Question

Thermo

(25) 7. A steam power plant operates on the Rankine cycle with steam entering the high pressure turbine at 1500 psi, 1000'C with a mass flow rate of 5x10lb/hr. The steam exits the high pressure turbine at 90 psi, 350°F where the steam is then sent back to the boiler and reheated to 900°F before expanding through the low pressure turbine to 1 psi. Cooling water enters the condenser from a lake at 65°F and exits at 90°F Assume that the low pressure turbine has an isentropic efficiency of 92% and that the pumps operate isentropically. Draw a T-s diagram of this cycle and determine the following: a. b. high pressure turbine isentropic efficiency net power output, MW low pressure turbine exit quality cycle thermal efficiency the mass flow rate of the cooling water from the lake d. e. BONUS: What is the rate of entropy production in the low pressure turbine? (Btu/hr 'R)

Answer 1

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