Question

Question Ref. No: 2925 A steam power plant operates on a simple ideal Rankine cycle between the pressure limits of 9 Mpa and 30 kPa. Superheated vapors enter the turbine at 650 oC. Determine for the cycle: a) The rate of heat supplied to the boiler per kg of steam flow. b) The net power output per kg of steam flow through the boiler. c) The thermal efficiency. If the cycle above is modified to include one open feedwater heater operating at 0.5 MPa with saturated liquid exits the feedwater heater at 0.5 Mpa, determine for the modified cycle: a) The rate of heat supplied to the boiler per kg of steam flow. b) The net power output per kg of steam flow through the boiler. c) The thermal efficiency. h1 (kJ/kg) Format: 282.58 wp.in (kJ/kg) Format : 3.82 h2 (kJ/kg) Format : 296.52 h3 (kJ/kg) Format: 6985.3 s3 (kJ/kg) Format : 2.0796 h4 (kJ/kg) Format : 9634.4 qin (kJ/kg) Format : 6587.85 wnet (kJ/kg) Format : 6356.49 nth (%) Format: 44.3 у Format : 0.3886 wnet (kJ/kg) Format : 3869.938306 nth% Format : 93.9

Answer 1

#2925 + I 90 bar (Rankine cycle). 4. 0.3 bar 3 S KJ/kg k - 0.3 bar P = P2 = 90 bar; z = Py T - 650°C. All data quoted below are taken from Steam Table. hy = 33752.05 kJ/kg ; S = 70885 for ideal cycle, Process. 2-3 will be isentropic S = = 1.0855 kJ/kgK GDL from steam Table we confirm that State 3 wet region. so, we need to find the dryness fraction (x). S = St + x liej in (Sg - Sf). >> 7.0885 = 0.944 4x (7.769 - 0-944). x = 0.9 so, ha = hf + z (.hg - hf) = (2 89.3 + oig so h = 2392-47 kJ /kg hy.= 289-3 kJ / kg m²/kg (2625-4- 289-3) ; y = 0.001022m

ump work = Vy (P - P4) = hy hu) -> 0.00 1022 * (9000 - 30) So, h, = 298.467 kJ/kg. (h, -289-32 a) Heat supplied in the boiler = Qona (3752.05-298467) (h₂ - h). so Qup. Net power = = 3953.58 kJ/kg (Ansera) Turbine work - pump pump work) Turbine work = (3-ha) = (3752.05 - 2392-47) 13.59.58 kJ/kg h-hu) (298.467–289-3) - 9. 167 k5/kg (359.58.9.167) KJ/kg - 13.59.58 Pump work So, Net power = or, Net Power = 1350.413 kJ/kg (Ans-b?

c) Thermal efficiency (th). Net power Heat supplied 11350.413 3453.58 8) x 100 11th 39.101/ (Ansare) qobar Steam power cycle with open feed water heater. 7 5 bar 5 0.3 bar S- properties from previous problem we have de termined following from Steam Table h = 3752.05 kJ/kg ; hy. = 2392.47 kJ /kg) hs = 289.3 kJ/kg ; 4 = 3 = $y = 7:0885 kkgk From steam Table see that State 3 lies in super heated region. = 207°C ; P = Py = 5 bar; = 2870 kJ /kg v = 0.001022 m / kg We

2 1 unit i unit Boiler Turbine y unita (1-4) whit (1 unit 7 4. feedpump. open feed water heater (1-9) unit 6 Condenser ose Feed pump. (1-4 unit from Steam Table hy = 640.1 kJ/kg Also, & (Pc - Ps) = (hoahs) -30.00 1022 (500-30) = (hs - 289.3) so b6 = 289.78. KJ / kg Energy Conservation around open feed water heater bg + y + (1-y the = hy => 2870 xy + (1-9) * 289.78 = 6.40.1 -Solving we gest, y = 0.136

Also, (hy-hy). - Vog (P-P) NE 2 0.601093 mol kg hi - 640.1 0.001093 (9000 - 500) So, h = 649.39 kJ/kg a) Heat supplied in boiler = 14 Cha-hy) 67 Qeupp. = (3752005-641:39) Jasupp KJ/kg (Ans. a) = 3102.66

b). Net power = 16: -6.)+ 8-9)(hy -ho. - [t-}}+ a - by); (hi ha). 52.05 – 2870) + (1-0-136)* (2 870 – 2392:47) [(1-0-136)*( 0-136)*(289.78 -289-3) + (649.39-640.1) So Net work output 1284.93 KJ/kg (Ansb) C) Thermal efficiency (th). 'Net work output Qsupp. = 1284 93 3102.66 18 or, 11th 41-4101 (Ans. e)

All answers are boxed and steps are shown clearly. NOTE: Value of enthalpy and entropy varies for different steam Table but the change in enthaly or change in entropy remains same for all Steam Table. Thank You and stay safe.