CH1: Introduction and Basic Concepts
Eng. Ismail Salim
Renewable Energy Engineering Department
Middle East University
Course: 0905325 – Heat Transfer
1
Objectives
 Understand how thermodynamics and heat transfer are
related.
 Distinguish thermal energy from other forms of energy, and
heat transfer from other forms of energy transfer.
 Perform general energy balances as well as surface
energy balances.
2
Objectives
 Understand the basic mechanisms of heat transfer
1. Conduction, Fourier's law of heat conduction.
2. Convection, Newton's law of cooling.
3. Radiation, Stefan–Boltzmann law of radiation.
3
Thermodynamics and Heat Transfer
 The science of thermodynamics deals with the amount of
heat transfer as a system undergoes a process from one
equilibrium state to another, and makes no reference to
how long the process will take.
 The science of heat transfer deals with the determination of
the rates of energy that can be transferred from one
system to another as a result of temperature difference.
4
Thermodynamics and Heat Transfer
 Thermodynamics deals with equilibrium states and
changes from one equilibrium state to another.
 Heat transfer, on the other hand, deals with systems that
lack thermal equilibrium, and thus it is a non equilibrium
phenomenon.
 Therefore, the study of heat transfer cannot be based on
the principles of thermodynamics alone. However, the laws
of thermodynamics lay the framework for the science of
heat transfer.
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Heat Transfer
 The basic requirement for heat transfer is the
presence of a temperature difference.
 The second law requires that heat be
transferred in the direction of decreasing temperature.
 The temperature difference is the driving force for heat transfer.
 The rate of heat transfer in a certain direction depends on the
magnitude of the temperature gradient in that direction.
 The larger the temperature gradient, the higher the rate of heat
transfer.
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Application Areas of Heat Transfer
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Heat and Other Forms of Energy
 Energy can exist in numerous forms such as:
 Thermal
 Kinetic
 Potential
 Electrical
 Chemical
 Nuclear.
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Heat and Other Forms of Energy
 Their sum constitutes the total energy E (or e on a unit
mass basis) of a system.
 The kinetic energy of the molecules is called sensible heat.
 The internal energy associated with the phase change of a
system is called latent heat.
9
Specific Heats of Gases,
Liquids, and Solids
 Specific heat is defined as the energy required to raise the
temperature of a unit mass of a substance by one degree.
 Two kinds of specific heats:
 Specific heat at constant volume Cv, and
 Specific heat at constant pressure Cp.
 The specific heats of a substance, in general, depend on
two independent properties such as temperature and
pressure.
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Specific Heats
 A substance whose specific volume (or density) does not
change with temperature or pressure is called an
incompressible substance.
 The constant-volume and constant-pressure specific heats
are identical for incompressible substances (e.g. solid).
 The specific heats of incompressible substances depend
on temperature only.
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Energy Balance
Closed systems:
 Stationary closed system, no
work, the heat transfer is:
𝑄 = 𝑚𝑐𝑣 Δ𝑇
(𝐽)
12
Energy Balance
Steady-Flow Systems:
 For system with one inlet and one exit, the mass flow rate:
𝑚𝑖𝑛 = 𝑚𝑜𝑢𝑡 = 𝑚
𝑚 = 𝜌𝑉 = 𝜌𝜈𝐴
(𝑘𝑔/𝑠)
 𝑉 is the Volume flow rate (m3/s)
13
Energy Balance
Steady-Flow Systems:
 When kinetic and potential energies are negligible the heat
transfer is:
𝑄 = 𝑚Δ𝑕 = 𝑚𝑐𝑃 Δ𝑇
(𝑊)
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Energy Transfer
 Energy can be transferred to or from a given mass by two
mechanisms:
 Heat transfer (Q).
 Work (W).
 The amount of heat transferred per unit time is called heat
transfer rate, and is denoted by 𝑄 .
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Energy Transfer
 The total amount of heat transfer Q during a time interval
Δt can be determined from:
 The rate of heat transfer per unit area normal to the
direction of heat transfer is called heat flux, and the
average heat flux is expressed as:
16
Energy Transfer
 Heat flux is heat transfer per
unit area, and is equal to
𝑄
24 𝑊
2
𝑞= =
=
4
𝑊
𝑚
𝐴
6 𝑚2
when 𝑄 is uniform over the
area A.
17
EXAMPLE
 A 10-cm diameter copper ball is to be heated from 100°C
to an average temperature of 150°C in 30 minutes.
(ρ=8950 kg/m3 and Cp= 0.395 kJ/kg·°C), Determine:
 (a) the total amount of heat transfer to the copper ball.
 (b) the average rate of heat transfer to the ball.
 (c) the average heat flux.
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EXAMPLE…Solution
 (a) the total amount of heat transfer to the copper ball.
𝑄 = 𝑚𝑐𝑝 (𝑇2 − 𝑇1 )
𝜋 3
𝑘𝑔 𝜋
𝑚 = 𝜌𝑉 = 𝜌 𝐷 = 8950 3
0.1𝑚
6
𝑚 6
𝑄 = 4.69𝑘𝑔
𝑘𝐽
0.395
𝑘𝑔. 𝑜𝐶
3
= 4.69 𝑘𝑔
150 − 100 𝑜𝐶 = 𝟗𝟐. 𝟔 𝒌𝑱
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EXAMPLE…Solution
 (b) the average rate of heat transfer to the ball.
𝑄
92.6 𝑘𝐽
𝑄=
=
= 𝟓𝟏. 𝟒 𝑾
Δ𝑡
(30 ∗ 60) 𝑠
 (c) the average heat flux.
𝑄
𝑄
51.4 𝑊
𝟐
𝑞= =
=
=
𝟏𝟔𝟑𝟔
𝑾
𝒎
𝐴 𝜋𝐷2
𝜋 0.1𝑚 2
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The First Law of Thermodynamics
 The first law of thermodynamics states that energy can neither be
created nor destroyed during a process; it can only change forms.
𝑇𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦
𝑒𝑛𝑡𝑒𝑟𝑖𝑛𝑔 𝑡𝑕𝑒
𝑠𝑦𝑠𝑡𝑒𝑚
𝑇𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦
− 𝑙𝑒𝑎𝑣𝑖𝑛𝑔 𝑡𝑕𝑒
𝑠𝑦𝑠𝑡𝑒𝑚
𝐶𝑕𝑎𝑛𝑔𝑒 𝑖𝑛 𝑡𝑕𝑒
= 𝑡𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑓
𝑡𝑕𝑒 𝑠𝑦𝑠𝑡𝑒𝑚
 The energy balance for any system can be expressed as (rate form)
 𝐸𝑖𝑛 − 𝐸𝑜𝑢𝑡

𝑑𝐸𝑠𝑦𝑠𝑡𝑒𝑚
𝑑𝑡
𝑑𝐸𝑠𝑦𝑠𝑡𝑒𝑚
𝐸𝑖𝑛 − 𝐸𝑜𝑢𝑡 =
(𝑊)
𝑑𝑡
: Rate of net energy transfer by heat, work, and mass.
: Rate of change in internal (kinetic, potential, etc.) energies.
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Energy Balance
 In heat transfer problems it is convenient to write a heat
balance and to treat the conversion of nuclear, chemical,
mechanical, and electrical energies into thermal energy as
heat generation.
 The energy balance in that case can be expressed as
𝑄𝑖𝑛 − 𝑄𝑜𝑢𝑡 + 𝐸𝑔𝑒𝑛 = Δ𝐸𝑇𝑕𝑒𝑟𝑚𝑎𝑙,𝑆𝑦𝑠𝑡𝑒𝑚 (𝐽)
 𝑄𝑖𝑛 − 𝑄𝑜𝑢𝑡 : Net heat Transfer.
 𝐸𝑔𝑒𝑛 : Heat generation.
 Δ𝐸𝑇𝑕𝑒𝑟𝑚𝑎𝑙,𝑆𝑦𝑠𝑡𝑒𝑚 : Change in thermal energy of the system22
Energy Balance
Surface Energy Balance:
 A surface contains no volume
or mass, and thus no energy.
𝐸𝑖𝑛 = 𝐸𝑜𝑢𝑡
 The energy balance for the
outer surface of the wall
𝑄1 = 𝑄2 + 𝑄3
23
Heat Transfer Mechanisms
 Heat can be transferred in three basic modes:
 Conduction
 Convection
 Radiation.
 All modes of heat transfer require the existence of a
temperature difference.
 All modes are from the high-temperature medium to a
lower-temperature one.
24
Conduction
 Conduction is the transfer of energy from the more
energetic particles of a substance to the adjacent less
energetic ones as a result of interactions between the
particles.
 Conduction can take place in solids, liquids, or gases
25
Conduction
 In gases and liquids
conduction is due to the
collisions and diffusion of the
molecules during their random
motion.
 In solids conduction is due to
the combination of vibrations of
the molecules in a lattice and
the energy transport by free
electrons.
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Conduction
 𝐶𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑜𝑛 ∝
𝑄𝑐𝑜𝑛𝑑
(Area)(Temperature difference)
(𝑇𝑕𝑖𝑐𝑘𝑛𝑒𝑠𝑠)
𝑇1 − 𝑇2
Δ𝑇
= 𝑘𝐴
= − 𝑘𝐴
Δ𝑥
Δ𝑥
(𝑊)
 Where the constant of proportionality k is
the thermal conductivity of the material.
 In differential form:
𝑄𝑐𝑜𝑛𝑑
𝑑𝑇
= − 𝑘𝐴
𝑑𝑥
 Which is called Fourier’s law of heat
conduction.
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Thermal Conductivity
 The thermal conductivity of a material is a measure of the
ability of the material to conduct heat.
 High value  good heat conductor
 Low value  poor heat conductor or insulator.
 Pure crystals and metals have the highest thermal
conductivities, and gases and insulating materials the
lowest
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Thermal Conductivities of Materials
29
Thermal Conductivities and
Temperature
 The thermal conductivities of materials vary with
temperature.
 The temperature dependence of thermal conductivity
causes considerable complexity in conduction analysis.
 Therefore, it is common practice to evaluate the thermal
conductivity k at the average temperature and treat it as a
constant in calculations.
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Thermal Conductivities and
Temperature
 The variation of the
thermal
conductivity with
temperature
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Thermal diffusivity
𝐻𝑒𝑎𝑡 𝐶𝑜𝑛𝑑𝑢𝑐𝑡𝑒𝑑
𝑘
𝛼=
=
𝐻𝑒𝑎𝑡 𝑆𝑡𝑜𝑟𝑒𝑑
𝜌𝑐
𝑚2
( )
𝑠
 The thermal diffusivity represents heat how fast diffuses
through a material, it appears in the transient heat
conduction analysis.
 A material that has a high thermal conductivity or a low
heat capacity will have a large thermal diffusivity.
 The larger the thermal diffusivity, the faster the propagation
of heat into the medium.
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Example
 An aluminum pan whose thermal conductivity is 237
W/m.°C has a flat bottom with diameter 20 cm and
thickness 0.4 cm. Heat is transferred steadily to boiling
water in the pan through its bottom at a rate of 800 W. If
the inner surface of the bottom of the pan is at 105°C.
Determine the temperature of the outer surface of the
bottom of the pan.
33
Example…Solution
𝐴 = 𝜋𝑟 2 = 𝜋 0.1𝑚
𝑄𝑐𝑜𝑛𝑑
2
= 0.0314 𝑚2
Δ𝑇
𝑇2 − 𝑇1
= − 𝑘𝐴
= − 𝑘𝐴
Δ𝑥
Δ𝑥
𝑊
2
800 𝑊 = −237
0.0314
𝑚
𝑚. 𝑜𝐶
105 − 𝑇1 𝑜𝐶
0.004 𝑚
 T1 = 105.43 oC
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Convection
 Convection = Conduction + Advection (fluid motion)
 Convection is the mode of energy transfer between a solid
surface and the adjacent liquid or gas that is in motion.
 Convection is commonly classified into three sub-modes:
 Forced convection,
 Natural (or free) convection,
 Change of phase (liquid/vapor,
solid/liquid)
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Convection
 The rate of convection heat transfer is expressed by
Newton’s law of cooling as:
𝑄𝑐𝑜𝑛𝑣 = 𝑕𝐴𝑆 (𝑇𝑠 − 𝑇∞ )
(𝑊)
 h is the convection heat transfer coefficient in W/m2°C.
 h depends on variables such as the surface geometry, the
nature of fluid motion, the properties of the fluid, and the
fluid velocity.
36
Convection
37
Heat Transfer Involve
Change of Phase
 They are considered to be convection because of the fluid
motion induced through the process.
 Evaporation:
𝑄 = 𝑚𝑕𝑓𝑔
𝐽
 hfg: latent heat of vaporization, Table A-2 (for water 2257
kJ/kg)
 Freezing:
𝑄 = 𝑚𝑕𝑖𝑓
𝐽
 hif: latent heat of fusion, Table A-2 (for water 333.7 kJ/kg)
38
Radiation
 Radiation is the energy emitted by matter in the form of
electromagnetic waves (or photons) as a result of the
changes in the electronic configurations of the atoms or
molecules.
 Heat transfer by radiation does not require the presence of
an intervening medium.
 In heat transfer studies we are interested in thermal
radiation (radiation emitted by bodies because of their
temperature).
39
Radiation
 Radiation is a volumetric phenomenon. However, radiation
is usually considered to be a surface phenomenon for
solids that are opaque to thermal radiation.
 The maximum rate of radiation that can be emitted from a
surface at a temperature Ts (in K) is given by the Stefan–
Boltzmann law as:
𝑄𝑒𝑚𝑖𝑡,𝑚𝑎𝑥 = 𝜍𝐴𝑆 𝑇𝑠4
(𝑊)
 σ =5.67X10-8 W/m2·K4 is the Stefan Boltzmann constant
40
Radiation
 The idealized surface that emits radiation at this maximum
rate is called a blackbody.
 The radiation emitted by all real surfaces is less than the
radiation emitted by a blackbody at the same temperature,
and is expressed as:
𝑄𝑒𝑚𝑖𝑡 = 𝜀𝜍𝐴𝑆 𝑇𝑠4
(𝑊)
 𝜀 is the emissivity of the surface (0 ≤ 𝜀 ≤ 1).
41
Radiation
 The fraction of the radiation energy incident on a surface
that is absorbed by the surface is termed the absorptivity
𝛼. (0 ≤ 𝛼 ≤ 1).
 Both 𝜀 and 𝛼 of a surface
depend on the temperature
and the wavelength of the
radiation.
42
Radiation between two surfaces
 When a surface of emissivity and surface area As at an
absolute temperature Ts is completely enclosed by a
much larger surface at absolute temperature Tsurr
separated by a gas (such as air) that does not intervene
with radiation, the net rate of radiation heat transfer
between these two surfaces is given by:
4
𝑄𝑅𝑎𝑑 = 𝜀𝜍𝐴𝑆 (𝑇𝑠4 − 𝑇𝑠𝑢𝑟𝑟
)
(𝑊)
43
Radiation between two surfaces
4
𝑄𝑅𝑎𝑑 = 𝜀𝜍𝐴𝑆 (𝑇𝑠4 − 𝑇𝑠𝑢𝑟𝑟
)
(𝑊)
44
Radiation between two surfaces
 For simplicity and convenience, by defining a combined
heat transfer coefficient hcombined that includes the effects
of both convection and radiation.
 Then the total heat transfer rate to or from a surface by
convection and radiation is expressed as
𝑄𝑡𝑜𝑡 = 𝑕𝑐𝑜𝑚𝑏𝑖𝑛𝑒𝑑 𝐴𝑆 (𝑇𝑠 − 𝑇∞ )
(𝑊)
45
Simultaneous Heat Transfer
Mechanisms
 Although there are three mechanisms of heat transfer, a
medium may involve only two of them simultaneously.
46
Simultaneous Heat Transfer
Mechanisms
47
Example
 Consider a person standing in a room at 23 oC. Determine
the total rate of heat transfer from this person if the
exposed surface area and the skin temperature of the
person are 1.7 m2 and 32 oC, respectively, and the
convection heat transfer coefficient is 5 W/m2.oC. Take the
emissivity of the skin and clothes to be 0.9, and assume
the temperature of the inner surfaces of the room to be as
the air temperature.
48
Example…Solution
 𝑄𝑐𝑜𝑛𝑣 = 𝑕𝐴𝑆 𝑇𝑠 − 𝑇∞ = 5
𝑊
𝑚2.𝑜𝐶
1.7 𝑚2 32 − 23 𝑜𝐶
𝑄𝑐𝑜𝑛𝑣 = 𝟕𝟔. 𝟓 𝑾
4
 𝑄𝑅𝑎𝑑 = 𝜀𝜍𝐴𝑆 𝑇𝑠4 − 𝑇𝑠𝑢𝑟𝑟
−8
= 0.9 5.67𝑥10
𝑊
𝑚2 𝐾 4
1.7 𝑚2 [ (32 + 273)4 − 23 + 273 4 ]
𝑄𝑅𝑎𝑑 = 𝟖𝟒. 𝟖 𝑾
 𝑄𝑡𝑜𝑡 = 𝑄𝑐𝑜𝑛𝑣 + 𝑄𝑅𝑎𝑑 = 𝟏𝟔𝟏. 𝟑 𝑾
49
End of Chapter
50