PC ME 501.pdf

PC ME 501
Heat Transfer
By
Dr Abhishek Samanta
Associate Professor
Department of Mechanical Engineering
HIT, Haldia

THERMODYNAMICS AND HEAT
TRANSFER
• The science of heat transfer is concerned with the
calculation of the rate at which heat flows within a
medium, across an inter face, or from one surface to
another, and the associated temperature distribution.
• Thermodynamics deals with systems in equilibrium
and calculates the energy transferred to change a system
from one equilibrium state to another.

For example, if 1 kg ingot of iron is quenched from
1000°C to 100°C in an oil bath, thermodynamics tells us
that the loss in internal energy of the ingot is
A closed system containing a fixed mass of a solid (Fig.)
has a volume V(m3) and density r(kg/m3).
There is heat transfer into the system at a rate Q(W), and
heat may be generated internally within the solid, say, by
nuclear fission or electrical current at a rate QG(W).
The principle of energy conservation requires that over a
time interval Δt(s).

HEAT TRANSFER
• Heat always moves from a warmer place to a
cooler place.
• Hot objects in a cooler room will cool to room
temperature.
• Cold objects in a warmer room will heat up to
room temperature.

• If a cup of coffee and a ice were left on the
table in this room what would happen to them?
Why?
• The cup of coffee will cool until it reaches
room temperature. The ice will melt and then
the liquid will warm to room temperature.

Heat Transfer Methods
• Heat transfers in three ways:
–Conduction
–Convection
–Radiation

Conduction
When you heat a metal strip at one end, the heat travels to the other end.
As you heat the metal, the particles vibrate, these vibrations make the adjacent
particles vibrate, and so on and so on, the vibrations are passed along the metal
and so is the heat. We call this?
Conduction

Metals are different
The outer e______ of metal atoms
drift, and are free to move.
When the metal is heated, this
‘sea of electrons’ gain k_____
energy and transfer it
throughout the metal.
Insulators, such as w___ and p____, do not have this ‘sea of
electrons’ which is why they do not conduct heat as well as
metals.
lectrons
inetic
ood lastic

Why does metal feel colder than wood, if they
are both at the same temperature?
Metal is a conductor, wood is an insulator. Metal
conducts the heat away from your hands. Wood does not
conduct the heat away from your hands as well as the
metal, so the wood feels warmer than the metal.

Convection
What happens to the particles in a liquid or a
gas when you heat them?
The particles spread out and
become less dense.
This effects fluid movement.
What is a fluid?
A liquid or gas.

Fluid movement
Cooler, more d____, fluids
sink through w_____, less
dense fluids.
In effect, warmer liquids and gases r___
up.
Cooler liquids and gases s___.
ense
armer
ise
ink

Water movement
Hot water
rises
Cooler
water sinks
Convection
current
Cools at the
surface

Why is it windy at the seaside?

Cold air sinks
Where is the
freezer
compartment
put in a fridge?
Freezer
compartment
It is put at the top,
because cool air
sinks, so it cools the
food on the way
down.
It is warmer at
the bottom, so
this warmer air
rises and a
convection
current is set up.

The third method of heat transfer
How does heat energy get
from the Sun to the Earth? There are no particles between
the Sun and the Earth so it
CANNOT travel by conduction or
by convection.
?
RADIATION

Radiation
Radiation travels in straight lines
True/False
Radiation can travel through a vacuum
True/False
Radiation requires particles to travel
True/False
Radiation travels at the speed of light
True/False

Emission experiment
Four containers were filled with warm water. Which
container would have the warmest water after ten minutes?
Shiny metal
Dull metal
Dull black
Shiny black
The __________ container would be the warmest after ten
minutes because its shiny surface reflects heat _______ back
into the container so less is lost. The ________ container
would be the coolest because it is the best at _______ heat
radiation.
shiny metal
radiation
dull black
emitting

Absorption experiment
Four containers were placed equidistant from a heater. Which
container would have the warmest water after ten minutes?
The __________ container would be the warmest after ten
minutes because its surface absorbs heat _______ the best.
The _________ container would be the coolest because it is
the poorest at __________ heat radiation.
dull black
radiation
shiny metal
absorbing
Shiny metal
Dull metal
Dull black
Shiny black

Convection questions
Why are boilers placed beneath hot water
tanks in people’s homes?
Hot water rises.
So when the boiler heats the water, and the hot water rises, the water tank
is filled with hot water.
Why does hot air rise and cold air sink?
Cool air is more dense than warm air, so the cool air ‘falls
through’ the warm air.

Radiation questions
Why are houses painted white in hot countries?
White reflects heat radiation and keeps the house cooler.
Why are shiny foil blankets wrapped around marathon runners at the end of a
race?
The shiny metal reflects the heat radiation from the runner back in, this stops
the runner getting cold.

1. Which of the following is not a
method of heat transfer?
A. Radiation
B. Insulation
C. Conduction
D. Convection

2. In which of the following are
the particles closest together?
A. Solid
B. Liquid
C. Gas
D. Fluid

3. How does heat energy reach
the Earth from the Sun?
A. Radiation
B. Conduction
C. Convection
D. Insulation

4. Which is the best surface for
reflecting heat radiation?
A. Shiny white
B. Dull white
C. Shiny black
D. Dull black

5. Which is the best surface for
absorbing heat radiation?
A. Shiny white
B. Dull white
C. Shiny black
D. Dull black

HEAT CONDUCTION
Conduction is the transfer of energy from the more energetic particles of a substance to the
adjacent less energetic ones as a result of interactions between the particles.
 Conduction can take place in solids, liquids, or gases.
 In gases and liquids, conduction is due to the collisions and diffusion of the molecules
during their random motion.
 In solids, it is due to the combination of vibrations of the molecules in a lattice and the
energy transport by free electrons.
 The rate of heat conduction through a medium depends on the geometry of the
medium, its thickness, and the material of the medium, as well as the temperature
difference across the medium.

We know that a hot water tank will lose heat at a higher rate when the
temperature of the room housing the tank is lowered.
Further, the larger the tank, the larger the surface area and thus the rate of heat
loss.
Consider steady heat conduction through a large plane wall of thickness Δx =L
and area A, as shown in Fig.
The temperature difference across the wall is ΔT = T2-T1.
Experiments have shown that the rate of heat transfer Q through the wall is
doubled when the temperature difference ΔT across the wall or the area A
normal to the direction of heat transfer is doubled, but is halved when the
wall thickness L is doubled.
Thus we conclude that the rate of heat conduction through a plane layer is
proportional to the temperature difference across the layer and the heat transfer
area, but is inversely proportional to the thickness of the layer.

k is the constant of proportionality, which is a property of the material. Which is a
measure of the ability of a material to conduct heat. This property of the material is
called thermal conductivity (W/m K).

Fourier’s law (after the French scientist J.B.J. Fourier who proposed it in 1822) of heat
conduction states that the rate of heat transfer is linearly proportional to the temperature
gradient. For one-dimensional or unidirectional heat conduction
where qk is the rate of heat flux (a vector) in W/m2,
dT/dx is the temperature gradient in the direction of heat flow x and
k is the constant of proportionality, which is a property of the material through which
heat propagates. This property of the material is called thermal conductivity (W/m K).
The negative sign is used because heat flows from a high to a low temperature and the
slope dT/dx is negative

The magnitude of the thermal conductivity k for a given substance very much
depends on its microscopic structure and also tends to vary some what with
temperature.

For the simple case of steady-state one-dimensional heat flow through a plane wall
(Fig.), the temperature gradient and the heat flow do not vary with time, so that from
Eq.
where the temperature at the left face (x = 0) is uniform at T1 and the temperature
at the right face (x = L) is uniform at T2.

Since dt/dx = – qk/k, for the same ,
if k is low (i.e. for an insulator), dT/dx will be large i.e., there will be a large
temperature difference across the wall, and
if k is high (i.e. for a conductor), dT/dx will be small, or there will be a small
temperature difference across the wall (Fig.).

Resistance Concept
Heat flow has an analogy in the flow of electricity.
Ohm’s law states that the current I (Ampere) flowing through a wire (Fig.) is
equal to the voltage potential E1 – E2 (V), divided by the electrical resistance Re
(W) or

where Rk = L/kA is the conductive thermal
resistance to heat flow offered by the wall

The ratio k/L is the thermal conductance per unit area.
where ζ (= 1/ρ) is the electrical conductivity [(Ω m)–1 or mho] and dE/dl is the potential
gradient (V/m).
The similarity of Eqs (1.5) and (1.9) can be noticed. The reciprocal of the thermal resistance
is referred to as the thermal conductance Kk, defined by
For many materials, thermal conductivity can be approximated as a linear function of
temperature over limited temperature ranges:
where βk is an empirical constant and k0 is the value of thermal conductivity at a
reference temperature.

Composite Walls
 For a composite wall, as shown in Fig., there are three resistances in series.
 The rate of heat conduction is the same through all sections. The slope of the temperature
profile in each depends on the thermal conductivity k of the material of that section.
 The lower the k, the more will be the slope and the higher is the temperature difference.
 The higher the k, the less will be the slope and the lower is the temperature difference.

where T2 and T3 are the interface temperatures. The walls are assumed to be in good
thermal contact, with no contact resistance.
The total thermal resistance

Conduction can occur in a wall with two different materials in parallel (Fig. ).
If the temperatures over the left and right faces are uniform at T1 and T2, the
equivalent thermal circuit is shown to the right of the physical system.
The total resistance R is given by

Heat is transferred through a more complex composite structure involving thermal
resistances in series and in parallel. The total resistance is

Contact-Resistance
Thermal contact resistance develops when two conducting surfaces do not fit
tightly together and a thin layer of fluid is trapped between them.

This resistance is primarily a function of
 Surface roughness,
 The pressure holding the two surfaces in contact,
 The interface fluid
 The interface temperature.
If the heat flux through the two solid surfaces in contact is and the temperature
difference across the fluid gap is ΔTi (= Ts1 – Ts2), the interface resistance Ri is
defined by

An increase in contact pressure can reduce the contact resistance
significantly.
The interfacial fluid also affects the thermal resistance
 Putting a viscous liquid like glycerin on the interface reduces the contact
resistance 10 times with respect to air at a given pressure.
 A thermally conducting liquid called a thermal grease such as silicone oil is
applied between the contact surfaces before they are pressed against each
other.
 It is commonly done when attaching electronic components such as power
transistors to heat sinks.

The variation of the thermal
conductivity of various solids,
liquids, and gases with
temperature.

Thermal Diffusivity
𝛼 =
𝐻𝑒𝑎𝑡 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑒𝑑
𝐻𝑒𝑎𝑡 𝑠𝑡𝑜𝑟𝑒𝑑
=
𝑘
𝜌𝐶𝑝
(m2/s)
 The product 𝜌Cp is called the heat capacity of a material.
 Both the specific heat Cp and the heat capacity 𝜌Cp represent the heat storage
capability of a material.
 But Cp expresses it per unit mass whereas 𝜌Cp expresses it per unit volume, as
can be noticed from their units J/kg · °C and J/m3 · °C, respectively.
 Another material property that appears in the transient heat conduction analysis is
the thermal diffusivity, which represents how fast heat diffuses through a material
and is defined as

Note:
 The thermal conductivity k represents how well a material conducts heat, and the heat
capacity 𝜌Cp represents how much energy a material stores per unit volume.
 Therefore, the thermal diffusivity of a material can be viewed as the ratio of the heat
conducted through the material to the heat stored per unit volume.
 A material that has a high thermal conductivity or a low heat capacity will obviously have
a large thermal diffusivity.
 The larger the thermal diffusivity, the faster the propagation of heat into the medium.
 A small value of thermal diffusivity means that heat is mostly absorbed by the material
and a small amount of heat will be conducted further.

Heat Conduction Through a Cylinder
 Assume that the inside and outside surfaces
of the cylinder temperatures are T1 and T2
respectively, and T1 is greater than T2.
 Heat will be assumed to be flowing under
steady state only in the radial direction, and
there is no heat conduction along the length
or the periphery of the cylinder.
 The rate of heat transfer through the thin
cylinder of thickness dr is given by

Heat Conduction Through a Sphere

CONVECTION HEAT TRANSFER
Convection is the mode of energy transfer between a solid surface and the adjacent
liquid or gas that is in motion, and it involves the combined effects of conduction and
fluid motion.
 In the absence of any bulk fluid motion, heat transfer between a solid surface and the
adjacent fluid is by pure conduction.
 The presence of bulk motion of the fluid enhances the heat transfer between the solid
surface and the fluid, but it also complicates the determination of heat transfer rates.

When a temperature difference produces a density difference which results in mass
movement (Fig.), the process is called free or natural convection.
When the mass motion of the fluid is caused by an external device like a pump,
compressor, blower or fan, the process is called forced convection (Fig.).

A thermal boundary layer thickness δt is defined in an exactly analogous manner to
the velocity boundary layer thickness δ.
 Within this layer, the temperature varies from Tw at the wall to 𝑇∞ in the
undisturbed flow.
 Since temperature within the boundary layer approaches 𝑇∞ asymptotically, δt is
defined as the thickness at which Tw – T = 0.99 (Tw – 𝑇∞). In general, δt is not equal
to δ.
Ludwig Prandtl suggested that the field of flow can be divided into two regions:
A thin layer next to the wall, which he called the boundary layer where the shear
stress is confined,
The region outside this layer, where the fluid is “ideal” i.e., nonviscous and
incompressible.
The boundary layer thickness δ is defined as the distance from the wall where u =
0.99𝑢∞ i.e., the fluid velocity is 99% of the free stream velocity. As a result of viscous
forces the velocity of the fluid is zero at the wall and increases to 𝑢∞
.

Velocity and temperature profiles for natural convection heat transfer

where kf is the thermal conductivity of
the fluid film.
The rate of convection heat transfer from
the wall to the fluid
Heat transfer through a stationary fluid film

 The rate of heat transfer Qc increases with the increase in the value of heat
transfer coefficient hc.
 The higher the value of kf and the lower the value of δt, the higher will be the
value of hc, and hence, Qc.
 As the velocity of the fluid increases, film thickness δt decreases and hc
increases.
 For gases kf is low, and so Qc will also be low, compared to a liquid.
The film or surface coefficient of heat transfer
hc may be defined as
The above equation is known as Newton’s law of cooling.

Strictly speaking, convection applies to fluid motion. The mechanism of heat transfer is by
conduction:

COMBINED CONDUCTION AND CONVECTION
Some common flows: (a) Forced flow in a pipe, Re @ 50,000. The flow is initially laminar
because of ‘bell-mouth’ entrance, but becomes turbulent downstream. (b) Laminar forced flow
over a cylinder, Re = 25. (c) Forced flow through a tube bank, (d) Laminar and turbulent natural
convection boundary layers on vertical walls. (e) Laminar natural convection about a heated
horizontal plate.

 The term forced convection is only applied to flows in which the effects of
these buoyancy forces are negligible.
 In some flows in which a forced velocity exists, the effects of these buoyancy
forces may not be negligible. Such flows are termed combined or mixed, free
and forced convection flows.
Convective flows can be classified as external and internal flows (Fig.).
 External flows involve a flow over the outer surface of a body and
 Internal flows involve the flow through a duct or channel.

The rate of heat transfer from the wall to the mean or bulk temperature of the fluid is given
by
where hc is the heat transfer coefficient (W/m2 K) and Tb is the bulk
temperature of the fluid.
local rate of heat transfer per unit area, the local heat
transfer coefficient, the local wall temperature and the
local fluid (bulk) temperature =qw, hc, Tw and Tb,
These vary with position on the wall or surface

a mean or average heat transfer coefficient, ℎ𝑐 such that if Q is the total heat transfer
rate from the entire surface of area A, then

OVERALL HEAT TRANSFER COEFFICIENT

RADIATION HEAT TRANSFER
Radiation is the energy emitted by matter in the form of electromagnetic waves
(or photons) as a result of the changes in the electronic configurations of the atoms
or molecules.
 The transfer of energy by radiation does not require the presence of an intervening
medium.
 Energy transfer by radiation is fastest (at the speed of light) and it suffers no
attenuation in a vacuum.
 Radiation differs from other forms of electromagnetic radiation such as x-rays,
gamma rays, microwaves, radio waves, and television waves that are not related to
temperature.
 All bodies at a temperature above absolute zero emit thermal radiation.

If Q is the total radiant energy incident upon the surface of a body, some part of it
(Qa) will be absorbed, some (Qr) will be reflected and some (Qt) will be
transmitted through the body.
where α is known as absorptivity, ρ as reflectivity and η as transmissivity.
For an opaque body, η = 0 and so α + ρ = 1. Most solids are opaque. By increasing ρ
with high surface polishing, α can be decreased.

The radiant heat exchange between two gray bodies at temperatures T1 and T2 depends on how
the two bodies view each other, and their emissivities, and it is given by
(1)
Black body

where ε1 and ε2 are the emissivities of the two bodies of surface areas A1 and A2, and F12 is the
view factor of two similar black bodies, or the fraction of energy that leaves the black surface
1 and is incident on the black surface 2. It can be shown that
which is known as the reciprocity theorem.

The T4 dependence of radiant heat transfer.
For the special case of surface 1 surrounded by surface 2, where either area A1 is small as
compared to area A2, or surface 2 is nearly black, F12 ≈ε1 and Eq. (1) becomes
When T1 and T2 are not too different, it is convenient to linearize the Eq. by factoring
the term (T4
1 – T4
2) to obtain

The radiation heat transfer coefficient at room temperature is about six times the surface
emissivity.
For T1 = 320 K and T2 = 300 K, the error incurred in using the approximation of Eq. (2) is
only 0.1%; for T1 = 400 K and T2 = 300 K, the error is 2%.
(2)

Combined Convection and Radiation
Heat is transferred from a hot body both by natural convection and radiation.
Rate of heat transfer by natural convection,
and that by radiation
where hr is known as the radiation heat transfer coefficient (W/m2K).

Combined Conduction, Convection
and Radiation
Tꚙ Tꚙ

THERMAL INSULATION
Apparent thermal conductivity of an insulating material accounting for
conduction through solid material, and conduction or convection through the
air space as well as radiation

Types of insulation materials:
1. Fibrous -- Mineral wool at temperatures below 700°C, and fibre
glass for temperatures below 200°C. Between 700°C
and 1700°C, one can use refractory fibres such as
alumina (Al2O3) or silica (SiO2).
2. Cellular----Polyurethane and expanded polystyrene foam
3. Granular---Perlite powder, diatomaceous silica, and vermiculite

Insulation material property
 Low thermal conductivity,
 Should have structural rigidity,
 Low density,
 Less degradation,
 Good chemical stability
 Low cost.

DIFFUSION AND MASS TRANSFER
Diffusional mass transfer occurs at a microscopic or molecular level which deals with the
transport of one constituent of a fluid solution or gas mixture from a region of higher
concentration to a region of lower concentration (mol/m3).
Heat is transferred in a direction which reduces an existing temperature gradient, and
mass is transferred in a direction which reduces an existing concentration gradient.

The rate of molecular diffusion is proportional to the concentration gradient
where NA/A is the diffusion rate per unit area (kg mol/m2s) of the diffusing species
A, and dCA/dy is the concentration gradient (kgmol/m4) in the direction of
diffusion Y, CA being the concentration of A in kg mol/m3
where D is the constant of proportionality, called the diffusivity (m2/s).
It is called the Fick’s law of diffusion, which is similar to Fourier’s law of heat
conduction.

The mass fraction of chemical species is defined as
Fick’s law of diffusion governing ordinary diffusion, states that the local mass flux of a
chemical species is proportional to the negative of local concentration gradient.

Fick’s law then gives the diffusion mass flux j1 (kg/m2 s) of species 1 in a
binary mixture of species 1 and 2 as
for one-dimensional diffusion in the x-direction.
where ρD1–2 is the constant of proportionality, ρ(kg/m3) is the local mixture
density and D1–2 is the binary diffusion coefficient or mass diffusivity (m2/s).

Mass convection is essentially identical to heat convection.
The flow may be forced or natural, internal or external, and laminar or
turbulent.
Newton’s law of cooling we may write
Here, Jm1 (kg/m2 s) is the mass transfer coefficient.
The mass transfer for laminar flow in a tube like heat convection
where D is the tube diameter.

The average mass transfer coefficient is

Q. A furnace wall has the inside surface temperature of 1100°C, while the ambient air
temperature is 25°C. The wall consists of 125 mm thick refractory bricks (k = 1.6
W/mK), 125 mm thick firebricks (k = 0.3 W/mK) and 12 mm thick plaster (k = 0.14
W/mK). There is an air gap which offers a thermal resistance of 0.16 K/W. The heat
transfer coefficient on the outside wall to the air is 17 W/m2K. To find: (a) The rate of
heat loss per unit area of wall surface, (b) the interface temperatures throughout the
wall, and (c) the temperature of the outside surface of the wall.

Q. A steam pipe made of steel (k = 58 W/mK) has i.d. of 160 mm and o.d. of 170
mm. The saturated steam flowing through it is at 300°C, while the ambient air is at
50°C. It has two layers of insulation, the inner layer (k = 0.17 W/mK) is 30 mm
thick and the outer layer (k = 0.023 W/mK) is 50 mm thick, the heat transfer coeffi
cients on the inside and outside walls are 30 and 5.8 W/m2K respectively. To find:
The rate of heat loss per unit length of the pipe.

Q. A plastic pipe (k = 0.5 W/mK) of i.d. 3 cm and o.d. 4 cm carries a fl uid of
average temperature 100°C and h = 300 W/m2K. The rate of heat transfer per
unit length is 500 W/m. To find: (i) The outside surface temperature of pipe, (ii)
the overall heat transfer coefficient based on outside area.

Q. The roof of an electrically heated home is 6 m long, 8 m wide, and 0.25 m
thick, and is made of a flat layer of concrete whose thermal conductivity is k 0.8
W/m · °C (Fig.). The temperatures of the inner and the outer surfaces of the roof
one night are measured to be 15°C and 4°C, respectively, for a period of 10 hours.
Determine (a) the rate of heat loss through the roof that night and (b) the cost of
that heat loss to the home owner if the cost of electricity is $0.08/kWh.

Q. A 2-m-long, 0.3-cm-diameter electrical wire extends across a room at 15°C, as
shown in Fig.. Heat is generated in the wire as a result of resistance heating, and the
surface temperature of the wire is measured to be 152°C in steady operation. Also, the
voltage drop and electric current through the wire are measured to be 60 V and 1.5 A,
respectively. Disregarding any heat transfer by radiation, determine the convection heat
transfer coefficient for heat transfer between the outer surface of the wire and the air in
the room.

Q. Consider a person standing in a breezy room at 20°C. Determine the total rate of
heat transfer from this person if the exposed surface area and the average outer
surface temperature of the person are 1.6 m2 and 29°C, respectively, and the
convection heat transfer coefficient is 6 W/m2 · °C (Fig.).

SOLUTION The total rate of heat transfer from a person by both convection and
radiation to the surrounding air and surfaces at specified temperatures is to be
determined.
Assumptions 1 Steady operating conditions exist. 2 The person is completely surrounded
by the interior surfaces of the room. 3 The surrounding surfaces are at the same
temperature as the air in the room. 4 Heat conduction to the floor through the feet is
negligible.
Properties The emissivity of a person is 0.95
Analysis The heat transfer between the person and the air in the room will be by
convection (instead of conduction) since it is conceivable that the air in the vicinity of the
skin or clothing will warm up and rise as a result of heat transfer from the body, initiating
natural convection currents. It appears that the experimentally determined value for the
rate of convection heat transfer in this case is 6 W per unit surface area (m2) per unit
temperature difference (in K or °C) between the person and the air away from the person.
Thus, the rate of convection heat transfer from the person to the air in the room is

The person will also lose heat by radiation to the surrounding wall surfaces. We
take the temperature of the surfaces of the walls, ceiling, and floor to be equal to
the air temperature in this case for simplicity, but we recognize that this does not
need to be the case. These surfaces may be at a higher or lower temperature than
the average temperature of the room air, depending on the outdoor conditions
and the structure of the walls. Considering that air does not intervene with
radiation and the person is completely enclosed by the surrounding surfaces, the
net rate of radiation heat transfer from the person to the surrounding walls,
ceiling, and floor is

Note that we must use absolute temperatures in radiation calculations. Also note
that we used the emissivity value for the skin and clothing at room temperature
since the emissivity is not expected to change significantly at a slightly higher
temperature.
Then the rate of total heat transfer from the body is determined by adding these two
quantities:
Discussion In these calculations, heat transfer through the feet to the floor by conduction,
which is usually very small, is neglected. Heat transfer from the skin by perspiration, which
is the dominant mode of heat transfer in hot environments, is not considered here.

Q. Two large aluminium (k = 240 W/mK), each 2 cm thick, with 10 µm surface
roughness are placed in contact at 105 N/m2 pressure with the outside surface
temperatures of 390 °C and 406°C. The thermal contact resistance is 2.75 X 10–4
m2K/W. To find: (i) The heat flux, (ii) the temperature drop due to contact
resistance, and (iii) the contact temperatures.

Q. A 0.8 m high, 1.5 m wide double pane window consists of
two 4 mm thick layers of glass (k = 78 W/mK) and is
separated by a 10 mm wide stagnant air space (k = 0.026
W/mK). The room is at 20°C and the outside air is at –10°C.
The heat transfer coefficients are hi = 10 and ho = 40 W/m2K.
To find: (i) The rate of heat transfer through the window, (ii)
the inside surface temperature.

Q. Steam at 350°C flowing in a pipe (k = 80 W/mK) 5 cm
i.d., 5.6 cm o.d. is covered with 3 cm thick insulation (k =
0.05 W/mK). Heat is lost to the surroundings at 5°C by
natural convection and radiation with combined h = 20
W/m2K and hi = 60 W/m2K. To find: (i) The rate of heat loss
from the pipe per unit length, (ii) the temperature drops
across the pipe and the insulation.

Q. A refrigerated container in the form of a cube with 2 m sides
and 5 mm thick aluminium walls (k = 204 W/mK) is insulated
with a 0.1 m layer of cork (k = 0.043 W/mK) and the surface
temperatures are Ti = –5°C and T0 = 20°C. To find: The cooling
load of the refrigerator.

Q. An electronic transistor capsule of 2 cm dia spherical
shape is kept in an evacuated space with black walls at 30°C.
Heat loss at the rate of 300 mW from the capsule to the case
walls takes place only by radiation. To find: The capsule
temperature if it is (i) bright aluminium (ε = 0.035) and (ii)
black anodized aluminium (ε = 0.80).

Q. A steam pipe (o.d. = 10 cm, Ts = 500 K, e = 0.8) passing
through a large room at 300 K. The pipe loses heat by natural
convection (h = 15 W/m2K) and radiation. To find: (i) The
surface emissive power of the pipe, (ii) the total radiation
falling upon the pipe, and (iii) the total rate of heat loss from
the pipe.

Steady and transient mode of HEAT CONDUCTION

Multidimensional HEAT CONDUCTION

GENERAL EQUATION OF HEAT CONDUCTION
The mathematical formulation of the law of heat conduction (Fourier’s law) as
interpreted in Fig. is expressed as

The heat flux Q/A or q flows along the normal n of the decreasing temperature. It is
a vector quantity. In Fig. are shown the isotherms T and T + dT in a body. The
normal to these isotherms is designated by the axis n, which is also normal to the
differential area dA. The heat flux can be calculated in the direction of the normal
and in the direction s shown as follows:

Thus, qs is a component of the heat flux vector qn. The greatest heat flux will
occur when it is calculated along the normal to the isothermal surface (θ = 0). If
the component fluxes are related to the planes containing the x, y, z coordinate
system, the fluxes are
For an isotropic solid kx = ky = kz = k. The fluxes shown in last Eq. are components
of the heat flux vector so that

Let us consider an infinitesimal volume element of sides dx, dy and dz (Fig.). The
considerations here will include the nonsteady condition of temperature variation with
time t.
According to the Fourier heat conduction law, the
heat flowing into the left-most face of the element
in the x-direction

The value of the heat flow out of the right face of the element can be obtained by
expanding dQx in a Taylor series and retaining only the first two terms as an
approximation:
The net heat flow by conduction in the x-direction is therefore

Here, the solid has been assumed to be isotropic and homogeneous with properties
uniform in all directions. Let us consider that there is some heat source within the
solid, and heat is produced internally as a result of the flow of electrical current or
nuclear or chemical reactions. Let qG is the rate at which heat is generated internally
per unit volume (W/m3). Then the total rate of heat generation in the elemental volume
is

The net heat flow owing to conduction and the heat generated within the element
together will increase the internal energy of the volume element. The rate of
accumulation of internal energy within the control volume is
where c is the specific heat and ρ is the density of the solid.
An energy balance can be achieved on the volume element as:
Rate of energy storage within the solid = Rate of heat influx – Rate of heat outflux
+ Rate of heat generation

where α is the thermal diffusivity of the solid given by k/ρc.
This equation is also called Fourier-Biot equation

If the temperature of a material is not a function of time, the system is in the
steady state and does not store any energy. The steady state form of a three-
dimensional conduction equation in rectangular coordinates is
It is known as Poisson’s equation.
If the system is in steady state and no heat is generated internally, the
conduction equation further simplifies to
Equation is known as the Laplace equation.

It occurs in a number of areas in addition to heat transfer, for example, in diffusion
of mass or in electromagnetic fields. The operation of taking the second derivatives
of the potential in a field has therefore been given a short symbol, 𝛻2, called the
Laplacean operator. For the rectangular coordinate system Eq. becomes

Vector Method
Let us consider a control surface S enclosing a volume V (Fig.).
The net rate of heat outflow across the surface S
is given by 𝑞. 𝑛𝑑𝑆 where n is the normal
direction. Converting the surface integral to
volume integral

Cylindrical Coordinates
For a general transient three-dimensional heat conduction problem in the
cylindrical coordinates with T = T(r, θ, z, t), let us consider an elementary volume
dV = drr dθ d z

Spherical Coordinates
For spherical coordinates, as shown in Fig. the
temperature is a function of the three space
coordinates r, q and f and time t, i.e., T = T (r,
θ, ϕ, t). The general form of the conduction
equation in spherical coordinates can be found
as

where the Laplacian includes the first three terms of the above equation in
spherical coordinates.
Let us consider an elemental volume having the coordinates (r, ϕ, θ ) for three
dimensional heat conduction analysis [Fig.].
The volume of the element = dr . rdθ. r sin θ dϕ.

(A) Net heat accumulated in the element due to conduction of heat from all the
coordinate directions

Combined One-Dimensional Heat Conduction
Equation
one-dimensional transient heat conduction equations for the plane wall,
cylinder, and sphere reveals that all three equations can be expressed in a
compact form as
where n = 0 for a plane wall, n = 1 for a cylinder, and n = 2 for a sphere. In the
case of a plane wall, it is customary to replace the variable r by x. This equation
can be simplified for steady-state or no heat generation cases as described before.

STEADY HEAT CONDUCTION IN SIMPLE
GEOMETRIES
Plane Wall Without Heat Generation
(a)

Cylindrical Wall
(a) Without heat Generation
Heat transfer through a hollow pipe with a fluid flowing inside (Fig.).
Heat is assumed to flow only radially.
We want to determine the temperature distribution and the heat transfer rate in a long
hollow cylinder of length L if the inside and outside surface temperatures are Ti and To
respectively and there is no internal heat generation.

With heat Generation
Let us consider a long solid cylinder of radius R
with internal heat generation (Fig.), such as an
electric coil in which heat is generated as a
result of the electric current in the wire or a
cylindrical nuclear fuel element in which heat
is generated by nuclear fission. The one-
dimensional heat conduction equation in
cylindrical coordinates is

Spherical Wall
For a hollow sphere with uniform temperature
at the inner and outer surfaces (Fig.), the
temperature distribution without heat
generation in the steady state can be obtained
by simplifying general heat conduction
equation in spherical coordinates. Under these
conditions the temperature is only a function of
the radius r, and the conduction equation in
spherical coordinates is

CRITICAL RADIUS OF INSULATION
 Let the tube of radius rt and at temperature Tt be covered with insulation.
 At the outside radius of the insulation ro,
 a surface coefficient ha is assumed for heat transfer to the atmosphere at
temperature Ta (Fig.).
 As ro increases, xw (= ro – rt) increases and Q decreases.
 Again, as ro increases, Ao increases and Q increases.

Since Tt, Ta, ki, rt and ha are all constant, the heat loss Q depends only on ro.
As ro increases, (1/ki )ln (ro/rt) increases, but 1/(ha ro) decreases.
Differentiating Q with ro,
There is, thus, an optimum ro at which Q is the maximum.

Therefore, at ro = rcr, the heat loss will be maximum.
If rt < (ro)cr, as ro increases, Q increases till ro = (ro)cr (Fig.).
If ro > (ro)cr, as ro increases, Q decreases.
If rt > (ro)cr, any increase of insulation will decrease the rate of heat trans fer.
If rt < (ro)cr, the increase of insulation will increase Q till Q = Qmax.

The results given above for the critical radius do not include the effects of thermal
radiation. The heat transfer coefficient ha at the outer surface of insulation is
approximated by the sum of a convection component hc and a radiation component hr in
the form
(a) (b)
sphere

Let us consider the plane wall of Fig (a). If Tw is fixed, there are three ways in which the
heat transfer rate may be increased.
1. The convection coefficient h could be increased by increasing the fluid velocity or/and
the fluid temperature Tꚙ could be reduced. However, increasing h even to the maximum
possible value is often insufficient to obtain the desired heat transfer rate or the costs
related to blower or pump power required to increase h may be prohibitive.
2. The second option of reducing Tꚙ is often impractical.
3. The heat transfer rate may be increased by increasing the surface area across which
convection occurs. This may be done by using fins that extend from the wall into the
surrounding fluid [Fig. (b)]. The thermal conductivity of the fin material has a strong
effect on the temperature distribution along the fin and thus the degree to which the heat
transfer rate is enhanced.

Fins of Uniform Cross-sectional Area
Each fin is attached to the base surface
of To and extends into a fluid of
temperature Tꚙ.
The perimeter of the fin, P, which is
uniform is 2(L + b) or, P ≈ 2L, if the
thickness of the fin is small (b << L). If
the fin is very thin and its length l is
long, it can be assumed that there is no
radial temperature variation and heat
gets conducted axially along the length.
This heat is then dissipated to the
surroundings by convection.

Let us consider a small volume
element at a distance x from the
base or root of the fin of thickness
dx.
The rate at which heat enters the
element is Qx and the heat leaving
the element is Qx + dx.
In that small distance dx, let QΔ be
the heat transferred by convection.
By energy balance.

(h)
(g) (h)
(i)
(j)
(k)
(l)
(m)
(n)
(i)
(j) (m)

Fin Performance
For the infinite fin (Case 4) the result is
The fin effectiveness, ef.

A fin resistance may be defined as
fin efficiency, ηf

The heat loss from the fin tip can be taken into account approximately by increasing l by
b/2 and assuming that the tip is insulated.
This approximation keeps the surface area from which heat is lost the same as in the real
case, and fin efficiency then becomes
where the corrected fin length lc = l + b/2.

The total heat transfer will be the sum of the heat transfer from the two portions.

Limitation of an Extended Surface
For long fins the rate of heat loss from the fin is
given by
which is equal to the heat loss from the primary surface with no extended surface.
When
h/mk = 1, an extended surface will not increase the heat transfer rate from the primary
surface whatever be the length of the extended surface.
h/mk > 1, Q < hAθo and hence adding a secondary surface reduces the heat transfer and
the added surface will act as an insulation.
h/mk < 1, Q > hA θo, and the extended surface will increase the heat transfer. The heat
transfer would be more effective when h/k is low for a given geometry.

Rectangular Fin of Minimum Weight
Weight of one fin = b×l × L × ρ
A1 = b × l = area of fin cross-section normal to L.

The above equation is identical to the modified Bessel equation of zero
order (n = 0) and its general solution is

If the heat flow into the triangular fin [Fig.] is optimised in order to determine the best
ratio of height l to base b, the following expression as shown by Eckert and Drake [4]
results.

Q. 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, as shown in
Figure. 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.

Q. Consider the base plate of a 1200-W household iron that has a thickness of L =0.5 cm,
base area of A = 300 cm2, and thermal conductivity of k =15 W/m · °C. The inner surface of
the base plate is subjected to uniform heat flux generated by the resistance heaters inside,
and the outer surface loses heat to the surroundings at Tꚙ = 20°C by convection, as shown in
Figure. Taking the convection heat transfer coefficient to be h =80 W/m2 · °C and
disregarding heat loss by radiation, obtain an expression for the variation of temperature in
the base plate, and evaluate the temperatures at the inner and the outer surfaces.

Q. Consider a large plane wall of thickness L =0.06 m and thermal conductivity k =1.2
W/m · °C in space. The wall is covered with white porcelain tiles that have an emissivity
of ε=0.85 and a solar absorptivity of α= 0.26, as shown in Figure. The inner surface of
the wall is maintained at T1 =300 K at all times, while the outer surface is exposed to
solar radiation that is incident at a rate of q·solar = 800 W/m2. The outer surface is also
losing heat by radiation to deep space at 0 K. Determine the temperature of the outer
surface of the wall and the rate of heat transfer through the wall when steady operating
conditions are reached. What would your response be if no solar radiation was incident
on the surface?

Q. Consider a steam pipe of length L=20 m, inner radius r1=6 cm, outer radius
r2=8 cm, and thermal conductivity k=20 W/m · °C, as shown in Figure. 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.

Q. Consider a spherical container of inner radius r1 =8 cm, outer radius r2 =10 cm,
and thermal conductivity k = 45 W/m · °C, as shown in Figure. 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.

Q: A tube 2 cm o.d. maintained at uniform temperature of Ti is covered with insulation (k
= 0.20W/mK) to reduce heat loss to the ambient air at Tꚙ with ha = 15 W/m2K. To find:
(i) The critical thickness rc of insulation (ii) the ratio of heat loss from the tube with
insulation to that without insulation, (a) if the thickness of insulation is equal to rc, (b) if
the thickness of insulation is (rc + 2) cm.

Q. A 1 mm dia electric wire is covered with 2 mm thick layer of insulation (k = 0.5
W/mK). Air surrounding the wire is at 25ºC and h = 25 W/m2K. The wire
temperature is 100ºC. To find: (i) The rate of heat dissipation from the wire per unit
length with and without insulation, (ii) The critical radius of insulation, (iii) The
maximum value of heat dissipation.

Q. A 50 mm 50 mm iron bar 0.4 m long is connected to the walls of two heated
reservoirs, each at 120°C. The ambient air temperature is 35°C and the convective
heat transfer coefficient is 17.4 W/m2 K.
Calculate the rate of heat loss from the bar and the temperature of the bar midway
between the reservoirs. The thermal conductivity of iron is 52 W/mK.

Fins, 12 in number, having k = 75 W/m K and 0.75 mm thickness protrude 25 mm
from a cylindrical surface of 50 mm diameter and 1 m length placed in an
atmosphere of 40°C. If the cylindrical surface is maintained at 150°C and the heat
transfer coefficient is 23 W/m2K, calculate (a) the rate of heat transfer, (b) the
percentage increase in heat transfer due to fi ns, (c) the temperature at the centre of
fi ns and (d) the fi n efficiency and the fin effectiveness.

Transient Heat Conduction
Biot number Bi
where
h is the average heat transfer coefficient,
L is a characteristic dimension obtained by dividing the volume of the body
by its surface area
k is the thermal conductivity of the solid body.
For Bi < 0.1, i.e. when the internal resistance is less than 10% of the external
resistance, the internal resistance can be ignored, and there is only one temperature at
a certain instant that stands for the system as a whole, independent of space
coordinates x, y and z.

consider the cooling of a small billet or metal casting in a quenching bath or air
after it is removed from a hot furnace [Fig.]. Neglecting any temperature gradient
within the solid, an energy balance for the billet over a small time interval dt
gives.
where r = density of billet, kg/m3;
V = volume of billet, m3;
c = specific heat of billet, J/kg K;
T = average temperature of billet, K;
T• = surrounding fluid temperature, K;
h = average heat transfer coefficient, W/m2 K;
A = surface area of billet, m2
dT = temperature change, K, during time interval dt, s.

Initial temperature of the billet be Ti when t = 0. After time t has elapsed, the
temperature of the billet falls to T. On integrating last Eq. between these limits
Let Bi = hL/k and Fo = Fourier number =αt/ L2 ,
where α = thermal diffusivity (m2/s). The product

PLANE WALL WITH CONVECTION
Infinite Plate with no External Thermal Resistance (h = ꚙ)
Heat conduction equation in one dimension

Infinite Solid Plate with Both Internal and External
Resistances
Large plate uniform thickness = 2l
Uniform temperature Ti at t = 0
t > 0, with Tꚙ and h remaining uniform

from which temperature T at any distance x from the midplane can be estimated

INFINITE CYLINDER AND SPHERE WITH CONVECTION
Eq. (a)
(a)
(b)

TWO- AND THREE-DIMENSIONAL SOLUTIONS OF TRANSIENT HEAT
CONDUC TION

Time–temperature history of a finite cylinder of radius R and length 2L
is equivalent to the product solution for an infinite cylinder of radius R
and an infinite plate of thickness 2L

Q. A load of peas at a temperature of 25°C is to be cooled down in a room at a
constant air temperature of 1°C. (a) How long the peas will rquire to cool down to
2°C when the surface heat transfer coefficient of the peas is 5.81 W/m2K? (b)
What is the temperature of the peas after a lapse of 10 min from the start of cool
ing? (c) What air temperature must be used if the peas were to be cooled down to
5°C in 30 min? The peas are supposed to have an average diameter of 8 mm. Their
density is 750 kg/m3 and specific heat 3.35 kJ/kg K.

Q. A steel tube of length 20 cm with internal and external diameters of 10 and 12
cm is quenched from 500°C to 30°C in a large reservoir of water at 10°C. Below
100°C the heat transfer coefficient is 1.5 kW/m2K. Above 100°C it is less owing
to a fi lm of vapour being produced at the surface, and an effective mean value
between 500°C and 100°C is 0.5 kW/m2 K. The density of steel is 7800 kg/m3
and the specific heat is 0.47 kJ/kg K. Neglecting internal thermal resistance of
the steel tube, deter mine the quenching time.

Q. Steel ball bearings (k = 50 W/m K, α = 1.3 × 10–5 m2/s) having a diameter of 40 mm
are heated to a temperature of 650°C and then quenched in a tank of oil at 55°C. If the
heat transfer coefficient between the ball bearings and oil is 300 W/m2 K, determine (a)
the duration of time the bearings must remain in oil to reach a temperature of 200°C, (b)
the total amount of heat removed from each bearing during this time and (c) the
instantaneous heat transfer rate from the bearings when they are first immersed in oil and
when they reach 200°C.

Q. An average human body modeled as a 30 cm diameter, 170 cm long cylinder
has 72% water by mass, so that its properties may be taken as those of water at
room temperature: ρ = 1000 kg/m3, c = 4180 J/kg K and k = 0.608 W/m K. A
person is found dead at 5 am in a room the temperature of which is 20°C. The
temperature of the body is measured to be 25°C when found, and the heat transfer
coefficient is estimated to be 8 W/m2 K. Assuming the body temperature of a
living man is 37°C, estimate the time of death of the above person.

Q. An egg is being cooked in boiling water. Its initial temperature is 5°C. It is
dropped into boiling water at 95°C. Assumptions: 1. The egg is spherical in shape
with a radius of r0 = 0.025 m. 2. Heat conduction in the egg is one dimensional
(radial only) because of thermal symmetry about the centre. 3. The thermal
properties of the egg and the heat transfer coefficient are constant. 4. The Fourier
number F0 > 0.2 so that the one term approximate solutions are applicable. 5. The
thermal conductivity and diffusivity of eggs are approximated as those of water. k
= 0.6 W/mK, α = 0.14 × 10–6 m2/s. The heat transfer coefficient is 1200 W/m2K. To
find: The time for the centre of the egg to reach 70°C.

THERMAL RADIATION
Thermal radiation is generally described in terms of electromagnetic waves, all of
which travel at the velocity of light.
The wavelength and frequency of radiation propagating in a medium are related by

 The type of electromagnetic radiation that is pertinent to heat transfer is the
thermal radiation emitted as a result of vibration and rotational motions of
molecules, atoms and electrons of a substance, and higher the temperature,
higher is the rate of emission. It is the portion of electromagnetic spectrum that
extends from 0.1 µm to 100 µm wavelength range. Thus, it includes the entire
visible and infrared radiation as well as a portion of the ultra violet radiation.
 The visible portion of the electromagnetic spectrum lying between 0.4 µm and
0.76 µm is what we call light, which triggers the sensation of vision in the
human eye.
 Light consists of narrow bands of colour from violet (0.40 – 0.44 µm) to red
(0.63 – 0.76 µm).
 A surface that reflects radiation in the wavelength range 0.63 – 0.76 µm and
absorbs the rest of visible radiation appears red to the eye.
 A surface that reflects all of the light appears white, while a surface that absorbs
all the incident radiation appears black.

 A body that emits some radiation in the visible range is called a light source.
The sun is our primary light source.
 The electromagnetic radiation emitted by the sun is called the solar radiation
which falls in the wavelength band 0.1 – 3 µm.
 About half of solar radiation is light (visible range), while the rest is ultraviolet
and infrared.
 Radiation emitted by bodies at room temperature falls into the infrared
region extending from 0.76 to 100 µm.
 Bodies start emitting visible radiation at temperatures above 800 K.
 The tungsten filament in the light bulb heated to temperatures above 2000 K
emits radiation significantly in the visible range.
 The ultraviolet radiation in wavelengths varying from 0.01 to 0.40 µm forms
about 12% of solar radiation and is highly pernicious, killing microorganisms
and causing serious damage to humans and other living organisms.
 Fortunately, the ozone layer in the atmosphere absorbs most of this ultraviolet
radiation.

 Microwave ovens utilize electromagnetic radiation in the microwave region of
the spectrum, 102 – 105 µm, generated by microwave tubes called magnetrons.
 These microwaves are very suitable for use in cooking, since they are reflected
by metals, transmit ted by glass and plastics, and absorbed by food (especially
water) molecules.
 The electric energy converted to radiation in a microwave oven eventually
becomes part of the internal energy of the food.
 This speeds cooking which becomes more efficient and less power consuming.

 Radars and cordless telephones also use electromagnetic radiation in the
microwave region.
 The electromagnetic waves used in radio and TV broadcasting varying in
wavelengths between 1 µm and 1000 µm fall in the radio wave region of the
spectrum.
 Radiation emitted from heated bodies comprises rays with wavelengths 0.4 – 800
mm whereas the visible range is from 0.4 – 0.8 µm.
 The large part of thermal radiation between 0.8 and 800 µm which has waves
outside the visible range is called the infrared radiation.

The study of radiation energy transfer is thus separated into two distinct parts:
1. Radiation energy transfer in a nonparticipating medium where the medium is
either vacuum or does not interfere with the energy propagation.
2. Radiation energy transfer in a participating medium, which intervenes with its
propagation by absorbing, emitting or scattering energy. Layers of certain gases
and vapours which absorb and emit radiation at certain ranges of wavelengths
would be considered as the participating medium.

PREVOST’S THEORY
All bodies emit thermal radiation, unless the body is at absolute zero temperature.
 If a body is placed in a surrounding at the same temperature as itself, its
temperature does not change, since the rate at which energy is radiated is equal
to the rate at which energy is received from the surroundings.
 When a hot body is placed in cooler surroundings, the rate at which it radiates is
faster than the rate at which it absorbs the incident radiation, as a result of which
its temperature decreases till the state of thermal equilibrium is reached.
This is known as Prevost’s theory of heat exchange.

ABSORPTIVITY, REFLECTIVITY AND TRANSMISSIVITY
When radiation strikes a surface, part of it is absorbed, part of it is reflected,
and the remaining part, if any, is transmitted, as illustrated in Figure.
The fraction of irradiation absorbed by the surface is called the absorptivity α,
the fraction reflected by the surface is called the reflectivity ρ, and the fraction
transmitted is called the transmissivity τ

If Q is the total radiant energy incident upon the surface of a body, some part of it
(QA) will be absorbed, some part (QR) reflected and some part (QTr) transmitted
through the body (Fig.). By energy balance,
 A body is said to be opaque if η = 0 and α + ρ = 1. Most
solids do not transmit any radiation and are opaque. If ρ is
reduced, a increases.
 The reflectivity depends on the character of the surface.
Therefore, the absorptivity of an opaque body can be
increased or decreased by appropriate surface treatment.
 Roughening a solid surface enhances its absorptivity of
incident radiation.

where G is the radiation energy incident on the surface, and Gabs, Gref, and Gtr are the
absorbed, reflected, and transmitted portions of it, respectively.
The first law of thermodynamics requires that the sum of the absorbed, reflected, and
transmitted radiation energy be equal to the incident radiation.

 When the surface is highly polished, the angle of incidence θi is equal to the
angle of reflection θr
 Most gases have high values of τ and low values of α and ρ.
 Air at atmospheric pressure and temperature is transparent to thermal radiation
for which τ = 1 and α = τ = 0.
 Gases like CO2 and H2O vapour are highly absorptive at certain ranges of wave
lengths.

EMISSIVE POWER
The total emissive power of a body, E, is defined as the total radiant energy
emitted by the body at a certain temperature per unit time and per unit surface area
at all wavelengths.
The monochromatic emissive power of a body, Eλ, is defined as the radiant
energy emitted by the body per unit time and per unit surface area at a particular
wavelength and temperature.
The radiation energy emitted by a black body per unit time and per unit surface
area is given by

The total radiation, Eb, is distributed among wavelengths varying from 0 to ꚙ.
Therefore, at a particular temperature,
which is also called spectral or radiation intensity of a black body. Since the
area under each curve for a certain temperature represents the total emissive
power Eb, as the temperature in creases, the area increases and therefore Eb
increases.

Emissivity
The emissivity of a surface is defined as the ratio of radiation emitted by the surface
to the radiation emitted by a black body at the same temperature.
The emissivity of a surface is denoted by ε, and it varies between zero and one.
Emissivity is a measure of how closely a surface approximates a blackbody, for which
ε = 1.
 The emissivity of a real surface varies with the temperature of the surface as well as
the wavelength and direction of emitted radiation.
 The emissivity of a surface at a certain wavelength is called spectral emissivity ελ.
 The emissivity in a certain direction is called directional emissivity εθ, where θ is the
angle between the direction of radiation and the normal to the surface.
 The emissivity of a surface averaged over all directions is called the hemispherical
emissivity,
 The emissivity averaged over all wavelengths is called the total emissivity.

The total hemispherical emissivity of a surface is simply the average emissivity
over all directions and wavelengths, and can be expressed as
where E(T ) is the total emissive power of the real surface at temperature T and
Eb(T) is the same for a black body.
Thus

KIRCHHOFF’S LAW
This is Kirchhoff’s law which states that the emissivity of the surface of a body is equal to
its absorptivity when the body is in thermal equilibrium with its surroundings.

BLACKBODY RADIATION
A blackbody is defined as a perfect emitter and absorber of radiation.
 The amount of radiation energy emitted from a surface at a given wavelength
depends on the
 Material of the body
 The condition of its surface
 The surface temperature
 A blackbody absorbs all incident radiation, regardless of wavelength and
direction.
 A blackbody emits radiation energy uniformly in all directions per unit area
normal to direction of emission.
The radiation energy emitted by a blackbody per
unit time and per unit surface area was determined
experimentally by Joseph Stefan in 1879

Several observations can be made from this figure:
1. The emitted radiation is a continuous function of wavelength. At any specified
temperature, it increases with wavelength, reaches a peak, and then decreases
with increasing wavelength.
2. At any wavelength, the amount of emitted radiation increases with increasing
temperature.
3. As temperature increases, the curves shift to the left to the shorter wavelength
region. Consequently, a larger fraction of the radiation is emitted at shorter
wavelengths at higher temperatures.

BLACK BODY
The energy emitted from a portion of the surface in the cavity
is εEb.
ρεEb
ρ2εEb
ρ3εEb
ρ4εEb
Then the energy emitted
from the hole is

Planck’s law
It follows that the emissive power Eb characterized by individual isotherms passes
through a maximum, as shown in Fig. With wavelengths λ = 0 and λ = ꚙ it vanishes.

Rayleigh–Jeans’ Law
This equation is known as Rayleigh–Jeans law which is found to be valid for large
wavelengths.

Wien’s Law
This equation is known as Wien’s law, which is found to be valid for short
wavelengths.
The unity present in the denominator of Plank Eq. can be neglected

Wien’s Displacement Law
The maximum values of the emissive power Eb can be obtained by
differentiating Planck’s Eq. with respect to λ and equating it to zero.
Therefore,
This is known as Wien’s displacement law.
Here, λmax is the wavelength at which Ebλ is the maximum at a particular temperature.
The value of (Ebλ)max shifts towards the shorter wave lengths with increasing
temperature.

where the constant C3 = 1.287 x 10–5 W/m3 K5.
The maximum emissive power of a black body can be found from Planck’s law; λ =
λmax is replaced

Planck’s Law in Dimensionless Form

This is also equal to the ratio of area A1 to the total area under the curve at T.
Similarly, for the wavelength range between λ1 and λ2, the fraction of radiation at
temperature T, would be

Black Body Radiation in a Certain Range of
Wavelength
The fraction of total radiation from a black body in the wavelength range of 0 – λ at a
certain temperature T is given by

At T = 1 K, the curve represents Ebλ vs λ. The area under the curve is equal to the
Stefan–Boltzmann constant. The fractional energy in the range of λ1T and λ2T is
given by

INTENSITY OF RADIATION
The intensity of radiation, I is defined as the rate of heat radiation in a given direction from
a surface per unit solid angle per unit area of the projection of the surface on a plane normal
to the direction of radiation
A solid angle (steradian)
is defined as the ratio of
the spherical surface
enclosed by a cone, with
its vertex at the centre of
the sphere, to the square
of the radius of the sphere
zenith angle θ
azimuth angle Φ

Let us consider the radiation emanating from an elemental surface of area dA1 at the
temperature T1.
Energy will be radiated in all directions in the entire hemisphere.
The spherical strip area dA of radius rsin ϕ1 and thickness rdϕ1,
Solid angle of dA/r2.
The amount of radiation dQ1 directed towards this area is
π

RADIANT HEAT EXCHANGE BETWEEN TWO
BLACK BODIES SEPARATED BY A NON-
ABSORBING MEDIUM
The radiant heat exchange between two bodies
depends upon
1. The views the surfaces have of each other i.e.,
how they “see” each other.
2. Their emitting and absorbing characteristics and
3. The medium that intervenes the two bodies.

Let us assume that the two bodies are black and the medium is nonparticipating in
the energy exchange.
Let us consider the area elements dA1 and dA2 on the two surfaces.
The distance between them is r
The angles made by the normals to the two area elements with the line joining
them are ϕ 1 and ϕ 2 respectively.
The projected area of dA1 in the direction of radiation is dA1 cos ϕ1.

ELECTRICAL ANALOGY
The rate of radiant flux between two black surfaces may be written as
where (Eb1– Eb2) is the driving force or potential between the two nodes 1 and 2 for
radiative energy transfer and 1/A1 F12 is the resistance.
The corresponding network is given

For radiative flux between the four walls of a black enclosure

RADIANT HEAT TRANSFER BETWEEN TWO BLACK
SURFACES CONNECTED BY NONCONDUCTING AND
RERADIATING WALLS
An enclosure having a single source surface 1,
a single sink surface 2 and refractory surface R,
each of the surfaces maintained at uniform
temperatures T1, T2 and TR respectively

F1R = 1 – F12 and F2R = 1 – F21

RADIATION HEAT TRANSFER BETWEEN
GRAY BODIES
 Radiant Exchange between Two Small Gray Bodies
The energy emitted by the body 1 = ζε1 A1 T1
4,
Incident on the second body = F12 (ζε1 A1 T1
4)
Absorbed = α2 (F12 ζε1 A1 T1
4).
Since the bodies are gray, α2 = ε2
Energy transfer from 1 to 2 is

Radiant Exchange between Two
Infinite Parallel Gray Planes
The radiant interchange between two infinite
parallel gray planes involves no geometry
factor, since F12 = F21 = 1.0.
Let us consider two gray planes.
For gray surfaces, α = ε and ρ = 1 – ε.
Surface 1 emits ε1Eb1 per unit time and area.
Surface 2 absorbs the fraction α2ε1Eb1 or
ε2ε1Eb1
Surface 2 reflects ρ2ε1Eb1 or (1 – ε2) ε1Eb1 back
towards A1.
The net heat transferred per unit of surface 1 to
2 is the emission ε1Eb1 minus the fraction of
ε1Eb1 and ε2Eb2 which is ultimately absorbed by
surface 1 after successive reflections.

RADIOSITY AND IRRADIATION
Radiosity is the term used to indicate the total radiant energy leaving a surface per
unit time and per unit surface area.
It is denoted by the symbol J.
This quantity differs from the emissive power in that the radiosity includes reflected
energy as well as the original emission, regardless of any directional dependence or
spectral preference.
Irradiation is the term used to denote the total radiation incident on a surface per
unit time and per unit surface area. It is denoted by the symbol G.

The radiosity is the sum of the energy emitted and the energy reflected, when no energy
is transmitted
The net energy leaving the surface is the difference between the radiosity and
irradiation

RADIATION NETWORK FOR GRAY SURFACES EXCHANGING ENERGY
A network which represents two surfaces exchanging radiative energy with each
other
Surface resistance = (1 – ε)/Aε
space resistance = 1/(Ax Fxy)

The net heat transfer would be given by dividing the overall potential difference with
the sum of the resistances.

Node JR is not connected to any surface resistance, since the surface R has no source of
its own and it only reflects energy. It is called the floating node.

Let R’ is the sum of the resistances 1/[A1(1 – F12)] and [1/A2(1 – F12)], which are in
series,

Let Req be the equivalent resistance of the parallel resistances of R’ and 1/(A1
F12), so that

By the use of one radiation shield, the net radiant heat transfer is reduced by 50%.
σ

If N shields are placed between the two planes 1 and 2,
Surface resistances = (2N + 2)
two for each shield and one for each heat transfer surface,
space resistances = (N + 1) (which would all be unity).
The total resistance would thus be

RADIATION FROM CAVITIES
The conical cavity, as shown in Fig. (a), which is of diameter D, height H, lateral
length L, semi-vertex angle a and surface area A1. The temperature T1 of the
surface is uniform. Part of the radiation from the surface falls on itself, of which a
portion is absorbed and the remainder is reflected.

Of this, the amount falling on A1 and absorbed by it =

flat surface A2 closing the cavity. Since A1 and A2 together form an
enclosure,

Q. A domestic hot water tank (0.5 m diameter and 1 m high) is in stalled in a
large space. The ambient temperature is 25°C. If the tank surface is oxidised
copper with an emissivity of 0.8, fi nd the heat loss from the tank surface at
temperature 80°C by radiation. What would be the reduction in heat loss if a
coating of aluminium paint having an emissivity of 0.3 is given to the tank? What
would be the increase in heat loss if a white paint having an emissivity of 0.97 is
given to the tank?

Q. A black body emits radiation at 2000 K. Calculate (i) the monochromatic
emissive power at 1 mm wavelength, (ii) wavelength at which the emission is
maximum, and (iii) the maximum emissive power.

Q. Assuming the sun as a black body, it emits maximum radiation at 0.5 µm
wavelength. Calculate (i) the surface temperature of the sun, (ii) its emissive power,
(iii) the energy received by the surface of the earth and (iv) the energy received by a
2 m × 2 m solar collector whose normal is inclined at 60° to the sun. Take the
diameter of the sun as 1.4 × 109 m, diameter of the earth as 13 ¥ 106 m and the
distance of the earth from the sun as 15 × 1010 m.

Q. A long steel rod 20 mm in diameter is to be heated from 427°C to 538°C. It is
placed concentrically in a long cylindrical furnace which has an inside diameter of
160 mm. The inner surface of the furnace is at a temperature of 1093°C and has an
emissivity of 0.85. If the surface of the rod has an emissivity of 0.6, estimate the
time required for the heating operation. Take the density of steel as 7800 kg/m3
and its specific heat 0.67 kJ/kg K.

Q. An enclosure measures 1.5 m × 1.7 m with a height of 2 m. The walls and
ceiling are maintained at 250°C and the floor at 130°C. The walls and ceiling
have an emissivity of 0.82 and the floor 0.7. Determine the net radiation to the
floor.

Q. Two very large parallel planes with emissivities 0.3 and 0.8 exchange radiative
energy. Determine the percentage reduction in radiative energy transfer when a
polished aluminium radiation shield (ε = 0.04) is placed between them.

Heat Exchangers
Heat exchangers can be grouped into three broad
classes:
1. Transfer type heat exchangers or recuperators,
2. Storage type heat exchangers or regenerators,
3. Direct contact type heat exchangers or mixers.

Transfer type heat exchangers or recuperators

Flow Arrangements in Recuperative Heat
Exchangers

Storage type heat exchangers or regenerators
(i) stationary or (ii) rotating

COMPACT, SHELL-AND-TUBE AND PLATE HEAT
EXCHANGERS
A heat exchanger having a large surface area per unit volume is called a compact
heat exchanger.
The ratio of the heat transfer surface area to the volume is called the area density β.
A heat exchanger with β >700 m2/m3 is said to be compact,
e.g. car radiators (β ≈ 1000 m2/m3),
ceramic regenerators in gas turbines (β ≈ 6000 m2/m3),
Stirling engine regenerator (β ≈ 15,000 m2/m3)
Human lung ≈ 20000 m2/m3

Fouling Factor
The net effect of accumulations on heat transfer is represented by a fouling factor Rf,
which is a measure of the thermal resistance introduced by fouling.

COMPARISON OF PARALLEL FLOW AND COUNTER FLOW
HEAT EXCHANGERS

 For the same inlet and exit temperatures of the two fluids, it is found that
(ΔT )1m for counter flow is always greater than that for parallel flow.
 For the same heat transfer Q and the same overall heat transfer coefficient
U0, the surface area required for counter flow operation is always less
than that for parallel flow operation.
 In a parallel flow heat exchanger, Th2> Tc2 , i.e. the hot fluid cannot be
cooled to a temperature lower than the cold fluid temperature.
 In a counter flow heat exchanger, Th2 can be less than Tc2 or Tc2 can be
higher than Th2, i.e. the hot fluid can be cooled below Tc2 or the cold fluid
can be heated above Th2

HEAT TRANSFER WITH PHASE CHANGE

Steam in the condenser of a power plant is to be condensed at a temperature of
30°C with cooling water from a nearby lake, which enters the tubes of the
condenser at 14°C and leaves at 22°C. The surface area of the tubes is 45 m2, and
the overall heat transfer coefficient is 2100 W/m2 · °C. Determine the mass flow
rate of the cooling water needed and the rate of condensation of the steam in the
condenser.

Q. Cold water enters a counter-flow heat exchanger at 10°C at a rate of 8 kg/s,
where it is heated by a hot water stream that enters the heat exchanger at 70°C at a
rate of 2 kg/s. Assuming the specific heat of water to remain constant at Cp = 4.18
kJ/kg · °C, determine the maximum heat transfer rate and the outlet temperatures of
the cold and the hot water streams for this limiting case.

Cold and hot water streams enter a heat exchanger at specified temperatures and
flow rates. The maximum rate of heat transfer in the heat exchanger is to be
determined.
Assumptions
1 Steady operating conditions exist.
2 The heat exchanger is well insulated so that heat loss to the surroundings is
negligible and thus heat transfer from the hot fluid is equal to heat transfer to the
cold fluid.
3 Changes in the kinetic and potential energies of fluid streams are negligible.
4 Heat transfer coefficients and fouling factors are constant and uniform.
5 The thermal resistance of the inner tube is negligible since the tube is thin-
walled and highly conductive.

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