Chapter One
Introduction
1.1 Definition:
Heat: is the energy transit as a result of the temperature difference.
Heat transfer: is that science which seeks to predict the energy
transfer that may take place between material bodes as a result of a
temperature difference
Heat flux: heat transfer flow in the direction per unit area (q”).
Steady state: Temperature does not vary with time (dT/dt) =0.
Unsteady state: temperature is depending on time
1.2 Modes of Heat Transfer
There are three fundamental modes of heat transfer
-conduction,
- convection, and
-radiation.
1.3 A Conduction Heat Transfer
Conduction may be viewed as the transfer of energy from
the more energetic to the less energetic particles of a substance due
to interactions between the particles. A temperature gradient within a
homogeneous substance results in an energy transfer rate within the
medium which can be calculated by Fourier's law
1
q= -KA
( 1-1)
Where q is the heat transfer rate (W or J/s) and k thermal
conductivity (W/m K)) is an experimental constant for the medium
involved, and it may depend upon other properties, such as
temperature and pressure.
:
Is the temperature gradient in the direction normal to the area
A.
Figure 1.1 Temperature distributions for steady state conduction
through a plate wall
The minus sign in Fourier's Law ((1.1)) is required by the second law
of thermodynamics: thermal energy transfer resulting from a thermal
gradient must be from a warmer to a colder region. If the temperature
profile within the medium is linear Fig. 1.1 it is permissible to replace the
temperature gradient (partial derivative) with
q= -KA
(1.2)
2
The quantity (L/kA) is equivalent to a thermal resistance Rth (K/W)
which is equal to the reciprocal of the conductance. As:
q=
, Rt =
(1-3)
Such linearity always exists in a homogeneous medium of fixed k during
steady state heat transfer occurs whenever the temperature at every
point within the body, including the surfaces, is independent of time.
Figure 1.2 Association off conduction heat transfer with diffusion of
energy due to molecular activity.
3
1-3-1 The Energy Balance
In this special case the control surface includes no mass or
volume and appears as shown in Figure 1.8.Accordingly, the
generation and storage terms of the Energy expression,
Ein –Eout -Storage + Generation= 0
Consequently, there can be no generation and storage. The
conservation requirement then becomes
Ein –Eout = 0
In Figure 1.8 three heat transfer terms are shown for the control
surface. On a unit area basis they are conduction from the medium
to the control surface q"cond convection from the surface to a
fluid q"conv, and net radiation exchange from the surface to the
surroundings q"rad. The energy balance then takes the Form and
we can express each of the terms according to the appropriate rate
equations.
4
Fig (1-3)) Application of conservation of energy law at a surface of a system
1-3-2 GENERAL EQUATION OF HEAT CONDACTION
We now set ourselves the problem of determining the basic equation that
governs the transfer of heat in a solid, using Equation ((1-1)) as a starting
point.
Consider the one-dimensional
dimensional system shown in Figure 1-4.. If the
system is in a steady state, i.e., if the temperature does not change with time,
then the problem is a simple one, and we need only integrate Equation ((1-1)
and substitute the appropriate values to solve for the desired quantity.
However, if the temperature of the solid is changing with time, or if there are
heat sources or sinks within the solid, the situation is more complex. We
consider the general case where the temperature may be changing with time
and heat sources may be present within the body. For the element of
thickness dx, the following energy balance may be made:
Energy conducted in left face
ace + heat generated within element= change in
internal energy + energy conducted out right face
5
These energy quantities are given as follows:
Energy in left face = q= −kA
Energy generated within element =
.
˙A dx
Figure 1--4 Elemental volume for onedimensional heat- conduction analysis.
6
7
8
where the quantity α = k/ρc is called the thermal diffusivity of the material.
The larger the value of α, the faster heat will diffuse through the material.
Thermal diffusivity α has units of square meters per second.
Equation (1-4a)
a) may be transformed into either cylindrical or spherical
coordinates by standard calculus techniques. The results are as follows:
it is worthwhile to show the reduced form of the general equations for
several cases of practical interest.
9
.
10
11
1.4 Convection Heat Transfer
Whenever a solid body is exposed to a moving fluid
having a temperature different from that of the body, energy is carried
12
or convected from or to the body by the fluid If the upstream
temperature of the fluid is T , and the surface temperature of the solid is
Ts the heat transfer per unit time is given by Newton s Law of cooling:
q= h A (
-
∞)
(1-9)
2
Where h is Convective Heat transfer coefficient (W/m K) as the
constant of proportionality relating the heat transfer per unit time and
area to the overall temperature difference. It is important to keep in
mind that the fundamental energy
Fig (1.4)) Velocity and temperature di
distribution on flat plate
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14
15
16
17
18
Example: A glass window of width W= 1 m and height H=2 m is 5 mm
thick and has a thermal conductivity of kg = 1.4 w/m .k . If the inner and
outer surface temperatures of the glass are 15⁰C and -20⁰C,
C, respectively, on
19
a cold winter day, what is the rate of heat loss through the glass? To reduce
heat loss through windows, it is customary to use a double pane construction
in which adjoining panes are separated by an air space. If the spacing is 10
mm and the glass surfaces in contact with the air have temperatures of 10⁰C
10
and -15 ⁰C,
C, what is the rate of heat loss from a 1 m x 2 m window? The
thermal conductivity of air is ka =
=0.024 W/m.k
20
Example:- what is the thickness required of a masonry wall having thermal
conductivity 0.75 w/m. k if the heat rate is to be 80% of the heat rate through
a composite structural wall having a thermal conductivity of 0.25 w/m. k and
a thickness of 100 mm? Both walls are subjected to the same surface
temperature difference.
21
Example: - At a given instant of time the temperature distribution within an
infinite homogenous body is given by the function
22
T(x, y, z) = x2 – 2 y2 + z2 – x y + 2 y z
Assuming constant properties and no internal heat generation, determine the
regions where the temperature changes with time.
23
(From Holman book)
Example 1-1 Conduction through copper plate
One face of a copper plate 3 cm thick is maintained at 400co , and the other
face is maintained at 100 c0. How much heat is transferred through the plate
?
Solution
Thermal conductivity for copper is 370 w/m.C0 at 250 Co.
From Fourir,s law
= −
24
= −
=
−(370)(100 − 400)
= 3.7
3 × 10
/
Example 1-2 convection calculation
Air at 20 C0 blows over a hot plate 50 by 75 cm maintained at 250 Co. The
convection heat – transfer coefficient is 25 W/m2 .C0 .calculate the heat
transfer.
Solution
From Newton,s law of cooling
q = hA (Tw -
∞
) = (25)(0.50)(0.75)(250 – 20 )= 2.156 kw
Example 1-3 Multimode heat transfer
Assuming that the plate in example 1-2 is made of carbon steel ( 1% ) 2
cm thick and that 300 W is lost from the plate surface by radiation. Calculate
the inside plate temperature.
Solution
The heat conducted through the plate must be equal to the sum of convection
and radiation heat losses:
qcond = qconv + qrad
-kA
∆
∆
= 2.156 + 0.3 = 2.456
(
)( .
( . )( . )(
)
)
°
25
The inside plate temperature is therefore
Ti =250 + 3.05 =253.05 C0
Example 1-4 Heat source and convection
An electric current is passed through a wire 1 mm in diameter and 10 cm
long. The wire is submerged in liquid water at atmospheric pressure ,and the
current is increased until the water boils. For this situation h = 5000 w/m2 .
C0. and the water temperature will be 100 C0. How much electric power
must be supplied to the wire to maintain the wire surface at 114 C0?
Solution
The total convection loss is given by
Q = h A (Tw -
∞
)
For this the surface area of the wire is
A = π d L = π ( 1× 10 )(10 × 10
The heat transfer is therefore
q =( 5000 w/m2. C0)( .142 × 10
) =3.142 × 10
m2
m2 )( 114 -100 ) = 21.99 W
and this is equal to the electric power which must be applied.
Example 1-5 Radiation heat transfer
Tow infinite black plates at 800 and 300 C0 exchange heat by radiation.
Calculate the heat transfer per unit area.
Solution
q/A =σ (T14 – T24 ) =( 5.669 × 10 ) (10734 -5734 ) = 69.03 KW/m3
26
Example 1-6 total heat loss by convection and radiation
A horizontal steel pipe having a diameter of 5 cm is maintained at a
temperature of 50 C0 in a large room where the air and wall temperature are
at 20 C0. The surface emissivity of the steel may be taken as 0.8. Using the
data of table 2 calculate the total heat lost by the pipe per unit length.
Solution The total heat loss is the sum of convection and radiation. From
table 2 we see that an estimate for the heat- transfer coefficient for free
convection with this geometry and air is h=6.5 w/m2 . C0. The surface area
is πdL, so the convection loss per unit length is
q/L]conv =h(πd )( Tw -
∞
)= (6.5)( π)(50-20) = 30.63 w/m
this pipe is a body surrounded by a large enclosure so the radiation heat
transfer can be calculated from this equation
q/L]rad =ε1(πd1)σ (T14 – T24 ) = (0.8 )( π )( 0.05) ( 5.669× 10 ) (3234
– 2934) = 25.04 w/m
the total heat loss is therefore
q/L]tot =
q/L]conv +
q/L]rad =30.63 +25.04 =55.67 w/m
In this example we see that the convection and radiation are about the same.
To neglect either would be a serious mistake.
Chapter 2
Steady- State Conduction
27
–OneiDimensionon
2-1 THE PLANEWALL
28
29
30
3. The time rate of change of the temperature at any point in the medium
may be determined from the heat equa on, Equa on 2.15, as
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32
1.7.1 Plane Walls in Series
33
In Fig. 1.9 for a three-layer
layer system, the temperature gradients in tthe
layers are different. The rate of heat conduction through each layer is qk,
and from Eq. (1.1) we get
Eliminating the intermediate temperatures T2 and T3 in Eq. qk can be
expressed in the form
Similarly, for N layers in series we have
surface temperature of layer 1 and TN+1 is the
where T1 is the outer-surface
outer- surface temperature of layer N. and ∆T is the overall
temperature difference, often called the temperature potential.
34
Figure 1.9 Conduc on Through a Three-Layer System in Series.
Example 1. 10 Calculate the rate of heat loss from a furnace wall per
unit area. The wall is constructed from an inner layer of 0.5 cm thick steel
(k : 40 W/m K) and an outer layer of 10 cm zirconium brick (k = 2.5 W/m
K) as shown in Fig. The inner-surface
surface temperature is 900 K and the
outside surface temperature is 460 K. What is the temperature at the
interface?
35
Solution
Assumptions:
• Assume that steady state exists,
• neglect effects at the corners and edges of the wall,
• the surface temperatures are uniform.
The rate of heat loss per unit area can be calculated from Eq:
36
Example 1. 11 Two large aluminum plates (k = 240 W/m K), each 1 cm
thick, with 10 µm. surface roughness the contact resistance Ri = 2.75 x 10
104 m2 K/W. The temperatures at the outside surfaces are 395°C and
405°C. Calculate (a) the heat flux (b) the temperature drop
p due to the
contact resistance.
Figure 1.11 Schema c Diagram of Interface Between Plates.
37
1.7.2 Plane Walls in Parallel
Conduction can occur in a section with two different materials in
parallel between the same poten al. Fig. 1.18 shows a slab with two
different materials of areas AA and AB in parallel. If the temperatures over
the left and right faces are uniform
orm at T1 and T2, the total rate of heat flow is
the sum of the flows through AA and AB:
38
Note that the total heat transfer area is the sum of AA and AB
and that the total resistance equals the product of the individual
resistancess divided by their sum, as in any parallel circuit. A more complex
applica on of the thermal network approach is illustrated in Fig. 1.19,
where heat is transferred through a composite structure involving
thermal resistances in series and in parallel. For this system the resistance
of the middle layer, R2 becomes and the rate of heat flow is
39
Figure 1.19 Conduc on Through a Wall Consis ng of Series and
Parallel Thermal Paths.
Example 1. 12
A layer of 2 in thick firebrick (kb = 1.0 Btu/hr °F) is placed between two ¼
in.-thick
thick steel plates (ks = 30 Btu/hr °F). The faces of the brick adjacent to
the plates are rough, having solid
solid-to-solid
solid contact over only 30 % of
the total area, with the average height of asperi es being L2=1/32 in. If
the surface temperatures of the steel plates are 200° and 800°F,
40
respec vely. The conduc vity of air ka is 0.02 Btu/hr
41
42
Example 1.13
43
A 0.1 m thick brick wall (k = 0.7 W/m K) is exposed to a cold wind at
270 K through a convec on heat transfer coefficient of 40 W/m2 K. On the
other side is air at 330 K, with a natural convec on heat transfer coefficient
of 10 W/m2 K. Calculate the rate
te of heat transfer per unit area. Solution
The three resistances are the rate of heat transfer per unit area is :
1.8.2 Convec on and Radia on in Parallel
In many engineering problems
roblems a surface loses or receives
thermal energy by convection
vection and radiation simultaneously.
Figure 1.23 illustrates the co current heat transfer from a
surface to its surroundings by convection and radiation.
44
45
46
47
An overall heat transfer coefficient U can be based on any chosen
area.
Example 1.16
In the design of a heat exchanger for aircraft application, the
maximum wall temperature in steady state is not to exceed 800 k.
For the conditions tabulated below, determine the maximum
permissible unit thermal resistance per square meter of the
metal wall that separates the hot gas Tgh = 1300 K from the
48
cold gas Tgc = 300 K. Combined heat transfer coefficient on
hot side h1= 200 W/m2 k combined heat heat transfer coefficient
on cold side h3 = 400 W/m2 k
49
Solving for R2 gives
R2 = 0.0025 m2 K/W
Example 1.11
The door for an industrial gas furnace is 2 m x 4 m in surface area
and is to beinsulated to reduce heat loss to no more than 1200
W/m2. The interior surface is a 3/8
3/8- in.-thick Inconel 600 sheet
(K= 25 W/m K), and the outer surface is a l/4 in.
in.-thick
thick sheet
of Stainless steel 316. Between these metal sheets a suitable
thickness of insulators material is to be placed. The effective gas
temperature inside the furnace is 1200°C, and the overall heat
transfer coefficient between the gas and the door is Ui = 20 W/m2
K. The heat transfer coefficient between the outer surface of the
door and the surroundings at 20°C is hc= 5 W/m2 K. calculate the
thickness of insulated should be use
50
51
52
53
-The composite wall of an oven consists of three materials, two of which are
of known thermal conductivity, kA= 20 w/m. k and kc = 50 w/m. k, and
known thickness, LA = 0.3 m and LC = 0.15 m. The third material , B, which
is sandwiched between materials A and C, is of known thickness , LB= 0.15
m, but unknown thermal conductivity kB . Under steady -state
state operations
conditions, measurements reveal an outer surface temperature of To,s =
20⁰C,
C, an inner surface temperature of Ti,s = 600⁰C, and an oven air
temperature of Tᴔ= 800⁰C.
C. The inside convection coefficient h is known to
2
be 25 w/m . k. what is the value of kB
54
2-4 RADIAL SYSTEMS
Cylinders
55
Consider a long cylinder of inside radius ri, outside radius ro, and length
L, such as the one shown in Figure 2-3.We expose this cylinder to a
temperature differential Ti −To and ask what the heat flow will
be.
Figure 2-3 One-dimensional
dimensional heat flow through a hollow cylinder
and electrical analog
Ar =2πrL
so that Fourier’s law is written
qr =−k Ar dT/dr
or
qr =−2π r L k dT/dr
56
[2-7]
For a general transient three
three-dimensional
dimensional in the cylindrical
coordinates T= T(r, φ ,z, t),, the general form of the conduction
equation incylindrical coordinates becomes
If the heat flow in a cylindrical shape is only in the radial direction and
for steady-state conditions with no heat generation the conduction
equation reduces to
57
A hot fluid flows through a tube that is covered by an insulating
58
material. The system loses heat to the surrounding air through an average
heat transfer coefficient hc,o. the thermal resistance of the two cylinders at
the inside of the tube and the outside of the insulation gives the thermal
network shown below the physical system where Th,∞ hot fluid temperature
Tc,∞ and the environmental air temperature
the rate of heat flow is
it is often convenient to define an overall heat transfer coefficient by the
equation
q = UAo (Thot-Tcold)
59
The area varies with radial distance. Thus, the numerical value of U will
depend on the area selected. Since the outermost diameter is the easiest to
measure in practice, Ao= 2л r3L is usually chosen as the base area.
Comparing between above Equations. we see that
Note that
UA=Ui Ai=Uo Ao
Figure 2-4 One-dimensional heat flow through multiple cylindrical sections
and electrical analog.
60
Figure 2-6 Resistance analogy for hollow cylinder with convection
boundaries.
61
For a hollow cylinder exposed to a convection environment on its inner
and outer surfaces,the electric-resistance
resistance analogy would appear as in Figure
2-6 where, again, TA and TB are the two fluid temperatures. Note that the
area for convection is not the sam
same for both fluids in this case, these areas
depending on the inside tube diameter and wall thickness. The overall heat
transfer would be expressed by
62
63
64
where Ta is the interface temperature, which may be obtained as Ta =595.8◦C
The largest thermal resistance clearly results from the insulation, and thus the major
portion of the temperature drop is through that material.
65
66
67
68
69
2-6 CRITICAL THICKNESS OF INSULATION
us consider a layer of insulation which might be installed around a circular
pipe, as shown in Figure 2-7.. The inner temperature of the insulation is fixed
at Ti, and the outer
Let
70
71
72
73
74
75
Heat Generation
A common thermal energy generation process involves
• The conversion from electrical to thermal energy in a current
current-carrying
carrying medium
Eg=I2R.
• The deceleration and absorption of neutrons in the fuel element of a nuclear reactor
• Exothermic chemical reactions occurring within a medium. Endothermic reacti
reactions
ons
would, of course, have the inverse effect
• A conversion from electro magnetic to thermal energy may occur due to the
absorption of radiation within the medium.
Plane Wall with Heat Generation
76
(a) Asymmetrical plane wall (b)Symmetrical plane wall (c) Adiabatic surface
at midline
Assumptions
• Uniform heat generation per unit volume q =Const.
• For constant thermal conductivity k=Const.
• One dimension and steady state heat transfer.
The appropriate form of the heat equation, is
The equation may be integrated twice to obtain the general solution
77
To obtain the constants of integration, C1 and C2 boundary conditions must
be introduced.
The Symmetrical Plane Wall
when both surfaces are maintained at a common temperature, Ts1= Ts2= Ts.
The temperature distribution is given by
The maximum temperature (T=To) exists at the midline (x=0).
78
for (case b) at x = L or the insulated plane wall case c). The energy balance
given by:
The surface temperature is
Example
A plane wall is a composite of two materials, A and B. The wall of material
A (k = 75 W/m K) has uniform heat generation 1.5 X 106 W/m3, and
thickness 50 mm. The wall material B has no generation with (k = 150 W/m
K) and thickness 20 mm. The inner surface of material A is well insulated,
while the outer surface of material B is cooled by a water stream with 30°C
and heat transfer coefficient 1000 W/m2 K.
1. Sketch the temperature distribution that exists in the composite
under steady-state conditions.
79
2. Determine the maximum temperature To of the insulated surface and
the temperature of the cooled surface Ts.
Solution
Assumptions:
1. Steady-state conditions.
2. One-dimensional conduction in x direction.
3. Negligible contact resistance between walls.
4. Inner surface of A adiabatic.
5. Constant properties for materials A and B.
80
2-8 CYLINDER WITH HEAT SOURCES
81
82
83
Example
84
2-9 CONDUCTION-CONVECTION
CONVECTION SYSTEMS
85
The heat that is conducted through a body must frequently be
removed (or delivered) by some convection process. Obviously, an analysis
of combined conduction-convection systems is very important from a
practical standpoint. Extended surfaces have wide industrial application as
fins attached to the walls of heat transfer equipment in order to increase the
rate of heating or cooling q = h As (Ts- T∞). Fins come in many shapes and
forms, some of which are shown below:
86
Consider a pin fin having the shape of a rod whose base is attached to a
wall at surface temperature Ts. The fin is cooled along its surface by a fluid at
temperature T∞ To derive an equation for temperature distribution, we make
a heat balance for a small element of the fin. Heat flows by conduction into
the left face of the element, while heat flows out of the element by
conduction through the right face and by convection from the surface.
Assumptions
1. The fin has a uniform cross-sectional area
2. The fin is made of a material having uniform conductivity (k = constant)
3. The heat transfer coefficient between the fin and the fluid is constant
(h=constant).
4. One dimensional steady state condition only.
5. Non heat generation (q=0).
6. Radiation is negligible.
87
Where
P is the perimeter of the fin
Pdx is the fin surface area between x and x+dx.
A Cross section area of fin
If k and h are uniform, Eq. 2.34 simplifies to the form
If
θ(x) = [T(x) - T∞],
88
89
Representations of Four Boundary Conditions at the Tip of a Fin
90
Differentiating
91
92
Substituting the above relations for C1 and C2 into Eq.(2.37)
93
The temperature distribution is:
The heat loss from the fin can be found by substituting the temperature
gradient at the root into Eq.(2.37), we get
94
Substituting above equations in B.C.2
95
Substituting the above relations for C1 and C2 into Eq.(2.37)
The temperature distribution is:
96
The heat loss from the fin can be found by substituting the temperature
gradient at the root into Eq.(2.37), we get
Temperature distribution and rate of heat transfer for fins
97
Fin Performance
The heat transfer effectiveness of a fin is measured by a parameter called fin
effectiveness and the fin efficiency, which is defined as
I. Fin Effectiveness ε. A ratio of the fin heat transfer rate to the heat transfer
rate that would exist without the fin.
98
where Ac is the fin cross-sectional area at the base. the use of fins may rarely
be justified unless ε >= 2.
II. Fin Efficiency η
Where as for a fin of rectangular cross section (length L & thickness t) and
an adiabatic end (Case 2) is
a corrected fin length of the form Lc = L + (t/2).
99
A fin efficiency for a circular pin fin (Diameter D & Length L) and an
adiabatic end (Case 2) is
where Af is the total surface area of the fin and Ab is the base area. For the
insulated-tipfin described by Equation ((2-36),
Af = PL
Ab = A and the heat ratio would become
This term is sometimes called the fin effectiveness.
100
101
102
103
Example
Example
104
The cylinder barrel of a motorcycle is constructed of 2024-T6 aluminum
alloy (k = 186 W/m K) and is of height H = 0.15 m and OD = 50 mm. Under
typical operating conditions the outer surface of the cylinder is at a
temperature of 500 K and is exposed to ambient air at 300 K, with a
convection coefficient of 50 W/m2 K. Annular fins of rectangular profile are
typically added to increase heat transfer to the surroundings. Assume that
five (N=5) such fins, which are of thickness t = 6 mm, length L = 20 mm and
equally spaced, are added. What is the increase in heat transfer due to
addition of the fins?
Solution
Assumptions:
1. Steady-state conditions.
2. One-dimensional radial conduction in fins.
3. Constant properties.
4. No internal heat generation.
5. Negligible radiation exchange with surroundings.
6. Uniform convection coefficient over outer surface (with or without fins).
With the fins in place, the heat transfer rate is q=qf+qb
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106
107
108
109
110
111
112
113
114
115
116
117
ANSWER SHEET
QUESTION 1
118
QUESTION 2
119
QUESTION 3
120
QUESTION 4
121
QUESTION 5
122
QUESTION 6
123
QUESTION 7
124
125
QUESTION 8
126
QUESTION 9
127
QUESTION 10
128
Chapter -3- Steady State Conduction – Multiple
Dimensions
Introduction
D steady state without heat source in Cart. Coor., the Laplace equation
2-D
The total heat flow at any point in the material is he resultant of the qx and qy at that point. Thus the
total heat-flow
flow vector is directed so that it is perpendicular to the lines of constant temperature in the
material in Fig. 3-1.
Separation of variables in the steady state problems
Here we consider an example. (Fig. 3-2)
129
130
3-3 GRAPHICALANALYSIS
Consider the two-dimensional
dimensional system shown in Figure 3-3. The inside surface is maintained at some
temperature T1, and the outer surface is maintained at T2.We wish to calculate the heat transfer.
Isotherms and heat-flow
flow lanes have been sketched to aid in this calculation. The isotherms and heat
flow lanes form groupings of curvilinear figures like that shown in Figure 3-3b.. The heat flow across
this curvilinear section is given by Fourier’s law, assuming unit depth of material:
Figure 3-3 Sketch showing element used for curvilinear
curvilinear-square
analysis of two-dimensional
dimensional heat flow.
131
This heat flow will be the same through each section within this heat
heat-flow lane, and the total
heat flow will be the sum of the heat flows through all the lanes. If the sketch is drawn so
that ∆x ≈ ∆y,, the heat flow is proportional to the ∆T across the element and, since this heat
flow is constant, the ∆T across each element must be the same within the same heat
heat-flow
lane. Thus the ∆T across an element is given by
by:
where N is the number of temperature increments betwee
between the inner and outer surfaces.
Furthermore, the heat flow through each lane is the same since it is independent of the
dimensions ∆x and ∆y when they are constructed equal. Thus we write for the total heat
transfer
where M is the number of heat-flow lanes. So, to calculate the heat transfer, we need only construct
these curvilinear-square
square plots and count the number of temperature increments and heat-flow
flow lanes.
Care must be taken to construct the plot so that ∆x ≈ ∆y and the lines are perpendicular. For the corner
section shown in Figure 3-3a the number of temperature increments between the inner and outer
surfaces is about N = 4,, while the number of heat flow lanes for the corner section may be estimated as
M =8.2. The total number of heat-flow lanes is four times this value, or 4×8.2=32.8.. The ratio M/N is
thus 32.8/4=8.2 for the whole wall section. This ratio will be called the conduction shape factor in
subsequent discussions.
3-4 THE CONDUCTION SHAPE FACTOR
The values of S have been worked out for several geometries and are summarized in
Table 3-1
Table 3-1 Conduction shape factors, summarized from References 6 and 7.
Note: For buried objects the temperature difference is ∆T =Tobject −Tfar field. The far-field
temperature is taken the same as the isothermal surface temperature for semi
semi-infinite
infinite media.
132
133
134
For a three-dimensional
dimensional wall, as in a furnace, separate shape factors are used to calculate the heat flow
through the edge and corner sections, with the dimensions shown in Figure 3-4. When all the interior
dimensions are greater than one-fifth
fifth of the wall thickness,
135
where
A= area of wall
L= wall thickness
D= length of edge
Figure 3-4 Sketch illustrating dimensions for use in calculating three
dimensional shape factors
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NUMERICAL METHOD OFANALYSIS
In many practical situations the geometry or boundary conditions are such that an analytical
solution has not been obtained at all, or if the solution has been developed, it involves such a complex
series solution that numerical evaluation becomes exceedingly difficult. For such situations the most
fruitful approach to the problem is one based on finite-difference techniques, the basic
principles of which we shall outline in this section.
Consider a two-dimensional
dimensional body that is to be divided into equal increments in both the
x and y directions, as shown in Figure 3-5.
The nodal points are designated as shown, the m locations indicating the x increment and
the n locations indicating the y increment. We wish to establish the temperatures at any of
these nodal points within the body, using Equation ((3-1)) as a governing condition. Finite differences
are used to approximate differential increments in the temperature and space coordinates; and the
smaller we choose these finite increments, the more closely the true temperature
distribution will be approximated
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Equation (3-24) states very simply that the net heat flow into any node is zero at steady-state
conditions.
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Chapter 4
Unsteady- State Condition
To analyze a transient heat-transfer problem, we could proceed by solving the general heat conduction
equation by the separation-of-variables method, similar to the analytical
treatment used for the two-dimensional steady-state problem
Consider the infinite plate of thickness 2L shown in Figure 4-1. Initially the plate is at a uniform
temperature Ti, and at time zero the surfaces are suddenly lowered to T =T1. The differential equation
is
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This solution will be presented in graphical form for calculation purposes.
LUMPED-HEAT-CAPACITY SYSTEM
We continue our discussion of transient heat conduction by analyzing systems that may be
Considered
- uniform in temperature. This type of analysis is called the lumped-heat-capacity method.
Such systems are obviously idealized because a temperature gradient must exist in a material if heat is
to be conducted into or out of the material. In general,
- the smaller the physical size of the body, the more realistic the assumption of a uniform
temperature throughout.
-If the resistance to heat transfer by conduction were small compared with the convection
resistance at the surface, so that the major temperature gradient would occur through the fluid layer
at the surface. The lumped-heat-capacity analysis, then, is one that assumes that the internal resistance
of the body is negligible in comparison with the external resistance. The convection heat loss from the
body is evidenced as a decrease in the internal energy of the body, as shown in Figure 4-2. Thus,
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…………………1
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where Te is the temperature of the convection environment. The thermal network for the singlecapacity system is shown in Figure 4-2b. In this network we notice that the thermal capacity of the
system is “charged” initially at the potential T0 by closing the switch S.
Then, when the switch is opened, the energy stored in the thermal capacitance is dissipated through the
resistance 1/hA. The analogy between this thermal system and an electric system is apparent, and we
could easily construct an electric system that would behave exactly like the thermal system as long as
we made the ratio
The quantity c ρV/hA is called the time constant of the system because it
has the dimensions of time. When
it is noted that the temperature difference T −T∞ has a value of 36.8 percent of the initial difference
T0 −T∞.
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Substitute eq. 1
Example
Quenching of a Steel Plate
A steel plate 1 cm thick is taken from a furnace at 600°Cand
Cand quenched in a bath of oil at 30
°C.
C. If the heat transfercoefficient is estimated to be 400 W/m2 K, how long will
it take for the plate to cool to 100°C?
C? Take k, ρ, and c for the steel as 50 W/m K, 7800
kg/m3, and 450 J/kg K, respectively.
Given: Steel plate quenched in an oil bath.
Required: Time to cool from 600°C to 100
100°C.
Assumptions: Lumped thermal capacity model valid.
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The temperature of a gas stream is to be measured by a thermocouple whose junction
can be approximated as a 1.2-mm-diameter sphere. The properties of the junction are k
= 35 W/m · °C, ρ=8500 kg/m3, and Cp = 320 J/kg · °C, andthe heat transfer coefficient
between the junction and the gas is h = 65 W/m2 · °C. Determine how long it will take
for the thermocouple to read 99 percent of the initial temperature difference.
Answer: 38.5 s
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In a manufacturing facility, 2-in.-diameter brass balls (k = 64.1 Btu/h
· ft · °F, ρ = 532 lbm/ft3, and Cp = 0.092 Btu/lbm · °F) initially at 250°F
are quenched in a water bath at 120°F for a period of 2 min at a rate of 120
balls per minute. If the convection heat transfer coefficient is 42 Btu/h · ft2
· °F, determine (a) the temperature of the balls after quenching and (b) the
rate at which heat needs to be removed from the water in order to keep its
temperature constant at 120°F
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To warm up some milk for a baby, a mother pours milk into a thin-walled
glass whose diameter is 6 cm. The height of the milk in the glass is 7 cm.
She then places the glass into a large pan filled with hot water at 60°C. The
milk is stirred constantly, so that its temperature is uniform at all times. If
the heat transfer coefficient between the water and the glass is 120 W/m2 ·
°C, determine how long it will take for the milk to warm up from 3°C to
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38°C. Take the properties of the milk to be the same as those of water. Can
the milk in this case be treated as a lumped system? Why?
Answer: 5.8 min
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