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MAE204_F11_HW5_Solutions

MAE204_F11_HW5_Solutions - P ‘[00\FPC COMXY‘Mb...

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Unformatted text preview: P ‘ : [00 \FPC. COMXY‘Mb p$\;tfb/BIZ_ (D A 5 \ :HOO \LP .3 ' FEM;— vabm ; a mi '33P] 310° Has A%3~JN‘L bkEZbP/ESO ( QUSM (Leash/33¢ spec, L h \1 e7U/[I IbP,U/Y]/ .LCSCC\ CTCIX Un‘gg5 osvhmk: 35,6513 (T ~1\\) A P 0.3 a: “’75 : CJQ'EE; / A =LlumL \‘M 1 1606 HluuHL-z 7% k. CW“ ’ 0 W“ {XL (MA A .1) 2:5 QMM; : grotfim¥(uw_;qb) “NCO. “ l \ \ 9 b QWIQ ILJZ (DJ\7'ECV>;L fA’ Cnéb—s bounfi ’ P \"n [flu/J; «fly RU“: 0‘ \9801 ”/ng (,NAL A1) RV6_~ m RT; :3 2 MR1}. _\ Q LUdSo‘JGI)C L109”) 3 _ 5“ Pl H023 m 2 \\ ‘3’y’l hfifS '3 “953 1.9.8 3 l 0 \aVA & P1, V3 :7 V3: <PA W"; ) A'l” Hmcl.133bl)\'%>\/""° P3 \ Lad ’4 8,23%; m3 :7 [05,“: W543: VA _‘ Q00)QZ,EH%)—(H0o)(l.l’31‘l) “F‘fi— - W 1 923,? L? wmg: H ”911,? L? Lam; 99+ (QINA'S ’* mug‘“mma’v “(3,31 2 MCUM CE‘T;§*LJ5 (3-1 :(gucmm) (T3433 Arumfi‘) LT, 11: now, T3 '2— P’; v1 V LVJO >L$.’$L{&) Q3 2 ~ , L m R cmwsc‘fig? ‘ 830% 2 =7 me :0, We; (8%‘2— \aoe) +Hgsr) 143‘ ; !—\'>9J.71crgggfw 4-11 Refrigerant-134a in a cylinder is heated at constant pressure until its temperature rises to a specified value. The boundary work done during this process is to be determined. Assumptions The process is quasi-equilibrium. Properties Noting that the pressure remains constant during this process, the specific volumes at the initial and the final states are (Table A—ll through A—13) P = 500 kP ' , , a V1 [email protected] 1d,, =0.0008059 m3/kg P Sat. hquld (kPa) P = 500 kP 2 a v2 =0.052427 m3/kg 1 2 T2 = 70°C 500 Analysis The boundary work is determined from its definition to be V1 0.05 m3 U V] 0.0008059 m /kg and 2 Wm... = I] Pdv = P(Vz — «4) =mP<v2 — v1) = (62.04 kg)(500 kPa)(0.052427 — 0.0008059)m3/kg is 1 kPa - m = 1600 RJ Discussion The positive sign indicates that work is done by the system (work output). 4-21 C02 gas in a cylinder is compressed until the volume drops to a specified value. The pressure changes during the process with volume as P = a l/_2 . The boundary work done during this process is to be determined. Assumptions The process is quasi-equilibrium. P Analysis The boundary work done during this process is determined from 2 2 a 1 1 Wm —IlPdV— L [WJdV_—a[V_2—71] 1 =_(3kpa.m6) 1 3_ 1 3 —kJ3 0.1m 0.3m lkPa-m = —53.3 kJ Discussion The negative sign indicates that work is done on the system (work input). 4-24 A piston-cylinder device contains air gas at a specified state. The air undergoes a cycle with three processes. The boundary work for each process and the net work of the cycle are to be determined. Properties The properties of air are R = 0.287 kJ/kg.K , k = 1.4 (Table A-2a). Analysis For the isothermal expansion process: v, = mRT = (0.15 kg)(0.287 kJ/kg.K)(350 +273 K) = 0.01341m3 p] (2000 kPa) Air 2 MPa v2 = M” = W = 0.05364 m3 3500c P2 (500 kPa) 0.05364 m3 0.01341 m3 WM2 = PM ln[%] = (2000 kPa)(0.01341m3)ln[ =37.18kJ 1 For the polytropic compression process: 1021/2" = P3V3” ——>(500 kPa)(0.05364 m3 )‘~2 = (2000 kPa)V3‘2 ——>V3 = 0.01690 m3 PV—PV . 3— . 3 Wb,2—3 = 3 31 2 2 = (2000 kPa)(0 01690 m1 )1 2500 kPa)(0 05364 In ) = _34_86 N —n — . For the constant pressure compression process: Wb,3_1 = 133(1/1 — 1/3 ) = (2000 kPa)(0.01341— 0.01690)m3 = -6.97 kJ The net work for the cycle is the sum of the works for each process Wnet = W,,,1_2 + Wb,2_3 + Wb,3_] = 37.18 + (—34.86) + (—6.97) = -4.65 kJ 4-38 A cylinder is initially filled with saturated liquid water at a specified pressure. The water is heated electrically as it is stirred by a paddle-wheel at constant pressure. The voltage of the current source is to be determined, and the process is to be shown on a P—Vdiagram. Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 The cylinder is well- insulated and thus heat transfer is negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 The compression or expansion process is quasi-equilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as E in ‘ Eout = AEsystem W ad Net energy transfer Change in internal, kinetic, by heat, “/0119 and mass potential, etc. energies Wedn + WPW,in _ Wb,out = AU (Slnce Q = K-E = PE = 0) We,in +pr,in = ”1012 _hl) (VIAt) + pr,in = m(h2 —hl) since AU + W, = AH during a constant pressure quasi-equilibrium process. The properties of water are (Tables A-4 through A-6) sat.liquid u, = [email protected] m = 0.001057 m3/kg P P = 175 kP 2 0 5 a } h2 = hf + xzhfg = 487.01+ (0.5 x 2213.1): 1593.6 kJ/kg x2 = . 3 m=fi=—0'005m3 =4.731kg 1 2 v1 0.001057 m /kg Substituting, VIAt+(400kJ) = (4.731 kg)(1593.6— 487.01)kJ/kg V VIAt = 4835 k] 4835 kJ [1000 VA = — = 223.9 v (8 A)(45>< 60 s) 1 kJ/s 4-46 Two tanks initially separated by a partition contain steam at different states. Now the partition is removed and they are allowed to mix until equilibrium is established. The temperature and quality of the steam at the final state and the amount of heat 10st from the tanks are to be determined. Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero. 2 There are no work interactions. Analysis (a) We take the contents of both tanks as the system. This is a closed system since no mass enters or leaves. Noting that the volume of the system is constant and thus there is no boundary work, the energy balance for this stationary closed system can be expressed as E in — E out = AE system ‘—v—’ H—‘ Net energy 'mee' Change in internal, kinetic, Q by heat, work, and mass potential, etc. energies — Qout = AUA + AUB = [m(u2 — u,)]A + [m(u2 — u,)]B (since W = KE = PE = 0) The properties of steam in both tanks at the initial state are (Tables A—4 through A-6) PM = 1000 kPa um = 0.25799 m3/kg TM = 300°C um = 2793.7 kJ/kg TU, =150°C uf =0.001091, ug =0.39248 m3/kg x1 = 0.50 u f = 631.66, u f8 = 1927.4 kJ/kg a”, = uf +xlu 12: = 0.001091+[0.50 x(0.39248— 0001091)]: 0.19679 m3/kg um = u f +x1ufg = 631.66+(0.50><1927.4)= 1595.4 kJ/kg The total volume and total mass of the system are u = VA +1/B = mAUM +mBuU, = (2 kg)(0.25799 m3/kg) +(3 kg)(0.19679 m3/kg) = 1.106 m3 m=mA +mB =3+2=5kg Now, the specific volume at the final state may be determined 1.1 3 u2 =2 =fl= 0.22127 m3/kg m 5kg which fixes the final state and we can determine other properties T2 = [email protected] =133-5oc 122 =300 kPa }x _ u2 —uf _ 0.22127—0001073 _ 2 v2 = 0.22127 m3/kg ' 0g —uf ' 0.60582—0.001073 ' u2 = uf +x2ufg = 561.1 1 + (0.3641x 1982.1) = 1282.8 kJ/kg 0.3641 (b) Substituting, _Qout =AUA +AUB = [m(u2 —“1)l,4 + [m(u2 —“1)l3 = (2 kg)(1282.8 — 2793.7)kJ/kg + (3 kg)(1282.8 —1595.4)kJ/kg = —3959 U 01‘ Qom = 3959 RJ 4-57 The total internal energy changes for neon and argon are to be determined for a given temperature change. Assumptions At specified conditions, neon and argon behave as an ideal gas. Properties The constant-volume specific heats of neon and argon are 0.6179 kJ/kg-K and 0.3122 kJ/kg-K, respectively (Table A—2a). Analysis The total internal energy changes are AUneon = mcu AT = (2 kg)(0.6l79 kJ/kg . K)(180—20)K = 129.7 kJ wagon = quT = (2 kg)(0.3122 kJ/kg - K)(180—20)K = 99.9 kJ 4-67 Nitrogen in a piston-cylinder device undergoes an isobaric process. The final pressure and the heat transfer are to be determined. Assumptions 1 Nitrogen is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 126.2 K (227.1 R) and 3.39 MPa (492 psia). 2 The kinetic and potential energy changes are negligible, Ake —:— Ape —:— 0 . 3 Constant specific heats at room temperature can be used for nitrogen. Properties For nitrogen, cp = 1.039 kJ/kg-K at room temperature (Table A-2a). Analysis Since this is an isobaric pressure, the pressure remains constant during the process and thus, P2=Pl=1MPa We take the nitrogen as the system. This is a closed system since no mass crosses the boundaries of the system. The energy balance for this system can be expressed as E in E out _ AE system W HF; Net energy transfer Changein internal, kinetic, by heat, work, 311d mass potential, etc. energies —qom —w,,,°u, =Au=(u2 —u1) (sinceKE =PE =0) _qout = wb,ou! +(u2 _ul) _qout =hZ _hl qout =mcp(Tl_T2) since Au + wb = Ah during a constant pressure (isobaric) quasi-equilibrium process. Substituting, qout = cp (T1 — T2) = (1.039 kJ/kg - °F)(427— 27)°F = 416 lekg 4-77 A cylinder initially contains nitrogen gas at a specified state. The gas is compressed polytropically until the volume is reduced by one-half. The work done and the heat transfer are to be determined. Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 The N2 is an ideal gas with constant specific heats. 3 The thermal energy stored in the cylinder itself is negligible. 4 The compression or expansion process is quasi-equilibrium. Properties The gas constant of N2 is R = 0.2968 kPa.m3/kg.K (Table A- 1). The cuvalue of N; at the anticipated average temperature of 350 K is 0.744 kJ/kg.K (Table A-2b). Analysis We take the contents of the cylinder as the system. This is a closed system since no mass crosses the system boundary. The energy balance for this closed system can be expressed as Ein _ Eout _ AEsystem ‘ér—’ H—l Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies ”1min _ Qout 2 AU = m(u2 _ "1) Wan _ Qout = mcv (TZ _ Tl) The final pressure and temperature of nitrogen are Q V 1.3 1921/;3 = Flt/113 —> P2 =[V—‘J Pl =21-3(100 kPa)=246.2 kPa 2 v PV v 4 . —P‘ ‘= 2 2 —>T2=5—2Tl =flx0.5x(290K)=357.0K T1 T2 Pl V1 lOOkPa Then the boundary work for this polytropic process can be determined from W. jmflm b,m l l—n 1_n “W495“ 1—1.3 ' Substituting into the energy balance gives Qout = Wb,in _mcv(T2 _T1) = 99.5 k] — (1 .5 kg)(0.744 kJ/kg.K)(357.0 — 290)K = 24.7 kJ ...
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