Numerical Heat Transfer ( 数值传热学 ) Chapter 4 Numerical Solution of Diffusion Equation and its Applications(2)

Transcription

1 Numerical Heat ransfer ( 数值传热学 ) Chapter 4 Numerical Solution of Diffusion Equation and its Applications(2) Instructor ao, Wen-Quan CFD-NH-EH Center Key Laboratory of hermo-fluid Science & Engineering Xi an Jiaotong University Xi an, 207-Oct- /56

2 数值传热学 第四章扩散方程的数值解及其应用 (2) 主讲陶文铨 西安交通大学能源与动力工程学院热流科学与工程教育部重点实验室 207 年 0 月 日, 西安 2/56

3 Contents 4. -D Heat Conduction Equation 4.2 Fully Implicit Scheme of Multi-dimensional Heat Conduction Equation 4.3 reatments of Source erm and B.C. 4.4 DMA & ADI Methods for Solving ABEs 4.5 Fully Developed H in Circular ubes 4.6 Fully Developed H in Rectangle Ducts 3/56

4 4.4 DMA & ADI Methods for Solving ABEs 4.4. DMA algorithm ( 算法 ) for -D conduction problem.general form of algebraic equations of -D conduction problems 2.homas algorithm 3.reatment of st kind boundary condition ADI method for solving multidimensional problem. Introduction to the matrix of 2-D problem 2. ADI iteration of Peaceman-Rachford 4/56

5 4.4 DMA & ADI Methods for Solving ABEqs 4.4. DMA algorithm for -D conduction problem.general form of algebraic equations. of -D conduction problems he ABEqs for steady and unsteady (f>0) problems take the form app aee aw W b he matrix ( 矩阵 ) of the coefficients is a tridiagonal ( 三对角 ) one. hree unknons 5/56

6 2. homas algorithm( 算法 ) he numbering method of W-P-E is humanized ( 人性化 ), but it can not be accepted by a computer! Rerite above equation: A i i B i i C i i Di, i,2,... M (a) End conditions:i=, C i =0; i=m, B i =0 () Elimination ( 消元 )-Reducing the unknons at each line from 3 to 2 Assuming the eq. after elimination as P Q i i i i (b) Coefficient has been treated to. 6/56

7 he purpose of the elimination procedure is to find the relationship beteen P i, Q i ith A i, B i, C i, D i: Multiplying Eq.(b) by C i, and adding to Eq.(a): A i i i i i A B C D (a) i i i i i i i C CP CQ i i i i i i i C P B D C Q i i i i i (b) Yielding Comparing ith i B D C Q ( ) A C P A C P i i i i i i i i i i i P Q i i i i 7/56

8 P i A B i C P i i i ; Q i D A C Q i i i C P i i i he above equations are recursive -i.e., In order to get P i, Q i, P and Q must be knon. In order to get P,Q,use Eq.(a) A i i B i i C i i Di, i,2,... M (a) End condition:i=, C i =0; i=m, B i =0 Applying Eq.(a) to i=, and comparing it ith Eq. (b), the expressions of P,Q can be ; 8/56

9 obtained: B D 2 A A i, C 0, A B 2 D (2) Back substitution( 回代 )-Starting from M via Eq.(b) to get i sequentially( 顺序地 ), B M PM M Q i M Pi ; Ai CiPi End condition: P 0 i=m, B i =0 M P B A ; Q D A Q M M P Q i i i i M,...,. to get: 2 9/56

10 3. Implementation of homas algorithm for st kind B.C. For st kind B.C., the solution region is from i=2,.to M-=M2. Applying Eq.(b) to i= ith given,given : P 2 Q P 0; Q,given Because M is knon,back substitution should be started from M 2 : P Q M 2 M 2 M 2 When the ASM is adopted to deal ith B.C. of 2 nd, and 3 rd kind, the numerical B.C. for all cases is regarded as st kind, and the above treatment should be adopted. 0/56

11 4.4.2 ADI method for solving multi-dimensional problem. Introduction to the matrix of 2-D problem E N W N P E P S W S -D storage ( 一维存储 )of variables and its relation to matrix coefficients /56

12 Numerical methods for solving ABEqs. of 2-D problems. () Penta-diagonal algorithm(pdma, 五对角阵算法 ) (2) Alternative( 交替的 )-direction Implicit (ADI, 交替方向隐式方法 ) 2. 3-D Peaceman-Rachford ADI method t 2-D ADI Dividing In the st /3 t into three uniform parts t implicit in x direction, and explicit in y, z directions; In the 2 nd and 3 rd t /3 y, z direction, respectively. implicit in 2/56

13 Set u i,j,k, v i,j,k the temporary( 临时的 ) solutions at to sub-time levels 2 n x i, j, k st subtime level u 2 nd sub- v time level: -CD for 2 nd derivative at n time level in x direction i, j, k i, j, k t t /3 u /3 n i, j, k 2 2 n 2 n a x ui, j, k yi, j, k z i, j, k ( ) n i, j, k a x ui, j, k yvi, j, k z ui, j, k ( ) 3 rd subtime level n n i, j, k i, j, k 2 2 n 2 n a xvi, j, k y vi, j, k zi, j, k t v /3 ( ) 3/56

14 Stability condition by von Neumann method: at( ) x y z Is the alloed maximum time step three times of -D case? Actually, No! For 2-D case P-R method is absolutely stable. 3. o ADI methods: ADI-implicit( 交替方向隐式 ) for transient problems and ADI-iteration( 交替方向迭代 ) multi-dimensional problems. hey are very similar. 4/56

15 5 FDH in Circular ubes 4.5. Introduction to FDH in tubes and ducts Physical and Mathematical Models Governing equations and their nondimensional forms Conditions for unique solution Numerical solution method reatment of numerical results Discussion on numerical results 5/56

16 4.5 Fully Developed H in Circular ubes 4.5. Introduction to FDH in tubes and ducts. Simple fully developed heat transfer Physically:Velocity components normal to flo direction equal zero; Fluid dimensionless temperature distribution is independent on( 无关 ) the position in the flo direction Mathematically:Both dimensionless momentum and energy equations are of diffusion type. Present chapter is limited to simple cases. 6/56

17 FDH in straight duct is an example of simple cases. x m, ( ) 0, m 2. Complicated FDH b In the cross section normal to flo direction there exist velocity components, and the dimensionless temperature depends on the axial position, often exhibits periodic ( 周期的 )character. he full Navier- Stokes equations must be solved his subject is discussed in Chapter of the textbook. 7/56

18 Examples of complicated FDH ube bundle (bank) ( 管束 ) Fin-and-tube heat exchanger Louver fin ( 百叶窗翅片 ) 8/56

19 3. Collection of partial examples able 4-5 Numerical examples of simple FDH No Cross section B. Condition Refs See pp for details 9/56

20 4.5.2 Physical and mathematical models A laminar flo in a long tube is cooled (heated) by an external fluid ith temp. and heat transfer coefficient h e. Determine the heat transfer coefficient and Nusselt number in the FDH region. 20/56

21 .Simplification ( 简化 ) assumptions () hermo-physical properties are constant ; (2) Axial heat conduction in the fluid is neglected (3) Viscous dissipation ( 耗散 )is neglected; (4) Natural convection is neglected; (5) Wall thermal resistance is neglected; (6) he flo is fully developed: u u m r R 2 2[ ( ) ]; v 0 2/56

22 2. Mathematical formulation ( 描述 ) ()Energy equation Cylindrical coordinate, symmetric temp. distribution, and no natural convection (A4): cp( u v ) ( r ) ( ) S x r r r r x x FD flo (A6) c u p ( r ) x r r r 2-D parabolic eq.! No axial cond. (A2) No dissipation (A3) ype of eq.? 22/56

23 (2)Boundary condition r r 0, 0 r R, he ( ) r (Symmetric condition); (External convective condition!) Internal fluid thermal conductivity External ( 外部 )convective heat transfer No all thermal resistance(a5), tube outer radius =R; 23/56

24 4.5.3 Governing eqs. and dimensionless forms From FD condition a dimensionless temp. can be introduced, transforming the PDE to ordinary eq.. Defining b b b db hen: ( b ) ; x x dx Defining dimensionless space and time coordinates: x 2Rcpum 2Rum ; X Pe a r R R Pe Constant properties(a ) 24/56

25 Energy eq. can be reritten as: db / dx d d u ( ) / ( ) d d 2 u b = 0 Dependent on X only Dependent on only m is called eigenvalue ( 特征值 ) 25/56

26 Folloing ordinary PDE for the dimensionless temperature. eq. can be obtained d d u ( ) /( ) d d u 2 m he original to B.Cs. are transformed ( 转换成 ) into: d 0, 0; d (a) (b) d( ) b he R d, ( ) r d( ) b d R (c) Question:hether from Eqs.(a)-(c) a unique ( 唯一的 ) solution can be obtained? ) Bi 26/56

27 4.5.4 Analysis of condition for unique solution Because of the homogeneous ( 齐次性 ) character : Every term in the differential equation contains a linear part of dependent variable or its st /2nd derivative. d ( d ) /( u ) d d u 2 m d d u ( ) ( ) d d u 2 m In addition, the given B.Cs. are also homogeneous: d d 0, 0; ) Bi d d For the above mathematical formulation there exists an uncertainty ( 不确定性 )of being able to be multiplied by a constant for its solution. 27/56

28 While in order to solve the problem, the value of the formulation has to be determined. In order to get a unique solution and to specify the eigenvalue, e need to supply one more condition! We examine the definition of dimenionless temperature: in ( ) b Physically, the averaged temp. is defined by b b 0 R 2 rudr 2 Rum b b b.0 r u r 2 d( ) 0 R u R m = 28/56

29 hus the complete formulation is: d d u ( ) ( ) 0 (a) d d u 2 m d 0, 0; (b) d d d 0 ) Bi u u m d / 2 Non-homogeneous term! (c) (d) 29/56

30 4.5.5 Numerical solution method his is a -D conduction equation ith a source term! u 2 um ()Let d ( d ) ( u ) 0 d d u,hose value should be determined during the solution process iteratively. Patankar-Sparro proposed folloing numerical solution method: 2 m Because of the homogeneous character, the form of the equation is not changed only replacing by. 30/56

31 d d u ( ) ( ) 0 d d u (a) 2 m d 0, 0; (b) d d ) Bi d (c) 0 u u m d / 2 Non-homogeneous term (d) u /(2 d) It can be used to iteratively 0 u determine the eigenvalue. m 3/56

32 (2)Assuming an initial field *, get (3)Solving an ordinary differential eq. ith a source term to get an improved (4)Repeating the above procedure until: *, 0 ~ * his iterative procedure is easy to approach convergence: S 2 u ( u/ u ( ) u m) 2 4 ( u / u ) d 4 ( ) d 2 m u /(2 d) 0 u m 0 m 0 32/56

33 exists in both numerator and denominator, thus only the distribution, rather than absolute value ill affect the source term reatment of numerical results o ays for obtaining heat transfer coefficient:. From solved temp. distribution using Fourier s la of heat conduction and Neton s la of cooling: r R, h( b) h r r For inner fluid Different from Boundary condition r ) r R b R, he ( ) r 33/56

34 34/56 2. From the eigenvalue( 特征值 ): From heat balance beteen inner and external heat transfer ( ) ( ) b e h h Inner External Get: e b h h e b h e b h e e b h h h e b h h

35 Nu h h h = e e h e R h e 2Bi 2 From the specified values Bi eigenvalues, 2Rh, the corresponding, can be obtained. hus it is not necessary to find the st derivative at the all of function for determining Nusselt number Discussion on numerical results 35/56

36 able 4-6 Numerical results of FDH in tubes ( Nu) q ( Nu) 36/56

37 . Bi effect: From definition Bi, h e h e 0 Bi Rh e External heat transfer is very strong,the all temp. approaches fluid temp. his is corresponding to constant all temp condition, hus Nu=3.66 Bi 0, Is this adiabatic? No! Product of very small H coefficient and very large temp. difference makes heat flux almost constant. q h const e 37/56

38 2. Computer implementation of Bi and Bi 0 Bi by progressively ( 逐渐地 ) increasing Bi: Bi= 5 0, 6 0 7,0... Nu Bi=0 by progressively decreasing Bi: Bi= 0., 0.0, 0.00, 0.000, , Double decision ( 双精度 )must be used for Computation: 2Bi, Bi 0, 0, /56

39 4.6 Fully Developed H in Rectangle Ducts 4.6. Physical and mathematical models Governing eqs. and their dimensionless forms Condition for unique solution reatment of numerical results Other cases(2070) 39/56

40 Brief revie of lecture key points. DMA solution algorithm for -D problem P () Elimination ( 消元 )-Reducing the unknons at each line from 3 to 2 A B C D (a) i i i i i i i i A B i C P i i i ; Q i D A C Q i i i C P i i i 2. ADI method for solving 2-D unsteady problem ; P Q i i i i P recursive (2) Back substitution( 回代 )-Starting from the last node via Eq.(b) to get i sequentially Dividing t into to uniform parts;in the st t /2 B A ; Q (b) D A 40/56

41 implicit in x direction, and explicit in y direction; t /2 In the 2 nd implicit in y direction, and explicit in x direction. By implementing to times of DMA the algebraic equations for forarding one time step is solved. 3. Homogeneous problems Every term in the differential equation and boundary conditions only contains a linear part of dependent variable or its st or 2nd derivative. For such a mathematical formulation there exists an uncertainty of being able to be multiplied by a constant for its solution. 4/56

42 4.6 Fully Developed H in Rectangle Ducts 4.6. Physical and mathematical models Fluid ith constant properties flos in a long rectangle duct ith a constant all temp. Determine the friction factor and H coefficient in the fully developed region for laminar flo.. Momentum eq. For the fully developed flo u=v=0, only the velocity component in z-direction is not zero. Its governing equation: 42/56

43 p ( u v ) ( ) x y z z x y z Neglecting cross section variation of p 2 2 dp ( ) x y dz p ( ) 0 x y z aking ¼ region as the computational domain because of symmetry. Boundary conditions are: At the all,=0; At center line, First order normal derivative equals zero: n 0 43/56

44 Defining a dimensionless velocity as : W dp D 2 dz here D is the referenced length, say: D=a, or D=b. Defining dimensionless coordinates:x=x/d, Y=y/D, then: dp ( ) x y dz It is a heat conduction problem ith a source term and a constant diffusivity! W W X Y At all,w=0; W At center lines, 0 n 44/56

45 2. Energy equation 0 0 ( ) ( ) ( ) ( c ) p u v 0 x y z x x y y z z hus: c p ( ) ( ) z x x y y Boundary conditions: At the all,= ; At the center line, 0 n Neglecting axial heat conduction ype of equation? Parabolic!Z is a one-ay coordinate like time! 45/56

46 4.6.2 Dimensionless governing equation We should define an appropriate dimensionless temperature such that the dimension of the problem can be reduced from 3 to 2: Separating the one-ay coordinate z from the to-ay coordinates x,y hen ( ) b b ( b ) z z Defining: X x/ D, Y y / D, Z z /( DPe) b Pe c D p One-ay coordinate! m 46/56

47 Dimensionless governing eq. hus: 2 2 ( ) 2 2 b X Y Z W b W Dependent on Z only W 2 2 0; 2 2 X Y W At the all 0 m n At center line, 0 m Dependent on X,Y only d( ) b dz Heat conduction ith an inner source! b 0 47/56

48 4.6.3 Analysis on the unique solution condition Because of the homogeneous character,these also exists an uncertainty of being magnifying by any times! Introducing average temperature (difference): ( ) da da A b b A b da A A A A da b m b m W ( ) da A W It is the additional condition for the unique solution. Numerical solution method is the same as that for a circular tube. A m 48/56

49 4.6.4 reatment of numerical results* After receiving converged velocity and temperature fields, friction factor and Nusselt number can be obtained as follos:.fre- for laminar problems fre =constant: dp De D Definition of W m e f Re [ dz ]( ) 2 De 2 f Re ( ) m W Wm D 2 dp 2. Nu- Making an energy balance : d 2 D dz b cpm A qp,p is the duct circumference length dz 49/56 2

50 d( b ) i.e., db db DPe ( b) dz b dz dz db ( b ) Substituting in dz DPe db cpm A qp dz Ac pm d Ac b pm yields q ( b) P dz P DPe Nu A yields: q ( ) 2 b PD hd q D e e b b D A e PD 2 ( ) D e Pe 4A P b c D p m D ( e ) 2 4 D 50/56

51 f 2 D Re ( ) W D e 2 D e 2 m Nu= ( ) 4 D Other cases 5/56

52 Home Work ( =50, f =25), 4-4, 4-2, 4-4, 4-8 Due in October 23 52/56

53 Problem 4-2: As shon in Fig. 4-22, in -D steady heat conduction problem, knon conditions are: =50, Lambda=5, S=50, f =25, h=5, the units in every term are consistent. ry to determine the values of 2, 3 ; Prove that the solution meet the overall conservation requirement even though only three nodes are used. Problem 4-4: A large plate ith thickness of 0. m, uniform 3 3 source S=50 0 W/m, 0 W / (m C) ;One of its all is kept at 75 C,hile the other all is cooled by a fluid ith f 25 C and heat 2 transfer coefficient h 50 W/m C. Adopt Practice B, divide the plate thickness into three uniform CVs, determine the inner node temperature. ake 2 nd order accuracy for the inner node, adopt the additional source term method for the right boundary node. 53/56

54 Problem 4-2: Problem 4-4: 54/56

55 Problem 4-8: Shon in Fig.4-25 is a laminar fully developed heat transfer in a duct of half circular cross. ry: () Write the mathematical formulation of the heat transfer problem; (2) Make the formulation dimensionless by introducing some dimensionless parameters; (3) Derive the expressions for fre and Nu from numerical solutions, here the characteristic length for Re and Nu is the equivalent diameter D e. 55/56

56 同舟共济渡彼岸! People in the same boat help each other to cross to the other bank, here. 56/56