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PS6 Solutions - l MAE 105A Introduction to Engineering...

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Unformatted text preview: l. MAE 105A Introduction to Engineering Thermodynamics Winter 2008 Problem Set 6 In a steam power plant 1 MW is added in the boiler, 0.58 MW is taken out in the condenser, and the pump work is 0.02 MW. Find the thermal efficiency of the power plant. The refrigerator in a kitchen receives an electrical input power of 150 W to drive the system, and it rejects 400 Wto the kitchen air. Find the rate of energy taken out of the cold space and the coefficient of performance of the refrigerator. A heat pump is used to meet the heating requirements of a house and maintain it at 20°C. 011 a day when the outdoor air temperature drops to —2°C, the house is estimated to lose heat at a rate of 80,000 kJ/h. If the heat pump under these conditions has a coefficient of performance of 2.5, determine: (a) the power input required, in kJ/h and kW. (b) the rate at which heat is absorbed from the cold outdoor air, in kJ/h and kW. Due Tuesday, February 19, 2008 (. Known: m: Pisa/d: 6%Iic2ienafl fl 5%{fl rig; MuJJJ'. 1. Con-hml volume com-sisfiw? 01L film-Hm Plant. 9., Sfiadg, Cam-{L SchEMgfic 43 GI'VeA Da‘l‘a 2 QM 54cm POM/fir F lanai. @h‘foou-r: Wt-fl/l)? SubS‘f'f‘HL'Hna) («w-Hum? Chi/£1610”? 7'12 51"?)1 ant flack Harm: +1Mvd no.58MM/ 2 wt - 0.01MW .. wt: mass 19:202. no.4.4Mw' w¢=o.+4+w 71x4 WYM’ enLfl'cl‘el‘ 71.0%. 1‘5 . , DzSI‘v—ed 0w.+2“+ __ WM? _ Wt'l%l : 3M: 0.4.2 Q: @73qu Infwf “ (aim _‘ 62"” 1. Commmis I Vi: 0'42 , 1. The sold-ion refimimcl use. 0-? Find” Law.) as we!) as 4A; WM'HDM 04- mel {Gh'ciemccd, z_ Ra-Fexm‘. ‘Hvz. Wgngfim fiakma'fi'c, we comm also qu-i—l’c. C? = Q”- c __ --——-—-—-I~I0V5—B ’2 “HQ” — ; mam .‘g; a simple; calmlafi'an) bu)!- “ aPProarA 6:};ng above bpmw‘ics amaktr- {nfiiabd’ l’m-f-oflaaci-uak Plani‘oparafi'onsk E i k L [ 2.. H'DM rcgrlauafiw‘ 3.551.: QCHB W: l. S-Leaoba. Shh ‘ ' 2' CQMM VOW” cow‘sflnfr 04 {:ch Syskm. Schmafi’c “1° Gl’ven 193153 .' g t [ 11L codach mt fur—Fermena. I; _ M: £322: 3;: 25W: (.621?— E‘. Klaul‘htcl IW‘I' WC We. [saw Mi [- Wsotwh'oh Maine! LL51 (97‘ Ma Firsf Law as We” as 14“ CILHML‘HUA «91‘- 7‘70. ugfiq'cgud- 0+ radon/n MC!— -E:»r- a rt-Fwi‘gzraw‘vw. 2.. Pt Valu-L :91!- #:1-6? [s 10v Far H—LCJCFfi'edznf . 01‘ lavatormwez. 91¢ a rafiifiem'fow. 3, 2.94;”; ~[o Kg mFbt‘gzrav‘or Saki/mafi'c 3+ flu; N" } w/«E coda! a.st Wrik @_ .4 *4; = I.“ as More. l. S-l—eastfi $113+: , 2-13” WNW‘ WWW“ (a) Head? Pump 4 up) gmsya Amalia: Taking 7'14 how-.51 ’as a wn-h—o! Volum: 12:3: g’sw ‘Ew =9 Germs: 0 => Qr—QLmMGWM M ““W 1"“ “*W” “a 7“ “Ema?” r r 94' W 01(— Vlvé L? manq a FMMPLT 94:1._ ‘ _ 91:21.: "qu‘Joo-"F. a £9: 45% __ W‘: _> Wt.“ Y 2.8- ”" “321000 A “(36905 —-- vb; = «3% ow if: ~615qu Takl'I/Ifim Maj” Plump as a UOLLWLLg ‘ '7» Q¢+ QR '=' we => (3;: Wc‘QH '3 ‘32! 000-631)].000) = —-32..Lmao-[- 801000.: +5L Doro "LR:- K k a (5.3 law 36:09: Comm: . L Soiufion réfiuireal Frrfi" L‘W Mat QC“; 491w" T1'3-SEW 71C JJL-Hm'hah 3-!— W Mdficimt’f 0.? a; Reef ULMP. Fer-Fofm’aA/tm ‘ 2. NHL Mai 634 is oSin +aéu'n? 1'11 [nous—e ES 3 (:0an you”) wit—[1.2. 45,... 1'5 nefiah'vz hum? m Maffaump a,s A Can‘t???) {infant-2. 3, A- kw? 149 m Selma.“ Was +0 Magm‘y— +49»?! 6224:9510” £9? 7’10; Muse. 3f" 54—66443. SEA-s: madd'h'ous. ...
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PS6 Solutions - l MAE 105A Introduction to Engineering...

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