Thermo (25) 7. A steam power plant operates on the Rankine cycle with steam entering the...

T-s Diagram of cycle. q (1000 €) , gopsi M3 (900°F) 1500 Psi $2584s are Lesentropic exit) state s A 0.1326 igas (Btylem-R) (correction :- Tempercatwee at Inlet of High pressure turbine is 1000°F not 1000 Because it would have no meaning for a twebine having Entropy at Inlet higher than at outlet. Ť Q.a af state 1. P = 1 soo psi T=1000°F .

H = 1490.8 Btyebm , S, = 1.6 btyebrow or at state 2, Pa = 90 psi, T2 = 350°F elsing steam Property table h2 = 1203 Btyabm Now, at Isentropic outlet condition. Pa=90psi s Enthalpy would have ($29=S, = 1.6 sty been. has = 1176.45 Bty bbm [Note:- 2s will in 2-phase region. | - so, has = hf +(33) meses ?

Gsentropic Efficiency of high pressure turbine, nt = h,-h2 - 1490.8 – 1203 ni-has 1490.8-1176.45 i 17, = 91.55 @ 924 by at state 3, (Ps = gopsi & Tz = 900°F) hz - 1481 Btu/ebm S3 = 118955 Bt%bor at Isentropic state 4s. Sa = 52= 1,8955 BEY bur so, ideal Quality (*4s) = 54s-Sf = 18955-0.13262 Isso 1.34495 X4s = 0.9555 so, ideal enthalpy, has = hf +45 Thefg).

= 69.72 + (0-95554 1035.7) € 1059.331 BtWebmu Actual Enthalpy , hq = hz - M (hz_has) = 1481- 0.92 (1481-1059.331) = 1093.06 Btrebne at states. hs = hf = 69.72 BEY/b.. Pump work = (Po-Ps) x Vs = (1500-1) X0.01614 = 24.19 Psi - Agleme = 4.47 Bty ebon Nef Power output = m x 5 (2,-hz) +(hg_hq) – Pump work = 5xlo® x [287.8 +387.94 – 4.477 - 932319,44 3600 Bt4/s

Hence, | Net Net poy Powele Output = 983.6 MW low pressure furbine Exit quality X4 = ha-hf = 1093.06–69.72 = 0.98 hfg 103507 cycle theremal Efficiency, 7th = Wnet - 983.6 d) 983.6+1499,477 1th = 39.6% © Using Energy balance, ricooling xCwx (OT) = Qrejected = mx [ha-hs] arhie x 1 x 90-65) = 5x106 x [1093.06-69.72] Cy. mass flow leate of cooling watere in cooling = 5116700000 - 204.66 X100 l 6 / 25