A numerical study of heat and mass transfer in porous fluid coupled domains

... Ochoa-Tapia and Whitaker’s stress jump conditions for the flow, heat and mass transfer in porous/ fluid coupled domains, and to investigate their effects on local and global heat and mass transfer; ... incorporated with heat or mass transfer There are also limited numerical experiments and systematic results for heat and mass transfer in porous/ fluid coupled domains for certain numerical applications ... Interface Conditions For heat transfer interface conditions, usually continuities of temperature and heat flux are required (Neale and Nader, 1974; Vafai and Thiyagaraja, 1987; OchoaTapia and Whitaker,...

MÔ HìNH HOá QUá TRìNH TRAO ĐổI NHIệT ẩM TRONG MáY ấP TRứNG GIA CầM Modeling of heat and mass transfer in chicken egg incubators

... 3.4 Mô hình toán học trình trao đổi nhiệt ẩm vùng chứa trứng Từ mô hình động học (6) phơng trình nhiệt độ độ ẩm dòng khí trứng (35), (36) (38) ta xác định đợc hệ phơng trình mô tả trạng thái nhiệt ... lợng trứng: 224 (1 ) drT h fT t = hcTa Av (Ta TT ) + (1 ) drT hP + (1 ) drT hD (TT ) X t (34) 3.3 Đơn giản hoá mô hình nhiệt ẩm vùng chứa trứng Mô hình trao đổi nhiệt ẩm vùng chứa trứng ... (1988), Modeling Simultaneous Heat and Mass Transfer in Isotropic Sphere, Trans ASEA, pp 31 Husain A., Chen C.S., Clayton J T, Whitney L.F (1972), Mathematical Simulation of Mass and Heat Transfer in...

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fundamentals of heat and mass transfer solutions manual phần 2 docx

... reductions in q and m are achieved with r2 = 0 .26 -3 m With increasing r2, q and m decrease from values of 17 02 W and 8×10 kg/s at r2 = 0 .25 m to 0. 627 W and 2. 9×10-6 kg/s at r2 = 0.30 m COMMENTS: ... surface of the insulation, it follows that q′ cond = q′ conv + q′ rad Ts,1 − Ts ,2 ln ( r2 r1 ) 2 k ( = Ts ,2 − T∞ + Ts ,2 − Tsur ( 2 r2 h ) ( 2 r2 h r ) where h r = εσ Ts ,2 + Tsur 2 ) (Ts ,2 + Tsur ... r1 ) − (1 r2 )] , and, from Eq 2 ) (Ts ,2 + Tsur ) With t = 10 mm (r2 = 26 0 mm), ε = 0 .2 and T∞ = Tsur = 29 8 K, an iterative solution of the energy balance equation yields Ts ,2 ≈ 29 7.7 K, where...

fundamentals of heat and mass transfer solutions manual phần 3 pdf

... 0.6 939 T11 + 7.65 The initial estimates (k = 0) are carefully chosen to minimize calculation labor; let ε < 1.0 k T1 34 0 33 8.9 33 8 .3 338 .8 33 9.4 33 9.8 34 0.1 34 0 .3 T2 33 0 33 6 .3 337 .4 33 8.4 33 8.8 33 9.2 ... 282.4 31 0.9 34 0.0 25 180.2 2 03. 6 228 .3 254.4 281.8 31 0.5 34 0.0 50 178.4 201.6 226.2 252.4 280.1 30 9 .3 339 .6 75 175.4 198.2 222.6 248.7 276.9 30 7.1 33 9.1 100 171.1 1 93. 3 217.2 2 43. 1 271.6 30 3.2 33 7.9 ... 33 9.2 33 9.4 33 9.5 T3 31 5 32 4 .3 328.0 32 8.2 32 8.9 32 9 .3 329.7 32 9.9 T4 250 237 .2 241.4 247.7 251.6 254.0 255.4 256.4 T5 225 232 .1 241.5 245.7 248.7 250.5 251.7 252.5 T6 205 225.4 226.6 230 .6 232 .9...

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fundamentals of heat and mass transfer solutions manual phần 4 docx

... 356.9 357 .4 357.9 358 .4 358.9 359 .4 T02 3 54. 9 355 .4 355.9 356 .4 356.9 357 .4 T03 351.6 352.1 352.6 353.1 353.6 3 54. 1 T 04 346 .9 347 .4 347 .9 348 .4 348 .9 349 .3 T05 340 .9 341 .4 341 .9 342 .3 342 .8 343 .2 ... 20 30 40 45 T0 85 85 85 85 76.9 76.9 68.8 68.8 61.7 61.7 T1 85 85 85 76.9 76.9 68.8 68.8 61.7 61.7 55.6 T2 85 85 68.8 68.8 60.7 60.7 54. 6 54. 6 49 .5 49 .5 T3 85 52.5 52.5 44 .4 44. 4 40 .4 40 .4 37.3 ... 125.00 137.50 143 .75 146 .88 148 .44 149 .22 149 .61 149 .80 T2 (°C) Implicit 50 83.33 105.55 120.37 130.25 136.83 141 .22 144 .15 146 .10 147 .40 100 100 100 100 100 100 100 100 100 100 < The heat flux at...

fundamentals of heat and mass transfer solutions manual phần 5 ppsx

... 75. 5 W m ⋅ K q′ = hπ D ( Ts − T∞ ) = 75. 5 W m ⋅ K π ( 0.01m ) (50 − 20 ) C = 71.1W m Fluid: Saturated Water (5 m s ) 0.01m VD = = 68, 55 2 Re D = ν 7 25 × 10−6 N ⋅ s m 994 kg m3 0.62 ( 68, 55 2 ... ≈ 0 .51 ( 9 25) ( 0.686) k 0.0373 W/m ⋅ K h = Nu D = 13 .5 = 50 .4 W/m2 ⋅ K Do 0.01 m 0 .5 0.37 1/4 0 .5 0.37 ×1 = 13 .5 Hence ( ) 1/2  50 .4 W/m2 ⋅ K 0.0314 m   m =  5 m   ( 35 W/m ⋅ K ) 5. 89 ... 600 55 0 50 0 450 400 350 300 50 100 150 200 250 300 Time, t(s) Without radiation With radiation, eps = 0.8 Continued The agreement between predictions with and without radiation for t < 50 s implies...

fundamentals of heat and mass transfer solutions manual phần 6 docx

... 8 .60 was used The effects of m are as follows Continued PROBLEM 8.30 (Cont.) 67 70 69 .8 Outlet temperature, Tmo(C) Outlet temperature, Tmo(C) 66 65 64 63 62 61 60 0.005 69 .6 69.4 69 .2 69 0.0 06 ... rate and properties of hot fluid Initial temperature, volume and properties of pharmaceutical Heat transfer coefficient at outer surface and dimensions of coil FIND: (a) Expressions for Tc(t) and ... the heat loss is q ′ = 3 46 W / m PROBLEM 8. 36 KNOWN: Inner and outer diameter of a steel pipe insulated on the outside and experiencing uniform heat generation Flow rate and inlet temperature of...

fundamentals of heat and mass transfer solutions manual phần 7 pot

... Node Tm(K) 391 .70 385.95 380 .70 375 .92 371 .56 10 3 67. 61 364.03 360.81 3 57. 91 355.32 11 12 13 14 15 353.02 351.00 349.25 3 47. 75 346.50 16 17 18 19 20 345.49 344 .70 344.15 343.82 343 .71 From these ... 12.2 343 − Ts,i = 0. 175 0.60 + 2.64 Ts,i − 273 )1/ } (Ts,i − 273 ) + 0.5 × 5. 67 × 10−8 Ts,i − ( 273 )4    Ts,i ≈ 314 .7 K ≈ 41 .7 C A trial -and- error solution gives < The heat loss per unit length ... significant if 0.1 37 ReD < 4 .7 GrD Pr ( ) where GrD = RaD/Pr, find using results from above and in part (a) for ω = 5000 rpm, 0.1 37 11, 076 ? < ? 4 .7 ( 27, 981 0 .70 28 ) 0 .70 18     = 383...

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fundamentals of heat and mass transfer solutions manual phần 8 ppsx

... 11.14 and 11.15 where ∆T1 = Th,i - Tc,o = 80 °C and ∆T2 = Th,o - Tc,i = 120°C Hence, ∆T1m = (120 80 )°C/ln(120 /80 ) = 98. 7°C and L= q 3000 W = = 0.9 68 m (π D ) U∆T1m (π × 0.02 m ) 500 W m2⋅ K × 98. 7 ... Cmin/Cmax = (5,000 × 41 78) /(10,000 × 4 188 ) = 0.499 Combining Eqs (1) and (2), find Th,o = 80 °C − 0.7 × 3 48. 2 × 103 W / (10,000/3600 ) k g / s × 4 188 J / k g ⋅ K ( Th,o = ( 80 − 21.0 ) ° C = 59°C ... Overall heat transfer coefficient, U, for the HXer, outlet temperature of cooling water, Tc,o, and condensation rate of the steam m h ; and (b) Compute and plot Tc,o and m h as a function of the...

fundamentals of heat and mass transfer solutions manual phần 9 ppt

... PROBLEM 12 .99 (Cont.) t= ( −11 2.4504 ×10 K −1 −1 ⋅s ) ( 292 8 K )  3  ln 292 8 + 2741 292 8 + 300 − ln 292 8 − 2741 292 8 − 300   2741  −1  300   +2  tan −1   − tan  292 8    292 8   ... 4.17E-5 0.0417 6.39E-5 0.702 600 5.179E-5 0.0446 7.39E-5 0.701 700 6.671E-5 0.0 499 9. 44E-5 0.706 800 8.29E-5 0.0548 0.000116 0.715 90 0 0.0001003 0.0 597 0.0001 39 0.721 1000 0.0001187 0.0647 0.000165 ... rate, M is the mass of PCM, and hfg is the heat of fusion of PCM Irradiation: -8 4 G = σTw = 5.67 × 10 W/m ⋅K (1200 K) = 117,573 W/m Emissive power: Eb = σ Tm = σ ( 87 +273 ) = 95 2 W / m λT =...

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