fundamentals of heat and mass transfer solutions manual phần 5 ppsx

... the Explore and Graph options of IHT. 5 10 15 20 25 Freestream velocity, uinf(m/s)0 50 01000 150 02000 250 03000 350 04000Power dissipation, q(W) Continued Dimensions of chip and pin fin. ... at x = 95 mm, 10x0.095mhh.==1/21/310xkh0. 453 RePrx= 5 x-62ux40 m/s0.0 95 mRe2.0 651 018.410 m/sν∞×===××()( )1/21/3 52 100.0282 W/mKh0. 453 2.0 651 00.70 354 .3W/mK0.0 95 ⋅=×=⋅( ... obtained.0 50 100 150 200 250 300Time, t(s)300 350 400 450 50 0 55 0600Temperature, T(K)Without radiationWith radiation, eps = 0.8Continued Diameter, thickness and thermal conductivity of steel...

fundamentals of heat and mass transfer solutions manual phần 8 ppsx

... values of P and R as given by Fig. 11.11 with()()()()P 953 5/300 350 .23R300196/120 351 .22=−−==−−=giving F ≈ 1. Based upon CF arrangement,()()()()m,CFT300 951 9635C/n300 95/ 196.2 351 82K.∆=−−−°−−=llUsing ... 210.45W/K C C 755 .25W K== ==.Hence, Cmin/Cmax = 0.279 and ()max min h,i c,iq C T T 210. 45 W K (100 K) 21,045W=−= =maxq q 14,732W 21,045W 0.700ε== =.Figure 11.18 yields NTU ≈ 1 .5, ... 13 ,50 0 W/m2 ⋅K.FIND: (a) Overall heat transfer coefficient, U, for the HXer, outlet temperature of cooling water, Tc,o, and condensation rate of the steam hm; and (b) Compute and...

fundamentals of heat and mass transfer solutions manual phần 2 docx

... system.PROBLEM 3 .50 (Cont.)0.0 35 0.0 45 0. 055 0.0 65 0.0 75 Outer radius of insulation, r3(m)0400800120016002000 Heat loss, qprime(W/m)q1 0.0 35 0.0 45 0. 055 0.0 65 0.0 75 Outer radius of insulation, ... trial -and- error solution, Ts,o ≈ 50 2K. Hence the heat loss is()()44oo s,o ,o o s,o surq = D h T T D T Tπεπσ∞′−+ −() ( )()2844424Wq = 0.075m 25 W/m K 50 2-293 0.8 0.075m 5. 67 10 50 2 ... 0.38 0.44 0 .5 Outer radius of insulation, m4080120160200240Outer surface temperature, CTs,o0.2 0.26 0.32 0.38 0.44 0 .5 Ou ter radius of ins ulation, m0 50 01000 150 02000 250 0 Heat rates...

fundamentals of heat and mass transfer solutions manual phần 3 pdf

... 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 50 15 50 63 .58 67.73 63 .58 50 30 50 56 .27 58 .58 56 .27 50 45 50 52 .91 54 .07 52 .91 50 60 50 51 .32 51 .86 ... 52 .91 50 60 50 51 .32 51 .86 51 .32 50 75 50 50 .51 50 .72 50 .51 50 90 50 50 50 50 50 The improved prediction of the temperature field has a significant influence on the heat rate, where,accounting ... where x and y are in mm and the temperatures are in °C.y\x 0 25 50 75 100 1 25 150 1 75 200 2 25 250 2 75 3000 180.7 180.2 178.4 1 75. 4 171.1 1 65. 3 158 .1 149.6 140.1 129.9 119.4 108.7 98.0 25 204.2...

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fundamentals of heat and mass transfer solutions manual phần 4 docx

... 3 25 3 25 3 25 1 18 304.2 324.7 3 25 3 25 3 25 3 25 3 25 3 25 2 36 303.2 3 15. 3 324 .5 3 25 3 25 3 25 3 25 3 25 3 54 294.7 313.7 320.3 324 .5 3 25 3 25 3 25 3 25 4 72 293.0 307.8 318.9 322 .5 324 .5 3 25 3 25 3 25 ... p⋅∆t = 300 p (s) = 5 p (min).pt(min) T0T1T2T3TL(°C)0 0 85 85 85 85 201 85 85 85 52 .5 202 10 85 85 68.8 52 .5 203 85 76.9 68.8 44.4 204 20 76.9 76.9 60.7 44.4 20 5 76.9 68.8 60.7 ... 3 25 3 25 5 90 287.6 3 05. 8 3 15. 2 321 .5 323 .5 324 .5 3 25 3 25 6 108 2 85. 6 301.6 313 .5 319.3 322.7 324.0 324 .5 3 25 7 126 281.8 299 .5 310 .5 317.9 321.4 323.3 324.2 8 144 279.8 296.2 308.6 3 15. 8 320.4...

fundamentals of heat and mass transfer solutions manual phần 6 docx

... (K)(Assumed) Eq. (5) Eq. (6) Eq. (7) 1 1000 3.0 25 × 10-2348.6 9 15. 0 2 950 4.321 × 10-2463.7 898. 3 900 5. 041 × 10-2 52 4.6 891.0 4 890 5. 2 15 × 10-2 53 9.0 889 .5 Hence, we find2am,om5.2210 kg/s ... 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) ... effect of m on q and Tm,o was determined by accessing the Correlations Toolpad of IHT, and theresults are plotted as follows.0.1 0.2 0.3 0.4 0 .5 Mass flowrate, mdot(kg/s) 50 0010000 150 0020000 250 0030000Heat...

fundamentals of heat and mass transfer solutions manual phần 7 pot

... 3.03Table A-6, Water 0.497 650 0. 158 3. 15 0 .50 4Table A -5, Ethylene glycol 5. 15 260 0.0936 55 .0 0. 65 ANALYSIS: The electrical power (Pe) required to offset convection heat transfer is( ) ( )2convsssqhATThDTT.π∞∞=−=−(1)The ... )D21/688/279/160.3871.0 251 0Nu0.6 056 .9310 .55 9/0.703×=+=+D2DhNuk/D56.930.0285W/mK/0.300m5.4W/mK.==×⋅=⋅Continued … PROBLEM 9.72KNOWN: Velocity and temperature of air flowing ... find122t1 1 1 1 15 1ln 0.430 W m KR 2 0. 051 W m K 65 5.09 W m K 0.115m 11,000 W m K 0. 055 mπππ−=++ =⋅′×⋅⋅× ⋅ × and from Eq. (1), ()q 0.430 W m K 416 298 K 50 .8W m′=⋅−=<Continued...

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fundamentals of heat and mass transfer solutions manual phần 9 ppt

... for qsh - d/qtot =0 .5, 1 or 1 .5 % was computed and plotted as a function of the sample emissivity.0. 05 0. 15 0. 25 0. 35 Sample emissivity, epss 50 100 150 200 250 Shield temperature, Th ... 0 .55 9 0.7 15 =+ =⋅+ (5) With values of ε and α from part (a), the rate of temperature change with time is() ()824 2 8244s34 0 .53 5. 67 10 W m K 1300K 15. 1W ... M is the mass of PCM, and hfg is the heat of fusion of PCM.Irradiation: G = σTw = 5. 67 × 10-8 W/m2⋅K4(1200 K)4 = 117 ,57 3 W/m2Emissive power: Eb = ( )442mT87273 952 W/mσσ=+=Emissivity:...

fundamentals of heat and mass transfer solutions manual phần 10 doc

... 13 .5 or Table 13 .5, ()1/22214 4 1F 0 .5 S S 4 r /r=−− (5) ()1/22214F 0 .5 5.01 5. 01 4 15/ 150 0.001997=−− =2242211 R 1 0. 05 S 1 1 5. 010R0 .5 ++=+ ... 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 ... Dimensions and emissivity of double pane window. Thickness of air gap. Temperatures of room and ambient air and the related surroundings.FIND: (a) Temperatures of glass panes and rate of heat transfer...

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