lectures on heat transfer - introduction - applications - fundamentals - governing laws
TRANSCRIPT
Lectures on Heat Transfer:Introduction - Fundamentals
by
Dr. M. Thirumaleshwarformerly:
Professor, Dept. of Mechanical Engineering,St. Joseph Engg. College, Vamanjoor,
Mangalore
Preface
• This file contains Introduction to Heat Transfer and Fundamental laws governing heat transfer.
• The slides were prepared while teaching • The slides were prepared while teaching Heat Transfer course to the M.Tech. students in Mechanical Engineering Dept. of St. Joseph Engineering College, Vamanjoor, Mangalore, India, during Sept. – Dec. 2010.
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• It is hoped that these Slides will be useful to teachers, students, researchers and professionals working in this field.
• For students, it should be particularly useful to study, quickly review the subject, useful to study, quickly review the subject, and to prepare for the examinations.
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References• 1. Cengel Y. A. Heat Transfer: A Practical
Approach, 2nd Ed. McGraw Hill Co., 2003• 2.Cengel, Y. A. and Ghajar, A. J., Heat and
Mass Transfer - Fundamentals and Applications, 5th Ed., McGraw-Hill, New York, NY, 2014.
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5th Ed., McGraw-Hill, New York, NY, 2014.• 3. Incropera , Dewitt, Bergman, Lavine:
Fundamentals of Heat and Mass Transfer, 6th
Ed., Wiley Intl.• 4. Necati Ozisik: Heat Transfer – A Basic
Approach, McGraw Hill
References… contd.
• 5. M. Thirumaleshwar: Fundamentals of Heat & Mass Transfer, Pearson Edu., 2006
• 6. M. Thirumaleshwar: Software Solutions to Problems on Heat Transfer – CONDUCTION-Problems on Heat Transfer – CONDUCTION-Part-I, Bookboon, 2013
• http://bookboon.com/en/software-solutions-to-problems-on-heat-transfer-ebook
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Heat TransferIntroduction - Fundamentals
• Applications - Modes of heat transfer-Fundamental laws – governing rate equations – concept of thermal resistance
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equations – concept of thermal resistance – general heat conduction equation –different boundary conditions.
Applications of Heat Transfer:
• Mechanical Engineering: Boilers, Heat Exchangers, Turbine systems, Internal combustion engines etc.
• Metallurgical Engineering: Furnaces,
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• Metallurgical Engineering: Furnaces, Heat treatment of components etc.
• Electrical Engineering: Cooling systems for electric motors, generators, transformers etc.
Applications of Heat Transfer(Contd.)..
• Chemical Engineering: Process equipments used in Refineries, Chemical plants etc.
• Nuclear Engineering: In removal of heat generated by nuclear fission using liquid metal
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generated by nuclear fission using liquid metal coolants, design of nuclear fuel rods against possible burn – out etc.
• Aerospace Engineering & Space Technology:In the design of aircraft systems and components, Rockets, Missiles etc.
Applications of Heat Transfer(Contd.)..
• Cryogenic Engineering: In the production, storage, transportation and utilization of cryogenic liquids (at very low temperatures ranging from 100 K to 4 K or even lower) for various Industrial, Research and Defence applications.
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applications.• Civil Engineering: In the design of Suspension
bridges, railway tracks, Airconditioning and Insulation of buildings etc.
• Principles and methods of Heat Transfer are widely applied in many, many areas that affect our lives.
Fundamental Laws governing Heat Transfer:
– 1. First Law of Thermodynamics – gives conservation of energy.
– 2. Second Law of Thermodynamics – gives direction of heat flow.
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direction of heat flow.– 3. Equation of continuity – gives
conservation of mass.– 4. Equation of flow – Newton’s Second Law
of motion—Navier Stokes’ Equations
Fundamental Laws governing Heat Transfer (contd.):
– 5. Rate equations governing the three modes of Heat Transfer:
– Conduction – Fourier’s Law of Conduction– Convection – Newton’s Law of cooling
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– Convection – Newton’s Law of cooling– Radiation – Stefan – Boltzmann’s Law– 6. Empirical relations for fluid properties
such as specific heat, thermal conductivity, viscosity etc.
– 7. Equation of State for the fluid.
Modes of Heat Transfer
• Three main modes:• Conduction …. heat flow by direct contact• Convection … heat carried by moving fluid• Radiation … heat flow does not need an • Radiation … heat flow does not need an
intervening medium• In practical cases, a combination of one or
more of the above modes are present
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Conduction:
• Governing ‘rate equation’ for conduction is: Fourier’s Law.
• Q = -k A (T2 – T1)/L
k
T1
Q
Slope = dT/dx
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• Qx = -k A (T2 – T1)/L...For a plane slab, in steady state
= k A (T1 – T2)/L(Watts)
T2
X
L
Assumptions behind Fourier’s Law:
• Fourier’s Law is an empirical law, derived from experimental observations and not from fundamental, theoretical considerations.
• Fourier’s Law is defined for steady
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• Fourier’s Law is defined for steady state, one dimensional heat flow.
• It is assumed that the bounding surfaces between which heat flows are isothermal and that the temperature gradient is constant i.e. the temperature profile is linear.
Assumptions behind Fourier’s Law (contd.):
• There is no internal heat generation in the material.
• The material is homogeneous (i.e. constant density) and isotropic (i.e. thermal conductivity
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density) and isotropic (i.e. thermal conductivity is the same in all directions).
• Fourier’s Law is applicable to all states of matter i.e. solid, liquid or gas.
• Fourier’s Law helps to define ‘thermal conductivity’
Rate of heat transfer by conduction, Q, is given by:L
T1 T2
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Rate of heat transfer by conduction, Q, is given by:
Q = k A �T/L (W)
where, k = thermal cond. (W/m.K)
A = Area of cross-section, (m^2)
L = Length, (m), �T= temp. difference, (K)
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Range of ‘k’ of various materials at room temperature
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Variation of ‘k’ with temperature
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Variation of ‘k’ with temperature for Cu and Al
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Convection
• In convection, heat is carried from place to place by bulk movement of fluidof fluid
• Convection currents are set up when a pan of water is heated
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Newton’s Law of Cooling:Q = h.A.�T
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Radiation
• energy is transferred by means of electromagnetic waves.
• can take place through vacuum.
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• can take place through vacuum.
• electromagnetic waves can propagate through empty space.
Radiationaverage frequency ∝ absolute temperature
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Radiation
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Greenhouse Effect
Incoming UV radiation from Sun easily passes through the glass walls of a greenhouse. Weaker IR radiation, however,
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IR radiation, however, has difficulty passing through the glass walls and is trapped inside, thus warming the greenhouse.
Black body
A good absorber like lampblack is also a good emitter
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good emitterAnd, a poor absorber like polished silver is also a poor emitter
The Stefan–Boltzmann Law of Radiation
Rate of radiant energy emission is proportional to the fourth power of its Absolute temperature.
Stefan’s law and is expressed as follows:
P = � .�. A. T4 (W)
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P = � .�. A. T4 (W)
� is the Stefan-Boltzmann constant, = 5.67 × 10-8 W/m2.K4.� is the emissivity, which is a number between 0 and 1.
T is the temp. in Kelvin
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Radiation Heat Transferbetween bodies
• If the surrounding is at TS then the net power radiated is:
P = � � A [ T4 - TS4]
• Assuming on a dark, dry, night, T = 3 K:
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• Assuming on a dark, dry, night, TS = 3 K:• Frost may form even if air temperature > 0 C since radiation cools the surface faster than conduction heat lost from the ground or air.
Concept of Thermal resistance:
• Conduction:• Q = kA(T1 – T2)/L
k
T1
T2
T(x)
Q
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T2
XL
QT1 T2
Rcond = L/kA
Q
Rth = L/(kA) is known as “Thermal resistance” of the slab for conduction.
Concept of Thermal resistance (contd.):
• It is seen that there is a clear analogy between the flow of heat and flow of electricity, as shown below:
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Concept of Thermal resistance (contd.):
• Convection heat transfer – thermal resistance :
U, Tf, h TsQ
Q = h A (Ts – Tf)
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Q QTs Tf
Rconv = 1/hA
Rconv = 1/(hA)
Note that the Units are : (C/W) or (K/W)
Concept of Thermal resistance (contd.):
• Radiation heat transfer – thermal resistance :• Q1 = F1 A1 σ (T1
4 – T24), W
• F1 is known as view factor, which includes the effects of orientation, emissivities and the distance between the surfaces
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Practical applications of Thermal resistance concept:
• To analyse the problems where one or more modes of heat transfer occursimultaneously
• To analyse the problems where multiple
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• To analyse the problems where multiple layers of materials of different thermal conductivities are used; ex: in furnace walls
Limitations for the use of thermal resistance concept:
• Thermal concept can be used only when all the following conditions are satisfied:
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• One dimensional conduction• Steady state conduction • No internal heat generation
Thermal diffusivity (αααα):• Often, while dealing with transient
conduction problems, we come across a quantity called ‘Thermal diffusivity’. Note that Unit of α is: m2/s
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that Unit of α is: m /s
Values of αααα for materials vary over a wide range:
For copper at room temperature, its value is approx. 113 * 10-6 m2/s
For glass it is about 0.34 * 10-6 m2/s.
Thermal diffusivity (αααα) for a few materials at room temperature:
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Thermal diffusivity (αααα) for a few materials at room temperature(contd.):
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GENERAL DIFFERENTIAL EQUATION FOR HEAT CONDUCTION
• This is also known as ‘heat diffusion equation’ or, simply ‘heat equation’.
• Consider a differential volume element (dx.dy.dz); • Making an energy balance on this differential element:
Q
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E
B
C
D
A
H
G
F
dxdy
dzy
z
x
Q x Q x+dx
Q y
Q y+dy
Q z+dz
Q z
GENERAL DIFFERENTIAL EQUATION FOR HEAT CONDUCTION (contd.)
B
CG
Fz
Q y+dy
Q z+dz
• Ein – Eout + Egen = Est …eqn. (1)
Energy, In
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E
D
A
H
dxdy
dzy
x
Q x Q x+dx
Q y
Q z
Energy, In
Energy, out
Energy, generated
Energy stored
GENERAL DIFFERENTIAL EQUATION FOR HEAT CONDUCTION (contd.)
Applying eqn (1):
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where qg is the heat gen. rate per unit volume, (W/m3)
Now, etc.
eqn. (2)
GENERAL DIFFERENTIAL EQUATION FOR HEAT CONDUCTION (contd.)
Then, from eqn. (2):
But, from Fourier’s Law:
eqn. (3)
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But, from Fourier’s Law:
GENERAL DIFFERENTIAL EQUATION FOR HEAT CONDUCTION (contd.)
Then, subst. in eqn. (3) and dividing by dx.dy.dz, we get:
eqn. (4)
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This is the general form of heat diffusion equation in Cartesian coordinates, for time dependent (i.e. unsteady state) heat conduction, with variable thermal conductivity and uniform heat generation within the body.
GENERAL DIFFERENTIAL EQUATION FOR HEAT CONDUCTION (contd.)
Now, if the material is isotropic i.e. the thermal conductivity is the same in all the three directions, i.e. kx = ky = kz = k say, :
eqn. (5)
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If k is constant and does not vary with temperaturei.e. k does not change with position, then:
eqn. (6)
GENERAL DIFFERENTIAL EQUATION FOR HEAT CONDUCTION (contd.)
i.e.eqn. (7)
where α = k/(ρcp) is thermal diffusivity, and
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where α = k/(ρcp) is thermal diffusivity, and∇ = Laplacian operator
Special cases:
1. Steady state: i.e.
GENERAL DIFFERENTIAL EQUATION FOR HEAT CONDUCTION (contd.)
eqn. (8)
Then, eqn. (7) becomes:
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This is known as ‘Poisson equation’.
2. With no internal heat generation, i.e. qg = 0:
Then, eqn. (7) becomes:
eqn. (9)
This is known as ‘Diffusion equation’.
GENERAL DIFFERENTIAL EQUATION FOR HEAT CONDUCTION (contd.)
3. Steady state, with no internal heat generation, i.e.
eqn. (10)
τ∂∂T
Therefore,
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This is known as ‘Laplace equation’.
4. One dimensional, steady state, with no internal heat generation,i.e.
Therefore,
GENERAL DIFFERENTIAL EQUATION FOR HEAT CONDUCTION (contd.)
eqn. (11)
So, equation (7) becomes:
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GENERAL DIFFERENTIAL EQUATION in cylindrical coordinates:
eqn. (12)
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For one dimensional conduction in r - direction only, we get:
eqn. (13)
GENERAL DIFFERENTIAL EQUATION in spherical coordinates:
eqn. (14)
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For one dimensional conduction in r - direction only, we get:
eqn. (15)
Boundary and Initial conditions:Commonly encountered Boundary Conditions (B.C’s) are:
• Prescribed temperature conditions at the boundaries – known as B.C. of the first kind or Dirichlet condition:
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Boundary and Initial conditions:Commonly encountered Boundary Conditions (B.C’s) are:
• Prescribed heat flux condition at the boundaries - known as B.C. of the second kind or Neumann condition
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Prescribed heat flux condition at the boundaries:
Two special cases:
1. Insulated boundary: 2. Thermal symmetry:
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Boundary and Initial conditions:Commonly encountered Boundary Conditions (B.C’s) are:
• Convection boundary condition - known as B.C. of the third kind:
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Boundary and Initial conditions:Commonly encountered Boundary Conditions (B.C’s) are:
• Interface boundary condition - known as B.C. of the fourth kind:
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• Example3.1: Temperature variation in a slab is given by: T(x) = 100 + 200 x –500 x2, where x is in metres; x = 0 at the left face and x = 0.3 m at the right face. Thermal conductivity of the material k = 45 W/(m.C). Also, cp = 4 kJ/(kg.K) and ρρρρ = 1600 kg/m3. Determine:
• Temperature at both surfaces• Heat transfer at left face and its direction• Heat transfer at right face and its direction
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• Heat transfer at right face and its direction• Is there any heat generation in the slab? If so, how
much?• Max. temperature in the slab and its location• Time rate of change of temperature at x = 0.1 m if
the heat generation rate is suddenly doubled• Draw the temperature profile in the slab
k, qg
Temp. Profile
QrightQleft
Data:
L 0.3 m
k 45 W/m.C
c p 4000 J/kg.K
ρ 1600 kg/m^3
α k
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L
α k
ρ c p.
α 7.031 10 6= m2/s
Fig. Ex. 3.1 (a)
T x( ) 100 200 x. 500 x2. Define T(x)...i.e. temp. as a function of x
Temp. at left face: i.e. at x = 0: T 0( ) 100= C....Ans.
Temp. at right face: i.e. at x = 0.3 m: T 0.3( ) 115= C....Ans.
To find max. temp.:
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Define the first derivative of T(x): T' x( )xT x( )d
d
Also, define the second derivative of T(x): T'' x( )xT' x( )d
d
--------------------------------------------------------------------------------------------------------------------------
• By hand calculation: • We get: T (̀x) = 200 –1000.x• We set T (̀x) equal to zero to get the
position xmax where temp. is max.:• i.e. 200 – 1000.x = 0. • This gives x = 0.2 m. Substitute this value
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• This gives x = 0.2 m. Substitute this value of xmax in T(x) to get the value of Tmax.
• So, Tmax = T(0.2) = 100 + 200 x 0.2 – 500 x (0.2)2. = 120 C.
Temp. distribution in the slab:
T x( )110
115
120Variation of T(x) with x for Slab
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x0 0.1 0.2 0.3
100
105
Fig. Ex. 3.1 (b)
Note from the graph that the max. temp. occurs at x = 0.2 m and its value is 120 C, as
already calculated.
To calculate the heat fluxes at the left and right faces :
Apply the Fourier's Law at x = 0 and at x = 0.3 m, remembering that temp. gradient is
given by T'(x), aready defined.
q left k T' 0( ). ..applying Fourier's Law at left face i.e. at x = 0
q left 9 103= ...Heat flux at the left face (W/m^2) ; note that -ve sign indicates heat flowing from right to left
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q right k T' 0.3( ). ..applying Fourier's Law at right face i.e. at x = 0
q right 4.5 103= ...Heat flux at the right face (W/m^2 ); note that +ve sign indicates heat flowing from left to right.
q total q left q right ...Total heat generated per m^2 of surface
q total 1.35 104= W/m^2...Total heat generated/m^2
• To calculate the time rate of change of temperature at x = 0.1 m when qg is suddenly doubled:
• We have the time dependent differential equation for
Therefore, qg, the volumetric heat gen. rate is given by Total heat gen. per unit volume:
q gq total
1 0.3.q g 4.5 104= W/m^3...vol. heat gen. rate in the slab....Ans .
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• We have the time dependent differential equation for heat conduction in Cartesian coordinates:
τατρ
∂∂=
∂∂=+
∂∂ TT
k
c
k
q
xT pg 12
2
• Therefore, αα
τ k
q
xTT g+
∂∂=
∂∂
2
2
From the given equation for temperature distribution, it is clear that does not depend on x, i.e.
depends only on qg:
2
2
xT
∂∂
τ∂∂T
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depends only on qg: τ∂
dtbyd τ x( ) α T'' x( ). α2 q g
.
k. ...define dT/d ττττ as a function x . Now, we can get dT/dτ
at any x by simply substituting that value of x in the function defined
dtbyd τ 0.1( ) 7.031 10 3= C/s....time rate of change of temp. ...Ans.
Note that this is true for all x since T''(x) does not depend on x for the temperature distribution given
• Example 1.3 : Electronic power devices are mounted to a heat sink having an exposed surface area of 0.045 m^2 and an emissivity of 0.8. When the devices dissipate a total power of 20 W and air and surroundings are at 27 C, the and surroundings are at 27 C, the average sink temperature is 42 C. What average temperature will the heat sink reach when the devices dissipate 30 W for the same environmental conditions?
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Variation of T2 with epsilon:
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