fundamentals of heat and mass transfer solutions manual phần 3 pdf

... 276.9 271.6 2 63. 3 250.5 232 .8 2 13. 5 125 31 0.9 31 0.5 30 9 .3 307.1 30 3.2 296.0 282.2 257.5 150 34 0.0 34 0.0 33 9.6 33 9.1 33 7.9 33 5 .3 324.7 Agreement between the temperature fields for the (a) and (b) grids ... ) 2 1 T 237 4.6 439 8.062.5K390.2K 2 =−×+×−= < Node 3 (to find T 3 ): 2 c2B3 TTT300K4Tqx/k0 +++−+∆= & ( ) 3 1 T348. 539 0. 237 4. 630 062.5K369.0K...

Ngày tải lên: 08/08/2014, 17:20

... C 0.8 0. 033 m 5.67 10 W/m K 31 3 2 93 K − =⋅ +× ×× ⋅ −  q 2.64 W 3. 33 W 5.97 W =+= < The heat loss is much less than the electrical power, and the assumption of negligible heat loss ... ) () 2 224444 ss 8 7 833 W 6 0 .30 m 200W / m K T 30 3K 0.8 5.67 10 W / m K T 30 3 K − =⋅−+××⋅−    A trial -and- error solution yields s T 37 3K 100 C≈=° <...

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... that () ()() () ( ) s,i s,o 44 3os,o ,o s,o sur 21 32 st ins 2T T 2r h T T T T nr/r nr/r kk π πεσ ∞ −  =−+−   +  () () () () () 2824444 3 3 284 832 3K 2 r 6 W / m K 32 3 30 0 K 0.20 5.67 10 W / m K 32 3 30 0 K nr/0.18n ... . ∞  =− −  (4) From Eqs. (1) and (3) , 3 1 i Cqr/3k.=−  From Eqs. (1), (2) and (4) 33 ii 2 2 o o 33 ii 2 2 o o qr qr h CT k3rk 3kr qr qr CT...

Ngày tải lên: 08/08/2014, 17:20

... 32 4.5 32 5 32 5 32 5 32 5 32 5 3 54 294.7 31 3.7 32 0 .3 324.5 32 5 32 5 32 5 32 5 4 72 2 93. 0 30 7.8 31 8.9 32 2.5 32 4.5 32 5 32 5 32 5 5 90 287.6 30 5.8 31 5.2 32 1.5 32 3.5 32 4.5 32 5 32 5 6 108 285.6 30 1.6 31 3.5 31 9 .3 ... (2) and (4). The results are tabulated below. p t(s) T 0 T 1 T 2 T 3 T 4 T 5 T 6 T 7 (K) 0 0 32 5 32 5 32 5...

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... general, () 1/2 1 /3 x x 1 /3 3/4 0 .33 2Re Pr Nu 1x ξ =  −   (1) ( ) () () 1 /3 1/2 1/2 x x x 1 /3 1 /3 3/4 3/ 4 0 .33 2k Pr Re Re h 0.00 832 W m K x1 x x1 x ξξ ==⋅   −−     . With Unheated Starting ... ) 3 2 1/2 1 /3 3/4 xdx q27.50 27.501.41 739 W 12.0/x − =∫=×=  −   < COMMENTS: Values of h with and without the unheated starting length are 3. 9 an...

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... from Eqs. 8.20 and 8.22, ( ) 1/4 1/ 43 D f0 .31 6Re0 .31 69.1 431 00. 032 3 − − ==×= 2 m u pfL 2D ρ ∆= ( )( ) 2 3 2 0.7740 kg/m200/101 .36 m/s2 m p0. 032 371.1 N/m. 20.025 m × ∆== × < The heat transfer rate ... RePr0.0 231 .33 3100.65 439 . 83 ==×= 2 hNuk/D39. 830 .30 4 W/mK/0.02 m605 W/mK. =⋅=×⋅=⋅ Hence, the surface temperature is ( ) 2 s -3 s 0.020 m0.780 m605 W/mK T1000 K ex...

Ngày tải lên: 08/08/2014, 17:20

... % 32 3 3 62 62 5 9 8 0 0 033 2 73 015 1589 10 22 5 10 30 8 10 and from Eq. 9 .34 , () () () 2 2 1/6 5 1/6 s,i D,i o 8/27 8/27 9/16 9/16 i 0 .38 7 3. 08 10 T T 0 .38 7Ra k 0.02 63 h 0.60 0.60 D0.15 1 0.559 ... correlation ( ) ( ) 4/50 .3 4/50 .3 D D Nu0.023RePr0.0 231 0,9062.9954 .3 === 2 D i Nuk54 .30 .653W/mK h1418W/mK. D0.025m ×⋅ ===⋅ To estimate o h, find ( ) ( ) ( ) ( ) 3...

Ngày tải lên: 08/08/2014, 17:20

... K 2 ⋅ "' 015 1787 38 .4 37 .6 0.20 2250 38 .4 37 .8 0.25 2690 38 .4 37 .9 0 .30 31 12 38 .4 37 .9 Note that while h i varies nearly 50%, there is a negligible effect on the value of U. COMMENTS: Note ... Fig. 11.11 with ( ) ( ) ( ) ( ) P9 535 /30 035 0.23R300196/12 035 1.22=−−==−−= giving F ≈ 1. Based upon CF arrangement, ( ) ( ) ( ) ( ) m,CF T3009519 635 C/n30095/1...

Ngày tải lên: 08/08/2014, 17:20

... V m W 0.2 732 0. 039 3 600K 0. 039 3 0.0010 1 36 0K µ εσ ε σ =⋅ − + − − where for λ 2 T sur = 5 µ m × 36 0 K = 1800 µ m ⋅ K, () 2sur 0T F λ − = 0. 039 3 and λ 1 T sur = 3 µ m × 36 0 K = 1080 µ m ⋅ K, ... values of ε and α from part (a), the rate of temperature change with time is () () 824 2 8244 s 3 4 0. 53 5.67 10 W m K 130 0K 15.1W m K 1000K 0.8 5.67 10 W m...

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... PROBLEM 13. 89 (Cont.) A 3 : () b3 3 3 1 3 2 3 4 3 3 33 1 33 2 33 4 E J JJ JJ JJ 1 /A 1/A F 1/A F 1/A F ε −−−− =++ − (5) Recognize that in the above equation set, there are three equations and three ... of heat transfer to tube without ash deposit, (b) Rate of heat transfer with an ash deposit of diameter D d = 0.06 m, (c) Effect of deposit diameter and conve...

Ngày tải lên: 08/08/2014, 17:20