Question

The figure shows a schematic of a power plant that has two steam turbines. The water (the working fluid for the plant) leaves the high pressure turbine (HPT) as saturated vapour at 1000 [kPa]. Part of this flow is diverted to an insulated heat exchanger, which we hot airstream to create superheated steam that enters the low pressure turbine (LPT). The air mas flow rate is 19.5 k/s and the air temperature drops from 1100 to 6001 as it goes through one side of the heat exchanger. The remaining portion of the exit flow from the high pressure turbine goes to a mixing chamber. This flow is mixed with the stream coming from the low pressure turbine branch after it has gone through the low pressure turbine, the condenser, and Pump 1. The flow leaves the mixing chamber as saturated liquid and then enters Pump 2. After leaving Pump 2 at a pressure of 8 MPal, the water enters the steam generator, where it exits at 8 MP and 450" before flowing into the high pressure turbine. Further details of the plant operating conditions are shown in Figure 2. The turbines and pumps are insulated (ie, adiabatic). Neglect changes in potential and kinetic energies (a) Determine the power output of the low pressure turbine, Wur, in kW). (1) Determine the power input to Pump 1, Wp, in kW). (c) Determine the power input to Pump 2, Wro, in kW). (d) Determine the thermal efficiency of the power plant (e) On a T (temperature specific volume) diagram, draw a proces representative for the working fluid in the power plant. On the diagram, clearly indicate the labelled state points, the process paths (use a dashed line if the path is unknown), and the constant pressure lines that we through the state points. Indicate state T and values. Do any additional work mary to label the diagram. YPT steam P8 MP Thu 450 [PC generator WPT mine = 19.5 kg/s] high prest T, = 1100 To = 600 g Pump 2 turbine saturated பம்பா saturated quid vapour HT ហហហហហ TMC THPT Mixing T: 350 Chamber water side P = 1000 [kPa] neat exchanger WT TLPT - PP pressure WP turbine Pump 1 THPT HULPT condenser saturated quid B = 40 kPa) 18

Answer 1

The problem would be easy if we start with T-S diagram. Read the problem thoroughly so that it would be easy point out the states. to ATA Pg=8000kPa 8 To = 450/00 MHPT Pr=P, _6=10 = 1000 kPa = ls Не / 72 =3000 Mme 40 kPa MLPT meg P₂= Put 5 met X:0.95 Sat: Lig Sat. Vop S7 P= P2 = Ps= P6 = 1000 kPa P₂ = P4 = 40 kPa Pq = P8 = 8000 kia from Saturated tables I kg = h kJ/kg At 1000 kPa => hy= 2797.1 1000 kPa - ho = hq = 762.51 At

From Superheated Table At P-1000kPa T= 3500 = ₂ = 3157.7 kJ/kg Consider Heat Exchanger 10 To 600°C. 9 Tg = 11000 ma 19.5 hi MLPT from Energy Balancs Ein = out mitch, -hy) a na (Ta-To) Pair Upt ( 3157.7–2777.1) - 195 (1100-600) (-os) mipt = 250 62 kkg / from saturated Table liquid At P = 40 kPa =) ha a hf 317.62 kJ lieg hfg kilkg since we know dryness fraction &z=0.95 = 2318.4

hz hut dg hfg hiz - 317.62 +0.95 ( 2318.4) h₂= 252001. kJ / kg Low Pressure Turbine (а Power output of ů LPT = M LPT Chaahz) (25:62) (3154.7 - 2520.1)(1000) = 16335.3x8kW WEPT = 16335.3x103 kW from Saturated Table At Pa 40 kPa =) S4 = 1.026 1 kJ/kg since Pumps are adiabatic Surso = 1.0261 KI/kg Ny = 16026x10 m/kg from sub cooled Table mkg csillag i do 6026 At P=40kPa and Ss = 1:0261 kJ/kg 7 hs = 317.62 kJ/kg

6 Power Input to pumpt Wpp mept " (Ps - Py) - (25.62X18° ) (1.026 X163) (1000 -40) TWA : 25-23x103 kW

from Saturated Table At P= 1000kPa =) T = Tg = 179.8 C - hi-hg = 2777.1 777.115/19 =) ho= bf = 76251 kJ/kg =) S6 = Sf = 2.138 kJ/kg » Vc = Uf = 1.127x103 m3 1149 Mixing Chamber rime (h.) + m Lp T (hs) an HPT Cho) ime (2777.1) + 0 25.62x18 ( 317.62) = 2007 (7625) and we know that I napra imme of MUT

MHPT = mme + 25062x103 manca mapr - 25.62 410 3 ( 1218 )( 2929.) + 25.62 (32.0.) no 1-5) MART ( 2797,1 - 76251) 25.67X10} (277.7.1-319.62) 3. ſ = 31027x10 kg / HPT = 5.65x103 kg Is in Mc LPT НРТ Mc = nHp W pz Power input to pump 2 MART Vo (Py-Po) = ( 31.27X10 ) (8000-1000) (6-127X163) 246.68X103 KW We

@ Thermal Efficiency of Power Plant Whet ; Whet = Pin-Pout = Wirul-Waime th Vin = What & Rout Tub -Wpump) & Pout What & Pout o (Wrub [linor tekst) teintha)] = mor (hy the - noor Che-ho) + (6 335.3XB) - (25•23x7° 42450m) $(252011 – 319.62) مرا + 25.62x10

he= 8272 rJlrg at p= 8MPa and T=450c. tinct =(3129 X 10 ) [3272-2977.] + 16335.381d? 3 - 271.91810 Wnet = 31538.9X10 kW Qin 31538.9x103 + 25.62x10 (252011-317662) din 8796604X103 KW - 0.35 Ath 31538.98703 87966.4x102 ie 35% efficiency