CHAPTER 1 INTRODUCTION 1.1 ROLE OF HEAT TRANSFER...
Transcript of CHAPTER 1 INTRODUCTION 1.1 ROLE OF HEAT TRANSFER...
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CHAPTER 1
INTRODUCTION
1.1 ROLE OF HEAT TRANSFER IN ENGINEERING
Heat transfer is one of the most important subjects for mechanical engineers,
as it occurs in a large number of engineering applications and by several
mechanisms.
Application of heat transfer :
a. Domestic application
Baking oven, bread toaster, boiling water.
b. Energy production
Steam power plant, gas turbine, and furnace.
c. Cooling of electronic equipment
Removal of heat that is generated due to flow as electric current
in electronic equipment.
d. Manufacturing / materials processing
Welding, brazing, soldering, foundry, machining, grinding.
e. Automobiles
Engines are cooled by radiator and passengers are cooled by
air condition.
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ii. Some industrial applications of heat transfer :
a. Petroleum refining
Preheating of crude oil and fractional distillation
Evaporation of the various hydrocarbons
b. Sugar industry
Steam generation in boiler
Concentration of sugar cane juice (heating by steam)
c. Paper industry
Steam generation
Soaking of wood logs
Drying of paper on steam heated drums
d. Iron making
Smelting of core
Solidification of casting
e. Tire industry
Steam generation
Vulcanizing of rubber
Forming of tire in moulds
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Example 1.1
The walls of furnace are covered with insulation. Heat is lost by convection from the
outer surface of insulation. The ambient is at 27 0C
The furnace is fired by natural gas having a calorific value of 30 MJ/m3 and a density
of 0.6 kg/m3. If gas costs RM1.00 /kg, calculate the cost of gas per kWh of heat loss
through the wall.
Solution:
1 kg natural gas ― RM 1
1 kWh ― RM ?
Assumptions:
1. Natural gas is assumed to be an ideal gas
2. There is no infiltration through the wall of furnace
Find heat released by burning 1kg of natural gas
Q = Cv = 30 x 103 kJ/m3
ρ 0.6 kg/m3
= 50000 kJ/kg
This means that 1kg of natural gas produces 50000 kJ of heat.
Conversion factors 1 kWh = 3600 kJ
Cv = 30 MJ/m3
ρ = 0.6 kg/m3
T∞ = 27 0C
Q
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Heat transfer rate, by burning 1kg of natural gas
= Q = 50000 kJ
t = 3600 h
= 13.89 kWh
Amount of natural gas used to release 1kWh heat
1kg 13.89 kWh
? 1 kWh
m = 1 kWh x 1kg
13.89 kWh
= 0.072 kg
Costs of natural gas per kWh
1kg RM 1.00
0.072 kg ?
Costs of natural gas per kWh = 0.072 x RM 1.00
1kg
= RM 0.072
≈ 7 cents
Q
Q
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1.2 RELATIONSHIP BETWEEN THERMODYNAMICS AND HEAT TRANSFER
1. Heat is the form of energy that can be transferred from one system to another
as a result of temperature difference.
2. Amount of heat, Q [J] can be determined by means of thermodynamic
analysis but no indication about how long the process will take.
Example 1.2
The initial conditions of gas are as follow:
Mass, m = 3.4 kg, volume, V = 0.92 m3 dan temperature, T = 17 0C. Gas is heated to
the temperarature of 147 0C. Calculate the amount of heat needed to heat up the gas
if the volume of the gas is constant.
Specific heat at constant volume, Cv = 0.72 kJ/kg.K
Solution:
Q = mCv [T2 – T1]
= 3.4 (0.72) [147 - 17]
= 318.24 kJ Only gives the idea of amount
heat being transferred.
318.24 kJ of heat needed to heat up the gas
Gas
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Example 1.3
The initial conditions of gas are as follow:
Pressure, P = 293 kN/m², volume, V = 0.082 m³, temperature, T = 1800C. Gas is cooled
at constant pressure to the temperature of 24oC. By assuming R = 0.287 kJ/kg.K dan
Cp = 1.005 kJ/kg.K, calculate amount of heat needed.
Solution:
PV = mRT
293 [0.082] = m (0.287) (180 + 273)
m = 0.185 kg
Q = mCp [T2 –T1]
= 0.185 [1.005] [24 - 180]
= -29kJ
However, in practice we are more concern about the rate of heat transfer [heat transfer
per unit time, J/s], than the amount of heat itself.
29 kJ out from the gas
Gas
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1.2.1 First Law Thermodynamic
1. Relate to conservation of energy.
2. What’s stated in First Law Thermodynamic?
“The rate of energy transfer into a system be equal to the rate of the
energy of that system”
Or
“Jumlah tenaga yang dibekalkan kepada sistem oleh sekeliling adalah
sama dengan jumlah tenaga yang dilakukan oleh sistem ke atas
sekeliling.”
1.2.2 Second Law of Thermodynamic
1. Related to the direction of heat transfer.
2. Heat is being transferred in the direction of decreasing temperature.
3. The basic requirement for heat transfer is the presence of a temperature
difference.
There can be no heat transfer between two bodies that are at the same
temperature.
4. Temperature difference is the driving force for heat transfer.
∑Q = -Ve ∑W= +Ve
System
Boundary
Environment ∑Q = +Ve ∑W= -Ve
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5. The rate of heat transfer depends on the magnitude of the temperature
gradient, [temperature difference per unit length]
Q
6. From 2nd Law of Thermodynamic,
“Heat can flow by itself only from a higher temperature to a lower
temperature”
7. Such flow of heat from higher to lower temperature is said to be
spontaneous.
8. The principle common to all of these is that a flow takes place from a higher
potential to a lower potential.
9. The potential may be temperature, level, pressure or voltage.
10. All these processes are said to be analogous - They all have the same
underlying principles.
11. A variety of processes undergo spontaneous change. Some examples:-
a) Radio-active decay
b) Flow of rivers downhill
c) Deflation of a balloon
d) Lightning strikes
12. Non-spontaneous process:-
a) The cup of coffee did not reach a high temperature spontaneously.
It has to be heated [ forced or driven]
b) The can of chilled coke warms and reaches equilibrium with the
surroundings spontaneously. But it has to be first chilled its
temperature down by consumption of electricity in a refrigerator.
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c) An electric iron is unplugged will cool spontaneously but when
power is switched on, it will heat up. This heating is a driven
process.
13. Equilibrium – if two systems are at the same temperature, they are in
thermal equilibrium.
14. Steady state – the parameter such as temperature is not change with
time at a specified location.
15. Uniform state – the parameter such as temperature is not change with
position throughout a surface or region at a specified time.
Example 1.4
Sketch the temperature history of an electric iron that is switch on at zero time and
switched off after steady state has been reached.
Solution:
Pin
Electric iron
Assumptions:
1. Iron is modeled without automatic temperature control
2. Heat transfer coefficient, h is constant.
T∞
stored
released
Q
Q
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Graph temperature versus time
1.3 HEAT AND OTHER FORMS OF ENERGY
1. Energy can exist in numerous forms such as thermal, mechanical, kinetic,
potential, electrical, magnetic, chemical and nuclear.
2. Microscopic energy – the forms of energy related to the molecular structure
of a system and the degree of the molecular activity. (i.e. kinetic and potential
energy)
3. The sum of all microscopic forms of energy [kinetic and potential energy] is
called the internal energy of a system, u [KJ/Kg]
i. The portion of internal energy of a system associated with the kinetic
energy is called as sensible energy or sensible heat.
“The average of velocity and the degree of activity of the molecules
proportional to the temperature”
Temperature ,velocity ,kinetic energy Internal Energy, U
ii. The internal energy is also associated with the intermolecular forces
between the molecules of the system.
Temperature, T
Driven Steady State Spontaneous
Heating Cooling Time, t
ON OFF
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There will be a phase change if sufficient energy added to the system. The
internal energy associated with the phase of a system is called latent energy
and latent heat.
u Gas >Liquid>Solid
4. The internal energy, u represents the microscopic energy of a non-flowing fluid.
Kinetic
Energy = u
Stationary fluid Potential
Microscopic energy of a non-flowing fluid
5. The enthalpy, h represents the microscopic energy of a flowing fluid.
Flowing Fluid Energy = h = u + PV
Microscopic energy of a flowing fluid
Where,
PV = flow energy of the fluid [flow work]
* Energy needed to push a fluid and to maintain flow
P = pressure [N/m2] and V = specific volume [m3/kg]
Solid Liquid Gas
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2. Specific heat represents the energy required to raise temperature of a unit
mass of a substance by one degree.
i. Specific heat at constant volume, Cv [kJ/kg.K]
ii. Specific heat at constant pressure, Cp [kJ/kg.K]
3. The specific heats, Cp and Cv are related to each other by :
Cp = Cv + R
1 kJ/kg °C = 1 kJ/kg.K [Identical]
since ΔT (°C) = ΔT (K)
4. The specific heat for an ideal gas depends only on temperature.
9. The differential changes in the internal energy, u and enthalpy, h of an ideal
gas can be expressed in terms of the specific heat as:
du = Cv dT and dh = Cp dT
constant volume constant pressure
Stationary fluid flowing fluid
Differential changes in the internal energy, u and enthalpy, h
10. The finite changes in the internal energy and enthalpy of an ideal gas during a
process can be expressed approximately by using specific heat values at the
average temperature.
Δu = Cv,avg ΔT and Δh = Cp,avg ΔT [kJ/kg]
1kJ
M= 1kg ∆T = 1°C
C = 1 kJ/kg.K
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or
ΔU = m Cv,avg ΔT and ΔH = m Cp,avg ΔT [kJ]
Where,
m = mass of the system [kg]
11. Incompressible substance – A substance whose specific volume (or
density) does not change with temperature or pressure [i.e. liquid & solid].
12. The Cv and Cp values of incompressible substances are identical
Cp = Cv = Cavg
13. Thus, the change of incompressible of solids and fluids can be expressed as:
Δu = m Cavg ΔT
Where Cavg is evaluated at average temperature
14. The specific heat of several common gases, liquids and solids are given in
the steam table.
Q = The amount of heat transferred during the process [kJ]
= Heat transfer rate [kJ/s or kW]
Where = Q / Δt
q = Heat flux, the rate of heat transfer per unit area normal to the
direction of heat transfer [W/m2]
q = / A = 10W
3m
q = 10/15 W/m2
Example of heat flux
5m
Q
Q
Q
Q
qQQ
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1.4 ENERGY BALANCE
Ein Eout
Heat Heat
Work Work
Mass Mass
– =
Ein – Eout = ΔEsystem [J]
In the rate form,
E in –
E out = dEsystem / dt [W]
In steady state condition, 0dt
dEsystem
E in –
E out = 0
E in =
E out
In heat transfer analysis, we are usually interested only in the forms of energy that can
be transferred as a result of temperature difference, that is, heat or thermal energy.
System
Net energy transfer by heat,
work, mass
Change in internal, potential, kinetic etc.
energies
Rate of net energy transfer by heat, work & mass
Rate of change in internal, kinetic, potential
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The energy balance in that case, can be expressed as:
Qin – Qout + Egen = ΔEthermal,system
Where,
Egen = heat generation – nuclear, chemical, mechanical, electrical energies etc
1. 4.1 Energy Balance For Closed System (Fixed Mass)
1. The total energy, E for most systems encountered in practice consists of the
internal energy, u.
2. This is especially the case for stationary systems since they don’t involve any
changes in their velocity or elevation during a process.
Stationary closed system: Ein – Eout = Δu = m Cv ΔT
3. When the systems involve heat transfer only and no work interactions across
its boundary, the energy balance relation further reduces to:
Stationary closed system, no work: Q = m Cv ΔT [J]
Where,
Q = the net amount of heat transfer to or from the system [Qin – Qout]
Net heat transfer
Heat generation
Change in thermal energy of the system
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1.4.2 Energy Balance for Steady Flow Systems
1. A large number of engineering devices involve mass flow in and out of system
water heater and car radiators.
2. These sorts of devices are modeled as control volume and analyzed under
steady operating conditions, ΔEcv = 0.
3. The flow of a fluid through a pipe or duct can often be approximated to be one
dimensional.
4. All properties are assumed to be uniform at any cross-section normal to the
flow direction, and properties are assumed to have bulk average values over
the entire cross-section.
5. The mass flow rates of a fluid flowing in a pipe or duct can be expressed as:
m = ρ v Ac [kg/s]
where,
ρ = fluid density [kg/m3]
v = average velocity of fluid [m/s]
Ac = cross-sectional area of the pipe or duct [m2]
6. The volume flow rate of a fluid flowing in a pipe or duct can be expressed as:
v = v Ac =
m / ρ [m3/s]
7. For a steady flow system with one inlet and one outlet, the rate of mass flow
into the control volume must be equal to the rate of mass flow out of it.
m in =
m out =
m
8. The energy balance for steady flow system is:
Q =
m Δh =
m Cp ΔT [kJ/s]
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Example 1.5
A 10cm diameter copper ball is to be heated from 100°C to an average temperature
of 150°C in 30 min. Taking the average density and specific heat of copper in this
temperature range to be ρ=8950 kg/m³ and Cp=0.395 KJ/kg°C, respectively,
determine :
a) The total amount of heat transfer to the copper ball, Q
b) The average rate of heat transfer to the ball,
c) The average heat flux.
Solution:
Copper ball
Assumption:
1. Heat is being transferred uniformly to the ball.
2. Constant properties can be used for the copper at average temperature.
Properties:
ρ=8950 kg/m³ and Cp= Cavg = 0.395 KJ/kg°C are taken at average temperature.
a) Q = ?
Q = Δu =m Cavg [T₂ –T₁]
m = ρV= 8950[π(0.1)³/6]
Q
D =10cm
T1 = 100°C
T2 = 150°C
t = 30 min =1800 sec
V = πD³/6 and A = πD²
Sphere
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Thus,
Q = 4.687(0.395)(150-100)
= 92.6 kJ
It means that 92.6 kJ of heat needs to be transferred to the copper ball to heat the
ball from 100°C to 150°C.
b)
avg = Q/Δt
= 92.6/1800
= 0.0514 KJ/s
= 51.4 W
c)
q
q avg = avg/A
= avg/πD²
= 51.4/π (0.1)²
= 1636 W/m²
Q
Q
Q
Q
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Example 1.6
1.2 kg of liquid water initially at 15°C is to be heated to 95°C in a teapot equipment
with a 1200W electric heating element inside. The teapot is 0.5 kg and has an average
specific heat of 0.7kJ/kg.K. Taking the specific heat of water to be 4.18 kJ/kg.K and
disregarding any heat loss from the teapot, determine how long will take for the water
to be heated.
Solution:
Assumptions:
1. Heat loss from teapot is negligible.
2. Steady state condition - constant properties.
Properties:
Cwater = 4.18 kJ/kg.k , Cteapot = 0.7 kJ/kg.K
From energy balance :
Ein – Eout = ΔEsystem ------------------------------------------( 1 )
No mass and work interactions, only heat transfer involves in the system, thus:
mwater = 1.3 kg mteapot = 0.5 kg
T1 = 15 °C Δt = ? T2 = 95°C
P = 1200W
Teapot
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Qin – Qout = ΔEsystem stationary fluid system
Q = ΔEsystem
Q = Δuwater + Δuteapot --------------------------------------( 2 )
Q = [ mCΔT ]water + [ mCΔT ]teapot
= [1.2 x 4.18 x (95-15)] + [0.5 x 0.7 x (95-15)]
= 401.28 + 28
= 429.3 kJ
It means that, 429.3 kJ of heat needed to raise the temperature of water from 15°C to
95°C.
The time needed to supply 429.3 kJ of heat is
P = 1.2 kW = 1.2 kJ/s
1.2 kJ 1 sec
429.3 kJ ?
Δt = ( 429.3 kJ x 1 sec )/ (1.2 kJ )
= 358 sec ≈ 6 min
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Example 1.7
A 5m long section of an air heating system of a house passes through an unheated
space in basement. The cross-section of the rectangular duct of the heating system is
20cm × 25cm. Hot air enters the duct at 100kPa and 60°C at an average velocity of
5m/s. The temperature of the air in the duct drops to 54°C as a result of heat loss to
the cool space in the basement. Determine the rate of heat loss from the air in the duct
to the basement under steady conditions. Also determine the cost of heat loss per hour
if the house is heated by a natural gas furnace that has an efficiency of 80%, and the
cost of the natural gas in that area is $1.60/therm. (1therm = 105500kJ)
Solution:
T2 = 54˚C
P = 100kPa
T1 = 60˚C εfurnace = 0.80
V = 5m/s
Rectangular duct
Assumptions:
1. Steady operating condition,
m in =
m out =
m
2. Constant properties can be used and taken at average temperature.
3. Natural gas is treated to be an ideal gas. PV = mRT.
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Properties:
Average temperature = = 57°C
From table A-15
Cp @ T = 57°C = 1.007kJ/kg.K
From Energy Balance,
Ein – Eout = ∆Esystem ………….(1)
No mass and work interactions, only heat transfer involves in the system, thus:-
Qin – Qout = ∆Esystem--------→ steady flow system
Net heat transfer rate, in =
m Cp∆T…………..(2)
Where;
m = ρAV……………..(3)
As natural gas,it is assumed to be an ideal gas;
PV = mRT
P =
P =ρRT
ρ =
=
= 1.046kg/m3
Substitute ρ into (3)
m = ρAv
= 1.046 (0.20×0.25) × 5
= 0.2616kg/s
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Thus,
=
m Cp∆T
= 0.2616 (1.007)(54-60)
= -1.58kJ/s @kW
This means in 1sec, 1.58kJ amount of heat loss through the ducting.
1sec -----------→ 1.58kJ
3600sec-------→ =?
=
= 5688kWh ≈ 5688kJ/h
1kWh = 3600kJ ------------→ conversion factor
Heat loss of using furnace with 80% efficiency
=
= 7110kWh
Thus, the cost of heat loss by using 80% efficiency of furnace;
Cost of natural gas = $1.60/therm
The cost of heat loss = Rate of heat loss × unit cost of energy input
= 7110kWh × ×
= $0.1078 ≈ 10.8 cents per hour
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Example 1.8
Consider a house that has a floor space of 200 m2 and an average height of 3 m at
1500 m evaluation where the standard atmospheric pressure is 84.6 kPa. Initially the
house is at a uniform temperature at 10oC .Now ,the electric heather is turned on, and
the heather runs until the temperature of air in the house to average value of 200C
Determine the amount of energy transferred to the air, assuming:-
(a) The house is air-tight and thus no air escape during the heating process---Fixed
mass system (Cv)
(b) Some air escape through the cracks as the heated air in the house expands at
constant pressure –---steady flow system (Cp)
Also determine the cost of this heat for each case if the cost of electricity in that area
is $0.075/kWh.
Solution:
Afloor =200m2
Patm= 84.6kPa H
H = 3m
T1=100C
T2=200C
House
Assumption:
1. Steady operating condition-Constant properties could be used and taken at
average temperature.
2. Air is assumed to be an ideal gas
3. No heat generation, work and mass transfer.
Properties:
Tavg = = 15oC
From Table A-15
Cp @ T =150C =1.007kJ/kg.K, R=0.287kJ/kg.K
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As air is an ideal gas ,
Cp = Cv + R
Cv = Cp – R
=1.007-0.287
=0.72kJ/kg.K
(a) Close system-Fixed mass
From Energy Balance .
Ein-Eout = Esystem
Qin-Qout = Usystem
Q = mCv T……………..(1)
Where
PV = mRT
m =
m =
m = 625 kg
Q = mCv T
= 625(0.72)[20-10]
= 4500kJ
This means 4500kJ of heat needed to rise the temperature of air in the house from
100C to 200C
1kWh 3600 kJ
? 4500 kJ
So,
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=
= 1.25 kWh
The cost of energy,
1 kWh $0.075
1.25 kWh $ ?
Cost of energy =
= $ 0.094
(b) Steady flow system
From energy balance
Ein-Eout = Esystem
Qin-Qout = Hsystem
Q = Hsystem =mCp T……………..(2)
Q = mCp T
= 625(1.007)[20-10]
= 6294kJ
This means 6294kJ of heat needed to rise the temperature of air in the house from
100C to 200C
1kWh 3600 kJ
? 6294 kJ
So,
=
= 1.75 kWh
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The cost of energy
1 kWh $0.075
1.75 kWh $ ?
Cost of energy =
= $ 0.131
1.5 HEAT TRANSFER MECHANISMS
1. Heat is the forms of energy that can be transferred from one system to another as
a result of temperature different
2. Heat is transferred from the higher temperature medium to the lower temperature,
and heat transfer stops when the two medium reach the same temperature
[Equilibrium].
3. Heat can be transferred in three different modes:-
a) Conduction
b) Convection
c) Radiation
1.5.1 Conduction
1. Conduction is the transfer of energy from the more energetic particles of a
substance to the adjacent less energetic ones as a result of interaction between
the particles.
2. Conduction can take place in solid, liquid and gases
Solid Liquid Gas
Mechanisms of Conduction
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Solid= Lattice vibration, Flow of free electrons
Liquid & Gas = Molecular collisions, Molecular diffusion
3. The rate of conduction depends on :-
a) Geometry of the medium ,A
b) The thickness of the medium, t
c) The temperature difference, T
d) Material of the medium, k
t
T1
Where
α
=k A
4. The rate of heat conduction through a plane layer is proportional to the
temperature difference across the layer and the heat transfer area, but is
inversely proportional to the thickness of the layer.
5. Fourier’s Law is used to analyze heat transfer problem which is due to
conduction. It is stated by Fourier’s Law;
[W]
T2 A
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Where;
k = the constant of proportionality, thermal conductivity of the
material [W/mk]
A = the area normal to the direction of heat transfer [m²]
= Temperature difference [K]
= Thickness [m]
6. In the limiting case of 0, the equation reduces to:
[W]
Where;
= Temperature gradient
-ve sign = show that heat is transferred in the direction of decreasing
Temperature [2ndLaw of Thermodynamics]
7. The purpose of conduction analysis is to find the temperature
distribution/gradient in the medium.
e.g. :
Temperature distribution,
Temperature gradient,
8. The major purpose of obtaining temperature distribution/gradient is to obtain
the more practical result i.e, the heat transfer rate.
1.5.1.1 Conduction Properties
Thermal conductivity, k [W/mk]
Thermal conductivity, k is a measure of a material’s ability to conduct heat.
*Specific heat is a measure of material’s ability to store heat.
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e.g: Find thermal conductivity, k and specific heat, for water and iron at
room temperature.
= 4.18 kJ/kg , = 0.45 kJ/kg ,
= 0.607 W/m = 80.2 W/m
Water can store heat greater than iron, but it is a poor conductor relative to
iron.
Thermal conductivity can be defined as the rate of heat transfer through a unit
thickness through a unit thickness of the material per unit area per unit
temperature different.
k good heat conductor
k poor heat conductor (insulator)
Thermal conductivities of materials vary with temperature.
The temperature depends of thermal conductivity cause considerable
complexity in conduction analysis.
In analysis, thermal conductivity k is always assumed to be constant and taken
at the average temperature.
Thermal Diffusivity, [m²/s]
Thermal Diffusivity, represent how fast heat diffuses through a material.
[m²/s]
k , - The larger the thermal diffusivity, the faster the propagation of
heat into a medium.
k , - A small value of the thermal diffusivity means that heat is
mostly absorbed by the material and a small amount of heat is conducted
further.
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1.5.2 Convection
Convection is the mode of energy transfer between a solid surface and the adjacent
liquid or gas that is in motion, and it involves the combined effects of conduction and
liquid motion.
Mechanism of convection
There are two types of convection:
i. Forced convection
The fluid is forced to flow over the surface by external means such
as at pump, blower, and etc.
ii. Natural convection
The fluids motion is caused by buoyancy forces that are induced by
density different due to the variation of temperature in the fluid.
Heat transfer process that involves change of phase of a fluid are also considered to
be convection because of the fluid motion induced during the process example Boiling
and condensation process.
Convection heat transfer can be expressed by Newton’s Law of cooling as;
[W]
1-32 Bachelor Degree in Mechanical Engineering
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Where
= Convection heat transfer coefficient [W/m² ]
Surface area though which convection heat transfer takes place [m²]
= Surface temperature [
= Fluid temperature [
Convection heat transfer coefficient is influenced by;
iii. Surface geometry
iv. The nature of the fluid
v. Properties of the fluid
vi. Bulk fluid velocity
is not a property of the fluid.[ i.e we cannot get the value of from the manual]. It
needs to be determined first before convection heat transfer problem could be
solved.
Typical values of convection heat transfer coefficient,
Type of convection [W/m² ]
Free convection of gases 2-25
Free convection of liquids 10-1,000
Forced convection of gases 25-250
Forced convection of liquids 50-20,000
Boiling & Condensation 2500-100,000
1.5.3 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.
1-33 Bachelor Degree in Mechanical Engineering
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The transfer of heat by radiation does not require the presence of an intervening
medium- e.g. sun emissions.
Thermal radiation - The form of radiation emitted by body because of their
temperature.
Stefan- Boltzman Law is used to solve thermal radiation problem.
[W]
where,
= Stefan- Boltzman constant, 5.67 x 108 W/m
42 K
= Emissivity of the surface, 0 1 , for Blackbody, 1
1.5.3.1 Radiation Properties
Emissivity, , 0 1
This property provides a measure of how efficiently a surface emits energy
relative to a blackbody.
for blackbody is 1 – Blackbody is a perfect emitter.
This property depends strongly on the surface material and finish; and represent
active value are provided in the manual.
Absorbtivity, , 0 1
The fraction of the radiation energy incident on a surface that is absorbed by
the surface.
Blackbody is a perfect absorber ( )1 absorbs the entire radiation incident
on it.
Thermal radiation incident upon a body (or medium) may be partially reflected,
partially absorbed and the remainder transmitted.
1-34 Bachelor Degree in Mechanical Engineering
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P + 1
Mechanism of radiation
When a surface of emissivity, and surface area, A s at a thermodynamic
temperature, T s is completely enclosed by a much larger (or black) surface at
thermodynamic temperature, T surr separately 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
[W]
Radiation heat transfer between a surface and the surfaces surrounding it
1-35 Bachelor Degree in Mechanical Engineering
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1.6 COMBINATION OF HEAT TRANSFER MODES
Even there are three modes of heat transfer, i.e. conduction, convection and radiation;
a medium may involve only two of them simultaneously.
1-36 Bachelor Degree in Mechanical Engineering
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Combination of heat transfer
Medium Combination of heat transfer
1. Opaque solids
2. Semitransparent solids
3. Solid exposed to a fluid or other
surfaces
4. Still fluid (no bulk fluid motion)
5. Flowing fluid
conduction
conduction and radiation
convection and/or radiation
conduction and radiation
convection and radiation
1.7 IMPORTANT TO BE REMEMBERED
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Example 1.9
The roof of an electrically heated home is 6m long, 8m wide and 0.25m thick, and is
made of a flat layer of concrete whose thermal conductivity is CmWk ./8.0 (Figure
1-9). The temperatures of the inner and the outer surfaces of the roof one night are
measured to be 15 C and 4 C , respectively, for a period of 10 hours. Determine:-
(a) The rate of heat loss through the roof that night and
(b) The cost of that heat loss to the home owner if the cost of electricity is $0.08/kWh.
Solution:
?
Assumptions:
1. Steady state conditions for whole night – constant properties could be used.
2. Heat is uniformly being transferred over the entire surface of the roof.
Properties:
CmWkroof
./8.0
(a)
6m
8m
0.25m
CT 151
CT 42
hrt
CT
CT
mt
mxAroof
10
4
15
25.0
4886
2
1
2
Q
Cost = ?
Q
t
TTkAQ 21
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(b) Cost of heat transfer
Heat loss within 10 hours
= 1.69(10)
= 16.9 kWh
Cost of heat transfer = 16.96kWh ($0.08/kWh)
= $1.35
25.0
415)48(8.0
kW69.1
6.1689
Q
1-39 Bachelor Degree in Mechanical Engineering
Heat Transfer (BDA 30603) N. Nordin, 2018
Example 1.10
The wall of an industrial furnace is constructed from 0.15m thick fireclay brick having
a thermal conductivity of 1.7W/m.K. measurements made during steady-state
operation reveal temperatures of 1400 and 1500k at the inner and outer surfaces,
respectively. What is the rate of heat loss through a wall that is 0.5m by 1.2m on a
side?
Solution:
Assumptions:
1. Steady state conditions.
2. Constant thermal conductivity.
3. One-dimensional conduction through the wall.
4. Heat is uniformly being transferred.
Properties:
k=1.7W/m.K
Analysis:
x
TkAQ
cond
Q
1.2m 0.5m
0.15
m
Brick
wall
0.15m
?
Q
KT 14001
KT 11502
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x
TTkA 12
x
TTkA 21
15.0
115014002.15.07.1 x
W1700
Example 1.11
A 2-m-long, 0.3-cm-diameter electrical wire extends across a room at 15 C , as shown
in below. Heat is generated in the wire as a result of resistance heating, and the surface
temperature of the wire is measured to be 152 C in steady operation. Also, the voltage
drop and electric current through the wire are measured to be 60V and 1.5A,
respectively. Disregarding any heat transfer by radiation, determine the convection
heat transfer coefficient for heat transfer between the outer surface of the wire and the
air in the room.
Solution:
Assumptions:
1. Steady operating conditions.
2. Radiation heat transfer is negligible.
Analysis:
radQ WxIVE generated 905.160
22 01885.0)2)(103.0( mxdLAs
AI
vV
md
mL
5.1
60
3.0
2
60
V
convQ
1.5A
CTs
152CT 15
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CmWh
h
TThAQ ssconv
./9.34
)15152)(01885.0(90
)(
2
Example 1.12
It is a common experience to feel “chilly” in winter and “warm” in summer in our homes
even when the thermostat setting is kept the same. This is due to the so called
“radiation effect” resulting from radiation heat exchanger between our bodies and the
surrounding surfaces of the walls and the ceiling.
Consider a person standing in a room maintained at 22 C at all times. The inner
surfaces of the walls, floors, and the ceiling of the house are observed to be at an
average temperature of 1 C in winter and 25 C in summer. Determine the rate of
radiation heat transfer between this person and the surrounding surfaces if the
exposed surface area and the average outer surface temperature of the person are
1.4 C and 30 C , respectively
Solution:
CT erwsurr
10int,
CT summersurr
25,
radQ ?
The rates of radiation heat transfer between a person and the surrounding surfaces at
specified temperature are to be determined in summer and winter.
32C
radQ
surrT
CT
mA
s
s
30
4.1 2
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Assumptions:
1. Steady operating conditions.
2. Heat transfer by convection is not considered.
3. The surrounding surfaces are at a uniform temperature.
Properties:
The emissivity of a person is = 0.95 (Table 1-6, Cengel, Y.A., 2006).
Analysis:
)( 44
int, surrsserwrad TTAQ
= (0.95)(5.67 810x )(1.4)[ (30+273)4
- (10+273)4
]
= 152W
)( 44
, surrsssummerrad TTAQ
= (0.95)(5.67810x )(1.4)[ (30+273)
4- (25+273)
4]
= 40.9W