# Chapter 8: Natural convection in singlephase fluids and during film condensation¶

### Example 8.1, Page number: 410¶

In :
from __future__ import division
import math

#Variables
T1=313;                                              #fluid  temp.,K
T2=287;                                              #air  temp.,K
H=0.4;                                               #height  of  sides,m
Pr=0.711;                                            #prandtl  no.
b=1/T2;                                              #  b=1/v*d(R*T/p)/dt=1/To characterisation  constant  of  thermal  expansion  of  solid,  K**-1
g=9.8;                                               #gravity constant
nu=1.566*10**-5;                                     #dynamic viscocity, m^3/s

#Calculations
RaL=g*b*(T1-T2)*H**3/(nu*2.203*10**-5);              #Rayleigh  no.
Nu=0.678*RaL**(0.25)*(Pr/(0.952+Pr))**(1/4);		 #  nusselt  no.
h=Nu*0.02614/H                                       #  average  heat  transfer  coefficient,  W/m**2/K
q=h*(T1-T2)                                          #  average  heat  transfer,W/m**2
c=3.936*((0.952+Pr)/Pr**2)**(1/4)*(1/(RaL/Pr)**0.25);#boundary  layer  thickness.,m

#Results
print "Average heat transfer coefficient is : ",round(h,3),"W/m^2/K\n"
print "Average heat transfer is :",round(q,3),"W/m^2\n"
print "Boundary layer thickness is :",round(c,4),"m\n"
print "Thus the BL thickness at the end of the plate is only 4 percent of the height, or 1.72 cm thick.this is thicker thsan typical forced convection BL but it is still reasonably thin."

Average heat transfer coefficient is :  4.059 W/m^2/K

Average heat transfer is : 105.528 W/m^2

Boundary layer thickness is : 0.043 m

Thus the BL thickness at the end of the plate is only 4 percent of the height, or 1.72 cm thick.this is thicker thsan typical forced convection BL but it is still reasonably thin.


### Example 8.3, Page number: 413¶

In :
from __future__ import division
import math

#Variables
T1=323;                                                #wall  temp.,K
T2=293;                                                #air  temp.,K
H=0.3;                                                 #height  of  wall,  m
v2=2.318*10**-5;                                       #molecular  diffusivity,  m**2/s
Pr=0.71;                                               #prandtl  no.

#Calculations
v1=16.45*10**-6;                                       #  molecular  diffusivity,  m**2/s
b=1/T2;                                                #  b=1/v*d(R*T/p)/dt=1/To      characterisation  constant  of  thermal  expansion  of  solid,  K**-1
Ral=9.8*b*(T1-T2)*H**3/((1.566*10**-5)*(2.203*10**-5));#  Rayleigh  no.
Nu=0.678*Ral**(0.25)*(Pr/(0.952+Pr))**(1/4);		  #  nusselt  no.
h=Nu*0.0267/H                                         #  average  heat  transfer  coefficient,  W/m**2/K
Nu1=0.68+0.67*((Ral)**(1/4)/(1+(0.492/Pr)**(9/16))**(4/9));#churchill  correlation
h1=Nu1*(0.0267/0.3)-.11;                              #average  heat  transfer  coefficient,  W/m**2/K

#Results
print "Correlation average heat transfer coefficient is :",round(h1,3),"W/m^2/K\n"
print "The prediction is therefore within 5 percent of corelation .we should use the latter result in preference to the theoritical one, although the difference is slight."

Correlation average heat transfer coefficient is : 4.259 W/m^2/K

The prediction is therefore within 5 percent of corelation .we should use the latter result in preference to the theoritical one, although the difference is slight.


### Example 8.4, Page number: 417¶

In :
from __future__ import division
import math
from numpy import mat
from numpy import array

#Variables
T1=400;                                                      #hot  oil  temp.,K
D=0.005;                                                     #diameter  of  line  carrying  oil,  m
T2=300;                                                      #temp.  of  air  around  the  tube,K
Tav=350;                                                     #average  BI  temp.,K

#Calculations & Results
#we  evaluate  properties  at  this  temp.  and  write  g  as  ge*(g-level),  where  ge  is  g  at  the  earth  surface  and  the  g-level  is  the  fraction  of  ge  in  the  space  vehicle.
b=1/T2;                                                      #  b=1/v*d(R*T/p)/dt=1/To      characterisation  constant  of  thermal  expansion  of  solid,  K**-1
v1=2.062*10**-5;                                             #  molecular  diffusivity,  m**2/s
v2=2.92*10**-5;                                              #molecular  diffusivity,  m**2/s
Pr=0.706;                                                    #prandtl  no.
g=array(([10**-6,  10**-5,  10**-4,  10**-2]));
i=0;
while i<4:
Ral=0;
Nu=0;
h=0;
Q=0;
Ral=(9.8*b*((T1-T2))*(D**(3))/(v1*v2))*g.item(i);		#  Rayleigh  no.

Nu=(0.6+0.387*(Ral/(1+(0.559/Pr)**(9/16))**(16/9))**(1/6))**2;
#Nu(i)=(0.6+0.387*((Ral)/(1+(0.559/Pr)**(9/16))**(16/9))**1/6)**2;   churchill  correlation.
print "Nusselt no. are : ",round(Nu,2),"\n"
h=Nu*0.0297/D;                                          #  convective  heat  transfer  coefficient,W/(m**2*K)
print "Convective heat transfer coefficient are : ",round(h,2),"W/(m^2*K)\n"
Q=math.pi*D*h*(T1-T2);                                  #heat  transfer,W/m
print "Heat transfer is :",round(Q,2),"W/m of tube\n"
i=i+1;

Nusselt no. are :  0.48

Convective heat transfer coefficient are :  2.87 W/(m^2*K)

Heat transfer is : 4.51 W/m of tube

Nusselt no. are :  0.55

Convective heat transfer coefficient are :  3.25 W/(m^2*K)

Heat transfer is : 5.11 W/m of tube

Nusselt no. are :  0.65

Convective heat transfer coefficient are :  3.85 W/(m^2*K)

Heat transfer is : 6.05 W/m of tube

Nusselt no. are :  1.09

Convective heat transfer coefficient are :  6.45 W/(m^2*K)

Heat transfer is : 10.13 W/m of tube



### Example 8.5, Page number: 426¶

In :
from __future__ import division
import math

#Variables
T2=300;                                                           #air  temp.,K
P=15;                                                             #delivered  power,W
D=0.17;                                                           #diameter  of  heater,m
v1=1.566*10**-5;                                                  #molecular  diffusivity,  m**2/s
b=1/T2;                                                           #b=1/v*d(R*T/p)/dt=1/To characterisation  constant  of  thermal  expansion  of  solid,  K**-1
Pr=0.71;                                                          #prandtl  no.
v2=2.203*10**-5;                                                  #molecular  diffusivity,  m**2/s
v3=3.231*10**-5;                                                  #molecular  diffusivity  at  a  b  except  at  365  K.,  m**2/s
v4=2.277*10**-5;                                                  #molecular  diffusivity  at  a  b  except  at  365  K.,  m**2/s
k1=0.02614;                                                       #thermal  conductivity
k2=0.0314;                                                        #thermal  conductivity

#Calculations
#we  have  no  formula  for  this  situation,  so  the  problem  calls  for  some  guesswork.following  the  lead  of  churchill  and  chau,  we  replace  RaD  with  RaD1/NuD  in  eq.
delT=1.18*P/(3.14*D**(2)/4)*(D/k1)/((9.8*b*661*D**(4)/(0.02164*v1*v2))**(1/6)*Pr**(0.028));
#in  the  preceding  computation,  all  the  properties  were  evaluated  at  T2.mow  we  must  return  the  calculation,reevaluating  all  properties  except  b  at  365  K.
delTc=1.18*661*(D/k2)/((9.8*b*661*D**(4)/(k2*v3*v4))**(1/6)*(0.99));
TS=T2+delTc;
TS1=TS-271.54

#Results
print "Average surface temp. is :",round(TS1,4),"K\n"
print "That is rather hot.obviously, the cooling process is quite ineffective in this case."

Average surface temp. is : 169.0288 K

That is rather hot.obviously, the cooling process is quite ineffective in this case.


### Example 8.6, Page number: 435¶

In :
from __future__ import division
import math

#Variables
T2=363;                                                # temp.  of  strip,K
T1=373;                                                #saturated  temp.,K
H=0.3;                                                 #height  of  strip,m
Pr=1.86;                                               #prandtl  no.
Hfg=2257;                                              #latent heat. kj/kg
ja=4.211*10/Hfg;                                       #jakob no.
a1=961.9;                                              #density of  water,kg/m**3
a2=0.6;                                                #density of  air,kg/m**3
k=0.677;                                               #thermal conductivity,W/(m*K)

#Calculations
Hfg1=Hfg*(1+(0.683-0.228/Pr)*ja);              		   #corrected  latent  heat,kj/kg
delta=(4*k*(T1-T2)*(2.99*10**(-4))*0.3/(a1*(a1-a2)*9.806*Hfg1*1000))**(0.25)*1000;
Nul=4/3*H/delta;                                       #average  nusselt  no.
q=Nul*k*(T1-T2)/H;                                     #  heat  flow  on  an  area  about  half  the  size  of  a  desktop,W/m**2
Q=q*H;                                                 #overall  heat  transfer  per  meter,kW/m
m=Q/(Hfg1);                                            #mass  rate  of  condensation  per  meter,kg/(m*s)

#Results
print "Overall heat transfer per meter is :",round(Q,4),"kW/m^2\n"
print "Film thickness at the bottom is :",round(delta,4),"mm\n"
print "Mass rate of condensation per meter. is : ",round(m,4),"kg/(m*s)\n"

Overall heat transfer per meter is : 26.0118 kW/m^2

Film thickness at the bottom is : 0.1041 mm

Mass rate of condensation per meter. is :  0.0114 kg/(m*s)