HW 9 solution ME 114 S08

HW 9 solution ME 114 S08 - \ I . ' PROBLEM92 chl...

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Unformatted text preview: \ I . ' PROBLEM92 chl §ollxlfim KNOWN: Object with specified characteristic length and temperature difi'erence above ambient fluid. ‘ ' FIND: Grashof number for air, hydrogen, water, ethylene glycol for a pressure of 1 atm. . ' Quiescen+ fluid a 7;,157‘721 ASSUWTIONS: (1) Thermophysical properties evaluated at Tf = 350K, (2) For gases, ,5 =‘1/Tf- PROPERTIES: Evaluate at 1 atm, Table 11-4, Air: U=20.92X10—6 m2 /s; Hydrogen: u= 143x10*6 In2 /s Table A-6, Water (Sat, liquid): u=pf vf =37.5><1o-6 m2 /s, a =0.624><10'3 K‘1 _ Table A-5, Ethylene glycol: U=3.17><10_6 rnz/s, fl=0.65x10‘3 K‘l. Tf 4—- 350K: ANALYSIS: The Grashof number is Gm = stirs-Tog)? _ V2 Substituting numerical values for the fluid air with ,6 =' l/Tf, find GrL _ 9.8m/s‘2 x (1/350K) (25K) (0.25m)3 ’ ' '— 3" (20;92x10 8rug/s)2 given by Eq. 9.12, Gm“, = 2.50x107 . <1 Performing similar calculations for the other fluids, find Gmyd = 5.35x105 . .<] Grimm, = 1.70m),6 <l — ’ ,9 . . . <1 COMNIENTS: Higher values of GrL impltr increased free convection flows. However, other properties affect the value of the heat trans fer coeflicients. Note that for the gases, fl = 1 /Tf, assuming perfect gas behavior. ' m W . raffle/L T/fo €39 ' ‘ _ p ' PROBLEM 9.92 KNOWN: Cross flow over a cylinder with'prescribed surface temperature and free stream conditions. ' - FIND: .Whether free convection will be significant if the fluid is water or air. SCHEMATIC: D=50mml Cylinder V=0.05m/s —+-i> L20 C 7;:55‘73 ASSUMPTIONS": (1) Uniform properties, (2) Combined free and. forced heat transfer situation. - . a . ' PROPERTIES: Table A-6, Water. (T§=(T.,+Ts)/2=300K): V v=ttvf=855><10‘6 N-s/m2x1.003x10-3 m3/lcg=8.576x10‘7.m2/s, =276.1><10',6 K‘l; TableA-‘4, Air (30032.1 arm): v=15.89x10‘5 mZ/s,a=1/r;=3.333x10- K-l. ~ . ANALYSIS: Following the discussion of Section 9.9, the general criterion for delineating the a relative significance of free and forced convection depends upon the value of Gr/Re3. If free convection is significant, _ ' ' _ » . ' ‘ ‘GrD/Re2D21-.' - . - 3 ' ' «(1) where t ‘ _ . .ch = g‘B (T..,—T..,)D3/v2 - and I RC]; = Vii/v. I -- (2,3) (a) When the surIOunding fluid is water, find ‘ ' GrD = 9.8 m/s2 x 276.1x10‘5 K:1 (35—20)K (0.05m)3/(8.576x10-7 1312/5)2 = 68., 980 to, 'ReD=0.OS m/sxo.oSm/8.576x10-7 mzls = 2915 “C Q S7 CLSZOHO , » ' {Big CED/Reg = sense/29152 ==9£08T2f.(9, g I - b/ WM 7% mad ‘ 4 We conclude that since GrD'lReIZJ << 1, free convection is -w‘significant. It is apparent that forced convection dominates the heat transfer process. ' (b) When the surrounding fluid is air, find' ‘ , GID -= 9.8 m/s2 S< 3.333x10’3 K‘l _(35—20)K(0.058803/(15.89x1o-6 mgA/s)2 = 242.558 RED = 0.05 m/s x O.O5m/15.89x10‘6 mz/s = 157 GrD/Rc% = 242,558/1572 = 9.8 . , - ' i <1 We conclude that, since GrD/Refi >> 1, free convection dominates the heat transfer process. COMMENTS: Note also that for the air flow situation, surface radiation exchange is likely to be significant. r - y . A? [1407 M90 Calm/(an [(A4 [9.07% am it ComfiMQU ' ‘ g PROBLEM 11.4 ' KNOWN: Dimensions of heat exchanger tube with or without fins. Cold and hot side convection coefficients. . . : V FIND: Cold side overall heat transfer coefficient without and with fins. SCHEMATIC: [es-ow ‘FiTl 00W/,,,a./< +.=2mm§, ‘N=16 ASSUMPTIONS: (1) Negligible fouling, (2) Negligible contact resistance between fins and tube wall, (3) hh is not affected by fins, (4) One-dimensional conduction in fins, (5) Adiabatic . fin tip. . a _ ANALYSIS: Froni Equation 11.1, L = 1 + DimCDn/Di) + - Ac (Home 2k ("PhAhl " Without fins: TIM =1‘\'(,‘}1 =1. ' . . . a _1_ 1 + (0.02 m)ln(26/20) + a 1 20 ' U. = 8000 W/mz-K 100 W/m-K . 200 W/mz-K 36— ‘ ‘ 1/Uc = (1.25x10‘4 + 5415x10-s Jr 3.85x10‘3) mZ-K/w =.4.oz><10-3 mz-K/W Uc=249WIm2-K. '. _ . " _- v .. V . 4 ‘ Withfins: 110,; = 1,110... = 1 - (Af/Ax1 - hf) Per unit length along the tube axis, A;=N(2L;+t)=16(30+2)mm=51in1m ’ ‘ ' '11,, = Af+ (11D. — 160 = (512+ 81.7 - 32)mm = 561.7 m I I With m = (21mm = (400 W/mz-K/SO whim-1:20.002 m)”2 = 53.3 m-1 t (63.3 m-1)(0.01s m) = 0.95 i ‘ ' and Eq. 11.4 yields ' x ' " , 'flr é tanh(er)/m-Lr = 0.739/0.95 =, 0.7713. - The overall surface efficiency. is then ' ‘ ‘11., = l — (As/Ath —- 115) = 1 — (512/561.7)(1 f 0.778) = 0.798. , .. 1 ‘ 1: ySéL é;l(/LJI0,VLS F 1 .4 . ‘ .5 - I 2 ‘ V 4 .._. = 2 + . __.___.__. - — . . Hence. UP. (1. 5x10 5 25x10 + 0.798(200)56L7 m m g73;.<10-4 m 1 UC _ Vilmzi'K. ' 1 ‘ . f I ‘ - < ‘ COMMENTS: The fins provide significant heat transfer enhancement. ' v ' ‘ ' , PROBLEMIIJO KNOWN: A shell and tube Ileer (two shells, four tube passes) heats 10,000 kg/h of pressurized water from 35 ° C to 120 ° C with 5,000 kg/h Water entering at 300 C. FIND: Required heat transfer area, A4,. _ scant/mirror ‘ , 777150012 Th - Ll=1500w/m2-/< ‘ rp'zaflzofc. L--f{>7i2f5000kg//1 A72 mc =10,000kg//1 l a . ASSUMPTIONS: (1) Negligible heat lossto surroundings, (2) Negligible kinetic and potential energy changes, (3) Constant properties. ' PROPERTIES: TabliA-B, Wat'e1"'('.[‘c =350K):‘ cp =419'5 J/kg-K; Table A-6, Water (Assume TM, {3150 ° C, Th 8500K): cp =4660 J/kg‘K." p A ' ‘ ANALYSIS: The rate equation, Eq. 11.14, can be written in the form - _ ‘ As = Q/U ATrm ‘ i i ' 3 (1) andf’rom Eq. 11.18, ' ' ' i. H AT1.— AT2 ATgm = F ATgm'GF Where I ATngCF = m . (2.3.) From an energy balance on the cold fluid, the heat rate is _ ‘ 'q= 1h, cp,c_(Tc,c,4Tc,i) = W x 4195'fi(120—35)K = 9.905x105 W . From an energy balance on the hot fluid, the outlet temperatureis 5000 kg' J ‘ . -—-—-—-—- 4660—— = 147 C . . . 3600 -s x, kg-K- - From Fig. 11.11, determine F from values 'of P and R Where'P =(120—35)°C /(300—35)°C =0.32, R=(300—-147)‘-°C /(120—35)°C =1;8 giving F~0.97. The log mean temperature based upon OF arrangement follows from Eq. (3); find Thin .= Th), _- q/Ihh cm}, = 300" C — 9.905x105 W/ 300—120) ' AT = .300—120 — 147—35 K 4? (———— = 3.3 . A,’= 9.90.5x105W/1500-W/m2‘K><0.97><143.3K =4.7.5m2 . _ . 0. <1 QOMMENTS: - (1) Check Thz500K used in: property ' _ determination; Th =(300+147) ° 0/2 =497K. (2) Using the NTU-E method, determine first the capacity rate ratio, Cm], /Cmax = 0.56 . Then ' , E: q = Cmax(Tc,o—Tc,i) = 1 X (120—35)°C =057 qmax Cmin (Th,i_Tc,i) 0'56 (300—35) a C ' I From Fig. 11.17, find that NTU = AU/Cmi,1 $31.1 giving As =4.7 m2. é _ ' PROBLEM 11.24 KNOWN: ConCentric tube heat exchanger operating in parallel flow (PF) conditions with a thin-walled ' separator tube of lOO-mm diameter; fluid conditions as specified in Problem 11.23. FIND: (a) Required length for the exchanger; (b) Convection coefficient for water flow, assumed to be fully developed; (0) Compute and plot the heat transfer rate, q, and fluid inlet temperatures, Tm, and Tm, as a function of the tube length for 60 s L S 400 m with the PF arrangement and overall coefficient (U = ZOOW/m2 -K), inlet temperatures (Th, = 225°C and Tc, = 30°C), and fluid flow rates from Problem 11.23; (d) Reduction in required length relative to the value found in part (a) if the exchanger were i ' operated in the counterflow (CF) arrangement; and (e) Compute and plot the effectiveness and fluid outlet temperatures as a function of tube length for 60 S L S 400 m for the CF arrangement of part (c). SCHEMATIC: ' u = 200 W/mz-K TM: 225 "C _ a ‘ T TN, -100 c a, /— \D‘ , A7,, “’ —» To): 30 °C Tm, = 50 °C (D ' ® ASSUMPTIONS: (1) No losses to surroundings, (2) Negligible kinetic and potential energy changes, (3) Separation tube has negligible thermal resistance, (4) Water flow is fully developed, (5) Constant properties, (6) Exhaust gas properties are those of atmospheric air. PROPERTIES: Table A-4, Hot fluid, Air (1 atm, T: (225 +100)°C /2 = 436 K): cP = 1019 J/kg-K; Table A-6, Cold fluid, Water T = (30 + sow: /2 = 328 K): p = 1/v, = 935.4 kg/mfi, 'c, = 4183 J/kg-K, k = ' 0.648 W/m-K, u = 505 x 10'6 N-‘s/ml, Pr ; 3.58. ' ANALYSIS: (a) From the-rate equation, Eq. 11.14, with A ; 1rDL‘, the length of the exchanger is L=q/U.nD-A'Tg,.pp- - . . " . (I) The heat rate follows from an energy balance oni'the cold fluid, using Eq, 11.7, find q = rhccc(Tm ~T,,) = 3'1cg/s >< 41831/kg- K(80— 30)K = 6275 x103W. Using an energy balance on the hot fluid, find r'nh for later use. I En, = q/ch (Tm — T“) = 627.5 >‘< 103 W/1019J/kgi K(225'—-100)K = 4.93 kg/s (2) For parallel flow, Eqs. 11.15 and 11.16, ' AT — AT (225— 30)°c —(1oo—so)°c, AT =——1—-—“"‘—-=——-—-—-———=763°CI “m” tnAT, /AT2 €n(225— 30)/(1oo— so) Substitutingnumerical values into Eq. (1), find ‘ I V L=6275x1o-‘w/200W/m2 -K(itx0.1m)76.8K=130rn. ' ' ‘ < . Continued... PROBLEM 11.24 (Cont) (b) Considering the water flow within the separator tube, from Eq. 8.6, ReD = 4th/1tDtL = 4 x 3kg/s/(1t X 0.1m x 505 X 10—5 N/s - m2) = 75,638. I Since Rep > 2300, the flow is turbulent and since flow is assumed to be fully developed, use the Dittus- ' 4 Boelter correlation .With n ; 0.4 for heating; ' ' ' ' NuD ; 0.023 Reg8 Pr“ = 0023054533)"-8 (358)"-4 = 306.4 h=NuD-g=3O6.4><O.648W/m--K/(0.1m)=1935w/m2,Kl ' ' < (c) Using the [Wm Concentric Tube.'Parczllel Flaw, Efi‘eciiveness relation, and'the Properties Tool for Water and Air, a model was developed for the PF arrangement. With U = 200 W/rnle and prescribed inlet temperatures, TM = 225°C and Tel = 30°C, the outlet temperatures, TI”, and _' Ten. and heat rate, q, werepcornputed as ‘a function of tube length L. Parallel llnw arrang'amanl . o aw P0300“ 115 90 65 Temperature (C) or Heal BIB (W‘10A-4) 4o . . V . BO ' 120 130 240 300 350 Tube length. L (m) ', --N- Cold outlet lemperalure.Tuo (C) -9—- Hot outlet tern erature. The (C) -—- Heat rate, q'1 (W) As the tube le‘ngth'increases, the outlet temperatures approach one another and eventually reach Tm = ‘ I H Tc‘g = SSIEOC-~ ' > ‘ . ‘ (d) If the exchanger as for part (a) is operated in T counterflow (rather than parallel flow), the log . mean temperature difference is , ' _ . p ' Th,/= 225 °C ‘ ' ‘ _ ATw, _ - 3;}: AT =——‘-——-—=— , , TI =80 °o £m.CF enATl /4T2 . t , ca (223—80)—(100'— 30) ATgmflF = “1(225 _.80)/100 - 3O = 103.0”C . 'Using Eq. (1). the required length is ' . ' L=627.5x103W/200W/m3-Kx1tx0.1mx103.OK=97m. The reduction in required length of relative to PF operation'is , - ' ‘ ' Continued... PROBLEM 11.24. (Cont) I? 0 \‘30. v - ‘ flea/er =(l195f97l/‘m=5£‘% ' < (e) Using the [H T Heat Exchanger Tool, Concentric Tube. Counterflow, Efiectiveness relation, and the ' Properties Tool for Water and Air, a model was developed for the CF arrangement. For the same conditions as part (c), but CF rather than PF, the effectiveness and fluid outlet temperatures were computed as a function of tube length L. ' ‘ J“ i :140 120 100 Countadlow arrangement Issesiiae...iiiuu so iliifllllllllllll \ e . _ to IIIIIIIIEE!!!IIII ~o)\l\‘ . 2° IIIIIIIIII ' 50 120 130 240 300 >\ I I . Tubs length. le) —>+— Gold outlet temperature. Ten to) --9— Hal outlet temperature. The (C) — Ettacllvanese.eps'1uu BO Tho, Tm (C) or eps‘mfl Note that as the length increases, the effectiveness tends toward unity, and the hot fluid outlet temperature tends toward Tu = 30°C. Remember the heat rate for an infinitely long CF heat exchanger is qm and the minimum fluid (hot in our case) experiences the temperature change, TM -‘Tc_i. COMMENTS: (1) As anticipated, the required length for CF operations was less than for PF operation. (2) Note that U is. substantially less than hi implying that the‘gas-side coefficient must be the controlling thermal resistance. r ~ ' ' ' t, - ’ ...
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This note was uploaded on 06/06/2008 for the course ME 114 taught by Professor Okamoto during the Spring '08 term at San Jose State University .

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HW 9 solution ME 114 S08 - \ I . ' PROBLEM92 chl...

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