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Heat Conduction EquationBMCG 2123 Faculty of Manufacturing Engineering, UTeM Taufik Week 21/28/2011 BMCG 2123 Heat Transfer 1
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.1/28/2011 BMCG 2123 Heat Transfer 2
Introduction Although heat transfer and temperature are closely related, they are of a different nature. Temperature has only magnitude it 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
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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, J, z, t)), spherical (T(r, J, U, t)).
Introduction
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Classification of conduction heat transfer problems: steady versus transient heat transfer, multidimensional heat transfer, heat generation.
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:& ! kA dT Qcond dx (W) (21)
Heat is conducted in the direction of decreasing temperature, and thus the temperature gradient is negative when heat is conducted in the positive xdirection.
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 & ! kA dT Qn (W) (22) dn
General Relation for Fourier s Law of Heat ConductionContinue In rectangular coordinates, the heat conduction vector can be expressed in terms of its components r r r r as (23) ! i j kn x y z
which can be determined from Fourier s law as & ! kA xT Q x x xx & ! kA xT (24) Qy y xy & ! kA xT Q z z xz
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 & & (25) E ! e dV (W)gen
gen
V
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, and b) The heat flux on the outer surface of the wire as result of this heat generation.
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Solution 1a) The rate of heat generation: e gen ! gen
!
wire
D / 4 (0.3cm) / 4 cm L T 80 T2
gen 2
!
1200 W
! 212 W/cm 3
b) The heat flux on the outer surface: ! s E gen1/28/2011
!
E genT
ire
1200 W ! ! 15.9 W / c T (o.3c )(80c )
2
BMCG 2123 Heat Transfer
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OneDimensional Heat Conduction Equation  Plane WallRate of Rate of Rate of heat  heat + generation heat conductio conduction inside the at x+(x n element at x Rate of change = of the energy content of the element
& & Qx Qx (x & ,element gen(26)
(Eelement ! (t
( & Q & & Qx x (x gen ,element !
element
(t
(26)
The change in the energy content and the rate of heat generation can be expressed as Eelement ! Et (t Et ! mc t (t ( & & & E gen,element ! egenVelement ! egen A(x t
! Vc
(x
t (t
t
(27)(28)
Substituting into Eq. 2 6, we get& &x
& egen A(x ! VcA(x x (x
t (t
(t
t
(29)
Dividing by A(x, taking the limit as (x 0 and (t 0, and from Fourier s law:1 x xT kA xx A xx
xT & egen ! V c xt
(211)
The area A is constant for a plane wall the one dimensional transient heat conduction equation in a plane wall is (213) Variable conductivity: x k x egen ! V c x & Constantxx xx & x 2T egen 1 xT ! conductivity: 2 xx k E xt xtk ; E! (214) Vc
The onedimensional conduction equation may be reduces to the following forms under special & d 2T egen conditions ! 0 (215) 2 1) Steadystate: dx k 2) Transient, no heat generation: d 2T 3) Steadystate, no heat generation: 2dx
x 2T 1 xT ! 2 xx E xt !0
(216) (217)
OneDimensional Heat Conduction Equation  Long CylinderRate of Rate of Rate of heat  heat + generation heat conductio conduction inside the at r+(r n element at r Rate of change = of the energy content of the element
& Q & ,element ! ( & gen r r (r
element
(t
(218)
( &Q & & Qr r (r gen,element !
element
(t
(218)
The change in the energy content and the rate of heat generation can be expressed as Eelement ! Et (t Et ! mc Tt (t Tt ! VcA(r Tt (t Tt (219) ( & & & E (220) gen,element ! egenVelement ! egen A(r
Substituting into Eq. 2 18, we get& &r r (r
& A(r ! VcA(r Tt (t Tt egen (t
(221)
Dividing by A(r, taking the limit as (r 0 and (t 0, and from Fourier s law: 1 x xT xT & (223) kA egen ! V cA xr
xr
xt
Noting that the area varies with the independent variable r according to A=2TrL, the one dimensional transient heat conduction equation in a plane wall becomes 1 x xT xT (225) & rk xr egen ! V c xt Variable conductivity: r xr Constant conductivity:1 x xT r xr r xr & egen 1 xT k ! E xt (226)
The onedimensional conduction equation may be reduces to the following forms under special & conditions 1 d dT egen (227) r dr k ! 0 1) Steadystate: r dr 2) Transient, no heat generation:1 x xT r xr r xr
1 xT ! E xt !0
(228) (229)
3) Steadystate, no heat generation:
d dT r dr dr
OneDimensional Heat Conduction Equation  Sphere
Variable conductivity: Constant conductivity:
1 x 2 xT r k xr 2 r xr
xT & ! Vc egen xt
(230)
& 1 x 2 xT egen 1 xT r xr k ! E xt 2 r xr
(231)
General Heat Conduction Equation
& & & Q Q Qx (x Qy (y Qz (z Egen ,element ! ( Qx & & & y z
Rate of Rate of Rate of heat Rate of  heat + heat generation = change conductio conduction inside the of the at x+(x, n element energy y+(y, and at x, y, content of z+(z and z the elementelement
(236)
(t
Repeating the mathematical approach used for the onedimensional heat conduction the threedimensional heat conduction equation is determined to be Twodimensional
Constant conductivity:
& x 2T x 2T x 2T egen 1 xT 2 2 ! 2 xx xy xz k E xtThreedimensional
(239)
1) Steadystate: 2) 3)
& x 2T x 2T x 2T egen 2 2 ! 0 (240) 2 xx xy xz k(241) (242)
x 2T x 2T x 2T 1 xT Transient, no heat generation: xx 2 xy 2 xz 2 ! E xt x 2T x 2T x 2T Steadystate, no heat generation: 2 2 2 ! 0 xx xy xz
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.
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AnalysisThe 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.1/28/2011 BMCG 2123 Heat Transfer 24
Solution 1The rate of heat generation in the wire per unit volume: e gen ! E gen
!
E gen T 2
V
ire
/4
0.004m / 40.5m T2
!
2000W
! 0.318 x10 9 W/m 3
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 e1 d dT r r dr drgen !0 k
Which steady onedimensional heat conduction equation in cylindrical coordinates for the case of constant thermal conductivity.1/28/2011 BMCG 2123 Heat Transfer 25
Cylindrical Coordinates
1 x xT rk r xr xr
1 xT xT x xT 2 k k r xJ xJ xz xz
xT & egen ! V c xt (243)
Spherical Coordinates
1 x 2 xT 1 1 x xT x xT xT & kr xr r 2 sin 2 U xJ k xJ r 2 sin U xU k sin U xU egen ! V c xt 2 r xr
(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 ConditionFor onedimensional heat transfer through a plane wall of thickness L, for example, the specified temperature boundary conditions can be expressed asT(0, t) = T1 T(L, t) = T2 (246)
The specified temperatures can be constant, which is the case for steady heat conduction, or may vary with time.
Specified Heat Flux Boundary ConditionThe heat flux in the positive xdirection anywhere in the medium, including the boundaries, can be expressed by Fourier s law of heat conduction asdT &! k q ! dxHeat flux in the positive xdirection
(247)
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.
Two Special CasesInsulated boundary Thermal symmetry
xT (0, t ) k !0 xx
or
xT (0, t ) !0 xx
xT L , t 2 !0 xx
(249)
(250)
Convection Boundary ConditionHeat conduction at the surface in a selected direction Heat convection at the surface in the same direction
=
and
xT (0, t ) k ! h1 ?Tg1 T (0, t ) A xx xT ( L, t ) k ! h2 ?T ( L, t ) Tg 2 A xx
(251a) (251b)
Radiation Boundary Condition
Heat conduction at the surface in a selected direction
=
Radiation exchange at the surface in the same direction
xT (0, t ) 4 k ! I1W Tsurr ,1 T (0, t ) 4 xxand
(252a)
xT ( L, t ) 4 k ! I 2W T ( L, t ) 4 Tsurr ,2 xx
(252b)
Interface Boundary ConditionsAt the interface the requirements are: (1) two bodies in contact must have the same temperatureat 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)
xTA ( x0 , t ) xTB ( x0 , t ) k A ! kB (254) xx xx
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 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.
Heat transfer from the surface In all modes
Heat Generation in Solids The Surface TemperatureRate of heat transfer from the solid Rate of energy generation within the generation solid within
=
(263)
For uniform heat &! e V & gen
the medium(264)
(W)
The heat transfer rate by convection can also be expressed from Newton s law of cooling as

& Q ! hAs Ts Tg Ts ! Tg & egenV hAs
(W)
(265) (266)
Heat Generation in Solids The Surface TemperatureFor a large plane wall of thickness 2L (As=2Awall and V=2LAwall) & egen L (267) Ts , plane wall ! Tg h For a long solid cylinder of radius r0 (As=2Tr0L and V=Tr02L) & egen r0 (268) Ts ,cylinder ! Tg 2h For a solid sphere of radius r0 (As=4Tr02 and V=4/3Tr03)
Ts , sphere ! Tg
& egen r0 3h
(269)
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.dT & kAr ! egenVr dr(270)
Substituting these expressions into the above equation and separating the variables, we get& egen dT 2 & k 2T rL ! egen T r L p dT ! rdr 2k dr
Integrating from r =0 where T(0) =T0 to r=ro(Tmax,cylinder ! T0 Ts ! & egen r02 4k(271)
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
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Solution 3The heat generation per unit volume: e gen ! E gen ! Vwire
0.002m 0.5m T 2
gen Tr0 2 L
!
2000W
! 0.318 x10 9 W/m 3
The center temperature of the wire using Eq.271T0 ! Ts e gen r02 4k ! 105o
.318 x10 W / m 0.002m 0 4 W / m. C 159 3 o
2
! 126 o C
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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 CasesWhen 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 Tkave
!
2
T1
k (T )dT
(275)
T2 T1
The variation in thermal conductivity of a material with can often be approximated as a linear function and expressed as
k (T ) ! k0 (1 F T )
(279)
F the temperature coefficient of thermal conductivity.
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+ T) where k0 = 38 W/m.K and = 9.21x104 K1. Disregarding the effects and assuming steady onedimensional heat transfer, determine the rate of heat conduction through the plate.
Figure 2.1
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Solution 4 The average thermal conductivity of the medium is simply the value at the average temperature:T T1 k avg ! k Tavg ! k 0 1 F 2 2
600 400K ! 55.5 W/m.K ! 38W / m.K 1 9.21x10 4 K 1 2
The rate of heat conduction through the plate using Eq. 276:! kavg A
T1 T 2 L
! 55.5W / m.K 2mx 0.7m 1/28/2011
600 400 K0.1m
! 155 kW45
BMCG 2123 Heat Transfer