In [1]:

```
#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"
```

In [2]:

```
#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"
```

In [3]:

```
#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"
```

In [4]:

```
#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"
```

In [5]:

```
#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"
```