Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. hl = IOkPa = 191.

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1 ~ Chapter 9 Vapor and Combined Power Cycles 9-16 A steam power plant that operates on a simple ideal Rankine cycle is considered. he quality of the steam at the turbine exit, the thermal efficiency of the cycle, and the mass flow rate of the steam are to be determined. Analysis (a) From the steam tables (ables A-4, A-S, and A-6), hl = hf@ IOkPa = kj/kg VI =Vf@IOkPa =O.OO101m/kg Wp.in = VI(P2 -PI) / = ~ m/kg} kpa{ ~ \I kpa.m = 10.09kJ/kg h2 = hl + wp.in = = kj/kg lkj... qin ~J, )~~ / 1 qoul ""' PJ = 10 MPa J =500 C hj = 7.7 kj/kg JsJ =6.5966kJ/kg.K s P4 = 10 kpa S4 = S ~ = = 0.79 X4 = Sfg h4 = h f + x4h fg = (0.79 X292.8 ) = kj/kg (b) qin = h -~ = = kj/kg qout = h4 -hl = = kj/kg wne( = qin -qou( = = kj /kg and Ulrt kj /kg = 4Q2>/o '1 th = -q:: kj /kg (c) m= = 165 kg/s,, t ;;, ~, i,! I ~ J

2 Chapter 9 Vapor and Combined Power Cycles 9-2 A steam power plant operates on a simple non ideal Rankine cycle between the specified pressure limits. he thermal efficiency of the cycle, the mass flow rate of the steam, and the temperature rise of the cooling water are to be determined. Analysis (a) From the steam tables (ables A-4, A-5, and A-6), hl = hf@ IOkPa = kj/kg VI =Vf@IOkPa = m/kg Wp,in =VI(P2 -PI)/1lp = ( m/kg)7, kpa { -1:!!, 1/(0.87) \ 1 kpa.m ) = 8.11 kj/kg h2 =hl +Wp,in = =199.94kJ/kg 2~/ /!V qin IOkPa J- qoul v 4s 4 PJ = 7 MPa } hj = kj/kg J = 500 C $ = kj/kg.k P4 = 10 kpa S4 = S ] =!!1::.!!!- ~ h4 = h -1l (h -h4., ) h -h4., = (0.87X ) = kj/kg hus, qin = h -h2 = = kj Ikg qout = h4 -hl = = kj Ikg Wnet = qin -qout = = kj Ikg and U1Et kj /kg 1]1h = -= = l.8:1/o l1in ~10.ffi kj /kg (b) fn- ~ - 45.(00 kj /s -Utrl kj /kg = 41.5 kg /s (c) he rate of heat rejection to the cooling water and its temperature rise are QOUI = mqoul = (41.5 kg/sx kj/kg)= 88,14 kj/s Ll = Qout = 88,14 kj/s = 1054 C coo1lngwater ( mc. cooling water 2 O kg/s 4.18 /kg. ) ( 00 X kj C) o. 9-1

$09 kj/kg \lkpa.m h2 = hl + W p.in = 191.8 + 10.09 = 201.92 kj/kg 2 / / P = 10 MPa h = 7.7 kj/kg s = 500 C 5 = 6.5966 kj/kg.k P6 = 10 kpa S6 = Ss (b) W,ou! = (h -h4 )+ (h5 -h6 ) = 7.7-2782.5 + 478.$

3 ~ Chapter 9 Vapor and Combined Power Cycles 9.0 A steam power plant that operates on the ideal reheat Rankine cycle is considered. he quality (or temperature, if superheated) of the steam at the turbine exit, the thermal efficiency of the cycle, and the mass flow rate of the steam are to be determined. Analysis (a) From the steam tables (ables A-4, A-S, and A-6), hl = h I@' 10 kpa = kj/kg VI =:vi@ 10kPa = m/kg 5 w p.in = VI (P2 -~ ), = ~ m/kg}10,000-10kpa{ ~-~ - = kj/kg \lkpa.m h2 = hl + W p.in = = kj/kg 2 / / P = 10 MPa h = 7.7 kj/kg s = 500 C 5 = kj/kg.k P6 = 10 kpa S6 = Ss (b) W,ou! = (h -h4 )+ (h5 -h6 ) = = kj/kg qin = (h -h2 )+ (h5 -h4 ) = = kj/kg Wnet = W.out -W p,in = = kj/kg.w: OO.(XX) kj /s m=~= UlEt IW.41 kj /kg = 50.0 kg Is 9-17

~ Chapter 9 Vapor and Combined Power Cycles 9-8 A steam power plant that operates on an ideal regenerative Rankine cycle with an open feed water heater is considered.

$40 kj/kg VI =vf@zokpa =0.001017 m /kg wpl.in = Vl(PZ -PJ) = ~.001017 m/kg}400-20 kpa { ~! 1 = 0.9 kj/kg \lkpa.m hz = hj + wpl,in = 251.40+0.9 = 251.79 kj/kg 4 MPa ~~ j( 20 kpa \~,,-~ II q.$

4 ~ Chapter 9 Vapor and Combined Power Cycles 9-8 A steam power plant that operates on an ideal regenerative Rankine cycle with an open feed water heater is considered. he net work output per kg of steam and the thermal efficiency of the cycle are to be determined. Analysis (a) From the steam tables (ables A-4, A-5, and A-6), hj = hf@ ZOkPa = kj/kg VI =vf@zokpa = m /kg wpl.in = Vl(PZ -PJ) = ~ m/kg} kpa { ~! 1 = 0.9 kj/kg \lkpa.m hz = hj + wpl,in = = kj/kg 4 MPa ~~ j( 20 kpa \~,,-~ II q.:;, 5 l-y 7 P = 0.4 MPa h = h f@ 04 MPa = kj/kg sat.liquid.v = V f@ 04 MPa = m/kg Wpl/jn = V(P4 -P) lkj = ( m/kg} kpa { ~ = 6.07 kj/kg \ kpa.m h4 = h + w pl/jn = = kj/kg P5 =6MPa h5 = 01.8 kj/kg 5 = 450 C S5 = kj/kg.k $6 -$ f P6 =0.4 MPalx6 === ~ $ fg $6 = $S h6 =hf +x6hfg = (0.9655X21.8)=2664.9kJ/kg S7 -S f P7 =20kPa } X7 =-;;:-= S7 = s5 h7 =hf +x7hfg = ( X 258. ) =221.kJ/kg he fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feed water heater. Noting that Q = W = Me = f1pe = 0,.Mh' " Ein -out E -A -ills)'stem 1:' ",v\sleaay) - -O Ein = Eout = Lni;hi = Lniehe ~ ni6h6 + ni2h2 = nih ~ yh6 + (1- y }112 = l(h) where y is the fraction of steam extracted from the turbine ( = ni6 / ni ). Solving for y, -- y-~-hz h -hz = hen, qin = h5 -h4 = = kj/kg qout = (1- y Xh7 -hi ) = ( X ) = kj/kg 9-21

5 ~ Chapter 9 Vapor and Combined Power Cycles 1 and Wnet = qin -qout = = kj I kg (b) he thermal efficiency is determined from Jth = 1-i2!!L = kj Ikg = 7.8% qin kj Ikg!!.

6 Chapter 9 Vapor and Combined Power Cycles 9-4 A steam power plant operates on an ideal reheat-regenerative Rankine cycle with an open feedwater heater. he mass flow rate of steam through the boiler and the thermal efficiency of the cycle are to be determined. Analysis (a) From the steam tables (ables A-4, A-S, and A-6),, hi = hf@ IOkPa = kj/kg / vi=vf@iokpa= m/kg : Wpl.in = Vi(P2 -PI)= ( m/kg)800-10kpa{ ~ h2 =hl = 0.80 kj/kg +Wpl,in = =192.6kJ/kg \1 kpa.m 1~ "/ 7 l-y PJ = 0.8 MPa 1 hj = hf@ O.8MPa = kj/kg sat.liquid I V =Vf@O.8MPa = m/kg 8 s Wpl/~n = V(P4 -P)= ~ m/kg)10, kpa { - lkj = k1/kg \ 1 kpa.m h4 = h + w pl/ ~n = = 71.7 k1/kg P5 = 10 MPa h5 = k1/kg 5 = 550 C s5 = kj/kg. K 6 5 P6 = 0.8 MPa S6 = S5 } h6 = kj/kg P7 = 0.8 MPa 1 - h7 = kj/kg 7 = 500 C Pg = 10 kpa Sg = S7.S7 = kj/kg. K X8 =~=!.867-Q.649=0.962 s fg ::--~ h8 = h f + x8h fg = (0. 962X292.8) = kj/kg ~~:n ~. -.::-J ~ '-l.:. 7 l-y I 8 he fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feed water heaters. Noting that Q = W = Me = Llpe = O. E. E.."O(sle.dy).. in -OUI =!J.Esyslem = 0 ~ Ein = Eoul Lniihi = Lniehe --+ ni6h6 + ni2h2 = nih ---+ yh6 + (1- y }JI2 = l(h) where y is the fraction of stearn extracted from the turbine ( = ~ / ~ ). Solving for y, =~= = y h6 -~ hen, qin = (h5 -h4)+ (1- y Xh7 -h6 ) = ( )+ ( X ) = kj/kg qoul = (1- y Xh8 -h1 )= ( X ) = kj/kg ~'net = qin -qout = = kj/kg.wm 8),00) ~ /s and m=-= kJ/kg = 54.1 kg /s lqa Wnet kj/kg (b) 77lh = -=- "" = 44.8% kj/kg 9-29

$he fraction of steam extracted for regeneration and the thermal efficiency of the cycle are to be determined. Analysis -5 I -J, 4-6 l-y y I 7 I 5 ~ ~ 6 /ii 0.$

7 Chapter 9 Vapor and Combined Power Cycles 9.84 An ideal reheat-regenerative Rankine cycle with one open feedwater heater is considered. he fraction of steam extracted for regeneration and the thermal efficiency of the cycle are to be determined. Analysis -5 I -J, 4-6 l-y y I 7 I 5 ~ ~ 6 /ii 0.6 MPa / 7 ~ 2 -&~ (a) From the steam tables (ables A-4, A-5, and A-6), 9 s hi = 15kPa = kj/kg VI =Vf@15kPa = m/kg PJ = 0.6 MPa sat.liquid h = h 0.6 MPa = kj/kg V =Vf@06MPa = mfkg P5 = 10 MPa } h5 = 7.7 kj/kg 5 = 500 C S5 = kj/kg.k P9 = 15 kpa S9 = S7 9-6

8 Chapter 9 Vapor and Combined Power Cycles he fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heaters. Noting that Q = W = Me = ~e = O....<:1'0 (steady).. Ein -Eout = Msystem = 0 -+ Ein = Eout Lmjhi =Lmehe ~m8h8 +m2h2 =mh ---+yhg +(I-y)h2 =1(h) where y is the fraction of steam extracted from the turbine ( = 1hs / m ). Solving for y. 67Qffi-~ y= ~ = -= O.l4«) f8-19. :m.5- ~ (b) he thermal efficiency is determined from qin = (h5 -h4)+(h7 -h6) = ( )+ ( ) = 88.5 kj/kg qout = (1- y Xh9 -h1 ) = ( X ) = kj/kg and 1] lh = 1- ~ = kj kg qin 88.5 kj / kg = 42.1 % 9-64

Analysis (a) Process 1-2: isentropic compression. p = 755K. R7] h87kpa.m 3/kg.K }30~ = O.906m 3/kg = V max

~ Chapter 8 Gas Power Cycles Analysis (a) Process 1-2: isentropic compression. p - f ii, f T2=1j ( - VI = (300KX8)0.4 = 689K k-l I V2 - ~..., I ~=~ T2 Process 2-3: v = constant heat addition. q23,in =U3