Question

Problem 1 (40 pts): A power plant runs on a steam cycle that closely approximates a Rankine cycle with one exception: the turbine is adiabatic but not isentropic, and has an isentropic efficiency of 85%. The cycle operates between 20 kPa and 20 MPa and steam achieves a maximum temperature of 600 °C. The cycle receives heat from a combustion chamber at 1500 K and rejects heat to the surrounding. a) Calculate the net work output and thermal efficiency of the cycle. (15 pts) b) Calculate the entropy generated in each component of the cycle and the total exergy destroyed by the cycle. (25 pts)

Answer 1

let us draw T- s diagram of Cycle (2) TN 716600c) 365 - ---- -- --- --- A Lompa Lokla 60 - -osa 6.8320 6.5075 65075 is CEEAL) Using Appendix (A-4,5,6) for property value of steam at various state At State 1 : 200 MPa 2 hi - 3539 kJ/kg - 600°c J si = 6.5075 kJ/kgk At state 2: Under I sentropic Condition Szs =S, = 6.5075 kJ/kgk 1. P = 20kPa (saturated water - vapour mixture) Dryness fraction Yes = 326-- 6-5075 -0-83.0 1 sfg 7.0152 cs Scanned with Xss = 0.8021 CamScanner

Sample Say = 163-7813) ; correlation coefficient is given Tr= Scy [ Sri Syy o Substitute above values in equation 10 r = 153-3207 J176-0093 81637893 n = 153.3207 28828-4514 r = 153-3207 1. - 169-789432 (r = 0-9030| Yes, truly it confirms impression from the scatter plot. From the scatter plot we got the the positive correlation between measured load Calculated load and from the calculation of correlation coefficient, we got [2=0.9030 which is also positive. CamScanner Scanned with

93575 Teal Enthelpy hos hpl Agu hly 1261-031 (091263) » hus 219.3-6.313139/19 So, actual Enthalpy at 2, h ₂ hin, (hi-ha.) hg = 3539 - 0,85 (3537 - 3142-5313) ha = 2352 15/19 Actual entropy, S2 = 8f + (hy hy). Sty S =0-820 + (1352-251.42 )~7.0152 S2 = 7,136 kJ/kgk At State 3 ; Saturated liquid water hz = hy at 20 kPa = 251-42 kJ / kg S3 = Sf at 20 kPa = 0·8320 kJ/kgk V3 = Of at 20kPa -o-odlo17 m²/kg - At state 4 : h4 = h3 +(P4 - P₃) x V3 h4 =251.42+(20000 – 20 )x0.001017 – 271.74 kJ - (a) Net work output, Wout, net = (h, hz)-(hy-hy) Wout.net - (3539 - 2352) -( 271-74 -251.42) Wout, net = 1166.68 W/kg) Scanned with CamScanner

= 0.357 Thermal Efficiency, nth - Wout net Whet, out Q in Chi-hy) nth - 1166.65 (3539-271.74) "th = 35.7%) (b) Entropy Generation in components : Boiler : Sagen - S; - Sq = 6.5075-0-8320 - Sgen = 5.6755 kJ/kgk Turbine i Sgen - S-5 = 7.136-6-5075 = 0.6285 kJ/xgK Conden by : Sgen = S3-S2 = 0-320 - 7-136 =-6-304 kJ/kak Pump: Igen = S4-S3 =0 Total Erlergy Destruction Maximum work out put for a cycle working between the temperature limit 1500k and 298k would be that of reversible engine (carnot engine) Scanned with U CamScanner

1. ma 1 500 So, Wait, mar - (1 - 12 ) x 8 (supplied) Mat, a. (-279 (A, - - Bà (353-37) Woul, max = 26181643 koleg So, Energy destroyed = Wart, mez - Wout, net = 2618-1643 -1166-68 = 1451.484 Scanned with CamScanner