# Chapter 2 Steady State Conduction One Dimension¶

## Exa 2.1¶

In [1]:
#Example Number 2.1
#Calculate thickness of insulation to reduce heat loss/gain by 80 %

#Variable declaration

dx1 = 0.1	      # [m] thickness of layer of common brick
k1 = 0.7	      # [W/m degree celsius] heat transfer coefficient of common brick
dx2 = 0.0375          # [m] thickness of layer of gypsum plaster
k2 = 0.48 	      # [W/m degree celsius] heat transfer coefficient gypsum plaster

#Calculation

Rb = dx1/k1	      # [sq m degree C /W] thermal resistance of brick
Rp = dx2/k2           # [sq m degree C /W] thermal resistance of gypsum plaster
R = Rb+Rp 	      # [sq m degree C /W] thermal resistance without insulation
R1 = R/0.2 	      # [sq m degree C /W] with insulation

# heat loss with the rock-wool insulation is 20 percent

Rrw = R1-R 	      # [square meter degree celsius /W]
k3 = 0.065 	      # [W/m degree celsius] heat transfer coefficient
dx3 = Rrw*k3          # [m]

#Result

print "Length of thickness is:",round(dx3*100,2)," cm"
Length of thickness is: 5.75  cm

## Exa 2.2¶

In [2]:
#Example Number 2.2
#Calculate heat loss/metre length and tube-insulation interface temperature

#Variable declaration

ID = 0.02 		# [m] inner diameter of steel
OD = 0.04 		#[m] outer diameter of steel
t = 0.03 		#[m] thickness of asbestos insulation
# system is like three concentric cylinders
T1 = 600 		# [degree celsius] inside wall temperature
T2 = 100 		# [degree celsius] outside insulation temperature
Ks = 19 		#[W/m degree celsius] heat transfer coefficient of steel
Ka = 0.2 		# [W/m degree celsius] heat transfer coefficient of asbestos

# heat flow is given by per unit length

#Calculation

import math
Q_l = ((2*math.pi*(T1-T2))/((math.log(OD/ID)/Ks)+(math.log(0.1/OD)/Ka)))    # [W/m]

# above calculated heat flow is used to calculate the interface temperature
# between the outside wall and the insulation

Ta = Q_l*(math.log(0.1/OD)/(2*math.pi*Ka))+T2
# [degree C] Ta is interface temperature

#Result

print "Heat flow is given by:",round(Q_l)," W/m"
print "Interface temperature is",round(Ta),"degree celsius "
Heat flow is given by: 680.0  W/m
Interface temperature is 596.0 degree celsius

## Exa 2.3¶

In [3]:
#Example Number 2.3
# Calculate overall heat transfer coefficient and R Value of wall

# 1. heat transfer through studs for unit depth

#Variable declaration

l = 0.0413 	  # [m] length of wood studs
b = 1.0 	  # [m] unit depth
A = l*b		  # [square meter] area of studs for unit depth
hi = 7.5	  # [W/sq meter/degr C] convectional heat transfer coefficient
ho = 15 	  # [W/sq m per deg C] convectional heat transfer coefficient
Kb = 0.69 	  # [W/m per deg celsius] heat transfer coefficient of brick
Kgi = 0.96 	  # [W/m per deg C] heat transfer coefficient of gypsum inner sheath
Ki = 0.04 	  # [W/m per deg C] heat transfer coefficient of insulation
Kws = 0.1	  # [W/m per deg C] heat transfer coefficient of wood stud
Kgo = 0.48 	  # [W/m per deg C] heat transfer coefficient of gypsum outer sheath
Rair = 1/(ho*A)   # [degree C /W] convection resistance outside of brick
dx_b = 0.08 	  # [m] thickness of brick
dx_os = 0.019 	  #[m] thickness of outer sheet
dx_ws = 0.0921    # [m] thickness of wood stud
dx_is = 0.019     # [m] thickness of inner sheet
Rb = dx_b/(Kb*A)  	# [degr C /W] conduction resistance in brick
Ros = dx_os/(Kgi*A) 	# [deg C /W] conduction resistance through outer sheet
Rws = dx_ws/(Kws*A) 	# [deg C/W] conduction resistance through wood stud
Ris = dx_is/(Kgo*A) 	# [deg C/W] conduction resistance through inner sheet

#Calculation

Ri = 1/(hi*A) 		# [degree celsius /W] convection resistance on inside

Rt = Rair+Rb+Ros+Rws+Ris+Ri
# [deg C/W] total thermal Res through the wood stud section

print "TOTAL THERMAL RESISTANCE THROUGH,"
print"wood stud is:",round(Rt,2),"degree C"

# 2. Heat transfer through insulation section

#Calculation

A1 = 0.406-A		 # [sq meter] area of insulation section for unit depth
dx_ins = 0.0921          # [m] thickness of insulation
Rins = dx_ins/(Ki*A1)    # [deg C /W] conduction resistance through insulation section

# five of the materials are same but resistance involve different area
# i.e. (40.6-4.13) cm instead of 4.13 cm
# so that each of the previous must be multiplied by a factor of 			#(4.13/(40.6-4.13)) = 0.113

#Calculation

Rt_ins = (Rair+Rb+Ros+Ris+Ri)*0.113+Rins
# [deg C/W] total resistance through insulation section

print"Insulation section is:",round(Rt_ins,3),"degree C/W"

R_overall = 1/((1/Rt)+(1/Rt_ins))
# [degree celsius /W] overall resistance for the section

# the value is related to overall heat transfer coefficient by
# Q = U*A*dt = dt/R_overall
# where A is area of total section

A_ = 0.406 		# [sq meter] area of total section
U = 1/(R_overall*A_)    # [W/sq meter deg C] overall heat transfer coefficient
# R value is somewhat different from thermal resistance and is 			#given by
R_value = 1/U 		# [degree celsius square meter/W] R value of system

#Results

print"Overall heat transfer coefficient is:",round(U,3),"W/sq m per deg C"

print"R value is:",round(R_value,3),"sq m/W"
TOTAL THERMAL RESISTANCE THROUGH,
wood stud is: 31.39 degree C
Insulation section is: 7.34 degree C/W
Overall heat transfer coefficient is: 0.414 W/sq m per deg C
R value is: 2.415 sq m/W

## Exa 2.4¶

In [4]:
#Example Number 2.4
# Calculate overall heat transfer coeff and heat loss/unit length at 20 deg C

#Variable declaration

ID = 0.025 		# [m] inner diameter of steel
OD = ID+2*0.0008 	#[m] outer diameter of steel
hi = 3500 		# [W/sq m per deg C] convectional heat transfer coefficient of 			#inside
ho = 7.6 		# [W/sq m per deg C] convectional heat transfer coefficient of 			#outside
L = 1.0 		# [m] tube length
import math
Ai = math.pi*ID*L 	# [sq meter] inside crossectional area
Ao = math.pi*OD*L 	# [sq meter] outside crossectional area
k = 16 			# [W/sq meter per deg C] thermal conductivity of tube

#Calculation

Ri = 1/(hi*Ai)			     # [degree C /W] convection resistance inside tube
Rt = math.log(OD/ID)/(2*math.pi*k*L) # [degree C /W] thermal resistance
Ro = 1/(ho*Ao) 			     # [deg C /W] convection resistance outside tube
R_total = Ri+Rt+Ro		     # [deg C/W] total thermal and convection  				     		     #resistance
Uo = 1/(Ao*R_total) 		     # [W/sq m deg C] overall heat transfer 				     		     #coefficient

Tw = 50				     # [degree C] water temperature
Ta = 20 			     # [degree C] surrounding air temperature
dt = Tw-Ta 			     # [degree C] temperature difference
q = Uo*Ao*dt			     # [W] heat transfer

#Results

print"Overall heat transfer coefficient is:",round(Uo,2)," W/sq m degree C"

print"Heat loss per unit length is:",round(q)," W(for 1m length)"
Overall heat transfer coefficient is: 7.58  W/sq m degree C
Heat loss per unit length is: 19.0  W(for 1m length)

## Exa 2.5¶

In [5]:
#Example Number 2.5
#Calculate the heat loss

#Variable declaration

k = 0.17 		# [W/m per deg C] heat transfer coefficient of asbestos
Tr = 20 		# [degree celsius] temperature of room air
h = 3 			# [W/sq m per deg C] convectional heat transfer coefficient
Tp = 200 		# [degree celsius] temperature of pipe
d = 0.05 		# [m] diameter of pipe

# from equation (2-18) we calculate r_o as

#Calculation

r_o = k/h 		# [m] critical radius of insulation
print"Critical radius of insulation for asbestos is:",round(r_o*100,2),"cm "

Ri = d/2	        # [m] inside radius of insulation
# heat transfer is calculated from equation (2-17)
import math
q_by_L = (2*math.pi*(Tp-Tr))/(((math.log(r_o/Ri))/0.17)+(1/(h*r_o)))
# [W/m] heat transfer per unit length

#Results

print"Heat loss when covered with critical radius of insulation is:",round(q_by_L,1)," W/m"

# without insulation the convection from the outer surface of pipe is

q_by_L1 = h*2*math.pi*Ri*(Tp-Tr)
#[W/m] convection from outer surface without insulation
print"Heat loss without  insulation is:",round(q_by_L1,1)," W/m"
per_inc = ((q_by_L-q_by_L1)/q_by_L1)*100 	# percentage increase in heat transfer

print"Addition of 3.17 of insulation actually increases the heat transfer by:",round(per_inc),"%"
Critical radius of insulation for asbestos is: 5.67 cm
Heat loss when covered with critical radius of insulation is: 105.7  W/m
Heat loss without  insulation is: 84.8  W/m
Addition of 3.17 of insulation actually increases the heat transfer by: 25.0 %

## Exa 2.6¶

In [6]:
#Example Number 2.6
#Calculate the centre temperature of wire

#Variable declaration

# all the power generated in the wire must be dissipated by convection 		to the liquid
# P = i**(2)*R = q = h*A*dt
L = 100 	# [cm] length of the wire
k = 19 		# [W/m per deg C] heat transfer coefficient of steel wire

#Calculations

import math
A = math.pi*(0.15)**(2) 	# [sq m] crossectional area of wire
rho = 70*10**(-6) 		# [micro ohm cm] resistivity of steel
R = rho*L/A 			# [ohm] resistance of wire
i = 200 			# [ampere] current in the wire
P = i**(2)*R 			# [W] power generated in the wire
Tl = 110 			# [degree celsius] liquid temperature
d = 0.003 			# [m] diameter of wire
l = 1 				# [m] length of wire
Tw = (P/(4000*3.14*d*l))+110 	# [degree celsius] wire temperature

# heat generated per unit V q_dot is calculated as
# P = q_dot*V = q_dot*3.14*r**(2)*l
r = d/2 			# [m] radius of wire
q_dot = P/(math.pi*r**(2)*l) 	# [W/m**(3)]
# finally the center temperature of the wire is 					calculated from equation (2-26)

To = ((q_dot*(r**(2)))/(4*k))+Tw	 # [degree celsius]

#Result

print "Center temperature of the wire is:",round(To,1) ,"degree celsius"
Center temperature of the wire is: 231.7 degree celsius

## Exa 2.8¶

In [7]:
#Example Number 2.8
# Calculate heat loss per unit depth of the material

#Variable declaration

t = 0.003 		# [m] thickness of fin
L = 0.075 		# [m] length of fin
Tb = 300 		# [degree celsius] base temperature
Tair = 50 		# [degree celsius] ambient temperature
k = 200 		# [W/m per deg C] heat transfer coefficient of aluminium fin
h = 10 			# [W/sq m per deg C] convectional heat transfer coefficient
# We Will use the approximate method of solution by extending 			the fin
# With a fictitious length t/2
# using equation(2-36)

#Calculation
Lc = L+t/2 		# [m] corrected length
z = 1 			# [m] unit depth
p = (2*z+2*t) 		# [m] perimeter of fin
A = z*t 		# [square meter] crossectional area of fin
m = ((h*p)/(k*A))**(0.5)

# from equation(2-36)
dt = Tb-Tair 					# [degree C] temperature difference
import math
q = math.tanh(m*Lc)*((h*p*k*A)**(0.5))*dt	# [W/m] heat transfer per unit length

#Results
print "Heat loss from the fin per unit length is",round(q),"W/m"
Heat loss from the fin per unit length is 360.0 W/m

## Exa 2.9¶

In [8]:
#Example Number 2.9
# Calculate the heat loss per fin

#Variable declaration

t = 0.001 			# [m] thickness of fin
L = 0.015 			# [m] length of fin
Ts = 170			# [degree celsius] surface temperature
Tfluid = 25 			# [degree celsius] fluid temperature
k = 200 			# [W/m per deg C] heat transfer coefficient of 						  aluminium fin
h = 130 			# [W/sq m per deg C] 				  convectional heat transfercoefficient
d = 0.025 			# [m] tube diameter
Lc = L+t/2			# [m] corrected length
r1 = d/2 			# [m] radius of tube
r2_c = r1+Lc 			# [m] corrected radius

#Calculation

Am = t*(r2_c-r1) 		# [sq m] profile area
c = r2_c/r1 			# constant to determine	efficiency of fin from curve

c1 = ((Lc)**(1.5))*((h/(k*Am))**(0.5)) 	# constant to determine efficiency of fin from 					curve

# using c and c1 to determine the efficiency 						of the fin from figure (2-12)
# we get nf = 82 percent
# heat would be transferred if the entire fin were at 					the base temperature
# both sides of fin exchanging heat
import math
q_max = 2*math.pi*(r2_c**(2)-r1**(2))*h*(Ts-Tfluid) 	# [W] maximum heat transfer
q_act = 0.82*q_max 					#[W] actual heat transfer

#Result
print"The actual heat transferred is:",round(q_act,2)," W"
The actual heat transferred is: 60.97  W

## Exa 2.10¶

In [10]:
#Example Number 2.10
#Obtain an expression for the temperature distribution in rod

#Variable declaration & Calculations

# q_dot is uniform heat source per unit volume
# h is convection coefficient
# k is heat transfer coefficient
# A is area of crossection
# P is perimeter
# Tinf is environment temperature
# we first make an energy balance on the element of the rod shown in 	figure(2-10)
# energy in left place + heat generated in element  =  energy out right face + 	energy lost by convection
# or
# -(k*A*dT_by_dx)+(q_dot*A*dx) = 	-(k*A(dT_by_dx+(d2T_by_dx2)*dx))+h*P*dx*(T-Tinf)
# simlifying we have
# d2T_by_dx2-((h*P)/(k*A))*(T-Tinf)+q_dot/k = 0
# replacing theta = (T-Tinf) and (square meter) = ((h*P)/(k*A))
# d2theta_by_dx2-(square meter)*theta+q_dot/k = 0
# we can make a further substitution as theta = theta-(q_dot/(k*(square 	meter)))
# so that our differential equation becomes
# d2theta_by_dx2-(square meter)*theta
# which has the general solution theta = C1*exp^(-m*x)+C2*exp^(m*x)
# the two end temperatures are used to establish the boundary conditions:
# theta = theta1 = T1-Tinf-q_dot/(k*(square meter)) = C1+C2
# theta = theta2 = T2-Tinf-q_dot/(k*(square meter)) = 	C1*exp^(-m*L)+C2*exp^(m*L)
# solving for the constants C1 and C2 gives
#((theta1*exp^(2*m*L)-theta2*exp^(m*L))*exp^(-m*x))+((theta2exp^(m*L)-theta1)exp^(m*x))/(exp^(2*m*L)-1))

#RESULTS

print"The expression for the temperature distribution in the rod is "

print"theta=(((theta1*exp^(2*m*L)-theta2*exp^(m*L))*exp^(-m*x))+((theta2exp^(m*L)-theta1)exp^(m*x))/(exp^(2*m*L)-1))"

print"for an infinitely long heat generating fin with the left end maintained at T1, the temperature distribution becomes "

print"theta/theta1 = exp^(-m*x)"
The expression for the temperature distribution in the rod is
theta=(((theta1*exp^(2*m*L)-theta2*exp^(m*L))*exp^(-m*x))+((theta2exp^(m*L)-theta1)exp^(m*x))/(exp^(2*m*L)-1))
for an infinitely long heat generating fin with the left end maintained at T1, the temperature distribution becomes
theta/theta1 = exp^(-m*x)

## Exa 2.11¶

In [11]:
#Example Number 2.11
# Calculate the axial heat flow and temperature drop across the contact surface

#Variable declaration

d = 0.03		 #[m] diameter of steel bar
l = 0.1                  #[m] length of steel bar
import math
A = (math.pi*d**(2))/4	 # [square meter] crossectional area of bar
k = 16.3 		 # [W/sq m per degree celsius] thermal conductivity of tube
hc = 1893.93		 # [W/sq m per degree celsius] contact coefficient
# the overall heat flow is subjected to three thermal resistances
# one conduction resistance for each bar
# contact resistance
#Calculation

Rth = l/(k*A)            # [degree celsius /W]

# from table(2-2) the contact resistance is
Rc = 1/(hc*A) 	         # [degree celsius /W]
Rt = 2*Rth+Rc		 # [degree celsius /W] total resistance
dt = 100		 # [degree celsius] temperature difference
q = dt/Rt 		 # [W] overall heat flow

#Results

print "Overall heat flow is:",round(q,2),"W"

# temperature drop across the contact is found by taking the ratio
# of the contact resistance to the total thermal resistance

dt_c = (Rc/(2*Rth))*dt			 # [degree celsius]

print "The temperature drop across the contact is:",round(dt_c,2),"degree celsius"
Overall heat flow is: 5.52 W
The temperature drop across the contact is: 4.3 degree celsius