Chapter 1 Heat conduction

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Chapter 1 Heat conduction
1.1 Heat flow through plane wall 1.2 Heat flow
through radial systems cylinder and sphere 1.3
The overall heat-transfer coefficient. 1.4
Transient conduction
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1.1 Heat flow through plane wall
First consider the plane wall where a direct
application of Fouriers law may be made.
Integration yields
?x
T1
Q
Q heat rate in direction normal to surface ?x
Wall thickness T1, T2 the wall face
temperature A surface area k thermal
conductivity
T2
A plane wall
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1.1 Heat flow through plane wall
The temperature gradients in the three materials
are shown, and the heat flow may be written
Solving these three equations simultaneously, the
heat flow is written
composite wall
Note the heat flow must be the same through all
sections.
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1.1 Heat flow through plane wall
A relation quite like Ohms law in
electric-circuit theory Rth the thermal
resistances of the various materials
composite wall
Note the heat flow must be the same through all
sections.
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Ex.1 A composite wall is formed of a 2.5-cm
copper plate, a 3.2-mm layer of asbestos, and a
5-cm layer of glass wool. The wall is subjected
to an overall temperature difference of 560ºC.
Calculate the heat flow per unit area through the
composite structure. kcopper 385
W/m.ºC kasbestos 0.166 W/m.ºC Kglass wool 2.22
W/m.ºC
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1.1 Heat flow through plane wall
q ?
Construct the electrical analog
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1.1 Heat flow through plane wall
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1.1 Heat flow through plane wall
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1.2 Heat flow through radial system
Fouriers law
Boundary conditions TTi at rri TT0 at rr0
cylinder
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1.2 Heat flow through radial system
Multilayer cylinder
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Ex.2 A thick-wall tube of stainless steel 18
Cr, 8 Ni, k19 W/m. ºC with 2-cm inner diameter
(ID) and 4-cm outer diameter (OD) is covered with
a 3-cm layer of asbestos insulation k0.2 W/m.
ºC. If the inside wall temperature of the pipe
is maintained at 600 ºC and the outside surface
temperature is 100 ºC, calculate the heat loss
per meter of length. Also calculate the
tube-insulation interface temperature.
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1.2 Heat flow through radial system
A?
r0
ri
Boundary conditions TTi at rri TT0 at rr0
Hollow sphere
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EX.3 A hollow sphere is constructed of aluminum
with an inner diameter of 4 cm and an outer
diameter of 8 cm. The inside temperature is 100
ºC and the outer temperature is 50 ºC. Calculate
the heat transfer. Kaluminum 206 W/mºC
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1.3 Overall heat transfer coefficient
The overall heat transfer by combined conduction
and convection is frequently expressed in terms
of an overall heat transfer coefficient U
q3 convection
q1 convection
q2 conduction
q q1 q2 q3
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1.3 Overall heat transfer coefficient
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1.3 Overall heat transfer coefficient
Overall heat transfer coefficient
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EX.4 A certain building wall consists of 6.0 in
of concrete k1.2 W/m.ºC, 2.0 in of
fiberglassk0.05 W/m. ºC insulation, and 3/8 in
of gypsum board k0.05 W/m. ºC. The inside and
outside convection coefficients are 2.0 and 7.0
Btu/h.ft2.ºF, respectively. The outside air
temperature is 20ºF, and the inside temperature
is 72ºF. Calculate the overall heat-transfer
coefficient for the wall, the R value, and the
heat loss per unit area.
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1.4 Transient Conduction
Water
q
Hot Solid Vvolume Cpspecific heat ?
density T(t) temperature T0initial
temperature t time
Ambient fluid Too, h
Transient conduction is the temperature within
the system is changed with time t.
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1.4 Unsteady State Conduction
1.4.1 LUMPED-HEAT-CAPACITY SYSTEM
Heat loss equation
Rate of heat flow into the solid of volume V
through boundary surfaces A
Rate of increase of internal energy of the solid
of volume V
Initial condition
If the Biot number lt 0.1, the internal resistance
is negligible.
Biot number
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Ex.5 A steel ball c0.46 kJ/kg.ºC, k35 W/m.ºC
5.0 cm in diameter and initially at a uniform
temperature of 450 ºC is suddenly placed in a
controlled environment in which the temperature
is maintained at 100 ºC. The convection
heat-transfer coefficient is 10 W/m2. ºC.
Calculate the time required for the ball to
attain a temperature of 150 ºC.
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Homework2

1. Find the heat transfer per unit area through the
   composite wall in Figure below. Assume
   one-dimensional heat flow.
2. A 2.0-in schedule 40 steel pipe (Outer diameter
   2.375 in and inner diameter 2.067 in) has k
   27 Btu/h.ft.F. The fluid inside the pipe has h
   30 But/h.ft2. F, and the outer surface of the
   pipe is covered with 0.5-in fiberglass insulation
   with k 0.023 Btu/h.ft. F. The convection
   coefficient on the outer insulation surface is
   2.0 Btu/h.ft2. F. The inner fluid temperature is
   320 F and the ambient temperature is 70 F.
   Calculate the heat loss per foot of length.

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1.4.2 Transient Heat Flow in a Semi-infinite solid
Figure 0 Transient Temperature distribution in a
semi-infinite solid for three surface conditions
case (1) constant surface temperature, case (2)
constant surface heat flux, and case (3) surface
convection
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Case I Constant Surface of a Semi-infinite solid
The semi-infinite solid maintained at some Ti.
The surface temperature is suddenly lowered and
maintained at a T0.
x
The differential equation for the temperature
distribution T(x, ) is
The boundary and initial conditions are
for gt 0
The boundary and initial conditions are
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Where the Gauss error function is defined as The
values of the error function are tabulated in the
table below.
0.00 0.00000 0.76
0.71754 1.52 0.96841
0.02 0.02256 0.78
0.73001 1.54 0.97059
0.04 0.04511 0.80
0.74210 1.56 0.97263
0.06 0.06762 0.82
0.75381 1.58 0.97455
0.08 0.09008 0.84
0.76514 1.60 0.97636
0.10 0.11246 0.86
0.77610 1.62 0.97804
0.12 0.13476 0.88
0.78669 1.64 0.97962
0.14 0.15695 0.90
0.79691 1.66 0.98110
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0.16 0.17901 0.92
0.80677 1.68 0.98249
0.18 0.20094 0.94
0.81627 1.70 0.98379
0.20 0.22270 0.96
0.82542 1.72 0.98500
0.22 0.24430 0.98
0.83423 1.74 0.98613
0.24 0.26570 1.00
0.84270 1.76 0.98719
0.26 0.28690 1.02
0.85084 1.78 0.98817
0.28 0.30788 1.04
0.85865 1.80 0.98909
0.30 0.32863 1.06
0.86614 1.82 0.98994
0.32 0.34913 1.08
0.87333 1.84 0.99074
0.34 0.36936 1.10
0.88020 1.86 0.99147
0.36 0.38933 1.12
0.88079 1.88 0.99216
0.38 0.40901 1.14
0.89308 1.90 0.99279
0.40 0.42839 1.16
0.89910 1.92 0.99338
0.42 0.44749 1.18
0.90484 1.94 0.99392
0.44 0.46622 1.20
0.91031 1.96 0.99443
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0.46 0.48466 1.22
0.91553 1.92 0.99489
0.48 0.50275 1.24
0.92050 1.94 0.995322
0.50 0.52050 1.26
0.92524 1.96 0.997020
0.52 0.53790 1.28
0.92973 1.98 0.998137
0.54 0.55494 1.30
0.93401 2.20 0.998857
0.56 0.57162 1.32
0.93806 2.30 0.999311
0.58 0.58792 1.34
0.94191 2.40 0.999593
0.60 0.60386 1.36
0.94556 2.50 0.999764
0.62 0.61941 1.38
0.94902 2.70 0.999866
0.64 0.63459 1.40
0.95228 2.80 0.999925
0.66 0.64938 1.42
0.95538 2.90 0.999959
0.68 0.66278 1.44
0.95830 3.00 0.999978
0.70 0.67780 1.46
0.96105 3.20 0.999994
0.72 0.69143 1.48
0.96365 3.40 0.999998
0.74 0.70468 1.50
0.96610 3.60 1.000000
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Ex.6 A large block of steel k45 W/mC, a 1.4
x 10-5 m2/s is initially at a uniform
temperature of 35C. The surface is exposed to a
heat flux (a) by suddenly raising the surface
temperature to 250C and (b) through a constant
surface heat flux of 3.2x105 W/m2. Calculate the
temperature at a depth of 2.5 cm after a time of
0.5 min.
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Case II Constant heat flux of a Semi-infinite
solid
Semi-infinite solids suddenly exposed to
convection environment at T8
x
Constant surface heat flux
The differential equation for the temperature
distribution T(x, ) is
The result is
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Case III Constant Surface of a Semi-infinite solid
h
Semi-infinite solids suddenly exposed to
convection environment at T8
x
Heat convected into surface heat conducted into
surface
The differential equation for the temperature
distribution T(x, ) is
The result is
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Where
Ti initial temperature of solid T8 environment
temperature
This solution is presented in graphical form in
figure on right hand side
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Ex.7 A large slab of aluminum at a uniform
temperature of 200 C is suddenly exposed to a
convection-surface environment of 70C with a
heat-transfer coefficient of 525 W/m2. C.
Calculate the time required for the temperature
to reach 120C at the depth of 4.0 cm for this
circumstance.
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For other geometries that finite thickness
exposed to convection
Infinite plate to thickness 2 L infinite
cylinder of radius r0 Sphere of radius r0
r0
T0 centerline temperature T0 centerline
temperature T0 centerline temperature

The convection environment temperature is
designated as T and the center temperature for x
0 or r 0 is T0. At T0, each soild is assumed
to have a uniform initial temperature Ti.
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Fig. 1 Midplane temperature as a function of time
for a plane wall of thickness 2L
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Fig. 2 Temperature distribution in a plane wall
of thickness 2L
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Fig. 3 Internal energy change as a function of
time for a plane wall of thickness 2L
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Fig. 4 Centerline temperature as a function of
time for an infinite cylinder of radius r0
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Fig. 5 Temperature distribution in an infinite
cylinder of radius r0
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Fig. 6 Internal energy change as a function of
time for an infinite cylinder of radius r0
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Fig. 7 Center temperature as a function of time
sphere of radius r0
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Fig. 8 Temperature distribution as a function of
time sphere of radius r0
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Fig. 9 Internal energy change as a function of
time for sphere of radius r0
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Temperature in the solid are given in Fig. 1-9 as
functions of time and spatial position. In these
charts we note the definitions
or
If the centerline temperature is desired, only
one chart is required to obtain a value for
and . To determine an off-center temperature
two charts are required to calculate the product.
For example, Fig.1 and 4 would be employed to
calculate an off-center temperature for an
infinite plate.
The heat losses can be calculated by using below
equation.
In these figures Q is the actual heat lost by the
body in time .
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A quick inspection of Fig. 1 to 9 indicates that
the dimensionless temperature profiles and heat
flow may all be expressed in terms of two
dimensionless parameters called the biot and
Fourier numbers
Biot number Fourier number
The parameter Ls (Ls V/A) designates some
characteristic dimensions of the body for the
plate it is the half-thickness, whereas for the
cylinders and sphere it is the radius.
A very low value of the Biot modulus means that
internal-conduction resistance is negligible in
comparison with surface-convection resistance.
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Ex.8 A long aluminum cylinder 5.0 cm in diameter
and initially at 200C is suddenly exposed to a
convection environment at 70 C and h 525
W/m2.C. Calculate the temperature at a radius of
1.25 cm and the heat lost per unit length 1 min
after the cylinder is exposed to the environment.
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Homework3

1. On a hot summer day a concrete driveway may reach
   a temperature of 50C. Suppose that a stream of
   water is directed on the driveway so that the
   surface temperature is suddenly lowered to 10C.
   How long will it take to cool the concrete to
   25C at a depth of 5 cm from the surface?
2. A large slab of aluminum has a thickness of 10 cm
   and is initially uniform in temperature at 400C.
   Suddenly it is exposed to a convection
   environment at 90?C with h 1400 W/m2. C. How
   long does it take the centerline temperature to
   drop to 180C? Also calculate the heat loss per
   unit length.