# Chapter 3 Steady State Conduction Multiple Dimension¶

## Exa 3.1¶

In [1]:
#Example Number 3.1
# Calculate the heat loss by the pipe

# Variable declaration

d = 0.15 			# [m] diameter of pipe
r = d/2 			# [m] radius of pipe
L = 4 				# [m] length of pipe
Tp = 75				# [degree celsius] pipe wall temperature
Tes = 5 			# [degree celsius] earth surface temperature
k = 0.8				# [W/m per deg C] thermal conductivity of earth
D = 0.20 			# [m] depth of pipe inside earth

# We may calculate the shape factor for this situation using equation given in 	table 3-1

# since D<3*r
#Calculation
import math
S = (2*math.pi*L)/math.acosh(D/r) 	# [m] shape factor
# the heat flow is calculated from
q = k*S*(Tp-Tes) 			# [W]

#Result
print"Heat lost by the pipe is",round(q,1),"W"

Heat lost by the pipe is 859.9 W


## Exa 3.2¶

In [2]:
#Example Number 3.2
# Calculate heat loss through the walls

# VARIABLE DECLARATION

a = 0.5 	     # [m] length of side of cubical furnace
Ti = 500 	     # [degree celsius] inside furnace temperature
To = 50 	     # [degree celsius] outside temperature
k = 1.04 	     # [W/m per degree celsius] thermal conductivity of fireclay brick
t = 0.10 	     # [m] wall thickness
A = a*a 	     # [square meter] area of one face
# we compute the total shape factor by adding the shape factors 		       	       for the walls, edges and corners

#Calculation
Sw = A/t	     # [m] shape factor for wall
Se = 0.54*a 	     # [m] shape factor for edges
Sc = 0.15*t	     # [m] shape factor for corners

# there are six wall sections, twelve edges and eight corners, so 			the total shape factor S is

S = 6*Sw+12*Se+8*Sc  	# [m]

# the heat flow is calculated as

q = k*S*(Ti-To) 	# [W]

#Result
print"Heat lost through the walls is:",round(q/1000,3),"kW"

Heat lost through the walls is: 8.592 kW


## Exa 3.3¶

In [4]:
#Example Number 3.3
# Calculate the heat loss by the disk

# Variable declaration

import math
d = 0.30 	# [m] diameter of disk
r = d/2 	# [m] radius of disk
Td = 95 	# [degree celsius] disk temperature
Ts = 20 	# [degree celsius] isothermal surface temperature
k = 2.1 	# [W/m per degree celsius] thermal conductivity of medium
D = 1.0 	# [m] depth of disk in a semi-infinite medium
# We have to calculate shape factor using relation given in table (3-1)
# We select the relation for the shape factor is for the case D/(2*r)>1

#Calculation
S = (4*math.pi*r)/((math.pi/2)-math.atan(r/(2*D)))	 # [m] shape factor
# heat lost by the disk is
q = k*S*(Td-Ts) 					 # [W]

#Result
print"Heat lost by disk is:",round(q,2),"W"

Heat lost by disk is: 198.46 W


## Exa 3.4¶

In [5]:
#Example Number 3.4
#Calculate the heat transfer betwwen the disks

# Variable declaration

d = 0.50	 # [m] diameter of both disk
r = d/2 	 # [m] radius of disk
Td1 = 80	 # [degree celsius] first disk temperature
Td2 = 20 	 # [degree celsius] second disk temperature
k = 2.3 	 # [W/m per degree celsius] thermal conductivity of medium
D = 1.5		 # [m] seperation of disk in a infinite medium
# We have to calculate shape factor using relation given in table (3-1)
# We select the relation for the shape factor is for the case D>5*r
#Calculation
import math

S = (4*math.pi*r)/((math.pi/2)-math.atan(r/D)) # [m] shape factor
q = k*S*(Td1-Td2) # [W]

#Result
print"Heat transfer between the disks is:",round(q,1),"W"

Heat transfer between the disks is: 308.4 W