Question

A regenerative Rankin cycle utilized the schematic of the figure below.

1. A regenerative Rankine cycle utilized the schematic of the figure below. . Conditions: T1-450oC, P1-3 MPa, T2-2500C, P,-0.4 MPa, T-150oC, P,-0.1 MPa, P.-0.01 MPa, T,-140°C,H,-592k/kg.npump=0.8 Boiler/ Superheate 6 Condenser 10 Pump (a.) Determine the pressures for streams 5, 6, 8,9, and 10 in MPa (b.) Determine m/m, (c.) Determine the enthalpies of streams 5 and 6 in kJ/kg. (d.) Determine the efficiency of turbine stage (e.) Determine the output of turbine stage lll per kg of stream 4 if the turbine is 80% efficient. (f.) Determine ma/m. (.) Determine the work output of the system per kg of stream 1 circulated.

Answer 1

re CE WH You-ma m3 17. OF WH Z Ym-na-na m, - na - M3 no Condenson

al Pressure for streams at different points 7 is = P4 = 0.01 MPa at 5 at 6 > 16 = P2 = 0.1 MPa at 8 → P = P = 3 Mia. = 3 Mea م له f = at 10 P10 = P = 0.4 MPa 1 st 12 = 0.4MPa and T2 = 250°C, h₂ = 2964.5 KJ/Kg. ho = h = 604.7 KJ/Kg. 1 hg = 592 KJ/kg (Given) Nowo, to find enthalpy at 8 - ha = hen _417.5 KJ/Kg Pump work between__7_to_8, Wepa vam (po 2,2 were = 0.00 1043 (3 - 0:1/ x 10

ON 8. We = 3.78 kJ / kg ho = hy + Weg - 8 = 417 5 + 3.78 § 42128 kJ/kg

Apply closed energy balance in the water heater we get ma Cha - hiol = m (hy-hol - hg - ha ₂-hie - - 592 – 421-28 2964.5 - 604-7 22 0.072 Line = 0.072 (na) hs = h = 1918 kJ/kg (bas) c) lump work from 5 to 6, walang = y 1 – Ps) = 0.001010 (ool-001) x 103 01136 kJ/kg

hol = hst Wed - 10 / 08 | ا 3 - 0 = 30 يو-ق kJ / kg d At T = 450°C and P = 3 Mia 3 - 44 33 = ط ا kJ / kg | 7008 = - وا ت A = 0.4 MB کد کا ک مول وه Ta . 1 Tsat / د وي. - 7. ) يول . + تردد 430 = 6 18k - گرگی 18 - 82 - hq = hy + = 2737. cp (Iz - Isat) + 2018 2018 - 143 )

- hy = 2818.62 kJ / kg Efficiency of turbine I, hi - he = 3344-35-2964.3 334 4.35 - 2818.62 = 0 -7225 = 72-25 % I Ans)