SIMPLE RANKINE CYCLE. 3 expander. boiler. pump. condenser 1 W Q. cycle cycle. net. shaft

Transcription

1 SIMPLE RANKINE CYCLE um boiler exander condener Steady Flow, Oen Sytem - region ace Steady Flow Energy Equation for Procee m (u Pum Proce,, Boiler Proce,, V ρg) 0, 0, Exanion Proce,, 0, Condener Proce,, 0, aft out out m m m ( ) ( ) ( ) ( ) m Firt Law cycle cycle δ cycle cycle cycle δ net for Cycle cycle

2 ) ( ) ( ientroicroce VmR R IdealGaModel PROPERIES m m m u m u ysytem Unteadu gz V u m OenSytem E CloedSytem FIRSLA boundary o

3 CARNO CYCLE IH AER. C Carnot net H S L H L C net maximum at 0 C : net Carnot net.9 % 0. kj/kg maximum area H

4 SUPERHEA RANKINE CYCLE boiler um condener REHEA RANKINE CYCLE um boiler condener turbe

5 U Firt Law dq du dw dq d ubitiutg for dq and dw, d d ( du) ( d) ubitiutg for du, ( d d d) ( d) d d d for an adiabaticroce, d 0 d dw d u d du d d d d Second Law Boundary ork roerty defition, i an exact differential Examle: water umed from ia to 0 ia w ( ) ( 0ia ia) i/f w.lb/ft lbf 0 ft lbf w ft., (ft of fluid) lbm. lbm ft ft ft lbf BU w..0 BU/lb lb ft lb m Examle: water umed from 00 kpa to 00 kpa w ( ) w.000 m /kg w.0 m kg kpa, f m ( 00 kpa 00 kpa) kj/kg

6 A team ower lant run on a reeat cycle and roduce 0 M. e turbe let condition are 0 MPA, 00 C and MPA, 00 C. e condener oerate at 0 kpa. e efficiency of te turbe i 0%. e efficiency of te um i 9%. Determe: a) te turbe exit condition b) te cycle efficiency and c) te ma flow rate of te team. boiler 9% 0 MPa MPa 0 kpa 0% 9 0 condener turbe ex

7 A team ower lant run on a reeat cycle and roduce 0 M. e turbe let condition are 0 MPA, 00 C and MPA, 00 C. e condener oerate at 0 kpa. e efficiency of te turbe i 0%. e efficiency of te um i 9%. Determe: a) te turbe exit condition, b) te cycle efficiency and c) te ma flow rate of te team. Pt kpa 0 MPa 0 MPa 0 MPa 0 MPa MPa MPa MPa 0 kpa 0 kpa % 0 MPa MPa 0 kpa 0% ex

8 9-9 ex. J/kg MPa, 0. kj/kg.9 0) (0, kj/kg. 9.. kj/kg kj/kg MPa able A -, water, comreed liquid um um umactural actual um ideal um actual um ( ) ( ) ( ) ( ) C) 90 9., (. kj/kg kj/kg x kpa, 90.0 kj/kg..... O at fg f 9

9 out d ( ) ( ) ( ) (. 0.9). kj/kg ( 0 ) (.9.) ( ) ( 0 ) um (. 90.0) ( 9..) d out net... net. cycle.0%. otal ork 0,000kJ/ec m Secific ork kj/kg kj/kg.. kg/ec 9-9 ex

10 EES Model

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12 and deendent on at d FIRS LA LA δ δ Cycle CLOSED SYSEM quantity of ma E Proce δq du d Procee and for c, c, c, adiabatic, olytroic OPEN SYSEM V m (u Procee comreion, region ace gz) exanion, eat excanger, trottlg, diffuer, nozzle UNSEADY SYSEM unequal ma flow m u m u (m m ) Firt Law i an Energy Balance o boundary

13 PROPERIES Idealal Ga Model R room temerature cand c unieral ga contant R (.metric,.engli) molecular weigt R c c te ame unit Ientroic roce k k cln cln k k Rln Rln contant, 0, 0 Real Gae Steam, R a ABLES ma ga x ma mixture f x fg (alo u,,) - f x (alo u,,) ub cooled f Ideal Ga wit emerature deendent cand c ablea Pr Pr ablea roblem

14 SECOND LA PROCESS EFFICIENCY actual Heat Enge exanion roce benefit ork ientroic ideal ientroic effort ideal comreion roce out actual by Firt Law function ( ) CYCLE EFFICIENCY H, L H H CYCLE L L δ deendent of at roerty? FIRS ANDSECOND LAS COMBINED δ d du d 0 for reerible rocee > 0 for irreerible rocee S m > 0 iolated ytem, irreerible rocee out H L CARNO reerible H L COPrefrigerator out H L out H out COPeat um out H L

15 ermodynamic Problem Solg ecnique. Problem Statement Carbon dioxide i contaed a cylder wit a iton. e carbon dioxide i comreed wit eat remoal from, to,. e ga i ten eated from, to, at contant olume and ten exanded witout eat tranfer to te origal tate ot.. Scematic. Select ermodynamic Sytem oen - cloed - control olume a cloed termodynamic ytem comoed to te ma of carbon dioxide te cylder. Proerty Diagram tate ot - rocee - cycle,,,. Proerty Determation, u Concet Sytem Proertie State Pot Proce Cycle. Law of ermodynamic?? E? material flow? CO