ME 304 Ch 7 Examples

ME 304 Ch 7 Examples - mm ‘ Baum-Lu 7.1 Air at a pressure...

Info iconThis preview shows pages 1–17. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 10
Background image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 12
Background image of page 13

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 14
Background image of page 15

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 16
Background image of page 17
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: mm ‘ Baum-Lu 7.1 Air at a pressure of 6 kam2 and a temperulure 01' JUNE films with a velocin of 1G W5 over :1 Hal plate {1.5 m long. Estimate the mmling rate per unit width of the plan: needcd tn maintain it at a Surface 1emperature n’r‘ 21°C. firm I'I‘II'IN Known.- Airfiow over an isnthumal fiat plant. Find: lelng raic per unit width of the plain q' [Win-i]. Srhmlmfl'r: ‘ T_ I SUDJC —II- a“ =10 mfs —l* p,=5 kNa‘rnz—JI- 1|:- 2m; i | L—-L ELLEirI'I —-H i L” Assumptions: 1. Steady-state conditions. 2. Negligible radialidn effects. Properties: Table n.4, air ti"I = 43? K. p = I mm}: P = 30.84 x it)"I it: = 36.4 x in", Wi'm ' K, Pr = 0.531 Properties such as It. Pr, and p. Ina}r assumed to be independent of pressure to an excellent approximation. However. a gas, the kinematic viscosity P = mp will vary with pressure through its denee on density. From the ideal gas law. ,0 = MT. it follows that the ratio at ' matte viscosities fur a gas at the same temperature but at different pressures, p. p2. is intuit m- (Mpg. Hence the kinematic viscosity of air at 43? It. and p, = ti 10-“ Him: is 1 2 LUIJJ X 112* Mm = 5m x "r; it IS x 10’ Him? m 5 v = 30.34 x it)"tn:ts x Analysis: Fer a plate of unit width. it follows from Newtfln's law of coating the rate of convection heat transfer to the plate is q' = Eur.” m T3 T0 determine the appropriate convection com-latter: for computing 3, the Reyn number must first be determined Hut-'3 iflmts x {1.5 m R = —---=-—=959'i‘ 6' P 5.21 x10—4mtis Hence the fiflw is laminar over the entire plate, and the appropriate correlation 'a fiiven ’0? Equation 73$. N—HL = damsel” Fri” m omotsssti'fltoosn'” = 514 The average convection coefficient is then _ —lt ' 1 PM? =514EW=4szimnK {15 m and the required cooling rate per unit width of plate is q’=4.13WIm2'K.XD.5ml3m-2'?}"C=5?DWIm <1 Ennuunnts: The results of Table A»! apply u: gases at ammopheric pressure Except for the kinematic viscosity. mass density. and thermal diffusivity. the}! may generally be used at tail-tor pressures without commotion. The kinematic viscositya-d thermal dil‘l'usivltjlr for pressures m! then ] stm may be obtained by dividing the tabulated value by the pressure (at-m}. Emma 12 A flat Flatt: uf widfl‘: w = l m is maintajmd at & unifurm surface temperaturc, '1'} = 230°C. by using independentty cuntrulled. electrical strip hcamrs. each at which is. 51} mm long. If annmphedc air at 25°C flows out: the plat: at a vtlatity of 60 1111's. 1!! what heater is the elecutcal input a maximum? What is the value oFthjs input? Snwrmn Known: Airflow uvtr a flat plate wilh aegmtnted heaters. F ind : Maximum heater pnwur rcquirement. Sthnmir: assumptions: 1. Steady—state conditions. I. Negligible radiation effects. 3. Bottom Surface of plate adiabatic. Properties: Table n.4, air ii"Jr = 44]} K. p = 1 atm}: I' = 215.4] X lfl'fi mzi'a. it = ilifllli Wim - K. Pr = 0.6911 Analysis: The location of the heater requiring the maximum electrical power may be determined by first finding the point of boundary layer transition. The Reynolds number based on the length L, of the first heater is "all 2 fit] mis X flflfi m =1_iai<idS " zeal x lfl'fimzfs Rm = If the transition Reynolds number is assumed to he Relic = 5 X ICP. it follows that transition will occur on the fifth heater. or more specifically at _ tr _ 26.4l X lfl'anfa r _ ,Irc — Hm Re”. — wmm 5 K II] — {1.22 TI': The heater reoairing the maximum eiemrical power is that for which the average convection coefficient is largest Knowing hoa-r the local convection coefficient Titan 1' .4 Constanta of Equation 1.53 for the circular cylinder in cross flow [lift] If” C m I-to tits on dentin cat on 103—2 x it? one on 2 x ire—tofl care at a cylinder in cross flow. it is simply a matter of rcplac'tng ED by ED and Fr by 5:. mass transfer problems, boundary layer property variations are typically small. when using the mass transfer analog of Equation 153, the property ratio. accounts for nonconstant property effects. may be neglected. | EWPLE 1' .4 Experimean have been conducted on a metallic cylinder 12.? mm in diameter 94 mm long. The cylinder is heated internally by an electrical heater and subjected to a cross flow of air in a tow-speed wind tunnel. Under a specific set operating conditions for which the upstream air velocity and temperature at _ maintained at V = II} mfs and 262°C. respectively. the heater power dissi ‘ was measured to be P = as W. while the average cylinder surface temperature it determined to be i"1 = HEAT. It is estimated that IS% of the power dissipatan lost through the cumulative effect of surface radiation and conduction throufii endpieces. Thermoc on pie for I‘I‘IEHHI'II'IE airstream temperature " Heated F Insulated cylinder endptece P'rtot tube for _ determining velocity Thmbflflh tends Potter leads to electrical heater 1. Determine the convection heat transfer coefficient from the experimenul observations. 2. Compare the experimental result with the convection coefficient computed 'rl‘fil'l'l n.“ enamels-liens fihfl‘allfil;hfl SOLUTION Known.- Operating conditions for a heated Cylinder. Find:. Convection coefficient associated with the operating conditions. 2. Convection coefficient from an appropriate conelation. Schematic: T, = 262°C T,=128.4°c- - q = 391w ‘— -——-I- P = we w I D = 12.? mm alaarurrptr'ons: l. Steady-state conditions. 2. Uniform cylinder surface temperature. Properties: Table n.4, air {Ta = 262°C 2 300 K): v: [5.39 X It)"El mitts, t = 215.3 X It)" mm - It. Pr = 0.701 Table n.4, air (T! a: 350 K): p = zoo: x ID" mas. I: = 30 x It)" th- K. Pr = moo. Table M. air it; 2 were 2 am K]: Pr = 0.590. Analysis.- I. The c0nuection heat transfer coefficient ma}r be determined from the data by ming Newton‘s law of cooling. Thai. is. — _ l? h ‘ Ate - m With a = 0.35.” and A = 'ITDL, itfollows that 9.35 X 46 W —— = 2 , K. H: 2. Working with the Zultauakas relation, Equation 153‘ _ Pr IM NHHECRE’HFI‘" P—r, all properties. except Pr“ are evaluated at T3. Accordingly. VD lfl‘fl'lfa X 0.012? m = — = _.. = H 15.39 x it)" m'fa Hence, from Table 14, C = DIE- and m == '16. Mao. since Pr <1' II}. a = It follows that Rep _ on at... = d.20t1992)”tfl.?01t“ “( fi) = 50.5 ~_-— 5: UflEfiJWIm'K: =_ a-auflfl ace—flflmm lflfiWIrn x Corrine-Little: 1. Using the lChurchill relation. Equation 154. 0.52M,“ Pr'” [I +( Re” )1“ F = 0.3 + — "D [t + {BMW-13”?“ With all properties evaluated at Tr Pr = {Hi} and ” 1’ 20.92:: lfl'fimlfs Hence the Nuaselt number and the convection coefficient are 0.62tfiW1}'”t0.?0}"’-‘ I + m“ are as = 4M [1 + (omnjof’irt‘ 202.000 ‘ m... = 0.3 + _—i= flflfifl‘t‘th'm-KE 1, H—Nanfl ace—W 95.0mm a Alternatively. from the Hilpcrt correlation. Equation 152. Hi?“ = C Rafi Pr'” Witt: all properties evaluated at the film temperature. Rep = fulfil and e- 010. Hence. from Tahie T2. (3' = {LIE} and m I: 0.613. The Nasselt nu and the convection coefficient are then m, = ammo? I Macedon-3" = 37.3 —_— lt_ D.flJflWIm'K_ 2 ft - Nltpfi - 313W — 33 WIRE ' K I. Uncertainties emaciated with measuring the air velocity. estimating the loss from cylinder ends. and averaging the cylinder surface tent which varies: axially and circumferentially. render the experimental accurate It} titt- better than Iii}. Accordingly. calculations based on each of three correlations are within the experimental uncertainty of the result. 3. Recogniee the importance of using the proper temperature when eval fluid properties. EXAMPLE Tuf- Thc dwomtiw: plastic film on a. copper sphere of lfl—mm diam:ch is cured in oven at HT. Upon removal From me man. the alarm: is subij [1} an ' at I am and 23°C having a velocity of IE! mfs. Esrjmate haw long it will take: cool the splat to 35°C. SOLUTION Known: Sphere cooling in an aimrtam. Find.- Time I required to can] from T. = “35°C m T“) = 35°C. Schemmic: 1rI . ?5°C.. m}: 35H: Ass tempt Erma: l. Negligible thermal resistance and capacitance fer the plastic film. 2. Spatially isothermal sphere. 3. Negligible radiation effects. Pauperfleee Table A. I. copper {T == 323 K}: p = 3933 kgl'm". .l't = 39"} “Elm ' K. CF = 333 Mtg - K. Table M, an tn. = 295 Ki: a = lfllb x In" N — 5th. :u = |5.53 X Ill—'5' mzals1 k = [1.025] me ' K113!" = DEBS. Table A31. air {T5 =3 328 K]: pi = 1913 x It)" N'ei'mz. Atlfllyeis: The time required te complete the ceniing process ma}.r he chtainetl from results for a lumped capacitance. in particular, from Equations 5-4 and 5.5 pvcp T:_Tm f=":'_—|l1 ha, T-Tm unwith v = when“, = “:qu. t=pr£inE—TH ea T-Tw Ftttm Equaticn 15E: Nun = 2 + [flAReH'z + {Jfltjflegfiprfl‘ ( in. H4 a) when: Hence the Nusselt number and the cunvccliun cuefficicni are if?” = 2 + {Mia-um“ + fi.fl6(fi44fl}“lifl.?flsl°" 1r _ z m x(|32.fix m N arm) :41] IQTEX IU'TN'sfmz J: E=EDD=47IIW : 2- fllmm HEme K The lime [equirad fur cooling is then ,_§Wln(?fiflj=v:ss fixnswimJ-K 35’23 ' Cum menu: I. The validity of the lumped Capacitance math may be determined by caiculal- ing the Bic-t number. From Equation 5.1!] .‘ .-_hirfil_nawmI-Kxunme ... B“k,‘ k, ' 399WIm'K “135‘”: and the crilerimi is satisfied. 2. Atfliough their definitions are similar. the Noosett number is defined in the thermal conductivity of the fluid. whereas the Biot number is de terms of the theme! muttivily of the solid. 3. Options for enhancing production rates include accelerating the cooling by increasing the fluid veiocity andtor using a different fluid. Applyi foregoing procedures. the cooling time is computed and plotted for air helinn't over the range of velocities. 5 E v 5 25 into. 125 1m El 50 its} 25 5 ID 15 ED 25 'la' tn'u’sl' Although Reynolds numbers for He are much smaller than those for air. thermal conductivity is much larger and, as shown below, convection transfer is enhanced. Pressurized water is often available at elevated temperatures and may be used for space heatin g or industrial process applications. in such cases it is customer},r to use a tube bundle in which the water is passed through the tubes. while air is passed in cross flow over the tubes. Consider a staggered arrangement for which the tube out- side diameter is lost mm and the longitudinal and transverse pitches are 5}, = 3-4.3 mm and ST = 3|.3 mm. There are seven rows of tubes in the airflow direction and eight tubes per row. Under typical operating conditions the cylindm surface tern perfi atore is at NFC while the air upstream temperature and velocity ate 15°C and t5 nits, respectively. Determine the air-side convection coefficient and the rate ol' heat transfer for the tube bundle. What is the air-side pressure drop? SoLLr'rton Known: Goometry and operating conditions of a tube bank. Find: 1. Air-side convection coefficient and beat rate. 2. Pressure drop. Schema! Etc: Assumptions:. Steadyvstate conditions. 2. Negligible radiatien effects. 3. Negligible effect at change in air temperature across tube bank en ‘ properties. Properties: Table a4. air it"... = 15°C}: a = 1.21? kgrmt, a, = let)? tag- a = 14.82 x “3'6 mzr's, a = eases wrm ~ K. Pr = tutti. Table as. air it: = P:- = mm. Table AA, air {a} = are}: r = 114 a: [ti—‘5 are, a = sum wrait— Pr = ates. Analysis:. Frflm Equations T154 and 1155, the air-side Husselt number is _ Ira Nun = Ca: aremam arm (g) Sinee SD = [SE + {Efflflm = 31? mm is greater than {ST + mm. the matti- mum velocity occurs on the transverse plane+ A... of Figure 1| ]. Henee item Equatien 162 _ Sr _ 31.3 mm _ pm“ ' sT— D 1L“Hula - 16.4]mm 5W5_12"5"”5 1|With tr” D Rap M = “f = —‘2‘5 "“5 X Ufllfifm = 13.943 ‘ 14.32 X It] m'r's and 3r _ 31.3 mm _ 3—; - 34.3 mm _ '19] a: 2 it follows from Tables "3.? and "LB that _ “5 C = U.35( = 0.34. m =fi.fit}. and C3 = 0.95 L Hence _ ‘ e25 Nun = are x 0.34t13.943}“'wtfl.T110-‘( %) = era and 2 a.” X 0.0253 wan - a .i.’ = :'I ,- 5 Dfllmm 135.15 me K all From Equation 15? 5-0545 ) PVNr-S'rfp mo.o154m}554135.5wrm1-K1 ) |.21?kgfm3[fimfs}8(fl.ll1313m}IDDTJIICg- K T. -F Tfi= IE?"I — T,-} exp(— 1"”‘ H T” = [55"C) exp-(- TJ - TE. 2 445°C Hence from Equations 166 and 163 = (T, — T.) — or. - m = .155 —44.55*C on,“ = 45.55:: I T‘ _ l “ T, — 5:, " 44.5 and q’ = 4454055.; = 5547 5: I355 wxm= + K x 0.5554 n1 5: 49.5%: q" = 15.4 kWIm <5 2. The plflfifilll't.‘ drop may ho ubtainod from Equation 16'}. IVE-ax With Ream = 13,943. Pr= {STE}; = I3]. and {PTIPL} = 0.9!. it follows from Figure 114 that; == 1.534 and f = 1135. Hence with NL = T _ 5 3 _ 1 121 kym24125ws}]mj sip = 245 Wm2 = 146' X 113—3 bars 5:] op=TX1.fl4[ 5 . Utflfllfl'fl H .T I. With properties evaluated at Tr. Rea m = 1 1.376. With ST-ID 2 2 and SJ!) 2 2, it follow: from Tables 15 and 115 that C] = 11432, H: = 11556. and C? = 13.91 From Equations loft and TEN. the Husselt number is then mp = 36.7". and h I ' th: ' K. Values of h obtained from Equations Inf-[l and 164 therefore ' within 'ttita, which is well within their uncertainties. . Had tit??- E I} — T, been used in lieu of til-TI.” in Equlation 7.63, the would have been overpredicted by | “it. . Since the air temperature is predicted to increase by only I'll.5”C. evflu -‘_ the air properties at T. = 1513 is a reasonable approximation. Heat improved accuracy is desired. the calculations could be repeated Wili't the etties reevaluated at {2-1 + Tram = NEST. An exception pertains to the sity p in the esponentia] term of Equation "not. As it appears iii denominator of this term, p is matched 1tvith the inlet velocity to provide a M {pm that is linked to the "less flow rate of air entering the lube -=! Hence. in this term. ,9 should be evaluated at T,-. I . The air outlet temperature and heat rate may be increased by increasing.- number of tube rows. and for a fitted number of rows, they may be van- adjuszing the air velocity. Fen-5 E t'tt'I '5 25 and ir’ = El mils. parametric c #- tions based on Equations 1154 through 7.63 yield the following results: TH cc] The air uutlct temperature would asymptotically approach the surface mm with increasing m. at which paint thc hear ran: appmachcs a cansram and there is no advantage tn adding more rub-c rows. Nutc that fip incramcs early with increasing Ni. For N,“ = 25 and I s V s 21) mix. we obtain ...
View Full Document

Page1 / 17

ME 304 Ch 7 Examples - mm ‘ Baum-Lu 7.1 Air at a pressure...

This preview shows document pages 1 - 17. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online