# Chapter 9 Condensation and Boiling Heat Transfer¶

## Exa 9.1¶

In :
#Example Number 9.1
# condensation on vertical plate

#Variable declaration

# we have to check the reynolds no. to that film is laminar or turbulent
Tf = (100+98)/2 			# [degree celsius]
Tw = 98 				# [degree celsius]
RHOf=960 				# [kg/cubic meter]
MUf=2.82*10**(-4) 			# [kg/m s]
Kf=0.68 				# [W/m degree celsius]
g=9.81 					# [m/s**(2)]
L=0.3 					# [m]
# RHOf(RHOf-RHOv)~RHOf**(2)
# let us assume laminar film condensate
Tsat=100 				# [degree celsius]
Tg=100 					# [degree celsius]
Hfg=2255*10**(3) 			# [J/kg]

#Calculation

hbar=0.943*((RHOf**(2)*g*Hfg*Kf**(3)/(L*MUf*(Tg-Tw)))**(0.25)) 	# [W/sq m deg celsius]
h=hbar 					# [W/square meter degree celsius]
# checking reynolds no. with equation(9-17)
Ref=4*h*L*(Tsat-Tw)/(Hfg*MUf)

print "Value of reynolds no. is",round(Ref,1),"so the laminar assumption was correct"
# the heat transfer is now calculated from
A=0.3*0.3 				# [square meter]
q=hbar*A*(Tsat-Tw) 			# [W]
mdot=q/Hfg 				# [kg/h]

#Result

print "The heat transfer is",round(q),"w"
mdot=mdot*3600 				# [kg/h]
print "Total mass flow condensate is",round(mdot,2),"kg/h"

Value of reynolds no. is 49.7 so the laminar assumption was correct
The heat transfer is 2368.0 w
Total mass flow condensate is 3.78 kg/h


## Exa 9.2¶

In :
#Example Number 9.2
# condensation on tube tank

# Variable declaration

# the condensate properties are obtained from previous example
# replacing L by n*d
Tw=98 				# [degree celsius]
RHOf=960 			# [kg/cubic meter]
MUf=2.82*10**(-4) 		# [kg/m s]
Kf=0.68 			# [W/m degree celsius]
g=9.81 				# [m/s^(2)]
Tsat=100 			# [degree celsius]
Tg=100 				# [degree celsius]
Hfg=2255*10**(3) 		# [J/kg]
d=0.0127 			# [m]
n=10

#Calculation

hbar=0.725*((RHOf**(2)*g*Hfg*Kf**(3)/(n*d*MUf*(Tg-Tw)))**(0.25)) # [W/sq m deg C]
# total surface area is
n=100
Al=n*22*d/7 			# [square meter]
print "Total surface area is",round(Al,2),"square meter/m"
# so the heat transfer is
Ql=hbar*Al*(Tg-Tw)		 # [W]

#Result

print "Heat transfer is",round(Ql/1000,2),"kW/m"
# total mass flow of condensate is then
mdotl=Ql/Hfg 			# [kg/h]
mdotl=mdotl*3600		# [kg/h]
print "Total mass flow of condensate is",round(mdotl,1),"kg/h"

Total surface area is 3.99 square meter/m
Heat transfer is 100.1 kW/m
Total mass flow of condensate is 159.8 kg/h


## Exa 9.4¶

In :
#Example Number 9.4
# Flow boiling
# Variable declaration

p =0.5066			# [MPa] pressure of water
d = 0.0254 			# [m] diameter of tube
Tw = 10.0 			# [degree celsius]
# for calculation we use equation (9-45), noting that
dT = 10.0 			# [degree celsius]
# the heat transfer coefficient is calculated as

import math
h = 2.54*Tw**(3)*math.exp(p/1.551) 	# [W/square meter degree celsius]

# the surface area for a 1-m length of tube is
L = 1 				# [m]
import math
A = math.pi*d*L 		# [square meter]

# so the heat transfer is
q = h*A*dT 			# [W/m]
print "Heat transfer in a 1.0 m length of tube is",round(q),"W/m"

Heat transfer in a 1.0 m length of tube is 2810.0 W/m


## Exa 9.5¶

In :
#Example Number 9.5
# water boiling in a pan

# Variable declaration

p = 0.101				# [MPa] pressure of water
dT_x = 8 				# [degree celsius]
p1 = 0.17 				# [MPa] given operating pressure
# we will use the simplified relation of table 9-13(page no.-506) for the 		estimates.we do not know the value of q_by_A and so must choose one of the two 	relation for a horizontal surface from the table
# we anticipate nucleate boiling, so choose
h = 5.56*dT_x**(3) 			# [W/square meter degree celsius]
# and the heat flux is
q_by_A = h*dT_x 			# [W/square meter]
# for operation as a pressure cooker we obtain the value of h from 		equation(9-44)

#Calculation

hp = h*(p1/p)**(0.4) 			# [W/square meter degree celsius]
# the corresponding heat flux is
q_by_A1 = hp*dT_x 			# [W/square meter]

#Result

print "Heat flux obtained is",round(q_by_A/1000,1),"kW/square meter"
per_inc = 100*(q_by_A1-q_by_A)/q_by_A

print "If the pan operates as a pressure cooker at 0.17 MPa the increase in heat flux is",round(per_inc),"percent"

Heat flux obtained is 22.8 kW/square meter
If the pan operates as a pressure cooker at 0.17 MPa the increase in heat flux is 23.0 percent


## Exa 9.6¶

In :
#Example Number 9.6
# heat-flux comparisons

# Variable declaration

Tw = 200.0 				# [degree celsius] water temperature
L = 0.08 				# [m] length of solid copper bar
dT = 100.0 				# [degree C] temp differential in copper bar
#using the data of table 9-4(page no.-508)
# the heat flux per unit area is expressed as q_by_A = -k*del_T/dx
# from table A-2(page no.-) the thermal conductivity of copper is
k = 374.0 				# [W/m degree celsius]

#Calculation

q_by_A = -k*(-dT)/L 			# [W/square meter]

# from table 9-4(page no.-508) the typical axial heat flux for a water heat 		flux for a water heat pipe is
q_by_A_axial = 0.67 			# [kW/csquare meter]
q_by_A = q_by_A/(1000*10**(4.0)) 	# [kW/csquare meter]
time=q_by_A_axial/q_by_A

#Result

print "Thus the heat transfers ",round(time),"times the heat of a pure copper rod with a substantial temperature gradient"

Thus the heat transfers  14.0 times the heat of a pure copper rod with a substantial temperature gradient