# Chapter 10 : Unsteady State And Multidimensional Heat Conduction¶

## Example 10.8 Page No : 444¶

In [1]:
# Variables
l = 0.05                       			#m,thickness of margarine  slab
ro = 990.                       			#Kg/m**3, density of  margarine slab
cp = 0.55                      			#Kcal/kg C, ddpecific heat of slab
k = 0.143                      			#kcal/h mC,   thermal conductivity of slab
Ti = 4.                         			#C, initial temp
To = 25.                        			#C, ambient temp.
t = 4.                          			#hours, time
h = 8.                          			#kcal/h m**2 C

#calculation
Fo = k*t/(ro*cp*l**2)          			#, fourier no.
Bi = h*l/k                     			#Biot no.
#from fig. 10.6 a
Tcbar = 0.7                    			#Tcbar = (Tc-To)/(Ti-To)
Tc = To+Tcbar*(Ti-To)          			#C, centre temp.
#from fig 10.6 b
#(T-To)/(Tc-To) = 0.382
T = 0.382*(Tc-To)+To           			#c,top surface temp.
#again from fig. 10.6 b
Tm = 0.842*(Tc-To)+To          			#, mid plane temp.

# Results
print "The bottom surface temperature of given slab is %.1f C"%(Tc);
print "The top surface temperature of given slab is %.1f C"%(T);
print "The  mid plane temperature of given slab is %.1f C"%(Tm);

The bottom surface temperature of given slab is 10.3 C
The top surface temperature of given slab is 19.4 C
The  mid plane temperature of given slab is 12.6 C


## Example 10.9 Page No : 449¶

In [2]:
import math
# Variables
Ti = 870.                             			#C, initial temp.
To = 30.                              			#C, ambient  temp.
Tc = 200.                             			#C, centre line temp.
h = 2000.                              			#W/m**2 C, surface heat transfer coefficient
a = 0.05                             			#m, radius of cylinder
k = 20.                               			#W/m C, thermal conductivity
ro = 7800.                            			#kg/m**3, density
cp = 0.46*10**3                      			#j/kg C,  specific heat

#calculation
#i
Bi = h*a/k                          			#Biot no.
alpha = k/(ro*cp)                     			#m**2/C, thermal diffusivity
Tcbar = (Tc-To)/(Ti-To)             			# dimensionless centre line temp.
#from fig 10.7 a
fo = 0.51                           			#fourier  no. fo = alpha*t/a**2
t = fo*a**2/alpha                    			#s, time

#ii
#at the half radius, r/a = 0.5 & Bi = 5
T = To+0.77*(Tc-To)                 			#from fig. 10.7 b

#iii
x = Bi**2*fo
#for x  = 12.75 & Bi = 5.0. fig.10.9 b gives
#q/qi = 0.83
qi =  math.pi*a**2*(1)*ro*cp*(Ti-To)   			#kj, initial amount of heat energy
#present in 1 m length of shaft
q = 0.83*qi                        			#j, amount of heat transfered

# Results
print "i) time required for the cantre-line temp.to drop down to 200 C is %.0f s"%(t);
print "ii)the temp. at half radius  at that moment is %.0f C "%(T);
print "iii)the amount of heat that has been  transfered to the liquid is  %d Kj"%(q*10**-3)

i) time required for the cantre-line temp.to drop down to 200 C is 229 s
ii)the temp. at half radius  at that moment is 161 C
iii)the amount of heat that has been  transfered to the liquid is  19647 Kj