Chapter : Heat Conduction Equation Dr Ali Jawarneh Department of Mechanical Engineering, Hashemite University
Objectives When you finish studying this chapter, you should be able to: Understand multidimensionality and time dependence of heat transfer, and the conditions under which a heat transfer problem can be approximated as being one-dimensional, Obtain the differential equation of heat conduction in various coordinate systems, and simplify it for steady one-dimensional i case, Identify the thermal conditions on surfaces, and express them mathematically as boundary and initial conditions, Solve one-dimensional heat conduction problems and obtain the temperature distributions within a medium and the heat flux, Analyze one-dimensional heat conduction in solids that involve heat generation, and Evaluate heat conduction in solids with temperature-dependent thermal conductivity. cti it
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 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, φ, z, t)), spherical (T(r, φ, θ, t)).
Introduction Continue 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: one-dimensional, two dimensional, three-dimensional. In the most general case, heat transfer through a medium is three-dimensional. However, some problems can be classified as two- or one-dimensional depending di 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 x-direction) is expressed by Fourier s law of heat conduction for one-dimensional heat conduction as: dt Q & = ka (W) (-1) cond dx Heat is conducted in the direction of fdecreasing temperature, t and dthus the temperature gradient is negative when heat is conducted in the positive x- direction.
General Relation for Fourier s Law of Heat tconduction 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 t point can be expressed by Fourier s law as dt Qn = ka dn & (-) (W)
General Relation for Fourier s Law of Heat Conduction-Continue In rectangular coordinates, the heat conduction vector can be expressed in terms of its components as r r r r Q& = Q& i + Q& j + Q& k (-3) n x y z which can be determined from Fourier s law as TT Q& x = kax x & T Q y = ka y y T Q & z = kaz z & (-4)
Examples: Heat Generation electrical energy being converted to heat at a rate of I R, fuel elements of nuclear reactors, exothermic chemical reactions. Heat generation is a volumetric phenomenon. The rate of heat generation units : W/m 3 or Btu/h ft 3. 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 E& gen = e& gendv (W) (-5) V
One-Dimensional Heat Conduction Equation - Plane Wall Rate of heat Rate of heat Rate of heat Rate of change of conduction - conduction + generation inside the energy content = at x at x+δx the element of the element Q Q x +E x +Δ x gen, element Q & Q & + E & (-6) = ΔEelement Δt
ΔE & element x & x+δx + & gen, element = (-6) Q Q E The change in the energy content and the rate of heat generation can be expressed as Δ t ( ) ρ ( ) Δ Eelement = Et +Δ t Et = mc Tt +Δ t Tt = caδx Tt +Δt T (-7) t E & gen, element = e& genvelement = e& genaδx (-8) Substituting into Eq. 6, we get Q& x Q & x +Δ + e& T (-9) x gen AΔx t+δt Tt = ρ caδx Δt t Dividing by AΔx, taking the limit as Δx 0 and Δt 0, and from Fourier s law: 1 T T ka egen ρc + = A x x t & (-11)
The area A is constant for a plane wall the one dimensional transient heat conduction equation in a plane wall llis T T k e& gen = ρc x x + t Variable conductivity: (-13) Constant conductivity: y T e& gen 1 T k + = ; α = x k α t ρc (-14) The one-dimensional conduction equation may be reduces to the following forms under special conditions 1) Steady-state: dt + e gen = 0 dx & k (-15) ) Transient, no heat generation: T x 1 T = α t (-16) 3) Steady-state, no heat generation: dt 0 dx = (-17)
One-Dimensional Heat Conduction Equation - Long Cylinder Rate of heat conduction at r Rate of heat conduction at r+δr Rate of heat generation inside the element - + = Rate of change of the energy content of the element r ΔE + E & E Q & element gen element = r +Δ r Δt (-18) Q &,
ΔE Δ t & element r & r+δr + & gen, element = (-18) Q Q E The change in the energy content and the rate of heat generation can be expressed as ( ) ρ ( ) Δ Eelement = Et +Δ t Et = mc Tt +Δ t Tt = caδr Tt +Δt T (-19) t E & gen, element = e& genvelement = e& genaδr (-0) Substituting into Eq. 18, we get Q& r Q & r +Δ + e& T (-1) r gen AΔr t+δt Tt = ρ caδr Δt t Dividing by AΔr, taking the limit as Δr 0 and Δt 0, and from Fourier s law: 1 T T ka egen ρc + = A r r t & (-3)
Noting that the area varies with the independent variable r according to A=πrL, the one dimensional transient heat conduction equation in a long cylinder becomes Variable conductivity: 1 T T (-5) rk + e& = ρ gen c r r r t 1 T e gen 1 T & r = r r r + k α t Constant conductivity: (-6) The one-dimensional conduction equation may be reduces to the following forms under special conditions 1) Steady-state: 1 d dt e gen r & + = 0 rdr dr k 1 T 1 T r = r r r α t (-7) ) Transient, no heat generation: (-8) d dr dt r = dr 3) Steady-state, no heat generation: (-9) 0
One-Dimensional Heat Conduction Equation - Sphere 1 T T rk e gen ρc + & = r r r t Variable conductivity: (-30) 1 & T e gen 1 T r + = r r r k α t Constant conductivity: (-31)
Large Plane Wall: T Q= KAx x T Q= KA, A = πrl r Long Cylinder: r r Sphere: T Q = KA r, Ar = 4πr r
General Heat Conduction Equation Rate of heat conduction at x, y, and z Rate of heat conduction at x+δx, y+δy, Rate of heat generation inside the Rate of change of the energy content of the and z+δz element element - + = Q& + Q& + Q& x y z Q& Q& Q& x+δ x y+δ y z+δz ΔE E Δt element + E gen, element = (-36)
Repeating the mathematical approach used for the one- dimensional heat conduction the three-dimensional heat conduction equation is determined to be Two-dimensional T T T & T e gen 1 + + + = x y z k α t Constant conductivity: (-39) Three-dimensional 1) Steady-state: & x y z k T T T e gen + + + = 0 (-40) 1 + + = x y z α t T T T T ) Transient, no heat generation: (-41) T T T x y z + + = 0 3) Steady-state, no heat generation: (-4)
Cylindrical Coordinates 1 T 1 T T T rk + k k e T + + & gen = ρc r r r r φ φ z z t (-43)
Spherical Coordinates 1 T 1 T 1 kr k k sin T e T gen c r r r r sin r sin θ + + + & = ρ θ φ φ θ θ θ t (-44)
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 one-dimensional i heat transfer through a plane wall of thickness L, for example, the specified temperature boundary conditions can be expressed as T(0, t) = T 1 T(L, t) = T (-46) The specified temperatures can be constant, which is the case for steady heat conduction, or may vary with time.
Specified Heat Flux Boundary Condition The heat flux in the positive x- direction anywhere in the medium, including the boundaries, can be expressed by Fourier slawof of heat conduction as dt Heat flux in the & = k = positive x- dx direction q (-47) The sign of the specified heat flux is determined d 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 Cases Insulated boundary Thermal symmetry k ( L, t) T(0, t) T(0, t) T = 0 or = 0 x x = 0 x x (-49) (-50)
Cylindrical Thermal Symmetry: Spherical Thermal symmetry:
Convection Boundary Condition Heat conduction at the surface in a selected direction = Heat convection at the surface in the same direction and T(0, t) k = h1 T 1 T t xx [ (0, )] T( L, t) k = h T L t T xx [ (, ) ] (-51a) (-51b)
Radiation Boundary Condition Heat conduction at the surface in a selected direction = Radiation exchange at tthe surface in the same direction and T(0, t) 4 4 k = εσ 1 Tsurr,1 T(0, t) (-5a) x T T ( L, t ) k = ε σ T( L, t) Tsurr x 4 4, (-5b)
Interface Boundary Conditions At the interface the requirements are: (1) two bodies in contact must have the same temperature at the area of contact, () 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. T A (x 0, t) = T B (x 0, t) and T ( x 0, t A ) T B ( x 0, t ) ka = kb x x (-53) (-54)
Generalized Boundary Conditions In 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 transfer to the surface in all modes = Heat transfer from the surface In all modes
Important Comments: 1- The positive direction of heat transfer to be in the positive x or r direction - Set the conduction boundary condition equal to kdt/dx dt dt qo = k (0) dx dt qo = k (0) qo = k (0) dx dt dx dt q o = k (0) q o = k (0) dx dx dt dt ql = k (L) dt ql = k (L) k (L) = h[t(l) T ] dx dx dx dt dt q L = k (L) dt ql = k (L) k (L) = h[t T(L)] dx dx dx
Example: Consider a large plane wall of thickness L= 0.4 m, thermal conductivity k=.3 W/m C, and surface area A= 0 m.theleft side of the wall is maintained at a constant temperature of T 1 =80 C while the right side loses heat by convection to the surrounding air at T = 15 C with a heat transfer coefficient of h = 4 W/m C. Assuming constant thermal conductivity and no heat generation in the wall, (a) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the wall, (b) obtain a relation for the variation of temperature in the wall by solving the differential equation, and (c) evaluate the rate of heat transfer through the wall. k T 1 =80 C A=0 m L=0.4 m T =15 C h=4 W/m. C x
Assumptions: 1 Heat conduction is steady and one-dimensional. i Thermal conductivity is constant. 3 There is no heat generation. Properties The thermal conductivity is given to be k =.3 W/m C. Analysis (a) Taking the direction normal to the surface of the wall to be the x direction with x = 0 at the left surface, the mathematical formulation of this problem can be expressed as dt 0 dx = T( 0) = T = 80 C k dt ( L ) = h[ T( L) T ] dx 1
(b) Integrating the differential equation twice with respect to x yields dt dx = C 1 T( x) = C x+ C 1 x = 0: T( 0) = C 0+ C C = T x = L: kc = h[( C L + C ) T ] C = 1 1 1 1 1 hc ( T ) ht ( 1 T ) C1 = k + hl k + hl T ( ht T x ) ( 1 ) = k + hl x + T 1 ( 4 W/m C)( 80 15) C = ( 3. W/m C ) + ( 4W/m C )( 04. m) = 80 1311. x x + 80 C
(c) The rate of heat conduction through the wall is Q & wall = ka dt dx = kac = (.3 W/m C)(0 m = 6030 W 1 h = ka ( T 1 T k + hl ) (4 W/m C)(80 15) C ) (.3 W/m C) + (4 W/m C)(0.4 m)
EXAMPLE: Aspherical container of inner radius r 1 =m, outer radius r =.1 m, and thermal conductivity k = 30 W/m C is filled with iced water at 0 C. The container is gaining heat by convection from the surrounding air at T = 5 C with a heat transfer coefficient i of h = 18 W/m C. Assuming the inner surface temperature of the container to be 0 C, (a) express the differential equation and the boundary conditions for steady one-dimensional i heat conduction throughh the container, (b) obtain a relation for the variation of temperature in the container by solving the differential equation, and (c) evaluate the rate of heat gain to the iced water. k T 1 T h r 1 r
Assumptions: 1 Heat conduction is steady and one-dimensional since there is no change with time and there is thermal symmetry about the midpoint. Thermal conductivity is constant. 3 There is no heat generation. Properties: The thermal conductivity is given to be k = 30 W/m C. (a) d dr r dt dr = 0 Tr ( 1) = T1 = 0 C k dt ( r ) = ht [ ( r ) T ] dr
(b) Integrating the differential equation once with respect to r gives dt dt C r = C1 = 1 C1 dr Tr ()= +C dr r r C1 r = r1: Tr ( 1) = + C = T1 r r = r: 1 C1 C = 1 k h + C r r T C 1 = r ( T1 T ) C1 T1 T r and C = T1+ = T1 + r k r r k 1 r 1 1 r hr r hr 1 1 1
(c) The rate of heat conduction through the wall is dt C1 r (T1 T ) Q& = ka = k(4π r ) = 4π kc 1 = 4πk dr r r k 1 r hr 1 (.1 m)(0 5) C = 4 π (30 W/m C) = 3,460 W.1 30 W/m C 1 (18 W/m C)(.1 m)
Heat Generation in Solids The quantities of major interest in a medium with heat generation are the surface temperature T s and the maximum temperature T max that occurs in the medium in steady operation.
Heat Generation in Solids -The Surface Temperature Rate of heat transfer from the solid = Rate of energy generation within the solid (-63) For uniform heat generation within the medium Q& = e& V (W) (-64) gen The heat transfer rate by convection can also be expressed from Newton s law of cooling as Q& has Ts T - Q= ha ( T T ) (W) (-65) T s e V & gen = T + (-66) ha s
Heat Generation in Solids -The Surface Temperature For a large plane wall of thickness L (A s =A wall and V=LA wall ) egenl T = T + & (-67) s, plane wall h For a long solid cylinder of radius r 0 (A s =πr 0 L and V=πr 0 L) egenr0 Ts, cylinder = T + & (-68) h For a solid sphere of radius r 0 (A s =4πr 0 and V= 4 / 3 πr 03 ) T s, sphere & 0 egen r 0 = T + (-69) 3h
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 ka = e& V dr r gen r (-70) Substituting these expressions into the above equation and separating the variables, we get dt e gen k( π rl) = e& ( ) = & gen πr L dt rdr dr k Integrating from r =0 where T(0) =T 0 to r=r o egenr0 Δ Tmax, cylinder = T0 Ts = & (-71) 4k
The maximum temperature T o in: Plane Wall of thickness L Solid Sphere of radius r o Where: Δ Tmax = To Ts
Assumptions 1 Heat transfer is steady since there is no change with time. Heat transfer is one-dimensional since there is thermal symmetry about the center line and no change in the axial direction. 3 Thermal conductivity is constant. 4 Heat generation in the heater is uniform Properties The thermal conductivity is given to be k = 0 W/m C. Analysis The resistance heater converts electric energy into heat at a rate of kw. The rate of heat generation per unit volume of the wire is g& = Q& V gen wire = Q& πr gen o L = 000 W π (0.005 m) The center temperature of the wire is: (0.7 m) = 1.455 10 8 W/m 3 T o = T s + gr & o 4k = 110 C + (1.455 10 8 W/m 3 4(0 W/m. C) )(0.005 m) = 11.4 C
Analysis This insulated plate whose thickness is L is equivalent to one- half of an uninsulated plate whose thickness is L since the midplane of the uninsulated plate can be treated as insulated surface. The highest temperature will occur at the insulated surface while the lowest temperature will occur at the surface which is exposed to the environment. Note that L in the following relations is the full thickness of the given plate since the insulated side represents the center surface of a plate whose thickness is doubled. The desired values are determined directly from T s gl & ( 10 W/m )(0.05 m) = T + = 5 C + h 44 W/m. C 5 3 = 5.3 CC T o = T s + gl & k = 5.3 C + ( 10 5 W/m 3 )(0.05 m) (111 W/m. C) = 54.5 C
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 One-Dimensional Cases When the variation of thermal conductivity with temperature k(t) is known, the average value of the thermal conductivity in the temperature range between T 1 and T can be determined from k ave = T T 1 ktdt ( ) T T 1 (-75) The variation in thermal conductivity of a material with can often be approximated as a linear function and expressed as kt ( ) = k(1 + βt ) (-79) β the temperature coefficient of thermal conductivity. 0
Variable Thermal Conductivity For a plane wall the temperature varies linearly during steady onedimensional heat conduction when the thermal conductivity is constant. t This is no longer the case when the thermal conductivity it changes with temperature (even linearly).
k(t) T 1 T L
Properties The thermal conductivity is given to be kt ( ) = k0( 1 + β T). Analysis k ave = = = k( T ave ) = k 0 T + T 1+ β (5 W/m K) 1+ (87 (8.7 10 34.4 W/m K 1-4 K -1 ) (500 + 350) K Q & = k ave T A 1 T L = (34.4 W/m K)(1.5 m 0.6 m) (500 350)K 0.15 m = 30,80 W