Chapter 5. Transient Conduction. Islamic Azad University

Transcription

1 Chater 5 Transient Conduction Islamic Azad University Karaj Branch 1 Transient Conduction Many heat transfer roblems are time deendent Changes in oerating conditions in a system cause temerature variation with time, as well as location within a solid, until a new steady state (thermal equilibrium) is obtained. In this chater we will develo rocedures for determining the time deendence of the temerature distribution Real roblems may include finite and semi-infinite solids, or comlex geometries, as well as two and three dimensional conduction Solution techniques involve the lumed caacitance method, exact and aroximate solutions, and finite difference methods.! We will focus on the Lumed Caacitance Method, which can be used for solids within which temerature gradients are negligible (Sections ) 2

2 Lumed Caacitance Method Consider a solid that is initially at a uniform temerature, T i, and at t=0 is quenched by immersion in a cool liquid, of lower temerature! The temerature of the solid will decrease for time t>0, due to convection heat transfer at the solid-liquid interface, until it reaches T T! T! t=0 T ( x,0) = T i x 3 Lumed Caacitance Method If the thermal conductivity of the solid is very high, resistance to conduction within the solid will be small comared to resistance to heat transfer between solid and surroundings. Temerature gradients within the solid will be negligible, i.e. the temerature of the solid is satially uniform at any instant. T T ( x,0) = T i x 4

3 Lumed Caacitance Method Starting from an overall energy balance on the solid: dt ha s ( T T) =! Vc dt! = T T + % = ex ha (. s - ' * t! i T i T, Vc) /! Let s define a thermal time constant 0 (5.1) where! E! out = E! st = T T! = T i T i!! t = 1 % '((Vc) = R ha t C t s (5.2) R t is the resistance to convection heat transfer C t is the lumed thermal caacitance of the solid 5 Transient Temerature Resonse Based on eq. (5.1) the temerature difference between solid and fluid decays exonentially. 6

4 Transient Temerature Resonse From eq. (5.1) the time Vc! i required for the solid to reach = a temerature T is: ha s! t ln (5.3) The total energy transfer, Q, occurring u to some time t is: t t Q = q dt = ha S! dt = (Vc)! i 1 ex t /% t 0 0 [ ( )] (5.4) = T T = T i T i!! 7 Validity of Lumed Caacitance Method Surface energy balance: T s,1 q cond q conv T s,2 T! q cond = q conv ka ( Ts 1 Ts, 2 ) = ha( Ts, L T. 2! (5.5) ) 8

5 Validity of Lumed Caacitance Method (Rearranging 5.5) T T s, 1 s, 2 Ts, 2 Tsolid (due toconduction) ( L / ka) = = T Tsolid / liquid (due toconvection) ( 1/ ha) = R R cond conv = hl k! Bi? What is the relative magnitude of T solid versus T solid/liquid for the lumed caacitance method to be valid? 9 Biot and Fourier Numbers! The lumed caacitance method is valid when where the characteristic length: L hl c =V/ c A s =Volume of solid/surface area Bi = k < 0.1 We can also define a dimensionless time, the Fourier number: Fo =! t 2 L c where = k! c Eq. (5.1) becomes:! = T T = ex [Bi Fo] (5.6)! i T i T 10

6 True or False? A hot solid will cool down faster when it is cooled by forced convection in water rather than in air. For the same solid, the lumed caacitance method is likely more alicable when it is being cooled by forced convection in air than in water. The lumed caacitance method is likely more alicable for cooling of a hot solid made of aluminum (k=237 W/m.K) than coer (k=400 W/m.K) The transient resonse is accelerated by a decrease in the secific heat of the solid. The hysical meaning of the Biot number is that it reresents the relative magnitude of resistance due to conduction and resistance due to convection. 11 Examle (Problem 5.7 Textbook) The heat transfer coefficient for air flowing over a shere is to be determined by observing the temerature-time history of a shere fabricated from ure coer. The shere, which is 12.7 mm in diameter, is at 66 C before it is inserted into an air stream having a temerature of 27 C. A thermocoule on the outer surface of the shere indicates 55 C, 69 s after the shere is inserted in the air stream.! Calculate the heat transfer coefficient, assuming that the shere behaves as a sacewise isothermal object. Is your assumtion reasonable? 12

7 What if? What haens to the rate of cooling if h increases? What haens to the rate of cooling if the diameter of the shere increases? What haens if we have a huge shere? 13 General Lumed Caacitance Analysis In the general case we may have convection, radiation, internal energy generation and an alied heat flux. The energy balance becomes: E!,! gen Est T sur qs As h + E!, g ( qconv + qrad ) As ( c, r ) =! Vc dt dt q s q rad T!, h! Numerical solutions are generally required! Simlified solutions exist for no imosed heat flux or generation (see equations (5.19, 5.25) textbook). A s,h q conv A s(c,r) 14

8 Examle 5.2 Calculation of the steady state temerature of the thermocoule junction. How much time is needed for the temerature to increase from 25 C to within 1 C from its steady state value? 15 Examle

9 Examle (5.33) Microwave ovens oerate by raidly aligning and reversing water molecules within the food, resulting in volumetric energy generation. (a) Consider a frozen 1-kg sherical iece of ground beef at an initial temerature of Ti=-20 C. Determine how long it will take the beef to reach a uniform temerature of T=0 C, with all the water in the form of ice. Assume that 3% of the oven ower (P=1kW total) is absorbed by the food. (b) After all the ice is converted to liquid, determine how long it will take to heat the beef to T f =80 C, if 95% of the oven ower is absorbed. 17 Problem 5.12: Charging a thermal energy storage system consisting of a acked bed of aluminum sheres. KNOWN: Diameter, density, secific heat and thermal conductivity of aluminum sheres used in acked bed thermal energy storage system. Convection coefficient and inlet gas temerature. FIND: Time required for shere at inlet to acquire 90% of maximum ossible thermal energy and the corresonding center temerature. Schematic: 18

10 ASSUMPTIONS: (1) Negligible heat transfer to or from a shere by radiation or conduction due to contact with other sheres, (2) Constant roerties. ANALYSIS: To determine whether a lumed caacitance analysis can be used, first comute Bi = h(r o /3)/k = 75 W/m 2!K (0.025m)/150 W/m!K = <<1. Hence, the lumed caacitance aroximation may be made, and a uniform temerature may be assumed to exist in the shere at any time. From Eq. 5.8a, achievement of 90% of the maximum ossible thermal energy storage corresonds to! E st = 0.90 = 1! ex (!t /!!cv! t ) i! t = Vc / ha s = Dc / 6h = 2700kg / m3! 0.075m! 950J / kg K 6! 75W / m 2 = 427s. K t =!! t ln( 0.1) = 427s 2.30 = 984s From Eq. (5.6), the corresonding temerature at any location in the shere is ( ) = T g,i + T i! T g,i ex (!6ht /!Dc) T( 984s) = 300 C! 275 C ex!675w/m 2 K984s/2700kg/m m950J/kgK ( ) = C T 984s T 984s ( ) ( ) If the roduct of the density and secific heat of coer is (!c) Cu 8900 kg/m J/kgK = J/m 3 K, is there any advantage to using coer sheres of equivalent diameter in lieu of aluminum sheres? Does the time required for a shere to reach a rescribed state of thermal energy storage change with increasing distance from the bed inlet? If so, how and why? 19 Problem 5.16: Heating of coated furnace wall during start-u. KNOWN: Thickness and roerties of furnace wall. Thermal resistance of ceramic coating on surface of wall exosed to furnace gases. Initial wall temerature. FIND: (a) Time required for surface of wall to reach a rescribed temerature, (b) Corresonding value of coating surface temerature. Schematic: 20

11 ASSUMPTIONS: (1) Constant roerties, (2) Negligible coating thermal caacitance, (3) Negligible radiation. PROPERTIES: Carbon steel:! = 7850 kg/m 3, c = 430 J/kgK, k = 60 W/mK. ANALYSIS: Heat transfer to the wall is determined by the total resistance to heat transfer from the gas to the surface of the steel, and not simly by the convection resistance. Hence, with ( ) 1 = 1 h +!! U = R!! tot % 1 R 1 f ' ( = 25 W/m m 2 % ) K/W( ) K ' 1 = 20 W/m 2 ) K. Bi = UL k = 20 W/m2! K 0.01 m = << 1 60 W/m! K and the lumed caacitance method can be used. (a) From Eqs. (5.6) and (5.7), T! T = ex (!t/! T i! T t ) = ex (!t/r t C t ) = ex (!Ut/!Lc) ( )430 J/kg K t =!!Lc U ln T! T 7850 kg/m m =! T i! T 20 W/m 2 K t = 3886s = 1.08h. 1200!1300 ln 300! (b) Performing an energy balance at the outer surface (s,o), h( T! T s,o ) = ( T s,o T s,i ) / R f T s,o = ht! + T s,i / R f h + ( 1/ R f ) T s,o = 1220 K. = 25 W/m2 K 1300 K K/10-2 m 2 K/W ( )W/m 2 K How does the coating affect the thermal time constant? 22

12 Other transient roblems When the lumed caacitance analysis is not valid, we must solve the artial differential equations analytically or numerically Exact and aroximate solutions may be used Tabulated values of coefficients used in the solutions of these equations are available Transient temerature distributions for commonly encountered roblems involving semi-infinite solids can be found in the literature 23 The Finite-Difference Method An aroximate method for determining temeratures at discrete (nodal) oints of the hysical system and at discrete times during the transient rocess. Procedure:! Reresent the hysical system by a nodal network, with an m, n notation used to designate the location of discrete oints in the network, and discretize the roblem in time by designating a time increment!t and exressing the time as t =!t, where assumes integer values, ( = 0, 1, 2, ).! Use the energy balance method to obtain a finite-difference equation for each node of unknown temerature.! Solve the resulting set of equations for the nodal temeratures at t =!t, 2!t, 3!t,, until steady-state is reached. What is reresented by the temerature, T m,n? 24

13 Energy Balance and Finite-Difference Aroximation for the Storage Term For any nodal region, the energy balance is E i in + E i g = E i st (5.81) where, according to convention, all heat flow is assumed to be into the region. Discretization of temerature variation with time:!t T +1 m,n T m,n!t t (5.74) m,n Finite-difference form of the storage term: E i st ( m,n ) =!!c T +1 m,n T m,n t Existence of two otions for the time at which all other terms in the energy balance are evaluated: or The Exlicit Method of Solution All other terms in the energy balance are evaluated at the receding time corresonding to. Equation (5.74) is then termed a forward-difference aroximation. Examle: Two-dimensional conduction for an interior node with!x=!y. +1 = Fo T m+1,n + T m!1,n + T ( m,n +1 + T m,n!1 ) + ( 1! 4Fo)T m,n T m,n (5.76) Fo =!!t finite-difference form of Fourier number 2!x ( ) Unknown nodal temeratures at the new time, t = (+1)!t, are determined exclusively by known nodal temeratures at the receding time, t =!t, hence the term exlicit solution. 26

14 How is solution accuracy affected by the choice of!x and!t? Do other factors influence the choice of!t? What is the nature of an unstable solution? Stability criterion: Determined by requiring the coefficient for the node of interest at the revious time to be greater than or equal to zero. For a finite-difference equation of the form, T +1 m,n =... + AT m,n A! 0 Hence, for the two-dimensional interior node: ( 1! 4Fo) 0 Fo! 1 4 ( ) 2!t!x 4! Table 5.3 finite-difference equations for other common nodal regions. 27 The Imlicit Method of Solution All other terms in the energy balance are evaluated at the new time corresonding to +1. Equation (5.74) is then termed a backward-difference aroximation. Examle: Two-dimensional conduction for an interior node with!x=!y. ( ) = T m,n ( 1+ 4Fo)T +1 m,n! Fo T +1 m+1,n + T +1 + T +1 + T +1 m!1,n m,n +1 m,n!1 System of N finite-difference equations for N unknown nodal temeratures may be solved by matrix inversion or Gauss-Seidel iteration. Solution is unconditionally stable. i Table 5.3! finite-difference equations for other common nodal regions. (5.92) 28

15 Marching Solution Transient temerature distribution is determined by a marching solution. beginning with known initial conditions. Known t T 1 T 2 T 3.. T N 0 0 T 1,i T 2,i T 3,i. T N,i 1 t t t Steady-state Problem 5.93: Derivation of exlicit form of finite-difference equation for a nodal oint in a thin, electrically conducting rod confined by a vacuum enclosure. KNOWN: Thin rod of diameter D, initially in equilibrium with its surroundings, T sur, suddenly asses a current I; rod is in vacuum enclosure and has rescribed electrical resistivity,! e, and other thermohysical roerties. FIND: Transient, finite-difference equation for node m. SCHEMATIC: 30

16 ASSUMPTIONS: (1) One-dimensional, transient conduction in rod, (2) Surroundings are much larger than rod, (3) Constant roerties. ANALYSIS: Alying conservation of energy to a nodal region of volume! = Ac x, where A 2 c =! D /4,!E in!! E out +! E g =! E st Hence, with E! g = I 2 R e, where R e =! e!x/a c, and use of the forward-difference reresentation for the time derivative, q a + q b! q rad + I 2 R e =! cv T m +1! T m t T ka m-1! T m T c + ka m+1! T m x c!!dx % T x m % Dividing each term by!ca c x/t and solving for ( ) 4! T 4 sur +1 T m, 4 ( '( T +1 k m =! c! t x 2 T m-1 + T % m+1 ' ( 2! k! c! t x 2 '1 + * -T m ),!!P t T A c c ( m ) 4 '! T 4 sur ) % () + I 2! e A 2 t! c. c + I 2! e x A c =! ca c x T +1 m! Tm. t 31 or, with Fo =! t/x 2, T +1 m = Fo! Tm-1 + T m+1 + ( 1% 2 Fo!Px 2 )T m % ( ' Fo* T ka m c )* ( ) 4 % T 4 sur + -,- + I 2!e x 2 ' Fo. ka 2 c Basing the stability criterion on the coefficient of the T m term, it would follow that Fo!!. However, stability is also affected by the nonlinear term, T m be needed to insure its existence. ( ) 4, and smaller values of Fo may 32

17 Problem 5.127: Use of imlicit finite-difference method with a time interval of t = 0.1s to determine transient resonse of a water-cooled cold late attached to IBM multi-chi thermal conduction module. Features: Cold late is at a uniform temerature, T i =15 C, when a uniform heat flux of q!! = 10 5 W/ m 2 o is alied to its base due to activation of chis. During the transient rocess, heat transfer into the cold late ( q in ) increases its thermal energy while roviding for heat transfer by convection to the water ( q conv ). Steady state is reached when q = q conv in. 33 ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constant roerties. 34

18 ANALYSIS: Nodes 1 and 5: Nodes 2, 3, 4:!x 2 + 2!!t %!y 2 ' T 1 +1 ( 2!!t!x 2 T 2 +1 ( 2!!t!y 2 T 6 +1 = T 1!x 2 + 2!!t %!y 2 ' T 5 +1 ( 2!!t!x 2 T 4 +1 ( 2!!t!y 2 T = T 5!x 2 + 2!!t %!y 2 ' T +1!!t m,n (!x 2 T +1!!t m-1,n (!x 2 T +1 m+1,n Nodes 6 and 14: ( 2!!t!y 2 T +1 m,n-1 = Tm,n!x 2 + 2!!t!y 2 + 2h!!t % k!y ' T 6 +1 ( 2!!t!y 2 T 1 +1 ( 2!!t!x 2 T 7 +1 = 2h!!t k!y T ) +T 6!x 2 + 2!!t!y 2 + 2h!!t % k!y ' T ( 2!!t!x 2 T ( 2!!t!y 2 T = 2h!!t k!y T ) +T Nodes 7 and 15:!x 2 + 2!!t!y 2 + 2h!!t % k!y ' T 7 +1 ( 2!!t!y 2 T 2 +1 (!!t!x 2 T 6 +1 (!!t k!x 2 T 8 +1 = 2h!!t k!y T ) +T 7!x 2 + 2!!t!y 2 + 2h!!t % k!y ' T (!!t!x 2 T (!!t!x 2 T ( 2!!t!y 2 T = 2h!!t k!y T ) +T 15 Nodes 8 and 16:!x 2 + 2!!t!y ( 4!!t 3!x 2 T 9 +1 ( 2 3 h!!t % k!y ' T 8 +1 ( 4!!t 3!y 2 T 3 +1 ( = 2 h!!t 1 3 k!x + 1 %!y ' T ) + T 8 h!!t k!x + 2 3!!t!y 2 T 11!!t!x 2 T 7 +1!x 2 + 2!!t!y h!!t k!x + 2 h!!t % 3 k!y ' T ( 2!!t 3!y 2 T ( 2!!t 3!x 2 T ( 4!!t 3!x 2 T ( 4!!t 3!y 2 T = 2 h!!t 1 3 k!x + 1 %!y ' T ) + T 16 36

19 Node 11:!x 2 + 2!!t!y 2 + 2h!!t % k!x Nodes 9, 12, 17, 20, 21, 22: ' T (!!t!y 2 T 8 +1 ( 2!!t!x 2 T (!!t!y 2 T = 2h!!t k!x T ) +T 11!x 2 + 2!!t %!y 2 ' T +1!!t m,n (!y 2 T m,n+1 + T % m,n-1 (!!t!x 2 T +1 m-1,n + +1 Tm+1,n % = T m,n Nodes 10, 13, 18, 23: Node 19: Nodes 24, 28:!x 2 + 2!!t %!y 2 ' T +1!!t m,n (!y 2 T m,n+1 + T % m,n-1 ( 2!!t!x 2 T +1 m-1,n!x 2 + 2!!t %!y 2 ' T (!!t!y 2 T T % 24 ( 2!!t!x 2 T = T 19!x 2 + 2!!t %!y 2 ' T ( 2!!t!y 2 T ( 2!!t!x 2 T = 2 q o ))!!t k!y +T 24 = T m,n!x 2 + 2!!t %!y 2 ' T ( 2!!t!y 2 T ( 2!!t!x 2 T = 2 q o ))!!t k!y +T Nodes 25, 26, 27:!x 2 + 2!!t %!y 2 ' T +1 2!!t m,n (!y 2 T +1 m,n+1 (!!t!x 2 T +1 m-1,n + +1 Tm+1,n % = 2 q o ))!!t k!y +T +1 m,n The convection heat rate er unit length is q! conv = h % (x/2)( T 6 T )+x( T 7 T )+(x+y)( T 8 T )/2+y( T 11 T )+(x +y) ( T 16 T ) / 2 + x( T 15 T ) + (x/2)( T 14 T ) = q! out. The heat inut er unit length is ( ) q in! = q o!! 4x On a ercentage basis, the ratio of convection to heat in is ( ) 100. n! q conv / q in 38

20 Results of the calculations (in C) are as follows: Time: 5.00 s; n = 60.57% Time: s; n = 98.16% Time: s; n = 85.80% Time: s; n = 99.00% Time: s; n = 94.89% Temeratures at t = 23 s are everywhere within 0.13 C of the final steady-state values. 39 Summary The lumed caacitance analysis can be used when the temerature of the solid is satially uniform at any instant during a transient rocess! Temerature gradients within the solid are negligible! Resistance to conduction within the solid is small comared to the resistance to heat transfer between the solid and the surroundings The Biot number must be less 0.1 for the lumed caacitance analysis to be valid. Transient conduction roblems are characterized by the Biot and the Fourier numbers. 40