fundamentals of heat and mass transfer solutions manual phần 1 docx

fundamentals of heat and mass transfer solutions manual phần 2 docx

... lens:()t,wo232-33 311 1Rm4 0.35 W/m K 10 .2 12 .7 10 12 W/m K4 10 .2 10 mππ−=+−×⋅⋅×()22-33 19 1.2 K/W +13 .2 K/W+246.7 K/W=4 51. 1 K/W6 W/m K4 12 .7 10 mπ+=⋅×With lens:t,w3 311 1R 19 1.2 K/W +13 .2 ... With the insulation,s ,1 2w1 2 i2 33 11 1 11 1 1 TTq4k r r 4k r r4rhπππ∞=+ −+ −+()()4s ,1 2 11 1 1KT 25 C 489W 1. 84 10 4 0.04 0. 51 0.53 W4 0.53 6ππ−=°+ ... insulation()()()s ,1 g222w1 22TT50 25 CEq 11 1 1 111 1 417 W/mK 0.50m0.51m4k r r40.51m6W/mK4rhππππ∞−−°== =−+−+⋅⋅  ()g4225 CE q 489W 1. 84 10 5 .10 10 K / W−−°==...

fundamentals of heat and mass transfer solutions manual phần 4 docx

... 12 6 2 81. 8 299.5 310 .5 317 .9 3 21. 4 323.3 324.2 8 14 4 279.8 296.2 308.6 315 .8 320.4 322.5 9 16 2 276.7 294 .1 306.0 314 .3 319 .0 10 18 0 274.8 2 91. 3 304 .1 312 .4Hence, find( ) ( ) 10 1003T0, 18 0sT275C ... solution?PROBLEM 5 .11 2 (Cont.)() ()()()()()p +1 p +1 p +1 p +1 mmm -1 m +1 p +1 pp +1 mmompTT TTkd1 kd1xxTT q x 1 2h x 1 T T x d 1 ctρ∞−−⋅+⋅∆∆−′′+ ∆⋅ + ∆⋅ − = ∆⋅⋅∆()pp +1 mmT12Fo2FoBiT=+ ... +××()p1 p p pmmm1 m1T 0.029T 0.440 T T 2.3+−+=+ ++ (14 )Nodal Point 1: ()p1 p p p p 11 2 12 T 0.029T 0.440 200 T 2.3 0.029T 0.440T 90.3+=+ ++=++ (15 )Nodal Point 10 :()p1 p p p 10 10 ...

fundamentals of heat and mass transfer solutions manual phần 6 docx

... condition), and the following results were obtained.0.05 0.09 0 .13 0 .17 0. 21 0.25 Mass flowrate, mdot(kg/s) 10 00 14 00 18 00220026003000 Heat rate, q(W) 0.05 0.09 0 .13 0 .17 0. 21 0.25 Mass flowrate, ... from Eqs. 8.20 and 8.22,() 1/ 4 1/ 43Df0. 316 Re0. 316 9 .14 310 0.0323−−==×=2mupfL2Dρ∆=( )( )2320.7740 kg/m200 /10 1.36 m/s2 mp0.0323 71. 1 N/m.20.025 m×∆==×<The heat transfer rate ... Eq. (6) Eq. (7) 1 1000 3.025 × 10 -2348.6 915 .0 2 950 4.3 21 × 10 -2463.7 898. 3 900 5.0 41 × 10 -2524.6 8 91. 0 4 890 5. 215 × 10 -2539.0 889.5Hence, we find2am,om5.2 210 kg/s T890 K.−=×=&<COMMENTS:...

fundamentals of heat and mass transfer solutions manual phần 3 pdf

... 75 10 0 12 5 15 0 17 5 200 225 250 275 3000 18 0.7 18 0.2 17 8.4 17 5.4 17 1 .1 165.3 15 8 .1 149.6 14 0 .1 129.9 11 9.4 10 8.7 98.025 204.2 203.6 2 01. 6 19 8.2 19 3.3 18 6.7 17 8.3 16 8.4 15 7.4 14 5.6 13 3.4 12 1.050 ... isotherms.- 10 0 10 0 10 0 10 0 10 0 -50 86.0 10 5.6 11 9 13 1.7 15 1.6 20050 88.2 11 7.4 13 8.7 15 6 .1 174.6 20050 99.6 13 7 .1 162.5 17 9.2 19 0.8 20050 12 3.0 16 8.9 19 4.9 207.6 209.4 20050 17 3.4 220.7 ... T 1 (°C) T2(°C) T3(°C) T4(°C)0 18 5 18 5 18 5 18 5 ← initial estimate 1 186.3 18 6.6 18 6.6 10 5.82 18 7 .1 187.2 16 7.0 96.03 18 7.4 18 2.3 16 3.3 94.24 18 4.9 18 0.8 16 2.5 93.85 18 4.2 18 0.4 16 2.3...

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fundamentals of heat and mass transfer solutions manual phần 5 ppsx

... flow()( )L 1/ 2 1/ 3 1/ 21/ 35LNu0.664 Re Pr0.66 41. 3 710 0. 712 19.2==×=( )()L2ssk0.02 51 W/mKqNuATT 219 .22m5C55 W.L1m∞⋅=−==o<(b) The Reynolds number for the roof and collector of length ... KD==⋅=⋅06250 01 347 21 6902.., ′=qWm20 438,Fluid: Engine Oil()D625m s 0.01mVDRe 14 7340 10 m sν−== =×()( )()4/5 1/ 2 1/ 35/8D 1/ 42/30.62 14 7 4000 14 7Nu 0.3 1 120282,000 1 0.4 ... general,() 1/ 2 1/ 3xx 1/ 33/40.332Re PrNu1xξ=− (1) ()() () 1/ 3 1/ 2 1/ 2xxx 1/ 3 1/ 33/4 3/40.332k Pr ReReh 0.00832W m Kx1 x x1 xξξ==⋅ −−  .With Unheated...

fundamentals of heat and mass transfer solutions manual phần 7 pot

... 8 0 0033 273 015 15 89 10 22 5 10 308 10 and from Eq. 9.34,()()()22 1/ 65 1/ 6s,iD,io8/27 8/279 /16 9 /16 i0.387 3.08 10 T T0.387Rak 0.0263h 0.60 0.60D0 .15 1 0.559 Pr 1 0.559 0.707∞×−=+ ... )D2 1/ 68/279 /16 0.38 710 , 410 Nu0.604.40 10 .559/0.690=+=+D23k0.0338W/mKhNu4.4 011 .9W/mK.D 12 . 510 m−⋅==×=⋅×The heat rate is()( )23q 11. 9W/mK12. 510 m24020K103W/m.π−′ ... m Kh Nu 21. 4 5.09W m KD 0 .11 5m⋅== ×= ⋅.Substituting numerical values into Eq. (2), find 1 22t 1 1 1 115 1 ln 0.430 W m KR 2 0.051W m K 655.09 W m K 0 .11 5m 11 ,000 W m K 0.055mπππ−=++...

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fundamentals of heat and mass transfer solutions manual phần 8 ppsx

... and Eqs. 11 .22 and 11 .36a yield()c,o c,ih,i c,iTT 1 exp NTUTTε−==−−− (1) From Eq. 11 .1, () ()toochR 11 1 UA hA L hAηη′=++(2)With ()72DiRe 4m/ D 0 .12 kg /s/ 0.05m 19 6.4 10 ... = 4 .16 × 10 4 W/K( 410 K) = 1. 71 × 10 7 W. Hence, ε = (q/qmax) = (1. 254 × 10 7 W /1. 71 × 10 7 W) =0.735. With Cmin mixed and Cmax unmixed, Eq. 11 .35b gives NTU = 1. 54 and ()()422hminhA ... 19 0 W/K and Cmax = Cc = 420 W/K.Hence()()maxminh,ic,iqCTT190W/K10030K13,300W=−=−= and ε=q/qmax = 0.5 71. With Cr = Cmin/Cmax = 0.452 and using Eq. 11 .30b,minrrUA 111 0.5 711 NTUlnln1.00CC1C10.45 210 .5 710 .45 21 εε−−====−−−×−so...

fundamentals of heat and mass transfer solutions manual phần 9 ppt

... ),bb00Gd/G1E ,10 00Kd/E1000Kλλλλααλρλλ∞∞==−∫∫()( )()( ) 11 ,1, 2001F11F.λλλλαρρ→→=−+−−Using Table 12 .1 for λ 1 Tf = 4 × 10 00 = 4000 µm⋅K, F(0-λ) = 0.4 91 giving()()() 10 .20.4 911 0. 810 .4 910 .49.α=−×+−×−=<(b) ... 8244s34 0.53 5.67 10 W m K 13 00K 15 .1W m K 10 00K 0.8 5.67 10 W m K 300 KdT dt8933kg m 385J kg K 0.01m−−×× ⋅× + ⋅× −×× ⋅×=×⋅×  []4sdT dt 1. 16 10 85,829 15 ,10 0 3767 K s 11 .7K s−=× + ... Radiation Tool - Band emission factor:eps = eps1 * FL1T + eps2 * ( 1 - FL1T )/* The blackbody band emission factor, Figure 12 .14 and Table 12 .1, is */FL1T = F_lambda_T(lambda1,T) // Eq 12 .30// where...

fundamentals of heat and mass transfer solutions manual phần 10 doc

... rates,()()b1 1 in cv ,1 1 1 1 11 1EJqq qhATT 1/ Aεε∞−=+= −+−(2) and the radiosity J 1 can be determined from the radiation energy balance, Eq. 13 . 21, ()b1 1 1 312 11 1 11 2 11 3EJ JJJJ 1 / A 1/ A ... A2 and A3,respectively,A 1 :()b1 1 1 312 14 1 1 11 2 11 3 11 4EJ JJJJ JJ 1 /A 1/ A F 1/ A F 1/ A Fε−−−−=++−(3)A2:()b2 1 2 3 21 242 2 2 21 223 224EJ JJJJ JJ 1 /A 1/ A F 1/ A F 1/ A Fε−−−−=++−(4)Continued ... Section 13 .3.5 and Eq. 13 .30, the net radiation leaving theconductor surface A 1 is()( )[]b1 b2 1 3 1 1 11 33 11 3 11 2 332EEq 1 11 AAAF 1/ AF 1/ AFεεεε−−=−−++++ (1) where 4b1 1 ETσ=...

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