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
) ( ) ( ientroicroce VmR R IdealGaModel PROPERIES m m m u m u ysytem Unteadu gz V u m OenSytem E CloedSytem FIRSLA boundary o
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
SUPERHEA RANKINE CYCLE boiler um condener REHEA RANKINE CYCLE um boiler condener turbe
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
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 0 9-9 ex
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 9 0. 00 00 0 kpa 0 MPa 0 MPa 0 MPa 0 MPa MPa MPa MPa 0 kpa 0 kpa 9.. 9..9.9.99.99...0000 9% 0 MPa MPa 0 kpa 0% 9 0 9-9 ex
9-9 ex. J/kg.0. 9..99.9. 00 @ MPa, 0. kj/kg.9 0) (0,000.0000 0.9 kj/kg. 9.. kj/kg.9 9. 0.9.9.9 0.9 kj/kg.. 0.9.0 9. 0 0MPa able A -, water, comreed liquid um um umactural actual um ideal um actual um ( ) ( ) ( ) ( ) C) 90 9., (. kj/kg 0. 9.. 9... kj/kg 0. 9. x 9..9.99.9. x @0 kpa, 90.0 kj/kg..... O at 0 0 9 0 9 0 9 fg f 9
out d ( ) ( ) ( 9. 90.0) (. 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
EES Model
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
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,,) fg @ ub cooled f Ideal Ga wit emerature deendent cand c ablea Pr Pr ablea roblem
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
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