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Water vapor at 10 MPa, 600°C enters a turbine operating at steady state with a volumetric flow rate of 0.36 m3/s and exits at 0.1 bar and a quality of 92%. Stray heat transfer and kinetic and potential energy effects are negligible. Determine for the turbine (a) the mass flow rate, in kg/s, (b) the power developed by the turbine, in MW, (c) the rate at which entropy is produced, in kW/K, and (d) the isentropic turbine efficiency

Respuesta :

The mass flow rate when the Water vapor at 10 MPa, 600°C enters a turbine operating at steady state is 9.38kg/s.

How to calculate the mass?

At T1 = 600°C

P1 = 10MPa

h1 = 3625.8 kh/kg

At P2 = 0.1, h2 will be:

= 191.81 + 0.92 × 2392.1

= 2392.54kj/kg

The mass flow rate will be:

= q/v

= 0.36/0.038378

= 9.38kg/s

Applying the steady rate equation, the power will be:

= 9.38(7.5488 - 6.9045)

= 6.0435 kw/k

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