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Heat Conduction Equation
BMCG 2123Faculty of Manufacturing Engineering,
UTeMTaufik
Week 204/11/23 BMCG 2123 Heat Transfer 1
04/11/23 BMCG 2123 Heat Transfer 2
Learning Objectives At the end of this chapter, students
should be able to: Explain multidimensionality and time dependence of
heat transfer. Obtain the differential equation of heat conduction
in various coordinate systems. Solve onedimensional heat conduction problems
and obtain the temperature distributions within a medium and the heat flux,
Analyze onedimensional heat conduction in solids that involve heat generation.
Introduction
• Although heat transfer and temperature are closely related, they are of a different nature.
• Temperature has only magnitudeit is a scalar quantity.
• Heat transfer has direction as well as magnitude it is a vector quantity.
• We work with a coordinate system and indicate direction with plus or minus signs.
Introduction ─ Continue• The driving force for any form of heat transfer is the
temperature difference.• The larger the temperature difference, the larger the
rate of heat transfer.• Three prime coordinate systems:– rectangular (T(x, y, z, t)) ,– cylindrical (T(r, , z, t)),– spherical (T(r, , , t)).
Classification of conduction heat transfer problems:• steady versus transient heat transfer,• multidimensional heat transfer,• heat generation.
Introduction ─ Continue
Steady versus Transient Heat Transfer
• Steady implies no change with time at any point within the medium
• Transient implies variation with time or time dependence
Multidimensional Heat Transfer
• Heat transfer problems are also classified as being:– onedimensional,– two dimensional,– threedimensional.
• In the most general case, heat transfer through a medium is threedimensional. However, some problems can be classified as two or onedimensional depending on the relative magnitudes of heat transfer rates in different directions and the level of accuracy desired.
• The rate of heat conduction through a medium in a specified direction (say, in the xdirection) is expressed by Fourier’s law of heat conduction for onedimensional heat conduction as:
• Heat is conducted in the direction of decreasing temperature, and thus the temperature gradient is negative when heat is conducted in the positive xdirection.
(W)cond
dTQ kA
dx (21)
General Relation for Fourier’s Law of Heat Conduction
• The heat flux vector at a point P on the surface of the figure must be perpendicular to the surface, and it must point in the direction of decreasing temperature
• If n is the normal of the isothermal surface at point P, the rate of heat conduction at that point can be expressed by Fourier’s law as
(W)n
dTQ kA
dn (22)
General Relation for Fourier’s Law of Heat ConductionContinue
• In rectangular coordinates, the heat conduction vector can be expressed in terms of its components as
• which can be determined from Fourier’s law as
n x y zQ Q i Q j Q k
x x
y y
z z
TQ kA
xT
Q kAy
TQ kA
z
(23)
(24)
Heat Generation• Examples:– electrical energy being converted to heat at a rate of I2R,– fuel elements of nuclear reactors,– exothermic chemical reactions.
• Heat generation is a volumetric phenomenon.• The rate of heat generation units : W/m3 or Btu/h · ft3.• The rate of heat generation in a medium may vary with
time as well as position within the medium. • The total rate of heat generation in a medium of
volume V can be determined from
(W)gen gen
V
E e dV (25)
Exercise 1• The resistance wire of a 1200W hair dryer is 80 cm
long and has diameter of D = 0.3 cm. Determine: a) the rate of heat generation in the wire per unit
volume, in W/cm3, andb) The heat flux on the outer surface of the wire as
result of this heat generation.
04/11/23 BMCG 2123 Heat Transfer 12
Solution 1
a) The rate of heat generation:
b) The heat flux on the outer surface:
04/11/23 BMCG 2123 Heat Transfer 13
3
22 W/cm212
804/)3.0(
1200
4/
cmcm
W
LD
E
V
Ee
gen
wire
gengen
2/ 9.15)80)(3.(
W1200cmW
cmcmoDL
E
A
EQ gen
wire
gens
OneDimensional Heat Conduction Equation  Plane Wall
xQ
Rate of heat
conduction
at x
Rate of heat
conduction
at x+x
Rate of heat
generation inside the element
Rate of change of the energy content of
the element
 + =
,gen elementE x xQ elementE
t
(26)
• The change in the energy content and the rate of heat generation can be expressed as
• Substituting into Eq. 2–6, we get
,
element t t t t t t t t t
gen element gen element gen
E E E mc T T cA x T T
E e V e A x
,element
x x x gen element
EQ Q E
t
(26)
(27)
(28)
x x xQ Q (29)gene A x t t tT T
cA xt
1gen
T TkA e c
A x x t
(211)
• Dividing by Ax, taking the limit as x 0 and t 0, and from Fourier’s law:
The area A is constant for a plane wall the one dimensional transient heat conduction equation in a plane wall is
gen
T Tk e c
x x t
Variable conductivity:
Constant conductivity:2
2
1 ; geneT T k
x k t c
1) Steadystate:
2) Transient, no heat generation:
3) Steadystate, no heat generation:
2
20gened T
dx k
2
2
1T T
x t
2
20
d T
dx
The onedimensional conduction equation may be reduces to the following forms under special conditions
(213)
(214)
(215)
(216)
(217)
OneDimensional Heat Conduction Equation  Long
Cylinder
rQ
Rate of heat
conductionat r
Rate of heat
conduction
at r+r
Rate of heat
generation inside the element
Rate of change of the energy content of
the element
 + =
,gen elementE elementE
t
r rQ
(218)
• The change in the energy content and the rate of heat generation can be expressed as
• Substituting into Eq. 2–18, we get
,
element t t t t t t t t t
gen element gen element gen
E E E mc T T cA r T T
E e V e A r
,element
r r r gen element
EQ Q E
t
(218)
(219)(220)
r r rQ Q (221)gene A r t t tT T
cA rt
1gen
T TkA e c
A r r t
(223)
• Dividing by Ar, taking the limit as r 0 and t 0, and from Fourier’s law:
Noting that the area varies with the independent variable r according to A=2rL, the one dimensional transient heat conduction equation in a plane wall becomes1
gen
T Trk e c
r r r t
10gened dT
rr dr dr k
The onedimensional conduction equation may be reduces to the following forms under special conditions
1 1geneT Tr
r r r k t
1 1T Tr
r r r t
0d dT
rdr dr
Variable conductivity:
Constant conductivity:
1) Steadystate:
2) Transient, no heat generation:
3) Steadystate, no heat generation:
(225)
(226)
(227)
(228)
(229)
OneDimensional Heat Conduction Equation  Sphere
22
1gen
T Tr k e c
r r r t
22
1 1geneT Tr
r r r k t
Variable conductivity:
Constant conductivity:
(230)
(231)
General Heat Conduction Equation
x y zQ Q Q
Rate of heat
conduction
at x, y, and z
Rate of heat
conduction
at x+x, y+y, and
z+z
Rate of heat
generationinside theelement
Rate of changeof the energy
content of the
element
 + =
x x y y z zQ Q Q ,gen elementE elementE
t
(236)
Repeating the mathematical approach used for the onedimensional heat conduction the threedimensional heat conduction equation is determined to be
2 2 2
2 2 2
1geneT T T T
x y z k t
2 2 2
2 2 20geneT T T
x y z k
2 2 2
2 2 2
1T T T T
x y z t
2 2 2
2 2 20
T T T
x y z
Twodimensional
Threedimensional
1) Steadystate:
2) Transient, no heat generation:
3) Steadystate, no heat generation:
Constant conductivity: (239)
(240)
(241)
(242)
Exercise 2• A 2kW resistance heater with thermal conductivity
k = 15 W/m.K, diameter D = 0.4 cm, and length L = 50 cm is used to boil water by immersing it in water. Assuming the variation of thermal conductivity of the wire with temperature to be negligible, obtain the differential equation that describes the variation of the temperature in the wire during steady state operation.
04/11/23 BMCG 2123 Heat Transfer 23
Analysis
The resistance wire of a water heater is considered to be a very long cylinder since its length is more than 100 times its diameter. Also heat is generated uniformly in the wire and the conditions on the outer surface of the wire are uniform. Therefore, it is reasonable to expect the temperature in the wire to vary in the radial r direction only and thus the heat transfer to be onedimensional. Then we have T=T(r) during steady operation since the temperature in this case depends on r only.
04/11/23 BMCG 2123 Heat Transfer 24
Solution 1
The rate of heat generation in the wire per unit volume:
Nothing that the thermal conductivity is given to be constant, the differential equation that governs the variation of temperature in the wire is simply Eq.227
Which steady onedimensional heat conduction equation in cylindrical coordinates for the case of constant thermal conductivity.
04/11/23 BMCG 2123 Heat Transfer 25
39
22 W/m10318.0
5.04/004.0
2000
4/x
mm
W
LD
E
V
Ee
gen
wire
gengen
01
k
e
dr
dTr
dr
d
rgen
Cylindrical Coordinates
2
1 1gen
T T T T Trk k k e c
r r r r z z t
(243)
Spherical Coordinates
22 2 2 2
1 1 1sin
sin sin gen
T T T Tkr k k e c
r r r r r t
(244)
Boundary and Initial Conditions
• Specified Temperature Boundary Condition• Specified Heat Flux Boundary Condition• Convection Boundary Condition• Radiation Boundary Condition• Interface Boundary Conditions• Generalized Boundary Conditions
Specified Temperature Boundary Condition
For onedimensional heat transfer through a plane wall of thickness L, for example, the specified temperature boundary conditions can be expressed as
T(0, t) = T1
T(L, t) = T2
The specified temperatures can be constant, which is the case for steady heat conduction, or may vary with time.
(246)
Specified Heat Flux Boundary Condition
dTq k
dx
Heat flux in the positive xdirection
The sign of the specified heat flux is determined by inspection: positive if the heat flux is in the positive direction of the coordinate axis, and negative if it is in the opposite direction.
The heat flux in the positive xdirection anywhere in the medium, including the boundaries, can be expressed by Fourier’s law of heat conduction as
(247)
Two Special Cases
Insulated boundary Thermal symmetry
(0, ) (0, )0 or 0
T t T tk
x x
,2 0LT t
x
(249) (250)
Convection Boundary Condition
1 1
(0, )(0, )
T tk h T T t
x
2 2
( , )( , )
T L tk h T L t T
x
Heat conduction
at the surface in a
selected direction
Heat convection
at the surface in the same direction
=
and
(251a)
(251b)
Radiation Boundary Condition
Heat conductionat the surface in aselected direction
Radiation exchange at the surface
inthe same direction
=
4 41 ,1
(0, )(0, )surr
T tk T T t
x
4 42 ,2
( , )( , ) surr
T L tk T L t T
x
and
(252a)
(252b)
Interface Boundary Conditions
0 0( , ) ( , )A BA B
T x t T x tk k
x x
At the interface the requirements are:(1) two bodies in contact must have the same
temperature at the area of contact,(2) an interface (which is a
surface) cannot store any energy, and thus the heat flux on the two sides of an interface must be the same.
TA(x0, t) = TB(x0, t)
and(253)
(254)
Generalized Boundary ConditionsIn general a surface may involve convection, radiation, and specified heat flux simultaneously. The boundary condition in such cases is again obtained from a surface energy balance, expressed as
Heat transferto the
surfacein all
modes
Heat transferfrom the surface
In all modes
=
Heat Generation in SolidsThe quantities of major interest in a medium with heat generation are the surface temperature Ts and the maximum temperature Tmax that occurs in the medium in steady operation.
The heat transfer rate by convection can also be expressed from Newton’s law of cooling as (W)s sQ hA T T
gens
s
e VT T
hA
Rate ofheat
transferfrom the
solid
Rate ofenergy
generationwithin the
solid
=
For uniform heat generation within the medium (W)genQ e V

Heat Generation in Solids The Surface Temperature
(264)
(265)
(266)
(263)
Heat Generation in Solids The Surface Temperature
For a large plane wall of thickness 2L (As=2Awall and V=2LAwall)
, gen
s plane wall
e LT T
h
For a long solid cylinder of radius r0 (As=2r0L and V=r0
2L) 0, 2
gens cylinder
e rT T
h
For a solid sphere of radius r0 (As=4r02 and V=4/3r0
3)
0, 3
gens sphere
e rT T
h
(268)
(269)
(267)
Heat Generation in Solids The maximum Temperature in a Cylinder
(the Centerline)The heat generated within an inner cylinder must be equal to the heat conducted through its outer surface.
r gen r
dTkA e V
dr
Substituting these expressions into the above equation and separating the variables, we get 22
2gen
gen
edTk rL e r L dT rdr
dr k
Integrating from r =0 where T(0) =T0 to r=ro2
0max, 0 4
gencylinder s
e rT T T
k
(271)
(270)
Exercise 3• A 2kW resistance heater with thermal conductivity k
= 15 W/m.K, diameter D = 4 mm, and length L = 0.5 m is used to boil water. If the outer surface temperature of the resistance wire is Ts = 105oC, determine the temperature at the center of the wire.
• Assumption:– Heat transfer is steady since there is no change with time– Heat transfer is onedimensional since there is thermal symmetry
about the center line and no change in the axial direction.– Thermal conductivity is constant– Heat generation in the heater is uniform
04/11/23 BMCG 2123 Heat Transfer 39
Solution 3The heat generation per unit volume:
The center temperature of the wire using Eq.271
04/11/23 BMCG 2123 Heat Transfer 40
39
220
W/m10318.05.0002.0
2000x
mm
W
Lr
E
V
Ee gen
wire
gengen
C
CmW
mmWx
k
reTT o
oogen
s 126./154
002.0/10318.0105
4
23920
0
Variable Thermal Conductivity, k(T)
• The thermal conductivity of a material, in general, varies with temperature.
• An average value for the thermal conductivity is commonly used when the variation is mild.
• This is also common practice for other temperaturedependent properties such as the density and specific heat.
Variable Thermal Conductivity for OneDimensional Cases
2
1
2 1
( )T
Tave
k T dTk
T T
When the variation of thermal conductivity with temperature k(T) is known, the average value of the thermal conductivity in the temperature range between T1 and T2 can be determined from
The variation in thermal conductivity of a material with can often be approximated as a linear function and expressed as
0( ) (1 )k T k T the temperature coefficient of thermal conductivity.
(275)
(279)
Variable Thermal Conductivity
• For a plane wall the temperature varies linearly during steady onedimensional heat conduction when the thermal conductivity is constant.
• This is no longer the case when the thermal conductivity changes with temperature (even linearly).
Exercise 4• A 2mhigh and 0.7mwide bronze plate whose
thickness is 0.1 m. One side of the plate is maintained at a constant temperature of 600 K while the other side is maintained at400 K, as shown in Figure 2.1. Thermal conductivity of the bronze plate can be assumed to vary linearly in the temperature range as K(T)=k0(1+βT) where k0 = 38 W/m.K and β = 9.21x104 K1. Disregarding the effects and assuming steady onedimensional heat transfer, determine the rate of heat conduction through the plate.
04/11/23 BMCG 2123 Heat Transfer 44
Figure 2.1
Solution 4• The average thermal conductivity of the medium is
simply the value at the average temperature:
• The rate of heat conduction through the plate using Eq. 276:
04/11/23 BMCG 2123 Heat Transfer 45
W/m.K5.55
2
4006001021.91./38
21
14
120
KKxKmW
TTkTkk avgavg
kW 155
1.0
4006007.02./5.55
21
m
KmmxKmW
L
TTAkQ avg