Analysis of transient heat conduction in pyinkado (xylia xylocarpa) using finite element solutions

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This paper describes the one-dimensional, transient heat conduction in a rectangular piece of Pyinkado (Xylia xylocarpa) with cross grain and in an orthotropic wooden cylinder. Computerized solutions of a generalized, nonlinear heat equation are derived by discretizing the time domain using finite element techniques.

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Nội dung Text: Analysis of transient heat conduction in pyinkado (xylia xylocarpa) using finite element solutions

AGU International Journal of Sciences – 2019, Vol 7 (4), 91 – 99 ANALYSIS OF TRANSIENT HEAT CONDUCTION IN PYINKADO (Xylia xylocarpa) USING FINITE-ELEMENT SOLUTIONS Bui Thi Thien Kim1, Hoang Thi Thanh Huong1, Ho Xuan Cac2 1 Nong Lam University 2 Associate Technology And Science Sylvicultre Information: ABSTRACT Received: 10/09/2018 This paper describes the one-dimensional, transient heat conduction in a Accepted: 11/02/2019 rectangular piece of Pyinkado (Xylia xylocarpa) with cross grain and in an Published: 11/2019 orthotropic wooden cylinder. Computerized solutions of a generalized, Keywords: nonlinear heat equation are derived by discretizing the time domain using Finite element method, finite element techniques. A simplified example of linear heat conduction distribution of temperature on in cylindrical coordinates illustrates how to apply the finite element wood. solutions. The accuracy of the solutions for this special case is evaluated via comparing them with a well-known exact solution. The results could give significant information for fire-resistant solutions in house architecture and design. 1. INTRODUCTION The knowledge of transient temperature profiles In order to analyze an engineering system, a plays an important role in some wood uses, for mathematical model is developed to describe the instance, for simulating heat conditioning of logs behavior of the system. The mathematical in veneer and plywood mills and temperature expression usually consists of differential response of lumber in dry kilns. Since wood is a equations and given conditions. These hygroscopic and porous medium, heat transfer differential equations are usually very difficult to may occur by means of several modes: solve if handled analytically. The alternative conduction, convection and radiation. However, way to solve the differential equation is using all transfers can be accounted for by using only numerical methods. The numerical method has Fourier's law of heat conduction associated with some advantages, not only can it solve an effective thermal conductivity. complicated equations but also can it reduce the The values of these components can be cost needed to make experiments. Moreover, determined by knowing the values of the numerical methods are able to predict the principal conductivities and of the rotational physical phenomena so it can be studied and angles between the geometric and the implemented to make some devices. orthotropic axes. The likelihood of obtaining an Considering the advantage of the numerical exact solution to the problem of heat conduction method, this paper analyzes physics phenomena in wood is negligible. Therefore, approximated in a heat transfer problem on Pyinkado materials. solutions using numerical methods have been 91 AGU International Journal of Sciences – 2019, Vol 7 (4), 91 – 99 developed for special cases. The purpose of this discretization. In each element, a convenient work is to develop a more general, three- approximate solution is assumed and the dimensional finite difference solution for heat conditions of overall equilibrium of the structure conduction in wood that is shaped as a are derived. The derivation is using several rectangular piece with cross grain or as an mathematical analyses, likewise integration by orthotropic cylinder. The derivation of a one- or parts, weighted residual. Once this analyses two-dimensional solution, then, is straight done, each element will form matrix equation, forward. which resulting in global matrix equations. 2. MATERIALS AND METHODS These global matrix equations solved by MATLAB to get the solution for each node. All 2.1 Materials this processes are the basic procedures for the Specimens of green Pyinkado (Xylia Xylocarpa) finite element method. Systematically, each of (20x40x150mm), which are applied a series of the basic procedures will be discussed separately experimental runs, were taken from Dong Nai, for conduction and convection problem. Vietnam. We use finite element method to determine the 2.2 Methods transient distribution of temperature on wood The basic idea in the finite element method is to (Pyinkado). We research of a temperature- find the solution of a complicated problem by dependent T coefficient with the thermal replacing it by a simpler one. Since the actual conductivity coefficient k as show in Fig. 1 , problem is replaced by a simpler one; the with the t-thickness and z direction (constant), solution is only an approximation rather than the the amount of heat generated in the element is exact one. In the finite element method the Q (W/m3). As the amount of heat transferred to solution domain is divided into small domain the differential volume plus the amount of heat called elements; these elements connected each generated must be equal to the amount of heat other by nodes, therefore this method is called transferred, we have (Nguyễn Lương Dũng finite element. This subdivision process is called (1993) and Phan Anh Vũ (1994))  q   q y  qx dy t + q y dx t + Qdxdy t + Qsti =  qx + x dx dy t +  q y + dy dx t + Qsti+1 Eq.(2.1) [1,2]  x   y  qx q y From equation (2.1), to find out: + −Q = 0 Eq.(2.2) x y To change q x = − k T x ; q y = −k T y , Qsti=Qsti+1 because steady state so we have heat conduction equation:   T    T  k  + k +Q = 0 x  x  y  y  Eq.(2.3)[3,4,5] For the two-dimensional case as shown in Figure 1, heat move along the x y axis. Under constant pressure condition, Luikov (1966) equations 92 AGU International Journal of Sciences – 2019, Vol 7 (4), 91 – 99 Differential equation described two dimensional conductive process (2.3) was special case of general equation for Helmholtz’s conductive process. (O.C. Zienkiewicz (2000);Kwon, Y.W. and Bang, H. (2000);Zienkiewicz, O. C. and Taylor, R.L.(1989) where T is temperature, t time, K thermal conductivity coefficients, respectively. Boundary conditions Equation (2.3) must be solved with certain boundary conditions. We have T = T0 on wallside ST; We will be used triangular elements to solve conductive problem (Figure 1) We expressed heating field in element through: T = T1 N 1 + T2 N 2 + T 3 N 3 Eq. (2.4) [3,4,5] or: T = NTe  T3 y 3 =1 3 2 T(x,y)  T1 1 1 =1 T2 2 Reference element x Figure 1. first clas triangular elements in this problem where: N = 1 −  −    was function and T e = T1 T2 T3  was temperature at dots T which was finding unknown with triangular elements, we have A A qx dy Q y A-A qy dx (b) (a) x Figure 2. (a). 2D conductive problem model (b) conductive volume differential 93 AGU International Journal of Sciences – 2019, Vol 7 (4), 91 – 99 x = N1 x1 + N2 x2+ N3 x3 y = N1 y1 + N2 y2 + N3 y3 Eq. (2.5) [3,4,5] To find out T T x T y T T x T y = + = + Eq. (2.6) [3-5]  x  y   x  y  Or  T   T      x   T  = J  T  Eq. (2.7)[6,7]         y  J was Jacobian matrix which was defined : x y 21  J =  21  x31 y31  Where: xij = xi - xj; yij = yi - yj và det J = 2 Ae , here Ae is the area of the triangle element inverse (1.7), to get:  T   T   x    −1    1  y31 − y 21  − 1 1 0 e  T  = J  T  = − x  T Eq. (2.8)[5,8]     det J  31 x21  − 1 0 1  y     Or :  T   x   T  = BT T e Eq. (2.9)[8]    y  Where : 1  y21 − y31 y31 − y21  1  y23 y31 y12  BT = = det J  x31 − x21 − x31 x21  det J  x32 x13 x21  Eq. (2.10) [6-8] 94 AGU International Journal of Sciences – 2019, Vol 7 (4), 91 – 99 Results:  T  T   x  2  T   T   T 2 eT   +   =   T  = T BT BT T T e  Eq. (2.11) [6-8]  x   y   x y     y  Set up functions As reported, we need to solve equation (2.3) with boundary conditions (i). T = T0 on ST. We solve this equation similar minimum this function : 1   T   2  T  2  =  k   + k   − 2QT tdA Eq. (2.12) [3-8] 2 A   x   y    So that T = T0 on ST. To change equation (2.11) in two first term in first integral of , we will be had : 1   T   T  2  =  k   + k   tdA 2 A   x   y  1 =  T T kT e BT BT T e tdA Eq. (2.13) [3-8] e 2e 1 T 1 T 1 T =  T e ktAe BT BT T e = T e kT T e = T e K T T e T e 2 e 2 2 Where conductive matrix of element was defined by: kT = ktAe BT BT T Eq. (2.14) [6] And conductive matrix of system: K T =  kT Eq. (2.15) [3] e When Q =Qe was contans in element; thick of element t = const.       Qe NtdA T e = − r e − QTtdA = − T T Eq. (2.16) [3]   Q A e  e  e 1 Vì  N dA = 3 A , so that calory vector : e i e 95 AGU International Journal of Sciences – 2019, Vol 7 (4), 91 – 99 rQ = Qe tAe 1 1 1T Eq. (2.17) [6] 3 Final, we have this function (2.12) in the form: 1  = T T KT − R T T Eq. (2.18) [6,7] 2 Where : K =  kT Eq. (2.19) [6-8] R =  (rQ ) Eq. (2.20) [6-8] e To minimum functions  must perform so that 3. RESULTS AND DISCUSSION satisfy with conditions T = T0 on all the note on The objective of this paper is finding the ST. From there, to build problem apply the finite temperature distribution in wood as shown in element method to define heating elements on figure 4. The wood considered natural material, wood following the footsteps such as: inside the wood consist a solid substance. The • Elemental meshing initial temperature inside the wood is considered to 30 . the wood heated on the top sidewall while • Numbering each elements the bottom side is keeps at 30 temperature, • Establish the coordinate system and meanwhile the left and right of the wood is determine the coordinates of the nodes in remain insulated. Wood board with a thermal each element conductivity k = 0.17W / m.K have the boundary • Calculate the thermal conductivity of each conditions. The two opposite sides kept at a element temperature of 1000C = 373 K and below kept at 300C = 303K; Left side must be insulated. + Compute the Jacobi matrix of the element Determine the temperature distribution on the + Calculates the matrix B of the element wood section. The face of wood which was + Calculate thermal conductance ke surveyed to divide 8 elements with 10 dots is • Calculate the thermal conductivity of the illustrated as Figure 4. coefficient K • Solve the equation for determining the temperature vector at the Te node in the system KTe = R 96 AGU International Journal of Sciences – 2019, Vol 7 (4), 91 – 99 T=1000 T=1000 9 10 8 7 8 7 6 5 0,04m 5 6 4 3 4 3 k=0,17W/m. 2 1 K 1 2 0,02m T=300 Figure 4. Conductive problem model Following the diagram with element divided, we conducted to join many elements Table 1: Diagram with element divided Freedom level 1 2 3 Elements 1 1 2 3 2 4 1 3 3 5 4 3 4 5 3 6 5 5 6 7 6 8 5 7 7 9 8 7 8 9 7 10 Following the equation we have: 1  y 23 y31 y12  BT = x det J  32 x13 x21  We calculated: − 50 50 0  2 − 50 0 50  3  0 − 50 50  B1 =   B =  B =   0 − 100 100   100 − 100 0  100 − 100 0  − 50 0 50  5 − 50 50 0  6 − 50 0 50  B4 =   B =  B =   0 − 100 100   0 − 100 100   100 − 100 0   0 − 50 50  − 50 0 50  B7 =  B8 =  100 − 100 0   0 − 100 100  97 AGU International Journal of Sciences – 2019, Vol 7 (4), 91 – 99 Apply the equation kT = kAe BT BT (t=1 unit), we calculated heating matrix of element T  0,0425 − 0,0425 0   0,2125 − 0,17 − 0,0425  kT = − 0,0425  − 0,17  kT =  − 0,17 0  1 2 0,2125 0,17  0 − 0,17 0,17  − 0,0425 0 0,0425   0,17 − 0,17 0   0,0425 0 − 0,0425  kT 3  = − 0,17 0,2125 − 0,0425  kT =  0  4 0,17 − 0,17   0 − 0,0425 0,0425  − 0,0425 − 0,17 0,2125   0,0425 − 0,0425 0   0,2125 − 0,17 − 0,0425  = − 0,0425 − 0,17  kT =  − 0,17  0  5 6 kT 0,2125 0,17  0 − 0,17 0,17  − 0,0425 0 0,0425   0,17 − 0,17 0   0,0425 0 − 0,0425  kT 7  = − 0,17 0,2125 − 0,0425  kT =  0  8 0,17 − 0,17   0 − 0,0425 0,0425  − 0,0425 − 0,17 0,2125  Conductive matrix of system (dimensions 10x10) will be constructed based on join table of this element. Boundary condition T = 1000C at dot 9 and 10, T = 300C at dot 1 and 2 we have matrix Table 2. Matrix KT=R We use Matlab programs to solve matrix (KT = R) with results: T1 = 300C = 303K; T2 = 300C = 303K; T3 = 47,450C = 320,45K ; T4= 47,450C = 320,45K; T5= 64,970C = 337,97K; T6 = 64,970C = 337,97K; T7 = 82,480C = 355,48K; T8 = 82,480C = 355,48K; T9 = 1000C = 373K; T10 = 1000C = 373K. 98 AGU International Journal of Sciences – 2019, Vol 7 (4), 91 – 99 Figure 5. Conductive problem model 4. CONCLUSIONS REFERENCES This paper presents the study and Nguyen Luong Dung, (1993). Finite element implementation of finite element method to find method in mechanics. Ho Chi Minh City the temperature distribution inside wood University of Technology, Vietnam. material (Pyinkado). The temperature Phan Anh Vu (1994), Finite element method in distribution were analyzed by the help of the structural calculations, Ho Chi Minh City contour chart. It was shown that the temperature University of Technology. Ho Chi Minh, parameter governs the conduction on the heated Youth Publishing House. sidewall. As the temperature increase, the Dehghan, M. (2004). On Numerical Solution of temperature in the sidewall becomes increase the One-Dimensional Convection-Diffusion too. The results of the heated sidewall Equation. Mathematical Problems in conduction yield to the boundary condition for Engineering 2005. the convection diffusion inside the wood. It was shown in contour map for each layer inside of O.C. Zienkiewicz (2000), The finite element wood. The temperature distribution inside the method, McGraw - Hill wood was dominated by the temperature Kwon, Y.W. and Bang, H. (2000), The Finite conditions on the heated sidewall. The study will Element Method using MATLAB contribute to predicting the heat transfer process (2nded.)Florida: CRC Press LLC. in the logs, which will be useful for the drying Rannacher, R. (1999). Finite Element Method and thermal processing of wood in the wood for the Incompressible Navier-Stokes processing industry. Equations. Heidelberg: Institute of Applied 5. ACKNOWLEDGEMENTS Mathematics University of Heidelberg. I would like to express my endless thanks and Rao, S. (2005). The Finite Element Method in gratefulness to my supervisors and my students. Engineering (4thed.).Oxford: Elsevier Inc Zienkiewicz, O. C. and Taylor, R.L.(1989), The Finite Element Method, Vol1 Basic Formulation and Linier Problems, 4thEd., McGraw Hill, London 99