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practice - PROBLEM 13.94 KNOWN Diameter temperature and...

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Unformatted text preview: PROBLEM 13.94 KNOWN: Diameter, temperature and emissivity of boiler tube. Thermal conductivity and emissivity of ash deposit. Convection coefl'tcient and temperature of gas flow over the tube. Temperature of surroundings. PM: [a] Rate of heat transfer to tnbe without ash deposit, (b) Rate of heat transfer with an ash deposit of diameter Dd = 0.05 m, (c) Effect of deposit diameter and convection coefficient: on heat rate and comributions due to convection and radiation. SCHILNIAHC : Ash deposit. It *1 Wlm-K o,=o.05m Tt=600 K 8‘ -U.B ASSUMPTIONS: (1}Difi’uset'gray surface behavior, (2) Surroundings form a large enclosure about the tube and ma}.r be approximated as a blackbody, (3) One-dimensional conduction in ash, {4) Steady—state. ANALYSIS: (a) Without an ash deposit, the heat rate per unit tube length may be calculated directly. (1’ = ETDt (Too ‘Tt )+ EtWDt (Tails — T?) 4: 4 {=chwring.1:(at-)oosm(1300—500)K+n.3x5.67x10 wrm2.1c4(x)(o.osm)(lsno _6m4)1c q’=(1asso+35,150)wtm=54,om Wa’m < (b) Performing an energyI balance for a control surface about the outer surface ofthe ash deposit, GLmv +qiad = '1an- 01 = Hkfia 43:) 111(Dd IfDr) Hence, canoehngtt and considering an ash deposit for which Dd: [1.06 m, lDDWI'mz {(0.015 m)(180t]—Td)K+0.9x5.fi?x10—3me2 -K4 (0.06m)(lsoo“ —T§')K"' 3313:1000 -Tal+€dWDd (Tails —Ti) _ 2(1 WIm-K](Td—6QD)K Mommas) A trialanderror sohition yields Tdsu 1346 K, fi'omwhich it follows that q’ 2 Eerd (Tm —Td)+sdon'Dd (1;:1r 45‘] _g q’=1:}owsm2.K(:r)o.usm(lsoo—1345)K+ t}.9x5.fi?xll} th2 (34000364151104 —13464]K4 PROBLEM 13.94 (Cont) =(856D+1?,I4D)Wa’m225,?00an1 < (c) The foregoing energy balance 1was entered into the JET workspace andparametrie calculations were performed to explore the effects of _h and Dd on the heat rates. Hill I“. WWI emu-m nun-1mm +Gflll|n|hn +Wflm +Mu1¢n Fode= 0.06 mend 10 Eh E1000 mez -K, theheat rate to thetuhe, qgomi, as well as the contribution due to convection, 11;)an increase with increasing h. However, because the outer sudace temperature Td also increasa with h, the contribution due to radiation decreases and honours negative {heattransfer fromthe surface] wheanexceeds ISODKat h=54fl Win212 -K. Boththecomrection and radiation heat rates and hence the conduction heat: rate increase 1with decreasing Dd, as Td decreases andapproachesTt=lSDOK However,evenfode=fl_051n1[adepositt]:|icknmsoffl.5mrn],Td=W3 Kandflieashprmridesasignificantresistancetaoheat transfer. Cflhfl'rIENTS: Boiler operation in an energy efficient manner dictates that ash deposits be minimized. PRUBLEIH 13.96 KND‘WN: Dimensions= enJissivities and taeniperamres ef heated and cured surfaces at opposite ends of a cylindrical cavity. External conditions. KIND: Required heater power and Outside convection coefficient. SCHIL'IATIC: ASSUWTIDN‘S: (l) Steady-state cnnditions, (2] Opaque, I:l.i.‘Efi.1.se-gra},r surfaces, (3) Negligible cenvectien within cavity, {4) Isothermal disk and heater surfaca, [5) One—dimensional conductien in base. (6) Negligible centact resistance between heater and base, [Tr'] Sidewall is reradiating. ANALYSIS: The equivalent circuit is ELI—I ELI-ad .5. £5 a“, ‘31 day. lql,eand 2E fill] 93'; From an energyr balance on the heater surface, 111,313.: = ‘11,ch + (uni. q1=elec=kb (ND214)T1_TIJ+ “(T 4'24] 1—9 1 1-9 ' Lb 1 I + 2,! s A 1 s -A 1 1 Allis +[(UA1F1R )+(“A2ER )l— 2-1 3 when: 1311:1512 = @214 21:13. 12m]21'4 = 0.0113 in2 and fiern Fig 13.5, with 14:1 2 3.33 31:11ch 2 0.3 find F12 = F21 = 0.0??; hence, FIR = FQR = 0.923. The required heater power is sun—3:111 K 111 m :20 Wim-Kxomu “12% = 0.025 m 11.11113 m2 xsnhm‘swifm?‘ -K4 (31104 —4004)K4 + 1—0.9+ 1 +1415 0-9 0.1m +[|(1i11.923)+(1iu:1.323)]‘1 0-5 gum = 4521 W+32.9 w = 4604 w. '31 Anenergybalanee ferthe disk yields, .1de 2 gm” 21101112 (r2 —1;,.,)+le1120 (1’; 4;“). 32.9 w—o.9xo.n113 n12 xiii? x10_sWi"n12 -K‘* (4110‘ —3oo“)1<:“ h =—=54w1m2.1{ s: D 2 0.0113 n1 x100 K COMMENTS: Conduction flireugh the ceramic base represents an enerrneus system less. The base shnuld be insulated to greatly reduce this less and hence the electric pewer input. PROBLEM 13.5"?r KND‘WN: Electrical conductors in the form of parallel plates having one edge mounted to a ceramic insulated base. Plates exposed to large, isothermal surroundings, I'm. Operating temperature is T] = 500 K. KIND: (a) Electrical power dissipated in a conductor plate per unit length, qi, considering onlyr radiative exchangewith the sun'oundings; temperature ofthe ceramic insulatedbase T2; and, (h) qi ande when the surfaca experience CDflVBCliOflWifll anairstream at TDD: 300 K and a convection ceeflicient ofh = 24 me2-K SCHIJHATIC: .. . 1.: 5515'.) 'I "Fl-b1 'II" ‘ Iflqfiégfi ASSMIDNS: (1} Conductor surfaces are difiuse, gray, {2] Conductor and ceramic insulated base surfaces have uniform temperatura and radiositiesa (3) Surrcmndings are large, isothermal ANALYSIS: (a) Define the opening between the conductivities as the hypothetical area A3 at the temperature of the surroundings, T5“, with an emissivity £3 = 1 since all the radiation incident on the area will be absorbed. The conductor [l)—base (Zr-opening (3) form a three surface enclosure with one surface re—radiating (2). From Section 13.3.5 and Eq. 13.30, the net radiation leaving the conductorsurfaceAl is _¢ “1 ' 1—21 1 1—23 (I) +—+ l E11511 A1fi3+[[lfA1fi2)+(lfA3E32)]— 93133 where Ebl =ID'T14 and Eb] =D'T34. The View factors are evaluated as follows: F32: usethe relation for two aligned parallel rectangles, Table 13.2 orFig. 13.4, i=XIL=wIL=10MO=025 ‘?=YfL=oo F32 =o_1231 F13: appljringreciprocitybetweenAlandfig,whereA1=2Lf =2X0.040mf =D.DSD f andA3= wf =D.Dlfl f andf isthe length ofthe conductors nonnaltaothepage, f 3i5Lorw, A r1 = 3E1=OfllflfxflflTI'fiWOflSDE=0_1t}96 3 A1 where F31 can be obtainedbyusingthe summation rule onAg, F3121—F32 =l—U-lZ31=D.ETf-9 F12: bysymetry F12=F13=UJU§6 PROBLEM 13.97 (Cont) Substimting numerical values into Eq. (1], the net radiation leaving the conductor is 5.57x10‘3wm2-K4[5m4 400“)? [11: 1—0.3 1 —+—+n “SW-”30'? uosoix0.1096+[(1:0.0309xo_1096)+(1mnlugxuizsfl’l , 3544—4593 w q1=q1H= ( ) 2295me < 3.125l]+101.55?+fl [b] Consider now convection ocesses Declining at the conductor (1) and base (2) smfaces, and perform energy.r balances as fliiistrated in the schematic below. Ebi—Jl (1—91)JIIEIA1 and the radiosity J1 can be determined fi'omthe radiation energy balance, Eq. 13.21, Ehl—Il _ 11-h + J1-13 Qin=ch,l+q.l=hA1(Tl_Tm)+ {2) _ 3 (l—EIFEIAI “Alt—12 “Alt—13 0 where 13 =E1,3 =aT34 since A3 is black. Euphxs 2 : Since the sin-face is insulath (adiabatic), the energy balance 1135 the form EM 42 — 4 l—Egi'rEEAZ [) D: (lei-32 +512 2 1152 (T2 ‘TM)+ and the radiositjr J2 can be determined fromthe radiation energy balance, Eq. 13.21, . 12 43 (1—92)i92A2 1:11.2le 1:11.253 There are 4 equations, Eqs. (2—5), with 4 mknowns: J2, 12, T2 and q_1_ Substimting mimetical values, the simultaneous solution to the set yields EM 42 _ I2 41 {5) 11:3411r Win12 12 =1745 Win12 T2 =352 K qgn =441wm < COMIENTS: $1.3) The effiect of convection is substantial, increasing the heat removal rate Rom 29.5 W to 441 W for combined modes. :2) With the convection process, the current carrying capaci of the cenchictaors can be increased. ther advanta is that, 1With the presence ofconvection, e ceramic base operates at a cooler teniperauire: 35 K vs. 483 K. PROBLEM 13.106 KNOWN: Ifimmsirms and surface conditions of a cylindrical thermos bottle filled with hot coffee and lying horizontally. FIND: Heat loss. SCHEL'IATIC: . D900?» ’- 51=a25, 3175?: Q=aagm Airspace (pzfaf'm), L = 00053:! ASSIMTIDNS: (l) Steady-state conditions, [2) Negligible heat loss from ends (long infinite cylinders) (3) Ififfiise—gray sufiee behavior. PROPERTIES: “1" *5 ”H, Air (If: (T1 + mo 2 323 K, 1 film): 1: = 0.0234 Wim-K, v = 23.?4 x 1045 m2ts, u = 26.0 x 1045 11120, Pr = 0.0?033 B = 3.05 x 10‘3 1:4. ANALYSIS: The heat transfer across the air space is '1 2 Quad +‘10onv- From Eq. 13.25 for concentric cylinders o I(:rI::1L)(1‘fL — 13') seawa—g storm2 -1r“:r(0.0?x0.3)m2 (3434 — 3034 )K“ I310M! i+l—sz i _ 4+3(0.035I0.04) E1 92 f2 qml = 3.20 w. FromEq.9.25, 2 —3 —l 3 REL =gfl(T1—T2)L3 =9.3n1.l's [3.05x10 K )(40 K)(0.005 m) :36]- W 26.6XltII—6n12 J'sx 23.?4x1tT6 m2 is Heneefi'omEq.9.60 In D t Ra ln 0.0mm? 23-5.? R; =W= [ ( ”4 =;_35_ '5 5 5 L3 (DI—0'6 + DEM] (0.005 m)3 (0.0?“}-5 +0.03‘”) m‘3 However, the implication of such a small value of Ra: is that free convection eEects are negligible. Heat transfer across the airspace is therefore by conduction (km: 1:). From Eq. 3.2? mum —T2) _ 2::x03 mxt].t]234 W!m-K(?S—35)K =16.04 w. ln(r2 £11) ln(0.04!0.035) gonad: Hence the total heat loss is q=qnd+qmd=1924 w. < COLBEYIS: (1) End effects could be considered in a more detailed analysis, (2] Conthetion losses could be eliminated by evacuating the annulus. PEBBLE!“ 13.116 HOW: Diameter and surface temperature ofa fire tube. Gas low rate and temperature. Emissivity oftnbe and partition. FIND: (a) Heat transfer per unit tube length, (1’, without: the partition, (b) Partition temperature. TP, and heat rate with the partition, (c) Effect of flow rate and emissivity on (1’ and T1,. Effect: of emissivity on radiative and convective connibutions to q’. SCHELiATIC: I 900 K Tgm. mg=0..02 0.05. 0.03MB . Partition. T”. up = s, o = 0.01r m —|-—-| ASSURIPTIDNS: {l} Fully—developed flow in duct, {2] Ififi'usefgray surface behavior, (3) Negligible gas radiation. PROPERTIES: “1“" H nrmeg=900K)z p: 300x 10 710-0012 1;: 0.002 me-K Pr: 0. u; airfI's: 335K): u: 224x10 lN-.s.-’m2 ANALYSIS: (a) Without the partition, heat transfer to the tube wall is only by convection. With 11': = 0.05 kgf's and RBI}: 4 ring mm = 4 (0.05 kgrs)nr(0.0? m)300x10‘7 N-sfmz = 22,350, the flowis turbulent. FromEq. {3. 61], Nun: 0.033%” ’“3 (mm-14 =0.027(22,350)“5(032)1’3(393I224)0-14 =305 D 0.0".r m '12 -Tmh“'D( m.g _Ts): T1.3me2-K{:r)0.0? m(900—385)=BU?5 me <2 (b) The temperatne ofthe partition is determined fiom an energy balance which equates net radiation exchange with the tube wallto convection Eromthe gas. Hence, qiarl = (1:01“, where from Eq. 13.23, 0 4—T“ and——_(Tlp a) _1 A? + — +1 81:» 1:115 ea As whereF13=landApng=DfflmDfljl= 2hr: H.611? 'I'beflowisnowinanonurcflarductforwhich Db: 406m: 4(0D2f8}f[11:D0'2+D)= 0mm”): II}.=611D 0.0423n1anflfilly2=nngf2=fl.025 kga’s. Hence, Reg = n'il .9 2 thficu = In“ 2 Dy‘(nD2fB)p. = M0025 kg’s) (0.0428 n1}fn(0.0? In)2 398 x 10"“r 10-5er = 13,500 and Nun = 0.-::r27{13,000)‘5H 5 (0.02)“ 3 (393t224)fl'” =54} 1: 0.002 W! -K h=—NuD =—ms4.3=?0.? ani2 -K nh 0.0423 In PROBLEI’VI 13.116 (Cunt) Hencebwithss= Sp: 0.5mdqmv=h(T mg —Tp), 5.6?XID_8me2 -K4 (I; —3354)K4 =13;- Win12 -K(9fl]D—TP]K 1+1+n_:533r memo—gr; +mrrl, 41,302 = o A trial-and—error solution yields TP 2 T96 K <3: The heatrate tnone—halfofthe tube is then Do (T; 45“] '1le {liae +qconv =—+h(flD’r2)(TnLg _T‘S-] 1-81: +L 1—25 A1: 813 FPS £5 AS 0.0? n1(5.6?xlD_SWEm2 ~K4)(?96_ 4 4—335“)K4 2 git: =—+ 73.7 Wa'm -K((I_llfl m)(9m— 335)K 2.537 Ilia =5?2 me+4458 me=5030 Wfrn 'I'heheatrate formeentirembeis q'=2q’m=1o,osn warm {1 [e] The foregonig nndel was entered into the In T workspace, and pmnrtic calculations were perfumed to obtain the following results. =_---- -='--- I"'.--:---- - ---- -—-_g== __---- 0.1 I12 0.: 0.4 0.5 M M M 0.! cm I12 0.: an on M 41.7 0.3 as Emmew-m EM.W-Im ins-gem :m-s-sm +mamiaioem +m:uiua$ Radiation transfer from the partition increases with increasing SP = Es, ll:uereb}r reducing T13 while increasing (1’. Since hinereases with increasing in, T1] and q’ alsoincrease with n'1 PROBLEI‘UI 13.116 {Cunt} Ham ma. q'i'u'u'll'ml Although the radiative cmtrihution tn the heat rate increases with increasing SP = 25, it still remains small relative to convectiun. COLEEWS: Contrasting the heat rate mediewd fur part (b) with that for part (a): it is clear that use ufthepsrtitiunenhanoes heattransfermflie tube. Hflwever, the efi'eet is dueprimarily to an increase in h and secondarily to the addition of radiatim. ...
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