GATE Mechanical Engineering notes on Heat Transfer. Use these notes as a preparation for GATE Mechanical Engineering and other engineering competitive exams. For full course visit https://mindvis.in/courses/gate-2018-mechanical-engineering-online-course or call 9779434433.

1. 1 | www.mindvis.in
HEAT TRANSFER
Steady state conduction
Steady state conduction process means, that the
temperature does not vary with respect to time.
Therefore, the heat flux remains unchanged with time
during steady state heat transfer, through a medium at
any location.
For example, heat transfer through the walls of a house
will be steady when the conditions inside the house and
the outdoors remain constant for several hours.
Transient heat transfer, on the other hand is one in which
conditions change w.r.t. time. During such process
temperature and heat flux normally vary with time as well
as position.
In special cases, the variation is w.r.t. time and not with
position, such heat transfer systems are called lumped
systems.
Steady state conduction through a plane wall
Consider a thin element of thickness L in a large plane
wall, k as the thermal conductivity, T1 and T2 as the wall
face temperatures (T1 > T2).
When the thermal conductivity is a constant value,
Fourier law yields,
𝑞 = −
𝑘𝐴
𝐿
(𝑇2 − 𝑇1)
If the thermal conductivity varies w.r.t temperature as,
𝑘 = 𝑘0(1 + 𝛽𝑇)
The resultant heat flow equation becomes,
𝑞 = −
𝑘0 𝐴
𝐿
[(𝑇2 − 𝑇1) +
𝛽
2
(𝑇2
2
− 𝑇1
2
)]
Thermal Resistance concept
Rewriting the conduction equation for a plane wall,
𝑞 =
𝑇1 − 𝑇2
𝑅𝑡ℎ
Where,
𝑅𝑡ℎ =
𝐿
𝑘𝐴
Is known as the thermal resistance of the wall material
against heat conduction.
 Thermal resistance depends upon the
geometry and thermal properties of the
material
The heat conduction equation can be considered to be
analogous to flow of electric current in a circuit. As per
Ohm’s law,
𝐼 =
𝑉
𝑅
Where I = electric current (the flow quantity)
V = potential difference (driving force for the flow
quantity)
R = electrical resistance
Similarly, putting up the same concept for Fourier Law,
ℎ𝑒𝑎𝑡 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 =
∆𝑇(𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑜𝑟 𝑑𝑟𝑖𝑣𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒)
𝑡ℎ𝑒𝑟𝑚𝑎𝑙 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒
The figures given below, illustrate the above mentioned
concept,

2. 2 | www.mindvis.in
Consider a case of convection heat transfer across a
surface of area AS, and temperature TS to a fluid with
temperature 𝑇∞, convective heat transfer coefficient h.
Newton’s law of cooling gives us,
𝑞 = ℎ𝐴 𝑆(𝑇𝑆 − 𝑇∞)
𝑞 =
𝑇𝑆 − 𝑇∞
𝑅𝑡ℎ
Where,
𝑅𝑡ℎ =
1
ℎ𝐴 𝑆
Is the thermal resistance against convection heat
transfer.
 When the value of convection heat transfer
coefficient becomes too large, i.e. 𝒉 → ∞ the
convection resistance becomes almost zero
For a multilayered wall, as shown in the figure below, the
heat flow through each material is assumed to be the
same,
𝑞 = −𝑘1 𝐴
𝑇2 − 𝑇1
𝑥1
= −𝑘2 𝐴
𝑇3 − 𝑇2
𝑥2
= −𝑘3 𝐴
𝑇4 − 𝑇3
𝑥3
Upon solving all sections, the heat flow through the
complete system is written as,
𝑞 =
𝑇1 − 𝑇4
(𝑥1 𝑘1 𝐴⁄ ) + (𝑥2 𝑘2 𝐴⁄ ) + (𝑥3 𝑘3 𝐴)⁄
=
∆𝑇𝑜𝑣𝑒𝑟𝑎𝑙𝑙
∑ 𝑅𝑡ℎ𝑒𝑟𝑚𝑎𝑙
 In the above illustration, the multi layers are
joined in series with each other, so to find out
the overall thermal resistance, apply the series
combination law of electrical resistances
Heat conduction flow through a cylindrical wall
Consider a long cylinder of inside radius r1, outside radius
r2 and length L. The cylinder is exposed to a temperature
difference T1 – T2. The heat flow occurs in the radial
direction.
The following boundary conditions exist in the cylinder,
𝑇 = 𝑇1 𝑎𝑡 𝑟 = 𝑟1
𝑇 = 𝑇2 𝑎𝑡 𝑟 = 𝑟2
Invoking the heat conduction equation we get the
following expression,
𝑞 =
2𝜋𝑘𝐿(𝑇1 − 𝑇2)
ln(𝑟2 𝑟1⁄ )
=
∆𝑇𝑜𝑣𝑒𝑟𝑎𝑙𝑙
∑ 𝑅𝑡ℎ𝑒𝑟𝑚𝑎𝑙
Where 𝑅𝑡ℎ𝑒𝑟𝑚𝑎𝑙 =
𝑙𝑛 (𝑟2 𝑟1⁄ )
2𝜋𝑘𝐿
for a cylinder
Heat conduction for a spherical body
The above mentioned analysis can be run for a spherical
body subjected to same temperature differences.
𝑞 =
𝑇1 − 𝑇2
∑ 𝑅𝑡ℎ𝑒𝑟𝑚𝑎𝑙(𝑠𝑝ℎ𝑒𝑟𝑒)
Where,
𝑅𝑡ℎ𝑒𝑟𝑚𝑎𝑙(𝑠𝑝ℎ𝑒𝑟𝑒) =
𝑟2 − 𝑟1
4𝜋𝑟1 𝑟2 𝑘
Consider a 1-dimensional heat flow through a cylindrical
or spherical layer that is exposed to convection on both
sides due to fluids at temperatures 𝑇∞1 and 𝑇∞2 with
heat transfer coefficients h1 and h2, as shown in the figure
below,

3. 3 | www.mindvis.in
The thermal resistance network in this case is a
combination of one conduction and two convection
resistances in series. The heat flow through this system
can be expressed as,
𝑞 =
𝑇∞1 − 𝑇∞2
𝑅𝑡𝑜𝑡𝑎𝑙
Thermal contact resistance
When we analyse heat conduction through multi layered
solids, we assume perfect contact at the interface, which
results in NO temperature drop at the interface. This is
possible only when the surfaces in contact are perfectly
smooth and free from any sort of irregularity.
In reality, even high smooth surfaces are microscopically
rough. This is illustrated with the help of the figure below,
Due to the gaps visible in the actual thermal contact,
there will be air entrapped in them which acts as an
insulator, thereby increasing the thermal resistance at
the interface. The thermal resistance at the interface is
called thermal contact resistance (RC).
Critical thickness of insulation
By adding more insulation to a wall decreases the heat
transfer across it. The thicker the insulation, the lower is
the heat transfer rate. This happens because the heat
transfer area A remains constant, and by adding more
insulation will result in increase in the thermal resistance
of the wall without increasing the convection resistance.
But, adding more insulation to a cylindrical pipe does not
always result in decrease in heat transfer. Additional
insulation increases the thermal resistance of the
insulation layer but decreases the convection resistance
of the surface because of the increase in the outer surface
area for convection.
Consider a cylindrical pipe of outer radius r1 and outer
surface temperature T1. The pipe is now insulated with
an insulating material of thermal conductivity ‘k’ and
outer radius r2. Heat is lost to the surroundings
maintained at temperature 𝑇∞ and convection heat
transfer coefficient ‘h’. the heat flow through the pipe is
expressed as,
𝑞 =
𝑇1 − 𝑇∞
𝑅 𝑐𝑜𝑛𝑑 + 𝑅 𝑐𝑜𝑛𝑣
=
𝑇1 − 𝑇∞
𝑙𝑛(𝑟2 𝑟1⁄ )
2𝜋𝐿𝑘
+
1
ℎ(2𝜋𝑟2 𝐿)
The variation of ‘q’ with the outer radius is plotted, and
the value of r2 at which the ‘q’ reaches a maximum value
is called as critical thickness of insulation.
𝑟𝑐 (𝑐𝑦𝑙) =
𝑘
ℎ
 The value of critical radius will be largest when
‘k’ is large and ‘h’ is small.

4. 4 | www.mindvis.in
 The lowest value of ‘h’ encountered in practice
is about 5 W/m2-0C (in the case of convection of
gases)
 Thermal conductivity of commonly used
insulating materials is about
0.05 W/m-0C
 The largest value of critical radius is about 1 cm
For the spherical body, the value of critical value is given
as,
𝑟𝑐 (𝑠𝑝ℎ) =
2𝑘
ℎ
Heat source systems
In a number of engineering applications heat is generated
within a system.
Plane wall with heat source (internal heat generation)
Consider a plane wall with uniformly distributed heat
sources. The thickness of the wall is ‘2L’, and the heat
conduction is assumed to be in one dimension only.
Let the heat generated per unit volume is qg, and the
thermal conductivity does not vary with temperature.
The differential equation for such a condition is,
𝑑2
𝑇
𝑑𝑥2
+
𝑞 𝑔
𝑘
= 0
The boundary conditions can be specified as follows,
𝑇 = 𝑇 𝑤 𝑎𝑡 𝑥 = ±𝐿
The general solution of the differential equation then
becomes,
𝑇 = −
𝑞 𝑔
2𝑘
𝑥2
+ 𝐶1 𝑥 + 𝐶2
On applying the boundary conditions, the temperature
distribution can be found to be a parabolic expression as
follows,
𝑇 − 𝑇𝑜
𝑇 𝑤 − 𝑇0
= (
𝑥
𝐿
)
2
 At steady state conditions, the total heat
generated must equal the heat lost at the faces.
Relation between, Tw and To is,
𝑇0 =
𝑞 𝑔 𝐿2
2𝑘
+ 𝑇 𝑤
The maximum temperature in a symmetrical solid with
uniform heat generation occurs at its centre, as
illustrated in the figure,
Heat transfer through finned surfaces
The heat conducted through a body must be frequently
removed by some convection process. For this purpose,
we make use of finned surfaces. The heat transfer from
internally flowing fluid to fin wall is via convection. Heat
is conducted through the finned surface and finally
dissipated to the surroundings via convection.
To increase the heat transfer via convection:-
 Increase the convective heat transfer
coefficient ‘h’, or
 Increase the surface area ‘AS’
Increasing ‘h’ is not always a practical approach, so we
increase the surface area by using finned surfaces.
Fins are extended surfaces, made of highly conductive
materials like aluminium, and are manufactured with the
help of extrusion, welding etc. The finned surfaces are
commonly used to enhance the heat transfer rate, and
they often increase the heat transfer from a surface
severalfold.
In analysis of fins, we assume steady state operation with
no heat generation in the fin, constant thermal
conductivity ‘k’ of the fin material.
The value of convective heat transfer coefficient varies
along the fin length as well as its circumference, and its
value at a point is a function of the fluid motion at that
point. The value of ‘h’ is much lower at the fin base that
at its tip. This is because that fluid is surrounded by solid
surfaces near the fin base, this disrupts its motion, while
fluid near the fin tip has little contact with the solid
surface and thus encounters less resistance to flow.

5. 5 | www.mindvis.in
 Keeping in mind the above point, adding too
many fins on a surface may actually decrease
the overall heat transfer when the decrease in
‘h’ offsets any gain resulting from increase in
surface area
Fin equation
For the volume element shown in the figure, illustrating
the fin, under steady state conditions, the energy balance
is,
(
𝑟𝑎𝑡𝑒 𝑜𝑓 ℎ𝑒𝑎𝑡
𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑖𝑛𝑡𝑜
𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡
)
= (
𝑟𝑎𝑡𝑒 𝑜𝑓 ℎ𝑒𝑎𝑡
𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒
𝑒𝑙𝑒𝑚𝑒𝑛𝑡
)
+ (
𝑟𝑎𝑡𝑒 𝑜𝑓 ℎ𝑒𝑎𝑡
𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑜𝑛 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒
𝑒𝑙𝑒𝑚𝑒𝑛𝑡
)
The differential equation governing the heat flow from a
fin is given as,
𝑑2
𝜃
𝑑𝑥2
− 𝑎2
𝜃 = 0
Where,
𝜃
= 𝑇
− 𝑇∞ 𝑖𝑠 𝑘𝑛𝑜𝑤𝑛 𝑎𝑠 𝑡ℎ𝑒 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑒𝑥𝑐𝑒𝑠𝑠, 𝑎𝑛𝑑 𝑎2
=
ℎ𝑝
𝑘𝐴 𝑐
At the fin base,
𝜃 𝑏 = 𝑇𝑏 − 𝑇∞
There are four different boundary conditions at the fin
base and the fin tip,
1. Infinitely long fin (𝑻 𝒇𝒊𝒏 𝒕𝒊𝒑 = 𝑻∞)
The boundary condition at the tip is
𝜃(𝐿) = 0 𝑎𝑠 𝐿 → ∞
The expression for the heat transfer for an
infinitely long fin is given as,
𝑞𝑙𝑜𝑛𝑔 𝑓𝑖𝑛 = √ℎ𝑝𝑘𝐴 𝑐(𝑇𝑏 − 𝑇∞)
Where p = perimeter
AC = cross section area of the fin
The temperature along the fin in this case
decreases exponentially from 𝑇𝑏 to 𝑇∞.

6. 6 | www.mindvis.in
2. Fin tip with negligible heat loss (𝒒 𝒇𝒊𝒏 𝒕𝒊𝒑 = 𝟎)
The fins should not be so long, so as to make
their temperature approach the surrounding
temperature at the tip. A more realistic
situation can be realised in this case where the
heat transfer from the fin tip is negligible,
because the heat transfer from the fin is
proportional to its surface area, and the surface
area of the fin tip is taken to be negligible as
compared to the total area of the fin.
The mathematical expression for heat transfer
from an insulated fin tip is,
𝑞𝑖𝑛𝑠𝑢𝑙𝑎𝑡𝑒𝑑 𝑡𝑖𝑝 𝑓𝑖𝑛 = √ℎ𝑝𝑘𝐴 𝑐(𝑇𝑏
− 𝑇∞) 𝑡𝑎𝑛ℎ (𝑎𝐿)
The temperature distribution for an insulated
tip fin is expressed as,
𝑇(𝑥) − 𝑇∞
𝑇𝑏 − 𝑇∞
=
𝑐𝑜𝑠ℎ 𝑎(𝐿 − 𝑥)
𝑐𝑜𝑠ℎ (𝑎𝐿)
3. Fin of finite length and loses heat by convection
from its end
The fin tips, in actual, are exposed to
surroundings, which include convection and
also effects of radiation. In this case, the
temperature distribution can be expressed via
an expression,
𝑇(𝑥) − 𝑇∞
𝑇𝑏 − 𝑇∞
=
𝑐𝑜𝑠ℎ 𝑎(𝐿 − 𝑥) + (ℎ 𝑎𝑘)⁄ 𝑠𝑖𝑛ℎ 𝑎(𝐿 − 𝑥)
𝑐𝑜𝑠ℎ 𝑎𝐿 + (ℎ 𝑎𝑘)⁄ 𝑠𝑖𝑛ℎ (𝑎𝐿)
And the heat flow from such a fin is given as,
𝑞 𝑓𝑖𝑛
= √ℎ𝑝𝑘𝐴 𝑐(𝑇𝑏
− 𝑇∞)
𝑠𝑖𝑛ℎ (𝑎𝐿) + (ℎ 𝑎𝑘)⁄ 𝑐𝑜𝑠ℎ (𝑎𝐿)
𝑐𝑜𝑠ℎ (𝑎𝐿) + (ℎ/𝑎𝑘) 𝑠𝑖𝑛ℎ (𝑎𝐿)
Corrected fin length
It is a better way to determine the heat loss from the fin
tip. We replace the fin length ‘L’, in case of an insulated
tip, by a corrected length ‘LC’, defined as
𝐿 𝐶 = 𝐿 +
𝐴 𝐶
𝑝
The corrected length approximation gives us good results
when the variation of temperature near the fin tip is
small, and the heat transfer coefficient at the fin tip is
almost same as that at the lateral surface of the fin.
 Fins subjected to convection at the tip are
equivalent to fins with insulated tips by
replacing the actual length by corrected length
𝐿 𝐶,𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑢𝑙𝑎𝑟 = 𝐿 +
𝑡
2
, 𝑤ℎ𝑒𝑟𝑒 𝑡
= 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑓𝑖𝑛
𝐿 𝐶,𝑐𝑦𝑙𝑖𝑛𝑑𝑟𝑖𝑐𝑎𝑙 = 𝐿 +
𝐷
4
, 𝑤ℎ𝑒𝑟𝑒 𝐷
= 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑐𝑦𝑙𝑖𝑛𝑑𝑟𝑖𝑐𝑎𝑙 𝑓𝑖𝑛
Fin efficiency
Fin efficiency is mathematically defined as,
𝜂 𝑓𝑖𝑛
=
𝑎𝑐𝑡𝑢𝑎𝑙 ℎ𝑒𝑎𝑡 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑟𝑎𝑡𝑒 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑓𝑖𝑛
𝑖𝑑𝑒𝑎𝑙 ℎ𝑒𝑎𝑡 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑟𝑎𝑡𝑒 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑓𝑖𝑛 𝑖𝑓 𝑡ℎ𝑒 𝑒𝑛𝑡𝑖𝑟𝑒 𝑓𝑖𝑛 𝑤𝑒𝑟𝑒 𝑎𝑡 𝑏𝑎𝑠𝑒
For insulated fin tips,
𝜂 𝑓𝑖𝑛 =
√ℎ𝑝𝑘𝐴 𝐶 (𝑇𝑏 − 𝑇∞) 𝑡𝑎𝑛ℎ (𝑎𝐿)
ℎ𝑝𝐿(𝑇0 − 𝑇∞)
=
𝑡𝑎𝑛ℎ (𝑎𝐿)
𝑎𝐿
 Fins with triangular and parabolic profiles
contain less material and are more efficient
than ones with rectangular profiles, thus they
are more suitable for weight sensitive
applications
 An important design consideration for fins is
the fin length L. usually, a longer fin means
larger heat transfer area and higher heat
transfer rate from the fin. But a longer fin
means, more material, and more mass

7. 7 | www.mindvis.in
Fin effectiveness
Fins are used to enhance heat transfer from a surface,
and the use of fins on a surface cannot be justified, if the
enhanced heat transfer from the surface justifies the
added cost and design complexity associated with fins.
The performance of the fin is judged on the basis of
enhancement in heat transfer w.r.t. no-fin case.
Therefore, the fin performance is expressed in terms of
fin effectiveness,
𝜖 𝑓𝑖𝑛 =
𝑞 𝑓𝑖𝑛
𝑞 𝑛𝑜 𝑓𝑖𝑛
 𝝐 𝒇𝒊𝒏 = 𝟏 then the addition of fins has no effect
on heat transfer
 𝝐 𝒇𝒊𝒏 < 1 then the fins acts as an insulation, i.e.
slowing down the heat transfer
 𝝐 𝒇𝒊𝒏 > 1 then the fins are enhancing the heat
transfer
The use of fin is only justified, if the effectiveness is way
more than 1, as finned surfaces are designed on the basis
of maximizing effectiveness for a specified cost or
minimizing cost for a specified effectiveness.
 Relation between fin efficiency, and fin
effectiveness
𝜖 𝑓𝑖𝑛 =
𝐴 𝑓𝑖𝑛
𝐴 𝑏
𝜂 𝑓𝑖𝑛
Where Afin = pL
Ab = surface area with no fin
The effectiveness for a long fin can be determined by
using the relation,
𝝐𝒍𝒐𝒏𝒈 𝒇𝒊𝒏 = √
𝒌𝒑
𝒉𝑨 𝒄
Some important conclusions regarding the design and
selection of fins:
 The thermal conductivity of the fin material
should be as high as possible. More commonly
used materials for fins are copper, aluminium
and iron. The most widely used is aluminium as
it is of low cost and has high corrosion
resistance
 The ratio of perimeter to the cross section area
of the fin should be as high as possible. This can
be satisfied by thin plate fins and slender pin
fins
 The fins are most effective in use involving low
convective heat transfer coefficient, i.e. when
we use a gas instead of a liquid and heat
transfer occurs via free convection and not
forced convection.
It is due to the above reason, why, in liquid-to-
gas heat exchangers, e.g. car radiators, fins are
placed on the gas side
1. Unsteady State Conduction
Lumped system analysis
In many heat transfer applications, some bodies are said
to behave like a lump, i.e. their interior temperatures
remain essentially uniform at all times, during a given
heat transfer process. Thus, for such bodies, the
temperature can be said to be a part of time only, i.e. T =
f(t).
In lumped system analysis we utilise the simplification
provided to analyse such cases with considerable amount
of accuracy.
Consider a copper ball heated to a certain temperature.
Measurements indicate that the temperature of the
copper ball changes with time, but it doesn’t change with
position. This way the temperature of the ball remains
uniform at all times, and it behaves as a lumped system.
Consider an arbitrary shaped body with the following
parameters,
During a small time interval ‘dt’, the temperature of the
body rises by a small amount ‘dT’.
ℎ𝑒𝑎𝑡 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑖𝑛𝑡𝑜 𝑡ℎ𝑒 𝑏𝑜𝑑𝑦 𝑑𝑢𝑟𝑖𝑛𝑔 𝑑𝑡
= 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑓 𝑏𝑜𝑑𝑦 𝑑𝑢𝑟𝑖𝑛𝑔 𝑑𝑡
or,
ℎ𝐴 𝑆(𝑇∞ − 𝑇)𝑑𝑡 = 𝑚𝐶 𝑝 𝑑𝑇 (1)
where,
𝑚 = 𝜌𝑉, 𝑎𝑛𝑑 𝑑𝑇 = 𝑑(𝑇 − 𝑇∞)
Therefore, rearranging equation (1)
𝑑(𝑇 − 𝑇∞)
(𝑇 − 𝑇∞)
= −
ℎ𝐴 𝑆
𝜌𝑉𝐶 𝑝
𝑑𝑡 (2)

8. 8 | www.mindvis.in
Integrating equation (2) from t=0, at which T = Ti
𝑙𝑛
𝑇(𝑡) − 𝑇∞
𝑇𝑖 − 𝑇∞
= −
ℎ𝐴 𝑆
𝜌𝑉𝐶 𝑝
𝑡 (3)
Taking exponent on both sides of equation (3)
𝑇(𝑡) − 𝑇∞
𝑇𝑖 − 𝑇∞
= 𝑒−𝑏𝑡
Where
𝑏 =
ℎ𝐴 𝑆
𝜌𝑉𝐶 𝑝
The term ‘b’ is called time constant, a positive quantity,
whose units are (time)-1.
The figure below, shows a plot of equation (3) for
different values of ‘b’,
The temperature of a body reaches the ambient
temperature 𝑇∞ exponentially. A larger value of ‘b’
indicates that the body will reach the ambient
temperature in a short time, i.e. higher temperature
decay rate.
 Time constant is proportional to surface area
and inversely proportional to the mass and the
specific heat of the body.
 This indicates that it takes longer to cool or heat
a larger mass
The lumped system analysis provides convenient heat
transfer analysis, but it is important to know when to
apply it.
To establish a criterion to apply the lumped system
analysis, let us define a parameter, characteristic length
as,
𝐿 𝐶 =
𝑉
𝐴 𝑆
And another parameter, Biot number (Bi) as,
𝐵𝑖 = ℎ𝐿 𝐶/𝑘
Biot number can also expressed as,
𝐵𝑖 =
𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑡 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑜𝑓 𝑎 𝑏𝑜𝑑𝑦
𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑤𝑖𝑡ℎ𝑖𝑛 𝑡ℎ𝑒 𝑏𝑜𝑑𝑦
=
𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑜𝑛 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒
=
𝐿 𝐶/𝑘
1/ℎ
As per the above expression, a small Biot number
represents small resistance to heat conduction, and thus
small temperature gradients within the body. Lumped
system analysis assumes uniform temperature
distribution within the body, and it is possible only, when
conduction resistance is zero.
 Lumped system analysis is exact when Bi = 0,
and it is approximate when Bi > 0
 For a highly accurate lumped system analysis,
the value of Bi should be as low as possible
 Keeping in mind the uncertainties in the
convection process, the generally acceptable
value of Bi, to make lumped system analysis
applicable is
𝐵𝑖 ≤ 0.1
 The first step when applying lumped system
analysis, is to calculate Biot number and to
check the criteria for its applicability
 Small bodies, with high thermal conductivity
are best suited for application of lumped
system analysis, when they are in a medium
which is not a good conductor of heat
The hot small copper ball shown in the diagram, satisfies
the criterion for the application of lumped system
analysis,
Fourier Number (Fo)
Mathematically, it is expressed as,

9. 9 | www.mindvis.in
𝐹𝑜 =
𝛼𝑡
𝐿2
Fourier number is used in calculations, when heat
conduction and heat energy storage concurrently.
2. Convection
Introduction
The phenomenon of conduction and convection are
similar to each other in the context, that both require a
material medium for their occurrence. But convection
requires the presence of fluid motion.
In a solid, the main mode of heat transfer is conduction,
but in liquids or gases, the mode can be either conduction
or convection, depending upon the presence of bulk fluid
motion. The fluid motion enhances the heat transfer, as
it brings together hot and cool streams of fluid in contact
with each other. The heat transfer rate, via convection is
higher than the heat transfer rate via conduction, and
with a higher fluid velocity, the heat transfer rate is also
high.
Convection depends upon fluid properties like dynamic
viscosity (µ), thermal conductivity (k), density (ρ), specific
heat (CP) and fluid velocity (v). In addition to these
properties, it also depends upon geometry, roughness of
the solid surface, over which the fluid is flowing and the
type of fluid flow (i.e. laminar or turbulent).
The governing mathematical expression for convection
heat transfer process is called Newton’s law of cooling
and is expressed as,
𝑞 𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑜𝑛 = ℎ𝐴 𝑆(𝑇𝑆 − 𝑇∞)
Where, h = convection heat transfer coefficient (W/m2 –
0C)
AS = surface area for heat transfer (m2)
TS = surface temperature
𝑇∞ = temperature of the fluid sufficiently far from
the surface
Convection process is of two types:
a) Forced convection
b) Natural/ Free convection
The diagram below illustrates the two types of
convection processes named above,
When a fluid is made to flow over
a nonporous solid surface, it is observed that the fluid

10. 10 | www.mindvis.in
velocity at the solid surface is zero, i.e. the fluid layer in
immediate contact with the sold surface is stagnant. This
condition is known as no slip condition, and is mainly due
to the viscosity of the fluid. The no slip condition is
responsible for the development of the velocity profile of
the flow, which is a parabolic profile, due to friction
between adjacent layers of the fluid and viscosity.
A phenomenon, similar to no slip condition occurs, when
a fluid and solid surface will have the same temperature
at the point of contact. This is known as the no-
temperature-jump condition.
An implication of no-slip and no-temperature-jump
condition is that the heat transfer from the solid surface
is via pure conduction. This heat is then carried away by
the fluid motion, via convection.
Nusselt Number (Nu)
Nusselt number is defined as,
𝑁𝑢 =
ℎ𝐿 𝐶
𝑘
=
𝑞 𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑜𝑛
𝑞 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑜𝑛
Nusselt number represents the enhancement of heat
transfer as a result of convection relative to conduction
across a layer of fluid.
For effective convection, Nu should be as large as
possible. If Nu =1, then it represents a condition of heat
transfer by pure conduction.
Velocity Boundary Layer
Consider a parallel flow of a fluid over a flat plate, as
shown in the figure.
The fluid approaches the plate in the x-direction with a
uniform upstream velocity ‘V’, and can be put equal to
free stream velocity 𝑢∞ over the plate, away from the
surface.
The no slip condition exists in the vicinity of the plate
surface. Due to this, the adjacent layers of fluid slow
down. Therefore, the presence of the plate is
experienced upto a height 𝛿 from the plate surface.
Beyond this height𝛿, the effect of the plate, slowing down
the adjacent layers is no longer there, and the free stream
velocity is achieved.
The variation of fluid velocity w.r.t. y is as follows,
𝑢 = 0 𝑎𝑡 𝑦 = 0
𝑢 = 𝑢∞ 𝑎𝑡 𝑦 = 𝛿
The region of the flow above the plate bounded in 𝛿 in
which the effects of the viscous shearing effects are felt
is called the velocity boundary layer.
The boundary layer thickness, 𝛿, is usually defined as the
distance ‘y’ from the surface at which 𝑢 = 0.99 𝑢∞.
Thermal Boundary Layer
A velocity boundary layer develops when a fluid flows
over a surface as a result of fluid adjacent to the surface
exhibiting no-slip condition. Likewise, a thermal
boundary layer develops when a fluid at a specified
temperature flows over a surface that is at a different
temperature, as shown in the figure,
The thickness of the thermal boundary layer increases in
the flow direction.
The convection heat transfer rate, along the surface is
directly proportional to the temperature gradient. Thus,

11. 11 | www.mindvis.in
the temperature profile shape in the thermal boundary
layer dictates the convective heat transfer between solid
surface and the fluid flowing over it.
Prandtl Number (Pr)
Prandtl number is defined as,
𝑃𝑟 =
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑣𝑖𝑡𝑦 𝑜𝑓 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑣𝑖𝑡𝑦 𝑜𝑓 ℎ𝑒𝑎𝑡
=
𝜇𝐶 𝑃
𝑘
The value of Prandtl numbers for fluids range from less
than 0.01 for liquid metals and more than 10,000 for
heavy oils.
Prandtl number for gases is about 1, which indicates that
momentum and heat dissipate at the same rate through
it.
Reynolds Number (Re)
Reynolds number is defined as,
𝑅𝑒 =
𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑓𝑜𝑟𝑐𝑒𝑠
𝑣𝑖𝑠𝑐𝑜𝑢𝑠 𝑓𝑜𝑟𝑐𝑒𝑠
=
𝜌𝑉𝐿 𝐶
𝜇
At large values of Re, the inertia forces dominate the
viscous forces, and the viscous forces cannot prevent the
large and random fluctuations of the fluid, thus the flow
is termed as turbulent.
At low values of Re, the viscous forces dominate over the
inertia forces, which help in keeping the fluid flow in line,
thus the flow in this case is termed as laminar.
The value of Re, at which the transition occurs from
laminar flow to turbulent flow takes place is called critical
Reynolds number.
 The transition from laminar to turbulent flow
depends upon the surface geometry, surface
roughness, free stream velocity, and the type
of fluid
Physical significance of Dimensionless parameters
Consider the Reynold’s number (Re), which is defined as
the ratio of inertia to viscous forces in a region of
characteristic dimension L. Inertia forces are associated
with an increase in the momentum of a moving fluid. The
inertia forces dominate for large values of Re and the
viscous forces dominate for small values of Re. Reynolds
number is also used to determine the existence of
laminar or turbulent flow. The small disturbances present
in any kind of flow can be amplified to produce turbulent
flow conditions. For small values of Re, the viscous forces
are very high as compared to the inertia forces to
produce this kind of amplification, hence maintaining
laminar flow conditions. As the value of Re increases the
impact of viscous forces become less and the dominance
of inertia forces set in, thereby making small disturbances
amplified to create a transition to turbulent flow.
Prandtl number (Pr) is the ratio of momentum diffusivity
to the thermal diffusivity, and this number provides the
measure of the relative effectiveness of momentum and
energy transport by diffusion in the velocity and thermal
boundary layers. The value of Pr for gases is almost unity,
so the momentum and energy transfer by diffusion are
comparable. For liquid metals, Pr<<1, which indicate that
the energy diffusion rate exceeds the momentum
diffusion rate. For oils Pr>>1, which indicates the
opposite as compared to liquid metals.
Differential equations for Convection equations
Consider a flat plate with parallel flow of a fluid over its
surface. The flow directions along the surface is x and in
a direction normal to the surface is y. The fluid flow has
uniform free stream velocity 𝑢∞.
There are three differential equations for laminar flow in
boundary layers, namely:
1. Conservation of mass equation
2. Conservation of momentum equation
3. Conservation of energy equation
Conservation of Mass equation
According to this principle, the mass cannot be destroyed
or created. In a steady flow the amount of mass flowing
through the control volume remains constant,
(𝑟𝑎𝑡𝑒 𝑜𝑓 𝑚𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑖𝑛) = (𝑟𝑎𝑡𝑒 𝑜𝑓 𝑚𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑜𝑢𝑡)
Mathematically the conservation of mass equation is
given as,
𝜕𝑢
𝜕𝑥
+
𝜕𝑣
𝜕𝑦
= 0
Where u = flow velocity in x-direction and v =flow velocity
in y-direction
Conservation of Momentum equation
For this we make use of Newton’s second law, which
states, the net force acting on the control volume is equal
to the mass times the acceleration of the fluid element
within the control volume, which is also equal to the net
rate of momentum outflow from the control volume.
(𝑚𝑎𝑠𝑠)(𝑎𝑐𝑐𝑒𝑙𝑒𝑎𝑟𝑖𝑜𝑛 𝑖𝑛 𝑎 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑒𝑑 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛)
= (𝑛𝑒𝑡 𝑓𝑜𝑟𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔 𝑖𝑛 𝑡ℎ𝑎𝑡 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛)
The momentum conservation in x-direction is
mathematically expressed as,
𝜌 (𝑢
𝜕𝑢
𝜕𝑥
+ 𝑣
𝜕𝑢
𝜕𝑦
) = 𝜇
𝜕2
𝑢
𝜕𝑦2
−
𝜕𝑃
𝜕𝑥
, 𝑤ℎ𝑒𝑟𝑒 𝑃 𝑖𝑠 𝑡ℎ𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒
Conservation of Energy equation

12. 12 | www.mindvis.in
The energy equation for a steady two-dimensional flow
of a fluid with constant properties and negligible shear
stresses is expressed as,
𝜌𝐶 𝑃 (𝑢
𝜕𝑇
𝜕𝑥
+ 𝑣
𝜕𝑇
𝜕𝑦
) = 𝑘 (
𝜕2
𝑇
𝜕𝑥2
+
𝜕2
𝑇
𝜕𝑦2
)
The above equation states that the net energy convected
by the fluid out of the control volume is equal to the net
energy transferred into the control volume by heat
conduction.
Thermal Boundary Layer
Thermal boundary layer may be defined as that region
where temperature gradients are present in the flow.
These temperature gradients result from a heat exchange
process between the fluid and the stationary surface.
Let the temperature of the surface is TS and the
temperature of the fluid outside the thermal boundary
layer is 𝑇∞.
The thickness of the thermal boundary layer is designated
as 𝛿𝑡. At the wall, the velocity of the fluid is zero, and the
heat transfer into the fluid takes place via conduction,
𝑞 = −𝑘𝐴
𝑑𝑇
𝑑𝑦
|
𝑠𝑢𝑟𝑓𝑎𝑐𝑒
From Newton’s law of cooling,
𝑞 𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑜𝑛 = ℎ𝐴 𝑆(𝑇𝑆 − 𝑇∞)
Equating the above heat transfer rates, we get,
ℎ =
(−𝑘
𝑑𝑇
𝑑𝑦
|
𝑠𝑢𝑟𝑓𝑎𝑐𝑒
)
𝑇𝑆 − 𝑇∞
The following boundary conditions apply,
𝑇 = 𝑇𝑆 𝑎𝑡 𝑦 = 0
𝑑𝑇
𝑑𝑦
= 0 𝑎𝑡 𝑦 = 𝛿𝑡
𝑇 = 𝑇∞ 𝑎𝑡 𝑦 = 𝛿𝑡
After mathematical manipulation and substitution, we
get the following relation for the temperature profile,
which indicates a cubic curve,
𝑇 − 𝑇𝑆
𝑇𝑆 − 𝑇∞
=
𝜃
𝜃∞
=
3
2
(
𝑦
𝛿𝑡
) −
1
2
(
𝑦
𝛿𝑡
)
3
To find out the value of thermal boundary layer thickness,
use the following formula,
𝛿𝑡
𝛿
=
1
1.026
𝑃𝑟−1/3
Where 𝛿 = velocity boundary layer thickness
Pr = Prandtl Number
 The above analysis is based on the assumption,
that the fluid properties are constant
throughout the flow
 When there exists, an appreciable variation
between wall and free stream conditions, we
use a term film temperature to evaluate the
properties. The film temperature is defined as
the arithmetic mean of surface temperature
and free stream temperature,
𝑇𝑓 =
𝑇𝑆 + 𝑇∞
2

13. 13 | www.mindvis.in
3. Radiation
Introduction
Consider a hot object that is suspended in a vacuum
chamber, with its walls at room temperature.
The hot object will cool down and attain thermal
equilibrium with its surroundings, i.e. the object will lose
heat until its temperature becomes equal to the
temperature of the temperature of the walls of the
chamber. In a vacuum, conduction and convection
cannot bring about heat transfer, so the process which
comes into action in a vacuum, is radiation.
As it is clear from the above example, that radiation does
not require any material medium for its occurrence.
Radiation takes place across all three states of matter.
 Radiation heat transfer can occur between two
bodies separated by a medium colder than
both bodies, e.g. solar radiations reach the
surface of the earth after passing through cold
layers at high altitiudes
We associate thermal radiation with the rate at which
energy is emitted by the matter as a result of its finite
temperature. The energy emission mechanism is related
to the energy released as a result of oscillations of infinite
number of electrons, comprised in a matter. These
oscillations then in turn increase the internal energy of
the matter, which means a rise in its temperature.
 All forms of matter emit radiation
 For gases it is a volume phenomenon
 For opaque solids, like metals, woods etc.
radiation is considered to be a surface
phenomena, since the radiation emitted only
by the molecules at the surface can escape the
solid
Thermal radiation is viewed as electromagnetic waves
propagation. For thermal radiation propagation through
a medium, the wavelength (λ) and frequency (ν) are
related as,
𝜆 =
𝑐
𝜈
Where c = speed of light in the given medium
For vacuum, speed of light is given as cO = 2.998 X 108 m/s
Electromagnetic radiation is the propagation of a
collection of discrete packets of energy called photons or
quanta. Each photon of frequency ν is considered to have
an energy of,
𝑒 = ℎ𝜈 =
ℎ𝑐
𝜆
Where h = 6.6256 X 10-34 J.sec is the Planck’s constant
Although all EM waves have the same general features,
waves with different wavelength differ significantly in
their behaviour. The electromagnetic spectrum lists all
the EM radiation encountered in daily practice, i.e
gamma rays, X-rays, UV radiations and radio waves.

14. 14 | www.mindvis.in
Blackbody Radiations
A body at a temperature above the absolute zero emits
radiation in all directions over a wide range of
wavelengths, and the amount of radiation energy
emitted from a surface at a particular wavelength
depends upon the material of the body and its surface
temperature.
A blackbody is an object that emits maximum amount of
radiation, and is defined as a perfect emitter and
absorber of radiation. At a given value of temperature
and wavelength, no surface can emit more energy than a
blackbody. It also absorbs all incident energy on it,
regardless of t he wavelength and direction.
 A blackbody is a diffuse emitter, which means
emissions from a blackbody are independent of
direction
The energy emitted by a blackbody per unit time and per
unit surface area is given by Stefan-Boltzmann Law,
𝐸 𝑏(𝑇) = 𝜎𝑇4
Where σ = 5.67 X 10-8 W/m2 K4 is known as Stefan-
Boltzmann constant.
T = absolute temperature of the surface in K
Eb = blackbody emissive power and is a function of
temperature
 Any surface that absorbs the visible portion of
the spectrum would appear black to the eye
 Any surface that reflects the visible portion of
the spectrum would appear white to the eye
As per Stefan-Boltzmann’s law, blackbody emissive
power is ONLY a function of absolute temperature of the
surface, but as per Planck’s law, the blackbody emissive
power is a function of wavelength and temperature, as
follows,
𝐸 𝑏𝜆(𝜆, 𝑇) =
𝐶1
𝜆5 [exp (
𝐶2
𝜆𝑇
) − 1]
𝑊/𝑚2
. 𝜇𝑚
Where,
𝐶1 = 2𝜋ℎ𝑐 𝑜
2
= 3.742 𝑋 108
𝑊. 𝜇𝑚4
/𝑚2
𝐶2 =
ℎ𝑐 𝑜
𝑘
= 1.439 𝑋 108 𝜇𝑚. 𝐾
Where T = absolute temperature of the surface
λ = wavelength of the radiation emitted
k = Boltzmann’s constant = 1.38065 X 10-23 J/K
The relation discussed is valid for a surface in a vacuum
or a gas, for other mediums it needs to be changed by
replacing,
𝐶1 𝑏𝑦
𝐶1
𝑛2
Where n = index of refraction of the medium
The variation of the spectral emissive power of blackbody
with wavelength is illustrated in the figure below,

15. 15 | www.mindvis.in
Some important points to note regarding the above
figure are as follows:
 The emitted radiation is a continuous function
of wavelength, and it increases with
temperature, reaches a maximum value and
then starts to decrease with an increasing
wavelength
 At any wavelength, the quantity of emitted
radiation increases with increasing
temperature
 A larger fraction of the radiation is emitted at
shorter wavelengths at higher temperature,
and this is indicated on the curve with a shift of
the curve towards shorter wavelength region
as temperature increases. This shift can be
specified by Wein’s displacement law as,
(𝜆𝑇)max 𝑝𝑜𝑤𝑒𝑟 = 2897.8 𝜇 𝑚. 𝐾
 Surfaces at T≤ 800 K emit radiations in Infrared
region which is not visible to our eye, unless it
is reflected from some surface
The colour of an object is not dependent upon its
emission, which is primarily in the infrared zone, unless
the temperature is above 1000 K value. Actually the
colour of the object depends upon the absorption and
reflection characteristics of the surface, i.e. selective
absorption and selective reflection. In the figure given
below, an incident light with red, yellow, green and blue
colours is incident upon a surface, which absorbs all the
colours except the red one. Red colour is reflected by the
surface, thus the surface appears red to the human eye.
The total radiation energy emitted by a blackbody at a
particular temperature is given by the area under Ebλ and
λ chart as shown below in the figure,
Radiation Intensity
Radiation is emitted by parts of a plane surface in all
directions into the hemisphere above the surface, and
the directional distribution of emitted radiation is usually
not uniform. So, we need a quantity that describes the
magnitude of radiation emitted in a particular direction
in space. This quantity is known as radiation intensity (I).
The direction of radiation passing through a point is best
described in spherical coordinates in terms of the zenith
angle θ and azimuth angle φ, ass illustrated in the figure
below

16. 16 | www.mindvis.in
 If all surfaces emitted radiation uniformly in all
directions, the emissive power would be
sufficient to quantify radiation
Radiosity
Surfaces emit and reflect radiation, so the radiation
leaving the surface consists of emitted and reflected
components, as illustrated in the figure below,
Therefore, radiosity J, is defined as the total radiation
leaving the surface, without any dependence of origin. Its
units are W/m2.
 For a black body, radiosity is equivalent to the
emissive power Eb
Emissivity
The emissivity of a surface represents the ratio of the
radiation emitted by the surface at a given temperature
to the radiation emitted by a black body at the same
temperature. It is denoted by ‘ε’ and its value varies
between 0 and 1.
It is a measure of how closely a surface approximates a
blackbody.
 For a blackbody, ε = 1
 Incident radiation on a surface is called
IRRADIATION and is denoted by ‘G’
Absorptivity, Reflectivity and Transmissivity
When radiation strikes a surface a part of it is absorbed,
a part is reflected and the remaining part is transmitted.
The fraction of incident radiation which is absorbed by
the surface is called absorptivity (α),
𝛼 =
𝑎𝑏𝑠𝑜𝑟𝑏𝑒𝑑 𝑟𝑎𝑑𝑖𝑎𝑡𝑖𝑜𝑛
𝑡𝑜𝑡𝑎𝑙 𝑖𝑛𝑐𝑖𝑑𝑒𝑛𝑡 𝑟𝑎𝑑𝑖𝑎𝑡𝑖𝑜𝑛
, 0 ≤ 𝛼 ≤ 1
The fraction of incident radiation which is reflected is
called reflectivity (ρ),
𝜌 =
𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑒𝑑 𝑟𝑎𝑑𝑖𝑎𝑡𝑖𝑜𝑛
𝑡𝑜𝑡𝑎𝑙 𝑖𝑛𝑐𝑖𝑑𝑒𝑛𝑡 𝑟𝑎𝑑𝑖𝑎𝑡𝑖𝑜𝑛
, 0 ≤ 𝜌 ≤ 1
The fraction of incident radiation which is transmitted is
called transmissivity (τ),
𝜏 =
𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑡𝑡𝑒𝑑 𝑟𝑎𝑑𝑖𝑎𝑡𝑖𝑜𝑛
𝑡𝑜𝑡𝑎𝑙 𝑖𝑛𝑐𝑖𝑑𝑒𝑛𝑡 𝑟𝑎𝑑𝑖𝑎𝑡𝑖𝑜𝑛
, 0 ≤ 𝜏 ≤ 1
𝛼 + 𝜌 + 𝜏 = 1
For opaque surface, 𝜏 = 0
View factor
Radiation heat transfer between surfaces depends upon
the orientation of the surfaces w.r.t. each other as well as
their radiation properties and temperatures, as
illustrated by the figure below,

17. 17 | www.mindvis.in
This dependence on orientation is accounted for by the
view factor, also called shape factor and configuration
factor. This is a purely geometric quantity and is
independent of the surface properties and temperature.
The view factor from a surface i to j is denoted by 𝐹𝑖→𝑗 or
just 𝐹𝑖𝑗 and is defined as the fraction of the radiation
leaving surface i that strikes surface j directly.
 The radiation that strikes a surface does not
need to be absorbed by that surface, and the
radiation that strikes a surface after being
reflected by other surfaces is not considered in
the evaluation of view factors
The figure below shows the view factors for different
surface configurations,
4. Heat Exchangers
Introduction
Heat exchangers are devices that help in exchange of
heat energy between two fluids that are at different
temperatures, which are kept from mixing with each
other.
These devices use convection in each fluid and
conduction through the wall separating the two fluids.
While analysing heat exchangers we consider overall heat
transfer coefficient, U which encompasses all the effects
in a heat exchanger.
Types of heat exchangers
The simplest type of heat exchanger consists of two pipes
of different diameters, called the double pipe heat
exchanger. One fluid flows through the inner pipe while
the other fluid flows through the outer pipe, and
depending upon the direction of flow we classify the
double pipe heat exchanger as:
 Parallel flow: In this type, both the fluids flow
in the same direction
 Counter flow: In this type, both the fluids flow
in opposite directions
The figures below illustrate the two types of heat
exchangers described above,
 The ratio of heat transfer surface area of a heat
exchanger to its volume is called the area
density (β)
When two fluids move perpendicular to each other, the
heat exchanger is termed as cross flow heat exchanger.
This type of heat exchanger is further classified as:-
 Unmixed
 Mixed flow

18. 18 | www.mindvis.in
The most common type of heat exchanger used in
industrial applications is a shell & tube heat exchanger, as
illustrated in the figure below,
The figure shown above is a one-pass shell and tube heat
exchanger.
Baffles are commonly placed in the shell to force the
shell-side fluid to flow across the shell to enhance heat
transfer and to maintain uniform spacing between the
tubes. These types of heat exchangers are classified
according to the number of shell and tube passes
involved:
 One shell pass and two tube passes
 Two shell passes and four tube passes
Overall Heat Transfer Coefficient
Heat in a heat exchanger is first transferred from hot fluid
to the wall by convection, through the wall by conduction
and from the wall to the cold fluid via convection.
 Any radiation effects are usually taken care off
in the convective heat transfer coefficient
The figure below illustrates the thermal resistance
network associated with heat transfer in a double pipe
heat exchanger,
The total thermal resistance of the network is given as,
𝑅𝑡ℎ𝑒𝑟𝑚𝑎𝑙 = 𝑅𝑖 + 𝑅 𝑤𝑎𝑙𝑙 + 𝑅 𝑜 =
1
ℎ𝑖 𝐴𝑖
+
[ln (
𝐷 𝑜
𝐷𝑖
)]
2𝜋𝑘𝐿
+
1
ℎ 𝑜 𝐴 𝑜
Where Ai = inner surface area of the wall = 𝜋𝐷𝑖 𝐿
Ao = outer surface area of the wall = 𝜋𝐷 𝑜 𝐿
L = length of the tube
The heat transfer rate between two fluids is expressed as,
𝑄̇ =
∆𝑇
𝑅𝑡ℎ𝑒𝑟𝑚𝑎𝑙
= 𝑈𝐴∆𝑇
𝑅𝑡ℎ𝑒𝑟𝑚𝑎𝑙 =
1
𝑈𝐴
Where U = overall heat transfer coefficient in W/m2
 When the wall thickness is small and the
thermal conductivity of the tube material is
high, this means that the thermal resistance of
the wall, Rwall = 0 and inner and outer surface

19. 19 | www.mindvis.in
areas are almost equal, Ai =Ao, then the overall
heat transfer coefficient is,
1
𝑈
≈
1
ℎ𝑖
+
1
ℎ 𝑜
 The value of U is dominated by the smaller
convection coefficient
Fouling factor
The performance of heat exchangers deteriorates with
time due to accumulation of deposits on heat transfer
surfaces. This additional layer of deposits creates extra
thermal resistance, and decreases the heat transfer rate
in the heat exchanger.
The total effect of this accumulation is represented by Rf,
which is basically a measure of the thermal resistance
introduced by fouling.
 The most common type of fouling is the
precipitation of solid deposits in a fluid on the
heat transfer surfaces
 The other common type of fouling is chemical
fouling
The fouling factor depends upon the operating
temperature and the velocity of the fluids, and the length
of the service.
The value of fouling factor increases with increasing
temperature and decreasing velocity.
For an unfinned shell-and-tube heat exchanger, it can be
expressed as,
1
𝑈𝐴
= 𝑅𝑡ℎ𝑒𝑟𝑚𝑎𝑙 =
1
ℎ𝑖 𝐴𝑖
+
𝑅𝑓,𝑖
𝐴𝑖
+
[ln (
𝐷 𝑜
𝐷𝑖
)]
2𝜋𝑘𝐿
+
𝑅𝑓,𝑜
𝐴 𝑜
+
1
ℎ 𝑜 𝐴 𝑜
Where Rf,i = fouling factor at inner surfaces
Rf,o = fouling factor at outer surfaces
Log Mean Temperature Difference (LMTD)
As per this method, heat transfer rate in a heat exchanger
is given as,
𝑄̇ = 𝑈𝐴∆𝑇𝑙𝑛
Where, ΔTln = LMTD and it is expressed as,
∆𝑇𝑙𝑛 =
∆𝑇1 − ∆𝑇2
[ln (
∆𝑇1
∆𝑇2
)]
The values of ∆𝑇1 and ∆𝑇2 depends upon the flow type in
the double pipe heat exchanger, i.e. whether it is parallel
flow or counter flow heat exchanger.
The figure illustrates the above mentioned point,
The arithmetic mean temperature is given as,
∆𝑇 𝑚𝑒𝑎𝑛 =
∆𝑇1 + ∆𝑇2
2
 It should be noted that LMTD is always less then
arithmetic mean temperature
Effectiveness-NTU
The LMTD method discussed above is easy to use in a
heat exchanger analysis where the inlet and the outlet
temperatures of the hot and cold fluids are known or can
be determined. This means, that the LMTD method is
useful for determining the size of a heat exchanger when
the mass flow rates and the inlet and outlet temperatures
of the hot and cold fluids are given.
Another type of problem encountered in heat exchanger
analysis is that of determination of the ‘heat transfer rate’
and the ‘outlet temperatures’ of the hot and cold fluids
for the given fluid mass flow rates and inlet temperatures
when the type and size of the heat exchange are given. In
this situation the heat transfer surface area A of the heat
exchanger is known but not the outlet temperatures.
To solve the heat exchanger analysis in the above
condition, we will use the effectiveness-NTU method. In

20. 20 | www.mindvis.in
this method, we define a dimensionless parameter, heat
transfer effectiveness as,
𝜀 =
𝑎𝑐𝑡𝑢𝑎𝑙 ℎ𝑒𝑎𝑡 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑟𝑎𝑡𝑒
𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 ℎ𝑒𝑎𝑡 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑟𝑎𝑡𝑒
=
𝑄̇
𝑄 𝑚𝑎𝑥
̇
𝑄̇ = 𝐶 𝐶(𝑇𝑐,𝑜𝑢𝑡 − 𝑇𝑐,𝑖𝑛) = 𝐶ℎ(𝑇ℎ,𝑖𝑛 − 𝑇ℎ,𝑜𝑢𝑡)
𝐶 𝐶 = 𝑚 𝐶̇ 𝑐 𝑝𝑐 𝑎𝑛𝑑 𝐶ℎ = 𝑚ℎ̇ 𝑐 𝑝ℎ
To calculate the maximum possible heat transfer rate, we
have to consider the minimum value out of CC and Ch, and
substitute as Cmin in the relation,
𝑄 𝑚𝑎𝑥
̇ = 𝐶 𝑚𝑖𝑛(𝑇ℎ,𝑖𝑛 − 𝑇𝑐,𝑖𝑛)
Going by the above given relations, we can determine the
actual heat transfer rate, if the heat exchanger
effectiveness is given, as,
𝑄̇ = 𝜀 𝑄 𝑚𝑎𝑥
̇
This way, the effectiveness of a heat exchanger enables
us to determine the heat transfer rate without knowing
the outlet temperatures of the fluids handled.
 The effectiveness of a heat exchanger depends
upon the geometry of the exchanger and the
flow arrangement
For a parallel flow heat exchanger,
𝜀 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑓𝑙𝑜𝑤 =
1 − exp [
−𝑈𝐴 𝑆
𝐶 𝑚𝑖𝑛
(1 + (
𝐶 𝑚𝑖𝑛
𝐶 𝑚𝑎𝑥
))]
1 + (
𝐶 𝑚𝑖𝑛
𝐶 𝑚𝑎𝑥
)
The quantity
−𝑈𝐴 𝑆
𝐶 𝑚𝑖𝑛
is a dimensionless parameter called the
number of transfer units (NTU),
𝑁𝑇𝑈 =
−𝑈𝐴 𝑆
𝐶 𝑚𝑖𝑛
Where, U = overall heat transfer coefficient
AS = heat transfer surface area
Cmin = minimum heat capacity
 NTU is proportional to the surface area
 The larger the value of NTU, the larger will be
the heat exchanger
Another dimensionless parameter is defined, capacity
ratio (c),
𝑐 =
𝐶 𝑚𝑖𝑛
𝐶 𝑚𝑎𝑥
 Effectiveness is a function of the NTU and the
capacity ratio (c) as,
𝜀 = 𝑓 (
𝑈𝐴 𝑆
𝐶 𝑚𝑖𝑛
,
𝐶 𝑚𝑖𝑛
𝐶 𝑚𝑎𝑥
) = 𝑓(𝑁𝑇𝑈, 𝑐)
 The value of effectiveness ranges from 0 to 1,
and it increases rapidly with NTU for small
values but slowly for large values, so the use of
a heat exchanger with a large value of NTU
cannot be justified economically
 For a given value of NTU and capacity ratio, the
counter-flow heat exchanger has the highest
effectiveness
 The effectiveness of a heat exchanger is
independent of the capacity ratio for NTU
values less than 0.3
 The value of capacity ratio ranges between 0
and 1. For a given NTU, the effectiveness
becomes maximum for c = 0 and a minimum for
c = 1. If c = 0, it indicates a phase change
process, as in a condenser or a boiler.
Questions:
1. In descending order of magnitude, the thermal
conductivity of (a) pure iron, (b) liquid water, (c)
saturated water vapour and (d) aluminum can be
arranged as
(A) abcd (B) bcad
(C) dabc (D) dcba
2. For the circular tube of equal length and diameter
shown below, the view factor 𝐹13 is 0.17. The view
factor 𝐹12 in this case will be
𝑟 𝐷
(A) 0.17 (B) 0.21
(C) 0.79 (D) 0.83
3. For the same inlet and outlet temperatures of hot
and cold fluids, the Log mean Temperature
Difference (LMTD) is
(A) greater for parallel flow heat exchanger than for
counter flow heat exchanger
(B) greater for counter flow heat exchanger than for
parallel flow heat exchanger
(C) same for both parallel and counter flow heat
exchangers
(D) dependent on the properties of the fluids.
Common Data For 𝑄.4 and 𝑄.5
Heat is being transferred by convection 𝑓𝑖: 𝑜𝑚 water at
48 𝑜
𝐶 to a glass plate whose surface that is exposed to the
water is at 40 𝑜
C. The thermal conductivity of water is 0.6
𝑊/𝑚𝐾 and the thermal conductivity of glass is 1.2
𝑊/𝑚𝐾. The spatial gradient of temperature in the water
at the water‐ glass interface is 𝑑𝑇𝑙𝑑𝑦 = 1 × 104 𝐾/𝑚.

21. 21 | www.mindvis.in
4. The value of the temperature gradient in the glass at
the water‐glass interface in 𝐾/𝑚 is
(A) −2 × 104 (B) 0.0 × 104
(C) 0. 5 × 104
(𝐷) 2 × 104
5. The heat transfer coefficient ℎ in 𝑊/𝑚2 𝐾 is
(A) 0.0 (B) 4.8
(C) 6 (D) 750
6. Consider a laminar boundary layer over a heated flat
plate. The free stream velocity is 𝑈∞. At some
distance 𝑥 from the leading edge the velocity
boundary layer thickness is 𝛿 𝑣 and the thermal
boundary layer thickness is 𝛿 𝒯. If the Prandtl number
is greater than 1, then
(A) 𝛿 𝑣 > 𝛿 𝑇 (B) 𝛿 𝑇 > 𝛿 𝑣
(C) 𝛿 𝑣 ≈ 𝛿𝑖 ∼ (𝑈∞ 𝑥)−1/2
(D) 𝛿 𝑣 ≈ 𝛿𝑖 ∼ 𝑥−1/2
7. In a counter flow heat exchanger, for the hot fluid
the heat capacity = 2𝑘𝐽/𝑘𝑔𝐾, mass flow rate =
5𝑘𝑔/𝑠, inlet temperature = 150 𝑜 𝐶, outlet
temperature = 100 𝑜
C. For the cold fluid, heat
capacity = 4𝑘𝐽/𝑘𝑔𝐾, mass flow rate = 10𝑘𝑔/𝑠,
inlet temperature = 20 𝑜 C. Neglecting heat transfer
to the surroundings, the outlet temperature of the
cold fluid in 0 𝐶 is
(A) 7.5 (B) 32.5
(C) 45.5 (D) 70.0
8. A plate having 10 𝑐𝑚2
area each side is hanging in
the middle of a room of 100 𝑚2
𝑡 ot al surface area.
𝑇 he plat et emp erature and emissivity are
respectively 800 𝐾 and 0.6. The temperature and
emissivity values for the surfaces of the room are
300 𝐾 and 0.3 respectively. Boltzmann’s constant
𝜎 = 5.67 × 10−8
𝑊/𝑚2
𝐾4
The total heat loss ffom
the two surfaces of the plate is
(A) 13.66 𝑊 (B) 27.32 𝑊
(C) 27.87 𝑊 (D) 13.66 MW
9. In a condenser, water enters at 30 𝑜
𝐶 and flows at
the rate 1500 𝑘𝑔/ℎ𝑟. The condensing steam is at a
temperature of 120 𝑜
𝐶 and cooling water leaves the
condenser at 80 𝑜
C. Specific heat of water is 4. 187
𝑘𝐽/𝑘𝑔𝐾. If the overall heat transfer coefficient is
2000 𝑊/𝑚2
𝐾, then heat transfer area is
(A) 0.707 𝑚2
(B) 7.07 𝑚2
(C) 70.7 𝑚2
(D) 141.4 𝑚2
10. A spherical thermocouple junction of diameter
0.706 mm is to be used for the measurement of
temperature of a gas stream. The convective heat
transfer co‐efficient on the 𝑏 ead surface is 400
𝑊/𝑚2
𝐾. 𝑇 hermo‐physical properties of
thermocouple material are 𝑘 = 20𝑊/𝑚𝐾, 𝑐 =
400𝐽/ kg 𝐾 and 𝜌 = 8500𝑘𝑔/𝑚3 If the
thermocouple initially at 30 𝑜 𝐶 is placed in a hot
stream of 300 𝑜
𝐶, then time taken by the bead to
reach 298 𝑜 𝐶, is
(A) 2.35 𝑠 (B) 4.9 𝑠
(C) 14.7 𝑠 (D) 29.4 𝑠
11. A stainless steel tube (𝑘 𝑠 = 19𝑊/𝑚𝐾) of 2 cm 𝐼𝐷
and 5 cm 𝑂𝐷 is insulated with 3 cm thick asbestos
(𝑘 𝑎 = 0.2𝑊/𝑚𝐾) . If the temperature difference
between the innermost and outermost surfaces is
600 𝑜
𝐶, the heat transfer rate per unit length is
(A) 0.94
𝑊
𝑚
(B) 9.44 𝑊/𝑚
(C) 944.72
𝑊
𝑚
(D) 9447.21 𝑊/𝑚
12. One dimensional unsteady state heat transfer
equation for a sphere with heat generation at the
rate of 𝑞’ can be written as
(A)
1
𝑟
𝜕
𝜕𝑟
(𝑟
𝜃𝑇
𝜕𝑟
) +
𝑞
𝑘
=
1𝜕𝛤
𝛼𝜕𝑡
(B)
1𝜕
𝑟2 𝜕𝑟
(𝑟2 𝜃𝑇
𝜕𝑟
) +
𝑞
𝑘
=
1𝑓𝑓
𝛼𝜕𝑡
(C)
𝜕2 𝑇
𝜕𝑟2
+
𝑞
𝑘
=
1𝑓𝑓
𝛼𝜕𝑡
(D)
𝜕2
𝜕𝑟2 (𝑟𝑇) +
𝑞
𝑘
=
1𝑓𝑓
𝛼𝜕𝑡
Common Data For 𝑄.13 and 𝑄.14
An uninsulated air conditioning duct of rectangular cross
section 1 × 0.5𝑚 , carrying air at 20 𝑜
𝐶 with a velocity of
10 𝑚/𝑠, is exposed to an ambient of 30 𝑜 C. Neglect the
effect of duct construction material. For air in the range
of 20 − 30 𝑜
𝐶, data are as follows; thermal conductivity
= 0.025𝑊/𝑚𝐾 ; viscosity = 18𝜇𝑃 as, 𝑃 randtl number
= 0.73; density = 1.2𝑘𝑔/𝑚3
𝑇 he laminar flow Nusselt
number is 3.4 for constant wall temperature conditions
and for turbulent flow, Nu = 0.023Re08
𝑃𝑟033
13. The Reynolds number for the flow is
(A) 444 (B) 890
(C) 4. 44 × 105 (𝐷) 5. 33 × 105
14. The heat transfer per meter length of the duct, in
watts is
(A) 3.8 (B) 5.3
(C) 89 (D) 769
15. Hot oil is cooled 𝑓𝑖: 𝑜𝑚80 to 50 𝑜
𝐶 in an oil cooler
which uses air as the coolant. The air temperature
rises ffom 30 to 40 𝑜
C. The designer uses a LMTD
value of 26 𝑜
C. The type of heat exchange is
(A) parallel flow (B) double pipe
(C) counter flow (D) cross flow
16. A solid cylinder (surface 2) is located at the centre of
a hollow sphere (surface 1). The diameter of the
sphere is 1 𝑚, while the cylinder has a diameter and
length of 0.5 𝑚 each. The radiation configuration
factor 𝐹11 is
(A) 0.375 (B) 0.625

22. 22 | www.mindvis.in
(C) 0.75 (D) 1
17. A small copper ball of 5 mm diameter at 500 𝐾 is
dropped into an oil bath whose temperature is 300
K. The thermal conductivity of copper is 400 𝑊/𝑚𝐾,
its density 9000 𝑘𝑔/𝑚3
and its specific heat 385
𝐽/𝑘𝑔𝐾. If the heat transfer coefficient is 250
𝑊/𝑚2 𝐾 and lumped analysis is assumed to be valid,
the rate of fall of the temperature of the ball at the
beginning of cooling will be, in 𝐾/𝑠,
(A) 8.7 (B) 13.9
(C) 17.3 (D) 27.7
18. Heat flows through a composite slab, as shown
below. The depth ofthe slab is 1 𝑚. The 𝑘 values are
in 𝑊/𝑚𝐾. The overall thermal resistance in 𝐾/𝑊 is
(A) 17.2 (B) 21.9
(C) 28.6 (D) 39.2
19. The following figure was generated ffom
experimental data relating spectral black body
emissive power to wavelength at three temperature
𝑇1, 𝑇2 and 𝑇3(𝑇1 > 𝑇2 > 𝑇3) .
The conclusion is that the measurements are
(A) correct because the maxima in 𝐸 𝑏𝜆 show the correct
trend
(B) correct because Planck’s law is satisfied
(C) wrong because the Stefan Boltzmann law is not
satisfied
(D) wrong because Wien’s displacement law is not
satisfied
20. In a case of one dimensional heat conduction in a
medium with constant properties, 𝑇 is the
temperature at position 𝑥, at time 𝑡. Then
𝜃𝑇
𝜕𝑡
is
proportional to
(A)
𝑇
𝑥
(B)
𝜃𝑇
𝜕𝑥
(C)
𝜕2 𝑇
𝜕𝑥𝜕𝑡
(D)
𝜕2 𝑇
𝜕𝑥2
21. With an increase in the thickness of insulation
around a circular pipe, heat loss to surrounding due
to
(A) convection increase, while that the due to
conduction decreases
(B) convection decrease, while that due to
conduction increases
(C) convection and conduction decreases
(D)convection and conduction increases
22. A thin layer of water in a field is formed after a
farmer has watered it. The ambient air conditions
are: temperature 20 𝑜 𝐶 and relative humidity 5%. An
extract of steam tables is given below.
Temp
(∘ 𝐶)
−15 −10 −5 0.01 5 10 15 20
Saturati
on
Pressur
e (𝑘𝑃𝑎)
0.10 0.26 0.40 0.61 0.87 1.23 1.71 2.34
Neglecting the heat transfer between the water and the
ground, the water temperature in the field after
phase equilibrium is reached equals
(A) 10. 3 𝑜
𝐶
(B) −10. 3 𝑜
𝐶
(C) −14. 5 𝑜
𝐶
(D) 14. 5 𝑜
𝐶
23. 𝐴 100 𝑊 electric bulb was switched on in a 2.5 𝑚 ×
3𝑚 × 3𝑚 size thermally insulated room having a
temperature of 20 𝑜 C. The room temperature at the
end of 24 hours will be
(A) 321 𝑜
𝐶
(B) 341 𝑜
𝐶
(C) 450 𝑜
𝐶
(D) 470 𝑜
𝐶
24. In a composite slab, the temperature at the interface
(𝑇𝑖𝑛𝑡𝑒𝑟) between two material is equalto the
average ofthe temperature at the two ends.
Assuming steady one‐dimensional heat conduction,
which of the following statements is true about the
respective thermal conductivities?
(A) 2𝑘1 = 𝑘2 (B) 𝑘1 = 𝑘2
(C) 2𝑘1 = 3𝑘2 (D) 𝑘1 = 2𝑘2
0.5 m
1 m

23. 23 | www.mindvis.in
Common Data For 𝑄.25 and 𝑄.26
Consider steady one‐dimensional heat flow in a
plate of 20 mm thickness with a uniform heat
generation of 80 𝑀𝑊/𝑚3
The lefl and right faces are
kept at constant temperatures of 160 𝑜 𝐶 and
120 𝑜
𝐶 respectively. The plate has a constant
thermal conductivity of 200 𝑊/
𝑚𝐾.
25. The location of maximum temperature within the
plate 𝑓𝑟𝑜𝑚 its left face is
(A) 15 mm (B) 10 mm
(C) 5 mm (D) 0 mm
26. The maximum temperature within the plate in 0 𝐶 is
(A) 160 (B) 165
(C) 200 (D) 250
27. The average heat transfer co‐efficient on a thin hot
vertical plate suspended in still air can be
determined 𝑓𝑖: 𝑜𝑚 observations of the change in
plate temperature with time as it cools. Assume the
plate temperature to be uniform at any instant of
time and radiation heat exchange with the
surroundings negligible. The ambient temperature is
25 𝑜
𝐶, the plat has a total surface area of 0.1 𝑚2
and
a mass of 4 𝑘𝑔. The specific heat ofthe plate material
is 2.5 𝑘𝐽/𝑘𝑔𝐾. The convective heat transfer co‐
efficient in 𝑊/𝑚2
𝐾, at the instant when the plate
temperature is 225 𝑜 𝐶 and the change in plate
temperature with time 𝑑𝑇𝑙𝑑𝑡 = −0.02𝐾/𝑠, is
(A) 200 (B) 20
(C) 15 (D) 10
28. In a counter flow heat exchanger, hot fluid enters at
60 𝑜
𝐶 and cold fluid leaves at 30 𝑜
C. Mass flow rate
of the fluid is 1 𝑘𝑔/𝑠 and that of the cold fluid is 2
𝑘𝑔/𝑠. Specific heat of the hot fluid is 10 𝑘𝐽/𝑘𝑔𝐾 and
that of the cold fluid is 5 𝑘𝐽/𝑘𝑔𝐾. The Log Mean
Temperature Difference (LMTD) for the heat
exchanger in 0 𝐶 is
(A) 15 (B) 30
(C) 35 (D) 45
29. The temperature distribution within the thermal
boundary layer over a heated isothermal flat plate is
given by
𝑇 − 𝑇 𝑤
𝑇∞ − 𝑇 𝑤
=
3
2
(
𝑦
𝛿𝑡
) −
1
2
(
𝑦
𝛿𝑡
)3
,
where 𝑇 𝑤 and 𝑇∞ are the temperature ofplate
and 𝑓𝑖: 𝑒𝑒 stream respectively, and 𝑦 is the normal
distance measured ffom the plate. The local Nusselt
number based on the thermal boundary layer
thickness 𝛿𝑡 is given by
(A) 1.33 (B) 1.50
(C) 2.0 (D) 4.64
30. Steady two‐dimensional heat conduction takes
place in the body shown in the figure below. The
normal temperature gradients over surfaces 𝑃 and
𝑄 can be considered to be uniform. The temperature
gradient 𝜕𝑇𝑙𝜕𝑥 at surface 𝑄 is equal to 10 𝐾/𝑚.
Surfaces 𝑃 and 𝑄 are maintained at constant
temperature as shown in the figure, while the
remaining part ofthe boundary is insulated. The
body has a constant thermal conductivity of 0.1
𝑊/𝑚𝐾. The values of
𝜕𝑇
𝜕𝑥
and
𝜃𝑇
𝜕𝑦
at surface 𝑃 are
𝑦
(A)
𝜕𝑇
𝜕𝑥
= 20𝐾/𝑚,
𝜕𝑇
𝜕𝑦
= 0𝐾/𝑚
(B)
𝜃𝑇
𝜕𝑥
= 0𝐾/𝑚,
𝜃𝑇
𝜕𝑦
= 10𝐾/𝑚
(C)
𝜕𝑇
𝜕𝑥
= 10𝐾/𝑚,
𝜕𝑇
𝜕𝑦
= 10𝐾/𝑚
(D)
𝜕𝑇
𝜕𝑥
= 0𝐾/𝑚, 𝑓 = 20𝐾/𝑚
31. A hollow enclosure is formed between two infinitely
long concentric cylinders of radii 1 𝑚 and 2 𝑚,
respectively. Radiative heat exchange takes place
between the inner surface of the larger cylinder
(surface‐2) and the outer surface ofthe smaller
cylinder (surface‐l). The radiating surfaces are
difffise and the medium in the enclosure is non‐
participating. The fiaction of the thermal radiation
leaving the larger surface and striking itselfis
(A) 0.25 (B) 0.5
(C) 0.75 (D) 1
32. For the three‐dimensional object shown in the figure
below, five faces are insulated. The sixth face
(𝑃𝑄𝑅𝑆), which is not insulated, interacts thermally
with the ambient, with a convective heat transfer
coefficient of 10 𝑊/𝑚2
𝐾 The ambient temperature
is 30 𝑜
C. Heat is uniformly generated inside the
object at the rate of 100 𝑊/𝑚3
Assuming the face
𝑃𝑄𝑅𝑆 to be at uniform temperature, its steady state
temperature is
Surface ‐ 2

24. 24 | www.mindvis.in
𝑆
||
|1𝑚|
𝐻
(A) 10 𝑜
𝐶 (B) 20 𝑜
𝐶
(C) 30 𝑜
𝐶 (D) 40 𝑜
𝐶
33. The logarithmic mean temperature difference
(LMTD) of a counter flow heat exchanger is 20 𝑜 C.
The cold fluid enters at 20 𝑜 𝐶 and the hot fluid enters
at 100 𝑜
C. Mass flow rate of the cold fluid is twice
that of the hot fluid. Specific heat at constant
pressure of the hot fluid is twice that of the cold
fluid. The exit temperature of the cold fluid
(A) is 40 𝑜
𝐶
(B) is 60 𝑜
𝐶
(C) is 80 𝑜 𝐶
(D) cannot be determined
34. For flow of fluid over a heated plate, the following
fluid properties are known
Viscosity = 0.001 𝑃𝑎 − 𝑠;
Specific heat at constant pressure = 1𝑘𝐽/ kg.𝐾;
Thermal conductivity = l𝑊/m‐𝐾
The hydrodynamic boundary layer thickness at a
specified location on the plate is 1 mm. The thermal
boundary layer thickness at the same location is
(A) 0.001 mm (B) 0.01 mm
(C) 1 mm (D) 1000 mm
Common Data For 𝑄.35 and 𝑄.36
Radiative heat transfer is intended between the
inner surfaces of two very large isothermal
parallel metal plates. While the upper plate
(designated as plate 1) is a black surface and is
the warmer one being maintained at 727 𝑜
𝐶 ,
the lower plate (plate 2) is a diffise and gray
surface with an emissivity of 0.7 and is kept at
227 𝑜
𝐶.
Assume that the surfaces are sufficiently large
to form a two‐surface enclosure and steady‐
state conditions to exits. Stefan‐Boltzmann
constant is given as 5.67 × 10−8
𝑊/𝑚2
𝐾4
35. The irradiation (𝑖𝑛 𝑘𝑊/𝑚2
) for the plate (plate 1) is
(A) 2.5 (B) 3.6
(C) 17.0 (D) 19.5
36. If plate 1 is also diffuse and grey surface with an
emissivity value of 0.8, the net radiation heat
exchange (𝑖𝑛 𝑘𝑊/𝑚2
) between plate 1 and plate 2
is
(A) 17.0 (B) 19.5
(C) 23.0 (D) 31.7
37. Consider steady‐state conduction across the
thickness in a plane composite wall (as shown in the
figure) exposed to convection conditions on both
sides.
ℎ𝑖, 𝑇∞𝑖
Given: ℎ𝑖 = 20𝑊/𝑚2
𝐾, ℎ 𝑜 = 50𝑊/𝑚2
𝐾;, 𝑇∞𝑗 =
20 𝑜
𝐶; 𝑇∞,0 = −2 𝑜
𝐶, 𝑘1 = 20𝑊/𝑚𝐾; 𝑘2 = 50𝑊/
𝑚𝐾; 𝐿1 = 0.30𝑚 and 𝐿2 = 0.15𝑚.
Assuming negligible contact resistance between the wall
surfaces, the interface temperature, 𝑇 (in 0 𝐶), of the two
walls will be
(A) −0.50 (B) 2.75
(C) 3.75 (D) 4.50
38. In a parallel flow heat exchanger operating under
steady state, the heat capacity rates (product of
specific heat at constant pressure and mass flow
rate) of the hot and cold fluid are equal. The hot
fluid, flowing at 1 𝑘𝑔/𝑠 with 𝑐 𝑝 = 4𝑘𝐽/ kg 𝐾, enters
the heat exchanger at 102 𝑜
𝐶 while the cold fluid has
an inlet temperature of 15 𝑜
C. The overall heat
transfer coefficient for the heat exchanger is
estimated to be 1 𝑘𝑊/𝑚2
𝐾 and the corresponding
heat transfer surface area is 5 𝑚2
Neglect heat
transfer between the heat exchanger and the
ambient. The heat exchanger is characterized by the
following relations:
2𝜀 = − exp ( −2 NTU)
The exit temperature (in 0 𝐶) for the cold fluid is
(A) 45 (B) 55
(C) 65 (D) 75
39. A coolant fluid at 30 𝑜
𝐶 flows over a heated flat plate
maintained at constant temperature of 100 𝑜
C. The
boundary layer temperature distribution at a given
location on the plate may be approximated as 𝑇 =
30 + 70 exp (−𝑦) where 𝑦 (in m) is the distance
normal to the plate and 𝑇 is in ∘ C. Ifthermal
conductivity of the fluid is 1. 0𝑊/𝑚𝐾, the local
convective heat transfer coefficient (𝑖𝑛 𝑊/𝑚2
𝐾) at
that location will be
(A) 0.2 (B) 1
(C) 5 (D) 10
40. A fin has 5 mm diameter and 100 mm length. The
thermal conductivity of fin material is 400 Wm
−1𝐾 − 1 One end ofthe fm is maintained at 130 𝑜 𝐶
and its remaining surface is exposed to ambient air
at 30 𝑜
C. If the convective heat transfer coefficient
is 40 Wm −2𝐾 − 1, the heat loss (in 𝑊) from the fm
is
(A) 0.08 (B) 5.0
(C) 7.0 (D) 7.8
1 2
L2L1
ℎ𝑖, 𝑇∞,𝑖

25. 25 | www.mindvis.in
41. The ratios of the laminar hydrodynamic boundary
layer thickness to thermal boundary layer thickness
of flows of two fluids 𝑃 and 𝑄 on a flat plate are 1/2
and 2 respectively. The Reynolds number based on
the plate length for both the flows is 104
The Prandtl
and Nusselt numbers for 𝑃 are 1/8 and 35
respectively. The Prandtl and Nusselt numbers for 𝑄
are respectively
(A) 8 and 140 (B) 8 and 70
(C) 4 and 40 (D) 4 and 35
42. A spherical steel ball of 12 mm diameter is initially at
1000 K. It is slowly cooled in surrounding of 300 K.
The heat transfer coefficient between the steel ball
and the surrounding is 5 𝑊/𝑚2
K. The thermal
conductivity of steel is 20 𝑊/𝑚𝐾. The temperature
difference between the centre and the surface of
the steel ball is
(A) large because conduction resistance is far higher
than the convective resistance.
(B) large because conduction resistance is far less
than the convective resistance.
(C) small because conduction resistance is far higher
than the convective resistance.
(D) small because conduction resistance is far less
than the convective resistance.
43. A pipe of 25 mm outer diameter carries steam. The
heat transfer coefficient between the cylinder and
surroundings is 5 𝑊/𝑚2
K. It is proposed to reduce
the heat loss ffom the pipe by adding insulation
having a thermal conductivity of 0.05 𝑊/𝑚 K. Which
one of the following statements is TRUE?
(A) The outer radius of the pipe is equal to the critical
radius.
(B) The outer radius of the pipe is less than the
critical radius.
(C) Adding the insulation will reduce the heat loss.
(D) Adding the insulation will increases the heat loss.
44. In a condenser of a power plant, the steam
condenses at a temperatures of 60 𝑜
C. The cooling
water enters at 30 𝑜
𝐶 and leaves at 45 𝑜
C. The
logarithmic mean temperature difference (LMTD) of
the condenser is
(A) 16. 2 𝑜
𝐶 (B) 21. 6 𝑜
𝐶
(C) 30 𝑜
𝐶 (D) 37. 5 𝑜
𝐶
45. Water (𝑐 𝑝 = 4.18𝑘𝐽/𝑘𝑔𝐾) at 80 𝑜
𝐶 enters a
counter flow heat exchanger with a mass flow rate
of 0.5 𝑘𝑔/𝑠. Air (𝑐 𝑝 = 1𝑘𝐽/𝑘𝑔𝐾) enters at 30 𝑜
𝐶
with a mass flow rate of 2.09 𝑘𝑔/𝑠. If the
effectiveness of the heat exchanger is 0.8, the LMTD
(in𝐶) is
(A) 40 (C) 10 (B) 20 (D) 5
46. Consider two infinitely long thin concentric tubes of
circular cross section as shown in the figure. If 𝐷1
and 𝐷2 are the diameters of the inner and outer
tubes respectively, then the view factor 𝐹22 is give
by
(A) (
𝐷2
𝐷1
) − 1 (B) zero
(C) (
𝐷1
𝐷2
) (𝐷) 1 − (
𝐷1
𝐷2
)
47. Which one of the following configurations has the
highest fm effectiveness?
(A) Thin, closely spaced fins
(B) Thin, widely spaced fins
(C) Thick, widely spaced fins
(D) Thick, closely spaced fins
48. For an opaque surface, the absorptivity (𝛼) ,
transmissivity (𝜏) and reflectivity (𝜌) are related by
the equation:
(A) 𝛼 + 𝜌 = 𝜏
(B) 𝜌 + 𝛼 + 𝜏 = 0
(C) 𝛼 + 𝜌 = 1
(D) 𝛼 + 𝜌 = 0
Answers:
1 C 31 B
2 D 32 D
3 C 33 C
4 C 34 C
5 D 35 D
6 A 36 D
7 B 37 C
8 B 38 B
9 A 39 B
10 B 40 B
11 C 41 A
12 B 42 D
13 C 43 C
14 D 44 B
15 D 45 C
16 C 46 D
17 C 47 A
18 C 48 C
19 D
20 D
21 B
22 C
23 D
24 D
25 C
26 B
27 D
28 B
29 B
30 D