Question

6. Water is the working fluid in an ideal Rankine cycle. Saturated vapor enters the turbine at 16 MPa, and the condenser pressure is 8 kPa. The mass flow rate of steam entering the turbine is 120 kg/s. Determine: (a) Draw the ideal Rankine cycle in T-S diagram (b) The net power developed, in kW. (b) The rate of heat transfer to the steam passing through the boiler, in kW. (c) The thermal efficiency. 2 P (kPa) 16000 16000 TC) 41.39 347.36 41.39 0.609 v (m3/kg) 0.0010084 0.009312 h (kJ/kg) 173.362 2581 1636.56 s (kJ/kg.K) 5.2466 S3-S4 1 3 4 8 8 х 0

Answer 1

Ideal Rankine cycle Qin Boiler Turbine Wout Condenser Win Pump Qout Given Pi=8Kia 12 = 16 MPa = 16,000kPa - 120 kg/sec 3. solution : Working fluid Sat. curve ? -P2 T3 - 2 Temp Pi Tz Ty = Ti Si=S2 S3=54 Entropy T-s diagram of ideal Ranking cycle Process 1-2 : Iso entropic Compression in a pump Process 2-3 : Heat addition at constant Pressure Process 3-4 :> Iso entronie Expansion in Torbice Process 41 : Heal Rejection at constant Pressure in a condenser.

b) Net Power developed , KW work done by the - Change in Enthoply during turbine Expams ion hz-hu 2581 - 1636-58 C Given in Table) 944.42 K5/kg 120 X 946.42 = 113330.4 KW WT .mxWT work done on Pump tie P2 il -Svde PU Vf (P2-B1) Constant Volume Process value of Vf at Point given in Table = 0.0010084 m3/kg :: Wp Wp = -0.0010084x (16000 -8) =-16.13 KS/kg :: Net Power = 244.42-16-13 output 928.3 KJ/kg c) Rate of heat transfer to Steam in Boiler qin = hz-hz -wp hz-hi +16.13 hz-173-362 : h2= 189.49 KJ/kg Q in = 2581 in = 2581 - 189.49 18 =.2391.5 * 5/kg

As mass flow rate of steam is not mentioned, considering mass How of steam entering turbine = mass flow of steam leaving boiler = 120 kg/sec Q in = Winx m = 2391.5 X 120 = Ato 286980.96 kWt ansc d) Efficifency Thermal Efficiency net work done X loo Heat supplied WTtWp Qin X100 928.13 Xloo 239.115 38.87 Hence Thermal Efficiency = 38.81. * ans d >