fundamentals of heat and mass transfer solutions manual phần 8 ppsx

... 79 .8 Cε=+ − =°+ −°=°<The rate of heat transfer to the air is()pc,o c,iq mc T T 0.03kg/s 1008J/ kg K 62 .8 C 1900 W=−=×⋅×°= and the rate of condensation is4cond6fgq 1900Wm 8. 70 ... = Cmin/Cmax = (5,000 × 41 78) /(10,000 × 4 188 ) = 0.499. Combining Eqs. (1) and (2), find()( )3h,oT80C0.73 48. 210W/10,000/3600kg/s4 188 J/kgK=°−×××⋅()h,oT8021.0C59C.=−°=°<COMMENTS: ... (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 plot Tc,o and hm as...

fundamentals of heat and mass transfer solutions manual phần 5 ppsx

... emissivity of Nichrome wire. Electricalcurrent. Temperature of air flow and surroundings. Velocity of air flow. (a) Surface and centerline temperatures of the wire, (b) Effect of flow velocity and ... stream velocity and temperature of air coolant. (a) Average pin convection coefficient, (b) Pin heat transfer rate, (c) Total heat rate, (d) Effect of velocity and pin diameter on total heat rate. ... /s 18, 880 ,ν−==× × = application of theChurchill-Bernstein correlation yields()()()D4/51/2 1/35 /8 1/42/30.62 18, 800 0.707 18, 880 Nu 0.3 1 76.6 282 ,0001 0.4/0.707=+ + =+D2aok0.0263W...

fundamentals of heat and mass transfer solutions manual phần 2 docx

... thickness and thermal conductivity of steel tubes. Temperature of steamflowing through the tubes. Thermal conductivity of insulation and emissivity of aluminum sheath.Temperature of ambient air and ... Temperature and volume of hot water heater. Nature of heater insulating material. Ambientair temperature and convection coefficient. Unit cost of electric power.FIND: Heater dimensions and insulation ... 214mm=−=<The heat rate is then()()()2 84 8 323 Kq 420W / mn 0. 18/ 0.15 n 0.394/ 0. 18 35W / m K 0.10W / m Kπ−′==+⋅⋅<(b) The effects of r3 on Ts,o and q′ have been computed and...

fundamentals of heat and mass transfer solutions manual phần 3 pdf

... 185 185 185 185 ← initial estimate1 186 .3 186 .6 186 .6 105 .8 2 187 .1 187 .2 167.0 96.03 187 .4 182 .3 163.3 94.24 184 .9 180 .8 162.5 93 .8 5 184 .2 180 .4 162.3 93.76 184 .0 180 .3 162.3 93.67 183 .9 ... 250 275 3000 180 .7 180 .2 1 78. 4 175.4 171.1 165.3 1 58. 1 149.6 140.1 129.9 119.4 1 08. 7 98. 025 204.2 203.6 201.6 1 98. 2 193.3 186 .7 1 78. 3 1 68. 4 157.4 145.6 133.4 121.050 2 28. 9 2 28. 3 226.2 222.6 ... 200.1 188 .4 175.4 161.6 147.575 255.0 254.4 252.4 2 48. 7 243.1 235.0 223.9 209 .8 194.1 177 .8 100 282 .4 281 .8 280 .1 276.9 271.6 263.3 250.5 232 .8 213.5125 310.9 310.5 309.3 307.1 303.2 296.0 282 .2...

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fundamentals of heat and mass transfer solutions manual phần 4 docx

... 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 44.4 205 76.9 68. 8 60.7 40.4 206 30 68. 8 68. 8 54.6 40.4 207 68. 8 61.7 54.6 37.3 20 8 ... 90 287 .6 305 .8 315.2 321.5 323.5 324.5 325 325 6 1 08 285 .6 301.6 313.5 319.3 322.7 324.0 324.5 325 7 126 281 .8 299.5 310.5 317.9 321.4 323.3 324.2 8 144 279 .8 296.2 3 08. 6 315 .8 320.4 322.5 ... 120 s) and T(0.15 m, 120s) using the finite-element software FEHTfor a surface emissivity of 0.94 and (b) Plot the temperature histories for x = 0, 150 and 600 mm, and explain key features of your...

fundamentals of heat and mass transfer solutions manual phần 6 docx

... PROBLEM 8. 28 KNOWN: Inlet temperature, flow rate and properties of hot fluid. Initial temperature, volume and properties of pharmaceutical. Heat transfer coefficient at outer surface and dimensions ... ,π−==×mcum/A2 .83 m/s,ρ== and with DRe 38, 980 ,= Fig. 8. 3 yields f ≈ 0.022. Hence,()23242 .83 m /s 5mp 0.022 5.391kg /m 95N /m 9.4 10 atm2 0.05m−∆≈ × = = ××<The rate of heat transfer ... properties.PROBLEM 8. 34KNOWN: Flow rate and inlet temperature of water passing through a tube of prescribed length,diameter and surface temperature.FIND: (a) Outlet water temperature and rate of heat transfer...

fundamentals of heat and mass transfer solutions manual phần 7 pot

... per unit length of the tube; effect onquality, x, at outlet of 30 m length of tube; (b) Effect of radiation on heat transfer and quality of outletflow; (c) Effect of emissivity and insulation ... %# %# %323362 6259 8 0 0033 273 0151 589 10 22 5 103 08 10 and from Eq. 9.34,()()()221/651/6s,iD,io 8/ 27 8/ 279/16 9/16i0. 387 3. 08 10 T T0. 387 Rak 0.0263h 0.60 0.60D0.151 ... )32413sD6 282 9.8m/s810K5527.4K0.025mgTTDRa510m/s8 .85 10m/sβνα−−∞−−×−−==×××6DRa7.6410=× and using the correlation of Eq. 9.34,( )D21/6D 8/ 279/160. 387 RaNu0.6035.010.559/Pr=+=+D2ok0.15W/mKhNu35.0210W/mK.D0.025m⋅===⋅Alternatively,...

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fundamentals of heat and mass transfer solutions manual phần 9 ppt

... T = 289 8 µm⋅K, it follows that λmax(e) = 289 8 µm⋅K/400 K = 7.25 µm and λmax(r) = 289 8µm⋅K/273 K = 10.62 µm. <λ = 7.25 µm: Table 12.1 → Iλ,b(400 K) = 0.722 × 10-4 (5.67 × 10 -8 )(400)5 ... dissipationrate, 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 = 117,573 W/m2Emissive power: Eb = ( )442mT87273952W/mσσ=+=Emissivity: ... is() () 82 4 2 82 44s34 0.53 5.67 10 W m K 1300K 15.1W m K 1000K 0 .8 5.67 10 W m K 300 KdT dt 89 33kg m 385 J kg K 0.01m−−×× ⋅× + ⋅× −×× ⋅×=×⋅×  []4sdT dt 1.16 10 85 ,82 9 15,100...

fundamentals of heat and mass transfer solutions manual phần 10 doc

... 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 ... of ash deposit. Convection coefficient and temperature of gas flow over the tube. Temperature of surroundings.FIND: (a) Rate of heat transfer to tube without ash deposit, (b) Rate of heat transfer ... )rad,hc222 82 412210 .85 1 0.01R0.01 m 0.002 m 5.67 10 W / m K 288 K 2 68 K 288 2 68 K−−++=−×× ⋅ + +where, again, it is assumed that T1 = 15°C and T2 =...

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