HMT Notes.pptx

1. T2
k
MODULE 1
INTRODUCTORY CONCEPTS AND BASIC LAWS OF HEAT TRANSFER
1.Introduction: According to the law of conservation of energy, the energy can be changed
from one form to another form but the energy is neither be created nor be destroyed. The
science of Thermodynamics deals with the relation between heat and other forms of energy. But
the science of Heat Transfer is concerned with the analysis of rate of heat transfer taking place in a
system. The energy transfer by heat flow cannot be measured directly but the concept can be
analyzed by measuring the physical quantity called ‘Temperature’. It is observed that when there
is temperature difference in a system, heat flows from high temperature region to that of low
temperature region. Since the heat flow takes place whenever there is a temperature gradient in a
system, a knowledge of the temperature distribution in a system is essential in heat transfer studies.
Once the temperature distribution id known, the heat flux, which is the amount of heat transfer per
unit area per unit time, is readily calculated from the law relating the heat flux to the temperature
gradient.
The problem of determining temperature distribution and heat flow is of interest in many
branches of science and engineering. In designing of heat exchangers such as boilers, condensers,
radiators etc for sizing such equipments. In heating and air conditioning applications for buildings,
a proper heat transfer analysis is necessary to estimate the amount of insulation to be provided to
prevent excessive heat losses or gains.
2.Heat Transfer Mechanisms: There are three mechanisms by which heat transfer can
take place. All the three modes require the existence of temperature difference. In the study of
heat transfer, there are three main modes of heat transfer, namely:
(i) Conduction (ii) Convection and (iii) Radiation
1.Conduction: It is the energy transfer that takes place at molecular levels. Conduction is
the mode of heat transfer in which energy exchange takes place from the region of high
temperature to that of low temperature from one molecule to another molecule without
appreciable motion of the molecules. In the case of liquids and gases, conduction is due to
collisions and diffusion of the molecules during their random motion. In solids, it is due to the
vibrations of the molecules in a lattice and motion of free electrons. Heat transfer in solids is
generally referred to Conduction mode.
The rate of heat transfer by conduction is calculated by using ‘Fourier law of heat
conduction equation’. Fourier law of heat conduction states that ‘the rate of heat flow by
conduction in a given direction is proportional to the area normal to the heat flow and the
temperature gradient in that direction’.
For heat flow in x-direction (say), the Fourier law is given by
Q  A
dT T1
x
dx Q
or
Qx  QCond
 k A
dT
dx
(1.1)
Where k = proportionality constant
= Thermal conductivity of the material, W/m-K
Qx = Heat flow in x-direction, W
A = Area normal to the heat flow, m2
dT
= Temperature gradient in x-direction, 0
C/m
dx
The negative sign indicates that the heat flows in the direction of decreasing temperature and
serves to make the heat flux positive in the positive x-direction. Fig.1.1 shows the heat flow due
to conduction through the infinite slab.
L
x
Fig.1.1 Heat conduction
in an infinite slab

2. 2
If eqn.(1.1) is divided by the area, then
x
A
q 
Qx
 k
dT
dx
W/m2
(1.2)
Where qx is called as the heat flux.
Thermal Conductivity: - The constant of proportionality in the equation of Fourier’s law of
conduction is a material property called the thermal conductivity. The units of thermal conductivity
can be obtained from equation (1.2) as follows:
Solving for k from Eq. (1.2) we have, k = − qx / (dT/dx)
Therefore, units of k = W/(m–K) or W/(m–0
C). Thermal conductivity is a measure of a
material’s ability to conduct heat.
Table-1 shows the typical values of thermal conductivity for different material. We can observe
from the Table-1 that the pure metals have the highest values of thermal conductivities while the
gases and vapours have the lowest, and insulating materials and inorganic liquids have thermal
conductivities that lie in between those of metals and gases.
Thermal conductivity is also a function of temperature; for pure metals it decreases with
increasing temperature, where as for gases and insulating materials it increases with increasing
temperature.
Table-1: Thermal conductivities of common substances at 200
C
Substance K (W/m-K)
Silver, pure 407.0
Copper, pure 386.0
Aluminium, pure 175.6
Mild steel 37.2
Lead 29.8
Stainless steel 19.3
Wood 0.15
Asbestos, fibre 0.095
Water 0.51
Air 0.022
If a solid is of good electrical conductor, then generally it will also serve as a good heat
conductor. Ex: Silver, gold, copper etc.
Thermal Diffusivity: - This is a property which is very helpful in analyzing
transient heat conduction problem and is normally denoted by the symbol α . It is defined as
follows:
α 
Heat stored per unit
volume

Heat conducted k
ρ Cp
m2
/s ……
(1.3)
It can be seen from the definition of thermal diffusivity that the numerator represents the
ability of the material to conduct heat across its layers and the denominator represents the ability of
the material to store heat per unit volume. Hence we can conclude that larger the value of the
thermal diffusivity, faster will be the propagation of heat into the medium. A small value of
thermal diffusivity indicates that heat is mostly absorbed by the material and only a small quantity
of heat

3. will be conducted across the material.

4. 3
1.2.2 Convection:
When the fluid flows over a solid body or inside a channel while temperatures of the fluid and
the solid surface are different, the heat transfer between the fluid and solid surface takes place as a
consequence of the motion of fluid relative to the solid surface. This mode of heat transfer is called
‘Convection’.
Convective mode of heat transfer is calculated by using Newton’s law of cooling, which states
that “the rate of heat transfer is directly proportional to the area normal to the flow and the
temperature difference between the solid body and the fluid flow”. It is given by the equation:
Qconv  A T
or Qconv  h A T
(1.4)
Where h = proportionality constant or also called as convective heat transfer coefficient or
surface/film coefficient, W/m2
-K
Q = Rate of heat transfer, W
A = Area normal to the heat flow, m2
T = Temperature difference, Tw - Tf, 0
C
Fig.1.2 shows the fluid flow over the solid
surface during which the heat transfer between the
solid surface and flowing fluid takes place by
convection.
If the fluid motion is artificially induced,
say
Tf
Fluid flow
Fluid temperature
profile
by using pump or fan that forces the fluid flow
over the solid surface, the heat transfer is said to Solid body TW > Tf
TW
be by Forced convection, and if the fluid motion
is setup by buoyancy effects resulting from
Fig.1.2 Heat flow by convection
density difference caused by the temperature difference in the fluid, the heat transfer is said to be
by Natural or Free convection.
The heat transfer coefficient h depends on: (i) the type of flow (i.e. whether the flow is laminar
or turbulent), (ii) the geometry of the body and flow passage area, (iii) the thermo- physical
properties of the fluid namely the density ρ, viscosity μ, specific heat at constant pressure Cp and
the thermal conductivity of the fluid k and (iv) whether the mechanism of convection is forced
convection or free convection. The heat transfer coefficient for free convection will be generally
lower than that for forced convection as the fluid velocities in free convection are much lower than
those in forced convection.
Table-2 shows the approximate range of convective heat transfer coefficient for forced and free
convection encountered in typical engineering applications.
Table-2 Typical values of convective heat transfer coefficient
Types of fluid/flow h (W/m2
-K)
Free convection, air 5 - 15
Forced convection, air 10 - 500
Water 100 – 17,000
Boiling water 2500 – 57,000

5. Condensation of steam 5000 – 1,00,000

6. 4
Radiation:
All bodies at all temperature levels emit thermal radiation. Unlike conduction and convection,
this mode of heat transfer does not require any material media, the propagation of energy being
carried out by electromagnetic waves emitted from the surface of the body. When the
electromagnetic waves strikes the surface of another body, a part of the energy is reflected, a part
transmitted through the body and the remainder absorbed by the body.
The most commonly used law of thermal radiation is the Stefan-Boltzmann law, which states
that ‘the thermal radiation energy emitted by the body per unit area is proportional to the fourth
power of its absolute temperature’. It is given by,
Qrad  E   AT 4
(1.5)
b
where  is called the Stefan-Boltzmann constant whose value is 5.67x10-8
W/m2
-K4
, and T
is the surface temperature of the body in degree Kelvin.
For a non black surface the emissive power is given by
Qrad  Eg   AT 4
(1.6)
where ε is called the emissivity of the surface (0 ≤ ε ≤ 1). The emissivity provides a measure of
how efficiently a surface emits radiation relative to a black body. The emissivity strongly
depends on the surface material and finish.
The net radiant heat exchange between the two black bodies, which are at different temperatures
T1 and T2 respectively is given by
Qrad   A T1
4
 T2
4
 (1.7)
The real surfaces, like polished metal plate, do not radiate as much energy as a black body (is
one which absorb all the radiant energy falling on it and also emits all the absorbed energy).
Case (i): The radiation heat transfer between the two gray surfaces, which are at
different temperatures and having different emissivities, is given by:
(1.8)
(1.9)
Case (ii): Consider two finite surfaces A1 and A2, as shown in Fig.1.3, are maintained at absolute temperatures
T1 and T2 respectively, and have emissivities ε1 and ε2.
Q 12  1 A1T1
4
  2 A2T2
4

If the area and emissivity of the both surfaces are same, then
Q12   A T1
4
 T2
4


7. Fig.1.3: Radiation exchange between surfaces A1 and A2
Surroundings
A2, ε2, T2
A1, ε1, T1

8. 5
Part of the radiation leaving A1 reaches A2, while the remaining energy is lost to the
surroundings. Similar considerations apply for the radiation leaving A2. If it is assumed that the radiation
from the surroundings is negligible when compared to the radiation from the surfaces A1 and A2 then we can
write the expression for the radiation emitted by A1 and reaching A2 as
4
Q1→2 = F1− 2 A1ε1σ T1
(1.10)
where F1 – 2 is defined as the fraction of radiation energy emitted by A1 and reaching A2. Similarly, the
radiation energy emitted by A2 and reaching A1 is given by
4
Q2→1 = F2− 1 A2 ε2 σ T2
(1.11)
where F2 – 1 is the fraction of radiation energy leaving A2 and reaching A1.
Hence the net radiation energy transfer from A1 to A2 is given by
Q1 – 2 = Q1→2 − Q2→1
4
4
= [F1− 2 A1ε1σ T1 ] − [F2− 1 A2
ε2 σ T2 ]
(1.12)
F1-2 is called the view factor (or geometric shape factor or configuration factor) of A2 with respect to
A1 and F2 - 1 is the view factor of A1 with respect to A2. It will be shown in the Radiation
relative
chapter that the view factor is purely a geometric property which depends on the
orientations of A1 and A2 satisfying the reciprocity relation,
A1 F1 – 2 = A2 F2 – 1.
Therefore, Q1 – 2 = A1F1 – 2 σ [ε1 T14 − ε2 T24] (1.13)
Radiation Heat Transfer Coefficient:- Under certain restrictive conditions it is possible to simplify
the radiation heat transfer calculations by defining a radiation heat transfer coefficient hr analogous to
convective heat transfer coefficient as
Qrad = hr A ΔT
(1.14)
For the example of radiation exchange between a surface and the surroundings [Eq.(1.9)] using
the concept of radiation heat transfer coefficient we can write
Qrad = hr A[T1 – T2] = A ε σ [T1
4
– T2
4
]
 T 4
 T 4
  T 2
 T 2
 T  T T  T 
h r


1 2

1 2
T  T 
1 2 1 2 1 2
T1  T2 
Or h 
 
2
2
T  T   T  T  (1.15)
r 1 2 1 2

9. Objectives
 In this chapter, the governing basic equations for conduction in Cartesian
coordinate system will be derived.
The corresponding equations in cylindrical and spherical coordinate
systems will also be presented.
 Mathematical representations of different types of boundary conditions
and the initial condition required to solve conduction problems will also be
discussed.
Learn to evaluate the heat flow through a 1-D, SS system with no heat
generation for rectangular, cylindrical and spherical geometries for constant
thermal conductivity.
Understand the use of equivalent thermal circuits and with the expressions
for the conduction resistances that pertain to each of the three common
geometrics.
Study of composite thermal resistances for 1-D, Steady state heat transfer
with no heat sources placed in parallel or in series using electrical analogy.

10. Q˙ g
q'''
x.y.
z

First law equation for a control volume:
“The rate at which thermal and mechanical energy enters a control volume, plus
the rate at which thermal energy is generated within the control volume, minus
the rate at which thermal and mechanical energy leaves the control volume must
be equal to the rate of increase of stored energy within the control volume”.
Q˙
in  Q˙
out  Q˙
g  E˙ st
Rate of change of
stored energy
Net rate of heat
transfer
Rate of heat
generation
=Rate of heat generation is a Volumetric
phenomenon
First Law of Thermodynamics (Law of conservation of energy)
as applied to Heat Transfer Problems
Introduction

12. 



Three dimensional conduction equation in
Cartesian coordinate system

x x
Rate of heat flow into the element in X-
direction is,
Q 

 k
T

y z
x


Rate of heat flow from the element in X-direction is ,
T   T 

y z   k x y
z
Q  
xdx 

x
x
x
x x




Q  Q

k
Net heat flow due to conductio
n in X
-direction
is, 
T 
x
y z
x xdx x
x

x

T
Similarly, Net heat flow due to condu

c

tion

in Y

and Z-
directions is,
k

13. 

Qy
 Qy dy 
y
 ky
x.
y.z
y 
Q  Q 
 
k
T

x.y.z
z
z dz z
z

z


14. (Net heat conducted in x, y and z direction per unit time) + (Heat generated internally
in the volume element) = (Increase of internal energy in the volume element)
Hence the net rate of heat conduction into the element from all the three
directions,
Q
x
x
   T 

  T    T

y
 
z

kz

net    kx

z x.y.z …
(1)



y
 ky



Rate of heat thermal energy generation in the
element,
Qgen

q'''
x.y.
z
…(2)
E
  C
Rate of increase of internal energy within the element,
According to 1st law of TD for the volume element,
p
T
(x.y.
z)
t t
…(3)
Qnet + Qg = ∂E / ∂t …(4)

15. After substitution of eqn.(1), (2) and (3) into the eqn.(4) simplification, we
get,
  T    T    T 
'''
T
x
 kx
x
 
y
 ky
y
 
z
 kz
z  
q

  C
p
t
…(5)





Above equation is the most general form of conduction equation in Cartesian coordinate
system.

16. 1-D Heat Conduction
equation.
(i) For isotropic solids, thermal conductivity is independent of direction; i.e., kx = ky = kz = k,
then the equation
becomes, 2T 2T
x2

y2

z2
2T 
q'''
k 

Cp
T 1
T
k t


t
(6)
Where 

k

Thermal conductivity Is called as Thermal diffusivity

Cp
Thermal capacity
Eq (6) is as the “Fourier–Biot
equation”
(ii) Steady State conduction [i.e., (∂T/∂t) = 0]
2
T 2
T 2
T
x2
y2
z2
  
q'''
k
q'''
k

0
…(7) is called “Poisson's equation”.
(iii) No thermal Heat generation [i.e. q’’’ =
0]:
2T 
2T 
2T
x2
y2 z
2


t
1 T
…(8) is called as “Diffusion equation”.
 0  2
T 

17. (iv) Steady State conduction without heat generation [i.e., (∂T/∂t) = 0 and q’’’=0]:
2
T

2
T

2
T  0  2
T 
0
…(9) is called as “Laplace equation”.
x2
y2
z2

18. 

 1

q'''
It is convenient to express the governing conduction equation in cylindrical
coordinate system when we want to analyze conduction in cylinders and hallow
pipes.
2
 
T
 
2

1 T
 2
1 T
1
 T
r r r
r 2

2

2
T 
  q'''
z2 
k
 t


r r

 r

r 2

2
 z2
2
T

 


k t
1 T
Transformation
equations:
Three dimensional conduction equation
in cylindrical coordinate system

r
T   1 2
T

19. x = r cos θ
;
y = r sin
θ ;
z = z

20. 
  T  q''' 1 T
It is convenient to express the governing conduction equation in Spherical
coordinate system when we want to analyze conduction in Solid/Hallow Spheres.
2
 T
r2

2 T
r r  1 2
T

1
r 2
sin2
  2
r2
sin


 sin T 

q'''

1
T

  k 
t
1 

2 T  1 2
T
1
Three dimensional conduction equation
in Spherical coordinate system

21. 2
r r


r


r

r 2
sin2

2
 2 sin   
 r sin     k 
t

22. Boundary and Initial Conditions
The temperature distribution within any solid is obtained by integrating the
conduction equation with respect to the space variable and time
 The solution thus obtained is called the “general solution” involving arbitrary
constants of integration.
 The solution to a particular conduction problem is arrived by obtaining these
constants which depends on the conditions at the bounding surfaces of the solid as
well as the initial condition
The thermal conditions at the boundary surfaces are called the “boundary
conditions”
Boundary conditions normally encountered in practice are:
(i) Specified temperature, T (also called as B C of the 1st kind),
(ii) Specified heat flux, q” (also known as B C of the 2nd kind),
(iii)Convective boundary condition, h (also known as B C of the 3rd kind)

23. Specified Temperatures at the Boundary ( B C of 1st kind)
BC’s are:
(i) At x=0, T=T1
(ii)At x=L, T=T2
Ψ(y)
BC’s are:
(i) at x = 0, T(0,y) = Ψ(y) for all values of
y;
(ii) at y = 0, T(x,0) = T1 for all values of x
(iii)at x = a, T(a,y) = T2 for all values of y;
(iv) at y = b, T(x,b) = φ(x) for all values of
x
y
T(x,y)
T = φ(x)
T2
b
a
x
T1
Fig. Boundary conditions of first
kind for a rectangular plate

24. supply
Specified heat flux (B C of 2nd kind)
Heat
 k
dT
dr
(b) Hallow cylinder or sphere
r 
r
i
ri
 q
0
 qr
r r0
 k
dT
dr
Heat supply

25. Convective boundary condition (B C of 3rd kind)
Surface in contact with fluid at
Ti with surface heat transfer
coefficient h i L
T(x)
Surface in contact with fluid at
T0 with surface heat transfer
coefficient h0
Ti > T(x) > To
x
Fig. Boundaries subjected to convective heat transfer on a plane
wall
qconv = q cond;
BC’s are:
(i) at x = 0, i.e., hi[Ti − T|x = 0 ] = − k(dT / dx)|x = 0
(ii) at x = L, , q cond; = qconv , i.e, − k(dT / dx)|x = L = h0 [T|x = L − T0]

26.  k
dT
dr
x L
 h (T T )
Fluid flow
T2, h2
ri r0
h (T T r ri
)   k
dT
dr
r ri
Fluid flow
T1, h1
dr
Convective boundary condition (B C of 3rd kind)
Fluid flow
T1, h1
Convection
Conduction
Conduction
dT
Convection
1 1 x0
h (T T )   k
dx
x0 Fluid flow
T2, h2
(a) Slab (b) Hallow cylinder or sphere
BC’s:
(i) At x=0, h1(T1 -Tx=0) = - k dT/dx
BC’s:
(i) At
r=ri,
1 1
h (T T r ri )   k
dT
dr r ri
(ii) At x=L, h2(T2 -Tx=L) = +k dT/dx
0
(i) At
r=r ,
 k
dT r ro
2 2 r ro
 h (T  T )
x L  h2 (T2  T xL
 k
dx
dT

27. Radiation Boundary
Condition
x
Fig. :Boundary surface at x = L subjected to radiation heat
transfer for a plane wall
BC:
(i) At x = L, qcond = qrad; i.e., − k (dT/dx)| x = L = σ ε [(T| x = L)4 − T  4]
T(x,t)
Surface with emissivity ε is
radiating heat to the
surroundings at T
0K
L

28. Special Case: Insulated Boundary
A well-insulated surface can be
modeled as a surface with a specified
heat flux of zero. Then the boundary
condition on a perfectly insulated
surface (at x = 0, for example) can be
expressed as
On an insulated surface, the first
derivative of temperature with respect
to the space variable (the temperature
gradient) in the direction normal to the
insulated surface is zero.

29. Another Special Case: Thermal Symmetry
Some heat transfer problems possess thermal
symmetry
as a result of the symmetry in imposed thermal
conditions.
For example, the two surfaces of a large hot plate of
thickness L suspended vertically in air is subjected to
the same thermal conditions, and thus the temperature
distribution in one half of the plate is the same as that
in the other half.
That is, the heat transfer problem in this plate
possesses thermal symmetry about the center plane
at x = L/2.
Therefore, the center plane can be viewed as an
insulated surface, and the thermal condition at this
plane of symmetry can be expressed as
which resembles the insulation or zero heat
flux
boundary condition.

30. Special case: Interface Boundary Conditions
The boundary conditions at an interface are based
on the requirements that
1. two bodies in contact must have the
same temperature at the area of contact
and
2. an interface (which is a surface) cannot store
any energy, and thus the heat flux on the two
sides of an interface must be the same.
The boundary conditions at the interface of two
bodies A and B in perfect contact at x = x0 can be
expressed as

31. Conduction Equation in compact form
 The general form of one – dimensional conduction equation for
plane walls, cylinders and spheres can be written in a compact
form
as
follows: 1   T  q''' 1 T
 rn
  
rn
r  r  k  t
Where n = 0 for plane walls, with r = x
n = 1 for radial conduction in cylinders
n = 2 for radial conduction in spheres,
 General form of conduction equation for 1-D, Steady-State
with no Heat generation for plane walls, cylinders and
spheres in a compact form is as follows:
rn r

1  
rn T 
 0

  r

32. Study of
One-Dimensional, Steady-State
Conduction without Heat Generation for
uniform thermal conductivity
in plane walls, cylinders and spheres

33. • Derive the expressions for heat flow rate and
temperature distribution across the slab for 1-D,
Steady- State conduction with constant k and no heat
generation.
The Plane Wall
 Implications:
Heat flux qx  is independent of x.
Heat rate qx  is independent of x.
• General H2
eat Equation:
 T
x2

0
   (2.1)
• Boundary Conditions:
(i) At x =0, T=T1
(ii) At x=L, T = T2
Integration of eqn(2.1), once we get dT  C
   (2.2)

34. dx 1
Integration of eqn(2.2), once we get T (x)  C1x  C2    (2.3)

35. T (x)  C1x  C2
The Plane Wall
   (2.3)
• With Boundary Condition, : (i) At x =0, T=T1
From eqn.(2.3), T(0)  T1  C1x0  C2  C2  T1
Now eqn.(2.3) becomes
T(x)  C1x  T1    (2.4)
• With Boundary Condition, :(ii) At x =L, T=T2
In eqn.(2.4), T(L)  T2  C1x L  T1 1
 C

T2  T1
L
• From eqn(2.4), Temperature Distribution for Constant k is:
T2  T1
L
T(x) 
x  T
• Heat Rate:
1
OR T1 - T(x) 
x
T1  T2 L
dT
   (2.5)
T  T  T  T 

36. From Fourier’s law of conduction Q˙  kA dx
 kA 2
L
1

kA
1
L
2
   (2.6)
q 
Q˙
 k
T1  T2 

• Heat Flux: W/m 2
   (2.7)

37. Electrical Circuit Theory of Heat Transfer
• A resistance can be defined as the ratio of a driving potential to a
corresponding
transfer
rate.
Analogy:
R 
V
I
For Heat Transfer
th
Q   Rth 
R
T
T
Electrical Thermal
Current, I Heat flow, Q
Voltage
difference, Δ
V
Temperature
difference,
ΔT=(T1-T2)
Electrical resistance, REle Thermal resistance, RTh

38. k
Q
L

1
Thermal Resistances in Heat Transfer
Q  kA
T 
T

L
Q
kA
 R
th
 R
th,Cond

L
T1
T2
Q
TS
Flui
d h
T
L kA
• Thermal resistance due to convection
Q 
hAT

T

1
Q
hA
 Rth  Rth,Conv
hA
Conduction Convection
T , 
,
 4 4

• Thermal resistance due to
R2adia2tion    
  A, F1-2 T2
Qr    AF12 T1
 T2    AF12
T1
 T2 T1  T2
T1  T2  hr A T1  T2
1 1
 h r
k
L
• Thermal resistance due to conduction
through slab/plane wall

39. 
1 T 
1
1
 R
Qr r
h A
 

F12 T 2

T2
2
T

T2

R
a
d
i
a
t
i
o
n
Q
r
Now Q  h
AT
r
r
 R

1
th th,Rad

  FAT 2
 T 2
T  T 

40.  kA
R 0.01111
Heat Conduction in a Plane Wall
Example-1: Consider a large plane wall of thickness L = 0.2 m, thermal conductivity k = 1.2 W/(m-
°C), and surface area A=15 m2. The two sides of the wall are maintained at constant temperatures
of T1=120°C and T2 = 50°C, respectively. Determine: (a) the variation of temperature within the
wall and the value of temperature at x = 0.1 m and (b) the rate of heat conduction through the
wall under steady conditions.
Solution: Given:- A=15m2, L=0.2m, k=1.2 W/(m-K), T1=1200C; T2=500C;
To find: (i) T(x) at x=0.1m, (ii) Q
(i) Temperature Distribution for Constant k is:
T1 - T(x) 
x
T1  T2 L
 120 - T(x) 
0.1
 T(x)  850 C
120  50 0.2
Ans
(ii) Rate of Heat conduction : Alternatively:
From Fourier’s law of conduction Rth 
L

kA
0.2
1.2x15  0.01111 0
C/W
Q  T
1

T

41. 2 
L  1.2x15
120  50  6300 W
0.2
Ans  Q 
T

120  50
 6300
W
Ans

42. r dr

Cylindrical
Pipe
• Derive the expressions for heat flow rate and
temperature distribution across the cylindrical pipe
for 1-D, Steady- State conduction with constant k and
no heat generation.
 Implications:
Heat rate Q is in radial direction and is function
of r
• General Heat Equation:
L
Q
r2
T2
r1
T1
d  dT

 
0
   (2.8) T2
dr

R
Q
•
Boundary Conditions: (i) At r =r1, T=T1 (ii) At r =r2, T = T2
T1 T2
Integration of eqn(2.8), once we
get dr
r
dT
 C1 
dT 
C1
dr
r
   (2.9)
Integration of eqn(2.9) again we get T (r)  C1 ln r  C2
   (2.10)

43. T2
 C1 ln r2  C2
 1 2

Using the boundary conditons T  T1 at r 
r1
T  T
at r  r
2 2
we get
T1  C1 ln r1  C2  T1  T2
T1  T2  C1 lnr1 r2  1
 C 
 lnr r 



44. ln r  T1  T2 ln r

T
Similarly for C2
Cylindrical Pipe
L

1
T   T1  T2

1 2
lnr ln r  C 
C
2
1
 T   T1 
T2
lnr ln
r
1
r 
r 
 1 2   1 2

Q
Substituting the values of C1 and C2 in eqn.(2.10), we
get the temperature distribution as r2
T2
r1
T1
T (r)  C ln r  C

T1  T2
T2
1 2
lnr r  lnr1 r2  R
T1 T2
Q
 T (r)  T  T1  T2
ln r  ln r

1

T (r)  T 1
 T1  T (r)     (2.11)
1
1 2
lnr r 
• Heat Rate:
From Fourier’s law of
conduction
T  T
1 2 T  T
1 2
1
 lnr r
1 2
1
lnr r
2 1
1 2
1 1

45. ln(r r )
 2 1
lnr r
2 1

Thermal resistance due to conduction through pipe
wall
dT
Q  kA  k (2r
L)
dT
 k(2r L)
C1 
1
 k (2r L) .  2L
k
 T1  T2
 T1 
T2

   (2.12)
dr
dr
r  ln(r1 r2 )
 r

ln(r2
r1)

 T1  T2
T1  T2
Q  2L
k



    (2.13)
ln(r
r )
 Rth  Rth

46. 1
r 
Heat Loss through a Steam Pipe
Example-1: Consider a steam pipe of length L=20 m, inner radius r1= 6cm, outer
radius r2= 8cm, and thermal conductivity k= 20 W/(m-°C). The inner and outer
surfaces of the pipe are maintained at average temperatures of T1=150°C and
T2=60°C, respectively. Obtain a general relation for the temperature distribution
inside the pipe under steady conditions, and determine the rate of heat loss from the
steam through the pipe. Find also the temperature at radius of 7cm.
Solution: Given:- L= 20m, r1=0.06m, r2=0.08m, k=20 W/(m-K), T1=1500C; T2=600C;
• Heat Rate: Q˙ 
2Lk
To find: (i) T(r) & Q (ii)T(r) at r = 0.07
m

 2  150  60  786 kW Ans
T1  T2
 ln(rr )

x20x20 ln(0.08 / 0.06)



 2 1

OR R

th

ln(0.08 0.06)
2x20x20
1.145x104 0
C/W
 Q˙ 
T1  T2
R  1.145x10
150  60
4  786 kW Ans
th T  T (r)
2
1
2L
k
ln(r
r )
ln7

47. 150  T (r)
• Temperature at r =
0.07m:
1

   T (r)  101.8 0C
Ans
150  60 ln 8 /
6
 

48. dr

 Implications:
Heat rate Q is in radial direction and is function
of r
• General Heat Equation:
d  r 2 dT 
 0
   (2.15)
 
dr

• Boundary Conditions: (i) At r =r1, T=T1 (ii) At r =r2, T = T2
Integration of eqn(2.15), once we get 2 dT
r 
C


dT C
1
   (2.16)
dr 1 dr r 2
Integration of eqn(2.16) again we get T (r)  
C1
r
 C 2    (2.17)
• Derive the expressions for heat flow rate and
temperature distribution across a spherical shell for
1-D, Steady-State conduction with constant k and no
heat generation.
Heat Conduction through a Spherical Shell

49. 1 r 1 r  1 




1 2
T  T 1
 C2 T
Using the boundary conditons :T 
T1
at r  r1 and T  T2 at r  r2
 1 1 
T  T
 T1  T2 
1
T1  
1 r 
  C2
we get T   C r  C  2 1 r1 
r1
T  T  C  

1
 r r

1 2
1
 C  1 2
Similarly for C
1 1 1
2
T   C r  C
2

50. Heat Conduction through a Spherical
Shell
Substituting the values of C1 and C2 in eqn.(2.17), we
get the temperature distribution as
T (r)  
C1
r
 C 2

T (r)  
 T1  T2  11

T  T 
1
1 2


 T1  
T  T 
1
1 2

 T (r) T1  
1
r

1 r2 1 r1 
r
1 r2 1 r1 
r1
1 r 
r
r

 2 1  1

 1 r1 1 r


T (r) T1  1 r1 1 r

T  T
1 r 1 r
 

 T1 T (r)
T  T
1 r 1 r
 

   (2.18)
1 2  2 1

1 2  1 2

• Heat
Rate:
From Fourier’s law of
conduction 2
dT
dT
2 C1
 T1  T2
2
)


.
1
Q  kA dr  k (4r ) dr  k (4r
)
r 2
 k (4r 1 r 1 r r 2

51. • Thermal resistance due to conduction through pipe
wall
4 k r
Q˙  4
k 
 T1  T2

 T1  T2

 2 1

 Q  4 k 1 r 1 r   4 k 1 r 1 r

   (2.19)
 2 1   1 2


T1  T2 1
th th
 R

 

 1 1

   (2.20)
  T  T 1 r 1 r

1 2 1 2
1 r 1 r Q
4
k
r

52. Rth
 1 r1 1 r

1 r 1 r
4.421x10
200 T (r)
200  80
Heat Conduction through a Spherical Shell
Example-1: Consider a spherical container of inner radius r1= 8cm, outer radius r2=
10cm, and thermal conductivity k= 45 W/(m-°C). The inner and outer surfaces of the
container are maintained at constant temperatures of T1=200°C and T2=80°C,
respectively as a result of some chemical reactions occurring inside. Obtain a
general relation for the temperature distribution inside the shell under steady
conditions, and determine the rate of heat loss from the container. Find also the
temperature at radius of 9cm.
Solution: Given:- r1=0.08m, r2=0.1m, k=45 W/(m-K), T1=2000C; T2=800C;
To find: (i) T(r) & Q (ii) T(r) at r = 0.09 m
• Heat
Rate:
From Fourier’s law of
conduction 
˙
Q  4
k
 T1  T2

 200  80

1 r 1 r
  200  80

x45

 4 1 0.08 1 0.1


4 1 0.08 1 0.1
x45 

27.143 kW Ans
OR R

 1 2 
1  1

1

 
3 0
 
T1  T2 200  80
˙
C/W 
Q 
th


4 x45 
0.08
 
4.421x10
0.1
  27.143
kW
3
Ans
1 0.08 1 0.07

0
T1 T (r)
T  T

53. • Temperature at r =
0.07m:






Ans
  T (r)  53.33
C 1 0.08 1 0.1 
1 2 1

54. Ti, hi
k3
k2
k1
Q Ri
R0
L L
Heat Transfer through Composite wall
• Derive the expressions for heat flow rate and Overall heat transfer coefficient
across a composite slab for 1-D, Steady-State conduction with constant k and no
heat generation.
 Consider a composite slab of thickness L1, L2 and L3
having thermal conductivity k1, k2 and k3 respectively. Ti
and T0 be the hot and cold gas temperature at inside
and outside of the composite wall, whose heat transfer
coefficient on hot and cold side of the surface at hi and
ho. Respectively. A is the surface area through which
heat flows.
Q
T0, h0
Ti T1
T2 T3
T4
T0
 By energy balance, we can write for rate of heat transfer,
Q˙  h AT  T 
k1 A
T  T 
k2 A
T  T 
k3 A
T  T  h AT  T     (2.21)
i i 1 1 2
1
2 3
2 3
3 4 0 4 0
Q
˙
 T  T ,
L1
Q˙  T  T ,
L2
Q˙  T  T ,
L3
Q˙  T  T ,
Q˙
 T  T     (2.22)
L1
R1
L2
R2
L3
R3
L

55. k A k A k A h A
i 1 1 2 2 3 3 4 4 0
h A

56. L2
L1
Q Ri
L3
R3
R1
R0
R2
1 
˙
0

Overall temperature difference can be
as,
Ti T0   
T4 T0
Ti T1  T1 T2  T2 T3  T3 T4   
From eqn.(2.22), we can
write
k1
T2
k2
T3
k3
h


1 
 T  T  1
T h
i, i
T  T 
 Q˙

i 0
T4
i 0
3
i 0
2 T h
T  T 
 1
T1
Q
2 3

L1

L2

L3 

i

h A k A k A k A h
AQ
2
1

Q  1

L1

L2

L3
k3
0

A 
hik1 k
  
A h  
k k  k h
1  1L L L  1


0, 0
 i 1 2
3 0

 Q˙ 
Ti
R
T0
 (2.24)
Where RTot  
1  1 L L L
1
2 3
   
Ti T1 T2
T3 T4
T0
Tot
For ' n' composite
layers
A 
h
i 1
k k2 k3 h
1 
  (2.25)
0 
 q

i 0
T T 
1 1
n
L  (2.26)

 
i

57. 
hi h0
i1
ki
 Overall heat transfer coefficient (U): A modified form of Newton’s Law of Cooling to
encompass
multiple resistances to heat
transfer.
T  T  1 1 1  1 L1 L2 L3 1  1  1 L1 L2 L3 1
    
 RTot        (2.27)
Q˙

i 0
 UATi  T0  
 UA 
  


58. • Contact
Resistance:
R 
TA  TB
Resistances in Parallel: Resistances in Series & Parallel:
t,c
q
x
Values depend on:
Materials A and B,
surface finishes,
interstitial
conditions, and
contact pressure
• Thermal Resistances: Series – Parallel for Composite Wall

59. Heat Transfer through Composite Cylinder/Tube
• Derive the expressions for heat flow rate and Overall heat transfer coefficient across a
composite cylinder for 1-D, Steady-State conduction with constant k and no heat
generation. L
Air
T0, h0 r3
r2
 Consider a composite cylinder of radius r1, r2 and r3
having thermal conductivity k1 and k2 respectively. Ti
and T0 be the hot and cold gas temperature at inside
and outside of the composite tube with the heat
transfer coefficient hi and ho respectively.
A is the heat flow surface area at any radius  2 r L
• Rate of heat flow from composite cylinder (Q):
Gas
T1 T 2
T 3
r1 ATir
h
0, 0
Q˙  Ti  T0   (2.28) Wher
e
 R  R  R  R Ti, hi
k1
k
R
Tot
RTot
1
Where R  , R  ln (r2
r1)
, R 
ln
(r3
i 1 2 0
2
r ) 1
, R 
2
Q Ri R1 R2 R0
i
2Lr
h
1 i
1
2Lk
1
2
2L
k
0
2Lr
h
3 0
T T
2 Ti T1
2 3 T0

60.   
 eqn.(2.28) can be written
as
˙ Ti T0 
Q
 1 
1
1

 (2.29) 
For ' n' composite layers
2 LT  T

Q˙ 
i 0
1 1
 (2.30)
 ln(r2 r1 )  ln(r3 r2 ) 
ln
r
i1 ri
n


61. 1. Overall heat transfer coefficient
based on inner surface area
(Ai=A1=2 r1 L): Ui
Heat Transfer through Composite Cylinder/Tube
 Over

all heat transfer coefficient (U):
Q˙

i 0
T T  UAT  T  1 1
  UA  
R
RTot
i 0 RTot UA
Tot
1 1 1 1

1

 R   

 Tot

1  r  1  r

 ln 2
  
ln 3
 
1 

 r2 
r3h0 
Ui Ai Ui Ai Ui (2ri
L)
2L 
rihi
k1
 r1

k2
 (2.31)
U h

1   1 
r  r  r  r


r 1 
r
2  2

h
0

k1 ln r2   k1 ln 3

r1
i i
 1  1

3
• Heat Rate: Q˙  U A
T
i i overall
Ti T1 T2 T3 T0
Air
T h
0, 0
Gas
Ti, hi k1
Air
T0, h0
k2
2. Overall heat transfer coefficient based
on

62. 1
1
 RTot
1 1 1  r2  1  r3

 1 1


r3
1 1 r3  r2  r3  r3

1

k r h
 
U (2r L) 2

 ln   ln       ln   ln  

r k r
1  1  2  2  3 0  U0  r1 hi k1  r1  k2
 r2 
U0 A0 0 3 L 
rihi
h0

• Heat
Rate:
Q˙  U
0
A0
Toverall
-1 is a constant and But, U alone depends on area. (2.32)
Note:
UA= (Rtot)

63. k1
k2
Heat Transfer through Composite Sphere
• Derive the expressions for heat flow rate and Overall heat transfer coefficient across a
composite Sphere for 1-D, Steady-State conduction with constant k and no heat
generation.
 Consider a composite spherer of radius r1, r2 and r3
having thermal conductivity k1 and k2 respectively. Ti
and T0 be the hot and cold gas temperature at inside
and outside of the composite tube with the heat
transfer coefficient hi and ho
respectively.
A is the heat flow surface area at any radius  4 r 2
Air
Air
T3
T h
0 0
Th
0 0
•
Rate of heat flow from composite sphere
(Q):
Q˙  Ti  T0   (2.33)
Where
R  R  R  R  R R R R
Wher
e
RTot
R 
1
, R 
1
1  1

1

,
1  1

i 1
1 2 0
, R

r
 0
1 Q Ri
Ti T
1
1 2 0
T2 T3 T0
i
4r
2
h
4k
 r
4k 
r
Tot
R 
r 
2
1  1 2 
4r 2
h
3
0
1 i 2  2 3


64.   r

 eqn.(2.33) can be written
as
For ' n' composite
layers
Q˙

Ti  T0 4 T  T 
1  1 1
 
1  (2.34)
Q˙

i n  (2.35)
2   r  
k r k
2  1 1
1  1 1  1
1
4

r
 

65. 1. Overall heat transfer coefficient
based on inner surface area
(Ai=A1=4ri
2
): Ui
1
k

Heat Transfer through Composite sphere
 Over

all heat transfer coefficient (U):
Q˙

i 0
T T  UAT  T  1 1
  UA  
R
Air
RTot
i 0 RTot UA Tot
Th
0 0
Air
T3

RTot
1
Ui Ai  U (4 r 2 ) 
4
1 1 
1
r 2h k r
 1  1 1


r
1  11

r
1 

T h
0 0
i 1  1
i
r 2  1 1

1  1 2  2 
2
r 2  1 1  r 2
3

3 0

1 
1

h
1

U


 1
k r  r

r  r
 1
r 2
h

3 0

 (2.36)
i
• Heat
Rate:
i 1  1 2  2 
2
3


˙
Ti T1
T2 T3 T0
Q  Ui Ai
Toverall


1 1

1

 R  1

1  1

1 

1  1

1 

1 
k1
k2
  k r    r 2h
2. Overall heat transfer coefficient based
on
outer surface area (A0=A3=4r 2):
U0
0


66. U r 2
U A
0
0
Tot
U (4 r
2 ) 4

r
2h
k r 2h

i 3
 r
2

1
i 3 0

1
1 r 2  1 1  r 2 
1
k  r r  
r r

1  1 2  2  2 3

1 
1 
˙
Heat Rate: Q  U A
T
0 0 overall
 
 
3
 3
   3

     (2.37) •
h k r r k r r h

67. Study of
One-Dimensional, Steady-State
Conduction
Without Heat Generation for
Variable thermal conductivity
in Plane walls, Cylinders and
Spheres

68. The Plane Wall
• Expressions for: (i) heat flow rate and
(ii) temperature distribution across the
slab
• Fourier Heat Conduction Equation:
0
Q˙  k 1 T A
dT
dx
   (2.1)
• Integrate eqn.(2.1) in the limits
of L T2
Q˙
 dx   A k0 1T
dT
   (2.2)
 Q˙ 
Ak0
x0 T1

T  T 

T 2
 T 2

L 

1 2
2
1 2 

wher
e
Ak 


 Q˙

m
T  T  ……(2.3)
2
km  k0 1 T1  T2 
L
1 2 


69. x
 
0
dx   A k  1T dT
• Temperature Distribution:
Integrate eqn.(2.1) in the
limits of
Q̇
T

70. 

x T
Q˙
 dx   A k0 1T
dT
x0 T1
 Q˙ 
Ak
0

T  T 

 
T 2  T 2
1
(2.8)


1
x
2



Q˙ x  
T T 

T 2
T 2

Ak0
2
1 1


 T 2 

T  2  Q˙ x 

T 2


0
 T
(2.9)
2  
Ak
 0 2
1 1
Eqn.(2.9) is a quadratic equation for unknown
temperature T
• Temperature Distribution in terms of heat rate or heat flux:

71. 
 T (x)  
1

(2.10)

73. MODULE-3
1-D Steady State Conduction- Heat
Transfer

74. Numerical on Basic Laws of
HT
1. Conduction:
Fourier’s law of conduction is given
by
QCond
T1
dT
 kA
dx
 QCond
/ A  k
dT
qCond dx
Q
 T  T


T

On integration, q
 Cond 
k
1 2
 k

W/m2
Cond
A  L  L

2. Convection:
Newton’s law of cooling or heating is given
by
QConv  h A T  hA[TS  T ]
T2
Q
k
L

75. 1
Net radiation energy transfer from A1 to A2 is given
by
Q1 – 2 = A1F1 – 2 σ [ε1 T 4 − ε2 T 4]
1
2
Radiation Heat Transfer Coefficient:-
hr = ε σ [T 2 + T 2 ][T + T
]
s sur
s
sur
3. Radiation:
Stefan-Boltzmann law is given
by
Qrad
  
AT
4
 T 2 4

Qr = hrA[Ts – Tsur] = A ε σ [T 4 – T 4 ]
s sur
Formulae
Numerical on Basic Laws of HT

76. 1-D Heat Conduction
equation.
(i) For isotropic solids, thermal conductivity is independent of direction; i.e., kx = ky = kz = k,
then the equation
becomes, 2T 2T
x2

y2

z2
2T 
q'''
k 

Cp
T 1
T
k t


t
(6)
Where 

k

Thermal conductivity Is called as Thermal diffusivity

Cp
Thermal capacity
Eq (6) is as the “Fourier–Biot
equation”
(ii) Steady State conduction [i.e., (∂T/∂t) = 0]
2
T 2
T 2
T
x2
y2
z2
  
q'''
k
q'''
k

0
…(7) is called “Poisson's equation”.
(iii) No thermal Heat generation [i.e. q’’’ =
0]:
2T 
2T 
2T
x2
y2 z
2


t
1 T
…(8) is called as “Diffusion equation”.
 0  2
T 

77. (iv) Steady State conduction without heat generation [i.e., (∂T/∂t) = 0 and q’’’=0]:
2
T

2
T

2
T  0  2
T 
0
…(9) is called as “Laplace equation”.
x2
y2
z2

78. Conduction Equation in compact form
 The general form of one – dimensional conduction equation for
plane walls, cylinders and spheres can be written in a compact
form
as
follows: 1   T  q''' 1 T
 rn
  
rn
r  r  k  t
Where n = 0 for plane walls, with r = x
n = 1 for radial conduction in cylinders
n = 2 for radial conduction in spheres,
 General form of conduction equation for 1-D, Steady-State
with no Heat generation for plane walls, cylinders and
spheres in a compact form is as follows:
rn r

1  
rn T 
 0

  r

79. lnr 1
Temperature distribution, rate of heat transfer, thermal resistance etc

T1 - T(x) x

T1  T2
L
L

kA
Q˙ 
kA
1 2
L Rth,Cond
T  T (r)  T  T

ln(r r )
2 1
2Lk
1
T  T

lnr r

Q 
2Lk
1 2
 Rth 
 ln(r r )

1 2 2 1  2 1

T T (r)  1 r 1 r

Q˙  4

1 2
k T  T

1


1
  1 r 1 r

Rth 


1  1 1



 
T1  T2
Temp. Distribution Rate of heat flow
T  T 
Thermal resistance
Hallow Sphere
Hallow Cylinder
Slab

80. 1
r1
1 r2   4 k 
r1
1
2

r2


81. k
Q
L

1
Q  kA
T 
T 
L
Q
kA
 R
th
 R
th,Cond

L
T1
T2
Q
TS
Flui
d h
T
L kA
• Thermal resistance due to convection
Q 
hAT

T

1
Q
hA
 Rth  Rth,Conv
hA
Conduction Convection
T , 
,
 4 4

• Thermal resistance due to
R2adia2tion    
  A, F1-2 T2
Qr    AF12 T1  T2    AF12
T1
 T2 T1  T2
T1  T2  hr A T1  T2
1 1
 h
r
k
L
• Thermal resistance due to conduction
through slab/plane wall
Thermal Resistances in Heat Transfer

82. 
1 T 
1
1
 R
Qr r
h A
 

F12 T 2

T2
2
T

T2

R
a
d
i
a
t
i
o
n
Q
r
Now Q  h
AT
r
r
 R

1
th th,Rad

  FAT 2
 T 2
T  T 

83. Ti, hi
k3
k2
k1
Q Ri
R0

Heat Transfer through Composite wall
Q˙  Ti
 T0  Q
RTot
R
Tot

1  1 
L1

1 

L2

L3 k
k
A  h
k
h

3 0

 i
1
2
 Overall Heat Transfer Coefficient (U): T h
0, 0

˙ 
Q
T  T 
i
0
RTot
i
 

UA T T
0
Ti T1 T2 T3 T4
T0
1 
1
U

h
L1
R1
L2
R2
L3
R3
Total resistance
Rate of heat flow

84. 
L1
k
 3
k

L2 L
1
k 

1 2 
h3

i
0


85. Heat Transfer through Composite Cylinder/Tube
T  T  L
Q˙  i 0
RTot
1 
1
ln(r r ) ln(r r ) 1

3
Tot
R 
 2 1  3 2  
Air
T h r
0, 0
r

2L
hir1
k1
k2
h0r3

2
T
1 T T Air
 Overall Heat Transfer Coefficient (U):
Q˙  Ti T0 UAT  T 
Gas k T0, h0
1 r  r  r  r r
 1 1

RTot
i 0
Ti, hi
Q
Ti
˙
1
2
Ri R1 R2 R0
T1
T2 T3 T0


 1
 ln 2 1 3
 ln  1

• Heat
Rate:
Q  U A
T
U h k 

r k 

r r h i i overall
i i 1  1

2  2 3 0

2 3
1. Overall heat transfer coefficient based on inner surface
area (Ai=A1=2 r1 L): Ui
Total resistance
Rate of heat flow
r1
k

86. h0

r
1
 


1
r3  • Heat Rate: Q˙  U A
T
0 0 overall
2. Overall heat transfer coefficient based on outer
surface area (A0=A3=2 r3 L): U0
 r3
 r   r 1

ln 2   3 ln 3
 
U0  r1 hi k1  r1

k2
 r2


87.  r  r
1
k

Heat Transfer through Composite Spheres
Q˙  Ti  T0 
RTot
1

1
1  1 1  1 
1
1

Air
1  Air

RTot 4

r
2
h
  
k r r 
k
 
r 
r 2
h

 r
 1 i 1  1 2  2
 2
T h
3

3 0

0 0
T  T  T0h0
Q˙

0
i
RTot
UATi T0 


1 1 r 1 1

2  2



2
r

1 1

r
1

 

U
h
 k1

r  r
 12
r h
 • Heat
Rate:
Q̇ 
U
i i
A
T
overall
1. Overall heat transfer coefficient based on inner surface
area 1
(Ai=A1=4 r 2):
Ui
 Overall Heat Transfer Coefficient (U):
Total resistance
Rate of heat flow

88. r 2
3
i  i 1  1 2  2  2 3  0

 r
2
1 1 r 2  1 1  r 2 
1
1

1

 
 
3
 3
   3

    • Heat
Rate:
Q˙  U0
A0Toverall
2. Overall heat transfer coefficient based on outer
surface
area (A0=A3=4 r 2):
U0
3
U h k r r k r r h

89. Q Ri
R1 R2 R3
R0
Example-1:- Calculate the heat loss per square meter area from a composite furnace
wall made up of the following materials: 200mm thick firebrick with k1 = 1.16 W/(m-
K), 150mm thick insulating brick with k2 = 0.1 W/(m-K), 100mm of red brick k3 = 1.74
W/(m-K). The inside and outside film coefficient of heat transfer is 200 and 20 W/(m2-
K) respectively. The inside gas temperature is 1200°C and temperature of room air is
50°C. Find also temperature drop across the insulating brick
Solution: Q
Rate of heat
flow/m2 1 2 3
 1 1

L L L 
1
RTot        

 K/W
k3
0.2 0.15 0.1 1 
      1.785
m2
1.16 0.1 1.74 20 
 hi k1 k2
h0  
200
T  T 
1200  50
 Q˙

i
0
RTot
1.785
  644.3
W/m2
Ans
OR  Overall Heat Transfer Coefficient (U):
1 T0, h0
U

R
Tot
T1
Ti, hi
T4
k3
T3
k2
T2
k1

90. insul brick 2
3
1.785  U  W/(m2
 K)
0.56
q  Q˙ / A  U Ti  T0   0.561200  50  644.3 W/m2
Ans Ti T1
T2 T3
T4
T0
T = (T -T ):
L2
 644.3x
0.15
 966.45 0
C Ans
L1 L2 L3
 T  T   q˙R
2 3

q˙
2

91. ln(r r )
Example-2:- A saturated refrigerant at -300C flows through a copper pipe of 10mm inside
diameter and 2mm wall thickness. A layer of 40mm thick thermocole is provided on the outer
surface of the pipe to reduce the heat flow. Determine the heat leakage in to the refrigerant
per meter length of pipe. It the ambient temperature is 300C. Assuming the internal and
external heat transfer coefficient to be 500 and 5 W/(m2-K) respectively. Find the amount of
refrigerant vaporized per hour per meter length of pipe when the pipe is covered and the pipe
is bare. Take k(copper) = 400W/(m-K) and k(thermocole)
=0.03W/(m-K) . Take hfg of the refrigerant at -300C = 266.75 kJ/kg
Solution: Given Data: r1=0.005m, r2=7mm=0.007m, r3=47mm=0.047m and kcu=k1=400W/mK, ktc=k2=0.03W/mk,
hi=500W/m2K, h0=5W/m2K, Ti= -300C and T0=300C.
(a) Whe1n pip1e is covered
 ln(r r )
R   1

Tot

k2 1
3
2
k
h r


2L h r
 i 1
 1  1
1

2
0 3

ln(7 5)  ln(47 7) 
1
 
10.84
0
m C/W
2 x1
500x0.005
400 0.03 5x0.047

Rate of heat flow with insulationQ˙ 
Ti  T0 

 30  30  5.54 W/m Ans

92. OR
1  1 r1 
r2 
    ln  
r1  r3 r1
ln  
1
1

RTot

1
10.84
0.005
 7 
ln 

0.005
 47 
ln 

5
1
x
2 0
   

93. ln(rk2
r1 )
ln(7 5)
 Q˙ 1  5.54 W/mAns
Q˙1  m˙ 1 h fg  m˙1  Q˙ h  5.54
266.75x103
1 fg
 m˙1  2.077x105 kg/s  0.075 kg/hr
(b) When pipe is bare
Ans
1

R  
1
Tot

    4.61 m 0
C/W
2L 
h r

i 1
400 5x0.007

1 

1  1
1
h r  2x1 500x0.005
0 2 
1
Rate of heat flow without insulation
Q˙2 
Ti  T0 

RTot
 30  30 13.01 W/m Ans
4.61
Remarks: More heat leaks into the refrigerant when pipe is
bare
Rate of vaporization (m1)
Rate of vaporization (m2)

94. Q˙ 2
 m˙
2hfg
 m˙ 2 
Q˙2
hfg  13.01 266.75x103  4.88x105 kg/s
 0.176 kg/hr Ans

95. Example-3:- A steel tube is covered with a layer of asbestos insulation material. This tube is
used for the flowof hot gases. The following data is given: I.D. of steel tube = 75 mm, O.D. of
steel tube
= 100 mm, Thickness of asbestos layer = 30 mm, Temperature of hot gas = 350°C.
Temperature of outside ambient air = 40°C. ksteel = 50 W/(m-K), kAsbestos = 0.15 W/(m-K), hstell tube
inside wall-hot gas film= 300 W/(m2-K), hasbestos-ambient air film= 20 W/(m2-K), Estimate: i) Overall heat transfer
coefficient based on outside surface area. ii) Loss of heat per meter length of pipe, and iii)
Temperature drop radially in steel and asbestos.
Solution: Given Data: r1=37.5x10-3
m, r2=50x10-3
m, r3=80x10-3
m and ksteel=k1=50W/mK, kAsbe=k2=0.15W/mk,
hi=300W/m2K, h0=20W/m2K, Ti= 3500C and T0=400C.
1

1  
 r3 1 r3  r2 
r3  r3  r h

U r h  k ln r   k ln
 
0

1
i 1  1  2  2

0

x 
1 0.08
ln  ln
 50  0.08
 80 



80
1

37.5 300 50 37.5 0.15 50


20


0
 U  3.244 W/(m 2  0 C)





(ii) Rate of heat flow


Q˙  U A
T
 U 2r LT  T   3.2442x0.08x1350  40  505.5 W/m
0 3 i 0
Ans
0 0 overall
(i) Overall Heat Transfer Coefficient based on inner surface area (U0):

96. Solution: Given Data: r1=37.5x10-3
m, r2=50x10-3
m, r3=80x10-3
m and ksteel=k1=50W/mK, kAsbe=k2=0.15W/mk,
hi=300W/m2K, h0=20W/m2K, Ti= 3500C and T0=400C.
(iii) Temperature drop across the tube and Asbestos
Resistance due tube:
R 
T across steel tube:
T  T  T  QR  505.5x9.157x104

ln(50 37.5)
1 1 1 2
1
0
 4
(m 0
C)/W  T1  0.46 C Ans
T across Asbestos
layer:
Res
R

due to Asbestos:

ln(80 50)
 0.5 (m 0
C)/W T  T  T  QR  505.5x0.5  252.6 0
C
Ans
2
2L
k
2x1x0.1
5
2 2 3 2


1 1 r1  r2  r1  r3 r1 1  1 0.0375  50 
0.0375  80 
37.5 1

2 0
2x1x50
ln(r2 r1 )
2Lk
1
9.157x10
istance
ln(r3 r2 )
 Overall Heat Transfer Coefficient based on inner surface area (Ui):
Check for Q:
2

97. U h k
    ln   ln  
r k r
r
1  1  2  2 
h
  300

ln 

50
37.5
0.15
ln 

50 80 i
x
20 
 U  6.92 W/(m 
C)
i i 3 0 






(ii) Rate of heat flow
˙ 


U
2r
L
T  T
 6.92 2x0.0375x1 350  40
 505.5 W/m Ans

98. 
Example-4:- A spherical vessel of ID 0.3m and thickness of 20mm is made of steel with
conductivity of 40 W/(m-K). The vessel is insulated with two layers of 60mm thickness of conductivity
0.05 and 0.15 W/(m-K). The inside surface is at –196°C. The outside is exposed to air at 30°C with
convection coefficient of 35 W/(m2-K). Determine the heat gain and also the surface temperatures
and the overall heat transfer coefficient based on the outside surface area of the metallic vessel.
Solution: Given Data: r1=0.15m, r2=0.17m, r3=0.23m, r4=0.29m, k1=40W/mK, k2=0.05W/mK, k3=0.15W/mK ,
h0=35W/m2K, Ti= -1960C and T0= 300C.
RTot
4 r 2
h k r r k r r k r r

1  1 
1  1 
1  
1  1 
1  
1
1 
1  r
1
h2


R

1

 1 i
1 
1
1  
1 
1
 
1  1 2  2  2 3  3  3 4 
4
1
 
1

1
0

 1 

1 
Tot
4 
40 0.15 0.17  0.05 0.17
0.23 0.15 0.23 0.29 

 2.948 0
C/W
2
0.29 x35

(i) Rate of heat flow
RTot 2.948
Q˙ 
T1  T 

196  30  76.66 W
Ans
r 2  1 1

1

1  r 2  11  r 2  1 1

  4
       

U0
4
k2
   
 r2 r3

4
k3
 k1  r1 r2

 r3 r4  h0

(ii) Overall heat transfer coefficient based on outer surface area: U0

99. 
0.292
1 
1
 
0.292
 1 
1
 
0.292
 1  1  
1
Ans


0

100. Given Data: r1=0.15m, r2=0.17m, r3=0.23m, r4=0.29m, k1=40W/mK,
k2=0.05W/mK, k3=0.15W/mK , h0=35W/m2K, Ti= -1960C and T0= 300C.
Check for Q:
Q˙  U A
T
U 4r 2
T T  0.3214x0.292
196  30 76.66 W Ans
0 0 overall 0 4 1 
(iii) Surface temperatures
Thermal Resistances:
1
Surface temperatures



1

 1.56x103 0
C/W
R   1 1  1 1
1 4k
 r
4x40
0.15
0.17

R 
2
1 
1
   2.44
0
C/W
T  T  QR  8.67  76.66x2.44  195.70
C
2 3 2
4k 
r
r 
1  1 2 
 1 

1  1
1
r  4x0.05
0.17
0.23

Ans
2  2 3 
1 
1
1  1 1 1  0

101. 4k
R3  
r

r
 
    0.477
0
C/W
T3  T4  QR3
 27.9  76.66x0.477 8.67 C Ans
3 
3
4

4x0.15
0.23
0.29

R 
0
1

1
 0.027 0
C/W T  T  QR  30  76.66x0.027 27.90
C
4  0
Ans
4r
2h
4
0
4x0.292
x35

102. 2
Example-5:- A hollow aluminum sphere (k =204 W/(m-K)) with an electrical heater in the
centre is used to determine the thermal conductivity of insulating materials. The inner and
outer radii of the sphere are 15 cm and 18 cm respectively and testing is done under steady
state conditions with the inner surface of the aluminum maintained at 2500C. In a particular
test, a spherical shell of insulation is cast on the outer surface of the aluminum sphere to a
thickness of 12cm. The system is in a room where the air temperature is 200C and the
convection coefficient is 30 W/(m2 – K). If 80 W are dissipated by the heater under steady
state conditions, what is the thermal conductivity of the insulating material?
Solution: Given Data: Q = 80W, r1=0.15m, r2=0.18m, r3=0.3m, k1=204 W/mK, h0=30W/m2K,
T1= 2500C and T0= 200C. To find: k2
Thermal Resistances: (i) Rate of heat flow
1  1 1
 1
R1   


1 1
 4 0
  4.33x10
C/W
˙ T  T  
T1 T0 
R  R  R
  4


Q  1 0
RTot
4k1  r1
r2

x204 
0.15
0.18

1 2 0
R2  4k r r
1  1 
1 
1 
1
1


0.177 0 C/W
 4xk 0.18  0.3 
 k
2
 80 
250  20
4.33x104  R  0.0295
R0  2  2 3

1
2
1
2


103. C
/
0W.02950
 R2 
2.845
0C/W
.
0
1
7
4r3
h0
4x0.3
x30
2
But R

2
0
C/W
Ans
2
 2.854 
0.177
 k 2  0.0622 W/(m  K)
k

104. K1
K2
2
(ii) When superior insulator (k ) next to pipe:

Example-6:- A thin hollow sphere of diameter 50cm contains a hot fluid. To reduce loss of
heat it is lagged by two layer of insulation of different thermal conductivity. The thickness of
each insulator being 30mm. The average thermal conductivity of one of them is four times
that of the other. Compare the heat transfer when:
1. The inferior insulator is next to the sphere
2. The superior insulator is next to the sphere. Find the % heat reduction in the
second.
Solution: Given Data: r1=0.25m, r2=0.28m, r3=0.31m,
Assuming k2=4k1 i.e, Material of thermal conductivity k2 is superior insulator (i.e, conducts more).
Q(˙i)When inferTi1
orT3
insulator (k1) next to pipe:T1 T3 
1
1  1  1 1  1  1 1  1  1  1 1
 1
 1 1 
 1.944k1T1 T3

T3
  
4 k 
r r  k 
r r
4
 


 
4k   

1  1 2  2 
2
3



k 
1 0.25
0.28

1 
0.28
0.31

T  T  T  T 

105. % of reduction:  2 1
˙
2
Q  
1 1 
1
1 3
1  1  1 1


1  1 
1
1 3
1  1 1
1

1
1 3
  2.214k T  T

  
 



 
   
0.28

k1
0.28



0.31

T3
4  k2  r1 r2 
k1  r2
r3


4 4k1
0.25
˙ ˙
Q  Q  2.211.94

x100 12.2%
Ans

106. hi
110
k
k2
Example-7: A furnace wall is of three layers, first layer of insulation brick of 12 cm
thickness of conductivity 0.6 W/mK. The face is exposed to gases at 870°C with a convection
coefficient of 110 W/m2K. This layer is backed by a 10cm layer of firebrick of conductivity 0.8
W/mK. There is a contact resistance between the layers of 2.6×10–4 m2-°C/W. The third layer
is the plate backing of 10mm thickness of conductivity 49 W/mK. The contact resistance
between the second and third layers is
1.5×10–4 m2-°C/W. The plate is exposed to air at 30°C with a convection coefficient of 15
W/m2K. Determine the heat flow, temperature drop due to contact resistances, the surface
temperatures and the overall heat transfer coefficient.
Solution: Given Data: L1=0.12m, k1=0.6 W/mK, L2=0.1m, k2=0.8W/mK,
Rs1= 2.6×10–4 m2-°C/W, L3=0.01m, k3=49 W/mK, Rs2= 1.5×10–4 m2-°C/W,
hi=110W/m2K, Ti= 8700C, h0=15W/m2K, and T0=300C.
Thermal Resistances: Let assume A = 1m2
R 
1 
1
 9.09x103
(m2
 K)/W, R 
L1
1
L1
3

0.12  0.2, R = 2.6×10–4
i s1
R


L2
0.1

0.125,
R = 1.5×10–4,   2.04x104 , R 
1 
0.067

1
h0
15
2
0.8 s2
k3
49
0.6
R 
0.01
3
0

107. 870  30 2091.1W/m2
RTot  Ri
 R1  Rs1  R2  Rs 2
 R0  0.402 (m2
 K)/W
(i) Rate of heat flow
Q˙ 
T1 T0 

108. Solution: Given Data: L1=0.12m, k1=0.6 W/mK, L2=0.1m, k2=0.8W/mK,
Rs1= 2.6×10–4 m2-°C/W, L3=0.01m, k3=49 W/mK, Rs2= 1.5×10–4 m2-°C/W,
hi=110W/m2K, Ti= 8700C, h0=15W/m2K, and T0=300C.
(ii) Surface temperatures
T1  Ti  QRi  870  2091.1x9.09x103
 851.00
C
T2  T1  QR1  851.0  2091.1x0.2
 432.80
C
Ans
Ans
Ts1  T2  TA   QRs1  2091.1x2.6x104
 0.113 0
C Ans
 T  T  T   432.8  0.113  432.70
C
Ans
A 2 s1
T3  TA  QR2  432.7  2091.1x0.125
 171.30
C Ans
Ts 2  T3  TB  QRs2  2091.1x1.5x104
 0.03 0
C Ans
  171.3  0.03  171.3
0C
 T  T  T
B 3
s 2
Ans
4
 170.80
C Ans Check for T : T  T  QR  30  2091.1x0.067  170.10
C

109. R 0.401
T4  T3  QR3  171.3  2091.1x2.04x10 4 4 0 0
(iii) Overall heat transfer coefficient: U
U  1

1
 2.494 W/(m2
 K) Ans

110. Example-8: Determine the heat transfer through the composite wall show in Fig.1. Take the
conductivities of A, B, C, D & E as 50, 10, 8, 20 & 30 W/m-K respectively, and inside and outside
convective heat-transfer coefficients as 200 and 20 W/m2-K. Assume one dimensional heat
transfer.
Solution: Given Data: LA=0.05m=LD, kA=50 W/mK, LB=LB=0.15m,
kB=10W/mK, kC=8W/mK, kD=20 W/mK, LE=0.08m, kE=30
W/mK, hi=200W/m2K, Ti= 8000C, h0=20W/m2K, and T0=1000C.
Let assume A = 1m2
Surface area for A, D and E are AA = AD=AE = A = 1m2
And Surface area for B and C are AB = AC= A/2 = 0.5m2
Thermal Resistances:
R 
i
1

1
 5x103
(m2
 K)/W, R

LA

0.05
 1x103
,
50
i
h A 200x1 A k
A A
LB
0.15

0.5x10
LC
0.15
 
0.0375,
0.5x8
1
1 1 R R
RB 
ABkB
LD

0.03,
C
R
 A k
C C
LE
Req
RB RC RB  RC
   Req  B C 
0.0167
R
 
0.05
3 R
 
0.08  2.67x103, R0 
1

1
A k
D D 1x20
 2.5x10 , E
A k
E E
1x30 0
h A 20x1
 0.05
A
D

111. RTot
 Ri  RA  Req  RD  RE  R0
 0.078 (m2
 K)/W
(i) Rate of heat flow Q
˙
 8989.3
W/m2
Ans

112. Example-9: A composite insulating wall has three layers of material held together by 3 cm diameter
aluminium rivet per 0.1 m2 of surface. The layers of material consists of 10 cm thick brick with hot
surface at 200°C, 1 cm thick wood with cold surface at 10°C. These two layers are interposed by
third layer of insulating material 25 cm thick. The conductivity of the materials are : kbrick = 0.93
W/m.K, kinsulation = 0.12 W/m.K, kwood = 0.175 W/m.K, kAluminium = 204 W/m.K. Assuming one
dimensional heat flow. Calculate: the percentage increase in heat transfer rate due to rivets.
Solution: Given Data: Aw = 0.1 m2, d = 3 cm = 0.03 m, L1 = 0.1 m, Ti = 200°C,
L2 = 0.25 m, L3 = 0.01 m, k1 = 0.93 W/m.K, T0 = 10°C, k2 = 0.12
W/m.K, k3 = 0.175 W/m.K, k4 = 204 W/m.K.
(i) Rate of heat flow without a rivet:
RTot
Q˙1 
Ti  T0 
RTot
(ii) Rate of heat flow with a rivet:
ARivet

d 2
4
x0.032
4  7.07x104
m2

113. Solution: Given Data: Aw = 0.1 m2, d = 3 cm = 0.03 m, L1 = 0.1 m, Ti = 200°C,
L2 = 0.25 m, L3 = 0.01 m, k1 = 0.93 W/m.K, T0 = 10°C, k2 = 0.12
W/m.K, k3 = 0.175 W/m.K, k4 = 204 W/m.K.
Thermal Resistance due to rivet:
With consideration of rivet, the net effective area of the
wall,
Now the wall resistance ;
RTot
and
(iii) % of increase of heat flow:

114. Example-10: A square plate heater (size : 15 cm × 15 cm) is inserted between two slabs. Slab A is
2 cm thick (k = 50 W/m.K) and slab B is 1 cm thick (k = 0.2 W/m.K). The outside heat transfer
coefficient on both sides of A and B are 200 and 50 W/m2.K, respectively. The temperature of
surrounding air is 25°C. If the rating of the heater is 1 kW, find : (i) Maximum temperature in the
system. (ii) Outer surface temperature of two slabs. Draw equivalent electrical circuit of the system.
Solution:
Given Data: A =0.15 m × 0.15 m = 0.0225 m2, LA = 0.02 m, LB = 0.01m,
kA = 50 W/m.K, kB = 0.2 W/m.K, hA = 200 W/m2.K, hB = 50 W/m2.K,
T∞ = 25°C, Q = 1 kW = 1000 W.
Thermal Resistances:
R

1

1
 0.22 K/W, R 
A
LA
 0.02
Conv,1
A
h A 200x0.0225 A k
A A 50x0.0225

0.0178,
R

LB

0.01
0.2x0.0225
 2.22, 1

1
A k
B B RConv,2

B
h A 50x0.0225

0.89
Left side equivalent
resistance:
B

115. Req,L  RA  RConv,1
 0.0178  0.22  0.2378 K/W
Right side equivalent resistance:
R  R  R
3.11K/W

116. 1
Req,R
 
0.2378 3.11
1 1
 T  245.9
C
eq,tot
 R  0.221
K/W
Req,tot
Total heat flow rate
1 1
Q 
˙ 1 

TT
 
Req,L
1000  1
0.221
T  25 Req,tot
Ans
0
Given Data: A =0.15 m × 0.15 m = 0.0225 m2, LA = 0.02 m, LB = 0.01m,
kA = 50 W/m.K, kB = 0.2 W/m.K, hA = 200 W/m2.K, hB = 50 W/m2.K,
T∞ = 25°C, Q = 1 kW = 1000 W.
(i) Rate of heat flow to left side (ii) Rate of heat flow to right side
Q˙1

T1 T  
T1  25 ....(i)
R 0.2378
Alternatively1
:
Q˙2  1 
R  1
T  25
T T   
3.11 ....(ii)
eq,L eq,R
2
But, Q˙  Q˙  Q˙  1000


0.2378
T  25 T  25
1 1
3.11
wher
e
1
 T  245.9 C
Ans
˙1
Now Q 
T T 
1 
Req,L

245.8  25
0.2378
 928.5
W
245.8  T 
L
0.0178
0 Q Q
Also Q˙1 
T  T 
1
L
RA
 928.5

 TL  229.3 C Ans

117. similarly Q˙2 
T1 T  
245.8  25
Req,R
3.11
T T 

71W
0
245.8  T
 

Ans

118. Q  kA dT
2
Example-11: A conical cylinder of length L and radius R1 and R2 (R1<R2) is fully insulated along the
outer surface. The surface of R1 is maintained at temperature T1 and the surface of R2 is
maintained at temperature T2 (T1>T2). Considering the heat flow only along the axis of the cylinder,
derive an expression for heat flow and also the temperature distribution along its length from its
radius smaller end. If L = 100 mm, R1=30 mm, R2= 50 mm, T1 = 2000C, T2=500C, k = 60 W/m-K, find
the rate of heat flow and also the temperature at a distance of 40 mm from the smaller radius end.
Solution:
Since the lateral surface is well insulated, the
conduction of heat is one-dimensional (x-direction),
using Fourier’s equation, we have,
x x
dx
....(i)
And from the geometry of the
figure,
R  R r  R L
tan  2 1

1
L x
an
d
dx 
(R2
dr
 R1)
Therefore eqn(i) can be written
as
dr R  R  k

119. L
Qxdx  k (r
2
)dT
 Qx
L
dr  k(r 2
)dT
 Qx   2 1
dT .........(ii)

120. x
r 2
 Q
dr
 
R2
 R1  k
dT
L
....(ii)
For total heat transfer, integrate the eqn(ii) from x=0 to L (i.e from r=R1 to R2) and T = T1 to T2, we
get
R
2
 1 1

R  R  k
Q dT  Qx    

2 1
T2 T1
x 
r 2
dr R  R  k
  2 1
L
T2
 
R R  L
r
R
1
x
 Q

 kR
R
1 2
L
1
2
T  T 
T
T
1
 1 2

....(iii)
 Temperature distribution:
For temperature distribution along the length of the
cone, integrate the eqn(ii) from x=0 to x (i.e from r=R1
to r) and
1
T = T to T, we
get,
r
dr R  R  k T

 1 1

R  R  k
2 1
Qx   2 1
L
dT  Q     
T  T 
1

r 2
x

R r  L

121. 1 1
r
R
1
T T1
 1

 r  R1

R2  R1  k 
kR1R2
 r  R1

R2  R1  k
 Q

x


rR L
1  
1 2
  T  T   T T    T  T 
 rR1 L
1
T T R R  r  R

T  T 


 
1 R1 1 r


 1 2  1   1
L

....(iv)

122. 1 0.03 1 0.038
 1
0
Given: If L = 100 mm, R1=30 mm, R2= 50 mm, T1 = 2000C, T2=500C, k = 60 W/m-K, find the rate of
heat flow and also the temperature at a distance of 40 mm from the smaller radius end.
Radius at distance x=40mm from the smaller end is,
R2  R1

r  R1

50  30 
r  30
L x 100
40
 r  38 mm or 0.038m
(i) Rate of heat flow
Q˙ 

kR1R2
1 2
T  T  
x60x0.03x0.05
L 0.1
200  50  424.1W Ans
(ii) Temperature distribution:
 1 R 1 r

T  T  200  T  







1
 
T1  T2
 1 R1 1 R2

200  50  1 0.03 1 0.05

x40mm
 T  121.05
C
An

123. 1
0
0
Example-12:A hollow cylinder with inner radius 30 mm and outer radius 50 mm is heated at the
inner surface at a rate of 105 W/m2 and dissipated heat by convection from outer surface into a
fluid at 80°C with heat transfer coefficient of 400 W/m2.K. There is no energy generation and
thermal conductivity of the material is constant at 15 W/m.K. Calculate the temperatures of
inside and outside surfaces of the cylinder.
Solution
Given : A hollow cylinder is heated at its inner surface, r1 =0.03 m,
= 80°C, h = 400 W/m2.K, k = 15
r2 = 0.05 m, q = 105 W/m2, T∞
W/m.K. To Find: T1 and T2.
Thermal Resistances:
(i) Rate of heat flow
˙ ˙
Q  q
(2
1
r L)  T  T 
R
1  
th
 105
(2x0.03L)  T  80x2L 0.084  T1  332 C Ans
Also Q˙  q˙(2r1L)  T1 T  R2
 105
(2x0.03L)  T  80

x2 L 0.05 2
 T  230 C
Ans
2

124. 1. Conduction:
Fourier’s law of conduction is given
by
QCond
T1
dT
 kA
dx
dT
 QCond
/ A  k dx
Q
 T  T

qCond
 T

On integration, q
 Cond

k
1 2
 k

W/m2
Cond
A  L  L

L

125. 2. Convection:
Newton’s law of cooling or heating is given
by
QConv  h A T  hA[TS  T ]

126. 1
Net radiation energy transfer from A1 to A2 is given
by
Q1 – 2 = A1F1 – 2 σ [ε1 T 4 − ε2 T 4]
1
2
Radiation Heat Transfer Coefficient:-
hr = ε σ [T 2 + T 2 ][T + T
]
s sur
s
sur
3. Radiation:
Stefan-Boltzmann law is given
by
Qrad
  
AT
4
 T 2 4

Qr = hrA[Ts – Tsur] = A ε σ [T 4 – T 4 ]
s sur
Formulae
Numerical on Basic Laws of
HT

127. Example 1.1:- Heat flux through a wood slab 50 mm thick, whose inner and outer surface
temperatures are 400C and 200C respectively, has been determined to be 40 W/m2. What is
the thermal conductivity of the wood slab?
Solution: Given:- T1=400C; T2=200C; L=0.05m, q = Q/A = 40 W/m2. To find: k
. By Fourier’s law of
conduction,
q  Q / A  k
T1  T2 
T1
L
 40  k
40  20
0.05
 k 
0.1
W/(m-0C)
Ans
T2
q
k
L

128. Example 1.2:- A concrete wall, which has a surface area of 20m2 and thickness
30cm, separates conditioned room air from ambient air. The temperature of the inner
surface of the wall is 250C and the thermal conductivity of the wall is 1.5 W/(m-K).
Determine the heat loss through the wall for ambient temperature of: (i) ─150C
during winter and (ii) 380C during summer conditions. Comments on the results
Solution: Given:- A=20m2, L=0.3m, T1=250C; k=1.5 W/(m-K), T2=200C; To find: Q
By Fourier’s law of
conduction, Rate of heat
transfer,
Q  kA
T1  T2 
(ii) For Summer, T2 = 380C,
By Fourier’s law of
conduction, Rate of heat
transfer,
Q  kA
T1  T2 
T1
Inne
r
Surfac
e
T2
q
Ambien
t
L
 Q 
1.5x20x
25  (15) 
0.3
4000
W
L
 Q  1.5x20x
25  38
0.3
 1300 W
Heat flux, q  Q / A 4000 / 20  200 W/m2
Heat flux, q  Q / A  1300 / 20
 65 W/m2
(i) For Winter, T2 = -150C,
L
k

129. Comments: Comments:
Heat flows from ambient to inner surface

130. Example 1.3:- What is the thickness required of a masonry wall having a thermal
conductivity of 0.75 W/(m-K), if the heat transfer rate is to be 80% of the rate through
another wall having thermal conductivity of 0.25 W/(m-K) and a thickness of 100 mm? Both
walls are subjected to the same temperature difference.
Solution: Given:- For masonry wall, L1=?, k1=0.75 W/(m-K), Q1;
For unknown wall, L2=0.1m, k2=0.25 W/(m-K), Q2,
Also given, Q1 = 0.8Q2 To find: L1
(i) For masonry,
By Fourier’s law of
conduction, Rate of heat
transfer,
(ii) For Unknown ,
By Fourier’s law of
conduction, Rate of heat
transfer,
T1
Inne
r
Surfac
e
T2
Q
Ambien
t
1 2
T  T 
Q1  k1A
L1
T1  T2  L2
Q2  k 2 A
L
k

131. 
k1
x
L2
k
L
For the same temp. difference,
Q1
we can write Q
 0.8 
0.75
x
0.1
0.25
L
 L  0.375 m or 375 mm Ans
1

132. Example 1.4:- A large surface at 500C is exposed to air at 200C. If the heat transfer
coefficient between the surface and the air is 15 W/(m2-K), determine the heat transferred
from 5 m2 of the surface area in 7 hours.
Solution: Given:- Ts=500C, T=200C h=15 W/(m2-K), A = 5m2; time, t=7 hrs To find: Q
By Newton’s law of
cooling, Rate of heat
transfer,
Q˙  hATs  T 
 Q˙  15x550  20
 Q˙  2250 W or
J/s
Air
Q
Solid surface

133. Amount of heat transfer, Q  Q˙ xt 
2250x7x3600
 56.7
MJ
Ans

134. Example 1.5:- A 25 cm diameter sphere at 1200C is suspended in air at 200C. If the
convective heat transfer coefficient between the surface and air is 15 W/(m2-K), determine
the heat loss from the sphere. What will be the change in heat loss per unit length if it is a
cylinder of same diameter.
Solution: Given:- D=0.25m, Ts=1200C, T=200C h=15 W/(m2-K), To find: Q
(i) Sphere:
By Newton’s law of
cooling, Rate of heat
transfer,
Q˙  hAs Ts  T   h (4R 2
)T 
T 
s
 Q˙ 
15x(4xx0.1252 )120  20
 294.52 W
Air
Ans
(ii) Cylinder:
By Newton’s law of
cooling, Rate of heat
transfer,
Q

135.  T 
Q˙  hAs Ts  T   h (DL)Ts
˙
 


Ans

136. Example 1.6:- Heavy water flows inside a 6cm diameter 2m long pipe with its
surface temperature maintained at 1500C. Determine the heat transfer rate from the
tube to the water if the heat transfer co-efficient between the water and the tube is
2500 W/(m2-0C) and water temperature is 400C.
Solution: Given:- D=0.06m, L = 2m, Ts=1500C, T=400C, h=2500 W/(m2-K), To find: Q
Cylinder:
By Newton’s law of
cooling, Rate of heat
transfer,
Q˙  hAs Ts  T   h (DL)Ts 
T 
 Q˙ 
2500x(x0.06x2)150  40
Ts = 1500C
Water
Q
T∞ = 400C
h = 2500 W/(m2-K)
D
 103.7
kW
Ans L
Q
Water

137. s 1
2 2 -8 2 2
= ε σ [T1 + T2 ][T1 + T2] = 1x5.67x10 [473 + 303 ][473 +
303]
Example 1.7:- A surface with an area of 2m2 has a temperature of 2000C. It exchanges
heat with another surface at 300C by radiation. Determine the heat exchange
assuming that the surfaces are black bodies. Also find the value of thermal resistance
and equivalent convection co-efficient.
Solution: Given:- A=2m2, T1=2000C+273=473K, T2=300C+273=303K, Black bodies
1=2=1,
To find: Q, Rth and hr
By Stefan Boltzmann law of
radiation, Rate of heat transfer,
Q̇    A T
 T
2
4 4
  8
 5.67x10
x2x1
4 4
473  303   4720.4
W
Ans
Radiation heat transfer coefficient is
given by
hr
= 13.88 W/(m2-K)
Ans
Thermal resistance is given
by
Alternativel
y:
R

1

1
 0.036
K/W
R 
T

200  30  0.036 K/W
Ans

138. r
h A 13.88x2
th th
Q
˙
4720.4

139. (i) Q for 1=1
s 1
 4R 2
 4 (0.1/
2)2
1
Example 1.8:- A sphere 10 cm in diameter is suspended inside a large evacuated
chamber whose walls are kept at 300 K. If the surface of the sphere is black and
maintained at 500 K what would be the radiation heat loss from the sphere to the
walls of the chamber?. What would be the heat loss if the surface of the sphere has an
emissivity of 0.8?.
Solution: Given:- D=0.1m, T1=500K, T2=300K, To find: (i) Q for 1=1, (ii) Q for 1=0.8
Surface area of the sphere,
As
 0.0314
m2
By Stefan Boltzmann law of
radiation,
Rate of heat
transfer,
Q
˙
B
   A
T
4
 T 2
4
 8
 5.67x10 x1x0.0314 500 4 4
 300   96.85
W
Ans
Radiation heat transfer rate
is
(ii) Q for 1=0.8

140. Q˙ G A T 4
  
 T24
 5.67x108 x0.8x0.0314 5004  3004
  77.48 W
Ans
OR
Q˙
G
 Q˙ B 
0.8x96.85
 77.48
W
Ans

141. Example 1.9:- A long copper pipe of 100mm diameter with its surface temperature
of 700C is passing through a room which is maintained at a temperature of 250C.
Calculate the heat lost by the pipe per unit length assuming the heat transfer co-
efficient of air as 7 W/(m2- 0C) and emissivity of the pipe surface as 0.75.
Solution: Given: D =0.1m, TW=700C or 343K, T =250C or 298K, h= 7 W/(m2-K),  =
0.75, To find: Q
Surface area of the pipe, As DL x0.1x1  0.314 m2
Total rate of heat transfer at the surface of the pipe is,
Q˙
tot  Q˙
Conv
 Q˙ r
Q̇tot  hAs TS  T    A T  T
4 4 
s S 
 7x0.31470  25 0.75x5.67x108 x0.3143434  2984 
 178.4 W/m Ans

142. Energy balance equation for the top
surface of the plate is given by
Qsolar = Qr + Qconv
qsolar As = ε σ As (T 4 - T 4) + h As (Ts - T∞)
s ∞
Example 1.10:- A flat plate has one surface insulated and the other surface exposed to
the sun. The exposed surface absorbs the solar radiation at a rate of 800 W/m2 and
dissipates heat by both convection and radiation into the ambient at 300K. If the
emissivity of the surface is 0.9 and the surface heat transfer coefficient is 12
W/(m2-K), determine the surface temperature of the plate.
Solution: Given:- QSolar= 800 W/m2, T=300K, =0.9, h = 12 W/(m2-K). To find: (i) TS
— 8 4
4
 800 = 0.9 x 5.67 x 10 x (T – 300 ) + 12 x (T –
s s 300)
…(i)
T =342.5
K
By trial and error method, we
get,
Ans

143. Example 1.11:- The solar radiation incident on the outside surface of an aluminum
shading device is 1000 W/m2. Aluminum absorbs 12% of the incident solar energy and
dissipates it by convection from the back surface and by combined convection and
radiation from the outer surface. The emissivity of aluminum is 0.10 and the convective
heat transfer coefficient or both the surfaces is 15 W/(m2-K). The ambient temperature
of air may be taken as 200C. Determine the temperature of the shading device.
Solution: Given:- qsolar = 1000 W/m2; absorptivity of aluminum, α = 0.12; ε of aluminum = 0.10; h
= 15 W /(m2-K) ; T∞ = 20 + 273 = 293 K; Solar radiation flux absorbed by aluminum,
qa= α qsolar= 0.12 x1000 = 120 W/m2. To find: (i) TS
The energy absorbed by aluminum is dissipated by
convection from the back surface and by combined
convection and radiation from the outer surface.
Hence the energy balance equation can be written as
Outer
surface
Back
surface

144. Given:- qsolar = 1000 W/m2; absorptivity of aluminum, α = 0.12; ε of aluminum = 0.10;
h = 15 W /(m2-K) ; T∞ = 20 + 273 = 293 K; Solar radiation flux absorbed by
aluminum, qa= α qsolar= 0.12 x1000 = 120 W/m2. To find: (i) TS
Hence the energy balance equation can be
written as
…(i)
By trial and error method, we get, TS = 297 K
Ans
Outer
surface
Back surface

145. Example 1.12:- A hot plate made of stainless steel of size 600mm x 750mm and
thickness 20mm is maintained at 2500C. Atmospheric air at 270C blows over this plate.
Calculate the heat transfer and inside plate temperature if 300W is lost from the
surface of the plate by radiation. Assume convective heat transfer co-efficient of
30W/(m2-K) and thermal conductivity of the plate as 45 W/(m-0C).
Solution: Given: A = 600mm x 750mm =0.45m2 , L=0.02m, T =270C,
TW=2500C, Qr=300W, h= 30W/(m2-K), k=4500W/(m-0C), To find: T1
Rate of heat transfer by convection is,
Q˙ Conv  hAs TW  T  30x0.45250  27 3010.5 W
By the energy balance at CV, we can be written as
Q˙ Cond  Q˙
Conv
 Q˙ r
kA
T1  TW   Q˙
L
Conv  Q˙ r  45x0.45
1
T  250  3010.5  300
0.02

146.  T1 = 253.3 0C Ans

147. 1 = 2000C
 =1 T2 = 300C
Air T∞ = 250C
1
A=2m2

=1
1
A=2m2
Ts = 700C QConv
Qr
h = 7 W/(m2-K)
D
T1
k
TW h, T
Inne
r
QCond
Qr
Ambien
t
Surfac
e QConv
L

148. L

149. Heat Conduction
L
• Fourier Law
dT
Cross-
Sectional
Area
Thermal
Conductivity
[W/m-K]
Materials Property
’cold
Convection Thermal Radiation
Newton's law of
cooling
Stefan-Boltzmann
Law for Blackbody
Convective Heat
Transfer Coefficient
Flow dependent
• Heat Flux
= —k ——-k9T}
tW/n*t
• Natural
Convection
• Forced Convection
Q
’
— AmT
4
o=5.67x10 W/
m2K*
• Heat transfer
( ,4
i actor
Einissisitv of
F=1 for two
parallel plates