HEAT CONDUCTION. q A q T

Transcription

1 HEAT CONDUCTION When a temperature gradient eist in a material, heat flows from the high temperature region to the low temperature region. The heat transfer mehanism is referred to as ondution and the heat transfer rate has been found to be proportional to the temperature gradient normal to the heat transfer surfae. For a 1-dimensional Cartesian geometry this may be written as A T = T (1) where: = Thermal ondutivity A = urfae (heat transfer) area = Heat transfer rate = Heat flu Note: The minus sign aounts for the fat that positive heat transfer ours when the gradient is negative. Euation 1 is alled Fourier's Law of heat ondution. The thermal ondutivity is a material speifi property and in general is a funtion of temperature, and therefore a funtion of position. T Figure 1: Temperature Distribution and Heat Transfer ate in a Conduting Medium 65

2 THE HEAT CONDUCTION EQUATION Derivation of the Heat Condution Euation in Cartesian Geometry z+ Δz y z + Δ y+ Δy z y Figure 1: Control Volume for Deriving the Heat Condution Euation Consider the volume ΔV = ΔΔΔ y z illustrated above. A heat balane on ΔV gives The heat onduted into ΔV during a time interval Δt - the heat onduted out of ΔV during a time interval Δt + the amount of heat generated in ΔV during Δt = hange in the stored (internal) energy in ΔV during Δt Let U = Internal energy r v = Heat ondution rate at an arbitrary loation v r in the v r diretion. = Volumetri heat generation rate In terms of these variables, and the heat ondution rates indiated on the diagram, the heat balane an be epressed as ( + + ) Δt ( + + ) Δt y z + Δ y+ Δy z+ Δz t+ Δt t + ΔΔyΔz (, y, z, t) Δt = U U (1) where [, + Δ ], y [ y, y+ Δ y], and z [, z z + Δ z] represent the loation within the volume where the funtion is eual to its average value. Divide by Δt 66

3 ( + + ) ( + + ) y z + Δ y+ Δy z+ Δz U + ΔΔyΔz (, y, z, t) = t+ Δt t Δt U () and tae the limit as Δt. eall the definition of a derivative lim Δt U t+ Δt t Δt U U t (3) suh that the heat balane is now ( + + ) ( + + ) y z + Δ y+ Δy z+ Δz U + ΔΔyΔz (, y, z, t) = t (4) We an write the total internal energy U in terms of the speifi internal energy and mass as U = Mu = ΔΔyΔzρ u where ρ is the density of the material. If the density is assumed onstant ( + + ) ( + + ) y z + Δ y+ Δy z+ Δz u + ΔΔyΔz (, y, z, t) = ΔΔyΔzρ t (5) We an eliminate internal energy in favor of temperature by applying the hain rule from alulus u u T = = C T p (6) t T t t u where Cp T is the speifi heat. Euation 5 is now ( + + ) ( + + ) y z + Δ y+ Δy z+ Δz + ΔΔyΔz (, y, z, t) = ΔΔyΔzρC T p t (7) The heat transfer rates an be written in terms of temperature by appliation of Fourier's Law. eferring to Figure 1 = ΔyΔz T,, yz (8a) y = ΔΔz T y,, yz (8b) z = ΔyΔ T z,, yz (8) imilarly 67

4 + Δ = ΔΔ y z T,, + Δ y z (8d) y+ Δy = ΔΔz T y,, y+ Δyz (8e) z+ Δz = ΔΔ y T z,, yz+ Δz (8f) Note, the thermal ondutivity is evaluated at the same loation as the gradients, and in general an be different at eah surfae. ubstituting into Euation 7 and rearranging gives ΔΔ y z T T ΔΔ z T + y T y + Δ, y, z, y, z r y, + Δyz, yz,, + ΔΔ y T T ΔΔΔ y z y z t ΔΔΔ y z C T ρ p z z t yz+ Δz yz + (,,, ) = r,,,, (9) Divide by ΔΔyΔz 1 1 Δ T T Δy T y Δ + 1 Δz T z +, y, z, y, z r y, + Δyz, yz,, T z r yz,, + Δz yz,, T y y z t ρc T p + (,,, ) = t + (1) and tae the limit as Δ, Δy and Δz go to zero to give ρ T y T y z T (, y, z, t) = C T p (11) z t whih is the general form of the Heat Condution Euation in Cartesian geometry. Euation 11 may be written in a more onvenient form by introduing the gradient operator in Cartesian geometry suh that Euation 11 redues to v + + i $ $ y j z $ (1) r r v T ( T ) T + ( r, t) = ρc p t (13) It an be shown, that Euation 13 is valid not only for Cartesian geometries, but also ylindrial and spherial geometries by substituting the appropriate definition for the gradient operator v. olution of the heat ondution euation, subjet to appropriate boundary and initial onditions gives the temperature distribution within the region of interest. 68

5 Euation 13 is a nonlinear partial differential euation, and an only be solved analytially for speial ases. One ommon simplifiation is to assume the thermal ondutivity to be onstant. Under these onditions, Euation 13 redues to + ( r v, t) 1 T T = (14) α t where α is the thermal diffusivity and is the Laplaian Operator. The Laplaian is given below for ρ C p several geometries of interest. Geometry Laplaian 3-Dimensional Cartesian T T T T = + + y z 1-Dimensional Cartesian T T = -Dimensional Cylindrial 1 (, r z ) T = + r r r T T r z 1-Dimensional Cylindrial 1 (adial) T = r r r T r 1-Dimensional pherial 1 (adial) = T r r r T r Table 1: Laplaian Operator for eleted Geometries 69

6 TEADY TATE OLUTION OF THE ONE-DIMENIONAL HEAT CONDUCTION EQUATION We have shown, that in general the heat ondution euation is given by r r v T ( T ) T + ( r, t) = ρc p t At steady-state, the volumetri heat generation rate is independent of time, and T t Condution Euation is then r r v ( T ) T + ( r ) =. The steady-state Heat (1) In this setion we onentrate on solutions of Euation 1 for onditions typial of nulear power systems. olution of the Heat Condution Euation in Plate Type Fuel Elements Consider the plate type fuel element illustrated below. T T z = y Figure 1: Plate Fuel Element The steady-state ondution euation in Cartesian geometry is T y T y z T (, y, z) = () z The thiness of the plate in the diretion, is muh smaller than the dimensions in the y and z diretions. This implies 7

7 T y T << and T z T << suh that the ondution euation is approimately T + (, y, z) = (3) or d d dt + (, y, z) = (4) d where we have taen advantage of the fat that the partial derivatives redue to total derivatives for a 1-dimensional steady-state problem. Note, that this euation does not assume the temperature is invariant in the y and z diretions. It does however assume that any variation in the temperature in these diretions an by aommodated by simply varying the soure term. For this eample, we will further assume the volumetri heat generation rate to be uniform, suh that d d dt + = (5) d Euation 5 is valid whether the thermal ondutivity is onstant or varies with position, and will provide the basis for determining the temperature distribution in the fuel element. Euation 5 is a seond order differential euation, and its solution reuires boundary onditions. We assume the solution is symmetri about =, suh that olution in the Fuel egion [, ] dt = (6) d The temperature distribution is obtained by integrating Euation 5 twie. While a number of approahes are possible, the tehniue hosen here is to integrate the ondution euation from the left, or inner most boundary, outward. For this eample, we integrate from =, to some arbitrary ontained within the fuel material. Applying the boundary ondition at = d dt d + d = (7) d d dt d dt d + = (8) dt + = (9) d 71

8 Euation 9 is valid for any [, ] and provides information about the heat flu within the fuel. This is seen more learly by evaluating Euation 9 at = and rearranging to give dt d =. (1) Multiplying by surfae area A s gives A dt s d = = As (11) whih is a simple energy balane stating that the total heat transferred aross the fuel surfae is eual to the heat generated within the fuel. To obtain the temperature distribution within the fuel, we assume for this eample that the thermal ondutivity is a onstant. Dividing by, Euation 9 is integrated again from =, to some arbitrary ontained within the fuel material dt d d + d = (1) T( ) T( ) + = T( ) = T( ) (13) (14) The fuel surfae temperature is obtained by evaluating Euation 14 at = T( ) = T( ) (15) and the temperature drop aross the fuel is T( ) T( ) = (16) olution in the ladding region [, + ] Assuming onstant thermal ondutivity in the lad, the ondution euation in the ladding is d d dt = (17) d where it is assumed that the volumetri heat generation rate in the ladding is zero. Boundary onditions are reuired at the fuel/lad interfae. ine there is no heat sin at the fuel/lad interfae, we reuire that no heat is lost within the interfae. The heat transfer rate out of the fuel is then eual to the heat transfer rate into the ladding. 7

9 ine the surfae areas are eual at the interfae, this is euivalent to reuiring onservation of heat flu. By Fourier's Law this implies dt d s = dt d s (18) In addition, for this problem we will assume the ladding is bound to the fuel suh that good thermal ontat eists at the interfae and the interfae temperature is the same for the fuel and lad, i.e. T( ) = T ( ) We again integrate the ondution euation from the left, or inner most boundary, outward. For the ladding, the left or inner most boundary is at =. d dt d d d = (19) dt d dt d = () To apply the boundary ondition at =, note However, from Euation 1 dt d s = dt d s dt d = suh that dt = d s (1) ubstituting into Euation dt + = () d Euation is valid for any [, + ] and provides information about the heat flu within the lad. The temperature distribution within the lad is obtained by assuming the thermal ondutivity within the lad is onstant, and integrating Euation from =, to some arbitrary ontained within the lad material dt d d + d = (3) 73

10 The lad surfae temperature is obtained by evaluating Euation 5 at = + and the temperature drop aross the lad is Heat Transfer from the Clad to the Coolant = + T( ) T( ) + = (4) ( ) T( ) = T( ) (5) T( + ) = T( ) (6) T( ) T( + ) = (7) Heat transfer from a solid to a flowing fluid is said to be transferred by onvetion. The onvetive heat transfer rate to a fluid of temperature T, is given by Newton's Law of Cooling whih states v v ( rs) = h T( rs) T [ ] (8) where: v ( ) T h r s v Tr ( s ) = loal heat flu at a position r v s on the solid surfae = loal bul fluid temperature = Convetive heat transfer oeffiient = loal temperature at a position r v s on the solid surfae In our problem, r v v s is at the lad/oolant interfae, Tr ( s) = T( + ) and the heat flu at the lad/oolant interfae is given by Fourier's Law as v r = + = dt ( s) ( ) d + (9) giving dt = h[ T( + ) T ] (3) d + where the gradient is given diretly by Euation dt = d + = h[ T( + ) T ] (31) The lad temperature an then be written in terms of the oolant temperature as 74

11 and the temperature drop from the lad to the oolant is Eamine the temperature drops aross eah region 1) Fuel T( ) T( ) = ) Clad T( ) T( + ) = 3) Coolant T ( + ) T = h T ( + ) = T + (3) h T ( + ) T = (33) h In many appliations, the intermediate surfae temperatures are not needed, and these three euations an be added to yield the total drop from the enterline of the fuel plate to the oolant. T T 1 ( ) = + + (34) h or T T 1 ( ) = (35) h Euation 35 gives the fuel enterline temperature given the physial dimensions and harateristis of the fuel plate and the loal bul fluid temperature. Alternatively, given the enterline temperature, the reuired oolant temperature ould be determined. The onvetive heat transfer oeffiient is governed by the fluid veloity, and the physial properties of the fluid (e.g. density, visosity, speifi heat, et.). By setting onstraints on any of the above parameters, the others may be adjusted to give the desired result. Euation 35 may also be written as = A = s T( ) T As h (36) T = ( ) T (37) where 75

12 1 1 = + + As h (38) is alled the thermal resistane and represents the total resistane to heat transfer from the enterline of the fuel out to the oolant. Euation 37 suggests, that the smaller the thermal resistane, the smaller the temperature differene neessary to drive the heat transfer, and therefore lower enterline temperatures. From Euation 38 it should be obvious, that the thermal resistane an be redued by inreasing the thermal ondutivity of a onduting region, dereasing the thiness of a onduting region, or inreasing the onvetive heat transfer oeffiient. It should also be noted, that even if the onvetive heat transfer oeffiient were infinite, suh that 1/ h, the minimum thermal resistane is still 1 min = + As (39) whih implies that for a given oolant temperature, the minimum fuel enterline temperature is ditated by the fuel dimensions and material properties, regardless of how muh ooling is available. Non Uniform Heat Generation In reality, the volumetri heat generation rate within the fuel plate is not uniform, but as shown previously, is proportional to the fission rate. In thermal reators the volumetri heat generation rate is then proportional to the thermal flu distribution within the fuel material. We again onsider the plate type fuel element of the previous eample, but now impose a realisti spatial dependene on the heat generation rate. If we assume a thermal reator, then it is reasonable to assume there are no thermal neutron soures in the fuel as fission neutrons undergo relatively few interations in the fuel material, reahing thermal energies in the moderator. If we assume all thermal neutrons are produed in the moderator, then we an write the neutron diffusion euation for the thermal flu within the fuel as If we assume a one-dimensional, artesian geometry whih has solution D φ φσ = (4) or ine the volumetri heat generation rate is proportional to the flu The ondution euation in the fuel is now a φ φ/ L =. (41) d φ φ = (4) d L φ( ) = φ osh( / L) (43) ( ) = osh( / L) (44) d d dt + L = d osh( / ) (45) 76

13 We assume the same boundary onditions, i.e. dt = d The solution proedure is idential to that for a uniform volumetri heat generation rate. We integrate from =, to some arbitrary ontained within the fuel material d dt d + L d = d d osh( / ) (46) dt d dt d Evaluating the integrals at the limits, and applying the boundary ondition at = + Lsinh( / L) = (47) dt L L d + sinh( / ) = (48) Euation 48 is valid for any [, ]. The heat flu at the fuel/lad interfae is obtained by evaluating Euation 48 at = and rearranging to give dt = ) d L sinh( / L s. (49) To obtain the temperature distribution within the fuel, we again assume the thermal ondutivity to be onstant, and Euation 48 is integrated from =, to some arbitrary ontained within the fuel material dt d d Lsinh( / L) + d = L osh( / L) T( ) T( ) + = L T( ) = T( ) + 1 osh( / ) { L } (5) (51) (5) The fuel surfae temperature is obtained by evaluating Euation 5 at = L T ( ) = T( ) + 1 osh( / ) { L } (53) and the temperature drop aross the fuel is L T( ) T( ) = { 1 osh( / L) } (54) 77

14 It is relatively easy to show, that the temperature drops aross the ladding and the lad/oolant interfae an be written in terms of the heat flu s as Clad: Coolant: s T ( ) T ( + ) = T ( + ) T = s h 78

15 CYLINDICAL FUEL ELEMENT We net onsider a ylindrial fuel element as illustrated below. Cladding Coolant Coolant Fuel r= o Figure 1: Cylindrial Fuel Element We again begin by assuming steady-state heat ondution, with uniform heat generation. We further assume the thermal ondutivity is onstant. If the diameter of the fuel rod is small ompared to the length, then heat ondution is predominately in the radial diretion, and we an assume the one-dimensional ondution euation in ylindrial geometry is valid 1 r r r T + =. (1) r We assume similar boundary onditions to those for the plate fuel element, i.e. symmetry about the origin, 1) dt = dr onservation of energy (heat flu) at the fuel/lad interfae, ) dt = dt dr dr good thermal ontat between fuel and ladding, 79

16 3) T ( ) = T( ) and the fuel rod is onvetively ooled by a fluid of temperature T. 4) dt = h[ T( o) T ] dr o olution in the Fuel egion r [, ] olution for the temperature distribution follows diretly from our solution in the plate type fuel element. The temperature distribution is obtained by integrating Euation 1 twie, starting from the left or inner most boundary and woring outward. For this eample, Euation 1 must first be multiplied by r, suh that the left and right hand sides of the euation are diretly integrable. We integrate from r =, to some arbitrary r ontained within the fuel material. r r T + r = () r r r d r dt dr + r dr = (3) dr dr r dt dt r r r + = (4) dr dr Evaluating Euation 4 at the limits, and applying the boundary ondition at r = r r dt + r = (5) dr Euation 5 is valid for any r [, ] and an be used to determine the heat flu at the fuel/lad interfae. Evaluating Euation 5 at r = and rearranging gives r dt dr = (6) or dt = dr. (7) To obtain the temperature distribution within the fuel, we assume the thermal ondutivity is onstant and divide Euation 5 by r to obtain integrable terms on both sides of the euation. Integrate again from r =, to some arbitrary r ontained within the fuel material r dt dr dr + r r dr = (8) 8

17 r r Tr () T() + = 4 r Tr () = T() 4 (9) (1) The fuel surfae temperature is obtained by evaluating Euation 1 at r = T ( ) = T( ) 4 (11) and the temperature drop aross the fuel is T( ) T( ) = (1) 4 olution in the ladding region r [, o ] The ondution euation in the ladding is r r T = (13) r where it is again assumed that the volumetri heat generation rate in the ladding is zero. Integrate the ondution euation from the left, or inner most boundary, outward. For the ladding, the left or inner most boundary is at r =. r d r dt dr dr dr = (14) r dt dr r r dt dr = (15) To apply the boundary ondition at r = However, from Euation 6 suh that, note dt = dt r dt = dr dr dr r dt dr = r dt dr r dt = dr (16) 81

18 ubstituting into Euation 15 r dt + = (17) dr Euation 17 is valid for any r [, o ] and provides information about the heat flu within the lad. The heat flu at the lad/oolant interfae an be obtained by evaluating Euation 17 at r = o r dt dr o = (18) The temperature distribution within the lad is obtained by assuming the thermal ondutivity onstant, dividing Euation 17 by r and integrating from r =, to some arbitrary r ontained within the lad material r r dt dr dr + = (19) dr r T() r T( ) + ln( r ) = () r r T() r = T( ) ln (1) The lad surfae temperature an then be obtained by evaluating Euation 1 at r = o o T( o) = T( ) ln () suh that the temperature drop aross the lad is o T( ) T( o) = ln (3) Heat Transfer from the Clad to the Coolant r = Heat transfer from the ladding to the oolant is given by Newton's Law of Cooling o = dt ( o) = h[ T( o) T ] (4) dr o or multiplying by o dt ( ) = r o o = h o[ T ( o ) T ] dr o (5) 8

19 From Euation 18 r dt dr o = suh that = h o[ T ( o ) T ] (6) The lad temperature an then be written in terms of the oolant temperature as T( o) = T + h o (7) and the temperature drop from the lad to the oolant is T( o) T = h o (8) Eamine the temperature drops aross eah region 1) Fuel T( ) T( ) = 4 o ) Clad T( ) T( o) = ln 3) Coolant T( o) T = h o The total temperature drop from the fuel enterline to the oolant is obtained by adding these three euations to give or in terms of the fuel enterline temperature 1 1 o 1 T( ) T = + ln + (9) ho 1 1 o 1 T( ) = T + + ln + (3) ho Euation 3 gives the fuel enterline temperature given the physial dimensions and harateristis of the fuel rod and the loal bul fluid temperature. For this eample of a uniformly heated rod, = π H = H and A s = π H. Euation 3 may then be written in terms of the linear heat rate and the thermal resistane as o T () T H = 83

20 where 1 o = + As The orresponding minimum thermal resistane (1/ h ) is o o 1 ln + (31) h min 1 o = + As o o ln (3) whih again implies that for a given oolant temperature, the minimum fuel enterline temperature is ditated by the fuel dimensions and material properties, regardless of how muh ooling is available. In addition, if we assume the maimum fuel enterline temperature is eual to the fuel melt temperature, then the maimum linear heat rate is T H = T melt ma (33) 84