fundamentals of heat and mass transfer solutions manual phần 4 docx

... 203 85 76.9 68.8 44 .4 20 4 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 37.3 208 40 61.7 61.7 49 .5 37.3 209 45 61.7 55.6 49 .5 34. 8 20<The ... Size∆xpmT (45 s)xq′′ (0 ,45 s)xq′′ (L ,45 s)(mm)(°C)kW/m2kW/m210 149 .8 - 249 .8 + 149 .85 149 .3 - 249 .0 + 149 .02 149 .4 - 249 .1 + 149 .0COMMENTS: (1) The center temperature and boundary heat ... m1T T 1 2 0 .44 0 0.208 0 .44 0 0 .44 0 T T 0.208 0 .44 0 25+−+=−×−× + + +××()p1 p p pmmm1 m1T 0.029T 0 .44 0 T T 2.3+−+=+ ++( 14) Nodal Point 1:()p1 p p p p11 2 12T 0.029T 0 .44 0 200 T 2.3...

fundamentals of heat and mass transfer solutions manual phần 2 docx

... 1TTq4k r r 4k r r4rhπππ∞=+ −+ −+()() 4 s,12111 1KT 25 C 48 9W 1. 84 10 4 0. 04 0.51 0.53 W 4 0.53 6ππ−=°+ × + − + 4 s,1KT 25 C 48 9W ... 0.38 0 .44 0.5Outer radius of insulation, m 40 80120160200 240 Outer surface temperature, CTs,o0.2 0.26 0.32 0.38 0 .44 0.5Ou ter radius of ins ulation, m05001000150020002500 Heat rates ... 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...

fundamentals of heat and mass transfer solutions manual phần 6 docx

... mdot(kg/s)50001000015000200002500030000 Heat rate, q(W) 0.1 0.2 0.3 0 .4 0.5 Mass flowrate, mdot(kg/s) 42 43 44 45 46 Outlet temperature, Tmo(C)Although q increases with increasing m due to the attendant increase in ReD, and ... ) 4/ 50 .4 4/50 .43 DNu0.023RePr0.0239. 143 100.68229.12==×=2ihNuk/D29.120.0373 W/mK/0.025 m43 .4 W/mK.=⋅=×⋅=⋅Hence, the outlet temperature is( )( )2m,o-3212T0.025 m2 m43 .4 W/mKexp 4. 50110 ... W/m0.04m0.02mqDDTT40C48.2C 4 h D 41 840 W/mK0.02m−−=+=+=×⋅oo& and ( ) ( )6322w,max10 W/mT0.02m0.01m 41 5 W/mK=−−×⋅( )26310 W/m0.02m0.02n48.2C52.4C.215...

fundamentals of heat and mass transfer solutions manual phần 3 pdf

... 186.7 178.3 168 .4 157 .4 145 .6 133 .4 121.050 228.9 228.3 226.2 222.6 217.2 209.7 200.1 188 .4 175 .4 161.6 147 .575 255.0 2 54. 4 252 .4 248 .7 243 .1 235.0 223.9 209.8 1 94. 1 177.8100 282 .4 281.8 280.1 ... and ()2h x / k 5000W / m K 0.005m / 240 W / m K 0.1 042 ,∆= ⋅ ⋅= the solution to Eqs. (1) – (7) yields12 34 T 46 .61 C, T 45 .67 C, T 45 .44 C, T 49 .23 C=°=°=°=°567T 48 .46 C, T 48 .00 C, T 47 .86 ... relative amounts of heat transfer to the air and heat sinkare determined by the values of h and t,cR.′′PROBLEM 4. 54 (Cont.)The heat rate by convection between the bar and fluid is given...

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fundamentals of heat and mass transfer solutions manual phần 5 ppsx

... KD==⋅=⋅0625001 347 21 6902.., ′=qWm20 43 8,Fluid: Engine Oil()D625m s 0.01mVDRe 147 340 10 m sν−== =×()( )() 4/ 51/2 1/35/8D1 /4 2/30.62 147 40 00 147 Nu 0.3 1 120282,0001 0 .4 4000=+ ... 0.0001375 6. 24 1 .44 0.978 0.98 14 9 .47 11. 648 E4 0 *//* Correlation results and air thermophysical properties at TfNuLbarPr ReL Tf hLbar k nu uinf22 94 0.7035 1.63E6 325 53.82 0.02815 1. 841 E-5 25 ... conductivity and 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...

fundamentals of heat and mass transfer solutions manual phần 7 pot

... (a) Heat transfer rate by free convection to the room, 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 ... emissivity, and power dissipation of cylindrical heater. Temperature of ambientair and surroundings.FIND: Steady-state temperature of heater and time required to come within 10°C of this temperature.SCHEMATIC:ASSUMPTIONS: ... )D1 /4 51 /4 D 4/ 94/ 99/169/160.5892.93100.589RaNu2212.5510 .46 9/Pr10 .46 9/0.700×=+=+=++( )D2hNuk/D12.550.0298W/mK/0. 040 m9.36W/mK.==⋅=⋅Substituting numerical values, the heat loss from the bulb is,( ) ( ) ( ) ( ) 84 244 2 24 WWq0. 040 m9.3612525K0.805.671012527325273KmKmKπ−=−+××+−+⋅⋅()q4.703.92W8.62W.=+=<COMMENTS:...

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fundamentals of heat and mass transfer solutions manual phần 8 ppsx

... K3000W m K 279 W m K=++⋅⋅⋅() 45 32 32h1 4. 44 10 9.59 10 3.58 10 m K W 4. 12 10 m K WU−−− −=×+×+× ⋅=× ⋅2hU 243 WmK=⋅.With Ch = Cmin = 4. 160 × 10 4 W/K and Cc = Cmax = 2.09 × ... NTU = 1. 54 and ()() 42 2hminhA NTU C U 1. 54 4.160 10 W K 243 W m K 264m==× ⋅=.Hence,()()()2hL2oTA 264mN70DWN0.02 2 30 mππ== =<(b) Using the IHT Correlations, Heat Exchangers ... )()22minANTUC/U0.9786 ,40 0W/K/2000W/mK41.9m==⋅= <ccp,cmC/c86 ,40 0W/K /41 78J/kgK20.7kg/s.==⋅=&<(b) Using the final overall heat transfer coefficient, find()22minNTUUA/C1000W/mK41.9m/86 ,40 0W/K0 .48 5.==⋅=Since...

fundamentals of heat and mass transfer solutions manual phần 9 ppt

... or()()() 44 12c2wss ccTTTTLk Lkεσ−=−+()8 244 44 221500K T0.8 5.67 10 W m K T 90 K0.008m 0.005m25W m K 60W m K−−=× × ⋅ −+⋅⋅6 84 223.72 10 247 9T 4. 54 10 T 2.98−×− = × − 84 622 4. 54 ... 6161 (4) and Gw is equal to the blackbody emissive power at Tch. Using Eqs. (3) and (4) and substitutingnumerical values, findσεσ εσTT Tbp 4 ww 4 wch 4 =+−116800 07 03 40 0 44 K T Kw 4 1616=+ ... Km=−⋅⋅ () () 44 48 24 24 WW0.537 5.67 10 1500K 0 .4 5.67 10 300KmK mK−−+×× −××⋅⋅)[] 42 2idT dt 1.163 10 m K J 18 ,48 0 1 54, 140 180 W m 20K s−=× ⋅ + − =<Defining...

fundamentals of heat and mass transfer solutions manual phần 10 doc

... 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 ... 8170 K / W R 6 944 K / W R 747 1 K / W.===The equivalent resistance is then1eq111R 249 8 K / W8170 6 944 747 1−=++ = and the total resistance isR 1603 249 8 1603 57 04 K / W.=++=<COMMENTS: ... diffuse-gray surfaces.()() 44 psps conv spsTTqqhTT1/ 2/ 1σεε∞−′′ ′′===−+−()( )() 44 pspssTh/TT1/1/1Tσεε∞=−+−+()()2 4 4p p8 24 35 W / m K 20 K11T 1 320 K T 48 4 K.0.8 0.35.67 10...

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