam to the tube wall to the heat gained by the water, we have
#Solving by trial and error method, we get
L=5.216152; #in meter

#result
print"Temperature obtained from trial and error =",round(T,4),"oC";
print"hc =",round(hc,4),"W/m^2 K";
print"Length of the tube =",round(L,4),"m";
Temperature obtained from trial and error = 86.965 oC
hc = 11230.3034 W/m^2 K
Length of the tube = 5.2162 m

## Example 8.4 , Page no:322¶

In [4]:
import math
from __future__ import division
#Properties at (Tw+Ts)/2 = 100.5 degree celsius
deltaT1 = 1;                  #in degree celsius
p1 = 7.55*10**-4;           #[K^(-1) p1 is coefficient of cubical expansion
v1 = 0.294*10**-6;                #[m^2/sec]  viscosity at 100.5 degree celsius
k1 = 0.683;                  #[W/m-k]thermal conductivity
Pr1 = 1.74;                  #Prandtl number
g = 9.81;                    #acceleration due to gravity
L = 0.14*10**-2;                 #diameter in meters
#Properties at (Tw+Ts)/2 =102.5
deltaT2 = 5;                 #in degree celsius
p2 = 7.66*10**-4;            #[K^(-1) p1 is coefficient of cubical expansion
v2 = 0.289*10**-6;            #[m^2/sec]  viscosity at 102.5 degree celsius
k2 = 0.684;                 #[W/m-k]thermal conductivity
Pr2 = 1.71;                  #Prandtl number
#Properties at (Tw+Ts)/2 =105
deltaT3 = 10;                 #in degree celsius
p3 = 7.80*10**-4;            #[K^(-1) p1 is coefficient of cubical expansion
v3 = 0.284*10**-6;            #[m^2/sec]  viscosity at 105 degree celsius
k3 = 0.684;                 #[W/m-k]thermal conductivity
Pr3 = 1.68;                  #Prandtl number

#Calculations

Ra1 = ((p1*g*deltaT1*L**3)/(v1**2))*Pr1;
q1=(k1/L)*(deltaT1)*(0.36+(0.518*Ra1**(1/4))/(1+(0.559/Pr1)**(9/16))**(4/9))

Ra2 = ((p2*g*deltaT2*L**3)/(v2**2))*Pr2;
q2=(k2/L)*(deltaT2)*(0.36+(0.518*Ra2**(1/4))/(1+(0.559/Pr2)**(9/16))**(4/9))

Ra3 = ((p3*g*deltaT3*L**3)/(v3**2))*Pr3;
q3=(k3/L)*(deltaT3)*(0.36+(0.518*Ra3**(1/4))/(1+(0.559/Pr3)**(9/16))**(4/9))

#At 100 degree celsius
Cpl = 4.220;          #[kJ/kg]
lamda = 2257;         #[kJ/kg]
ul = 282.4*10**-6;        #viscosity is in kg/m-sec
sigma = 589*10**-4;       #Surface tension is in N/m
pl = 958.4;           #density in kg/m^3
pv =0.598;            #density of vapour in kg/m^3
deltap = pl-pv;
Prl = 1.75;           #Prandtl no. of liquid
Ksf = 0.013;
deltaT11=5;
deltaT12=10;
deltaT13=20;
q11=141.32*deltaT11**3
q12=141.32*deltaT12**3
q13=141.32*deltaT13**3

L1 = (L/2)*(g*(pl-pv)/sigma)**(1/2);
f_L = 0.89+2.27*math.exp(-3.44*L1**(0.5));
q2 = f_L*((3.14/24)*lamda*10**(3)*pv**(0.5)*(sigma*g*(pl-pv))**(0.25));

Tn=pow(q2/141.32,1/3)
q3 = 0.09*lamda*10**3*pv*(sigma*g*(pl-pv)/(pl+pv)**(2))**(0.25);
Ts1 = 140;          #surface temperature in degree celsius
Ts2 = 200;          #surface temperature in degree celsius
Ts3 = 600;          #surface temperature in degree celsius
Twm1 = (140+100)/2; #Mean film temperature
#properties of steam at 120 degree celsius and 1.013 bar
kv = 0.02558;       #thermal conductivity in W/mK
pv1 = 0.5654;       #vapor density in kg/m**3
uv=13.185*10**(-6);  #viscosity of vapour in kg/m sec
lamda1 = (2716.1-419.1)*10**(3);#Latent heat of fusion in J/kg
hc = 0.62*((kv**3)*pv*(pl-pv)*g*lamda1/(L*uv*(140-100)))**(0.25);
qrad = 5.67*10**(-8)*(413**4 - 373**4)/((1/0.9)+1-1);
h = hc + 0.75*hr;

hc_200 = 0.62*((kv**3)*pv*(pl-pv)*g*lamda1/(L*uv*(200-100)))**(0.25);
qrad1 = 5.67*10**(-8)*(473**4 - 373**4)/((1/0.9)+1-1);
h_200 = hc_200 +0.75*hr_200;
hc_600 = 0.62*((kv**3)*pv*(pl-pv)*g*lamda1/(L*uv*(600-100)))**(0.25);
qrad2 = 5.67*10**(-8)*(873**4 - 373**4)/((1/0.9)+1-1);

#Results

print "\n q/A = ",round(q1,2)," W/m^2 at (Tw-Ts)=1";
print "\n q/A = ",round(q2,2)," W/m^2 at (Tw-Ts)=5";
print "\n q/A = ",round(q3,2)," W/m^2 at (Tw-Ts)=10";
print "\n q/A at deltaT = 5 degree celsius = ",q11," W/m^2";
print "\nq/A at deltaT = 10 degree celsius = ",q12," W/m^2";
print "\n q/A at deltaT =20 degree celsius = ",q13," W/m^2";
print "\n Peak heat flux L =  ",round(L1,2);
print "\n f(l) = ",round(f_L,2);
print "\n q/A = ",q2," W/m^2";
print "Tw-Ts = ",Tn," degree celsius"
print "\n\n Minimum heat flux";
print "\n q/A ",q3, "W/m^2"
print "\n\n Stable film boiling"
print "\n hc = ",hc," W/m^2"
print "\n q/A due to radiation = ",qrad," W/m^2";
print "\n hr = ",hr," W/m^2 K ";
print "\n Since hr<hc ";
print "\n The total heat transfer coefficient ";
print " h = ",h," W/m^2 K";
print "\n Total heat flux ",h*(140-100)," W/m^2 K";
print "\n\n hc = ",hc_200," W/m^2";
print "\n hr = ",hr_200," W/m^2 K";
print "\n q/A due to radiation = ",qrad1," W/m^2";
print "\n Total heat flux = ",h_200*100," W/m^2";
print "\n\n hc = ",hc_600," W/m^2";
print "\n hr = ",hr_600," W/m^2 K";
print "\n q/A due to radiation = ",qrad2," W/m^2";

#Graph

import matplotlib.pyplot as plt
import numpy as np
from pylab import linspace
%matplotlib inline
q = [q11, q12, q13];
plt.plot ([1, 5, 10],q);
deltaT=linspace(1,10,10);
q1=141.32*deltaT**3;
plt.plot (deltaT,q1)
plt.title ("Boiling curve");
plt.xlabel(" (Tw - Ts)degree celsius ");
plt.ylabel(" Heat flux,(q/A)W/m^2 ");
q/A =  1116.99  W/m^2 at (Tw-Ts)=1

q/A =  1393519.91  W/m^2 at (Tw-Ts)=5

q/A =  19025.3  W/m^2 at (Tw-Ts)=10

q/A at deltaT = 5 degree celsius =  17665.0  W/m^2

q/A at deltaT = 10 degree celsius =  141320.0  W/m^2

q/A at deltaT =20 degree celsius =  1130560.0  W/m^2

Peak heat flux L =   0.28

f(l) =  1.26

q/A =  1393519.90741  W/m^2
Tw-Ts =  21.4438708455  degree celsius

Minimum heat flux

q/A  19025.295556 W/m^2

Stable film boiling

hc =  455.986290831  W/m^2

q/A due to radiation =  496.874268274  W/m^2

hr =  12.4218567068  W/m^2 K

Since hr<hc

The total heat transfer coefficient
h =  465.302683361  W/m^2 K

Total heat flux  18612.1073344  W/m^2 K

hc =  362.632549817  W/m^2

hr =  15.665080604  W/m^2 K

q/A due to radiation =  1566.5080604  W/m^2

Total heat flux =  37438.136027  W/m^2

hc =  242.507001959  W/m^2

hr =  3.13301612081  W/m^2 K

q/A due to radiation =  28652.514946  W/m^2

## Example 8.5 , Page no:337¶

In [5]:
import math
from __future__ import division

#Variable declaration
D = 0.02 ; #m
l = 0.15 ; #m
T = 500+273 ; #K
Tc = -196+273 ; #K
e = 0.4;
#Properties
k = 0.0349 ; #W/m K
rho = 0.80 ; #kg/m^3
Cpavg = 1.048 ; #kJ/kg J
rholiq = 800 ; #kg/m^3

#calculations
s = 5.670*10**-8;
#Film boiling will occur, hence eqn 8.7.9 is applicable
Tm = (T+Tc) /2; #Film boiling will occur
u = 23*10**-6 ; #kg/m s
latent = 201*10**3 ; #J/kg
hfg = (latent + Cpavg *(Tm -Tc) *1000); #Jk/g
hc = 0.62*((( k**3) *rho *799.2*9.81* hfg )/(D*u*(T-Tc)) )**(1/4) ; #W/m^2 K
#Taking the emissivity of liquid surface to be unity and using equation 3.9.1, the exchange of radiant heat flux
flux = s*(T**4- Tc**4) /(1/ e +1/1 -1) ; #W/m^2
hr = flux /(T-Tc);
#Since h_r < h_c, total heat transfer coefficient is determined from eqn 8.7.11
h = hc +3/4* hr ; #W/m^2 K
fluxi = h*(T-Tc);
Rate = fluxi *3.14*D*l; #W

#result
print"Initial heat flux =",round(fluxi,4),"W/m^2";
print"Initial heat transfer rate =",round(Rate,4),"W";
Initial heat flux = 69646.6128 W/m^2
Initial heat transfer rate = 656.0711 W