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Suppose a steam activates a turbine. Steam enters the turbine 36,000 kg / h, P =...

Suppose a steam activates a turbine. Steam enters the turbine 36,000 kg / h, P = 16.38 atm, T = 0ºC and v = 18 km / h. The inlet pipe is 120m above the ground and the outlet pipe is 20m below the ground. The outlet conditions are P = 5.12 atm, T = 0ºC and v = 100 m / s. The heat lost by the process is 10,000 BTU / h and the turbine performs an arrow job of 10kW. The system operates in a steady state.

What is the change in the enthalpy flow (kW)?

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Answer #1

Step: 1 leiven, * 36000 kg/h 2 Mass flow rate mi of steam, 3: 2 36000 kg * Ihr (60*60) & he 10 kals 3 Pi Initial pressure, P.Inlet Velocity, V, - 18 km/h 18 km * 1000 m * The < hr 1 km (60#bo) s VI 2 smls Elevation Elevation of Inlet pipe, 77 2 12Outlet Velocity, Veq 100 ms Heat Lest ; & 7 - 10,000 btu ( Negative Sign hr since Heat Loss 1 Btu 1.055 KI 2 * Q 2 -10,000 btEnergy Balance for an open System m[ AH + AV? + Ag+] - $ + W 8. m = Mass flow rate AH = change in Enthalpy AV? change in kine* [ 매 + Lity + g[31], 4 + d m 29 10 + no - s (os) + 4,81m), (-30-12 m 2부 1000 Jus 1000기 서 2 -2-13 Ks 10Ks 10 | 내 + 4-13 - 1.3> 10kg/s* - 4.103 kJ / kg 2 - 49.03 kJ ls 1 kils : 1 kW 2 Enthalpy flow - 49.03 KW Negative sign implies decrease in Enthalpy

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