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"
```

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 "
```

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"
```

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)"
```

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),"%"
```

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"
```

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"
```

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"
```

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))+((theta2`exp^(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))+((theta2`exp^(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)"
```

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"
```