Method of calculating the dimensions of greenhouse-type single slope watermaker by taking into account the accumulation of solar energy parnik tipidagi bir nishabli suv chuchutgichi o
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Statement of the problemConsider the static problem of thermoelasticity for isotropic bodies. It consists of the equilibrium equations [2] 3 ij j1 xi X i 0, i 1,3 (1) the constitutive relation of Dughamel-Neumann representing the relationship between stresses and strains taking into account temperature ij ij 2 ij 3 2 (T T0)ij (2) Cauchy relations ui u j ij ( ) (3) xj xi and boundary conditions ui 1 uio , ijnj 2 Sio (4) j1 where, corresponds to thermal expansion coefficient, ij stress tensor,ij strain tensor, ui displacement, Ttemperature, T0 initial temperature Xi volume force, , Lame constants, spherical part of strain tensor, nj - outward normal to surface 2 , S S S1, 2 , 3 - external load vector components, ij delta Kronecker symbol and (3 2 ). The boundary value problem of thermoelasticity (1–4) can be reduced to a system of three differential equations for displacements, i.e. 2u 2u 2u 2v 2w T (2)x2 y2 z2 ()xy xz x 0 2v 2v 2v 2u 2w T x2 z2 (2) y2 ()xy yz y 0 (5) x2w2 y2w2 ()x2uz y2vz (2)z2w2 Tz 0. Differential equations (5) can be considered with boundary conditions in displacements, stresses or mixed type. In our case, we consider a parallelepiped with a free boundary from loads, and the boundary conditions take the following form u v w 11 x0,l1 ( 2 ) x y z (T T0) x0,l1 0 12x o l ,1 uy xv x o l ,1 0 (6) 13x o l ,1 u w 0 z x x o l ,1 22 y0,l2 ux ( 2 ) yv wz (T T0) y0,l2 0 v u 21y o l ,2 x y y o l ,2 0 (7) v w 23y o l ,2 z y y o l ,2 0 33 z0,l3 ux yv ( 2 ) wz (T T0) z0,l3 0 31z o l ,3 wx uz z o l ,3 0 . (8) w v 32z o l ,3 y z z o l ,3 0 Fig. 1. where li , i 1,3–are the lengths of the edges of the considered parallelepiped. Numerical implementationReplacing the derivatives in the boundary value problem (5–8) with difference relations, one can find the basic difference equations [3] ui1, ,j k 2ui j k, , ui1, ,j k ui j, 1, k 2ui j k, , ui j, 1, k ui j k, , 1 2ui j k, , ui j k, , 1 ( 2 ) 2 h1 vi 1, 1,j k vi 1, 1,j k vi 1, 1,j k vi 1, 1,j k ( ) 4hh1 2 Ti1, ,j k Ti1, ,j k 0 2h1 2 2 h2 h3 wi1, ,j k1 wi1, ,j k1 wi1, ,j k1 wi1, ,j k1 (9) 4h1 3h vi j k, , 1 2vi j k, , vi j k, , 1 2 h3 w w vi j, 1, k 2vi j k, , vi j, 1, k ( 2 ) vi1, ,j k 2vi j k, , vi1, ,j k 2 2 h2 h1 u u u u w w i 1, 1,j k i 1, 1,j k i 1, 1,j k i 1, 1,j k 4hh1 2 Ti j, 1, k Ti j, 1, k 0 2h2 i j, 1, 1 k i j, 1, 1 k i j, 1, 1 k i j, 1, 1 k (10) 4h h2 3 wi j k, , 1 2wi j k, , wi j k, , 1 wi1, ,j k 2wi j k, , wi1, ,j k wi j, 1, k 2wi j k, , wi j, 1, k ( 2 ) 2 2 2 h3 h1 h2 ui1, ,j k1 ui1, ,j k1 ui1, ,j k1 ui1, ,j k1 vi j, 1, 1 k vi j, 1, 1 k vi j, 1, 1 k vi j, 1, 1 k (11) 4hh1 3 4hh1 2 Ti j k, , 1 Ti j k, , 1 0 2h3 boundary conditions (6) ( 2 ) u N( 1, , )j k u N( 1 1, ,j k) v N( 1, j 1,k)v N( 1, j 1,k) h1 2h 2 w N( 1, ,j k 1) w N( 1, ,j k 1) (TN1, j T0 ) 0 2h3 u N( 1, j 1,k)u N( 1, j 1,k) v N( 1, , )j k v N( 1 1, ,j k) (12) 0 2h2 h1 u N( 1, ,j k 1)u N( 1, ,j k 1) w N( 1, , )j k w N( 1 1, ,j k) 0 2h3 h1 The finite-difference equations for the boundary conditions (7) and (8) can be written in a similar way. Solving equations (9–12) with respect to ui j k, , , vi j k, , , wi j k, , we can construct the following iterative process (n1) ui(n1), ,j k ui(n1), ,j k ui j(,n)1,k ui j(,n)1,k ui j k(, ,n) 1 ui j k(, ,n) 1 ui j k, , ( 2 ) 2 2 2 h1 h2 h3
i 1, j 1,k i 1, j 1,k i 1, j 1,k i 1, j 1,k i j, 1,k 1 i j, 1,k 1 i j, 1,k 1 i j, 1,k 1 i1, ,j k1 i1, ,j k1 i1, ,j k1 i1, ,j k1 i j, 1,k 1 i j, 1,k 1 i j, 1,k1 i j, 1,k 1 4hh1 3 4hh1 2 (13) Ti j k, , 12hTi j k, , 1 / 2( 2 2 ) 2 2 2 2 h3 h1 h3
3 h v(n) (N , j 1,k)v(n) (N , j 1,k) (n1) (n) 1 1 1 (n1) (n) h1 (n) (n) w (N1, , )j k w (N1 1, ,j k) 2h3 u (N1, ,j k 1)u (N1, ,j k 1) Equations (14) are written for a face of a parallelepiped at x l1 . For the rest of the faces, the corresponding relations can be found in a similar way. Download 2.25 Mb. Do'stlaringiz bilan baham: |
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