Question

# Steam enters an adiabatic turbine at 1 MPa and 400 °C and leaves at 150 kPa...

Steam enters an adiabatic turbine at 1 MPa and 400 °C and leaves at 150 kPa with a quality of 80 percent. Neglecting the changes in kinetic and potential energies, determine the mass flow rate required for a power output of 10 MW

Initial state

At 1 MPa , saturation temperature Ts = 179.88 C

Ts < T1

(superheated steam)

Temperature T1 = 400 C

Pressure P1 = 1 MPa

Specific enthalpy h1 = 3264.385 kJ/kg

Final state (saturated steam)

Pressure P2 = 150 kPa

Specific enthalpy of fluid hf = 467.08 kJ/kg

Specific enthalpy of steam hg = 2693.11 kJ/kg

Quality of steam x = 0.80

Enthalpy balance

h2 = hf + x(hg - hf)

= 467.08 + 0.80*(2693.11 - 467.08)

= 2247.904 kJ/kg

From the first law of thermodynamics

Q - W = m(h2 - h1)

For adiabatic process Q = 0

Power output W = 10 MW

Mass flow rate

m = (10 MW x 1000 kW/MW) / (3264.385 - 2247.904) kJ/kg

= 9.84 kg/s

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