Heat Conduction Basic Research Part 1 potx

... to calculate q k , we use another formula (Beck et al, 19 85, Kurpisz &Nowak, 19 95), namely   *1 1, 1 1, 1 1, 1 * 1 2 1 1, 1 1 R kr kr kr kr r kk R kr kr r UTZ qq Z            ... Heat Conduction – Basic Research, Edited by Vyacheslav S. Vikhrenko p. cm. ISBN 978-953-307-404-7 Part 1 Inverse Heat Conduction Problems H...

Ngày tải lên: 21/06/2014, 02:20

... (10 ),  // jj Tx q xk and   /1 jcon Tx T  , j = 1, 2,…,J , (Beck et al., 19 85). Solving the system of equations (12 ) for the unknown heat flux gives 22 2 2 11 1 1 2 222 2 11 1 JJ ... q k , we use another formula (Beck et al, 19 85, Kurpisz &Nowak, 19 95), namely   *1 1, 1 1, 1 1, 1 * 1 2 1 1, 1 1 R kr kr kr kr r kk R kr kr r UTZ qq Z...

Ngày tải lên: 18/06/2014, 22:20

... 2.2(i) λ b 1 =  1 0 b(x)[ψ  1, b (x)] 2 dx  1 0 [ψ 1, b (x)] 2 dx >  1 0 a(x)[ψ  1, b (x)] 2 dx  1 0 [ψ 1, b (x)] 2 dx ≥ inf ψ∈H 1 0 (0 ,1)  1 0 a(x)[ψ  (x)] 2 dx  1 0 [ψ(x)] 2 dx = λ a 1 provided ... x 1 ∈ [0, 1] . Indeed, suppose that 0 ≤ x a 1 < x b 1 ≤ 1, that is a (x)=  a 1 ,0< x < x a 1 a 2 , x a 1 < x < 1 and b (x)= ...

Ngày tải lên: 18/06/2014, 22:20

... (3 .18 ) Heat Conduction – Basic Research 11 4 and it follows from (2.7) and (2.8), that 11 1 21 2 12 1 2 ''' , '' ' (1) ( 21) ( 2) '' (2 1) ( 1) iii m i i ii ...     (33) Heat Conduction – Basic Research 12 8 Case 4. 2 10 10 01 110 0 1 1, , 0, , , , 2 2 ba n aaaa bbb k c a a na        ...

Ngày tải lên: 18/06/2014, 22:20

... of Heat and Mass Transfer, Vol.46, No .1, (January 2003), pp. 10 1 -11 1, ISSN 0 017 -9 310 Ciałkowski, M. J. (20 01) , New type of basic functions of FEM in application to solution of inverse heat conduction ... problem in a square (Hon & Wei, 2005):     12 1 2 ,:0 1, 0 1xx x x   ,     12 1 2 ,: 1, 0 1 D Sxxx x,     12 1 2 ,:0 1, 1 N S...

Ngày tải lên: 18/06/2014, 22:20

... No. 3 , pp .19 7- 213 . Heat Conduction – Basic Research 54 Function Specification Method GA PSO RPSO CRPSO FMLP RBFN Solution Time (s) 14 06 8430 618 9 5907 613 6 73 21 2 316 Table 1. Comparison ... inverse heat conduction analysis. C 0 0.7 0.8 0.95 1. 1 1. 2 PSO RPSO CRPSO 8 .10 5e+4 7.532e+4 7.079e+4 6.823e+4 6.257e+4 7.577e+4 7.064e+4 6.685e+4 6.346e+4...

Ngày tải lên: 18/06/2014, 22:20

... Division, Vol. 11 1 pp. 11 3 -11 9. Heat Conduction – Basic Research 10 2 Q w [kW/m²] Time [s] 0 20 40 60 80 10 0 12 0 14 0 16 0 18 0 200 0 20 40 60 80 10 0 12 0 r/R = 0 r/R = 0 .11 8 r/R = 0.2352 r/R ... h 0 0 0 .1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.2 0.4 0.6 0.8 1 17 g/s 15 g/s 12 g/s 10 g/s r/R 0 0 .1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.2 0.4 0.6...

Ngày tải lên: 18/06/2014, 22:20

... 0,2570 17 9,9 0,2600 18 2,0 0, 312 4 218 ,6 0,29 91 209,4 4,59 0 ,17 01 119 ,1 0 ,17 20 12 0,4 0, 210 9 14 7,6 0,2 019 14 1,3 5,52 0 ,10 23 71, 6 0 ,10 37 72,4 0 ,12 92 90,4 0 ,12 37 86,6 6,49 0,0468 32,8 0,0473 33 ,1 0,0602 ... Kirchhoff’s variable can be given as Heat Conduction – Basic Research 14 2 1 1 , n bn          , (53)     11 1 () (...

Ngày tải lên: 18/06/2014, 22:20

... (59)] 1 (2π) n 1  d n 1 pe ipr si n [(p 2 −m 2 + i0) 1 2 x 0 ] (p 2 −m 2 + i0) 1 2 = 1 (2π) n 1 2 1 x n 1 2 1  ∞ 0 p n 1 2 sin(p 2 −m 2 ) 1 2 x 0 (p 2 −m 2 ) 1 2 J n 1 2 1 (xp) dp, (67) where p =  p 2 1 + ... can be identified as the reheating process in Linde’s cosmology model. Time is in arbitrary units. 0. 01 0.02 0.03 0.04 0.05 t 2 10 12 4 10 12 6 10 12...

Ngày tải lên: 18/06/2014, 22:20

... (23). 0 50 10 0 15 0 200 0.96 0.98 1. 00 1. 02 1. 04 1. 06 1. 08 1. 10 1. 12 1. 14 1. 16 Normaliz ed Signal Am plitude Modulation Frequency (Hz) 80 0 1 000 12 00 14 0 0 16 00 18 00 0. 99993 0. 99994 0. 99995 0. ... value T 1 = -19 6 0 C (e.g. liquid 012 34567 37 38 39 40 41 42 43 44 Diamond woodPVC Glass Pb K Ni Co Cu T C ( 0 C )  (x 10 4 J m -2 K -1 s -1/...

Ngày tải lên: 18/06/2014, 22:20