# Chapter 9 - Heat Transfer¶

## Example 1: pg 256¶

In :
#pg 256
print('Example 9.1');

#  aim : To determine
#  the heat loss per hour through the wall and interface temperature

#  Given values
x1 = .25;# thickness of brick,[m]
x2 = .05;# thickness of concrete,[m]
t1 = 30.;# brick face temperature,[C]
t3 = 5.;# concrete face temperature,[C]
l = 10.;# length of the wall, [m]
h = 5.;# height of the wall, [m]
k1 = .69;# thermal conductivity of brick,[W/m/K]
k2 = .93;# thermal conductivity of concrete,[W/m/K]

#  solution
A = l*h;# area of heat transfer,[m**2]
Q_dot = A*(t1-t3)/(x1/k1+x2/k2);# heat transferred, [J/s]

#  so heat loss per hour is
Q = Q_dot*3600*10**-3;# [kJ]
print ' The heat lost per hour is (kJ) = ',round(Q)

#  interface temperature calculation
#   for  the brick wall, Q_dot=k1*A*(t1-t2)/x1;
#  hence
t2 = t1-Q_dot*x1/k1/A;# [C]
print ' The interface temperature is (C) = ',round(t2,1)

#  End

Example 9.1
The heat lost per hour is (kJ) =  10815.0
The interface temperature is (C) =  8.2


## Example 2: pg 258¶

In :
#pg 258
print('Example 9.2');

#  aim : To determine
#  the minimum
#  thickness of the lagging required
import math
#  Given values
r1 = 75./2;# external radious of the pipe,[mm]
L = 80.;# length of the pipe,[m]
m_dot = 1000.;# flow of steam, [kg/h]
P = 2.;# pressure, [MN/m**2]
x1 = .98;# inlet dryness fraction
x2 = .96;# outlet dryness fraction
k = .08;# thermal conductivity of of pipe, [W/m/K]
t2 = 27.;# outside temperature,[C]

#  solution
#  using steam table  at 2 MN/m**2 the enthalpy of evaporation of steam is,
hfg = 1888.6;# [kJ/kg]
#  so heat loss through the pipe is
Q_dot = m_dot*(x1-x2)*hfg/3600;# [kJ]

# also from steam table saturation temperature of steam at 2 MN/m**2 is,
t1 = 212.4;# [C]
# and for thick pipe, Q_dot=k*2*%pi*L*(t1-t2)/log(r2/r1)
# hence
r2 = r1*math.exp(k*2*math.pi*L*(t1-t2)*10**-3/Q_dot);# [mm]

t = r2-r1;# thickness, [mm]
#results
print ' The minimum thickness of the lagging required is (mm) = ',round(t,1)

#  End

Example 9.2
The minimum thickness of the lagging required is (mm) =  38.8


## Example 3: pg 260¶

In :
#pg 260
print('Example 9.3');

#  aim : To determine the
#  (a) heat loss per hour
#   (b) interface temperature og lagging
import math
# Given values
r1 = 50.; # radious of steam main,[mm]
r2 = 90.;# radious with first lagging,[mm]
r3 = 115.;# outside radious os steam main with lagging,[mm]
k1 = .07;# thermal conductivity of 1st lagging,[W/m/K]
k2 = .1;# thermal conductivity of 2nd lagging, [W/m/K]
P = 1.7;# steam pressure,[MN/m^2]
t_superheat = 30.;# superheat of steam, [K]
t3 = 24.;# outside temperature of the lagging,[C]
L = 20.;# length of the steam main,[m]

#  solution
#  (a)
#  using steam table saturation temperature of steam at 1.7 MN/m^2 is
t_sat = 204.3;# [C]
# hence
t1 = t_sat+t_superheat;# temperature of steam,[C]

Q_dot = 2*math.pi*L*(t1-t3)/(math.log(r2/r1)/k1+math.log(r3/r2)/k2);# heat loss,[W]
#  heat loss in hour is
Q = Q_dot*3600*10**-3;# [kJ]

print ' (a) The heat lost per hour is (kJ) = ',round(Q)

# (b)
#  using Q_dot=2*%pi*k1*(t1-t1)/log(r2/r1)
t2 = t1-Q_dot*math.log(r2/r1)/(2*math.pi*k1*L);# interface temperature of lagging,[C]

print ' (b) The interface temperature of the lagging is (C) = ',round(t2,1)

print 'There is some rounding off error in the book, so answer is not matching'

#  End

Example 9.3
(a) The heat lost per hour is (kJ) =  8770.0
(b) The interface temperature of the lagging is (C) =  71.5
There is some rounding off error in the book, so answer is not matching


## Example 4: pg 265¶

In :
#pg 265
print('Example 9.4');

# aim : To determine
#  the energy emetted from the surface

#  Given values
h = 3.;# height of surface, [m]
b = 4.;# width of surface, [m]
epsilon_s = .9;# emissivity of the surface
T = 273.+600;# surface temperature ,[K]
sigma = 5.67*10**-8;# [W/m^2/K^4]

#  solution
As = h*b;# area of the surface, [m^2]

Q_dot = epsilon_s*sigma*As*T**4*10**-3;# energy emitted, [kW]
#results
print ' The energy emitted from the surface is (kW) = ',round(Q_dot,1)

#  End

Example 9.4
The energy emitted from the surface is (kW) =  355.7


## Example 5: pg 265¶

In :
#pg 265
print('Example 9.5');

#  aim : To determine
#  the rate of energy transfer between furnace and the sphere and its direction
import math
#  Given values
l = 1.25;# internal side of cubical furnace, [m]
ti = 800.+273;# internal surface temperature of the furnace,[K]
r = .2;# sphere radious, [m]
epsilon = .6;# emissivity of sphere
ts = 300.+273;# surface temperature of sphere, [K]
sigma = 5.67*10**-8;# [W/m**2/K**4]

#  Solution
Af = 6*l**2;# internal surface area of furnace, [m**2]
As =4 *math.pi*r**2;# surface area of sphere, [m**2]

#  considering internal furnace to be black
Qf = sigma*Af*ti**4*10**-3;# [kW]

#  radiation emitted by sphere is
Qs = epsilon*sigma*As*ts**4*10**-3; # [kW]

#  Hence transfer of energy is
Q = Qf-Qs;# [kW]
#results
print ' The transfer of energy will be from furnace to sphere and transfer rate is (kW) = ',round(Q)
print' There is some calculation mistake in the book, so answer is not matching'

#  End

Example 9.5
The transfer of energy will be from furnace to sphere and transfer rate is (kW) =  703.0
There is some calculation mistake in the book, so answer is not matching


## Example 6: pg 271¶

In :
#pg 271
print('Example 9.6');

#  aim : To determine
#  the overall transfer coefficient and the heat loss per hour

#  Given values
x1 = 25*10**-3;# Thickness of insulating board, [m]
x2 = 75*10**-3;# Thickness of fibreglass, [m]
x3 = 110*10**-3;# Thickness of brickwork, [m]
k1 = .06;# Thermal conductivity of insulating board, [W/m K]
k2 = .04;# Thermal conductivity of fibreglass, [W/m K]
k3 = .6;# Thermal conductivity of brickwork, [W/m K]
Us1 = 2.5;#  surface heat transfer coefficient of the inside wall,[W/m**2 K]
Us2 = 3.1;#  surface heat transfer coefficient of the outside wall,[W/m**2 K]
ta1 = 27.;# internal ambient temperature, [C]
ta2 = 10.;# external ambient temperature, [C]
h = 6.;# height of the wall, [m]
l = 10.;# length of the wall, [m]

#  solution
U = 1/(1/Us1+x1/k1+x2/k2+x3/k3+1/Us2);# overall heta transfer coefficient,[W/m**2 K]

A = l*h;# area ,[m**2]

Q_dot = U*A*(ta1-ta2);# heat loss [W]

#  so heat loss per hour is
Q = Q_dot*3600*10**-3;# [kJ]
#results
print ' The overall heat transfer coefficient for the wall is (W/m**2 K) = ',round(U,3)
print ' The heat loss per hour through the wall is (kJ) = ',round(Q)

#  End

Example 9.6
The overall heat transfer coefficient for the wall is (W/m**2 K) =  0.313
The heat loss per hour through the wall is (kJ) =  1148.0


## Example 7: pg 272¶

In :
#pg 272
print('Example 9.7');

#  aim : To determine
#  the heat loss per hour and the surface temperature of the lagging
import math
#  Given values
r1 = 75.*10**-3;# External radiou of the pipe, [m]
t_l1 = 40.*10**-3;# Thickness of lagging1, [m]
t_l2 = t_l1;
k1 = .07;# thermal conductivity of lagging1, [W/m K]
k2 = .1;# thermal conductivity of lagging2, [W/m K]
Us = 7;# surface transfer coefficient for outer surface, [W/m**2 K]
L = 50.;# length of the pipe, [m]
ta = 27.;# ambient temperature, [C]
P = 3.6;# wet steam pressure, [MN/m**2]

#  solution
#  from steam table saturation temperature of the steam at given pressure is,
t1 =  244.2;# [C]
r2 = r1+t_l1;# radious of pipe with lagging1,[m]
r3 = r2+t_l2;# radious of pipe with both the lagging, [m]

R1 = math.log(r2/r1)/(2*math.pi*L*k1);# resistance due to lagging1,[C/W]
R2 = math.log(r3/r2)/(2*math.pi*L*k2);# resistance due to lagging2,[C/W]
R3 = 1/(Us*2*math.pi*r3*L);# ambient resistance, [C/W]

#  hence overall resistance is,
Req = R1+R2+R3;# [C/W]
tdf = t1-ta;# temperature driving force, [C]
Q_dot = tdf/Req;# rate of heat loss, [W]
#  so heat loss per hour is,
Q = Q_dot*3600*10**-3;# heat loss per hour, [kJ]

#  using eqn 
t3 = ta+Q_dot*R3;# surface temperature of the lagging, [C]
#results
print ' The heat loss per hour is (kJ) = ',round(Q,0)
print ' The surface temperature of the lagging is (C) = ',round(t3,2)

print 'there is minor variation in the answer due to rounding off error in textbook'

#  End

Example 9.7
The heat loss per hour is (kJ) =  24533.0
The surface temperature of the lagging is (C) =  46.99
there is minor variation in the answer due to rounding off error in textbook