MEE 302 Final, Spring 2013

Energy Conversion — Final Exam Name: Date: School of Engineering — Dep. of Industrial & Mechanical Eng. Stat Key Tuesday, June 4" 2013; 04:00 PM Location: SCI 402 Instructor: Dr. Wassim Habchi :) Notes: No documents allowed 1 00 Value: 40% of Total Grade Time: 3 hours Problem I (30 points) An ideal vapor-compression refrigeration cycle operates with Refrigerant 134a as the working fluid. Saturated vapor enters the compressor at -10°C, and saturated liquid leaves the condenser at 28°C. The mass flow rate of refrigerant is 6 kg/min. a) Sketch a T-s diagram of this cycle (5_points). b) Determine the specific enthalpy A of the refrigerant at all points in the cycle and the compressor power, in kW (7 points). c) Determine the refrigerating capacity, in kW (3 points). d) Determine the coefficient of performance (3 points). e) Using the definition of the Joule-Thomson coefficient: wt --»-1(2) | And assuming that the equation of state of refrigerant 134a is: P(v+a)=RT Where a is a constant positive number, determine whether the refrigerant temperature will decrease, increase or remain constant during the throttling process (5 points). f) Using Clapeyron equation (4) - as , determine the enthalpy of vaporization of the refrigerant at -10°C. Compare the diicel value to the tabulated one (7 points). Solution: All 2 Lebanese American University MEE 302 — Final Exam he tere 2 Bs, enwen oma ¢ —_ hee 2kec = s a Roe rere. = 427.2) KR @ @. Maha F225 KR Co.rWe, # OS3AC6 KINI. K— = 240. Uy — Siz Siz O82 Kyle. @okrhe f 0.83366 Ky/14 KK A 233.26 ky], walle O. _ COREL core x (233 126 — 230.41) + 270.4 Ok «6.2 => Wop are (a. 0) = £ x (231.188 ~ 2uu.Sl) = [2eca ko] €) Wee wk (4, -4.) = & x ( 264.81 — 90.68) = HS 822 kas se 2.6678 P i Lebanese American University MEE 302 — Final Exam 1) Oa + (2 AT st @-10% bok es = &. Oy = 0.038f5(6 — 0. 0007825 = 0.088K2 am “/ hy “ (Brat (BL: Meee at AQ wre ST st Owe ~ Cote = 219-08 ~ S.8% = 7.5295 Khe u K = Qe ewe = ZECENSS 0.03362 Sx OR omg = [oe 031 eg 4 | . A blcted Uolne— ‘Be 208.36 Ky] 4 us idk, “n vety dere te Ra cdetets voles . Lebanese American University Problem II (30 points) Consider a reheat-regenerative vapor power cycle with two feedwater heaters, a closed feedwater heater and an open feedwater heater. The cycle and its corresponding 7-s diagram are shown below. Steam enters the first turbine at 8.0 MPa, 480°C and expands to 0.7 MPa. The steam is reheated to 440°C before entering the second turbine, where it expands to the condenser pressure of 0.01 MPa. Steam is extracted from the first turbine at 2 MPa and fed to the closed feedwater heater. Feedwater leaves the closed heater at 205°C and 8.0 MPa while the steam extracted from the turbine exits the closed feedwater heater as saturated liquid at 2MPa which is then throttled into the open feedwater heater. Steam extracted from the second turbine at 0.3MPa is also fed into the open feedwater heater, which operates at 0.3 MPa. The stream exiting the open feedwater heater is saturated liquid at 0.3 MPa. The net power output of the cycle is 100 MW. Determine: a) The specific enthalpy / of the water at states 1 to 13 (13 points). b) The fractions of steam y’ and y”’ extracted from the turbine (7 points). c) The thermal efficiency (7 points). d) The mass flow rate of the steam entering the first turbine, in kg/h (3_points). ! T, =480°C T, = 440°C 8.0 MPa i 12 D0 MPa 0.3 MPa 0.01 MPa WwW, PN // Condenser Sx / 0.01 MPa \y Lebanese American University MEE 302 — Final Exam Solution: Fist Ok's dikimraie He ensgetes Kk oD pert » CO. &= nea [+ Q.— 3348.02 Kyle | 2 sia 668812 Kgl K Ty = & 80° a ee poe = —a|h. = 2364.38 kz/ig. | Smeesee OES S12 Kylky.k . Ss=S = 6.08812 «rly «> “ Ay Qe; Ope — 6334 o.Ph8828y 2060. £ Al, = 244). 28 Ky] 8 8 EL i= 2st tal Y= 4hore ° 8 - Ps 20- 3NP i Bow Tol. 38S x / 2 Sec S — PiISIez Ky (le. k Set 7.98862 - POUT 0.948) _ = ———_—_——— 2©: fi om Mh l , = a § ae Sem See BBSRC? KG ° > Nee Lee Ne Res = A4\. Bl + oly 2292 .\ Ren 2053.% Kq/4 ®@. A= Age v5 (yi) AVitg 0.001010 x ( 0.3 - 0.01) xlo” = [492.4 kif | 6: =p aim aria] @. Moe Og e 05(P,_ 4) = ShU3 4 001093 x (000-300) =/565.68 ley [lk ae VK = 2oS°c \- bn 2 } (3 “oi am Stes : Lebanese American University MEE 302 — Final Exam k) Em erp y bals—ex « Bp Fou. 4" es a (A-4'- 5°) Qe #yr “as, ee Qoed Fate Mu Bo = yt (A. -bn) © ‘ Qu Qo _ 992-518 _ 568.68 On s'= Gite = Wee sse8 _ Gisay] we Z23CL.5S'., Sos: O— Wy Bok. 38s 4 (\- 0.4393 — 4") x AG2 1 ONS FZ x BORGF = $61.42 “Sl. 28Sy " — 262.0643 7“ = 0.088984 (R, - Bu) & Cle g!) (Qc - 3) = (gous Ort - 23cu.38), (i= 0.14842) ( 286u 33 — 29098 % F3Sh.US Bos 285) + (10.6992 ~ 0.0999 £9) ( Hol. 386L~ 2459.96 — A824 451.81) ~(S43-63_S61.43) (3168.02 — 42.513) + (1 0.14392) (2954.U3_ 274(. 38) = U9 + 401.9354 + GES .299 — B26 Y2e8e. 122, — = O.422F 2542 . 4648 a k2. 9h +. 247(.S0} + 520, $663 =~ rs ad Ab 0x lo? ——- = ——~= 78.859 ad {ust (hau K/e =|2£2 £20. % 4] | 6 — Lebanese American University MEE Problem III (25 points) A regenerative gas turbine with intercooling and reheat operates at steady state. The corresponding cycle and 7-s diagram are shown below: 19 || Regenerator |! py=100kPa | ======= — an aan _ 25 = px = 300 kPa Z 7 : nit = 5.807 kg/s Pome : is) 9 Qo a 195 eo a 4s 4 3 por ase we og ae 3 1 s Air enters the compressor at 100 kPa, 300 K with a mass flow rate of 5.807 kg/s. The pressure ratio across the two-stage compressor is 10. The pressure ratio across the two-stage turbine is also 10. The intercooler and reheater each operate at 300 kPa. At the inlets to the turbine stages, the temperature is 1400 K. The temperature at the inlet to the second compressor stage is 300 K. The isentropic efficiency of each compressor and turbine stage is 80%. The regenerator effectiveness is 80%. Determine: a) The specific enthalpy h of Air at states 1 to 10 (10 points). b) The thermal efficiency (5_points). c) The back work ratio (5 points). d) The net power developed, in kW (5 points). Solution: o) e @. ys t00K —S XQ. = 200.18 Kz [ey 3 FG, = AE Stee =e «=» Ges Sx Bo Betioe ket Lebanese American University MEE 302 — Final Exam ie \= Gis -% = og zea ee Bas Oy As ES OE ett 0.3 0.8 , ha - & Wi= 433.024 43/6 e 3 — |9,=- Yoo. L$ Ky | A Pr, = ABR = = = — Sages ——» ‘ =’ . &6 = G62 _— > — _— x —» Osa 423. 786 Keg [Ie . bar & = pis Os a Ap = W23-736- 200.18 + 300.14 o.8 SS ifs 454. 694s kglkeg “Tox. obfech ve~ers oh tas [eige —encttn Eun. Qe 2@. Tes Wook —s [Rea ASS. 42Kz/H] SR. kSos fr. 3 Fea = ri ve a 0.&x LSo.g— A3S.1F Pee R Woo = Das— 1095. &6¢ Kz /k ¢—- Wt = ASIS.42 -0.8(isis. (a lo Ss . 266) =o = 1179.3768 Ks /h = 1 2 ke _ 32 s Mec Me ~ USos _ ASo A 8 [hss = 1423.64 Key Lebanese American University MEE 302 — Final Exam Lex \= Ss %s or —» Aga he 08 (Ke_ Qs) Qe _ Ag, Qe = ASIS.42- 0. 8(\SI5.42 — W24..¢2) Ag = A205. 456 x gl& &) =>) ¥. = Se of yO yak, = O.&( A208 .1g6 _ &is= AoSS. 0963 Ky/ PA Lly ; Emegy bola rege er: Qe wiles = Och, = Res Qe Qe +8, = AS. 16 _ Hogs. O36? +h 56.6378 Qox Cou. 2992 KI b) Yu 22 Nye = att GBs) (Ge As) ~(€4-%,) -( ty -@) $$ ee \~ (4 ~ &e) + (he 43) +(Ais.42 — Atos. 136) — (454.6895. 3eo.18) —(48s.004 (USIS.42 — Aoss.0%62) + (ASIS.42 — A494, 7262) (]3s.co32.. Ro. 224) (AS45096 « 128.940 GUS 8492 — 282 216 WSo.3232 4 235.6422 7 > Ne. = OUL2~SES =lLL.2E5 7. a= Compe . @,-&4) + (%- 83) 283. a Se NE S042 als. ¢ tind (46.29) + (Gg ~ 43) GUS . 8U22 Gen mali he of. 1) Wd = * , = ™ro.t = mx (645. 9432 - 282. % IS) = S.€oF (CoS. 622-287, 346) USh CRP) 4 USL. 6998 = USI5.42 — 4479.96 4 | \ F8S.966 8 Lebanese American University inal Exam Problem IV (15 points) The combined gas turbine—vapor power plant shown below has a net power output of 45 MW. Air enters the compressor of the gas turbine at 100 kPa, 300 K, and is compressed to 1200 kPa. The isentropic efficiency of the compressor is 84%. The condition at the inlet to the turbine is 1200 kPa, 1400 K. Air expands through the turbine, which has an isentropic efficiency of 88%, to a pressure of 100 kPa. The air then passes through the interconnecting heat exchanger and is finally discharged at 400 K. Steam enters the turbine of the vapor power cycle at 8 MPa, 400°C, and expands to the condenser pressure of 8 kPa. Water enters the pump as saturated liquid at 8 kPa. The turbine and pump of the vapor cycle have isentropic efficiencies of 90 and 80%, respectively. Determine: a) The mass flow rates of the air and the steam, each in kg/s, and the net power developed by the gas turbine and vapor power cycle, each in MW (10 points). b) The thermal efficiency of this combined cycle (5 points). Gin Combustor T, = 1400 K 2 3 p3= pr = 1200 kPa N. = 84% 4 1 = 88% sie Gas turbine ] ‘Compressor i = 4 4 i Wass = a es is Wi te “A 7, =300K 1 | 1 p, = 100 kPa Air inlet 4 Exhaust <#————_+++} Ts=400K > —jo— = 400°C 7 : =8 MPa Heat exchanger a 1), = 90% Turbine 3 > +6 Vapor ie | Wyap= cycle a J We Wp Pump Condenser SWUM 0... (a Np = 80% Po = pg=8 kPa Solution: -) + Fi Beil ty yet Boots ug lig] f f= 42he Ff. Moo wy Fale = Wb 4.286 = (6.632 F, ‘a oo = = — Glo. iS Ky [ie . 10