Question

1) A steam power plant operates on an ideal reheat regenerative Rankine cycle with one reheater and two open feedwater heaters. Steam enters the high pressure turbine at 100 bar and 600°C and leaves the low-pressure turbine at 0.1 bar. Steam is extracted from the turbine at 17 and 3 bar, and it is reheated to 550°C at a pressure of 10 bar. Water leaves both feedwater heaters as a saturated liquid. Heat is transferred to the steam in the boiler at a rate of 422 MW. Show the cycle on a 7-8 diagram with respect to saturation lines, and Determine (a) the mass flow rate of steam through the boiler, (b) the net power output of the plant, and (c) the thermal efficiency of the cycle. High-P Low-P turbine turbine Boiler Reheater 7 (10) Open FWH Open FWH Condenser Page 2 of 7

I need the answer by mollier chart not tabels

Answer 1

Solution :- T-S Diagram a lobar co ban N 17 bar y V rey) 3bar 3 11 (4-4-3) - (1-43) of bar from Mollier chart 100 barn, 600C ! ha = 3625 kJ/kg [ Suberheated regim) Sy = 6.85 kJ/ kg K A27 bar Sq = 57 = 6-85 kJ/kg ok so the = 3040 Kilkg [Subex heated Region) At lorbar. Sg = S8=57= 6.85 kJ/kg-ok ha = 2990 kJ/kg [Super treated Region] Atjoban, 550C ho = 3600 kJ/kg Csuperheated Region] Sto= 7.84 KJ/tOK

At ontbar. Sig= 7.84 kJ / kgk [ Wet Region] hy2 = 2496 K Jl lg Deryness fraction = 0.962 At 3 bar. Siy= 7.84 kJ / kg ok ha 1= 3155 kJ 1 Kg K [ super heated Region] Also, he = 195.8 kJ 1 kg hoe hy + Vf ( P2-P1) - 191.8+ .00010(3-0.1) X 102 = 192.0920 hz = 562.5 kJ lkg hq=hz + Uf3 [Pq - P3) = 561.5+:001074 [17-3] x202 =563.0034 25 = 871. 0 kJ / kg h 6 = 45+ 0 [P6-P5) 0718+0.001163C100-177 x 102 - 881.452 Heat balance equation for FWH II y hat (1-4) ha = 1.45 Yx 3040 +(1-7) 563.00 36 = 871:8 y = 0.125

Heat balance equation for FWHI 3. hast (1-4-3) hq = 1.43 3 (3555) + (1-0-125-3) 192.0999 = 561.5 3= 0.1331 (a) mass flow rate of steam through the boiler Heat Tresferred in Chy-h6] + n(1-4) Eho-ho] 422x103 = riſ3625-881.4.529]+ro [1-0-125][3650 - 2920] 422x103 =3338.547 1x mi => "= 1.26.4 Kg 18 (b) Power output of the Plout P=lin (hy ho] + m [-y] [ho-hg] + m (8-4) (h10 hl) + m (1-4-3/4 (hags- ) / - [m (1-4-3) (ha-h1) + m (3-9) Cha-h3) + mithshs)] Pam (2625-30 40)+(-0.125) (3040-2920)+(1-0-125)(3600-3155) + (1-0.125-0.133) (3155-2496) – (1-0-125-0-133)(19.2.0929 - 191.0) – (1-0-125)(563.0036-561.5) -(081-4529-871-8) pain (25 68.353-11 . 186] p=126.4 x 1557.167 FP = 196.025 MW

o Cycle. Thermal Efficiency of the Power output th - Heat i uput 196.025 4220 h = 0.4 664 or 46.64%