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extraproblemsexam2 - PROBLEM 5.24 KNOWN: Diameter and...

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Unformatted text preview: PROBLEM 5.24 KNOWN: Diameter and thermophysical properties of alumina particles. Convection conditions associated with a two—step heating process. FIND: (a) 'l‘inte—in—flight (tn) required for complete melting. (b) Validity ofassuming negligible radiation. SCHEMATIC: Al203 sphere, Dp : 50 um, 7}: 300 K, kp=ill5 th-K. p = 39m kgrm3, cp = 1560 thg-K, = 2318 K, hsf= 3577 thkg mp n = 3x104 wrmZ-K "\ rm: 10,000 K q-bonv ASSUMPTIONS: {1) Particle behaves as a lumped capacitance, (2) Negligible radiation, (3) Constant properties. ANALYSIS: (a) The two-step process involves (i) the time t1 to heat the particle to its melting point and (ii) the time 1: required to achieve complete melting. Hence, ti_f = t. + t;, where from Eq. {5.5). :ppvcp 91 pprCp 1n Ti-Too t1 In — = _ hAS 9 6h TIn —TDC, 3970 kg/m3 (50x10_6m)156OJ/kg - K (300_10 000) t1=————2--— n—’—= 4X10_4S 6(30,000w/m -K) (231340-000) Performng an energy balance for the second step. we obtain t1+t2 Ll (Iconvdt = AEst where qcom = hA,(T..., - Tmp) and ABS, 2 ppVhsf. Hence, ,, 3 —6 £22,0pr hsf :3970kg/m (50x10 m)x 3'577X106J/kg =5X104S 6h (Toe—Tmp) 6(30.000w/m3-K) (10,000*2318)K Hence Ii_f :9X10_4s==lms < (b) Contrasting the smallest value of the convection heat flux, quw‘mm = 11 (TM —Tmp ) : 2.3x108 W/m3 to the largest radiation flux. qgfld’max = 50' ('lfim —T:,r) = 6.5 x 105 W/mz. we conclude that radiation is. in fact, negligible. COMMENTS: [1) Since Bi = [hrp/3)fk : 0.05, the lumped capacitance assumption is good. (2) In an actual application. the droplet should impact the substrate in a superheated condition (T )- T,,,,,), which would require a slightly larger t-Hc. PROBLEM 5.71 KN OWN: Asphalt pavement, initially at 50°C, is suddenly exposed to a rainstorm reducing the surface temperature to 20D C. . 2 . . FIND: Total amount 01 energy removed (Jan ) from the pavement for a 30 minute period. SC HEM AT] C: ASSUMPTIONS: ( I) Asphalt pavement can be treated as a semi-infinite solid, (2) Effect of rainstorm is to suddenly reduce the surface temperature to 20°C and is maintained at that level for the period of interest. PROPERTIES: Table 21-3, Asphalt (300K): [3 = 2115 kg/m3, e = 920 J/kg-K, k = 0.062 W/m-K. ANALYSIS: This solution corresponds to Case 1, Figure 5.7, and the surface heat flux is given by Eq. 5.58 as I a lf2 QS(t):k(Ts_Ti)’x(flal) (I) The energy into the pavement over a period of time is the integral of the surface heat flux expressed as Q”: l; q; [t)dt. (2) Note that q: (t) is into the solid and, hence, Q represents energy into the solid, Substituting Eq. (I) for q; (t) into Eq. (2) and integrating find ,. 1/2‘4/2 k " Q=k(Ts—Ti]/(:im) lot dt*—(—“§—><2 (3) Substituting numerical values into Eq. (3) with a_i_ 0.062 W/m-K .0 c 2115 kgrm3 x920 J/kg- K find that for the 30 minute period, A, 0.062 W/m-K [20—50)K Q _ 1:2 :3.isx10‘3m2/s x2(30x605)“2 2&4.99Xl05 erz. < (ax3.18X10'8m2/s] COMMENTS: Note that the sign for Q” is negative implying that energy is removed from the solid. PROBLEM 6.23 KNOWN: Shaft of diameter 100 mm rotating at 9000 rpm in a journal bearing of 70 mm length. Uniform gap of 1 mm separates the shaft and bearing filled with lubricant. Outer surface of bearing is water—cooled and maintained at Twe = 30°C. FIND: (a) Viscous dissipation in the lubricant. u¢(W/m'l}, (b) Heat transfer rate from the lubricant, assuming no heat lost through the shaft, and (c) Temperatures of the bearing and shaft, Tb and "1",. SCHEMATIC: Wmm) Bearing Bearing. Rb = 45 WIm-K l= ?0 mm .- Lubricant, u = 0.03 N-srm2 ‘ k = 0.15 me-K fibfl‘m mm Water-cooled surface : DC hf: 9000 rpm fl [)0 = 200 mm i.— Two 30 // ASSUMPTIONS: (l) Steady-state conditions, (2) Fully developed Couette flow. (3) lneompressible fluid with constant properties. and (4) Negligible heat lost through the shaft. ANALYSIS: (a) The viscous dissipation, uCD, Eq. 6.40, for Couette flow from Example 6.4, is 2 2 2 4 . 1‘ maze 93 =u E 20.03N-s/m2 13E :6.656><107 w/m3 < dy L 0.00m where the velocity distribution is linear and the tangential velocity of the shaft is U : JIDN = r: (0.100m)x9000rpm><(min/60s) = 47.1m/s. (b) The heat transfer rate from the lubricant volume V through the bearing is q zncp-v = gown-Lt): 6.65x107 W/m3 (Hx0.100m><0.001mx0.070m): 1462W < where E: 70 mm is the length of the bearing normal to the page. PROBLEM 6.23 (Cont) (c) From Fourier‘s law, the heat rate through the bearing material of inner and outer diameters, D,- and DD, and thermal conductivity kl, is, from Eq. (3.27), _ Mikb (Tb rTwc) qr [n (Do/Di) 1 D “D- szTwc+Qr “( o/ l) Zflfkb 1462Wl 200100 Tb=30°C+ n( / ) *81.2°C < 2rr><0.070m><45 W/m'K To determine the temperature of the shaft. T(0) = T5, first the temperature distribution must be found beginning with the general solution, Example 6.4. U 2 in 2 T =—fi— C +c (y) 2k[L]y+3y 4 The boundary conditions are. at y = 0. the surface is adiabatic —] =0 C3=0 dy 3:0 and at y = L, the temperature is that of the bearing, Th , IuU"3 M 2 TL=T =—-—— L+0+C C=T+—U H b 2K{L] 4 4 b 2k Hence. the temperature distribution is and the temperature at the shaft‘ y = 0, is 0.03N-s/m2 ~———(47.lm/s)2 =303°c < 2x0.15W/m-K TS =T(0):Tb Jrfiuz :81.3°C+ PROBLEM 6.32 KNOWN: Variation of hx with x for flow over a flat plate. FIND: Ratio of average Nusselt number for the entire plate to the local Nusselt number at x : L. SCI-lEMA’l‘IC: l/’—7719r'ma/ boundary Iayer; bx: Cx- [a where |—’x L. C is a consf'anf' ANALYSIS: The expressions for the local and average Nusselt numbers are -l/2 hLL _ (CL )L _ CLU2 NuL=——— EkL k k m =_t where , L L h] =1] hx dx=EJ x‘l’2dngLl’222 (31:19. ‘ L 0 L 0 L Hence, _ 2CL‘U2 (L) 2CL“2 NUL :f: k and NuL _2 < NUL — . COMMENTS: Note the manner in which EL is defined in terms of FL. Also note that Nu l L T L :EJU 1\uX dx. PROBLEM 7.23 KNOWN: Prevailing wind with prescribed speed blows past ten window panels. each of 17m length, on a penthouse tower. FIND: (a) Average convection coefficient for the first, third and tenth window panels when the wind speed is 5 Ms; evaluate thennophysical properties at 300 K, but determine suitability when ambient air temperature is in the range -15 5 Ta, 3 38°C; (b) Compute and plot the average coefficients for the same panels with wind speeds for the range 5 S um S 100 km/h; explain features and relative magnitudes. SCHEMATIC: Window panel WEI/f 1 m length in x-direetion ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties. (3) Wind over panels approximates parallel flow over a smooth flat plate, and (4) Transition Reynolds number is Re“ = 5 x 105. PROPERTIES: Table 11.4, Air m: 300 K. 1 alm): v = 15.89 x105 nil/s, k = 20.3 x 10'3 W/m~K, Pr = 0.707. ANALYSIS: (a) The average convection coefficients for the first, third and tenth panels are — q E X —E X3 - E X -H X hl h2_3 :_fl_2_~ hEHO =fl_10__9_9 (1.2.3) X3—X2 Klei—K9 where h2 = hz (212), etc. If ReM = 5 X 105, with properties evaluated at '1}: 300 K, transition occurs at —6 2 5. ,c :LRBX C =MX5X105 =1,ng um ‘ Sm/s The flow over the first panel is laminar, and hi can be estimated using Eq. (7.31). 313,1 = Ll)” : 0.664 Re“2 PrU3 k X - 11'2 111=(0.0203w/m-Kx0.ss4/1m)(5mli’sx1m/1ss9x10‘6nil/s) (0.707)U3=8.73W/m2‘l( < The flow over the third and tenth panels is mixed, and hz , h3, h9 and hm can be estimated using Eq. (7.41). For the third panel with x3 = 3 m and x3 : 2 m, N—ufl : “3"3 = 0.037 Rem—s71 PrU3 k X H3 = (0.0263W/m- K/sm) 4/5 x 0.037(Sm/sxsm/issgx10—6mz/s) —871](0.707)“3=10.6w/m2-1< PROBLEM 7.23 (Cont.) fig 2 (0.0263 W/m-K/zm) 4/5 x 0.037 5m"'s><2m 15.89><10‘6 mzfs e871 0.7071’32368w m2.K .4 J" From Eq. (2). 2 o — 10.6iw/m -Kx3m—s.68w/m~-Kx2m h2—3 = —"_*—r——-—(3 2) — m Following the same procedure for the tenth panel. find hm = 11.64 mez‘K and hg = 11_71 W/mz-K and :14.5W/m2-K < h9_10:ll.lW/m2-K < Assuming that the window pane] temperature will always be close to room temperature. TS = 23°C = 296 K. If Tm ranges from —15 to 38°C. the film temperature, '1}: {TS + TW)/2, will vary from 275 to 310 K. We'll explore the effect of 'I‘t- subsequently. (b) Using the_IHT Tool. Correlations. External Flow, Flat Plate, results were obtained for the average coefficients h . Using Eqs. (2) and (3), average coefficients for the panels as a function of wind speed were computed and plotted. Average coeflicie nl. hbar {WIIW‘ZKJ Wind speed. uinl (kmth) *7 First panel —K— Third panel —fi-- Tenth panel COMMENTS: (1) The behavior of the panel average coefficients as a function of wind speed can be explained from the behavior of the local coefficient as a function of distance for difference velocities as plotted below. Local coefliciem. hx [W-im’QM) Distance 1mm leading edge. x [m] -"-‘— uinl : 5 Math + Uinf = 15 km‘h + uinl -. 25 km‘h + ulnl 7. 50 krwh Continued... PROBLEM 7.23 (Cont.) For low wind speeds, transition occurs near the mid—panel, making El and h9_10 nearly equal and very high because of leading-edge and turbulence effects. respectively. As the wind speed increases, transition occurs closer to the leading edge. Notice how h2_3 increases rather abruptly, subsequently becoming greater than h9_10 . The abrupt increase in h, around 30 km/h is a consequence oftransition occurring with x < lm. (2) Using the IHT code developed for the foregoing analysis with new : 5 m/s. the effect of T. is tabulated below MK) 275 300 310 El (“z/mam 8.72 8.73 8.70 h2_3 (W/mZK) 15.1 14.5 143 H1340 (W/mE-K) 11.6 11.1 10.8 The overall effect of '1} on estimates for the average panel coefficient is slight. less than 5%. PROBLEM 7.54 KNOWN: Diameter and length of a copper rod, with fixed end temperatures, inserted in an airstream of prescribed veloctty and temperature. FIND: (a) Midplane temperature of rod. (b) Rate of heat transfer from the rod. SCHEMATIC: 7;: 90°C ASSUMPTIONS: (1) Stead -state conditions, (2) One—dimensional conduction in rod, (3) Negligible contact resrstance, (4) Negllgi le radiatlon, (5) Constant properties. PROPERTIES: Table A—!. Copper (T = 80C = 353 K): k = 398 me-K; Tobie A-4, Air (Toc = -6 2 25°C : 300K, 1 atm): v = 15.8 x 10 m /s, k = 0.0263 W/mK, Pr = 0.707; Table A-4, Air (T5 : 80°C z 350K, 1 atm): Prs = 0.700. coshm(Lix) _ TiTw ANALYSIS: (a) For case B of Table 3.4. i = fl 6}, cosh(rnL) Tb -— Tm where HQ in : (FPr'kAC) : (rtE’kD)1 “i2. Using the Zhukauskas correlation with n = 0.37, NuD =C Reg,-l Prn (131713101!4 VD ZSm/s 0.01m ReD=-—:~——#=15,823 V 15.8x10‘6 mz/s and C = 026,111 2 0.6 from Table 7—4. Hence MD : 0.26(15,823)fl'6 (0.707)“37 (0707:0700)‘ "'4 = 75.8 _ k — _ 0.0263 wrm.K D D 0.01m 1 4x199W/m2-K J 75.8) =199 wrm2 -1< :' 2 — =14.2 mi]. 398 Wr'm‘ KXUD] m T L —T ‘ -l 0 LL31: L03“ ) =_li=0.79 Hence, — Tb ‘Tco cosh(l4.2 m'I X005 m) 1'26 T(t.)=25°C+0.79(90—25) =76.6°C. < {b} From liq. 3.?6, q = qu = 2M tanh rnL, 112 M: HPkA ( c) m-,K m-K 4 M =20? w q =2(28.7 W) tanh (142111" x0.05m)=35 w. < If? 0b=[l99 2’" (“001111) [398 w F4001me 65°C COMMENTS: Note adiabatic condition associated with symmetry about midplane. ...
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extraproblemsexam2 - PROBLEM 5.24 KNOWN: Diameter and...

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