HW6+Solutions

HW6+Solutions - PROBLEM 5. 18 KNOWN» Dacha arg Prav'dea...

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Unformatted text preview: PROBLEM 5. 18 KNOWN» Dacha arg Prav'dea far & ___' - ' Gare/r c cl: ' hot and amid [‘E‘Ser‘vo-Warl P 7 OPEr‘n'hnj Def-ween Frwb? '1' EACH (4: _. m e d + ‘ h + I . J c ermne c9531“ 6"“ H“ Cycle apemfcs raw-3.5!»; Opera. 9 IFft’VfVLbe—i) or- :IS I'mdaOES’rbiC. J Smil‘nfk'tlc 1: (amen “DATA-5 (a) W 0 Lynn. m TN.“ chum J WEAVER _ We'lth __ 4933*“— vlr _ _ 0.5- m“ 6,0073% TM maiimum +WM’E fib‘m'ewg‘a ‘4 M __ _1_"_c__1_:12‘1€:a.ee?- Hm! I m ' :m‘}? S‘WC- Y] CHMAK/ +4.“ (11“ operaff! {rrrwrn'k}; . U0) (9H: 4008b] (QC : 3008+» Fray... HIE emu-m, ba- U‘H‘HL \chlch: QH' (ha 2 flea-'3W3‘9mfih‘ {.mva Cam (too Sm.“ :'f\ ow, Hucyde “PWaRr few/SIM , Y1 (Pitta) E; (C) chd¢z 600 3+“, (be :400734-u. WW4 W8»? bahnu. We.»th Gar-CDC. 2» (9H:W‘TW‘* ®c= “MW -; L000 13in KJJerMM—IA-m‘ms/ MWr-«M effiaenu—l t4; ’h: “’0 =. 0.90 {000 Sunni.- Y‘LY‘MAY J“’('\“"'ICS_L (quafl Opemi-es mmaw YL: qow/O S .W Y] > n-Nhi 20-626? (p‘dm); W‘S‘ en'w mun-47‘ DPW“HM Clank? 3- T“ ff“!- bl‘e‘ w [(NTOwN: .S-I-Eo-d-T-s-h'k Ope'ra‘h'nj (14% am: [Drovrdzd furquwerpkht disckua/l‘n) Energy by RAJ Wash C; :4 FIND! Dckrwne Hr“ IIUICVC‘LS' aflu'kweflmrb Why?" frqre‘etc- ,.u ruck Eacho6 M : SCHEMRTLC x'. or.sz 'Dfi'Tl‘i: L T.“ Pow" Fb‘“t ‘5 Mdeled duo 5-. Pat-Mgr cvcle “Fe/ra‘h'lflj GL1 PM)! " Yin-LL- 2, Ttu. “abusing-n Pmduc‘h’ Firemen, {Herg'a WWAnfif-a—g HUI Paw-Er agate. Energy in chm:qu b3 M 1w...“ f-«r +0 "H" M'Ver_ "Th-M4 ovum only Kivcrm‘mlbfixlflskg/s "bran: fir? - - 3' 11* (Iver-punk.” LS “Md «4 Anagrams; :WWMfL—un‘uhk “firm Cami—mt- . Tremfi'c h-nl: c,‘ Fur 4L £0w+fb| V‘D‘VWlfi Chciah'mj 0- SEC of. W; A1:er 11W 14;... an EKQV‘fij ME- balance MAUCES 3“““9— Q0“): HEM—Ht] aka.“ (Doug; U“,th “MEL... Chafing H... 4441:? chaner gym-Kapower chick 1-2 mm“ r“- UJH‘k 61.3.2019) Ms “Can-44 CiaJ= “NEE-Ta] (a) fiv't‘lhnrowa—flcJL-flj lumdolhlhbcvcle+ém‘ lH " Lb‘rcu = { 'L-1 L2) Y] aquamw e" “'1 ““L‘m ] Clowlatm'wa‘ 64':- (I)a..~é {2)} ' J. chclc [n4] (3 Crl—T‘): ——.—-——-—-—- ) 1m C mm (2: +1. pantry F"... ‘5“: I149- " - - T5 — - 3%]: 0302'. I“.an yam; £142} (a) q-“HAK— [I —- [l 5 9‘70 I _ it) (Tr-"1.): GIBMWH I] :l-oS‘K (21-0509 ‘; LLB? K tart-=5]; )(q...1 Kglgrg) Cb) 4—- (7go3tnosl[ofl§a‘1"]_ 1.“ K (22.11%2) d. CUMMCHT: A“ 0+ Irreverrrhl/afilei Lut‘i‘ka'n'flfl-L Pam‘s-plant H; +0 {mum “PM ' d. 'HM-ss “mun-4' M Lech.“ ct b M Mug-Fl.“ To 1"“ 7‘. Ver- an. increugl {tam-Hwy. rc d'fgf u-F 14.. r.w-hr'ace-..blr h: Suck. W Mnc{a(_ PROBLEM 5-6;! , :3 shown in Fig P559, an air conditioner operating at ENCR' r""DDE.=I'..-' w r . , ' . v ‘ . ‘-_————. gfgigesESIe maintain. :gwelhng at 70°F on a day when the t. The. tint“. it If“: defines 1-1,»; mpera ure 18 °F. If the rate of h = - V‘ ' 4'" " “'5 ' ‘ ' the dwelling throu Lat [muster “no u h m “J C! 5'5 e “a h “4 575“ b“)- gh the walls and roof is 30,000 Btu/h, 2- TM «yr-ten a ..+ sit-cad)» shu‘e. might a net power input to the air cond' ' . u l _ moner compressor 3 M - HI ,5 0‘ 013 hp be sumuent! lt yes, determine the coefficient of per- ' To”; :2 “'L‘iufd and: a": 2:31“; .fl fin-Q an ‘LG V 9‘ / immantfi. .IIO determlne the: lllllllum I. t: It: WC It! I: h- “6‘7 ) m h 0 “Call [)0 r P C. . scHeMATLC é cwem DATA‘ —'——-‘—————.L_.________: 30,000 BIN/h Outside. 90" F (1 Inside, 70" F T :5‘3‘u‘ll K TH;55DDR. C Refrigenmt 100p /__ .___. Cundcuscr l Fig. P559 AMHLYSIS'V . _ mike—1:) :mntmhmfk: :eu'nj “1 ¥°°fi 1“" ‘91“ Candt'hquo Srskw. . " 3 23 n- u ' . , . Qm=3°tooo 6W5. 3 SF»— oat-at. M%30,oooBh/h, Than Alsa, é . fi QM ‘- 26.5” —D W; 7- 302900 lib/AI ihp MW WQ 7—0-5“ arms-mun, : 0'44” ‘"'P YES , CL Power- .‘nPu—l- o;— 3I~F Wain“ tn SLFFI‘Q'eni‘. TIM, urnflpnh'nfi Qcmh‘eud’ Q‘F. Pgrfgv‘mfltf L; W“ -—-— a :_ z gl‘LOdOBi'k/Il = > wwde 3k!) ZS‘H‘EE PROBLEM Sit-9 5.79 Two kilograms of air within a piston-cylinder assembly executes a Carnot power cycle with maximum and minimum temperatures of 750 K and 300 K, respectively. The heat transfer to the air during the isothermal expansion is 60 k]. At the end of the isothermal expansion, the pressure is 600 kPa and the volume is 0.4 m3. Assuming the ideal gas model for the air, determine (a) the thermal efficiency. (b) the pressure and volume at the beginning of the isothermal expansion, in kPa and m3, respectively. (0) the work and heat transfer for each of the four processes , in kJ. (d) Sketch the cycle on p - V coordinates. KNOWN: Two kg of air undergoes a Carnot cycle with known operating conditions. FIND: Determine the thermal efficiency; the pressure and volume at the beginning of the isothermal expansion in kPa and in}, respectively; and the work and heat transfer for each of the four processes, in k]. Sketch the cycle on p~V coordinates. SCHEMATIC AND GIVEN DATA: m =2 kg p (kPa) r1: :r2 :750 K n = T4 :300 K Q12 = 60 k1 p2 = 600 kPa V2 2 0.4 m3 V0113) ENGINEERING MODEL: (1) The system under consideration includes air modeled as an ideal gas within a piston- cylinder assembly as shown in accompanying diagram. (2) Volume change is the only work mode. (3) Kinetic and potential energy are ignored. ANALYSIS: mom-em 5. as (2—) (a) Using Eq. 5.9, the thermal efficiency is TH —TC _ 750—300 TH 750 (b) Determine p i, in bar and V1, in n13, where Q; = 60 k]. Use an energy balance for a closed system assuming negligible effects due to kinetic and potential energy. -Qi2‘_W12 =m(uz“ul) For an ideal gas, specific internal energy is a function of temperature only and therefore u; = u] and Q; = W32 = 60k}. = 0.60 = 60% ' 77: To fix state 1, use Eq. 2.17 and solve for an isothermal process involving an ideal gas, as follows: I" V 1 2 T W12 = J'pdV2 [M H dV=mRTHln 5 =60kJ V1 1/, V V. ‘ (2 kg 7" —- — (750 K) 28.97 kgK Vl = 0.348m3 Since T1 = T 2, the ideal gas equation of state gives: P‘sz = prl 3 pl z szz = (600kPa 0.:hn )=6901<Pa VJ 0.348m (c) To determine the work and heat transfer for each of the four processes, in k], analyze each process within the cycle. Process 1 to 2 [isothermal expansiou) QI2 = W,2 = 60 kJ ‘ Process 2 to 3 {adiabatic expansion) Recognize that the process is adiabatic and use a closed system energy balance and property data from Table A-22, as follows: Q23 z 0 W23 = m(ul —u3)= 2kg(551.99—214.07)£ = 676k] kg 4—— Process 3 to 4 {isothermal compression! For reversible cycles operating between two thermal reservoirs, Eq. 5.7 applies as follows: PROBLEM sea (3) [g] 2g 60%|]: 300K QH $213 TH Q12 750K 300K = 60kJ ———_ = 24 k] -IQ34l _ [750K] Then, since 143 = 114, Q34=W34:_24kj ‘— Proccss 4 to 1 [adiabatic compression! Recognize that the process is adiabatic and use a closed system energy balance and property data from Table A—22 for air. Q4120 +— W41=m(“4‘ul):—W33=—676KJ 1. Check the above calculations using chm = Qcyclc and 17 = chclc/Qm as follows: Qcycie = Q12 +Q23 +Q34 +Q4, =36 k] W =W12 +W23 +W34 +ngi =36kJ cycle Therefore, as required by every thermodynamic cycle, QM.6 = gym. W W, = 36 Id 2 60% Qi 60 kJ This is in agreement with part (a), as expected. n3 PROBLEM 5. 83 A system executes a power cycle while receiving 1000 k] by heat transfer at El temperature of 500 K and discharging energy by heat transfer at a temperature of 3UU KThere are 110 other heat transfers. Applying Eq. 5.13, determine time if the thermal efficiency is (a) 100%, (b) 40%, (e) 30%. Identify cases (if any} that are internally reversible or impossible. gnu-MN! Dan-Ea. are Prov-dad -For- a 17:49:“, etecu-h'nj a... Power cycle- dellfrmag Gyckc +or emh O-F +‘1f‘fl‘ was, scHsnrurnc 2; awn) DATA: EucmEERIuG MODELS ' 1. TM :yflem Show... M f“- takemmh'c. Underage: a. power cycle wk“ "PW-twinj CO“ at TH and cusclunrflrnj Gk; 41 Tc 2- C52“ and (Dc, arcpon'fwe m ‘l'M-n. ch'recffan our-I'M tau-rows. Thu-c are- M CAN-'3 k “L4 “franc-fin“. ANALYSIS: =t—35. =t> Qc=(|-VL)QH- 71‘9": 65'5"“ "‘4’ can (To-Ida = -§ :_ 9_*£_@_E. =— 3—'1--(“"‘G""l 'fil Tc. 73+ Tc. Hp..- hat-cycle, 7]. .L... -- O :- - Ka- G‘cthe: alOOO J 4______ Since ('1ch Can-«not hi? “'5‘ KM J +m‘s cm La ""P‘m'lbk' (b) 'n: O. 4(+D%] _I__ _ (1-0.4) _= .q._____. @ Get-file '5 "two‘s-[500C 300:. O Sine-e Tcyclff") J “PL-4.? Car: UVN’FW’W +0 ‘-"+'°m“”33 rev'rs‘ue OP'V‘Ih'M‘ can: e-at‘so‘m __{_“ __. [l‘o'3)]= 0.333 g +_.. K Gc1cle = "40°05: 5°05 “IMF— 5‘M-4 (Fe-1:14 >0J+HS cm: “(Hipondl‘ 4014* (“H-mu r; {VA-Ova“ fr raver “.17; “'11.?! . 1.7m Mun-19M +kefm~4 efficiency auxin ruck cycle C—Mn hm ab =04 (40%) ...
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HW6+Solutions - PROBLEM 5. 18 KNOWN» Dacha arg Prav'dea...

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