Question

Canvas Steam enters a turbine operating at steady state 2.2 MP-WC with a very 0.1 MPa and a velocity of 50 m/s. The elevation of the inletisierte e 15 kg/s, and the power developed is 7 mW.648 = 9.84m2. Deastice (a) the area at the inlet, in ? (3 points) (b) the enthalpy at the inlet, in kJ/kg (1.5 points) (c) the enthalpy at the outlet, in kJ/kg (1.5 points) (d) the rate of heat transfer between the turbine and its surroundings intarpainted Question 7 (a) the area at the inlet, in m2 (3 points) 11 Related Equations: Vippines)

Answer 1

(0) Given data Entry conditions of steam. Nessure n = 2NPA Temperature T = 360°c. velocity loomisec exit conditions of Steam m = 15 Kg/sec. Ayea pressure R Col mpa velocity V2 = 50m/sec. Saturated vapour at exit Rg = 100% mass How Yote power developed , p=7mW at the inlet mass How rate, r = Px Areax velocity 0 mi= A, Vi ng At 360°c Usup=? is Tsat = 212-4.2C < 360°c Superheated c. At 2mpa & 360°c vsup= 0.1411 m3 kg. :- 0 A, V A, X 100 29 → A,= 0.021165777 m = PxA, XVI ampa , Tsat = 212:4200. . Steam is » - → 15 0.141) 6 Avea at inlet, A, = 0.021165 m. enthalpy at inlet, hsup At 2 mpa & 360°c , h = hsup=3159.3 kJ/kg. ? from steam tables.

At 2mpor, 360°c, Sf = 1.3026 KI Sg 7 7.3594 KI kg k 6.9917= ². Enthalpy at outlet, ha = het lhfg2 asentroppe expansion Turbine follows '. S, - 52 360°c, S, = ssup 6.9917 KI/kgk At o impa , kg K Si=52 1.3026+ X(7.3594 – 1.3026) a X, = 0.939 .: At Oolupa hta 417046 kJ Kg htge = 2258 KJ/Kg h2 = 411.46 +0.939x2258 = 2538.37 kilks eathalpy at outlet, ho = 2538.37 kJ/kg. equation. Apply Steady flow energy U,” Chrt vin +922)+ P d mi +92) + Q = m(h24 2000 2000 15 ( 3159.3+ loor + 9.81x3) +8 = 15 (2538.31+504 +9(0) 2000 2000 41905.95 +9= 38094.3 + 1000 + 1000 ( 0 - 38094-3 + 7000 47905.95 -28 11.65 KW Rate of heat transfer between the dinge turbine and Surroundi 9 = 2811.65 kW Negative sign indicates heat transfer will takes to surround ings. place to