chapter2_homeworkexamples_2 - (‘1 l v, Thermodynamic...

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Unformatted text preview: (‘1 l v, Thermodynamic Cycles 2.71 The following table gives data, in k], for a system undergoing a thermodynamic cycle consisting of four pro— cesses in series. For the cycle, kinetic and potential energy effects can be neglected. Determine (a) the missing table entries, each in M. (b) whether the cycle is a power cycle or a refrigeration cycle. Process A U Q w— W 1~«2 — 610 2~3 670 230 3.4 0 ' 920 4—1 . —360 4 Q EM: (a) £0ka flee. hub/e oi? emerge, vague; 14W #3.»: aflwmmi 64a} oiWMIHOr whei‘iAW We ajcle 2s a. fewer afjda ar a mfiwigmf—Em ajcié. 5CH’5MAflfi g Q; e155,! 23m: Process AU Q W 1-2 0- C "610 2—3 670 d 230 3—4 h 0 ' 920 4—1 —360 5 O 5:5 .ngflgms: (a; The §V$§am is a dosed fiysi-gm.flb} bilge}ch W Pdemh‘ag eweng eH—wés am he vufi Led-(5‘. M93455: (1; 3%{Mm‘vx3 quJk 9c0£z$$ 344 b, 5U=Q~Wz AU:O—.on =~0t2c> => bro-0‘20 New) -Eor +Le aide) ZCA‘BF'O. “FMS c1. a+G10+£-QZO“;+(~3%O)=O f9 azbro EU; Q~V\J c. me = c:-(-(a\o\ =‘-> 6:0 01.010: (1. 230 —.=> cinema 2.~3éoo=e——c> =>a=-3(o@ (b) For «Hm aide . Wage/(c: = W'Z*W1‘§+ W3?1h W‘H =. {~éac)+(13,o)+(fizo)+—o = We 5M5?“ Wagdc 70.11%“? ajdé £5 a- h Far 44‘; wide, new. W ‘ : le+d13+ Q3q+ Q‘fl 2(0)“? (QCO)+(O)+(~3&O} : Sifo Thus lecgg =2 (Haida ) as expzefefi‘ FEOfiLEM 2.73 KNOQM: A gas undergoes a. +h6vMofidje/mmlc 5:36.56 whs'es-é'fvze} 5% ééwae prcééfis'gg‘ Emu}: ‘Ddrgvmghe Wm Lam-1‘ Mmkr M wart Qer pramg 2-«3 3am wkwfixer ‘HAE cgde M 0«— {WWW Calida Ora Fei—‘y‘igerafim made. Tim fiwllawiysg ficfia are gift/en {branch pmcem: Préccgs l—z: Com ressa‘m wiHA PV=aonst Car} v; 31.9 m3 +0 Vz=0.2m3 Pracess 2J3: Cmflmf- pmssure {a \Q‘Vj Pmcm awe: 6mm”? vaiumeJ IJ;~U3=-s:tfi w W: (I) The mm 23 males—ed s aka/k .62} IV fed Eggseflé. aluoé' , u if. 2? gokng'm! er;qu wagesfigg TM, Wremmffi'am 5% i +9 2.2: a, PM? w my“: prance“. fiMfiLjQisz “To +75% M work {aw Process 2—3) use; 2.1972 Ma WM [macaw Vs Wu :1”;an = pgvrw) V1. 555mg? 'Hifi FJV rem-Fm fafgaromas (“2; pi = Va)?! :2 giver/m. Téwsgw'lfk 3" w _ 8E Siluwlm 1m ZS—( m)(/.é-6.z)m tn:- ‘O3N.M :HZOEJ} W23 The energy bah/me. Qty Woccss 2’3 Vaduccs 40 Q25 = Luff-323+ W23 “2:: 5.2+ 22143;, (as): flag; {or +1”: W, ABA/mac, ms ‘ (“SJ-A ) 4” (Uiu‘f) *7 (UKUB) :0 $(L33’Uz): " Wrijg) 3’ Fawn?! 3 {331’ Wow“ 2/3 (21,3 = (am Mammy) =. aw m (323 Nexéi, 47m” Prodch {’2} AU =0} 50 ka=wlr 0mg ELF? -w -J’vmawrfl ’QMVw. 0.2.- Ems all- m." U? a P3"; ‘7’: (L6) £333 (i) AM; for yromgmgs Wm =0 1. mug QM : 'U; ~13; : ~33ch k3“; {Eggémfimfi mmmg; rgsufi-s - 0 WW : sz+wz3 +M =67 332«7)+(g:20] 2 +787? E3“ 3”)“ “fa/gale; >0.- W 90514 is a» (fewer aide. “33m £3“ 5‘ “‘3 M3 FROBiEM The rate of heat transfer betWeen a certain eleetriomotor and its surroundings varies with time as Q = -02“ ._ et—onso] 1 where r is in seconds and is in kilowatts. The shaft of the motor rotates at a constant speed of w = 100 rad/s (about 955 revolutions per minute, or RPM) and applies a constant torque of 97 = 18 N ' m to an external load. The motor draws a constant electric power input equal to 2.0 kW. Obtain an expression for the time rate of change of energy of the motor. SOLUTEON Known: A motor operates with constant electric power input, shaft speed, and applied torque. The time-varying rate of heat transfer between the motor and its surroundings is given. Find: Obtain an expression for the time rate of change of energy, Schematic and Given Doro.- §’= 18 N-m m -= 100 rad/s 3:: Q = -o‘zu _e(—0.05c)] kw Figure E2.S Assumption: The system shown in the accompanying sketch is a closed system. Analysis: The time rate of change of system energy is Wrepresents the net powerfrom the system: the sum of the power associated with the rotating shaft, {fit/Sm, and the power associated with theeiectricity fiow, Wm: W = Vyshnft + Wales The rate Wm is known from the problem statement: Wm = -2.Q kW; Where the negative sign is required because energy is carried into the system by electrical work. The term Wsm can be evaluated with Eq. 2.20 as WM = em = (18 N~m)(100 rad/s) = 1800 w = +1.8 kW Because energy exits the system along the rotating shaft, this energy transfer rate is positive. In summary w = Wm + WM = (—2.0 kW) + (+2.8 kW) = ~02 kW where the minus sign means than the eiectn’cai power input is greater than the power transferred out along the shaft. With the foregoing result for W and the given expression for Q, the energy rate balance becomes dE F t: -02” a 3(‘03501 .. (“not a 0 get-0.050 r a , . ...
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chapter2_homeworkexamples_2 - (‘1 l v, Thermodynamic...

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