fundamentals of heat and mass transfer solutions manual phần 2 docx

... radqqq′′′=+() ()()s,1 s ,2 s ,2 s ,2 sur 21 2 2rTT TTTTln r r 2 k 1 2 r h 1 2 r hππ π∞−−−=+where ()() 22 r s ,2 sur s ,2 surhTTTTεσ=+ +. Solving this equation for Ts ,2 , the heat rate may bedetermined ... 0−=Hence,()()()()[]s ,2 s,1 2 2s ,2 12 TTh4r T T1r 1r 4 kππ∞−−=−()()[]()()s ,2 s,1 2 21 2 s ,2 TTkr1r 1rhTT∞−−=−()()()()s ,2 s,1 22 2 11 1s ,2 TTr r k 1.5W m K 301r ... circuits.Without lens:()t,wo 23 2- 333111Rm4 0.35 W/m K 10 .2 12. 71012W/m K4 10 .2 10 mππ−=+−×⋅⋅×() 2 2-33191 .2 K/W+13 .2 K/W +24 6.7 K/W=451.1 K/W6 W/m K4 12. 7 10 mπ+=⋅×With lens:t,w33111R...

fundamentals of heat and mass transfer solutions manual phần 4 docx

... 315.3 324 .5 325 325 325 325 325 3 54 29 4.7 313.7 320 .3 324 .5 325 325 325 325 4 72 293.0 307.8 318.9 322 .5 324 .5 325 325 325 5 90 28 7.6 305.8 315 .2 321 .5 323 .5 324 .5 325 325 6 108 28 5.6 301.6 ... 319.3 322 .7 324 .0 324 .5 325 7 126 28 1.8 29 9.5 310.5 317.9 321 .4 323 .3 324 .2 8 144 27 9.8 29 6 .2 308.6 315.8 320 .4 322 .5 9 1 62 276.7 29 4.1 306.0 314.3 319.010 180 27 4.8 29 1.3 304.1 3 12. 4Hence, ... Eqs. (2) and (4). The results aretabulated below.p t(s) T0 T1 T 2 T3 T4 T5 T6 T7(K) 0 0 325 325 325 325 325 325 325 325 1 18 304 .2 324 .7 325 325 325 325 325 325 2 36 303 .2 315.3...

fundamentals of heat and mass transfer solutions manual phần 6 docx

... from Eqs. 8 .20 and 8 .22 ,()1/41/43Df0.316Re0.3169.143100.0 323 −−==×= 2 mupfL2Dρ∆=( )( ) 2 3 2 0.7740 kg/m200/101.36 m/s2 mp0.0 323 71.1 N/m. 20 . 025 m×∆==×<The heat transfer rate ... m) = 26 07 W/m 2 ⋅K. From Eqs. (2) and (3) the outlet mean and surface temperatures are()() 2 42 m,o0.012m 20 10 W mT 29 0K 29 0.7K 17.7 C0.01kg s 4184J kg K×=+ = =⋅ 42 s,o m,o 2 220 10WmT ... obtained.0.05 0.09 0.13 0.17 0 .21 0 .25 Mass flowrate, mdot(kg/s)100014001800 22 00 26 003000 Heat rate, q(W) 0.05 0.09 0.13 0.17 0 .21 0 .25 Mass flowrate, mdot(kg/s) 25 27 29 313335Outlet temperature,...

fundamentals of heat and mass transfer solutions manual phần 3 pdf

... 22 8.9 22 8.3 22 6 .2 222 .6 21 7 .2 209.7 20 0.1 188.4 175.4 161.6 147.575 25 5.0 25 4.4 25 2.4 24 8.7 24 3.1 23 5.0 22 3.9 20 9.8 194.1 177.8100 28 2.4 28 1.8 28 0.1 27 6.9 27 1.6 26 3.3 25 0.5 23 2.8 21 3.5 125 310.9 ... 0 600 29 2 .2 2 4 0 1 0 0 0 0 300 28 9 .2 1 0 4 1 1 0 0 0 300 27 9.70 1 2 4 0 1 0 0 0 27 2 .2 ACT0 0 1 0 4 1 1 0 300 25 4.50 0 0 1 2 4 0 1 0 24 0.10 0 0 0 2 0 6 1 500 198.10 0 0 0 0 2 2 6 20 0 179.4−−−−−−−===−−−−−−−From ... 125 150 175 20 0 22 5 25 0 27 5 3000 180.7 180 .2 178.4 175.4 171.1 165.3 158.1 149.6 140.1 129 .9 119.4 108.7 98.0 25 20 4 .2 203.6 20 1.6 198 .2 193.3 186.7 178.3 168.4 157.4 145.6 133.4 121 .050 22 8.9...

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1,261
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fundamentals of heat and mass transfer solutions manual phần 5 ppsx

... convection and ()() ()1 /2 1 /2 3 22 bM h Dk D 4 2 235W m K 0.002m 399W m K 50 K 2. 15Wππ θ π==⋅ ⋅=()()1 /2 1 /2 21cm hP kA 4 23 5W m K 399 W m K 0.002m 34.3m−==×⋅⋅×=()1mL 34.3m 0.012m ... W/m⋅K, ν = µvf = 3 52 ×10-6 N⋅s/m 2 × 1. 029 × 10-3 m3/kg = 3. 621 × 10-7 m 2 /s; Prs values for 20 ≤ Ts ≤ 80°C:T (K) 29 3 300 325 350 353Pr 7.00 5.83 3. 42 2 .29 2. 20ANALYSIS: Using ... Pr0.6641.37100.7 121 9 .2 ==×=( )()L 2 ssk0. 025 1 W/mKqNuATT219 .22 m5C55 W.L1m∞⋅=−==o<(b) The Reynolds number for the roof and collector of length L = 3m is5Lx,c- 62 2 m/s3mRe4.1110Re.14.610...

fundamentals of heat and mass transfer solutions manual phần 7 pot

... )( ) 22 22 300 K 1000 300 K 32. 1W m K++=⋅<(c) For the long stainless steel rod and the initial values of h and hr,()() 2 roBi h h r 2 k 42. 0W m K 0.0 125 m 25 W m K 0. 021 =+ ... m/ (20 .22 × 10-6 m 2 /s) = 2. 23 × 104, the internal flow is turbulent, and from the Dittus-Boelter correlation,()()4/50.34/5 0.3 4 2 iD,iik 0. 029 5W m Kh 0. 023 Re Pr 0. 023 2. 23 ... )D1/451/4D4/94/99/169/160.58 92. 93100.589RaNu 221 2.5510.469/Pr10.469/0.700×=+=+=++( )D 2 hNuk/D 12. 550. 029 8W/mK/0.040m9.36W/mK.==⋅=⋅Substituting numerical values, the heat loss from the bulb is,( ) ( ) ( ) ( )84 24 4 22 4WWq0.040m9.36 125 25K0.805.6710 125 27 325 273KmKmKπ−=−+××+−+⋅⋅()q4.703.92W8.62W.=+=<COMMENTS:...

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fundamentals of heat and mass transfer solutions manual phần 8 ppsx

... velocity is um = m/(ρπD 2 /4) so that() 2 22 254m DL2m Lpf f2DDρρπρπ∆= =()() 2 365 25 0.003kg s 1. 128 kg m L 2 p f 2. 320 6 10 f L DDπ−−∆= = ×( 12) Recall that the pressure ... 1UhD 2k h=+ + () 22 0.02mln 20 15 120 115 60W m K3000W m K 27 9 W m K=++⋅⋅⋅()45 32 32 h14.44 10 9.59 10 3.58 10 m K W 4. 12 10 m K WU−−− −=×+×+× ⋅=× ⋅ 2 hU243WmK=⋅.With Ch ... Cmin mixed and Cmax unmixed, Eq. 11.35b gives NTU = 1.54 and ()() 422 hminhA NTU C U 1.54 4.160 10 W K 24 3W m K 26 4m==× ⋅=.Hence,()()() 2 hL 2 oTA 26 4mN70DWN0. 02 2 30 mππ==...

fundamentals of heat and mass transfer solutions manual phần 9 ppt

... 0.005m 25 W m K 60W m K−−=× × ⋅ −+⋅⋅684 22 3. 72 10 24 79T 4.54 10 T 2. 98−×− = × −84 6 22 4.54 10 T 24 79T 3. 72 10−×+=×Solving, we obtainT 2 = 1 425 K<()()() 52 12 h 42 ss cc1500 1 425 ... )( )()( )4 824 4s 2 ss hsr0.95.6710W/mK273K1TGI90.2W/msrsrεσρπππ−×⋅−====⋅( )() 26 258sdq46 .29 0.2W/msr510m2 .22 10sr1.5110W−−−−=+⋅×××=×8 623 2dG1.5110W /21 0m7.5710W/m.−−−=××=× ... M is the mass of PCM, and hfg is the heat of fusion of PCM.Irradiation: G = σTw = 5.67 × 10-8 W/m 2 ⋅K4( 120 0 K)4 = 117,573 W/m 2 Emissive power: Eb = ( )4 42 mT8 727 3952W/mσσ=+=Emissivity:...

fundamentals of heat and mass transfer solutions manual phần 10 doc

... radiation energy balance, Eq. 13 .21 ,()b2 2 2 321 22 2 22 1 22 3EJ JJJJ1/A1/AF1/AFεε−−−=+−(5)There are 4 equations, Eqs. (2- 5), with 4 unknowns: J 2 , J 2 , T 2 and q1. Substituting numerical ... +=−++where ()()()()44 22 12 121 2 12 TT TTTTTT.−=+ + − Accordingly,( ) ()()( )rad,hc 22 2 824 12 210.85 1 0.01R0.01 m 0.0 02 m 5.67 10 W / m K 28 8 K 26 8 K 28 8 26 8 K−−++=−×× ⋅ + +where, ... balances, for A1, A 2 and A3,respectively,A1:()b1 1 1 3 12 141 1 1 12 113 114EJ JJJJ JJ1 /A 1/A F 1/A F 1/A Fε−−−−=++−(3)A 2 :()b2 1 2 321 24 2 2 22 1 22 3 22 4EJ JJJJ JJ1 /A 1/A...

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