# Chapter 6: Heat Transfer by Natural convection¶

## Example 6.1, Page no:258¶

In :
import math
from __future__ import division

#Variable declaration
H = 0.5 ; #m
Th = 100; #degree C
Tl = 40; #degree C
Pr = 0.694;
k = 0.0297; #W/m K

#calculations
v = 20.02*10**-6 ; #m/s
T = (Th+ Tl )/2 + 273 ; #K
B = 1/T;
Gr = 9.81* B *(( Th - Tl )*H**3) /(v**2) ;
Ra = Gr*Pr;
Nua = 0.64*( Gr**(1/4) )*( Pr**0.5) *((0.861+ Pr)**( -1/4) ); #Exact analysis
Nub = 0.68*( Gr**(1/4) )*( Pr**0.5) *((0.952+ Pr)**( -1/4) ); #Integral method
Nuc = 0.59*( Ra)**(1/4) ; #McAdams correlation
Nud = 0.68 + 0.670*( Ra**(1/4) ) /(1+(0.492/ Pr)**(9/16))**(4/9) ; #Churchill and Chu correlation

#result
print"Mean film temperature =",round(T,4),"K";
print" ( a ) ";
print"Exact analysis";
print"NuL =",round(Nua,4);
print"\n ( b ) ";
print"Integral method";
print"NuL =",round(Nub,4);
print"\n ( c ) ";
print"NuL =",round(Nuc,4);
print"\n ( d ) ";
print"Churchill and Chu correlation";
print"NuL =",round(Nud,4);

Mean film temperature = 343.0 K
( a )
Exact analysis
NuL = 72.6196

( b )
Integral method
NuL = 76.0691

( c )
NuL = 81.9066

( d )
Churchill and Chu correlation
NuL = 71.887


## Example 6.3, Page no:260¶

In :
import math
from __future__ import division

#Variable declaration
s = 0.2 ; #m
d = 0.005 ; #m
rho = 7900 ; #kg/m^3
Cp = 460 ; #J/kg K
Tair = 20 ; #C
Tavg = 380 ; #C
Pr = 0.680 ;
k = 0.0393 ; #W/m K
h2 = 7.348 ; #W/m^2 K
h3 = 6.780; #W/m^2 K

#calculations
Tm = ( Tavg + Tair )/2 ; #C
v = 34.85*10**-6 ; #m^2/s
Re = 9.81*1/(273+ Tm)*( Tavg - Tair )*(s**3) /(v**2) *Pr;
Nu = 0.68 + 0.670*( Re**(1/4) ) /(1+(0.492/ Pr)**(4/9))**(4/9) ;
h = Nu*k/s; #W/m^2 K
t1 = rho*s*s*d*Cp /(( s**2) *2*h)* math.log ((430 - Tair )/(330 -Tair )); #s
t2 = rho*s*s*d*Cp /(( s**2) *2* h2)* math.log ((330 - Tair )/(230 -Tair )); #s
t3 = rho*s*s*d*Cp /(( s**2) *2* h3)* math.log ((230 - Tair )/(130 -Tair )); #s
time = t1+t2+t3; #Total time
minute = time /60;

#result
print"Time required for the plate to cool from 430 C to 330 C is",round(t1,4),"s";
print"Time required for the plate to cool from 330 C to 230 C is",round(t2,4),"s";
print"Time required for the plate to cool from 230 C to 130 C is",round(t3,4),"s";
print"Hence, time required for the plate to cool from 430 C to 130 C";
print" =",round(time,4),"s";
print" =",round(minute,4),"min";

Time required for the plate to cool from 430 C to 330 C is 328.9673 s
Time required for the plate to cool from 330 C to 230 C is 481.5307 s
Time required for the plate to cool from 230 C to 130 C is 866.4613 s
Hence, time required for the plate to cool from 430 C to 130 C
= 1676.9593 s
= 27.9493 min


## Example 6.4, Page no:264¶

In :
import math
from __future__ import division

#Variable declaration
D = 0.006 ; #m
e = 0.1 ;
Ti = 800 ; #C
Ta = 1000 ; #C
k = 0.0763 ; #W/m K
Pr = 0.717 ;
Ra2 = 6.42 ;
Nu2 = 0.9841 ;
h2 = 12.15 ;
Ra3 = 6.93 ;
Nu3 = 0.9963 ;
h3 = 12.33 ;

#calculations
Tm = (Ta+Ti) /2;
v = 155.1*10**-6 ; #m^2/s
Gr = 9.81*1/1173*(200* D**3) /(v**2) ;
Ra = Gr*Pr ;
Nu = 0.36 + 0.518*( Ra**(1/4) ) /(1+(0.559/ Pr)**(9/16))**(4/9) ;
h = Nu*k/D;
x = h*(Ta -Ti); #W/m^2
x2 = h2 *(900 -800) ;
x3 = h3 *(910 -800) ;
T = 900 + (910 -900) *(1306 - x2)/(x3 -x2); #Interpolation

#result
print"Trial 1";
print"Let Ta =10000 C";
print"Value of h(Ta-800) =",round(x,4),"W/m^2, which is much larger than the required value of 1306 W/m^2";
print"\nTrial 2";
print"Let Ta = 900 C";
print"Value of h(Ta-800) =",round(x2,4),"W/m^2, which is a little less than the required value of 1306 W/m^2";
print"\nTrial 3";
print"Let Ta = 910 C";
print"Value of h(Ta-800) =",round(x3,4),"W/m^2 \nThis value is little more than the required value of 1306 W/m^2";
print"\nThe correct value of Ta obtained by interpolation is",round(T,4),"C";

Trial 1
Let Ta =10000 C
Value of h(Ta-800) = 2722.8585 W/m^2, which is much larger than the required value of 1306 W/m^2

Trial 2
Let Ta = 900 C
Value of h(Ta-800) = 1215.0 W/m^2, which is a little less than the required value of 1306 W/m^2

Trial 3
Let Ta = 910 C
Value of h(Ta-800) = 1356.3 W/m^2
This value is little more than the required value of 1306 W/m^2

The correct value of Ta obtained by interpolation is 906.4402 C


## Example 6.5, Page no:269¶

In :
import math
from __future__ import division

#Variable declaration
Tp = 75 ; #Temperature of absorber plate , degree C
Tc = 55 ; #Temperature of glass cover , degree C
L = 0.025 ; #m
H = 2 ; #m
Y = 70 ; #degree
k = 0.0294 ; #W/m K
Pr = 0.695 ;

#calculations
r = H/L ;
Tavg = ( Tp +Tc) /2+273 ; #K
v = 19.50*10**-6 ; #m^2/s
Ra = 9.81*(1/ Tavg )*( Tp -Tc)*(L**3) /(v**2) *Pr*math.cos(a);
Nu = 0.229*( Ra)**0.252;
h = Nu*k/L ; #W/m^2 K
Rate = h *2*1*( Tp -Tc); #W

#result
print"Heat transfer rate =",round(Rate,4),"W";

Heat transfer rate = 122.8843 W


## Example 6.6, Page no:270¶

In :
import math
from __future__ import division

#Variable declaration
Tair = 30 ;
D = 0.04 ;
Ts = 70 ;
V = 0.3 ;
Pr = 0.698 ;
k = 0.0283 ;

#calculations
Tm = ( Tair + Ts )/2 ;
v = 17.95*10**-6 ;
Gr = 9.81*1/323*( Ts - Tair )*(D**3) /v**2;
Re = V*D/v ;
X = Gr/Re**2 ;
Nuforced = 0.3 + 0.62*( Re**0.5) *( Pr**(1/3) ) /((1+(0.4/Pr)**(2/3) )**(1/4))*(1+( Re /282000)**(5/8))**(4/5) ;
Nu = Nuforced *(1+6.275*( X)**(7/4))**(1/7) ;
h = Nu *(k/D);

#result
print"Since Gr/Re^2 =",round(X,4),"is > 0.2, we have a combined convection situation.";
print"The Average heat transfer coefficient =",round(h,4),"W/m^2 K";

Since Gr/Re^2 = 0.5399 is > 0.2, we have a combined convection situation.
The Average heat transfer coefficient = 10.8276 W/m^2 K