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soln6A_08W_term1 - \r-l’tjmewW\¢ 6 A’(Q 1 ‘ Solve(2...

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Unformatted text preview: \r-l’tjmewW\¢ 6 A’ (Q 1 ‘ Solve: (2) Q is given as 1000 J. Using the energy transfer equation for the heat engine, QH ZQC +Wom : Q! 2Q: +Wout 3Q: :QI *Wom The thermal efficiency of a Carnot engine is 300K w qzl—gzl— =0.50= °"' TH 600 K Ql 3Q: ZQliflQl :Ql(177]):(1000 J)(lk0'50)=500‘l To determine Q] and Q4, we turn our attention to the Carnot refrigerator, which is driven by the output of the heat engine with Win 2 WM. The coefficient of performance is Tc : 400K =4‘0:_Q£=Q4_Q.. 71,4C 500K—4OOK K: “/in Wont — ”Q1 ::> Q‘ = KnQ, :(4‘0)(0.50)(1000 J) 2 2000} Using now the energy transfer equation Win + Q4 : Q3 , we have Q; = Wow +Q4 2 HQ. + Q“ : (0.50)(1000'.l) + 2000 J = 2500 J (1)) From part (a) Q3 22500.! and Q. = 1000 J , so Q3 >QI . (6) Although Q. = 1000 I and Q3 : 2500 J, the two devices together do not violate the second law of thermodynamics. This is because the hot and cold reservoirs are different for the heat engine and the refiigerator. (:2 Z n Model: For the closed cycle of the heat engine, process 1 —> 2 is adiabatic, process 2 w) 3 is isothermal, and process 3 —) i is isobaric. For a diatomic gas CV 2%R and y:-; . Solve: (a) From the graph V, =3000cm’ and 'p, :100 kPa . The number of moles ofgas is - my: (4.0xi05 Pa)(1000x10‘6 m3) ’1 : 2 RT, (8.31 meoi K)(400 K) : 0.1203 moi With pl, V.‘ and MR having been determined, we can find T. using the ideal-gas equation: T , PM 7(100x103 kpa)(3000x_io* m3) I _ = 300 K nR (8-3 IJ/mol K)(0.1203 moi) (b) For adiabatic process I —9 2, Q : 0 J and (4.0x105Pa)(1000xl04'm3)—(1-0><105 Pa](3000xi0"" m3) W5: p2V2_plvl 2 — 250] i— j! 1— L4 Because MI, : “W, + Q, A131, :4»; : 2501 . For isothermal process 2 —) 3, AEIh :0 J . From Equation 17.16, V3 w, :nRT, ln—z554J V2 . From the first law ofthermodynamics, Q : W, x 554] for process 2 —> 3. For isobaric process 3 —> 1, W5 : area under p—versus-V graph: w, :(100x103 Pa)(3000x10‘6 m3 —4000x10"‘ m3) : 400} 7 . anwmmw 5R (Tl-Lease] , 5 as, : nCVU] 4;): {3R}: —T,)=(o.:203 mol}[38.3l J/mol K](—100 ii) ME“, (J) W: U) Q (J) ‘ l a 2 250 —250 0 2 —> 3 -- 0 554 554 3 —) 1 —250 —iOO —350 Net 0 204 204 (c) The work per cycie is 204 J and the thermal efficiency of the engine is U2fl:_2°‘” 20.37 :37% QH 554} Assess: As expected, for a closed cycle (Pl/S)"cl 2 Qm and (Mn. )1“ :0 J_ Q3“ IDENTIFY and SET UP: First use the methods of Chapter 17 to calcuiate the final temperature T of the system. EXECUTE: 0.600 kg of water (cools from 450°C to T) Q = chT = (0.600 kg)(4190 Jl'kg - K)(T — 450°C) = (2514 JIK)T— 1.1313 x10S J 0.0500 kg of ice (warms to 07°C, melts, and water warms from 0°C to T) Q 2 mcbe(0°C — (—15.0°C)) + m. + mm (T — 0°C) Q =o.0500 kg[(2100 Mtg-K)(15.0"C)+334x103 Jikg+ (4190 J/kg- K)(T —0°C)] Q=1575 1+1 67x10“ J+(209. 5 J/K)T=l.828x10‘ J+(209 5 mgr ' gm :0 gives (2514 J/K)T— 1.1313x105 1+1. 828x104 J+(209.5 mgr— 0 (2. 724x103 1/102": 9485x104 J :r: (9 485 x104 J)/(2._'1'24x103 J/K): 34. 83°C: 303 K EVALUATE: The final temperature must lie between —.15 0°C and 45. 0°C. A final temperature of 34. 8°C 15 consistent with only liquid water being present at equilibrium. IDENTIFY and SET UP: Now we can calculate the entropy changes. Use AS = Q/ T for phase changes and the method of Example 20.6 to calculate AS for temperature changes. EXECUTE: ic_e: The process takes ice at —15°C and produces water at 343°C. Calculate AS for a reversible process between these two states, in which heat is added very slowly. AS is path independent, so AS for a reversible process is the same as AS for the actual (irreversible) process as long as the initial and final states are the same. AS = deQ/ T, where T must be in kelvins For a temperature change dQ = 111ch so AS = If (me/T) dTl= me 1110'} I 1}). For a phase change, since it occurs at constant T, as = fdQ/T = Q/T = imL/T. Therefore ASice = mes“ 111(273 1025 8 K) + mLf [273 K + mewamr 111(308 K1273 K) AS“: = (0-0500 kg)[(2100 J/kg -K)ln(273 K1258 K)+(334x103 J/kg)/273 K+ (4190 Jfkg ~ K)ln(308 K1273 K)} AS.“ = 5.93 J/K +61.17 J/K + 25.27 J/K =92,4 .l/K water: ASm, = me rum/1;) = (0.600 kg)(4190 J/kg - K) ln(308 Kf318 K) = —80.3 J/K For the system, AS = AS“ + ASW = 92.4 J/K — 80.3 J/K = +12 J/K EVALUATE: Our calculation gives AS > 0, as it must for an irreversible process of an isolated system. 1“: Model: The ciosed cycle of the heat engine invoives the following four processes: isothermal expansion, isochoric cooling, isothermal compression, and isochoric heating. For a monatomic gas CV h %R . Visualize: p (kPa) 500% i ‘5 600 K isotherm E 1" / 250 —'. 4 2 j 300 K isofl'ierm q (5 -%--—- ----- —r——--V‘ 'r—' V(cm3) 0 2000 4000 Solve: Using the ideal-gas law, _ “RTI _ (0.20 moi)(8.3l J/TOngXtSOO K) *4.986><]05 Pa V 2.0XEO" m 1’: At point 2, because of the isothermai conditions, '1"2 :Tl = 600 K and . 2_ '3 3 . pl =ply'r:(4.986>(103 Pa) M =2.493x10’ Pa V2 4.0x10 ’ m At point 3, because it is an isochoric process, V3 : V2 : 4900 cm3 and T, 5 [300K] 5 = f: 2.493x10 Pa :l.247x10 Pa P3 p27; ( ) 600K Likewise at point 4, "I; = T3 = 300 K and — £"(1247x105 Pa) M -2493x105 Pa , p“ p314 ' 2.0x10"-‘m3 ' Let us now caiculate Wm: 2 W,_,2 + W2_,3 + WH4 + W4_,.. For the isothermai processes, WH2 : um; 01%:(020 mol)(3.31 J/mol K)(600 K)ln(2) : 691.2 .f wH : "RT, 1111 = (0.20 moi)(3.31 J/moI K)(300 Ming) 2 —345.6 J V3 For the isochoric processes, W2_,3 : W44. = 0 1. Thus, Wm 2 345.6 J z 350 J . Because Q : W5 + 551m QH2 :wHz 44015,h )Hz : 691.2 .1 + 0 J = 691.2 J For the first isochoric process, Q2413 : "CVAT : (0'20 mol)(§R)(T3 —T2) = (0.20 moi)§(8.31 Jlmol K)(300 K — 600 K) = 447.9 K For the second isotherma! process QM 2 WM + (A5,, L; -345.6 1 + 0 .I = 7345.6 J F or the secoud isochoric process, Q44, :nCVAT = ”(g RMT, —fl) = (0.20 moi)(§)(3.31 J/mol K)(600 K 7300 K): 7479 K Thus, QH : QHi +QH, :1439.! J . The thermal efficiency ofthe engine is w,“ 345.6 1 ,7 _ z —— _ : 0.24 : 24% QH 1439.1 J Q 9 i Model: For the Diesel cycle, process I u) 2 is an adiabatic compression, process 2 a 3 is an isobaric expansion, process 3 —) 4 is an adiabatic expansion, and process 4 _> i is isochoric. Visualize: Please refer to 0319.72. Solve: (a) It will be useful to do some calculations using the compression ratio, which is V V max __ l _ 1050 cm} _ vm v2 50 cm] _ The number of moles ofgas is ply] (1.0mm? Pa)(1050x104' m3) ,1: _ c ‘00430 mol l, RTl (8.31 .r/mol K)[(25+273) K] i For an adiabatic process, V Y pr,’ = pZVZ’ :> p2 —[w‘}‘—] p, :r’pI :21"in atm :71.0 mm =7.l9XlO6 Pa 2 rel v 71W":7;VJ":>E :[4] T, : #47; =2I.°“x293 K=1007 K V2 ®§ 65A“: “4A Process 2 —> 3 is an isobaric heating with Q 2 i000 J. Constant pressure heating obeys Q Q : nCPAT :5 AT : RC? The gas has a specific heat ratio y = 1.40 : 7f5, thus CV — 800 K and then T; 2 T2 + AT: 1807 K. Finally, for an isobaric process we have %R and CP : % R. Knowing Cg), we can calculate first AT: v v, #2: , 3v, :Evz £807 K(50><10‘6 m3):89.7xl0"’ m3 T, T, T, i007i< Process 3 —> 4 is an adiabatic expansion to V,1 : V._ Thus, r _6 3 H .7 - V3] p3=[89 “0 m } (7.19x!0"Pa):2.30xlfl” Pa:2.27atm V’: V’: :— p“ p“ p“ [v 1050x1045m3 4 04 Y" - 45 3 - V 89.7Xl0 m TVY_|:TVY_':T : A T I 1807K 2675K 3 3 “ “ " v, ’ iosoxur‘ m3 ( ) Point -p V T i 1.00atm= 1.013x105 Pa 1050>< 104:1? 298K: 25°C 2 71.0 atm=7.l9x 10° Pa 50.0)(10‘Sm3 1007K: 734°C 3 7|.0atm:7.l9x I06 Pa s9.7><10*"m1 1307K=1534°C 4 2.27 atm : 2.30 X 105 Pa 1050 x 10"6 m1 675 K = 402°C (b) For adiabatic process 1 _> 2,7 _pV-pV_ 00.2 —2+_,,-'—‘-~633J For isobaric process 2 —> 3, (wish; : PZAV : p2 (V3 7V1) 1285 J For adiabatic process 3 a 4, V — V (“15)34:2;_2Ll:10093 "3’ For isochoric process 4 —9 l, (WSLl =0 J. Thus, (“l/JD)“: :Wom 205/5)” +(W5)23 +(“’5)34 +01%)“ 2661 J (c) The efficiency is W,m 661 J QH 1000 J :66.l%=66% 1?: (d) The power output of one cylinder is 66! J)(2400_cyclex 1 mm :26,440 1:264 kW cycle min 60 sec 5 For an 8-cyiinder engine the power wili be 2] 1 kW or 283 horsepower. ...
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