Solution_10.pdf - 7 Hot exhaust gases leaving an internal...

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Unformatted text preview: 7!. Hot exhaust gases leaving an internal combustion engine is to be used to obtain saturated steam in an adiabatic heat exchanger. The rate at which the steam is obtained, the rate of exergy destruction, and the second-law efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Air properties are used for exhaust gases. 4 Pressure drops in the heat exchanger are negligible. Properties The gas constant of air is R = 0.287 kag.K. The specific heat of air at the average temperature of exhaust gases (650 K) is cp = 1.063 kJ/kg.K (Table A-2). Analysis (a) We denote the inlet and exit states of exhaust gases by (1) and (2) and that of the water by (3) and (4). The properties of water are (Table A-4) T3 = 20°C} h3 = 83.91 kJ/kg Eggécgas x = 0 s = 0.29649 kJ/k .K 350°C 3 3 g 150 kPa Hmt T4 = 200°C 11,. = 2792.0 1(ng Exchanger X4 =1 S4 = 6.4302 kJ/kg.K Sat. yap. Water 200°C 20°C An energy balance on the heat exchanger gives mah, + thwh3 = mah, + rhwh4 macpm — T2) = ”1.0014 _ h3) (0.8 kg/s)(1.063 kJ/kg°C)(400 — 350)°C = mw(2792.o — 83.91)kJ/kg rhw = 0.01570 kg/s (b) The specific exergy changes of each stream as it flows in the heat exchanger is T Asa = c, 1n—2 = (0.8 kg/s)(1.063 kJ/kg.K)lnw T1 (400 + 273) K = —0.08206 kJ/kg.K Al/la : Cp(T2 — 7.l) _ TOASa = (1.063 kJ/kg.°C)(350 - 400)°C - (20 + 273 K)(-0.08206 kJ/kg.K) = —29. 106 kJ/kg All/w = h4 '“ ha ‘T0(S4 ‘53) = (2792.0 — 83.91)kJ/kg — (20 + 273 K)(6.4302 — 0.29649)kJ/kg.K = 910.913 kJ/kg The exergy destruction is determined from an exergy balance on the heat exchanger to be — xdest = WA 1/4, + n'1wA w. = (0.8 kg/s)(-29.106 kJ/kg) + (0.01570 kg/s)(910.913) kJ/kg = —8.98 kW or x,est = 8.98 kW (c) The second—law efficiency for a heat exchanger may be defined as the exergy increase of the cold fluid divided by the exergy decrease of the hot fluid. That is, _ mwww _ (0.01570kg/s)(910.913kJ/kg) —-,—- = 0.614 — maAl/la — (0.8 kg/s)(-29.106 kJ/kg) 2 , A modern compression ignition engine operates on the ideal dual cycle. The maximum temperature in the cycle, the net work output, the thermal efficiency, the mean effective pressure, the net power output, the second-law efficiency of the cycle, and the rate of exergy of the exhaust gases are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at 850 K are cp = 1.110 kJ/kg‘K, c., = 0.823 kJ/kg-K, R = 0.287 kJ/kg-K, and k = 1.349 (Table A-2b). Analysis ((1) The clearance volume and the total volume of the engine at the beginning of compression process (state 1) are 3 14=Vc+0.0028m —>vc = 0.0002154 m3 = v2 : vx V5 ‘4 1/1 = vc + vd = 0.0002154 +0.0028 = 0.003015 m3 = v4 Process 1-2: Isentropic compression k-l T2 = T1[%1—] = (328 K)(14)1‘349'1 = 823.9 K 2 ”—1 k P2 =P1[ ] = (95 kPaX14)1'349 =3341kPa “2 Process 2-x and x-3: Constant-volume and constant pressure heat addition processes: P Tx =T2 sz (823.9 K):—232—%= 2220K 2 q“ = cu (TX — T2) = (0.823 kJ/kg.K)(2220 — 823.9)K = 1149 kJ/kg q2_x : qx_3 = cp (T3 —Tx ) —>1149 kJ/kg = (0.823 kJ/kg.K)(T3 — 2220)K ——>T3 = 3254 K (b) qin = q2_x + qx_3 = 1149 + 1149 = 2298 kJ/kg 3254 K 2220 K T v3 = vx T—3 = (0.0002154 m3) x = 0.0003158 m3 Process 3-4: isentropic expansion. V H 00003158 3 13494 T4 =T3[—3] =(3254 Ki—l] =1481K V4 0.003015 m3 v k 0 0003158 3 1349 P4 = P3 A = (9000 kPa ———E3— = 428.9 kPa V4 0.003015 In Process 4-1: constant voume heat rejection. qout = cu (T4 — T1): (0.823 kJ/kg - K)(1481— 328)K = 948.7 kJ/kg The net work output and the thermal efficiency are net,out W ”m = net,out _ 1349 kJ/kg =°.587 qin _ 2298 kJ/kg (c) The mean effective pressure is determined to be _P1(/1_ (95kPa)(0.003015m3) _0003043kg RT1 (0.287 kPa -m3/kg-Ki328 K) m MEP = mwnew _ (0.003043 kg)(1349 kJ/kg) kPa-m3 Vl-Vz (0.003015—0.0002154)m3 kJ )2 1466 kPa (d) The power for engine speed of 3500 rpm is Wnet = mwnet % = (0.003043 kg)(1349 kJ/kg)————3500 (”V/mm) [1 mm )2120kW 605 (2 rev/cycle) Note that there are two revolutions in one cycle in four-stroke engines. (6) The second-law efficieny of the cycle is defined as the ratio of actual thermal efficiency to the maximum possible thermal efficiency (Carnot efficiency). We take the dead state temperature and pressure to be 25°C and 100 kPa. T0 _ (25 + 273) K — = 0.908 77““ T3 3254 K and — ’7‘“ =——0'587 =0.646 0.908 max The rate of exergy of the exhaust gases is determined as follows T P x4 = 144 —u0 -—T0(s4 —sO) = c‘,(T4 —T0)—T0[cp ln—4—Rln—i] T0 P0 = (0.823)(1481— 298)— (298) (1.110 kJ/kg.Kln 1481 —0.287 In 428'9 = 567.6 kag 298 100 X4 = mx4 1 = (0.003043 kg)(567.6 kJ/kgfisflOfl/m—m)£1 mm) = 50.4 kW 2 (2 rev/cycle) 60 s 5 ,1 The total exergy destruction associated with the Brayton cycle described in Prob. 9-119 and the exergy at the exhaust gases at the turbine exit are to be determined. Properties The gas constant of air is R = 0.287 kJ/kg-K (Table A-l). Analysis From Prob. 9-119, qin = 480.82, go“, = 372.73 kJ/kg, and T1 = 310 K —> s; =1.73498 kJ/kg-K h2 = 618.26 kJ/kg -—> s; = 2.42763 kJ/kg- K 1150 K T3 =1150 K —> s; = 3.12900 kJ/kg-K h4 = 803.14 kJ/kg —» s; = 2.69407 kJ/kg-K h5 = 738.43 kJ/kg ——> s; = 2.60815 kJ/kg-K 310 K and, from an energy balance on the heat exchanger, h5 —h2 = h4 —h6 ——>h6 = 803.14—(738.43—618.26) = 682.97 kJ/kg ——> s; = 2.52861 kJ/kg ~ K Thus, P _ _ _ o o 2 xdestroyed,12 — TOSgen,12 —T0(52 “S1)— To[52 ‘51 —R1n T, 1 = (310 K)(2.42763 —1.73498 — (0.287 kJ/kg . K)1n(7)) = 41.59 kI/kg P _ _ _ o o 4 xdestroyed,34 _ TOSgen,34 — T0 (S4 _S3)_ T0 [S4 _S3 —Rll’l P 3 = (310 K)(2.69407—3.12900 —(0.287kJ/kg - K)1n(1/7)) = 38.30 kJ/kg xdestroyed,regen = TOSgen,regen = T0 [(55 _ $2 )+ (S6 _ S4 )] : T0 [(SS‘3 _ 52 )+ (S; _ S2 )1 = (310 K)(2.60815 — 2.42763 + 2.52861— 2.69407) = 4.67 kJ/kg 4R 53 P3 &0 q- xdestroyed,53 = TOSgen,53 = To 53 ‘55 ” T, = To 53 ‘55 ‘Rh‘? ‘ Tm R 5 H 480.82 kJ/kg = 310 K 3.12900— 2.60815 — = 78.66 ( l 1800 K ] kJ/kg <90 5111,61 0 a P1 qout xdestroyed,61 = TOSgen,61 = T0[Sl _S6 + T 2 T0 51 _SG _Rln? + T R 6 L 372.73 kJ/kg : (310 K{1.73498— 2.52861+ 10 K J: 126.7 kJ/kg Noting that kg = [email protected] 310 K = 310.24 kJ/kg, the stream exergy at the exit of the regenerator (state 6) is determined from V2 &0 $0 ¢6 = (’16 “h0)—T0(56 ‘ 50)+—§— + 826 P (:90 where s6 —s0 = s6 —s1 = sg —sf 42m?6 = 2.52861—1.73498 = 0.79363 kJ/kg-K 1 Thus, 6 = 682.97—310.24—(310 K)(0.79363 kJ/kg - K): 126.7 kJ/kg A__ ..... Amnr’s m, T. = We, P1 = 2.0m, Ta,A=..._599:'sL_W _, P4 = 7.51%qu 151 = 2630 Sj/firL TH: 1500 K, TLffloo kl # 1:, = 300K ,7 ,, , _ iinnd: (ca dun am.“ (0 Wm): (00W (*9 Xd “-3 '11 A Solution _ , _fi A K;- anaim figi‘wrngzg gggfim:w*gv 53th 3mg? _ _____ _ ______ , ; _____ ”=1 Pifinflfujl::,,,5_99:EV=3_,5:f.....33?§;95"£3,553I S _= 6.za§____§J/eggx A __ “a"; 3’1 = 2.0 Wu, 3: = S. ==> R2 = 2999.45 53/53} _ “‘53 Pa = Pa, T3= 500‘c m:=> 83: 39-616 R3153, 53 = “an H/Rj-K “4» , P4 = 7.5 Rm, 5“ g 53 a 2‘, = 0393; 5‘, = 2317.13 133$qu y "5" 2» sat. 33'3“”?! 95 = P4: ==> ’55 = 158.99 Big, 55 595747 RJ/fz‘g-k ,, 175 = 0.0mm 1313/61 v, ,,,,, M ._.:§L?6= PI = 8 ”Pa ”Ti—“mi “5 *.,£’5$,‘,,’9:.m'§>ii]? 35 Mg; aim = '{nEifir $0105 - 31)]: o. 7306 I(ssqs.osV—I1e.3s}+ (3mg -.;k<n9.45)] mam: w) = if: = 03306 3313 ‘ 4, f M. - W _ i v ,- {1 m g {a ”m Aeuiw/ W “Ninja = mMRS) = 07306133” ~ “8-99) , - iikigm fig‘w WM ” ,, ,, ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, ‘ mm = dim ~ (flout = 1115.69 R»; #1? 39% “g” firm? fiwnggr 11%;.” , ““““““ W ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, bond: ”35.67 ”’14:, = 61;" 39,553 = 0.4176 {44.76%} , “23m fiwxesgiémmmxaéérmfiwm H 2d»; = 39,“. fa“, = 0 ‘ Renting: procesSes _____________________________________________________________ _, xdn-a = (1'43“)6‘223 ”V 7h(u;“$’3) wfiete: Wa'ué = (31‘33)~?o{32*33) = [g— ~%:-)(m—as) + ($2.53) - Tom—33)} m E = ~%{£,~Ba)-To(s,-sa)]mi ‘ = 86.93 SW 0 __ _ > = 61.03 fiw .- 1 u. A _ _ TH ' “ I 7777777777777777777777777777777777777 T ,, M94 ‘0’?)6164'4' m($6” 3*) = [("":3,;'Xfiu'fie) + (fie-S.)— $5“;ng = Hag-m - 1-006-st VVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVV .............. T : 813.83 aw $21»sz ~§ 39‘ La 43%: g 12: ‘gé' at a? {3} 1:29:31? 3’ I 7&1; = (1- E3643 = (I- 3” o.) 3615.53 7 52.27. I) ...
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