Question

2.

2. In the Rankine cycle with a feedwater heater as shown in the figure, the condensing temperature T1 is 25°C. The superheat temperature T3 is 370°C and the vaporization pressure P3 is 32 bar. Enough steam is bled from Turbine 1 at 7 bar (P2') to heat the feedwater 2 to a saturated liquid 2". Determine the thermal efficiency of this reversible cycle and compare it to a Carnot cycle operating between the same two temperature extremes. Show the cycle in a T-s diagram. Also compute the net work output per kilogram of steam. Pump 2 3 2" Boiler - superheater 2" Turbine 1 Turbine 2 W Feedwater heater W Pump 1 +W Pump 2 4' 2 4 Pump 1 1 Condenser

Answer 1

Eve TT T3 = 370°c 3 3.2 MPa let y fraction of steam extracted for feedwater y 41 211 0.7Mpa (1-4) 250 4 مک

from table = 3? bar (steam) nlow - first at Tz= 370°c & Pa = 3.2 MPa ha = 3158.4k 5 83 = 6.782 R5/kg-K rlow the expansion in turbine es isentropic (both turbine) Sy = 6.782 kJ/ky-K Syl P21) Lo nlow at feedweater pressure ( 7 bar = 0.7 MPa = properties from steam table hq = 697 kolky sf = = 1.9918 kJ/kgK hg : 2762.8k5ley Sg = 6.7071 kJ/ Rg-k hqg = 2065.8 kJ/kg $89€ 4.7153 kJ/kg-R Sy > 3g . point 4' lies in superheated state i at P = 0.2MPa & su 6.782 51 kg-R, from table hely = 2796.2 R 5) Ry & Ty = 178.8°C now hall =lato.smpat 697 kolky (steam) Uf = = 0.00100301 m mylly now at condensing temperature Tzas troferties from table ... Pond = 3.17 kPa hy = 104.83 R 51kg Sf = 0.36722 kJ/kg-k hy= 254605 kolky sg = 8.5566'Rolby - K - 2441.7 Rol Rey $69 = 8.1894 kJ/kg-k hfg =

now $y = 8z = 6.782 € 3) By-K at T=2sc sy- 88 + Sfg Sy- sf 819 a = 0.7833 --X= 6.782 - 0.36223 8.1894 & a quality of steam at tondenser inbt / turbine z excet. at T-25C: het a hag hy = 104.83 + 60-7833 X2441-7) hy = 2017.41 kJ/kg hy h, hartaasic = 104.83 kJ/kg Now -> energy balance equation at feedwater hall = y har + (1-4) ha nlow wothenbeut of pumpa Wp = Ubato.ampa x(Po-P) Yato.smm2 0.00110786 malkg .. Wpa = 0.0011079613.2-0. 2-0.7) x10 Bolig = 2.77 kJ/hg han = hall + Wp, ham - 697 +2.77 hall 699.77 holky

How considering Pump 1 : Wp = fattasch (P21 - P.) :: Wp, = 0.001003 (0.7 -0.00317) x103 pJ/kg Wp = 0.68 kJ/kg Wp, is very small in neglecting Wp, h, wh, Parroase 104.83 pJ/kg now eg = 1 of feed water han y (2796.2) + (1-y) 104.8 (1-9) ha oc hyl + 697 0.22 592.8 269104 -- fraction of steam extracted from turbine i = y -0.23. (hz - how) m now Heat input to boiler ma 1 kg = 3158.4 – 899:27 = 2458.63 kJ

now whock output - It is said that the work output from turbine T will be only used for pumth unprāt do, net work outfit from cycle will be from turbine Tz . We = n(1-9) (huicha) 11-0-28) ( 2796-2-201704) WI 2 = 667.464 kJ Wrz : Thermal efficiency = ln - Qs 602-404 2458.63 = 24.7% (370- 25) = 53.65% carnot cycle Von Ti - Tu TH (370+273) = 53.65% (2) (2) Tactual Carnot