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# assignchap7-47-87 - PROBLEM 7.47 KNOWN Dimensions ot‘chip...

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Unformatted text preview: PROBLEM 7.47 KNOWN: Dimensions ot‘chip and pin ﬁn. Chip temperature. Free stream velocity and temperature of air coolant. FIND: (:1) Average pin convection coefﬁcient. (b) Pin heat transfer rate, (c) Total heat rate. (d) Effect of velocity and pin diameter on totaI heat rate. SCHEMATIC: 2 ‘5 D S 4 mm 105Vs40rnf5 ! L:12mm no: 300 K —» W:4mm Tb=350K ASSUMPTIONS: (1) Steady—state conditions, (2) OneAdimcnsional conduction in pin, (3) Constant properties. (4] Convection coefﬁcients on pin surface [tip and side) and chip surface correspond to single cylinder in cross flow, (5) Negligible radiation. PROPERTIES: Table .41.], Copper (350 K): k = 399 W/m-K; Table .44. Air (Tr :325 K. 1 atm): V = 18.4l X 10’0 mzfs. k = 0.0282 W/m-K, P1" = 0.704. ANALYSIS: (:1) W’ith V 2 10 m/S and D r 0.002 m. VD i 10rn/s x0002 m v 18.41x10_61n2/!s ReD e 71087 Using the Churchill and Bernstein correlations, liq. (7.57). 1/2 in 5/8 — (HER I" R NuD 203+ ) CD I 1+ ED :l6.7 ,. 7p, “4 282.000 [1+(0.4...s-Pr)“ F : (NuDk/o): (16.7x0.0282 wf-“nr K/ooozm) = 235 w/m2 -K < (h) For the ﬁn with tip convection and — 3 U2 r, 1 ~02 M:(herk7rD“/4] 0h=(n/2)[235w/m—-1<(0.002m)-399W/m-KJ 50K:2.]5W — 1r: 7 .. 1x2 _1 ni=(hP/kAC) =(4xzsswmvn 39‘)Wl,.-’m-KXOUOEm) :34.3m mL 2 34.3m_l(0.012m)=0.41’2 — a 71 . (h/mk ) 2 (-35 w,- m“ ‘ K/34.3 m x399 w,»- m v K) x 0.0172. The fin heat rate is sinh mL +[hl,.-""111lt)cosh mL qt cosh m].+(h;"mk)sinh nlL =0sosw. < PROBLEM 7.47 (Cunt) (c) The total heat rate is that from the base and through the fin, q =qh +qf = H(w2 —noz/4)9b +qf = (0.151+0.868)W =1.019W. < (d) Using the IHT Extended Surface Model for a Pin Fin with the Correlations Tool Pad for :1 Cylinder in crossflow and Properties Tool Pad for Air. the following results were generated. 2 i 8 E E E? 16 - 9 iii 5 h H E ‘5 r.) g 1 4 -- AT:1 R .L 6 E ._ 12 - 1 _ _ 10 20 30 40 Pm diameter. Dtrnm'; Freestream velocity, V[n'L-‘s] + V -. 10 WE. iDugmm PV=40W5 Clearly, there is significant benefit associated with increasing V which increases the convection coefficient and the total heat rate. Although the convection coefficient decreases with increasing D, the increase in the total heat transfer surface area is sufficient to yield an increase in q with increasing I) The maximum heat rate is t] = 2.77 W for V = 40 nifs and D : 4 mm. COMMENTS: Radiation effects should be negligible although tip and base convection coefficients will differ from those calculated in parts (a) and (d). PROBLEM 7.87 KNOWN: An air duct heater consists of an aligned arrangement of electrical heating elements with S._ : ST = 24 mm. NL = 3 and NT 2 4. Atmospheric air with an upstream velocity of 12 m/s and temperature of 25°C moves in cross flow over the elements with a diameter of 12 mm and length of 250 mm maintained at a surface temperature of 350°C. FIND: (a) The total heat transfer to the air and the temperature ofthe air leaving the duct heater, (b) The pressure drop across the element bank and the fan power requirement, (c) Compare the average convection coefficient obtained in part (a) with the value for an isolated (single) element; explain the relative difference between the results; (d) What effect would increasing the longitudinal and transverse pitches to 30 mm have on the exit temperature of the air, the total heat rate. and the pressure drop? SCHEMATIC: 87:24 mm 6— Heating elements ;' """"""" " ; Ts=350°C N=NLxNT=12 "+4500 2 D=12mm NL=3 T50 Q oi L=250mm NT=4 Ti=25°C —i> a 2 ; 2 “ﬁr‘b' V212mi‘s O O 0: To —" O O O: —i> ASSUMPTIONS: [1) Steady—state conditions, (2) Negligible radiation effects, (3] Negligible effect of change in air temperature across tube bank on air properties. PROPERTIES: Table 7174. Air (T, = 298. 1 atm ): p = 1.171 kgfm], cp = 1007 J/kg-K; Air (Tm = (T, + T,,)i’2 = 309 K, 1 atm): p = 1.130 kg/m'i, c1, : 10071/kg~K,tt = 1.89 x 10'5 N-sfml. k = 0.021399 WitnK, Pr = 0.7057:.Air(1‘5 = 633 K, 1 atm): Prs =0.687: Air (Tf = (T; + T,,)/2 = 461 K, 1 atm): v = 3.373 X 10'5 mzfs, k = 0.03801 me-K, Pr = 0.686. ANALYSIS: (a) The total heat transfer to the air is determined from the rate equation, Eq. 7.71. q I N (EDJTD A’I‘hn) (l) where the log mean temperature difference. Eq. 7.09, is ‘ _ - T. —T— ATrm : T5 Tl/gm ( .5. . J). (2) Ts _ T0 (Ts _ To) and from the overall energy balance. Eq. 7.70. T -T. DNli s o 2 exp 7? D (3) TS H Ti pVNTSTcp The properties p and CD in EL}. (3) are evaluated at the inlet temperature '1}. The average convection coefficient using the Zhukaukus correlation, Eq. 7.67 and 7.68, where C : 0.27, m = 0.63 are determined from Table 7.7 for the aligned configuration with STISL : l > T 5 . . . 0.7 and 10‘ < Ream,“ S 10 . All properties except Pr5 are evaluated at the anthmctlc mean temperature Tm : (T, + TONE. The maximum Reynolds number. Eq. 7.62. is PROBLEM 7.87 (Cont.) Rchnax : toll/[1151,3tl3l H (5) where for the aligned arrangement. the maximum velocity occurs at the transverse plane, Eq. 7.65. ST V . : V (6) max ST _ D The results of the analyses for ST 2 SL : 24 mm are tabulated below. Vmax RCDJ‘HIIX N711 D hD AT? m qu To (mils) (WimzK) ( C) l ) (DC) 24 1,723x104 96.2 216 314 7671 47.6 < (b) The pressure drop across the tube bundle follows from Eq. 7.72, AP:NLZ(9V§1aX/2)f {7) where the friction factor, f. and correction factor, x, are determined from Fig. 7.13 using ReD‘nm : 1.723 x 104, f = 0.2 x = 1 Substituting numerical values. ap:3xi[t.171kg/m3x(24 m/s)2/2}<0.2 A132195 Nxm2 < The fan vaver requirement is P = VAP I V P =12 mfsx4x0.024mx0.250mx195 N/m2 P = so w < where V is the volumetric ﬂow rate. For this calculation. p in Eq. (7) was evaluated at Tm. (c) For a single element in cross ﬂew. the average convection coefficient can be estimated using the Clnirchill—Bernstein correlation, Eq. 7.57. 4/5 - 1/2 1/3 “SIS — I 0.62 Re Pr ‘ NuU = 113]) =0.3+ D 1+ RED {9) k an “4 282,000 [l+[().4/Pr)' where all properties are evaluated at the film temperature. Tr : (T, +1519. The results ofthe calculations are ReD :4269 NTD1 :33.4 Rm 2 l06W/mz -K < PROBLEM 7.87 (Cont.) For the isolated clement. hDJ = 106 W I n12 it compared to the average value for the array. hD : 2H3 W {m2 -K. Because the ﬁrst row of the array acts as a turbulence grid. the heat transfer coefficient for the second and third rows will be larger than for the ﬁrst row. Here. the array value is twice that for the isolated element. (CI) The effect of increasing the longitudinal and transverse pitches to 30 mm, should be to reduce the outlet temperature. heat rate. and pressure drop. The effect can he explained by recognizing that the maximum Reynolds number will be decreased. which in turn will result in lower values for the convection coefficient and pressure drop. Repeating the calculations of part (a) for SL : ST 2 30 ml‘l'l, find Vmax Renown: Nu D ED ATfm C] To (in/s) thm"-K> cc.) (w) PC) 12 1.4ox 104 86.7 193 3t? 6925 41.3 and part (b) for the pressure drop and fan power, find f=0.'l8 1:1 Ap=122 N/n12 P=44W ...
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## This note was uploaded on 08/08/2008 for the course ME 364 taught by Professor Rothamer during the Fall '08 term at University of Wisconsin.

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assignchap7-47-87 - PROBLEM 7.47 KNOWN Dimensions ot‘chip...

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