LIBRARY Mlchlgan State University PLACE It RETURN BOXto movothb chockout 1mm your ncord. TO AVOID FINES Mum on or More data duo. DATE DUE DATE DUE DATE DUE 1! I | MSU to An Affirmativ- ActtoNEquul Oppoctunlty Inctttmton Mala-9.1 7 -_ q. _ __ .. , __ _ _.gifi, .7 _,V~____ A COMPUTATIONAL MODEL FOR DETERMININGTHE EFFECTIVE THERMAL CONDUCTIVITY OF A POROUS MEDIUM By Daniel K. Lucas A THESIS Submitted to Michigan State University in partial fulfillment of the requirements for the degree of Master of Science Department of Mechanical Engineering 1996 ABSTRACT A COMPUTATIONAL MODEL FOR DETERMINING THE EFFECTIVE THERMAL CONDUCTIVITY OF POROUS MEDIUM By Daniel K. Lucas There are many systems of interest in engineering that involve heat transfer through a medium consisting of two or more'distinct materials. Such a medium is often considered to be a porous medium when one of the materials is in solid phase and another material is in fluid phase. Some examples of a porous medium are the packed bed of a cooling tower, building insulation, and the core of a catalytic converter. In evaluating these systems and others of interest in engineering thermodynamics, heat transfer, or fluid mechanics, the effective thermal conductivity of a porous medium must first be evaluated. It is difficult to predict the thermal conductivity of a medium consisting of two or more separate materials. There is no general model to predict the effective thermal conductivity of a porous medium. This paper proposes such a model starting with the two dimensional heat conduction equation. Finite differencing is employed to develop the numerical model. From this numerical model, a computer code is developed that predicts the effective thermal conductivity of a porous medium. To my wife and family: Susan, Lisa, Kerri, and Daniel. TABLE OF CONTENTS LIST OF FIGURES NOMENCLATURE Chapter APPENDIX INTRODUCTION Problem Statement Literature Review Proposal Statement Summary METHOD OF SOLUTION Governing Equations Boundary Conditions RESULTS AND DISCUSSION Porous Media Generation Results of Numerical Validation CONCLUSIONS AND RECOMMENDATIONS Conclusions Recommendations LIST OF REFERENCES Specific References General References 47 48 51 52 Figure 1. Figure 2. Figure 3. Figure 4. Figure 5. Figure 6. Figure 7. Figure 8. Figure 9. Figure 10. Figure 11. Figure 12. Figure 13. Figure 14. Figure 15. Figure 16. Figure 17. Figure 18. Figure 19. Figure 20. Figure 21. Figure 22. Figure 23. LIST OF FIGURES Conductive Heat Transfer in Porous Medium Parallel Resistance Model Parallel Resistance Circuit Series Resistance Model Series Resistance Circuit Parallel and Series Models of Effective Thermal Conductivity for y = 0.1 Loeb’s Model Thermal Circuit of Loeb’s Model Heat Conduction Model Array of Cylinders with Imposed Finite Element Grid Simulated Porous Media Where Shaded Area Represents Solid Finite Difference Element Normalized Porosity for Target Porostiy of 0.1 to 0.9 Frequency vs. 0.1 Normalized Porosity Frequency vs. 0.9 Normalized Porosity Calculated porosity vs. N x N Elements Calculated porosity vs. N x N Elements Effective Thermal Conductivity vs. N x N Elements Effective Thermal Conductivity vs. N x N Elements Effective Thermal Conductivity vs. Log (DT) Effective Thermal Conductivity vs. Target Porosity km vs. Generated Porosity km/kf vs. Target Porosity 32 33 36 37 39 4O 41 43 Nomenclature oQO>—I.or—x Symbols Subscripts h“"l'"13CD-IQO'O¢DC 3-~ Superscripts k Overbar Thermal Conductivity Length Heat Transfer Rate Temperature Area Thermal Resistance Diameter Average Temperature Gradient Time Step Porosity Solid Thermal Conductivity I Fluid Thermal conductivity Geometrical Factor Void Fraction Fluid Lower Boundary Effective Value Upper Boundary Solid Pore Continuous Phase Discontinuous Phase Top Cell Node Bottom Cell Node Right Cell Node Left Cell Node Index Transverse to Heat Flow Index Parallel to Heat Flow Index Direction Average Conditions CHAPTER I INTRODUCTION 1.1 Problem Statement There are many systems of interest in engineering that involve heat transfer through a medium consisting of two or more distinct materials. Such a medium is often considered to be a porous medium when one of the materials is in solid phase and the other material is in fluid phase. Some examples of a porous medium are the packed bed of a cooling tower, building insulation, and the core of a catalytic converter. In evaluating these systems and others of interest in engineering thermodynamics, heat transfer, or fluid mechanics, the effective thermal conductivity of a porous medium must first be evaluated. The thermal conductivity of a single material is defined as the amount of heat flowing by conduction through a unit area per unit time per unit temperature gradient. The evaluation of thermal conductivity for a single material is very straight forward. However, it is more difficult to predict the thermal conductivity of a medium consisting of two or more separate materials. There has been previous work to determine the thermal conductivity of a porous medium. However, this work has dealt with either specific systems or specific geometries, and there is no general model for evaluating the effective thermal conductivity of a two component system. 2 There are three modes of heat transfer in a porous medium: conduction, convection, and radiation. Conduction is heat transfer by molecular, atomic, and electronic motion, and is often the prominent mode of heat transfer. Convection is heat transfer by fluid motion and can be important as porosity increases. Finally, radiation is heat transfer by electromagnetic wave motion and becomes increasingly important as the temperature increases. The problem will be studied as a two-dimensional system similar to the system shown in Figure 1. This system consists of two or more anisotropic materials having different thermal characteristics and has fixed temperatures applied to the top and bottom surfaces and adiabatic sidewalls. The effective thermal conductivity is defined as km such that the heat transfer may modeled using the equation _ kmAT q: L. (1) Using the above equation, a control volume may be defined, that is small enough to have local thermal equilibrium yet large enough to retain the properties of the porous medium, over which Fourier’s law of heat conduction is valid, i.e., 9 = kaT. (2) 1.2 Literature Review Previous work by Baradat and Combarnous [1] has shown that the two simplest models for determining effective thermal conductivity also describe the V TL Figure 1. Conductive Heat Transfer in Porous Medium 4 upper and lower boundaries for the actual thermal conductivity of a porous medium. These are the parallel resistance model and the series model. In the parallel resistance model, the voids and solids hypothetically move in opposite directions transverse to the heat flow as shown in Figure 2. Working with the parallel thermal circuit shown in Figure 3, the thermal conductivity can be shown to be 51:841-.» (3) where e is the porosity and y is the ratio of solid thermal conductivity to the fluid thermal conductivity. In the series resistance, the voids and solids hypothetically move in opposite directions in the direction of the heat flow as in Figure 4. Now, working with the series thermal circuit shown in Figure 5, the thermal conductivity can be shown to be km (1 — s) 4 T(,— = [a + T] . (4) The current models are bounded by the series and parallel models as shown in the graph in Figure 6 as km/kf versus porosity with the solid to fluid thermal conductivity ratio, 7, set equal to 0.1. These two simple models represent a whole group of models that have simple solutions. In these simple models, either heat flow lines run parallel to the direction of the heat flux or the isotherms are perpendicular to the heat flow. The group of models based on these simple models have one common approach for solution. They use a thermal circuit network and allow the / Fluid Solid L / .IL |-— (1-e)A + 2A 4 Figure 2. Parallel Resistance Model m< (k or k + 1) il TB,ij ‘ TT,i - 1,] ka Figure 12. Finite Difference Element 24 The superscript i on the thermal conductivity indicates whether an element is to be either a solid (5) or a fluid (f) and the subscript indicates thermal conductivity as directionally dependent. The upper boundary is set at some temperature, Tu, and the lower boundary at some temperature, TL, where Tu > TL. Further, the sidewalls in the porous system will be considered to be adiabatic (Q = 0). Applying finite difference to Equation (18), the terms in Equation 18 become @21- (Tu; " 2Tij '1' TR,ij) ax2 = (sz) (19) and 62T (TB.ij — 2Tij + Tm.) _ = , 2o 63/2 (Ayz) ( ) Substituting Equations (19) and (20) into Equation (18), we can write Equation (18) in the finite difference form [TU] _ :Ti; + T”) + R91 x . k (i) x,ij Now, the Alternating Direction Implicit (ADI) method will be utilized to solve the set of equations represented in Equation (21). To employ the ADI method, we must first set a quasi-time derivative that corresponds to first stepping in the y-direction (k + 1I2 iterate) and then in the x-direction (k + 1 or k iterate). 25 The following equation is thus obtained: k (“)9 (k) _ X-l' (k) (k) (kl T, — Tij _ :(TL,j — 2Tij + TM) k ” + + + + AWE?“ V2) — 2T)" ’2’ + TI}, V9) (22) GT where 2 2 2 5T = __(Ax) = __(Ay) = i . (23) AtT AtT AtT Before the set of equations represented by Equation (22) can be solved, adjacent finite difference elements must be coupled. This is done by observing that TTJI' = Tam.) (24) T3."- = TU.” (25) TU]. = 1',"qu (26) TR."- = TLJJH (27) which represents the temperature continuity at the boundaries between adjacent elements. There is also a continuity of the heat flux at the boundaries between adjacent elements. Therefore, we can also write the following equation: T .. - T.. T. . - T . _ ky‘ij 1ij I] = ky'H M 1+ 1,) 9/ B,I+1,) . (28) 2 2 Substituting Equation (24) into Equation (28) yields the equation T .. - T. T. . - T .. ky,ij TJl ll = ky,i+1,j 1+1.J T,IJ ’ (29) 96 96 26 and solving for T1),- gives T - ky,i+1,j 12%;?) + ky "iTIK+}§I 30 _ . ( ) k' ‘+ k:.:1 YJ " 1-l Similarly, the terms T3,“, TL], and TR), can all be expressed in terms ofTi-1_,-, Ti,- - 1, and T1,). 1 such that 11M“ + ky" "TiIKTyI k . T8, = y,I-1j I-1.j (31) J ky,i-1,j + kyiiu k .. TI“) + k MT”) TLi. = XM" H 1 (32) J kx,i,j-1 T kxjju k .. TIKI+ k .. TI") TR, = X,l,j*1 i.j+1 XJI 'I (33) J kxjj + 1 + k X,lj respectively. Substituting Equations (30), (31), (32), and (33) into Equation (22) yields the equation TIIMM) _ TiIk I k k k = kx,iI kx,i-1.ITift.)I+ kx.iITiI ) _2T(k) + kx.i+1.j TigIIT- kX'lT'I ) 5 T kx,i-1.j + kxII 1) kx.i+1.1 + kxii (“)6 k+)6) (“)6 (“)9 +kyii [km- 1_T,,_ 1 I +k,,,,T,,( _2Tu(k+ )4) + ky.ij+1Ti.j+1 ) +kinTii ] . (34) GT ky,,,,-_, + kal J ky,i,j+1 T ky.ij By rewriting Equation (34) and collecting the coefficients on each element temperature, T, the coefficients can be written for TIE-W: B =- Mk” ‘ (35) ij cT(kY'I+kYJI 1) 27 k2.. k2 k M: c, :1. - +24% (36’ ‘ ‘ O’T(kY.iI +ky.i.I-1) °T(ky.ii +kaI+ 1) OT + k i' R 1'1' T515216): 00:" y" H 1 I (37) o-T(ky,ij+ky.i-I*1) k k 1,0,)” E,,- = x,ij X.i-1.I (38) ' O’T(kxij+kx,I- 1j) Ti(jk)2 Fij =1+ kiij + kid _zffl’ (39) OT(kx.ij+kx,i-1.I) OT(kx.iI+kx.i+1J) OT k k . 11(911' Gil: x“ H U (40) 61(kxij+kx,i+1,j) and the following equation can be written B, TOW) + c, T,"”Y2’ + D, To”) '1 1 i,j+1 = 5,119,, + F, T,"" + G, ,T,‘f,, (41) Further, Equation (41) can be written as a set of equations =0" )6) —(k+ )3) -(k) M) T) = S (42) =k+ 2) where M,- is represented by a tridiagonal matrix. By employing a Gaussian elimination routine called the Thomas algorithm , Equation (42) can be solved for TE“ ” . This routine has the advantage of being able to store the required coefficients in a compact 3 x N2 matrix. This is done as j is stepped form 1 to N. Similarly, an equation can be developed for the k + 1 iteration and solving for 28 k+1) T: as i is stepped from 1 to N. The solutions are obtained alternately in each direction iteratively until the solutions converge. 2.2 Boundam Conditions It is observed that at the left boundary, where the adiabatic walls exist, the heat flux equation becomes fl 8x ”0 = 0 (43) which becomes L—T— = o. Ax (44) Therefore, TL = TR. (45) Equation (45) is then substituted into Equation (22) to obtain Til-(Mg) _ Tillk) = &(2Téz) _ 213(k)) oT ' k i' + + + + Gil—(T3; 72) — 2T;k V2) + T3; 72)) (46) Similarly, the boundary condition at the right wall is E ax =0 x=N and the finite difference equation becomes 29 k .. (kW?) ('0 _ X-' (k) (k) T k .. + + + + firs: V2) - 2T9 y» + Tl}, Y»). (47) T The B,C,D,E,F, and G coefficients are then obtained as before. In addition, the initial condition at the bottom of the porous medium is T3,, = TL (48) and at the top, the initial condition is Tm = TU. (49) The B, C, D, E, F, and G coefficients are then again obtained as before. The effective thermal conductivity can be obtained by determining the average heat flux in the y-direction while x = O _ 1 " 6T and then applying the definition from Equation (1) and assuming L = 1, k = q , 48 CHAPTER III RESULTS AND DISCUSSION 3.1 Porous M Generation The porous media generation is accomplished by the use of the pseudo-random number generator function call available in Microsoft FORTRAN. Figure 13 shows the results of running the random number generator one hundred times for each porosity from 0.1 to 0.9. The values are then normalized by dividing the generated porosity by the target porosity. For example, with a target porosity of 0.1, the calculated porosities range from 0.04 to 0.18 with a population standard deviation of 0.027. This distribution is shown in Figure 14. This results in a normalized porosity of 0.4 to 1.8. Further, running the random number generator for a target porosity of 0.9 results in a range from 0.83 to 0.96 with a population standard deviation of 0.026 giving a normalized porosity of 0.92 to 1.07. This distribution is shown in Figure 15. As will be seen in the discussion on numerical testing, the randomness of the generated porous media will result in slight deviations as the different data are put into graphical form. Figure 16 shows calculated porosity as a function of grid size. A thermal conductivity of 0.3 was used. As before, the actual porosities tend toward the target porosities of 0.3, 0.5, and 0.8 as the grid size is increased. 30 31 Distribution of Cells for Porosity of 0.1 to 0.9 7 10.? no.2 no.3 no.4 .05 [0.6 . no.7 I08 L201 l i Solid Gels in Porous Nbdia Target Fbrosity Target Porosity g Figure 13. Normalized Porosity for Target Porostiy of 0.1 to 0.9 32 16 14 i» 12 .. 10 ~» 88.? 88.. $8.. 89; «83 5:3 . 8o: nomad 0.1 Target Pbrosity hemmd oomhd named um 56 ooovd Figure 14. Frequency vs. 0.1 Normalized Porosity 33 20 i 184» 16<~ 141» 12» l ~FREIBCY10«~ 3.. l 6 4T 2. o. N N 1") t") V to to IO N O N no s§§§sa§§s§s§§ O O O O O O O 1- !- v- v- ‘- W 0.9 Target Porosity Figure 15. Frequency vs. 0.9 Normalized Porosity 0.900 0.800 4 0.700 4» 0.600 1 0.500 . ~ Calculated Porosity 0.400 0.300 «» 0.200 0.100 0.000 l he.-.) 9 12 N x N Bernents 15 18 f-- L - . — Target W031 ....... Target Porosity = 0.5 Figure 16. Calculated porosity vs. N x N Elements - — — - Target Porosity = 0.8 35 Further, it is expected that as the grid size increases, the calculated porosity will converge around a single value as determined by the set target porosity. This is confirmed by the graph in Figure 17 which represents the calculated porosity as a function of grid size. Figure 18 shows that for a kl/ks = 0.5 (0.3/0.6) and 3 OT = 0.1, the effective thermal conductivity will approach a single value as the grid size is increased. 36 0.40 0.35 11 0.30 Calculated Porosity 0.25 0.20 - 0.15 1 95 105 115 ., 125 135 N x N Bements Figure 17. Calculated porosity vs. N x N Elements 36 0.40 035 ~ 0.30 1- Calculated Porosity 0.25 ., 0.20 1» 0.15 ,_ 0.10:1A.L.e.+1 IO ID ID 5 15 25 35 1 45 55 6 7 8 95 105 115 N x N Bements Figure 17. Calculated porosity vs. N x N Elements 125 » 135 0.50 0.45 1 0.40 0.35 0.30 Km 0.25 0.20 0.15 ,_ 0.101 0.05 < 0.00 37 l 1 ID ID In In In to to ID ID ID In In to I!) v- N ('0 1‘ In (D N G) O) o v- N co v- ‘- v- 1- N x N Bements Figure 18. Effective Thermal Conductivity vs. N x N Elements 38 3.2 Results of Numerical Validation For the following numerical testing, a target porosity of 0.3456 was used unless specifically stated otherwise. Thermal conductivity values of either 0.3 or 0.03 were used for the fluid and solid materials in the following numerical tests. Figure 19, showing effective thermal conductivity as a function of the grid size, indicates the relative range of thermal conductivities for kf/ks of 0.1, 1.0, and 10.0. As expected, these curves trend closer to one value as the grid size is increased. In Figure 20, the effective thermal conductivity is plotted as a function of the log1o of the time step. As before, thermal conductivities of either 0.3 and 0.03 were used for the fluid and solid materials. These graphs were expected to converge around one value as the time step is increased. The fluctuation that is seen is attributed to the randomness of the grid generation. However, these curves do tend toward a central value. Figure 21 shows the effective thermal conductivity as a function of the target porosity for thermal conductivity ratios of 0.5, 1.0, and 2.0 using thermal conductivities of 0.3 and 0.6. These curves also behave as expected. When kf/k,3 = 2.0 at a target porosity of 0.0 to 1.0, the curve is equal to 0.3 and 0.6 respectively, and when kf/ks = 0.5 with the target porosity once again running from 0.0 to 1.0, the curve runs from 0.6 to 0.3 respectively. This is also the expected result. Finally, with kf/ks =1.0 and the thermal conductivities set equal to 0.3, the curve is equal to 0.3 from a target porosity of 0.0 to 1.0. 39 0.35 . 0.30 “ /—\ d‘ ...................... .I \v” .I V 0.25 l 0201 . —.-—- kf/ks=0.1 I -.—--l