fundamentals of heat and mass transfer solutions manual phần 3 pdf

... 276.9 271.6 2 63. 3 250.5 232 .8 2 13. 5125 31 0.9 31 0.5 30 9 .3 307.1 30 3.2 296.0 282.2 257.5150 34 0.0 34 0.0 33 9.6 33 9.1 33 7.9 33 5 .3 324.7Agreement between the temperature fields for the (a) and (b) grids ... )21T 237 4.6 439 8.062.5K390.2K2=−×+×−= <Node 3 (to find T 3 ):2c2B3TTT300K4Tqx/k0+++−+∆=&( ) 3 1T348. 539 0. 237 4. 630 062.5K369.0K4=++++=<Node 1 (to find T1):2C21 30 02TT4Tqx/k0++−+∆=&( ... where x and y arein mm and the temperatures are in °C.y\x 0 15 30 45 600 50 80 .33 85.16 80 .33 5015 50 63. 58 67. 73 63. 58 50 30 50 56.27 58.58 56.27 5045 50 52.91 54.07 52.91 5060 50 51 .32 51.86...

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fundamentals of heat and mass transfer solutions manual phần 2 docx

... that()()()()()s,i s,o443os,o ,o s,o sur21 32 st ins2T T2r h T T T Tnr/r nr/rkkππεσ∞−=−+−+()()()()()2824444 3 3284 832 3K2 r 6 W / m K 32 3 30 0 K 0.20 5.67 10 W / m K 32 3 30 0 Knr/0.18n ... .∞=− −(4)From Eqs. (1) and (3) , 3 1iCqr/3k.=− From Eqs. (1), (2) and (4) 33 ii22oo 33 ii22ooqr qrh CTk3rk3krqr qrCT.3r k3hr∞∞−=− +−=−+Hence, ... lens:()t,wo 23 2 -3 331 11Rm4 0 .35 W/m K 10.2 12.71012W/m K4 10.2 10 mππ−=+−×⋅⋅×()22 -3 3191.2 K/W+ 13. 2 K/W+246.7 K/W=451.1 K/W6 W/m K4 12.7 10 mπ+=⋅×With lens:t,w 3 3111R...

fundamentals of heat and mass transfer solutions manual phần 4 docx

... 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 aretabulated below.p t(s) T0 T1 T2 T 3 T4 T5 T6 T7(K) 0 0 32 5 32 5 32 5 32 5 32 5 32 5 32 5 32 5 1 18 30 4.2 32 4.7 32 5 32 5 32 5 32 5 32 5 32 5 2 36 30 3.2 31 5 .3 324.5 ... 31 3.5 31 9 .3 322.7 32 4.0 32 4.5 32 5 7 126 281.8 299.5 31 0.5 31 7.9 32 1.4 32 3 .3 324.2 8 144 279.8 296.2 30 8.6 31 5.8 32 0.4 32 2.5 9 162 276.7 294.1 30 6.0 31 4 .3 319.010 180 274.8 291 .3 304.1 31 2.4Hence,...

fundamentals of heat and mass transfer solutions manual phần 5 ppsx

... general,()1/2 1 /3 xx1 /3 3/40 .33 2Re PrNu1xξ=−(1)()() ()1 /3 1/21/2xxx1 /3 1 /3 3/4 3/ 40 .33 2k Pr ReReh 0.00 832 W m Kx1 x x1 xξξ==⋅ −−  .With Unheated Starting ... ) 3 21/21 /3 3/4xdxq27.50 27.501.41 739 W12.0/x−=∫=×=−<COMMENTS: Values of h with and without the unheated starting length are 3. 9 and 5.5 W/m2⋅K.Prior development of ... 80°C:T (K) 2 93 300 32 5 35 0 35 3Pr 7.00 5. 83 3.42 2.29 2.20ANALYSIS: Using the Zhukauskus correlation for the range 40 < ReD < 4000 with C = 0.51 and m =0.5,D1/40.50 .37 DshDPrNu0.51RePr.kPr==with...

fundamentals of heat and mass transfer solutions manual phần 6 docx

... from Eqs. 8.20 and 8.22,()1/41/ 43 Df0 .31 6Re0 .31 69.1 431 00. 032 3−−==×=2mupfL2Dρ∆=( )( )2 3 20.7740 kg/m200/101 .36 m/s2 mp0. 032 371.1 N/m.20.025 m×∆==×<The heat transfer rate ... RePr0.0 231 .33 3100.65 439 . 83 ==×=2hNuk/D39. 830 .30 4 W/mK/0.02 m605 W/mK.=⋅=×⋅=⋅Hence, the surface temperature is( )2s -3 s0.020 m0.780 m605 W/mKT1000 Kexp0.4898T600 K810 kg/s51 93 J/kgKπ××⋅−=−=−××⋅sT 138 4 ... temperature, volume and properties of pharmaceutical. Heat transfer coefficient at outer surface and dimensions of coil.FIND: (a) Expressions for Tc(t) and Th,o(t), (b) Plots of Tc(t) and Th,o(t)...

fundamentals of heat and mass transfer solutions manual phần 7 pot

... % 32 3 362 6259 8 0 0 033 2 73 0151589 10 22 5 10 30 8 10 and from Eq. 9 .34 ,()()()221/651/6s,iD,io8/27 8/279/16 9/16i0 .38 7 3. 08 10 T T0 .38 7Rak 0.02 63 h 0.60 0.60D0.151 0.559 ... correlation( ) ( )4/50 .3 4/50 .3 DDNu0.023RePr0.0 231 0,9062.9954 .3 ===2DiNuk54 .30 .653W/mKh1418W/mK.D0.025m×⋅===⋅To estimate oh, find( )()( ) ( ) 3 241 3 sD62829.8m/s810K5527.4K0.025mgTTDRa510m/s8.8510m/sβνα−−∞−−×−−==×××6DRa7.6410=× and ... correlation of Eq. 9 .33 ,Continued … PROBLEM 9.80 (Cont.)Fluid()sTC°Tf(K) RaDNuDDh(W/m2⋅K)Air 57.5 31 4 3. 72×1048 .31 9.09Water 57.5 31 41.99x107 37 .1 940Numerical integration of...

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fundamentals of heat and mass transfer solutions manual phần 8 ppsx

... K2⋅"' 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 .9Note that while hi 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,CFT3009519 635 C/n30095/196. 235 182K.∆=−−−°−−=llUsing ... Heat Exchangers and Properties Toolpads to perform the parametriccalculations, we obtain the following results for NL = 90.20 24 28 32 36 40Gas flowrate, mdoth (kg/s) 30 0 31 0 32 0 33 0 34 0 35 0Water...

fundamentals of heat and mass transfer solutions manual phần 9 ppt

... V m W 0.2 732 0. 039 3 600K 0. 039 3 0.0010 1 36 0Kµεσ ε σ=⋅ − + − −where for λ2Tsur = 5 µm × 36 0 K = 1800 µm⋅K,()2sur0TFλ− = 0. 039 3 and λ1Tsur = 3 µm × 36 0 K = 1080µm⋅K, ... values of ε and α from part (a), the rate of temperature change with time is() ()824 2 8244s 3 4 0. 53 5.67 10 W m K 130 0K 15.1W m K 1000K 0.8 5.67 10 W m K 30 0 KdT dt8 933 kg m 38 5J kg ... and diffuse, (2) Spectral emissivity is independent of temperature, and (3) Furnace and cooling environment are large isothermal surroundings.ANALYSIS: (a) Using Eqs. 12 .38 and 12.46, ε and...

fundamentals of heat and mass transfer solutions manual phần 10 doc

... PROBLEM 13. 89 (Cont.)A 3 :()b3 3 3 1 3 2 3 4 3 3 33 1 33 2 33 4E J JJ JJ JJ1 /A 1/A F 1/A F 1/A Fε−−−−=++−(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 Dd = 0.06 m, (c) Effect of deposit diameter and convection coefficient on heat rate and contributions ... surface re-radiating (2). From Section 13. 3.5 and Eq. 13. 30, the net radiation leaving theconductor surface A1 is()( )[]b1 b21 3 1111 33 1 13 112 33 2EEq111AAAF 1/AF 1/AFεεεε−−=−−++++(1)where...

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