In [1]:

```
from __future__ import division
import math
#Variables
T2=363; # temp. of strip,K
T1=373; #saturated temp.,K
p=1.013*10**5; #pressure of water,N/m**2
psat=1.203*10**5; #saturated pressure at 108 C,N/m**2
psat1=1.769*10**5; #saturated pressure at 116 C,N/m**2
a=57.36*10**-3; #surface tension, N/mat Tsat=108 C
a1=55.78*10**-3; #surface tension, N/mat Tsat=116 C
#Calculations
Rb=2*a/(psat-p)*1000; #bulk radius at 108 C, mm
Rb1=2*a1/(psat1-p)*1000; # bulk radius at 116 C, mm
#Results
print "Bulk radius at 108 C is :",Rb," mm\n"
print "Bulk radius at 116 C is :",Rb1,"mm\n"
print "This means that the active nucleation sites would be holes with diameters very roughly on the order magnitude of 0.005 mm atleast on the heater .that is within the ransge of roughness of commercially finished surfaces."
```

In [27]:

```
from __future__ import division
import math
#Variables
q=800; #power delivered per unit area,KW/m**2
T1=373; #saturated temp.of water, K
delT=22; # temp. difference,K
Cp=4.22; #heat capacity of water,kj/(kg*K)
Pr=1.75; #prandtl no.
a=958; #desity difference,kg/m**3
s=0.0589; #surface tension,kg/s**2
Hfg=2257; #latent heat,kj/kg
drho=958; #
mu=0.0000282; #
g=980;
#Calculations
#by using rohensow correlation applied data for water boiling on 0.61 mm diameter platinum wire
RHS=mu*Cp**3/(Hfg**2*Pr**3)*math.sqrt(g*(drho)/s) #RHS of equation 9.4
Csf=(RHS*(delT)**3/(q))**(1/3); #surface correction factor of the heater surface
#Results
print "Surface correction factor of the heater surface is :",round(Csf,5),", this value compares favorably with Csf for a platinum or copper surface under water.\n"
```

In [3]:

```
from __future__ import division
import math
#Variables
p=1.013*10**5; #pressure of water,N/m**2
D=0.1; #inside diameter,m
l=0.04; #wavelength,m
a=0.0589; #surface tension,N/m
b=0.577; #density of gas, kg/m**3
#Calculations
u=math.sqrt(2*math.pi*a/(b*l)); #the flow will be helmholtz stable until the steam velocity reaches this value.
#Results
print "Steam velocity required to destablize the liquid flow is :",round(u,4),"m/s ,beyond that, the liquid will form whitecaps and be blown back upward.\n"
```

In [1]:

```
from __future__ import division
import math
#Variables
a=13600; #desity difference,kg/m**3
s=0.487; #surface tension,kg/s**2
#Calculations
L=2*math.pi*(3**0.5)*math.sqrt(s/(9.8*a))*100; #spacing wavelength,cm
#Results
print "Maximum spacing is :",round(L,4),"cm\n"
print "Actually this spacing would give the maximum rate of collapse.it can be shown that collapse would begin at 1/3^0.5 times this value or at 1.2 cm."
```

In [2]:

```
from __future__ import division
#Variables
T1=373; #saturated temp.of water, K
a=957.6; #desity difference,kg/m**3
s=0.0589; #surface tension,kg/s**2
Hfg=2257*1000; #latent heat,J/kg
a2=0.597; #density of gas, kg/m**3
g=9.8; #Gravitational constant, m/s^2
#Calculations
Qmax=0.149*math.sqrt(a2)*Hfg*(g*a*s)**0.25/1000000;
#Results
print "Peak heat flux is : ",round(Qmax,3),",from figure it can be shown that qmax =1.16 MW/m^2, which is less by only about 8 percent.\n"
```

In [7]:

```
from __future__ import division
#Variables
T1=628; #saturated temp.of water, K
a=13996; #desity difference,kg/m**3
s=0.418; #surface tension,kg/s**2
Hfg=292500; #latent heat,J/kg
a2=4; #density of mercury, kg/m**3
g=9.8; #Gravitational constant, m/s^2
#Calculations
Qmax=0.149*math.sqrt(a2)*Hfg*math.pow((g*a*s),0.25)/(10**6);#Peak heat flux in MW/m^2
#Results
print "Peak heat flux is :",round(Qmax,3),"MW/m^2\n"
print "The result is very close to that for water,the increase in density and surface tension have not been compensated by amuch lower latent heat."
```

In [11]:

```
from __future__ import division
import math
#Variables
T1=373; #saturated temp.of water, K
a=958 ; #desity difference,kg/m**3
s=0.0589; #surface tension,kg/s**2
Hfg=2257000; #latent heat,J/kg
a2=0.597; #density of gas, kg/m**3
A=400*10**-4; #area of mettalic body,m**2
V=0.0006; #volume of body, m**3
g=9.8; #Gravitational constant, m/s^2
#Calculations
Qmax=(0.131*math.sqrt(a2)*Hfg*math.pow((g*a*s),0.25))*0.9*A/1000 ;#large rate of energy removal, KW as the cooling process progresses,it goes through the boiling curve from film boiling,through qmin, up the transitional boiling regime,through qmax and down the3 nucleate boiling curve.
#R=V/A*(9.8*a/s)**0.5 since this value comes out to be 6.0, which is larger than the specified lower bound of about 4.
#to complete the calculation, it is necessary to check whether or not rate is large enough to justify the use.
R=V/A*math.sqrt(g*958/0.0589); #the most rapid rate of heat removal during the quench
#Results
print "The heat flow is :",round(Qmax,1),"KW\n"
print "The most rapid rate of heat removal during the quench is :",round(R),", this is larger than the specified lower bound of about 4.\n"
```

In [14]:

```
from __future__ import division
import math
#Variables
T1=373; #saturated temp.of water, K
a=958; #desity difference,kg/m**3
s=0.0589; #surface tension,kg/s**2
Hfg=2257*1000; #latent heat,J/kg
a2=0.597; #density of gas, kg/m**3
g=9.8; #Gravitational constant, m/s^2
#Calculations
Qmin=0.09*a2*Hfg*(g*a*s/(959**2))**0.25; #Using equation 9.34
#Results
print "peak heat flux is :",round(Qmin),"W/m^2 ,from the figure, we read 20000 W/m^2, which is the same, within the accuracy of graph.\n"
```

In [15]:

```
from __future__ import division
import math
#Variables
T1=480; #bulk temp.of water, K
m=0.6; #mass flow rate of saturated water,kg/s
D=0.05; #diameter of vertical tube,m
p=184000; #heating rate f tube, W/m**2
A=0.001964; #area of the pipe,m**2
Pr=0.892; #prandtl no.
x=0.2; #quality
a1=9.014; #density of gas,kg/m**3
a2=856.5 #density of water, kg/m**3
Hfg=1913*1000; #latent heat,J/kg
muf=1.297*10**-4 #
#Calculations
G=m/A; #superficial mass flux
Relo=G*D/(1.297*10**-4); #reynolds no. for liquid only.
f=1/(1.82/2.303*math.log(Relo)-1.64)**2; # formula for friction factor for smooth pipes
Nu=(f/8*Relo*Pr)/(1.07+12.7*math.sqrt(f/8)*(Pr**(2/3)-1));#formula for nusselt no.in fully developed flow in smooth pipes
hlo=0.659*Nu/D; #heat transfer coefficient,w/(m**2*K)
Co=((1-x)/x)**0.8*math.sqrt(a1/a2); # Convection no.
Bo=p/(G*Hfg); # boiling no.
Hfg1=(1-x)**0.8*(0.6683*Co**(-0.2)+1058*Bo**0.7)*hlo;#heat transfer coefficient for nucleate boiling dominant, w/(m**2*K)
Hfg2=(1-x)**0.8*(1.136*Co**(-0.9)+667.2*Bo**0.7)*hlo;#heat transfer coefficient for connective boiling dominant, w/(m**2*K)
#since the second value is larger,we will use it.
Tw=T1+p/Hfg2; #wall temperature ,K
Tw1=Tw-273; #wall temperature ,C
#Results
print "Wall temperature is :",round(Tw1,3),"C\n"
```