lecture notes heat transfer.pdf

Heat Transfer (MEng 3121) Lecture Notes
Tsegaye Getachew Alenka
Department of Mechanical Engineering
Wolaita Sodo University
tsegaye.getachew@wsu.edu.et
February 26, 2023
Contents
1 Course objectives and Learning Outcome 1
2 Introduction 1
2.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
2.2 Application of Heat/mass transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
2.3 Modes of Heat Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
2.4 Review Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
3 1D Steady Heat Conduction 4
3.0.1 Heat Transfer from Finned Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
4 2D Steady Heat Conduction 9
4.1 Analytic Method: The Method of Separation of Variables . . . . . . . . . . . . . . . . . . . . . . . . . 9
4.2 Graphical Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
4.3 Numerical Method: Finite-Difference Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
4.3.1 Nodal Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
4.3.2 Finite Difference Method FDM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
5 Unsteady-State Heat Transfer: Conductive and Convective BL 13
5.1 Lumped Capacitance Method: No Internal Resistance, . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
5.2 Significant Internal Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
5.2.1 Heat Transfer in an Infinite Geometries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
5.2.2 Heisler chart: Infinite (slab, cylinder and spherical) geometry . . . . . . . . . . . . . . . . . . . 15
5.3 Large Body: Finite and Semi-infinite Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
5.3.1 transient Heat Transfer in a Semi-Infinite Geometry . . . . . . . . . . . . . . . . . . . . . . . . 16
6 Convection Heat Transfer 18
6.1 Dimensionless numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
6.2 Forced Convection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
6.2.1 Types of Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
6.3 Natural Convection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
7 Radiation Heat Transfer 24
7.1 Cause of Radiation, Radiation as a Form of Electromagnetic Wave Energy . . . . . . . . . . . . . . . . 24
7.2 Absorptivity, Reflectivity, Emissivity, and Transmissivity . . . . . . . . . . . . . . . . . . . . . . . . . 25
7.3 Radiation Heat Transfer Between Two Infinite Plates and Radiation Shields . . . . . . . . . . . . . . . 26
7.4 View Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
7.4.1 View factor Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
I

8 Heat Exchangers 28
8.1 Fouling Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
8.2 Types of Heat Exchangers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
8.2.1 Double pipe – Parallel flow heat exchanger . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
8.2.2 Shell and Tube Heat Exchangers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
8.2.3 Plate Heat Exchangers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
8.3 Heat Exchanger Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
9 Condensation and Boiling Heat Transfer 31
9.1 Boiling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
9.1.1 Pool Boiling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
9.1.2 Pool Boiling Correlation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
9.1.3 External and Internal Forced Convection Boiling (Flow Boiling) . . . . . . . . . . . . . . . . . 32
9.2 Condensation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
10 Objective Questions 35
11 References 37
List of Figures
1 Some applications of Heat and Mass Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
2 Modes of Heat Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
3 Resistance Network b. Series b. Parallel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
4 a. Maximum temperature with heat generation b. Max.Temp. Thickness Symmetry with Imaginary
Wall . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
5 Boundary Condition Problem of Two dimensional Conduction and Convection . . . . . . . . . . . . . 10
6 Shape factor S for various selected geometries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
7 isothermal and heat flow lines in 2d conduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
8 Nodal Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
9 Finite Difference Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
10 FDM Steady State Cond. Worked Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
11 Lumped Capacitance Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
12 Hiesler Chart:Unsteady state diffusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
13 Semi-Infinite Graphical Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
14 Convection configuration in various flow scheme: a. various flow layer scheme b. boundary layer . . . 19
15 External flows boundary regions a. Flow over Plate b. Flow Over Cylinder . . . . . . . . . . . . . . . 20
16 various flow boundary regions in internal flow convection . . . . . . . . . . . . . . . . . . . . . . . . . . 21
17 Isothermal wall heating, exponential Temp. Decay a. Mean surface Temp. b. Log Mean Temp. . . . . 21
18 natural Convection Over surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
19 Electromagnetic Spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
20 Thermodynamics of Radiation a. Radiation Causes b. absorption, reflection, and transmission of
irradiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
21 Combined Radiation Convection Furnace Model a. Radiation Model b. Furnace C. Resistance model 26
22 View Factor for Various Geometries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
23 a. Comb. Radiation-Conv Heat Transfer between Parallel Plates b. Equivalent Resistance Network
Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
24 a. Double pipe Heat Exchanger b. Parallel & Counter Flow . . . . . . . . . . . . . . . . . . . . . . . 29
25 Shell & Tube - Counter flow heat exchanger a. single & double Shell b. baffles & flow configuration . 30
26 compact Plate Heat Exchanger . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
27 Pool Boiling with different forms depending on the temperature ∆Excess = Ts − Tsat . . . . . . . . . 32
28 Film Condensation Vert. Wall Model: a. Hydrodynamics b. Flow Regime . . . . . . . . . . . . . . . . 34
29 Film condensation in radial coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
List of Tables
1 Conduction, Convection & Radiation HT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2 α of various materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
3 Thermal Conductivity of Various Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
4 Heat Generation in Various geometries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
II

5 convective heat transfer coefficient h in W/m2
.K for various heat transfer types . . . . . . . . . . . . 18
6 Convection Controlling Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
7 Empirical formulation of heat transfer coefficients various flow regimes and geometries of Flow Over
Plates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
8 Convection Coeff. for Flow Over Cylinders and Spheres . . . . . . . . . . . . . . . . . . . . . . . . . . 21
9 Laminar Flow in Tube Isothermal UWT and Isoflux UWF . . . . . . . . . . . . . . . . . . . . . . . . 22
10 Natural Convection Heat Transfer Correlations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
11 Boiling Correlation Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
III

1 Course objectives and Learning Outcome
This module is meant to provide not as stand alone or sufficient reference for the national exit exam but as the
fundamental guide so that to enable students to have a review resource of the heat transfer under graduate course.
Ethiopian mechanical Engineering graduating students or those who are currently studying the course can have this
condensed comprehensive reference and prepare for the their upcoming national exam using the module by further
attempting exercises, examples and objective questions provided as a quick reference and comparing their response
to the question prompts by referring/consulting lecture notes provided here as well as check answer options for the
objective questions in the last portion of this module against the check answer options in the reference websites. The
course is designed to enable students to
ˆ Develop the fundamental heat transfer equations in Cartesian, cylindrical and spherical coordinates
ˆ Develop the students’ abilities to model and analyze thermal systems, and
ˆ Develop experience in the application of thermal analysis to elementary problems in engineering practice
The objectives of this course therefore are
1. To provide students with a clear and through presentation of the basic concepts of heat and mass transfer and
their applications.
2. To develop understanding of the coupling of fluid mechanics and thermodynamics
3. To provide an understanding of fundamental concepts of heat fluxes.
4. Apply principle of conservation of energy
5. Apply numerical techniques for spatial discretization: finite difference method.
2 Introduction
2.1 Definition
Heat transfer is the ability of the barrel heaters to transfer heat into the material being processed to create a uniform
temperature profile throughout the melt. Heat transfer methods are used in numerous disciplines, such as automotive
engineering, thermal management of electronic devices and systems, climate control, insulation, materials processing,
chemical engineering and power station engineering.
There is a distinct difference between the two. Temperature with symbol T is a measure of the amount of energy
possessed by the molecules of a substance. It is a relative measure of how hot or cold a substance is and can be used
to predict the direction of heat transfer. The scales for measuring temperature in SI units are the Celsius and Kelvin
temperature scales.
On the other hand, heat with symbol Q is energy in transit. The transfer of energy as heat occurs at the molecular
level as a result of a temperature difference. Heat is measured by the Joule and calorie in the SI system.
Furthermore, heat transfer is a complementary science to thermodynamics. Thermodynamics tells us: the amount
heat is transferred δQ, work is done δW and final state of the system alongside with the system overall performance
with the help of four thermodynamic laws.
Heat transfer on the other-hand studies mechanisms or modes δQ transferred, the rate of δQ transferred, and tem-
perature distribution inside the body.
One can plainly say heat transfer as study of energy/mass transfer as particular interest without considering an overall
system energy balance whereas thermodynamics as an overall study of energy equilibrium in a system.
Thermodynamics gives no indication of how long the process takes. In heat transfer, we are more concerned about
the rate of heat transfer.
2.2 Application of Heat/mass transfer
Thermal energy: Thermal energy is associated with the translation, rotation, vibration and electronic states of the
atoms and molecules that comprise matter. Thermal energy represents the cumulative effect of microscopic activities
and is directly linked to the temperature of matter. The application of heat/mass transfer can have a wide scope from
daily life application such as Heating, Cooling, and Cooking up to mechanical applications in systems such as:
heat exchangers, refrigeration and air conditioning, boilers, condensers etc. where processes considered include Heating
/ cooling, Vaporizing, and Energy conversion and advanced cooling applications such as electronic appliances such
as cellphones, computers and so on. The application areas of the science of heat and mass transfer shown below in
Fig. 1.
1

(a)
(b) (c)
Figure 1: Some applications of Heat and Mass Transfer
[1]
2.3 Modes of Heat Transfer
Heat transfer can be sought in three different mechanisms: conduction mode, convection mode and radiation mode.
Conduction: An energy transfer across a system boundary due to a temperature difference by the mechanism of
inter-molecular interactions. Conduction needs matter and does not require any bulk motion of matter. On the other
hand, Convection is an energy transfer across a system boundary due to a temperature difference by the combined
mechanisms of intermolecular interactions and bulk transport. Convection needs fluid matter. Radiation heat transfer
involves the transfer of heat by electromagnetic radiation that arises due to the temperature of the body. Radiation
does not need matter. The comparison of three modes of heat transfer presented in Fig. 1. It is important to
Figure 2: Modes of Heat Transfer
notice the introduction of the minus sign. In words, this simply means that the direction of heat flow is opposite
to the direction in which the temperature increases. Heat is transferred via solid material (conduction), liquids and
2

Table 1: Conduction, Convection & Radiation HT
[2]
Comparison of HT Modes
conduction Convection Radiation
Described by Fourier Law Described by Newton’s Law of Cooling Governed by Stefan Boltzman law
⃗
q = −kA∇T q = hAs∆T q = ϵσA(TS
4 − TSur
4)
⃗
q is heat flow vector q is heat flow scalar from the source ϵ is surface emissivity,
in W, k is thermal conductivity surface in W, h is HT coeff. A exposed surf. area in m2 & Ts
of material in W/mK, A is cross of material which also is influenced absolute temperature of
section in flow direction in m2 and by surface geometry, flow scheme of Surrounding in K
& material thermal prop. in W/m2K, As surface, Tsur is
∇T is temp. gradient is surface area in m2 and ∆T is temp. absolute temp.
∇T = ∂T
∂x
i + ∂T
∂y
j + ∂T
∂z
k difference in K, ∆T = Ts − T∞ ϵ = 1 for black body,
each direction heat flow should be Can be free convection induced σ = 5.67 × 10−8W/m2K4
computed separately & by buoyancy forces h = 2 − 25 for gases,
combined as h = 20 − 100 for liquid and forced
resultant convection that induced by external
forces like fan h = 20 − 250 gases
and h = 50 − 20, 000 for liquid
but h = 2, 500 − 100, 000 for
boiling and condensation
gases (convection), and electromagnetic waves (radiation). Heat is usually transferred in a combination of these three
types and randomly occurs on its own. As a result, it is important to understand those three phenomena taken
separately. Mass transfer by convection involves the transport of material between a boundary surface (such as solid
or liquid surface) and a moving fluid or between two relatively immiscible, moving fluids.
Another important thermal property parameter used in selection of heat storage in engineering is thermal diffusivity
designated by symbol α = k
ρCP
where k is thermal conductivity, ρ is density and CP is heat capacity. Fig. 2 shows
the diffusivity property (heat storage capacity) of various materials. Thermal diffusivity is heat propagation rate,
Table 2: α of various materials
[1]
Material silver gold cupper Aluminum Iron Mercury marble Ice
α in m2
/s 149 × 10−6
127 × 10−6
113 × 10−6
97.5 × 10−6
22.8 × 10−6
4.7 × 10−6
1.2 × 10−6
1.2 × 10−6
Matrials concrete brick dry heavy soil glass wool water beef wood
α in m2
/s 0.75 × 10−6
0.52 × 10−6
0.34 × 10−6
0.23 × 10−6
0.14 × 10−6
0.14 × 10−6
0.13 × 10−6
the ratio of thermal conductivity by the storage capacity which is the product of density and specific heat. Thermal
resistance is ratio of thickness by conductivity
q =
∆T
R
whereR =
∆x
kA
for Cond. & R =
1
hconv.A
for conv. or R =
1
hRad.A
for Rad. (1)
The thermal resistance network suits an analogy of electrical circuit. Heat transfer problems are often two or three
dimensional, but approximate solutions can be obtained by assuming one dimensional heat transfer with much more
ease using thermal resistance network.
Rtotal =
X
Ri for circuit in series & Rtotal =
X 1
Ri
for circuit in parallel (2)
2.4 Review Questions
1. What material among soil and brick would be preferred for thermal storage and why? consider table 2
2. Compare and contrast conduction, convection and radiation
3. Find the heat loss per square meter of surface through a brick wall (0.5m) thick, when the inner surface is at
(400k) and outside at (300K), the thermal conductivity of brick may be taken as (0.7w/m.k) ?
3

(a) (b)
Figure 3: Resistance Network b. Series b. Parallel
3 1D Steady Heat Conduction
Conduction is due to collision between molecules or atoms (fluid); lattice waves induced by atomic and translational
motion of the free electrons (solid). Thermal conductivity is a thermodynamic property of a material. From the State
Postulate given in thermodynamics, it may be recalled that thermodynamic properties of pure substances are functions
of two independent thermodynamic intensive properties, say temperature and pressure. Thermal conductivity of real
gases is largely independent of pressure and may be considered a function of temperature alone. For solids and liquids,
properties are largely independent of pressure and depend on temperature alone. It is important that the student gain
a basic perspective of the magnitude of thermal conductivity for various materials. Note that heat transfer is the only
energy interaction; the energy balance for the wall can be expressed:
˙
Qin − ˙
Qout =
dEsyst
dt
(3)
This is important because thermal design involves section of proper materials either for isolation/insulation purpose
Table 3: Thermal Conductivity of Various Materials
[3] Materials Copper Silver Gold Aluminum Steel Limestone Bakelite Water Air
k in W/m2
K 401 429 317 237 60.5 2.15 1.4 0.613 0.0263
in many applications such as refrigerator box design and conduction applications such as heat exchangers. Conversion
of some form of energy into heat energy in a medium is called heat generation. Heat generation leads to a temperature
rise throughout the medium. Some examples of heat generation are resistance heating in wires, exothermic chemical
reactions in solids, and nuclear reaction. Heat generation is usually expressed per unit volume (W/m3
). In most
applications, we are interested in maximum temperature Tmax and surface temperature Ts of solids which are involved
with heat generation. Heat generation is nothing but conversion of one form of energy (like electrical, chemical,
or nuclear energy) into thermal energy (heat energy) inside a body. Whereas one-dimensional steady state heat
conduction, without heat generation, in a plane wall; with boundary conditions as shown in the figure below. The
conductivity of the wall is given by k = k0 +bT; where k0 and b are positive constants, and T is temperature. maximum
temperature Tmax in a solid that involves uniform heat generation will occur at a location furthest away from the outer
surface when the outer surface is maintained at a constant temperature, Ts. Consider a solid medium of surface area
A, volume V , and constant thermal conductivity k, where heat is generated at a constant rate of ġ per unit volume.
Heat is transferred from the solid to the surroundings medium at T∞. Under steady conditions, the energy balance
for the solid can be expressed as:
Q̇ = ġV = hA(Ts − T∞) in W combining these Ts = T∞ +
ġV
A
(4)
Heat transfer problem involves either thermal resistance network, boundary condition, heat generation, or combination
of all.
3.0.1 Heat Transfer from Finned Surfaces
From the Newton’s law of cooling, dotQconv = hA(Ts − T∞), the rate of convective heat transfer from a surface at
a temperature Ts can be increased by two methods: 1. Increasing the convective heat transfer coefficient, h and 2.
4

(a) (b)
Figure 4: a. Maximum temperature with heat generation b. Max.Temp. Thickness Symmetry with Imaginary Wall
[4]
Table 4: Heat Generation in Various geometries
Geometry a plane wall of thickness 2L a long cylinder of radius r0 a sphere of radius r0
Surface Temp. Fourier Relation Ts = T∞ + ġL
h Ts = T∞ + ġr0
2h Ts = T∞ + ġr0
3h
Maximum temperature Diff ∆Tmax = ġL2
2k ∆Tmax =
ġr2
0
4k ∆Tmax =
ġr2
0
6k
Increasing the surface area A. Increasing the convective heat transfer coefficient may not be practical and/or adequate.
An increase in surface area by attaching extended surfaces called fins to the surface is more convenient. Finned surfaces
are commonly used in practice to enhance heat transfer. In the analysis of the fins, we consider steady operation with
no heat generation in the fin. We also assume that the convection heat transfer coefficient h to be constant and
uniform over the entire surface of the fin. In the limiting case of zero thermal resistance (k → ∞), the temperature
of the fin will be uniform at the base value of Tb. The heat transfer from the fin will be maximized in this case given
by Eq. 5. [4]
dotQconv = hAfins(Ts − T∞) (5)
Fin efficiency can be defined as given in Eq. 6
ηfins =
actual HT from fins
HT from fins if all fins were atTb
=
Q̇fins
Q̇fins,max
(6)
Therefore, Q̇fins = ηfinshAfins(Ts − T∞). Fin design considerations incorporates 1. the longer the fin, the larger the
heat transfer area and thus the higher the rate of heat transfer from the fin, 2. the larger the fin, the bigger the mass,
the higher the price, and larger the fluid friction and 3. the fin efficiency decreases with increasing fin length because
of the decrease in fin temperature with length. The performance of fins is judged on the basis of the enhancement in
heat transfer relative to the no-fin case, and expressed in terms of the fin effectiveness in Eq. 7
ϵfin =
Q̇fin
Q̇nofin
(7)
εfin =


< 1 fin acts as insulation
= 1 fin does not affect heat transfer
> 1 fin enhances heat transfer
For a sufficiently long fin of uniform cross-section Ac, the temperature at the tip of the fin will approach the environment
temperature, T∞. By writing energy balance and solving the differential equation, one finds Eq. 8 first two rows:
where Ac is the cross-sectional area, x is the distance from the base, and p is perimeter. The effectiveness becomes
the last row of Eq. 8:
T(x) − T∞
Tb − T∞
= exp −x
r
hp
kAc
!
Qlong fin =
p
hpkAc (Tb − T∞)
εlong fin =
r
kp
hAc
(8)
To increase fin effectiveness, one can conclude: [5]
ˆ the thermal conductivity of the fin material must be as high as possible
5

ˆ the ratio of perimeter to the cross-sectional area p/Ac should be as high as possible
ˆ the use of fin is most effective in applications that involve low convection heat transfer coefficient, i.e. natural
convection.
The critical radius of insulation is a counterintuitive concept within the study of heat transfer. The theory states
that adding insulation to a cylindrical or spherical object will increase the rate of heat loss rather than decrease it, if
the radius (thickness) of the insulation is at its “critical” value. It signifies an insulation thickness above which the
heat transfer is reduced.
The heat transfer problems of composite slab involves material properties: thermal resistance and capacitance, taking
in to account the effect of thermal resistance and the relation between thermal conductivity, thickness and area. Ther-
mal resistance (R) and thermal conductance (C) of the materials are reciprocals of one another and can be derived
from thermal conductivity (k) and the thickness of the materials. Example 1 1.2 kg of liquid water initially at 15o
C
is to be heated to 95o
C in a teapot equipped with a 1200−W electric heating element inside Fig . The teapot is 0.5 kg
and has an average specific heat of 0.7kJ/kgo
C. Taking the specific heat of water to be 4.18kJ/kgo
C and disregarding
any heat loss from the teapot, determine how long it will take for the water to be heated assuming negligible loss from
tea pot and constant properties.
Solution Using Eq. 3
Ein − Eout = Esystem
Ein = ∆U = ∆Uwater + ∆Uteapot
where ∆Uteapot ≈ 0
Ein = (mCP ∆T)water + (mCP ∆T)teapot
= 1.2kg × 4.18kJ/kg.o
C × (95o
C − 15o
C) + 0.5kg
×0.7kJ/kg.o
C × (95o
C − 15o
C) = 429.3kJ
∆t =
Esystem
˙
Esystem
=
429.3kJ
1200W
= 358sec ≡ 6minutes
Example 2 Consider the combined series-parallel composite arrangement shown in figure below. Assuming one–dimensional
heat transfer, determine the rate of heat transfer.
6

Solution The equivalent circuit diagram is shown below
Q̇ =
T1 − T∞
Rtotal
Rtotal = [(
1
R1
+
1
R2
)parallel + R3 + Rconv]series =
R1R2
R1 + R2
+ R3 + Rconv
Example 2: Thermal Network Problem Consider the combined series−parallel composite circular arrangement
shown in figure below. Assuming one–dimensional heat transfer, determine the rate of heat transfer. The cylinder has
radius r and length L.
Solution Since, there is no heat generation in the layer and thermal conductivity is constant, the Fourier law becomes:
Q̇cylCond = −kA
dT
dr
= −k(2πrL)
dT
dr
Z r2
r1
Q̇cylCond
2πrL
dr =
Z T2
T1
k dT =
T1 − T2
Rtotal
Q̇cylCond =
T1 − T2
Rtotal
= 2πkL
T1 − T2
lnr1/r2
Rcyl =
lnr1
r2
2πkL
Example 3 Thermal Composite System/Network Problem Steam at Tinfity, 1 = 320o
C flows in a cast iron
pipe [k = 80W/m.o
C] whose inner and outer diameter are D1 = 5cm and D2 = 5.5cm, respectively. The pipe is
covered with a 3cm thick glass wool insulation [k = 0.05W/m.o
C]. Heat is lost to the surroundings at Tinfity, 2 = 5o
C
by natural convection and radiation, with a combined heat transfer coefficient of h2 = 18W/m2
.o
C. Taking the heat
transfer coefficient inside the pipe to be h1 = 60W/m2
K, determine the rate of heat loss from the steam per unit length
of the pipe. Also determine the temperature drop across the pipe shell and the insulation. Assume: Steady-state and
one-dimensional heat transfer.
solution Taking L = 1m, the areas of the surfaces exposed to convection are:
A1 = 2πr1L = 0.157m2
A2 = 2πr2L = 0.361m2
Rcanv ,1 =
1
h1A1
=
1
(60 W/m2.o.C) (0.157m2)
= 0.106o
.C/W
R1 = Rpipe =
ln (r2/r1)
2πk1L
= 0.0002o
.C/W
R2 = Rinsulation =
ln (r3/r2)
2πk2L
= 2.35o
.C/W
Rcanv ,2 =
1
h2A2
= 0.154o
.C/W
Rtotal = Rconv ,1 + R1 + R2 + Rcanv ,2 = 2.61o
.C/W
7

The steady-state rate of heat loss from the steam becomes
Q̇ =
T∞,1 − T∞,2
Rtotal
= 120.7W (per m pipe length)
The total heat loss for a given length can be determined by multiplying the above quantity by the pipe length. The
temperature drop across the pipe and the insulation are:
∆Tpipe = Qo
.Rpipe = (120.7W) (0.00022o
.C/W) = 0.02o
.C
∆Tinsulation = Qo
.Rinsulation = (120.7W) (2.35o
.C/W) = 284o
.C
Note that the temperature difference (thermal resistance) across the pipe is too small relative to other resistances and
can be ignored.
Example 4 Boundary Condition Problem A steel pipe (K = 45.0W/m.K) having a 0.05m Outer Diameter
is covered with a 0.042 m thick layer of magnesia (K = 0.07W/m.K) which in turn covered with a 0.024 m layer
of fiberglass insulation (K = 0.048W/m.K). The pipe wall outside temperature is 370 K and the outer surface
temperature of the fiberglass is 305K. What is the interfacial temperature between the magnesia and fiberglass? Also
calculate the steady state heat transfer
Given
OD = 0.05m d1 = 0.05m r1 = 0.025m k1 = 45W/mK r2 = r1 + insulation thickness 1
r2 = 0.025 + 0.042 r2 = 0.067m k2 = 0.07W/mK k3 = 0.048W/mK r3 = r2 + tinsulation2 = 0.067 + 0.024
r3 = 0.091m T1 = 370K T3 = 305K Find (i)T2? (ii)Q?
Q =
∆Toverall
REquiv
=
370K − 305K
REquiv
=
65K
REquiv
REquiv = RT hrmalCercuit1 + RT hrmalCercuit2 =
ln r2
r1
2πk2L
+
ln r3
r2
2πk3L
REquiv = RT hrmalCercuit1 + RT hrmalCercuit2 =
ln 0.067
0.025
2π0.07 × 1
+
ln 0.091
0.067
2π0.07 × 1
= 2.24 + 1.02 = 3.26K/W
Q =
65K
3.26K/W
= 19.96W
To find T2, Q = T1−T2
Rth1
, T2 = T1 − (QRth1) = 370 − (19.96 × 2.24) = 325.26K
Example 5 Fins Problem A motor body is 360mm in diameter (outside) and 240mm long. Its surface temperature
should not exceed 55o
C when dissipating 340W. Longitudinal fins of 15mm thickness and 40mm height are proposed.
The convection coefficient is 40W/m2
.o
C. Determine the number of fins required. Atmospheric temperature is 30o
C.
thermal conductivity k = 40W/mo
eC.
Given and required
D = 0.36m L = 0.24m Tb = 55o
C Qg = 340W tfin = 0.015m hfin = 0.04m
h = 40W/m2
.o
C T∞ = 30o
C k = 40W/mo
C Find No. of fins N
Solution
N =
Qg
Qeachfin
=
340W
√
h2LkA(Tb − T∞) tanh(L
q
2hL
kA )
N =
340W
p
40 × 2 × 0.24 × 40 × (0.015 × 0.24)(55 − 30) tanh(0.24
q
2×40×0.24
40×(0.015×0.24) )
N = 72.06 ≈ 72fins
8

1. Which one of water and air better suited for insulation considering table 3 and why? what materials in the list
would you select for cooling or heating? Base your answer with reasons.
2. Given material with thermal conductivity k, suppose you are considering to make a heater. If you have an
available geometry option a plane wall of thickness L, long cylinder with radius r0 and sphere with radius r0,
what geometry would you prefer if L = r0,spere = r0,cylinder? refer table 4
3. Distinguish between fin efficiency and fin effectiveness
4. Describe electrical circuit analogy of thermal circuit
5. Thick-walled tube of stainless steel, K = 19W/m.o
.C, with (2cm) inner diameter and (4cm) outer diameter, If
the inside wall temperature of the pipe is (600oC), calculate the heat loss per meter of length, outside temp. Is
(100o
.C)?
6. A Furnace is constructed with (0.20m) of firebrick, (0.10m) insulating brick and (0.20m) of building brick, the
inside temperature is (1200K) and the outside temperature is (330K), if thermal conductivity (k) is 1.4, 0.21, 0.7
seriously, find the heat loss per unit area?
4 2D Steady Heat Conduction
There are many conduction problems in which the temperature gradient is significant in more than one coordinate
direction. In such cases, the multidimensional effects must be accounted for. In this chapter, the various techniques
for treating two-dimensional systems under steady-state conditions are considered. The objective of this chapter is
to mathematically model the way thermal energy moves through the plate. The heat equation for 2-D, steady-state
conditions with no generation and constant thermal conductivity given in Eq. 9
q(x, y) = −k∇T(x, y) = −k(
∂2
T
∂x2
+
∂2
T
∂y2
) = 0 (9)
Methods for solving equation 9 include the use of analytical, graphical, and numerical approaches.
4.1 Analytic Method: The Method of Separation of Variables
This is one of the analytical techniques employed to obtain exact solution to 2-D conduction problems. The major
drawbacks of analytical solution techniques are by applying mathematical superposition and seeking out solutions
with constant coefficients that are either sum or products of functions of one variable.
ˆ The solutions often involve complicated mathematical series and functions
ˆ Solutions may be obtained for only a restricted set of simple geometries and boundary conditions
Example 6 Given two dimensional problem of modified Eq. 9 ∂2
T
∂x2 + ∂2
T
∂y2 +
q′′′
0
k = 0 where the last term represent
heat generation. The boundary conditions are ∂T
∂x 0
= 0, ∂T
∂x L
=
q′′′
0
k , T(x̄, 0) = 1 and ∂T
∂y L
= −BiT̄(x̄, 0)
Solution Let’s introduce dimensionless parameters ¯
q” = q”0L
k(T1−T∞) , ¯
q”′ =
q”′
0L2
k(T1−T∞) and Bi = hL/K, this also known
as Biot number The superposition involve splitting the problem in two parts as shown in figure 5: Part A is adiabatic
condition at x̄ = 0 and 1, uniform temperature of 1 at ȳ = 0, convection at ȳ = 1 and uniform heat generation of Q”′
.
Part B is an adiabatic condition at x̄ = 0 and 1, uniform heat flux of ¯
q” at x̄ = 1, zero temp. at ȳ = 0, convection at
ȳ = 1 and no heat generation Otherwise the number of equation will be less than unknowns. The subsequent equation
for each condition can be written and substituted in the original equation to find the solution. It can be noted here
that the introduction of convention just increase the number of equations and unknowns that otherwise the method
is the same. In cylindrical coordinates, the superposition follows the same suit.The non dimensional temperature
is T̄ = T −T∞
TB−T∞
. Let L be the characteristic length, so that z = z/L and r = r/L. Again, it is good practice to
nondimensionalize all spacial variables by the same characteristic length. The governing steady state equation is
1
r
∂
∂r r∂2
T
∂r2 + ∂2
T
∂z2 = 0, and boundary value conditions T̄(r̄, 0) = 1, ∂T̄
∂r 1
= 0, T̄(z̄, 0) = ∞, and −∂T̄
r̄ = BiT̄(a, z̄).
Exact solutions have been obtained for many geometries and boundary conditions. Many 2 − D or 3 − D conduction
problems may be solved rapidly by utilizing these existing solution. These solutions have been presented in terms
of the shape factor, S. The heat transfer rate may be expressed in Eq. 10 where k is the thermal conductivity of
the medium and ∆T1−2 is the temperature difference between boundaries. It follows then that the 2 − D conduction
resistance is also in Eq. 10
q = Sk∆T1−2 where RCond2D =
1
Sk
(10)
9

Figure 5: Boundary Condition Problem of Two dimensional Conduction and Convection
For objects that are embedded within an infinite medium, a characteristic length may be defined in Lc =
q
As
4π or
steady-state dimensionless conduction heat rate, qss given
qss =
k
qLc
kAs(T1 − T2) (11)
Shape factor for various geometries presented in Table 4 alongside which shape factor that incorporated in Equation
10 presented in Figure 6.
4.2 Graphical Method
This method uses a graph called a heat flux plots. We already saw graphs in example 6 that simplify the calculation.
The constant temperature lines (shown in the body) are called isotherms, The directions of the heat flux vectors
are represented by the heat flow lines. They are perpendicular to the isotherms, and The heat flux components are
determined by applying the Fourier’s law. Sets of rules for graphical technique include
1. identify the symmetry lines of both geometry and thermal condition called adiabates
2. construct an isothermal lines at right angle to adiabtes
3. adiabates bisect isothermal corners
Example 7 Graphical Method Typical example is given here.consider a long prismatic thin rectangular solid plate
made of some thermally conductive material. Suppose the dimensions of the plate.
Solution: No heat can be conducted across a heat flow line. Therefore, heat flow lines are sometimes referred to as
adiabats. Adiabatic surfaces (or symmetry lines) are heat flow lines (since heat cannot be conducted across them).
4.3 Numerical Method: Finite-Difference Equations
Some 2-D conduction problems cannot be solved analytically due to the nature of the geometries and/or boundary
conditions. In such cases, the best alternative is the use of a numerical technique. Among the numerical techniques
available are the finite-difference, finite-element and boundary-element methods.
4.3.1 Nodal Analysis
A numerical solution enables determination of temperature at only discrete points. The first step in any numerical
analysis must therefore be to select these points. The medium is subdivided into a number of small regions. A discrete
point, where the temperature will be determined, is located at the center of each small region. This is frequently
termed a nodal point (or simply a node). The aggregate of points is termed a nodal network, grid, or mesh. The
temperature of a node is assumed to represent the average temperature of the region where it is located. Each node
is designated by its row and column numbers. The numerical accuracy of the calculations increases as the number
of nodal points selected increases. The nodal network identifies discrete points at which the temperature is to be
10

Figure 6: Shape factor S for various selected geometries
[1]
Figure 7: isothermal and heat flow lines in 2d conduction
[6]
determined and uses an m,n notation to designate their location as shown in Fig. 8. What is represented by the
temperature determined at a nodal point, as for example, Tm,n? How is the accuracy of the solution affected by
construction of the nodal network? What are the trade-offs between selection of a fine or a coarse mesh?
4.3.2 Finite Difference Method FDM
To determine the temperature distribution numerically, an appropriate conservation equation must be written for each
node. For any interior node of a 2-D system with no generation and uniform thermal conductivity, the exact form
of the energy conservation requirement is given by the heat equation. An approximate, or finite-difference, form of
equation 9 is developed for the m, n nodal point as in Eq.12
11

Figure 8: Nodal Analysis
Figure 9: Finite Difference Method
∂2
T
∂x m,n
≈
∂T/ ∂x|m+1/2,n − ∂T/ ∂x|m−1/2,n
∆x
∂T
∂x m+1/2,n
≈
Tm+1,n − Tm,n
∆x
∂T
∂x m−1/2,n
≈
Tm,n − Tm−1,n
∆x
T|m,n ≈
1
4
[Tm−1,n + Tm+1,n + Tm,n−1 + Tm,n+1 +
q̇(∆x)2
k
(12)
Represent the physical system by a nodal network i.e., discretization of problem. Use the energy balance method to
obtain a finite-difference equation for each node of unknown temperature. Solve the resulting set of algebraic equations
for the unknown nodal temperatures. Use the temperature field and Fourier’s Law to determine the heat transfer in
the medium.
Example 8 Numerical Method Consider the square plate plate temperature distribution with no heat generation
given in the Fig. 10 and determine T1, T2, T3, T4
Solution: Using the last row of equation 12, T1 ≈ 1
4 [100 + T3 + 50 + T2] and likewise,
T2 ≈
1
4
[100 + T1 + 200 + T4] T3 ≈
1
4
[300 + T1 + 50 + T4]
T4 ≈
1
4
[300 + T2 + 200 + T3] collecting cooefficients & results in
The resulting coefficient matrix and RHS array can be augmented and solved using gauss-seidel method.




[cccc|c] − 1 0.25 0.25 0 −37.5
0.25 −1 0 0.25 −75
0.25 0 −1 0.25 −87.5
0 0.25 0.25 −1 −125




Solving this yields T4 = 206.45o
C, T3 = 169o
C, T2 = 152.32o
C and T1 = 358.82o
C
1. What is the use of superposition and variable separation in mathematical formulation of 2D-steady conduction?
12

Figure 10: FDM Steady State Cond. Worked Example
2. How does various geometries contribute to an overall 2D steady conduction? Select a geometry from figure 6
and make it foundation for your discussion.
3. Consider Example 8 and solve it by using graphical method and attempt with an analytical superposi-
tion/variable separation method
5 Unsteady-State Heat Transfer: Conductive and Convective BL
Most engineering problems incorporate both conduction and convection. Energy balance is going to be used to solve
thermal analysis of such system. Conditions to be considered in this part include: i. temperature do not change
with position but time,(temperature is assumed to be spatially independent (uniform)). The analysis of time variant
temperature is called lumped system analysis. ii. Next analysis in this part involves temperature change with both
time and position for simple slab geometry, and iii. Finally, semi-infinite regions or near surface of a large body.
5.1 Lumped Capacitance Method: No Internal Resistance,
Lumped capacitance model is used in which no temperature gradient exists. Temperature is a function of time only
with Negligible Internal Resistance This means that the internal resistance of the body (conduction) is negligible in
comparison with the external resistance (convection). i.e small convective heat transfer coeff. h and large thermal
conductivity k. This decision is made depending on the magnitude of Biot number Bi with expression in Eq. 13
Bi =
Conductive Resistance
ConvectiveResistance
=
L
kA
1
hA
=
hL
k
(13)
Consider a solid with convection over its surface in Fig. 11
Employing energy balance,
Figure 11: Lumped Capacitance Model
[7]
13

− mCp∆T = hA(T − T∞)∆t θi = θ(t = 0) = T − T∞
−
∆T
∆t
=
hA
mCp
(T − T∞)
dθ
dt
= −
hA
mCp
θ
dT
dt
= −
hA
mCp
(T − T∞)
Z θ
θi
dθ
θ
= −
Z t
0
hA
mCp
dt
T(t = 0) = Ti ln
θ
θi
= −
hA
mCp
t
θ = T − T∞ Rearranging to exponential



























(14)
θ
θi
= e
− hA
mCp
t
= e
− t
hA
mCp (15)
For given A, V, ρ, CP , s = V/A are exposed surface area, volume, body density, specific heat capacity and characteristic
solid dimension respectively.
If Bi = hL
k = hV/A
k < 0.1 in Eq. 13, internal resistance can be ignored and lamped capacitance analysis yields
acceptable. The lumped system is exact if Bi = 0, and Bi = hL
k = hs
k where s = 1
2 Lfor plate, s = 1
2 rfor cylinder and
s = 1
3 rfor spere
Example 9 What is the temperature of the egg after 60 minutes?
Given Ti = 20o
C, Tair = 3820o
C, h = 5.2W/m2
.K, ρ = 1035kg/m3
, CP = 3350J/kg.K, k = 0.62W/m.k. V =
60 × 10−6
m3
, A = 7.85 × 10−3
m2
Assume egg is spherical, h is average value, and lumped capacitance analysis.
Solution For lumped capacitance analysis to be valid, Bi = hV/Ak = 5.2×60×10−6
7.85×10−3×0.62 = 0.07 < 0.1, the analysis is
therefore valid.
60minutes = 3600sec and except that all parameters given in SI units and therefore using equation 15, numbers can
be plugged in directly
T60minutes = T∞ + (Ti − T∞)e
− hA
mCP
t
= 38 + (20 − 38)e
− 5.2×7.85×10−3
1035×60×10−6×3350
3600sec
= 29.1o
C
5.2 Significant Internal Resistance
The energy balance equation 9 can incorporate heat conduction if the internal resistance is not neglible.
ρCP
∂T
∂t
| {z }
storage
+ u∂T
∂x
0
|{z}
bulk flow
= k
∂2
T
∂x2
| {z }
conduction
+ Q
0
|{z}
Generation
, neglect bulk& gen.
∂T
∂t
=
k
ρCP
∂2
T
∂x2
, with bound. Cond. for psymetry
∂T
∂t (x=0,t)
= 0 and T(L, t > 0) = Ts
θ
θi
=
∞
X
n=0
4(−1)n
(2π + 1)π
cos
(2n + 1)πx
2L
e−α((2n+1)π
2L )
2
t
, where
θ
θi
=
T − Ts
Ti − Ts
, & therm. diff. α =
k
ρCp

























(16)
5.2.1 Heat Transfer in an Infinite Geometries
Making n = 0 from Eq. 16. To understanding/feel T vs. x and t is vital, infinite cosine series should be simplified.
The following derivation involves temperature variation with size.
θ
θi
=
4
π
cos
πx
2L
e−α( π
2L )
2
t
remember n = 0 all n terms from Eq. 16 gone
ln
θ
θi
= ln
T − Ts
Ti − Ts
= ln

4
π
cos
πx
2L

− α
π
2L
2
t, introducing Tav =
1
L
Z L
0
Tdx in the left Eq.
ln
Tav − Ts
Ti − Ts
= ln
8
π2
− α
π
2L
2
t, Rearrenging, the T change with size,
αt
L2
=
4
π2
ln

π2
8
Tav − Ts
Ti − Ts


















(17)
New term in Eq. 17 can also be introduced for FO = αt
L2 known as Fourier Number
Example 10 Consider cylindrical food item with thickness L = 0.03m. Take thermal diffusivity of biomaterial
α = 1.44 × 10−7
m2
/s to be sterilized. Determine temperature variation within food sterilizer at t = 30s and for
t = 600s.
14

Solution The first step is to drop off time values Equation 16, for t = 30s and for t = 600s given.
Temp. drop time t = 30sec Temp. drop time t=600 sec
F0 =
αt
L2
=
1.44 × 10−7
30
0.032
F0 =
αt
L2
=
1.44 × 10−7
600
0.032
FO = 0.0048 =
4
π2
ln

π2
8
θav
θi

FO = 0.096 =
4
π2
ln

π2
8
θav
θi

Temp. decay slowly Temp. decay rapidly
5.2.2 Heisler chart: Infinite (slab, cylinder and spherical) geometry
If internal resistance is not negligible, (Tr=r ̸= Tr=0.5r ̸= Tr=0(i.e.Bi ≥ 0.1) for a cylindrical objects and n = x
L
for discretization (meaning numerical differentiation). This way eq. 16 can be simplified and solution charts can be
developed with the help of x− axis as x
L and y− axis as αt
L2 using: FO and n = x
L , m = B−1
for heat transfer, m = DAB
hmL
for mass transfer or film diffusion, and θ
θ∞
for a given problem, remember α = k
ρCp
Chart developed for n = 0 condition for Eq. 16 is also called heisler chart.Application of internal resistance
involving use of Hiesler chart with sets of assumptions for infinite geometries. It’s developed with calculations
made for the series (n = 0, 1, ...) using equation 16 for various applications. The assumptions include i.Uniform initial
Figure 12: Hiesler Chart:Unsteady state diffusion
15

temperatureii.Constant boundary fluid temperature iii.Perfect slab, cylinder or sphere iv. Far from edges v. No heat
generation (Q = 0) vi. Constant thermal properties (k, α, CP ) are constants vii. Typically for times long after initial
times, given by αt
L2 0.2
Example 11 Consider tuna sterilizer: a cylindrical can containing food to be sterilized with heating surface temper-
ature 121o
C. If Surface temperature of a slab of tuna is suddenly increased, determine the temperature from initial
point 40o
C of at the center of the slab after 30 min. Assume negligible side heating and constant thermal diffusion of
α = 2 × 10−7
m2
/s. The slab thickness is given 25 mm Solution L = thickness
2 = 0.0125m and t = 30min = 1800sec
n =
x
L
=
0
0.025
= 0 from the chart in Fig. 12,
m = Bi−1
=
k
hL
= 0 using Bi = ∞orm = 0
FO =
αt
L2
usingFO = 2.3,
θ
θi
= 0.0043
=
2 × 10−7
× 1800
0.0125
= 2.3
T − Ts
Ti − Ts
= 0.0043























T30min = 120.65o
C
Hiesler Chart can also be used for Convective Boundary Condition. The external fluid resistance considered in addition
to internal fluid resistance Near Surface of a. In convective boundary condition, surface temperature is not the
same as the bulk fluid temperature, T∞, signifying additional fluid resistance, where numerical method or chart used
to find solution. Eq. 18 is at the surface,
−k
∂T
∂x s
= h(Ts − T∞) (18)
5.3 Large Body: Finite and Semi-infinite Geometry
Large body here refers to semi-infinite regions or near surface of a large body. A finite geometry is considered as the
intersection of two or three infinite geometries

Txyz,t − Ts
Ti − Ts

| {z }
finite
=

Tx,t − Ts
Ti − Ts

| {z }
infinite x slab
+

Ty,t − Ts
Ti − Ts

| {z }
infinite y slab
+

Tz,t − Ts
Ti − Ts

| {z }
infinite z slab
finite slab

Tr,z,t − Ts
Ti − Ts

| {z }
finite
=

Tr,t − Ts
Ti − Ts

| {z }
infinite r slab
+

Tz,t − Ts
Ti − Ts

| {z }
infinite z slab
finite cylinder

















(19)
5.3.1 transient Heat Transfer in a Semi-Infinite Geometry
A semi-infinite region extends to infinity in two directions and a single identifiable surface in the other direction. A
typical example of semi-infinite geometry extends to infinity in the y and z directions and has an identifiable surface
16

at x = 0 It can be used practically in heat transfer for a relatively short time and/or in a relatively thick material
∂T
∂t
= α
∂2
T
∂x2
Governing Equation
T(t = 0) = Ti and T(x = 0) = Ts BC
T(x → ∞) = Ti IC
T − Ti
Ts − Ti
= erf

x
2
√
αt

Solution
erf(η) =
2
√
π
Z η
0
e−η2
dη Error function
Where η =
x
2
√
αt



































(20)
The temperature solution profile given in a long equation 21 so that you may appreciate.

T1−T
T1−To

= 4
π

e
−π2αt
4H2
sin πx
2H + 1
3 e
−32π2αt
4H2
sin 3πx
2H
+1
5 e
−52π2αt
4H2
sin 5πx
2H + · · ·





(21)
Heat flux at the surface of the semi-infinite region can be calculated with chain rule
−k
dT
dx x=0
= −k
dT
dη
dη
dx x=0
= −k(Ts − Ti)

−
2
√
π
e−η2

(
η = 0)
1
2
√
αt
=
k(Ts − Ti)
√
αt
x
2
√
αt
≥ 2, or ,x ≥ 4
√
αt, Semi-infinite Approximation Condition
Solution to an approximate semi-finite region with surface heat flux q”surface = q”s and convective boundary condi-
tion −k ∂T
∂x Surface
= h(TSurface − T∞) is given in Eq. 22
T − Ti =
2
k
q”s
r
αt
π
e− x2
4αt −
q”sx
k

1 − erf

x
2
√
αt

cond.BC
T − Ti
Ts − Ti
= 1 − erf

x
2
√
αt

− e
hx
k + h2αt
k2

1 − erf

x
2
√
αt
+
h
√
αt
k

conv.BC









(22)
1. What is a Lamped capacitance method?
2. Write and expression for ratio of temperature differences θ
θi
for insignificant and significant internal resistance
the later with an infinite geometries
3. List down assumptions for in using Heisler charts and time-space variation of temperature for an infinite geome-
tries
4. Describe reading procedures for Hiesler Chart for in the contexts of slab, cylinder and slab
5. Describe semi-infinite geometries in relation to finite and infinite geometries considering convective and convective
boundaries
6. The temperature has been 35o
.F for a while now, sufficient to chill the ground to this temperature for many
tens of feet below the surface. Suddenly the temperature drops to −20o
.F. How long will it take for freezing
temperatures (32o
.F) to reach my pipes, which are 8 ft under ground? Use the following physical properties:
h = 2.0 BT U
hft2o.F , αsoil = 0.018ft2
h , ksoil = 0.5 BT U
hfto.F
Use simplified version equation for Eq. 22 T −T0
T1−T0
= erfζ − eβ(2ζ+β)
erf(ζ + β) , ζ ≡ x
2
√
αt
both ζ and β depend
on time, β ≡ h
√
αt
k ,T0 =?, T1 =?, T =?, and T −T0
T1−T0
=? with mathcad graph Fig.13
17

Figure 13: Semi-Infinite Graphical Solution
[8]
6 Convection Heat Transfer
Convection is the mechanism of heat transfer through a fluid in the presence of bulk fluid motion. Convection
is classified as natural (or free) and forced convection depending on how the fluid motion is initiated. In natural
convection, any fluid motion is caused by natural means such as the buoyancy effect, i.e. the rise of warmer fluid and
fall the cooler fluid. Whereas in forced convection, the fluid is forced to flow over a surface or in a tube by external
means such as a pump or fan
The convection heat transfer without incorporating fluid motion analysed in heat transfer modes and finned surfaces
covered in Section 2.3 and 2.5. This section is about heat transfer any fluid motion occurring naturally due to causes
such as buoyancy and external drives such as propellers, pumps, fans..The former one is called Natural convection
and the later is known as Forced convection. convection can also be categorized based on channel of flow as external
and internal. The phenomena of convection governed by Newton’s law of cooling given in table 1 and appears once
again in Eq.23 taking in to account a convective heat transfer coefficient h in (W/m2
K).
QConv =
∆T
Rconv
= hA(Tw − T∞)where (23)
It is assumed that the velocity of the fluid is zero at the wall, called noslip condition. As a result, the heat transfer
Table 5: convective heat transfer coefficient h in W/m2.K for various heat transfer types
heat transfer Natural convention Forced convention Boiling Condensation
h in W/m2
K
gases:3 − 20
,water:60 − 90
gases:30 − 300
oils:60 − 1800
Water:100 − 1500
Water:300 − 105
Steam: 3000 − 105
from the solid surface to the fluid layer adjacent to the surface is by pure conduction,and other factors that affect
heat transfer via convection listed in the table Laminar flows are smooth and streamlined, whereas turbulent flows
are irregular and chaotic. A low Reynolds number indicates laminar flow while a high Reynolds number indicates
turbulent flow. All fluids have viscosity, meaning they are sticky, and stick to surfaces. Oil sticks to your hand. So
does water, and air as well. Air also sticks to a wing or turbine blade surface. The viscosity of a fluid is independent
of the speed of its flow, and more to do with its chemical makeup. The Reynolds Number (Re) is proportional to the
ratio between the momentum of a flowing fluid and its viscosity. The lower Re, the more viscosity governs the flow
behavior, so the more it sticks to a surface, and the flow is laminar. The higher the Re, the more its velocity governs,
so the less it sticks to surface, and the flow is turbulent. Laminar flow is straight lines. Turbulent flow is eddies and
swirling vortexes. The smaller Re range means that viscosity governs for more of that range, and hence there is more
aerodynamic drag for the amount of lift generated.
6.1 Dimensionless numbers
Other important dimensionless number other than Re relevant for convection heat transfer include Prandtl number,
Nusselt number, Rayleigh number, Grashof number, and Stanton number. Prandtl number (Pr) is defined as the
ratio of momentum diffusivity (kinematic viscosity) to thermal diffusivity. Fluids with small Prandtl numbers are
free-flowing liquids with high thermal conductivity and are therefore a good choice for heat conducting. Boundary
layer theory is used to describe the mechanism of heat transfer in fluids. In the heat transfer that occurs between
18

Table 6: Convection Controlling Factors
1.Factors 2.Geometry: 3.Flow Type: 4.Boundary: 5.Fluid Type:
Lists
shape, size, aspect
ratio, orientation
forced, natural,
laminar,turbulent,
internal, external:
TW = Constisothermal
or q̇ = Const.
viscous oil, water
, gases/liquid metals
6.Thermofluid Properties all properties determined at film temperature Tf = (Tw + T∞)/2
Note that: ρ and ν ∝ 1/Patm
Thermofluid Properties symbols (SI units) Thermofluid Properties symbols (SI units)
Density ρ(kg/m3
) dynamic viscosity µ(N.s/2
)
specific heat CP (J/kg.K) kinematic viscosity: ν = µ/ρ(m2
/s)
thermal conductivity: k(W/m.K) thermal diffusivity: α = k/(ρ.CP ), (m2
/s)
Prandtl number: ν/α volumetric compressibility: β, K−1
Reynold number Re =
UL
ν
|{z}
forced Con.
Nusselt Number Nu =
hL
k
|{z}
Forced/free Conv
Grashof Number
Gr =
buouyancy force
viscous force
| {z }
free conv
=
gβL3
ν2
(TW − T∞)
Rayleigh Number RaL
|{z}
for free conv
= GrLPr
a wall and a flowing fluid, heat is transported from the bulk of a fluid through a momentum boundary layer that
consists of the bulk fluid and a transition layer and a thermal boundary layer that consists of stagnant film in which
heat transport occurs by fluid conduction. The Prandtl number (Pr) of a fluid gives the relative importance of the
momentum boundary layer to the thermal boundary layer in the transfer of heat. Pr used along with Stanton number
to characterize heat transfer in forced convection flows. Stanton number is named after Thomas Stanton (engineer)
(1865–1931), the ratio of heat transferred into a fluid to the thermal capacity of fluid.
The Nusselt number is defined as the ratio of convection heat transfer to fluid conduction heat transfer under the
same conditions. Nusselt number is an important parameter that can contribute to a better rate of heat exchange.
It is basically a function of Reynolds and Prandtl number. Another dimensionless parameter regarded as a measure
of the driving forces of natural convectionis Rayleigh number, a measure of the instability of a layer of fluid due to
differences of temperature and density at the top and bottom during free convection. Rayleigh number used along with
Grashof number, the ratio between the buoyancy force due to spatial variation in fluid density (caused by temperature
differences) to the restraining force due to the viscosisty of the fluid.
6.2 Forced Convection
The region defined by the velocity gradient where the flow velocity is distributed among the different fluid layers is
called the velocity boundary layer. The thermal boundary layer is the region of fluid flow defined by the temperature
gradient formed due to the thermal energy exchange among the adjacent layers. The simplest forced convection
configuration to consider is the flow of mass and heat near a flat plate as shown in the figure 14. The flow region where
(a) (b)
Figure 14: Convection configuration in various flow scheme: a. various flow layer scheme b. boundary layer
the viscous effect dominate referred to as hydrodynamic or velocity boundary layer. The velocity at the surface of the
plate, y = 0, is set to zero, Uy=0 = 0m/s because of the no slip condition at the wall. The fluid velocity progressively
increase along height until it reaches 99 % U∞ that is denoted by δ. The region beyond the velocity boundary layer is
denoted as the inviscid flow region, where frictional effects are negligible and the velocity remains relatively constant
at U∞.
The thermal boundary layer is arbitrarily selected as the locus of points where T −TW
T∞−TW
= 0.99. For low Prandtl
19

number fluids the velocity boundary layer is fully contained within the thermal boundary layer. Conversely, for high
Prandtl number fluids the thermal boundary layer is contained within the velocity boundary layer.
6.2.1 Types of Flow
As seen in the introduction of this section, flows can be categorized as external and internal. External flows seen in
Fig. 15 can further be analysed as laminar, turbulent or combined in various geometries over which fluid flow as: flow
over plates, cylinders and spheres. Whereas laminar and turbulent internal flow see Fig. 16 analysis geometries can
either be duct or tubes.
(a) (b)
Figure 15: External flows boundary regions a. Flow over Plate b. Flow Over Cylinder
Table 7: Empirical formulation of heat transfer coefficients various flow regimes and geometries of Flow Over Plates
Lam. Turb. Uniform Wall Temperature Formulation UWT
Flow Regime Nusselt No. at x for various Pr Full extent Average NuL
Laminar Re 50, 000
Nux = 0.332Re1/2
x Pr1/3
forPr ≥ 0.6
Nux = 0.565Re1/2
x Pr1/3
for
low Pr such as liquid metalsPr ≤ 0.6
NuL = hLL
kf
Turbulent R50, 000 ≤ Re ≤ 107
Nux = 0.0.0296Re0.8
x Pr
1
3 for 0.6 Pr 60
NuL = 0.0.037Re0.8
L Pr
1
3 for
0.6 Pr 60
...
Combined Laminar-turbulent Uniform Wall Temp. UWT Formulation
Flow Regime Nusselt No. at L for various Pr
Lam. Turb. TW − T∞ = Const.
hL =
1
L
(Z xC r
0
hx
lam.
dx +
Z L
xC r
hx
lam.
dx
)
forPr ≤ 0.6
NuL = (0.037Re0.8
L − 871)Pr1/3
An average Combined for0.6 ≤ Pr ≤ 500, 000
...
Uniform Wall Flux UWF Formulation
Laminar Bound. Layer Isoflux Nux = 0.453Re
1/2
x Pr1/3
Local Laminar UWFPr ≥ 0.6
Turnulent Bound. Layer Isoflux Nux = 0.0308Re
4/5
x Pr1/3
Local Turbulent UWFPr ≥ 0.6
Internal Flow: The Reynolds number is given in 24 and the critical Reynolds number is ReDCr = 2300 :
ReD =
UmD
ν
for flow in tube
ReD 2300 laminar
2300 ReD 4000 transition
ReD 4000 turbulent
(24)
When the boundary layer grows to the tube radius, r, the boundary layers merge. The flow length is called the flow
entrance length, Lh. 0 ≤ x ≤ Lh and 0 ≤ x ≤ Lt is developing region for hydrodynamic/velocity and temperature
respectively, Lh ≤ x ≤ L is fully developed region. The boundary layer thickness approximated as
δ(x) ≈ 5x

Umx
nu
−1/2
thermal boundary layer thickness
hydrodynamic entry length can be approximated asLh ≈ 0.05ReDD
turbulent entry length can be approximated asLt ≈ 10D ≈ Lh
laminar temp. boundary layerLt ≈ 0.05ReDPrD = PrLh
20

Table 8: Convection Coeff. for Flow Over Cylinders and Spheres
...
Boundary Layer Flow Over Circular Cylinders, Isothermal. UWT Formulation
Geometry Average Nusselt No.
Cylinder L/D 100
NuD = 0.3 +
0.62Re
1/2
x Pr1/3
1 + ((0.4Pr2/3));1/4
[1 + (
ReD
282
)5/8
]4/5
at ReD 107
, 0 ≤ Pr ≤ ∞
all properties evaluated atT = (Tw + Tinfty)/2
NonCircular Cylinder NuD = CReD
m
Pr1/3
Spere
NuD = 2 + [0.4Re
1/2
D + 0.06Re
2/3
D ]Pr0.4
(
µ∞
µs
)1/4
at0.7 ≤ Pr ≤ 380
3.5 ReD 80, 000
Lets consider fluid flow in a duct bounded by a wall that is at a different temperature than the fluid. For simplicity
we will examine a round tube of diameter D as shown below. A thermal entrance region develops from 0 ≤ x ≤ Lt.
Figure 16: various flow boundary regions in internal flow convection
The thermal entry length can be approximated as Lt ≈ 0.05ReDDPr = PrLh for laminar flow and Lh ≈ Lt ≈ 10D
for turbulent. The total heat transfer from the wall to the fluid stream can be determined by performing an energy
balance over the tube. If we assume steady flow conditions, mass and energy balances
ṁ = ˙
minlet = ˙
moutflow
Q̇ = ˙
qwA = ṁ(hout − hin) = ṁCP (Tout − Tin)
Since the wall flux ˙
qW is uniform, the local mean temperature is linear with x. Tm,x = Tm,i + q̇wA
ṁCP
The surface
temperature can be determined from TW = Tm + q̇w
h The energy balance equation
(a) (b)
Figure 17: Isothermal wall heating, exponential Temp. Decay a. Mean surface Temp. b. Log Mean Temp.
Q̇ = hA (TW − Tm)
| {z }
avg∆T
= ṁCP dTm
21

by isolating temperature terms, integrating and expressing temp. difference in terms of log mean temperature
(TW − Tm)
dTm
| {z }
avg
=
ṁCP
hA
ln
TW − Tout
TW − Tin
= −
hA
ṁCP
δTln =
Tout − Tin
ln TW −Tout
TW −Tin
=
Tout − Tin
ln(∆Tout/∆Tin)
Therefore, the average rate of heat transfer to or from a fluid flowing in a tube
Q̇ = hA∆Tln (25)
For non-circular tubes the hydraulic diameter, Dh = 4A
P can be used in conjunction with Table 6 determine the
Table 9: Laminar Flow in Tube Isothermal UWT and Isoflux UWF
Developing Laminar NuD = 1.86 ReDP rD
L
1/3

µb
µs
0.14
...
Fully Developed Laminar NuD = 3.66 for UWT and L Lt, Lh
Fully Developed Laminar NuD = 3.66 for UWF and L Lt, Lh
Turbulent UWT UWF
NuD = 0.023Re0.8
D Prn
ReD 2, 300
n = 0.4heating
n = 0.3cooling
Mean fluid temperature Tmean = 1
2 (Tm,in + Tm,out)
Reynolds number and in turn the Nusselt number.
6.3 Natural Convection
Fluids tend to expand when heated and contract when cooled at constant pressure. Therefore a fluid layer adjacent
to a surface will become lighter if heated and heavier if cooled by the surface. A lighter fluid will flow upward and a
cooler fluid will flow downward. As the fluid sweeps the wall, heat transfer will occur in a similar manner to boundary
layer flow however in this case the bulk fluid is stationary as opposed to moving at a constant velocity in the case
of forced convection. In natural convection, the Grashof number is analogous to the Reynolds number see Table 6.
Natural convection heat transfer depends on geometry and orientation. natural convection heat transfer depends on
Figure 18: natural Convection Over surface
geometry and orientation. The velocity and temperature profiles within a boundary layer formed on a vertical plate
in a stationary fluid looks as shown in 18. The general form of the Nusselt number for natural convection is as follows:
Nu = f(Gr, Pr) ≡ CGrm
Prn
Ra = Gr.Pr
Tf =
1
2
(Tw − T∞)
β = 1/Tf













C depends on geometry, Boundary etc
m depends on laminar/turbulent
n depends on fluid type/flow
(26)
22

Example 12 Forced Con. Flow over plate: Consider Figure 15 (a), Hot engine oil with a bulk temperature of
60o
C flows over a horizontal, flat plate 5m long with a wall temperature of 20o
C. If the fluid has a free stream velocity
of 2m/s, determine the heat transfer rate from the oil to the plate if the plate is assumed to be of unit width.
Solution:Assume steady state; width W = 1m The film temperature is Tf = 1
2 (Tw + T∞) = 1
2 (20 + 60) = 40o
C
From Table 3, 1 and standard property table for unused engine oil ρ = 876kg/m3
, k = 0.1444W/(m.K), ν = 2.485 ×
10−4
m2
/s, Pr = 2962 The Reynolds number is
ReL =
U∞L
ν
=
2m/s × 5m
2.485 × 10−4m2/s
= 4.07 × 104 Recr = 500, 000
Therefore we are in the laminar regime over the entire length of the plate. The Nusselt number is given in Table 7 as
NuL =
havL
kf
= 0.664Re1/2LPr1/3 Average Nusselt
and the average heat transfer coefficient is hav =
k
L
0.664Re
1/2
L Pr1/3
=
0.1444W/(mK)
5m
× 0.664 × (4.07 × 104
)1/2
× 29621/3
= 55.6W/(m2
K)
The heat flow rate is
Q̇ = havW × L(T∞ − Tw) = 55.6W/(m2
K) × (1m × 5m) × (60 − 20)K = 11.112kWAnswer
Example 13 Forced Conv. Flow over Cylinder Consider Figure 15 (b) for this problem. An electric wire with a
1mm diameter and a wall temperature of 325K is cooled by air in cross flow with a free stream temperature of 275K.
Determine the air velocity required to maintain a steady heat loss per unit length of 70W/m.
Solution: Heat balance
Q = mCp(Tout − Tin) |viscosity correction factor.
Q = (6000/3600) × 4.93 × (65–15) = 411kW |Nu = 0.023(96441)0.8
(4.86)0.33
= 376
Cross-section of pipeA = (π/4)(50 × 10−3
)2
= 1.963 × 10−3
m2
|h = (0.385/50x10−3
) × 376 = 2895Wm−2
.o
C−1
Fluid velocity,u =
6000
3600
1
866
1
1.963 × 10−3
= 0.98m/s |Take the steam coefficient as 8000Wm−2
.o
C−1
Re =
866times0.98 × (50x10−3
)
0.44 × 10−3
= 96441 |
1
U0
=
1
8000
+
60 × 10−3
(60/50)
2 × 480
+
60
50
×
1
2895
Pr =
4.3 × 10−3
× 0.44 × 10−3
0.3895
= 4.86 |U0 = 1627Wm−2
.o
C−1
Liquid is not viscous and flow is turbulent,| ∆Tlm = (85–35)/ ln(85/35) = 56.4o
C
so use eqn in 9, with C = 0.023 and neglect the| Ao = (411 × 103
)/(1627 × 56.4) = 4.5m2
Ao = π × do × L, L = 4.5/(π × 60 × 10−3
) = 23.87m
Number of lengths = 23.87/3 = 8(rounded)
Check on viscosity correction
Heat flux, q̇ = 411/4.5 = 91.3kW/m2
∆Tacross boundary layer = q/h = 91, 300/2895 = 32o
C
Mean wall temperature = (15 + 65)/2 + 32 = 72o
C
From standard table, µw ≈ 300mNm−2
s
µ/µW = (0.44/0.3)0.14
= 1.055, so correction would increase
the coefficient and reducethe area required. Leave estimate at 8 lengths to allow for fouling.
1. Distinguish between natural and forced convection
2. Describe factors affecting convection heat transfer
3. Distinguish between natural and forced convection as well as boiling and condensation for various fluids
4. Write temperature at which the thermo-fluid properties are determined
23

Table 10: Natural Convection Heat Transfer Correlations
Laminar Flow Over a Vertical Plate, Isothermal (UWT)
General Form nusselt No. NuL = hL
kf
= C




gβ(TW − T∞)L3
ν2
| {z }
≡Gr




1/4
ν
α
1/4
= C Gr
1/4
L Pr1/4
| {z }
Ra1/4
...
Rayleigh no. RaL = GrLPr = gβ(TW −T∞)L3
να
Laminar Flow Over a Long Horizontal Circular Cylinder, Isothermal (UWT)
General Form nusselt No. NuD = hD
kf
= C




gβ(TW − T∞)D3
ν2
| {z }
≡Gr




1/4
ν
α
1/4
= C Gr
1/4
D Pr1/4
| {z }
Ra
1/4
D
...
Rayleigh no. RaD = GrDPr = gβ(TW −T∞)D3
να
Natural Convection From Plate Fin Heat Sinks
Nusselt Number
Isothermal fins have t s Average NuS = hS
kf
=
h
576
(RaS S/L)2 + 2.873
(RaS S/L)0.5
i−0.5
Q̇opt = hA(TW − T∞) Optimizing: As number of fins increases, so is heat transfer coefficient h
SOpt = 2.714

L
Ra
1/4
L

But this comes at expense of an area A = 2nHL increases see figure below
RaL = gβ(TW −T∞)L3
να
h = 1.307kf /Sopt
5. Which property in free convection is analogous to Reynolds number in forced convection?
6. Write an expression for Nusselt number for laminar and turbulent flows of internal and external flows considering
uniform wall temperature and uniform heat flux for various geometries
7. Consider Figure 18. Find the optimum fin spacing, Sopt and the rate of heat transfer, Q̇ for the following plate
fin heat sink cooled by natural convection. Given parameters:
W = 120mmL = 18mm H = 24mmt = 1mm
TW = 80o
CT∞ = 25C
P∞ = 1atmFluid = air
7 Radiation Heat Transfer
7.1 Cause of Radiation, Radiation as a Form of Electromagnetic Wave Energy
Radiation differs from Conduction and Convection heat transfer mechanisms that generally due to Brownian effect,
in the sense that it does not require the presence of a material medium to occur. Energy transfer by radiation occurs
at the speed of light through vacuum due to electromagnetic effect characterized by frequency v, wave length λ and
propagative or photon energy e. It can also occur between two bodies separated by a medium colder than both bodies.
A wide range of electromagnetic spectra shown in the figure 19. Einstein’s photon energy is postulated as in Eq. 27.
Thermal radiation emission is a direct result of vibrational and rotational motions of molecules, atoms, and electrons
of a substance.As Electromagnetic energy (incident radiation)hits surface and pushes some molecules into an excited
24

(a) (b)
Figure 19: Electromagnetic Spectra
state, they relax from the excited state emitting radiation. The incident radiation per unit area is an irradiation, G
e = hc/λ = hv
h = 6.625 × 10−34
J.s Plank Const
c0 = 2.99 × 108
m/s light speed
c = c0/n (where n is refraction index)
n=1.5 for water n= 1.5 for air















Thermal radiation is a volumetric
But,opaque solids like metals exhibit
Surface Rad. b/s it emitted in the
region never reach the surface.
(27)
7.2 Absorptivity, Reflectivity, Emissivity, and Transmissivity
The radiation interaction of bodies characterized by four phenomena: irradiation, absorptivity (α), the ratio of heat
absorbed to incident, emissivity (ϵ), the ratio of heat emitted by the body to that of black body, and transmittance,
τ, portion of irradiation pass through the body see Fig. 20. a. [9] b. [10] Absorptivity is a function of wavelength
(a) (b)
Figure 20: Thermodynamics of Radiation a. Radiation Causes b. absorption, reflection, and transmission of irradiation
(α = α(λ)). The α doesn’t depend on surface temperature but source temperature. Depending on absorptivity,
substances can be gray body or black body: gray body: a body for which α is constant does not depend on λ
black body: a body for which α = 1, i.e. absorbs all incident radiation. For opaque substances, τ = 0. Therefore,
Reflectivity, ρ = 1 − α for opaque surfaces.
α =
qabsorbed
qincident
qemitted, blackbody = σAT4
Stefan Boltzman’s Law
ϵ =
qemitted
qemitted, blackbody
qemitted, nonblackbody = ϵAσT4
σ = 5.676
W
m2K
0 ≤ α ≤ 1,0 ≤ ϵ ≤ 1 α = ϵ at the same temp., Kircholf’s law
(28)
Example 14 Consider a food body item in the furnace with an internal blower: where there is both convection and
radiation heat transfer in fig 21. Body is at temperature Tb and furnace surface at Ts, assume body and surface
exhibit the same emissivity ϵ∥Ts
, and wall acts as a black body, determine radiation heat transfer coefficient in terms
25

of convection heat transfer coefficient and temperature differences using energy conservation model and Kircholf’s law.
Solution This can be modelled in varieties of ways. Writing expressions using Eq. 23 and 28,
(a) (b) (c)
Figure 21: Combined Radiation Convection Furnace Model a. Radiation Model b. Furnace C. Resistance model
qtotal = qconv + qrad = (hconv + hrad)A(Ts − Tb) hrad(Ts − Tb) = ϵσf(Ts, Tb)
qabsorb,body = ασT4
s = ϵ|Ts
T4
s qemit,body = ϵ|Tb
σT4
s = ϵ|Tb
T4
b
qnet,body =
qabs,b − qemit,b
A
= ϵ|Ts
σT4
s − ϵ|Tb
T4
b Net flux to bulk object body
hrad(Ts − Tb) = ϵ|Ts
σ(T4
s − T4
b ) hrad =
ϵ|Ts
σ(T4
s − T4
b )
Ts − Tb

















Assuming ϵ|Ts
= ϵ|Tb
The solar energy reaching the edge of the earth’s atmosphere is called the solar constant: Gs = 1353W/m2
throughout
the year within +/- 3.4% variation. Oxygen absorbs narrow band of solar radiation λ = 0.76µm, ozone λ = 0.3−0.4µm.
Clear day solar radiation to the earth’s surface Gs = 950W/m2
in the wavelength range of λ = 0.3−2.5µm. The clear
sky has temperature of 230 K cold and 285 K warm, and considering the sky as black body, it’s fictitious radiation is
given by Gsky = σT4
s .
7.3 Radiation Heat Transfer Between Two Infinite Plates and Radiation Shields
Consider a quantity of radiation energy that is emitted from surface 1. And in column 3, With N heat shields present,
q falls by a factor of 1/N compared to no heat shield. Purpose of Heat Shields:is to reduce the amount of energy
transfer from (hotter) plate at T1 to second (cooler) plate at T3.
Scheme diag.
... ... ...
qRad, 1 to 2 q12 =
σT 4
1
1
ε1
+ 1
ε2
−1
q12 =
σ(T 4
1 −T 4
2
1
ε + 1
ε −1
... ... ...
qRad, A to B q21 =
σT 4
2
1
ε1
+ 1
ε2
−1
q23 =
σ(T 4
2 −T 4
3
1
ε + 1
ε −1
Net Radiation qnet,12 =
σ(T 4
1 −T 4
2
1
ε + 1
ε −1
qnet,1,viashieldto3 = 1
N+1
σ(T 4
1 −T 4
3
2
ε −1
... ... ...
7.4 View Factor
Radiation heat transfer between surfaces depends on the orientation of the surfaces relative to each other as well as
their radiation properties and temperatures. View factor (or shape factor) Fij is a purely geometrical parameter that
accounts for the effects of orientation on radiation between surfaces. In view factor calculations, we assume uniform
radiation in all directions throughout the surface, i.e., surfaces are isothermal and diffuse. It ranges from 0 up to 1,
26

0 ≤ Fij ≤ 1. Also the medium between two surfaces does not absorb, emit, or scatter radiation. Calculating view
factors between surfaces are usually very complex and difficult to perform.
7.4.1 View factor Relations
ˆ An enclosed N sided geometries require an N2
calculations of Fij
ˆ The reciprocals are equal, AiFij = AjFji
ˆ Sum = 1,
Pj=N
j=1 Fij = 1
ˆ Any two surfaces do not see each other such as opposite/back side of absorbing planes and convex surfaces have
Fij = 0
ˆ Two (or more) surfaces that possess symmetry about a third surface will have identical view factors from that
surface.
ˆ Geometries such as channels and ducts that are very long in one direction can be considered two-dimensional
(since radiation through end surfaces can be neglected). The view factor between their surfaces can be determined
by cross-string method developed by H. C. Hottel, as follows: Fij =
P
No. of Crossed strings from edges−
P
No. of uncrossed string
2×string on face
ˆ The view factor from a surface i to a surface j is equal to the sum of the view factors from surface i to the parts
of surface j. Fi,jk = Fi,j + Fi,k
ˆ The net radiation heat transfer Q̇ = Ebi−Ji
Ri =
Pj=N
j=1
Ji−Jj
Rij
where E is emitted radiation, J = ϵE + ρG is called
radiosity, sum of emitted and reflected radiations. Rij = 1
AiFij
is the resistance to radiation heat transfer.
Example 15 Determine the view factors F12 and F21 for the following geometries in Fig. 22: Solution
Figure 22: View Factor for Various Geometries
1. sphere in cube: 2. square duct: circular tube
By inspection, F12 = 1 summation r2/L = 0.5
Reciprocity summation F11 + F12 + F13 = 1 L/r1 = 2, F13 = 0.17
F21 =
A1
A2
F12 F21 =
A1
A2
F12 summation
πD2
6L2
× 1 =
π
6
F11 = 0, F11 = 0,
F21 + F22 = 1 → F22 =
5π
6
symetry F12 = F13 = 0.5 F12 = 1 − F13 = 0.83
=
√
2L
L
× 0.5 reciprocity, F21 =
A1
A2
F12F12
= 0.71 =
πD2
/4
πDL
× 0.83 = 0.21
1. Why does radiation heat transfer occur compared to conduction and convection?
2. Why is radiation heat transfer considered as surface phenomena for opaque substances like metals?
3. Describe Kircholf’s and Boltzman’s laws
4. A horizontal oxidized steel pipe carrying steam and having an OD of 0.1683m has a surface temperature of 374.9
K and is exposed to air at 297.1 K in a large enclosure. Calculate the heat loss for 0.305 m of pipe.
27

5. A horizontal oxidized steel pipe carrying steam and having an OD of 0.1683m has a surface temperature of 374.9
K and is exposed to air at 297.1 K in a large enclosure. Calculate the heat loss for 0.305 m of pipe.
6. Two parallel plates 0.5 by 1.0 m are spaced 0.5 m apart shown in the Fig 23. One plate is maintained at 1000o
C
and the other at 500o
C. The emissivities of the plates are 0.2 and 0.5, respectively. The plates are located in a
very large room, the walls of which are maintained at 27o
C. The plates exchange heat with each other and with
the room, but only the plate surfaces facing each other are to be considered in the analysis. Find the net heat
transfer rate to each plate and the room; neglect other modes of heat transfer, i.e., conduction and convection.
(a) (b)
Figure 23: a. Comb. Radiation-Conv Heat Transfer between Parallel Plates b. Equivalent Resistance Network Model
8 Heat Exchangers
Heat Exchanger: are used for heat transfer between two media, the media do not come into direct contact and there is
no mixing, heat is transport from the hot medium to cold medium by way of heat conducting partition. Heat Transfer
Equipments: are used for heat transfer between two media, the media do not come into direct contact and there is
no mixing, heat is transport from the hot medium to cold medium by way of heat conducting partition. Applications
of heat exchangers include the process, power, heat recovery, oil refinery, manufacturing industries, and the process
in the refining industry often require fluids to be heated or cooled with or without a change in phase during the
different operation. The fouling factor represents the theoretical resistance to heat flow due to a build-up of a layer
of dirt or other fouling substance on the tube surfaces of the heat exchanger. Some examples of heat exchangers are:
Car radiators [Media] (water/air), Oil coolers [Media] (oil/air or water), and Cooling Coils in refrigerators [Media]
(air/refrigerant).
There are many ways to increase the efficiency of heat exchangers
ˆ Remove the accumulated precipitated inside and outside the pipes.
ˆ Increase the surface area of contact between the two fluids by increase the number of pipes in (shell and tubes)
heat exchanger.
ˆ In (shell and tube) type we can install baffles to decrease the velocity of fluid in it.
8.1 Fouling Factor
During operation with liquids and gases a dirt film may build up on the heat exchanger surfaces . The deposit film is
referred to as fouling. Increased thermal resistance caused by the deposit can normally only be obtained from tests or
experience. The fouling factor can be determined as
Rd =
1
Ud
−
1
U
Ud =
1
Rd + (1/U)
(29)
where Rd is fouling factor - or unit thermal resistance of the deposit (m2
K/W), Ud, thermal conductance of heat
exchanger after fouling (W/m2K),U, thermal conductance of clean heat exchanger (W/m2
K). Typical Fouling Factors:
Alcohol vapors : Rd = 0.00009(m2
K/W), Boiler feed water, treated above 325 K : Rd = 0.0002(m2
K/W), Fuel oil :
Rd = 0.0009(m2
K/W, Industrial air : Rd = 0.0004(m2
K/W), Quenching oil : Rd = 0.0007(m2K/W), Refrigerating
liquid : Rd = 0.0002(m2
K/W), Seawater below 325 K : Rd = 0.00009(m2
K/W), Seawater above 325 K : Rd =
0.0002(m2K/W). The fouling factors to be used in the design of heat exchangers are normally specified by the client
based on his experience of running his plant or process to simulate dirt accumulation on the heat transfer surfaces,
but if these are not restricted to proper levels they can totally negate any benefits generated by skilful design. The
fouling factor represents the theoretical resistance to heat flow due to a build up of a layer of dirt or other fouling
substance on the tube surfaces of the heat exchanger but they are often overstated by the end user in an attempt to
28

minimise the frequency of cleaning. In reality they can, if badly chosen, lead to increased cleaning frequency. Fouling
mechanisms vary with the application but can be broadly classified into four common and readily identifiable types.
Common types of fouling: Chemical fouling; when chemical changes within the fluid cause a fouling layer to be
deposited onto the tube surface. A common example of this phenomenon is scaling in a kettle or boiler caused by
”hardness” salts depositing onto the heating elements as the solubility of the salts reduce with increasing temperature.
This is outside the control of the heat exchanger designer but can be minimised by careful control of the tube wall
temperature in contact with the fluid. When this type of fouling occurs it must be removed by either chemical treatment
or mechanical de-scaling processes (wire brushes or even drills to romove the scale or sometimes high pressure water
jets)
Biological fouling: this is caused by the growth of organisms within the fluid which deposit out onto the surfaces of
the heat exchanger. This is once again outside the direct control of the heat exchanger designer but it can be influenced
by the choice of materials as some, notably the non-ferrous brasses, are poisonous to some organisms. When this type
of fouling occurs it is normally removed by either chemical treatment or mechanical brushing processes. Deposition
fouling; this is when particles contained within the fluid settle out onto the surface when the fluid velocity falls below a
critical level. This is to a large extent within the control of the heat exchanger designer as the critical velocity for any
fluid/particle combination can be calculated to allow a design to be developed with minimum velocity levels higher
than the critical level. Mounting the heat exchanger vertically can also minimise the effect as gravity would tend to
pull the particles out of the heat exchanger away from the heat transfer surface even at low velocity levels. When this
type of fouling occurs it is normally removed by mechanical brushing processes.
Corrosion fouling: this is when a layer of corrosion products build up on the surfaces of the tube forming an extra
layer of, usually, high thermal resistance material. By careful choice of materials of construction the effects can be
minimised as a wide range of corrosion resistant materials based on stainless steel and other nickel based alloys are now
available to the heat exchanger manufacturer. One of the most misunderstood items in the heat exchanger schedule
is the fouling factor with different types of unit.
8.2 Types of Heat Exchangers
Heat exchangers classified in term of nature of used fluids which used for heating or cooling as: coolers- use water for
cooling, Condensers- Fluids circulate across the set-up and converted from vapour phase to liquid, heat exchangers- hot
fluid and cold fluid circulate around one-another where heat from hot fluid transferred to cold, and re-boilers: same as
heat exchanger but the heated fluid will change to gas phase (evaporated).
Types of heat exchangers depending on its design configuration: a. Double pipe heat exchange b. Shell and tube heat
exchanger. and c. Plate heat exchanger.
8.2.1 Double pipe – Parallel flow heat exchanger
It consists of central tube container within a larger tube, it is relativity cheap, flexible and hence, used in smaller
units, it is customary to operate with high pressure, high temperature and high density. Fins are used in double pipe
heat exchanger in the figure 24 to increase the efficiency of exchanger and increase the surface area of heat exchanger
tubing. Parallel flow heat exchangers offer the following advantages
(a) (b)
Figure 24: a. Double pipe Heat Exchanger b. Parallel Counter Flow
(a) The configuration gives a large surface area in small volume.
29

(b) Good mechanical layout (a good shape for pressure operation).
(c) Uses well- Established fabrication techniques.
(d) Can be constructed from a wide range of materials.
(e) Easily cleaned.
(f) Well- Established design procedures.
8.2.2 Shell and Tube Heat Exchangers
A type of heat exchanger widely used in chemical process industries is that of (shell tube) arrangement shown in
the figure below, one fluid flows on the inside of tubes while the other fluid is forced through the shell and over the
outside of tubes, to ensure that the shell side fluid will flow across the tubes and thus in duce higher heat transfer
“Baffles” are placed in the shown figure.
Baffles can be defined as metal plates cutted from one side 25% and drilled many holes where the pipes in the heat
exchangers put through it, these baffles arranged in a particular position and the distance between them are similar.
Aim of using Baffles: To ensure that the shell side fluid will flow across the tubes by reduce fluid velocity so that make
higher heat transfer and to maintain the fluid in the shell as long as possible time to increase the efficiency of the heat
exchanger.
(a) (b)
Figure 25: Shell Tube - Counter flow heat exchanger a. single double Shell b. baffles flow configuration
8.2.3 Plate Heat Exchangers
A plate hate exchanger consists of stack of closely spaced thin metal plates, clamped together in a frame, a thin
gasket seals the plates round their edges, the gap between the plates is normally between about (3 and 6) mm, corner
ports in plates direct the flow from plate to plate, the basic flow arrangements are shown in the figure below, various
combinations of these arrangements are used, plate are available in wide range of materials including (stainless steel,
plastics and asbestos) based gasket materials are used.
Heat transfer coefficients are generally higher in plate heat exchanger than shell tube exchangers and the units are
more compact, plate are available with effective areas from (0.03 to 1.3) m2
, and up to 400 plates can be contained in
a large frame, a plate is not good shape to resist pressure, and the maximum operating pressure is limited to about
(20 bar), the operating temperature is limited by performance of the available gasket materials to about (250 C), plate
heat exchangers are used extensively in food industries, we have two type of flow for plate.
8.3 Heat Exchanger Analysis
All Double pipe, plate and shell tube based on convection heat transfer the general equation 25 with correction
factor f = t2−t1
T2−T1
where 0.35 ≤ f ≤ 1 shown here again in Eq. 30where t and t are temperatures of tube and shell
respectively while 1 and 2 are inlet and exit: [11]
∆Tln =
Tout − Tin
ln TW −Tout
TW −Tin
=
Tout − Tin
ln(∆Tout/∆Tin)
where Q̇ = fhA∆Tln = ṁCP ∆T
A = πDL for Parallel Pipe
A = NπDL for Shell Tube
N is No. of Pipes
(30)
30

Figure 26: compact Plate Heat Exchanger
The effectiveness ϵ from Eq. 8 of can be written in terms of heat capacitance rate [W/K], C, and change in temperature
[K]. The heat capacitance rate is defined in terms of mass flow rate [kg/s], , and specific heat [kJ/(kgK)]. The Number
of Transfer Units (NTU) Method is used to calculate the rate of heat transfer in heat exchangers. Refer [1] for details
in NTU method for all types of heat exchangers.
εparallel flow =
Q̇Actual
Q̇Max.Possibe
=
1 − exp[− UAs
Cmin
(1 + Cmin
CMax
)
1 + Cmin
Cmax
ε = f(UA/Cmin, Ctextmin/Cmax) = f(NTU, c)
(31)
Example 16 Water at the rate of (68 Kg/min) is heated from (35to75o
C by an oil having a specific heat of
(1.9KJ/Kg.o
C), the fluids are used counter flow double pipe heat exchanger and the oil enters the exchanger at
(110o
C) and leave at (75o
C), the overall heat transfer coefficient is (320w/m2
.o
C), calculate the heat transfer area?
Solution Q̇ = ṁCP ∆T = 68 × 1.9 × (75 − 35) = 86.1kW and log mean temperature
∆Tlm =
Tout − Tin
ln(∆Tout/∆Tin)
=
Tout − Tin
ln TW −Tout
TW −Tin
=
(110 − 75) − (75 − 35)
ln 110−75
75−35
= 37.44o
C
A =
Q
∆Tlm
=
86100
320 × 37.44
= 7.186m2
| {z }
Answer
1. Distinguish between double pipe, shell tube and plate type heat exchanger construction with an aid of schematic
sketches
2. write log mean temperature expression for convection heat transfer analysis described in section 6.2.1 and section
8.3 in equation 30 with an aid of temprature vs L sketch
3. What is the effectiveness of a counter-flow heat exchanger that has a heat capacitance UA value of 24 kW/K
if the respective mass rates of flow and specific heats of the two fluids are 10 kg/s, 2 kJ/(kgK) and 4 kg/s, 4
kJ/(kgK)?
4. In a processing plant a material must be heated from 20 to 80o
C in order for the desired reaction to proceed,
whereupon the material is cooled in a regenerative heat exchanger, as shown in the figure below. The specific
heat of the material before and after the reaction is 3.0 kJ/ (kgK). If the UA of this counter-flow regenerative
heat exchanger is 2.1 kW/K and the flow rate is 1.2 kg/s, what is the temperature T leaving the heat exchanger?
9 Condensation and Boiling Heat Transfer
This last section aimed to differentiate between evaporation and boiling, and gain familiarity with different types of
boiling, develop an understanding of the boiling curve, and the different boiling regimes corresponding to different
regions of the boiling curve, calculate the heat flux and its critical value associated with nucleate boiling A − C in
Fig 27 [1], and examine the methods of boiling heat transfer enhancement, and derive a relation for the heat transfer
coefficient in laminar film condensation over a vertical plate. Furthermore to calculate the heat flux associated with
condensation on inclined and horizontal plates, vertical and horizontal cylinders or spheres, and tube bundles,and
examine dropwise condensation and understand the uncertainties associated with them.
Boiling and condensation are vital links in the heat transfer of heat from a hot to a colder region in countless
applications, e.g., thermal and nuclear power generation in steam plants, refrigeration, refining, heat transmission,
etc.
31

9.1 Boiling
The production of vapour bubbles during boiling gives rise to presence of surface tension. The vapour inside a bubble
must be at a higher pressure then the surrounding liquid. The pressure difference increases as the diameter of the
bubble decreases, and is insignificant when the bubble is large. However, when the bubble is minute, appreciable
pressure difference exists.
9.1.1 Pool Boiling
Figure 27: Pool Boiling with different forms depending on the temperature ∆Excess = Ts − Tsat
Pool boiling refers to boiling with insignificant fluid velocity can be categorized in two: sub-cooled: Bubbles form
on the heating surface but on release from the surface are absorbed by the mass of the liquid and saturated. When the
temperature of the main body of the liquid is below the saturation temperature during sub-cooled boiling. When the
temperature of the liquid is equal to the saturation temperature.
In pool boiling, in absence of bulk flow, Any motion of the fluid is due to natural convection currents and the motion
of the bubbles under the influence of buoyancy. Boiling is called flow boiling in the presence of bulk fluid flow. In flow
boiling, the fluid is forced to move in a heated pipe or over a surface by external means such as a pump.
Evaporation occurs at the liquid–vapor interface when the vapour pressure is less than the saturation pressure of
the liquid at a given temperature. Whereas Boiling occurs at the solid–liquid interface when a liquid is brought into
contact with a surface maintained at a temperature sufficiently above the saturation temperature of the liquid.
At Point A, Bubbles do not form on the heating surface until the liquid is heated a few degrees above the saturation
temperature. the liquid is slightly superheated in this case (metastable state). The fluid motion in this mode of
boiling is governed by natural convection currents. Heat transfer from the heating surface to the fluid is by natural
convection. In Point A − C, the bubbles form at an increasing rate at an increasing number of nucleation sites as
we move along the boiling curve toward point C. Region A–B :isolated bubbles; Region B–C : numerous continuous
columns of vapor in the liquid.After point B the heat flux increases at a lower rate with increasing DeltaTexcess, and
reaches a maximum at point C called critical heat flux. DeltaTexcess is increased past point C, the heat flux decreases
because a large fraction of the heater surface is covered by a vapor film, In the transition boiling regime, both nucleate
and film boiling partially occur. Beyond Point D the heater surface is completely covered by a continuous stable vapor
film. Point D, where the heat flux reaches a minimum is called the Leidenfrost point.
9.1.2 Pool Boiling Correlation
9.1.3 External and Internal Forced Convection Boiling (Flow Boiling)
In flow boiling, the fluid is forced to move by an externalsource such as a pump as it undergoes a phase-change process.
Flow boiling is classified as either external or internal flow boiling.
Typical flow regimes:
32

Table 11: Boiling Correlation Equations
Region Heat Transfer Correlation Note
Nucleate Boiling q”s = µlhfg
h
g(ρl−ρv)
σ
i1/2 h
CP,l∆Te
Cs,f hfgP rn
l
i3
Rohsenow 1952 Any Geometry
Critical Heat Flux CHF q”max = Chfgρv
h
σ g(ρl−ρv)
ρ2
v
i1/4
S. S. Kutateladze 1948 for any, µ, k, CP
Minimum Heat Flux MHF q”min = 0.09hfgρv
h
σg(ρl−ρv)
ρ2
v
i1/4
Zuber drived for Leidenfrost point
Film Boiling NuD = h̄D
kv
= C
h
g(ρl−ρv)h′
fg
νvkv(Ts−Tsat)
i1/4
Cylinder Sphere diam. D
above 300o
C,Radiation Significant
h̄rad =
εσ(T 4
s −T 4
sat)
Ts−Tsat
h̄4/3
= h̄
4/3
conv + h̄radh̄1/3
Enhancement of Heat Transf. Depend on Heating surface and at Nucleate Pt
↑ h bubble formation rate surface irregularities and roughness
10 × h ↑ 3 × h ↑ Factors achieved by enhancement during nucleate and CHF resp.
Types External Internal
Flow Regime
Description
In flow boiling, the fluid is forced to move by
an external source such as a pump as it
undergoes a phase change process. The boiling
in this case exhibits the combined effects
of convection pool boiling. The
nucleate boil. heat flux CHF ↑ with vel.
The two phase flow in a tube exhibits different
flow boiling regimes, depending on the relative
amounts of the liquid and the vapor phases.
Typical flow regimes: Liquid single-phase flow,
Bubbly flow, Slug flow, Annular flow,
Mist flow, Vapor single phase flow.
1. Liquid single-phase flow: In the inlet region the liquid is subcooled and heat transfer to the liquid is by forced
convection (assuming no subcooled boiling).
2. Bubbly flow: Individual bubbles and Low mass qualities
3. Slug flow: Bubbles coalesce into slugs of vapor. and Moderate mass qualities
4. Annular flow: Core of the flow consists of vapor only, and liquid adjacent to the walls. and Very high heat
transfer coefficients
5. Mist flow: a sharp decrease in the heat transfer coefficient
6. Vapor single-phase flow: The liquid phase is completely evaporated and vapor is superheated
9.2 Condensation
Condensation occurs when the temperature of a vapor is reduced below its saturation temperature. Only condensation
on solid surfaces is considered in this section. Two forms of condensation: film condensation and dropwise condensa-
33

tion.
In film condensation, the condensate wets the surface and forms a liquid film. The surface is blanketed by a liquid
film which serves as a resistance to heat transfer. liquid film starts forming at the top of the plate and flows downward
under the influence of gravity. The hydrodynamics of film condensation is analysed using Newton’s second law and
heat transfer assumed to be only conduction using Fourier law. The condensate Weight adds up Viscous shear force
and Buoyancy force.
In droplet condensation, the most effective mechanisms of heat transfer, and extremely large heat transfer coeffi-
cients can be achieved. Small droplets grow as a result of continued condensation, coalesce into large droplets, and
slide down when they reach a certain size. Large heat transfer coefficients enable designers to achieve a specified heat
transfer rate with a smaller surface area. The challenge in dropwise condensation is to sustain it for prolonged periods
of time. Dropwise condensation has been studied experimentally for a number of surface fluid combinations. The
waves at the liquid vapor interface tend to increase heat transfer. Film condensation model ASSUMPTIONS: 1. Both
the plate and the vapor are maintained at constant temperatures of Ts and Tsat, respectively, and the temperature
across the liquid film varies linearly. 2. Heat transfer across the liquid film is by pure conduction. 3. The velocity of
the vapor is low (or zero) so that it exerts no drag on the condensate (no viscous shear on the liquid vapor interface).
4. The flow of the condensate is laminar (Re 30) and the properties of the liquid are constant. 5. The acceleration
of the condensate layer is negligible.
(a) (b)
Figure 28: Film Condensation Vert. Wall Model: a. Hydrodynamics b. Flow Regime
du
dy
=
g(ρl − ρg)(δ − y)
µl
HEAT TRANSFER
obtained from writing Newton’s law in dq = hfg
˙
dm = k(bdx)
Tsat − Ts
δ
terms of viscous shear force and buoyance force ˙
qx = hx(Tsat − Ts) = kx
Tsat − Ts
δ
u(y) =
g(ρl − ρg)δ2
µl

y
δ
−
1
2
y
δ
2

hx =
kx
δ
mass flow rate ṁ =
gρl(ρl − ρv)δ3
3µl
, Nu =
¯
hLL
kl
→ h̄ =
4
3
hx=L δ =

4µlkl(Tsat − Ts)hfgx
gρl(ρl − ρv)
1/4
NuL = 0.943

gρl(ρl − ρv)hfgL3
µl(Tsat − Ts)
1/4
For turbulent flows, Nu is f(Re) as many empirical relations established for different flow regimes. A single smooth
tube or smooth sphere given in Fiq 29can have correlation shown in Eq. 32,
NuD =
¯
hDD
kl
= C

gρl(ρl − ρv)hfgL3
µl(Tsat − Ts)
1/4
(32)
34

lecture notes heat transfer.pdf

lecture notes heat transfer.pdf

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to lecture notes heat transfer.pdf

Similar to lecture notes heat transfer.pdf (20)

Recently uploaded

Recently uploaded (20)

lecture notes heat transfer.pdf