extraproblem11-83

Extraproblem11-83 - PROBLEM 11.83 KNOWN Cooling coil geometry Air tlow rate and inlet temperature Freon pressure and convection coefficient FIND

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Unformatted text preview: PROBLEM 11.83 KNOWN: Cooling coil geometry. Air tlow rate and inlet temperature. Freon pressure and convection coefficient. FIND: Air outlet temperature. SCHEMATIC: pzlav‘m hc=5000w/mz-K mh=1.5k_9/s 7;,z310K —'> —°‘7Z,o A; =O.16m2- surface CF—‘ZO-S/BJ, Di 113.8771777, N = 4 ASSUMPTIONS: (l) Negligible fouling. {2) Constant properties, (3) Negligible heat loss to surroundings. PROPER'I‘IES: Ruble A-4, Air (Th : 300 K. 1 arm): cp = 1007 li’kgvK, ti = 184.6 >< 10'T N-srmzj k = 0.0263 Wi’m-K, Pr = 0.707; Table [4-5. Sat. R-12 atm): Tsm : TC 7' 243 K, hfg = 165 lei/kg. ANALYSIS: To obtain the air outlet temperature, we must first obtain the heat rate from the e -NTU method. To find Ah, first find the heat exchanger length, L : (NL —1)sL + of = 3(0.0343m)+0.0285m 2 0.13m. Hence, v =20er =0.l6m2(0.l3lm]=0.021m3 Ah =orV =(269m2/m3)0.021m3 =5.651n2. The overall coefficient is 1 l L I _ mm + Uh he (Ac/Ah) n0,hhh where Ex. 11,6 yields ('AcfAh) : 0.143 and AhRw I 3.5] x 1075 m2-K/W. With mh 1.50 kg/s G: :—2:20.9kg/s-m 0A0 0.449x 0.16 m 2 ooh _ 20.9 kg/S-1n2X6.68XlO_3111 Re: H 184.6x10’7 N-sfmZ : 7563. Fig. 11.20 gives jH 2 0.0068. Hence, _ Gep hhzjh 2}} 20.0068 Pr (0.707 20.9 kgrs -m2x1007Jrkg.K —)2r3 iihzisowrmzx. PROBLEM 11.33 (Cont) _ "5 1112 With LL. = 6.18 mm and AP =15? x 10 6 m" from FX, 11.6, Li;2(hh f'kAp) = 0.338 and, from Fig. 3.19, nf = 0.89 for rgct’r] = 1.75. Hence, as in Ex. 11.6, mm = [1.91 and 1 _ 1 — =—-—12—+3_51><10 5tttZ-Kfl/V +———2—— Uh (sooowm .K)0.143 091(180W/m K) 2 Uh 2133 Wr‘m -K. With Cmin :Ch :n‘nh epsh :15 kg/s(10071fkg-K)=1511WfK UhAh _133wrm2.1<x5.65m2 Cmin 1511W/K NTU : : 0.4.97. With Cmiw/Cnrm,C = 0, liq. 11.3621 yields 8 = l~ exp ( —NTU) = l— exp(—0.497) I 0.392. Hence, q = Eqmax = ECmm(Th,i—Tc1j)=0.392[151IWz’K)67 K :1 239,685 W. The air outlet temperature is Th0=Thi_i:310K_mz283.7 K. < ' ' Ch 1511WfK COMMENTS: IfR-12 enters the tubes as saturated liquid, a flow rate of at least q m—39’685W 20.241kgls _@_165,000Jrkg me would be needed to maintain saturated conditions in the tubes. ...
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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 Wisconsin.

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Extraproblem11-83 - PROBLEM 11.83 KNOWN Cooling coil geometry Air tlow rate and inlet temperature Freon pressure and convection coefficient FIND

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