Question

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 Feedwater heater W W Pump 1 + W Pump 2 2 Pump 1 1 Condenser

Answer 1

It is stated in diagram that work output from turbine 1 will only be used to drive both pump, and work output from turbine 2 will be net work output. So I took only work output from turbine 2 to calculate thermal efficiency of cycle.

T T3 = 370°c 3 3.2 MPa let y fraction of steam extracted for feedwater No, y 4 0.7 MPa (-y) 25°C 4 ng

= 32 bar from table (steam) nlow first at Tz= 370°c & P3 = 3.2 MPa h₂ = 3158.4kJ/ky 83 = 6.782 R5)g-k rlow the expansion en turbine isentropic (both turbine) 83 = sy Sy 6.782 kJ/kg-K So now at feedwater pressure ( 7 bar = 0.7 MPa = 0.7 MPa = P21) properties from steam table hq = 697 kolky = 1.9918 3 ROK - 2764.8 k 31 Ring 3g = 6.7071 kJ/mg-k =2065.8 kJ 4.7153 kJ/beg-R hag J/kg Sys > By : point y lies in superheated slate : at P = 0.3MPa & Sui = 6.782 kJ / kg -K from table hy = 2796.2 kolhy & Tyne) 78.8°C now hall lato.ame = 697 kJ/kg = now at condensing temperature T-asc forofarties from table (steam) Pond = 3.17 kPa ~ = 0.00100301 ml Rey hy = kJ/kg sf- 0.36722 kJ/kg-K hy= 254605 kilky Sg = = 8.55bb kJ R.5 / by k hag = 2441-7 kolky = 8.1894 kg/kg-K = 104.83

Now $y = 83 = at T=25c 60782 ks) By K Sy- & sy= 8f + x sbg x = := 6.782 - 0.36223 8.1894 La = 0.7833 quality of steam at londenser inbet turbine 2 edit. at T=25c : hy hog het a .hu = 104.83 + (0.7833 x2441-7) hy – 2017.41 kJ/kg h, - - 104.83 kJ/kg But T=asc Now - energy balance equation at feedwater hall = j hai + (1-4) hai now work input of pumpa x(Ps - P3) Wea- bato.ampa Wat oor moet 0.00110796 malag 3 Bolig 1. Wp = 0.00110796 ( 3.2 (3.2-0.7) X10 -2.77 RJ l bg hall = hall + Wp, - ha = 697 +2.77 halli = 699.77 pol kg

slow considering Pump 1 - Wp, = 4 Vatrast (P₂1 -P) :. Wp, 0.001003 (0.7 -0.00317) x103 DJ l by : Wp = 0.68 kJ/kg ", Wp, is very small now in neglecting We, Bi hy = Hae toata 104.83 pJ kJ/kg 1 of feed water hall y hul + (1-9) hal y (2796.2) + (1-4) 10408 eq = 697 0.22 y 592.8 269104 - faction of steam extracted from turbine 1 = y -0.22 m ma now Heat input to boiler (13 – how) : 1 kg =3158.4 – 899:27 Qs = 2458.63 kJ

Now Work output & It is said that the work output from turbine T will be only used for pumps umped do, Net work outfit from cycle will be from turbine Tz . Wh2 = n(1-9) (hu-hu 41-0-28) ( 2296.2 -201704) Wy2 = 607.464 RJ : Thermal efficiency = Von Qs --You 6070404 2458.63 = 24.7% Carnot cycle Von Ti Tu = (370-25) = (370– 25) = 53.65% TH (370+273) time 53.65% (2m). < (96) carnot Tactual

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