Question

precise le quality 4) Determine the fraction m of the extracted steam from the turbine HPT 5) What is the overall thermal efficiency of this cycle? Exercise 13: A steam power plant operates on the reheat-regenerative Rankine cycle with closed feed-water heater. Steam enters the turbine at 12.5 MPa and 550°C at a rate of 24kg/s and is condensed in the condenser at a pressure of 20 kPa. Steam is reheated at 5 MPa to 550°C. some steam is extracted from the low-pressure turbine at 1.0MPa, is completely condensed in the closed feed-water heater, and pumped to 12.5 MPa before it mixes with the feed-water at the same pressure. Assuming an isentropic efficiency of 88% for both the turbine and the pump. 1) Draw a simplified schematic for this plant 2) Determine the temperature of the steam at the inlet of the closed feed-water heater. 3) Determine the mass flow rate of the steam extracted from the turbine for the closed feed-water heater. And the fraction y of bled steam. 4) Determine the net power output 5) Determine the thermal efficiency of this plant. 6) Depict the thermodynamic cycle on the p-h diagram.

Answer 1

Given data as per Question & entry pressure of steam, pr 12.5 mpa Steam Temp. while entering = 550c flow rate = 24 kg/s Isentropic efficiency = 88% Steam pressure in Condenser, Pr 20kPa calculating Pump work Wpump = Vf P2 - 1 1 - 0 Considering Saturated water pressure table at pressure 20kPa hi hf = 251.42 kJ/kg V = Vf = 0.001017 m®/kg wpump = ump = Vf P2-P, Vf ne Putting above values in equation wpump = 0.001017(12500-20) 6.98 - 14.422 kJ/kg calculating enthalpy at state 2. h2 = hi+ wpump h2 = 251.42 + 14.422 - 265.84 kJ /kg calculating Pump work at state 2, Wpump = Vs (Py - P3)

.89 considering saturated water pressure table for the Pressure 1 MPa. the = hy = 762.51 kJ/kg Uz = Uf = 0.0011 27 m²/kg wpump = 0.001127(12500-1000) = = 14.73 kJ/kg Enthalpy at state 11. hii = tzt wpumpe 762.51+ 14.73 = 777.25 kJ/kg 0. 88 2 2 Enthalpy equals at hisa hio ha For PS & Ts considering super heat water. super heat water - pr. table. = Ps=12.smpa considering Ts=55000 5-34765 kJ/kg Ss = 6.6317 kJ /kg. k to = 3476.5k/kg Process 5-6 is isentropic S6 = So obtaining enthalpy from Super heated water-pressure ta P6 = 5MPa S6 = 6.6317 kJ 1 kg .k has = 3185.6 kJ lkg

We know that isentropic ic efficiency of turbine = efficiency hs-ho hs-hos ho=hs-nx [hs-hos] 6 = 34765-0.88(3476:5-3185-6) ho = 3220.508 kJ/kg Ir & s at state 7 at Py = 5MPa and Tz = 550°c from steam table hz = 3434.7+(3666.9-3434.7) 550-500 660-500 = 3550.8 kJ/kg S7 = 6.9781+ (7.2605-6.97) x 55o-sao S7 = 7.1238 kJ/kg.k Entropy is constant for process 7-8. from super heated water -pr. tables. Pq = Impa, ho = 3111.1 kJ/kg is. To = 300+(350-300), 3111-1_3051.6 To = 327.90°C 6corso 3158.2-3051.6 outlet of low pr. turbine at stag S = Sqs

aring enthalpy for super heated water- pr. tables Pq = 20kPa Sas = 7.1238 kJ/kg.k Visentropic of turbine = ha ha hz- has hq = hy h+ (hz- has) hq=3550.9 -0.88/3550-9-2347.9) ha = 2492.26 kJ/kg Appling energy balance equation & on feed water heater (1-4) (h10 ha) = 7 (he-hz) (1-4) (777.25-205785) = 4(3.11.1-702.51) y = 0.1788 fraction of steam extracted from mass flow rate of steam turbine = 0.1788 for closed ing=yms feed water mg =0.1788x24 = 4.2912 kg/s

Heat input qin=hs-huthi-ho qin = 3476.5-777.25 +3550.9-3220.5 qin - 3029.65 kilkg Heat reject from condenser Yout = (1-4) (ha - hit Gout = (1-0.1788)( 2492.2-251.42) 1840-12kJ/kg Net Power output What = milain-Yout) = 24 (3029.65-1840.12) = 28548.72 kW thermal efficiency of the cycle 2. = 1 - Yout 9 in - - 1840,1 3029.7 Nth = 39.264%

HP Turbine Turbine Mixing Schamber closed FWII (Condenson feed water 1 2 Temperature of the steam at inlet of closed I heater = 327.9°C 1 ® mass flow rate of steam extracted from fraction y of bled steam = & Net Power out = 28548.72 kw 1o thermal n = 39.264% turbine =4.2012 Ka 0.1788