Question

Thermodynamics II Assignment 5 Instructor: Mohd AWAD 999999REPPI A steam power plant operates on an ideal reheat-regenerative Rankine cycle and has a net power output of 80 MW. Steam enters the high-pressure turbine at 10 MPa and 550°C and leaves at 0.8 MPa. Some steam is extracted at this pressure to heat the feedwater in an open feedwater heater. The rest of the steam is reheated to 500°C and is expanded in the low-pressure turbine to the condenser pressure of 10 kPa. Show the cycle on a T-s diagram with respect to saturation lines, and determine:- 2- The mass flow rate of steam through the boiler. b- The thermal efficiency of the cycle.

Thermodynamics

Answer 1

Giren Pr = Compa, Ti : 5500 From Super heated steam Table hi = 3580.9 KU S, 6:7561 kJ ܘܐ Process 1-2 is Isendoopic Hence SIE S2 = 6.7561 Now Sg at oompa = 6.6627 KJ :: Sz>sg hence process state 2 is in . super heated state at o.ampa. By Interpolation in super heated State at о. Вмра T2 = 109.46°c,.h2 = 28.11.84 kJ Now T3 = 500°C P3 = 0.om pa h3 = 34.80.6 S3 = 7.8672 KJ process 3.4. is gsentropic S4= 53 = 7.2672 hence P te Rok Sq at lokpa = 8.150 1 ks con is greater than sy hence d' State 4 is in wet ( liquid trapower) region. Let quality of steam at this state is nu his state is my then, ..

CS4 = (st + MySf9) 0 kpa 7.8692 0.6492 + R4 X 7.5010 24 = 0.96] Hence hd = 191.014 0.96X2392.0 = 2400.090 kJ. 191.01 kJ ..^5 = hf at jokpa =

Pump work (First pump) We, - Usdp = 0.00l0lo (900-10) = 0.7979. Kj poleaola Hence ho-h5+l5dp = 191.01 + 0.7979 = 192.61 ha = (ht) 0.0 MPa = 390.976 kJ second pump work We2 = 67. dp = 0.001030 (10000-000) = 9.5496 kJ Mence hochot 67dp = 390.976 + 9.5496= 400.53 ks arle

Energy balance. For open Feed water heater M- mass Flow rate in boiler (Kgls) m- bleeding of steam (kg(s) м . * 742 mhz + (M-m) ho = Mhz - MX390.976 mx 2011-84, + (M-m) x 192.61 = 269.23 m = MX 198.366 m= 0.076 M 7 Ned power output is Oo Mw Hence Doxio*w = M (ho-h2) + (M=m) (h3– hy) goooo = M (3500.g. – 2011.84) +(M-0.076M)* ( 3480.6-24 20000 = 689.06M + 0.924M X 991.702 80,000 = 1605.39 M. M = 49.0 kg/s] This is mass Flow rate of steam in haine i Steam in boiler

o Thermal efficiency nr pump MTE Wnet INTE. Wnel MChi-ho) + (M=m) (h3-42) 80,000 49.8 (3500.9 – 400-53) +(49.8 – 0.076x49.8) * ( 348016 - 2011.04) MT= 20,000 MT= 154390.426+ 30773.125 ME: 0.432 MT = 43.2%