Computational Fluid Mechanics and Heat Transfer Third Edition 1 docx

... = ρcV hA = ρc h πD 3 /6 πD 2 = ρcD 6h = (9300)(0 .18 )(0.0 01) 6(250) kg m 3 kJ kg·K m m 2 ·K W 10 00 W kJ/s = 1. 116 s Therefore, eqn. (1. 22) becomes T −200 ◦ C (20 −200) ◦ C = e −t /1. 116 or T = 200 18 0 e −t /1. 116 ◦ C This result is plotted in Fig. 1. 12, where we see that, ... 597 11 .2 Mixture compositions and species fluxes 600 11 .3 Diffusion fluxes and Fick’s law 608 11 ....

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... 17 3 Equation (4. 41) or Fig. 4.9, on the other hand, gives Θ tip = 1 1.3986 = 0. 715 0 so the approximate tip temperature is T tip = 26 + 0. 715 (15 0 − 26) = 11 4.66 ◦ C Thus the insulated-tip approximation ... + tanh mL 1 +(Bi ax /mL) tanh mL (4.49) 17 6 Analysis of heat conduction and some steady one-dimensional problems §4.5 Solution. From Example 4.8 we have mL = 0.8656 and...

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... 1. 14 41 0.9960 0.60 0.70507 1. 0 814 0.9940 1. 018 44 1. 1345 0.9936 1. 26440 1. 1 713 0.9944 0.70 0.75056 1. 0 918 0.9922 1. 08725 1. 1539 0.9 916 1. 35252 1. 1978 0.9925 0.80 0.7 910 3 1. 1 016 0.9903 1. 14897 1. 1724 ... 0.9839 1. 302 51 1.2232 0.9 815 1. 6 319 9 1. 2970 0.9828 1. 20 0. 917 85 1. 1344 0.9 817 1. 34558 1. 2387 0.9787 1. 68868 1. 32 01 0.9800 1. 3...

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... 0.00000 1. 00000 1. 10 0.880 21 0 .11 980 0.05 0.05637 0.94363 1. 20 0. 910 31 0.08969 0 .10 0 .11 246 0.88754 1. 30 0.934 01 0.06599 0 .15 0 .16 800 0.83200 1. 40 0.95229 0.047 71 0.20 0.22270 0.77730 1. 50 0.96 611 ... 0.565, and Bi 1 = Bi 2 = hL  k = 800(0.04/2)/76 = 0. 210 5, and we then write Θ  x L  1 = 0,  x L  2 = 1 2 , Fo 1 , Fo 2 , Bi 1 1 , Bi 1 2  =...

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... 0.00664 0.066 41 0.00332 0.3 319 9 0.40 0.02656 0 .13 277 0. 013 22 0.3 314 7 0.60 0.05974 0 .19 894 0.029 81 0.33008 0.80 0 .10 611 0.264 71 0.05283 0.32739 1. 00 0 .16 557 0.32979 0.08 211 0.323 01 2.00 0.65003 ... D.C., 19 87. [5 .15 ] V. S. Arpaci. Conduction Heat Transfer. Ginn Press/Pearson Cus- tom Publishing, Needham Heights, Mass., 19 91. [5 .16 ] E. Hahne and U. Grigull...

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... T w >T b , n = 0 .11 in eqn. (7.44). Thus, with eqn. (7. 41) we have Nu D = (0. 012 8/8)(5.74 × 10 5 )(2.47) 1. 07 + 12 .7  0. 012 8/8  2.47 2/3 1  (1. 74) 0 .11 = 16 17 or h = Nu D k D = 16 17 0.6 61 0 .12 = 8, ... =  1. 82 log 10 (19 , 011 ) − 1. 64  −2 = 0.02646 and the Nusselt number is found with Gnielinski’s equation, (7.43) Nu D h = (0.02646/8) (19 , 011 − 1...

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... surface 411 and Pr = 0. 711 , where the properties are evaluated at 300 K = 27 ◦ C. Then, from eqn. (8.26), Nu L = 0.678  1. 645 10 8  1/ 4  0. 711 0.952 +0. 711  1/ 4 = 62 .1 so h = 62.1k L = 62 .1( 0.02 614 ) 0.4 = ... 2. 318 10 −5 m 2 /s, and β = 1  (273+20) = 0.003 41 K 1 . Then Ra L = gβ∆TL 3 αν = 9.8(0.003 41) (30)(0.3) 3 (16 .45)(0.2 318 )10 10 = 7 .10 10 7 T...

Ngày tải lên: 21/06/2014, 22:20

... at 27 + (14 0/2) = 97 ◦ C: ∆T corrected = 1. 18 6 61( 0 .17 )/0.0 310 4  9.8 [15 /π(0.085) 2 ]0 .17 4 300(0.0 310 4)(3.2 31) (2.277 )10 10  1/ 6 (0.99) = 14 2 K so the surface temperature is 27 + 14 2 = 16 9 ◦ C. That ... J. Heat Mass Transfer, 15 (4): 755–767, 19 72. [8 .17 ] G. C. Vliet. Natural convection local heat transfer on constant heat transfer inclined surfac...

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... (3, 11 5) (1 −0.2) 0.8  0.6683 (0. 311 0) −0.2 (1) +10 58 (3 .14 7 10 −4 ) 0.7 (1)  = 11 , 950 W/m 2 K h fb    cbd = (3, 11 5) (1 −0.2) 0.8  1. 136 (0. 311 0) −0.9 (1) +667.2 (3 .14 7 10 −4 ) 0.7 (1)  = ... Additional values are given in [9.47]. Fluid F Fluid F Water 1. 0 R -12 4 1. 90 Propane 2 .15 R -12 5 1. 10 R -12 1. 50 R -13 4a 1. 63 R-22 2.20 R -15 2a...

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... using eqns. (11 .1) , (11 .4), and (11 .6) and (11 .5), (11 .4), and (11 .2), respectively, M may be written in terms of either mole or mass fraction M =  i x i M i or 1 M =  i m i M i (11 .8) Mole fraction ... pressures, 1 p =  i p i (11 .15 ) Finally, we combine eqns. (11 .6), (11 .13 ), and (11 .15 ) to obtain a very useful relationship between x i and p i : x i = c i c...

Ngày tải lên: 21/06/2014, 22:20