10-106 10-102 An ideal reheat-regenerative Rankine cycle with one open feedwater heater is considered. The fraction of steam extracted for regeneration and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 h f @ 15 kPa 225.94 kJ/kg v1 v f @ 15 kPa 0.001014 m 3 /kg wpI,in v 1 P2 P1 1 kJ 0.001014 m 3 /kg 600 15 kPa 1 kPa m 3 0.59 kJ/kg h2 h1 wpI,in 225.94 0.59 226.53 kJ/kg P3 0.6 MPa sat. liquid 5 7 P II h4 h3 wpII,in 670.38 10.35 680.73 kJ/kg 1 3 PI 1 MPa 7 5 T 10 MPa P6 1.0 MPa h6 2783.8 kJ/kg s6 s5 P8 0.6 MPa h8 3310.2 kJ/kg s8 s7 Condens. 2 P7 1.0 MPa h7 3479.1 kJ/kg T7 500C s7 7.7642 kJ/kg K 9 Open fwh 4 1 kJ 0.001101 m3/kg 10,000 600 kPa 1 kPa m3 10.35 kJ/kg P5 10 MPa h5 3375.1 kJ/kg T5 500C s5 6.5995 kJ/kg K 1-y 8 y h3 h f @ 0.6 MPa 670.38 kJ/kg v v 3 f @ 0.6 MPa 0.001101 m /kg 3 wpII,in v 3 P4 P3 Turbine 6 Boiler 4 3 6 8 0.6 MPa 2 15 kPa 1 9 s s9 s f 7.7642 0.7549 0.9665 P9 15 kPa x9 s 7.2522 fg s9 s7 h9 h f x9 h fg 225.94 0.96652372.3 2518.8 kJ/kg The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heaters. Noting that Q W Δke Δpe 0 , E in E out E system0 (steady) 0 E in E out m h m h i i e e m 8 h8 m 2 h2 m 3 h3 yh8 1 y h2 1h3 where y is the fraction of steam extracted from the turbine ( m 8 / m 3 ). Solving for y, y h3 h2 670.38 226.53 0.144 h8 h2 3310.2 226.53 (b) The thermal efficiency is determined from q in h5 h4 h7 h6 3375.1 680.73 3479.1 2783.8 3389.7 kJ/kg q out 1 y h9 h1 1 0.14402518.8 225.94 1962.7 kJ/kg and th 1 q out 1962.7 kJ/kg 1 42.1% 3389.7 kJ/kg q in PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.