# Boiling and Condensation¶

## Example 10.1 Page 632¶

In [1]:
import math
#Operating Conditions
Ts = 118+273.    				;#[K] Surface Temperature
Tsat = 100+273.    				;#[K] Saturated Temperature
D = .3            				;#[m] Diameter of pan
g = 9.81          				;#[m^2/s] gravitaional constant
#Table A.6 Saturated water Liquid Properties T = 373 K
rhol = 957.9            		;#[kg/m^3] Density
cp = 4.217*math.pow(10,3)       ;#[J/kg] Specific Heat
u = 279*math.pow(10,-6)         ;#[N.s/m^2] Viscosity
Pr = 1.76                		;# Prandtl Number
hfg = 2257*math.pow(10,3)       ;#[J/kg] Specific Heat
si = 58.9*math.pow(10,-3)      	;#[N/m]
#Table A.6 Saturated water Vapor Properties T = 373 K
rhov = .5956            		;#[kg/m^3] Density

Te = Ts-Tsat;
#calculations

#From Table 10.1
C = .0128;
n = 1.;
q = u*hfg*math.pow(g*(rhol-rhov)/si,.5)*math.pow((cp*Te/(C*hfg*math.pow(Pr,n))),3);
qs = q*math.pi*D*D/4.; 			#Boiling heat transfer rate

m = qs/hfg; 					#Rate of evaporation

qmax = .149*hfg*rhov*math.pow(si*g*(rhol-rhov)/(rhov*rhov),.25); 	#Critical heat flux
#results

print '%s %.2f %s' %("\n Boiling Heat transfer rate = ",qs/1000. ,"kW")
print '%s %d %s' %("\n Rate of water evaporation due to boiling =",m*3600 ,"kg/h")
print '%s %.2f %s' %("\n Critical Heat flux corresponding to the burnout point =",qmax/math.pow(10,6) ,"MW/m^2");
#END

 Boiling Heat transfer rate =  59.13 kW

Rate of water evaporation due to boiling = 94 kg/h

Critical Heat flux corresponding to the burnout point = 1.26 MW/m^2


## Example 10.2 Page 635¶

In [2]:
import math
#Operating Conditions
Ts = 255+273.    					;#[K] Surface Temperature
Tsat = 100+273.    					;#[K] Saturated Temperature
D = 6*math.pow(10,-3)            	;#[m] Diameter of pan
e = 1            					;# emissivity
stfncnstt=5.67*math.pow(10,(-8))    ;# [W/m^2.K^4] - Stefan Boltzmann Constant
g = 9.81          					;#[m^2/s] gravitaional constant
#Table A.6 Saturated water Liquid Properties T = 373 K
rhol = 957.9            			;#[kg/m^3] Density
hfg = 2257*math.pow(10,3)        	;#[J/kg] Specific Heat
#Table A.4 Water Vapor Properties T = 450 K
rhov = .4902            			;#[kg/m^3] Density
cpv = 1.98*math.pow(10,3)           ;#[J/kg.K] Specific Heat
kv = 0.0299                			;#[W/m.K] Conductivity
uv = 15.25*math.pow(10,-6)          ;#[N.s/m^2] Viscosity
#calculations

Te = Ts-Tsat;

hconv = .62*math.pow((kv*kv*kv*rhov*(rhol-rhov)*g*(hfg+.8*cpv*Te)/(uv*D*Te)),.25);

#From eqn 10.9 h^(4/3) = hconv^(4/3) + hrad*h^(1/3)
#Newton Raphson
h=250.;        						#Initial Assumption
while 1>0 :
f = math.pow(h,(4./3.)) - (math.pow(hconv,(4./3.)) + math.pow(hrad*h,(1./3.)));
hn=h-f/fd;
if z < .01:
break;
h=hn;

q = h*math.pi*D*Te; 				#power dissipation
#results

print '%s %d %s' %("\n Power Dissipation per unith length for the cylinder, qs= ",q,"W/m");
print '%s' %("The answer is a bit different due to rounding off error")
#END

 Power Dissipation per unith length for the cylinder, qs=  730 W/m
The answer is a bit different due to rounding off error


## Example 10.3 Page 648¶

In [3]:
import math
#Operating Conditions
Ts = 50+273.    			;#[K] Surface Temperature
Tsat = 100+273.    			;#[K] Saturated Temperature
D = .08            			;#[m] Diameter of pan
g = 9.81          			;#[m^2/s] gravitaional constant
L = 1                		#[m] Length
#Table A.6 Saturated Vapor Properties p = 1.0133 bars
rhov = .596            		;#[kg/m^3] Density
hfg = 2257*1000.        	;#[J/kg] Specific Heat
#Table A.6 Saturated water Liquid Properties T = 348 K
rhol = 975.            		;#[kg/m^3] Density
cpl = 4193.             	; #[J/kg.K] Specific Heat
kl = 0.668               	;#[W/m.K] Conductivity
ul = 375*math.pow(10,-6)    ;#[N.s/m^2] Viscosity
#calculations

uvl = ul/rhol           	;#[N.s.m/Kg] Kinematic viscosity
Ja = cpl*(Tsat-Ts)/hfg;
hfg2 = hfg*(1+.68*Ja);

#Equation 10.43
Re = math.pow((3.70*kl*L*(Tsat-Ts)/(ul*hfg2*math.pow((uvl*uvl/g),.33334))+4.8),.82); #Reynolds number

#From equation 10.41
hL = Re*ul*hfg2/(4*L*(Tsat-Ts)); 		#Transfer coefficient
q = hL*(math.pi*D*L)*(Tsat-Ts); 		#Heat transfer rate

m = q/hfg;								#Rate of condensation
#Using Equation 10.26
delta = math.pow((4*kl*ul*(Tsat-Ts)*L/(g*rhol*(rhol-rhov)*hfg2)),.25);
#results

print '%s %.2f %s %.4f %s' %("\n Heat Transfer Rate = ",q/1000.,"kW and Condensation Rates=",m," kg/s");
print '%s %.3f %s %.2f %s' %("\n And as del(L)", delta*1000,"<< (D/2)", D/2. ,"m use of vertical cylinder correlation is justified");
#END

 Heat Transfer Rate =  66.62 kW and Condensation Rates= 0.0295  kg/s

And as del(L) 0.218 << (D/2) 0.04 m use of vertical cylinder correlation is justified


## Example 10.4 Page 652¶

In [4]:
import math
#Operating Conditions
Ts = 25+273.    					;#[K] Surface Temperature
Tsat = 54+273.    					;#[K] Saturated Temperature
D = .006          					;  #[m] Diameter of pan
g = 9.81          					;#[m^2/s] gravitaional constant
N = 20                				# No of tubes

#Table A.6 Saturated Vapor Properties p = 1.015 bar
rhov = .098            				;#[kg/m^3] Density
hfg = 2373*1000.        			;#[J/kg] Specific Heat
#Table A.6 Saturated water Liquid Properties Tf = 312.5 K
rhol = 992.            				;#[kg/m^3] Density
cpl = 4178.              			;#[J/kg.K] Specific Heat
kl = 0.631              			; #[W/m.K] Conductivity
ul = 663*math.pow(10,-6)           	; #[N.s/m^2] Viscosity
#calculations

Ja = cpl*(Tsat-Ts)/hfg;
hfg2 = hfg*(1+.68*Ja); 				#Coefficient of condensation
#Equation 10.46
h = .729*math.pow((g*rhol*(rhol-rhov)*kl*kl*kl*hfg2/(N*ul*(Tsat-Ts)*D)),.25);
#Equation 10.34
m1 = h*(math.pi*D)*(Tsat-Ts)/hfg2;	#Average condensation rate

m = N*N*m1;							#Rate per unit length
#results

print '%s %.3f %s' %("\n For the complete array of tubes, the condensation per unit length is",m ," kg/s.m");
#END

EXAMPLE 10.4   Page 652

For the complete array of tubes, the condensation per unit length is 0.463  kg/s.m