fundamentals of heat and mass transfer solutions manual phần 2 docx

fundamentals of heat and mass transfer solutions manual phần 2 docx

fundamentals of heat and mass transfer solutions manual phần 2 docx

... rad qqq ′′′ =+ () ()() s,1 s ,2 s ,2 s ,2 sur 21 2 2r TT TTTT ln r r 2 k 1 2 r h 1 2 r h ππ π ∞ −−− =+ where () () 22 r s ,2 sur s ,2 sur hTTTT εσ =+ + . Solving this equation for T s ,2 , the heat rate may be determined ... 0 −= Hence, ( ) () ()() [] s ,2 s,1 2 2s ,2 12 TT h4r T T 1r 1r 4 k π π ∞ − −= − ()() [] () () s ,2 s,1 2 21 2 s ,2 TT k r1r 1r h TT ∞ − −=...

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220 656 4
fundamentals of heat and mass transfer solutions manual phần 1 docx

fundamentals of heat and mass transfer solutions manual phần 1 docx

... sur h4T TTT2 εσ ==+ For the condition of ε = 0.05, T s = T sur + 10 = 35°C = 308 K and T sur = 25 °C = 29 8 K, find that () ( ) 824 22 3 2 r h 0.05 5.67 10 W m K 308 29 8 308 29 8 K 0.32W m K − =×× ... () 2 A DL 0.1m 25 m 7.85m . ππ == ×= Hence, () () 22 824 444 q 7.85m 10 W/m K 150 25 K 0.8 5.67 10 W/m K 423 29 8 K −  =⋅−+××⋅−   ()() 22 q 7.85m...

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220 928 4
fundamentals of heat and mass transfer solutions manual phần 4 docx

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 3...

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fundamentals of heat and mass transfer solutions manual phần 6 docx

fundamentals of heat and mass transfer solutions manual phần 6 docx

... from Eqs. 8 .20 and 8 .22 , ( ) 1/4 1/43 D f0.316Re0.3169.143100.0 323 − − ==×= 2 m u pfL 2D ρ ∆= ( )( ) 2 3 2 0.7740 kg/m200/101.36 m/s2 m p0.0 323 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,o 0.012m 20 10 W m T 29 0K 29 0.7K 17.7 C 0.01kg s 4184J...

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fundamentals of heat and mass transfer solutions manual phần 3 pdf

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 0.1 188.4 175.4 161.6 147.5 75 25 5.0 25 4.4 25 2.4 24 8.7 24 3.1 23 5.0 22 3.9 20 9.8 194.1 177.8 100 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 9.7 0 1 2 4 0 1 0 0 0 27 2 .2 ACT 0 0 1 0 4 1 1 0 300 25 4.5 0 0 0 1...

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220 1,3K 1
fundamentals of heat and mass transfer solutions manual phần 5 ppsx

fundamentals of heat and mass transfer solutions manual phần 5 ppsx

... convection and () () () 1 /2 1 /2 3 22 b M h Dk D 4 2 235W m K 0.002m 399W m K 50 K 2. 15W ππ θ π ==⋅ ⋅=    () () 1 /2 1 /2 21 c m hP kA 4 23 5W m K 399 W m K 0.002m 34.3m − ==×⋅⋅×= () 1 mL 34.3m 0.012m ... W/m⋅K, ν = µv f = 3 52 × 10 -6 N⋅s/m 2 × 1. 029 × 10 -3 m 3 /kg = 3. 621 × 10 -7 m 2 /s; Pr s values for 20 ≤ T s ≤ 80°C: T (K) 29 3 300 325 350 353 Pr 7.00...

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fundamentals of heat and mass transfer solutions manual phần 7 pot

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 h r , () () 2 ro Bi 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 × 10 4 , the internal flow is turbulent, and from the Dittus-Boelter correlation, ( ) () 4/5 0.3 4/5 0.3 4 2 iD,i i k 0. 029 5W m...

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220 1,2K 0
fundamentals of heat and mass transfer solutions manual phần 8 ppsx

fundamentals of heat and mass transfer solutions manual phần 8 ppsx

... velocity is u m = m / ( ρπ D 2 / 4) so that ( ) 2 22 25 4m DL 2m L pf f 2D D ρρ π ρ π ∆= = () ( ) 2 3 65 25 0.003kg s 1. 128 kg m L 2 p f 2. 320 6 10 f L D D π −− ∆= = × ( 12) Recall that the pressure ... 1 UhD 2k h =+ + () 22 0.02mln 20 15 120 1 15 60W m K 3000W m K 27 9 W m K =++ ⋅ ⋅⋅ ( ) 45 32 32 h 1 4.44 10 9.59 10 3.58 10 m K W 4. 12 10 m K W U −−− − =×+×+× ⋅=×...

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220 690 1
fundamentals of heat and mass transfer solutions manual phần 9 ppt

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 obtain T 2 = 1 425 K < ()() () 52 12 h 42 ss cc 1500 1 425 ... ) ( ) ( ) ( ) 4 824 4 s 2 ss h sr 0.95.6710W/mK273K 1T G I90.2W/msr sr εσ ρ πππ − ×⋅ − ====⋅ ( ) ( ) 26 258 sd q46 .29 0.2W/msr510m2 .22 10sr1.5110W −−− − =+⋅×××=...

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220 483 0
fundamentals of heat and mass transfer solutions manual phần 10 doc

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 3 EJ JJ JJ 1/A1/AF1/AF εε −−− =+ − (5) There are 4 equations, Eqs. (2- 5), with 4 unknowns: J 2 , J 2 , T 2 and q 1 . Substituting numerical ... + = −++ where ()() ()() 44 22 12 121 2 12 TT TTTTTT. −=+ + − Accordingly, ( ) ()()( ) rad,hc 22 2 824 12 21 0.85 1 0.01 R 0.01 m 0.0 02 m 5.67 10 W / m K 28...

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