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Adiabatic steam turbine 70% isentropic efficiency

Short answer

The required mass flow rate is 4.43 kg/s. The actual exit state is superheated steam with 0% moisture, well below the 2% limit, so no excessive blade wear occurs.

An adiabatic steam turbine has an isentropic efficiency of 70% and operates between 2 MPa and 0.175 MPa. The inlet temperature is 400°C. What mass flow rate is required for the turbine output of 1750 kW? Excessive wear of the turbine may occur if the moisture content exceeds 2 percent. Does it occur in this case?

Steam table values used (IAPWS-IF97 / standard tables)

State 1: 2 MPa, 400 °C (superheated): h₁ = 3248.2 kJ/kg, s₁ = 7.1290 kJ/kg·K.

At P₂ = 0.175 MPa (saturated, T_sat = 116.04 °C): h_f = 487.0, h_g = 2700.1 kJ/kg; s_f = 1.4849, s_g = 7.1714 kJ/kg·K.

Step 1: Isentropic exit state (2s)

Here s₁ = 7.1290 < s_g = 7.1714, so the isentropic exit is a wet mixture:

x₂s = (s₁ − s_f)/(s_g − s_f) = (7.1290 − 1.4849)/(7.1714 − 1.4849) = 0.9925

h₂s = h_f + x₂s h_fg = 487.0 + 0.9925 × (2700.1 − 487.0) = 2683.6 kJ/kg

Step 2: Actual exit state

η_T = (h₁ − h₂)/(h₁ − h₂s) = 0.70

h₁ − h₂ = 0.70 × (3248.2 − 2683.6) = 0.70 × 564.6 = 395.2 kJ/kg h₂ = 3248.2 − 395.2 = 2853.0 kJ/kg

Step 3: Mass flow rate

Adiabatic turbine, kinetic and potential energy neglected: W = ṁ (h₁ − h₂)

ṁ = 1750/395.2 = 4.43 kg/s

Step 4: Moisture at exit

h₂ = 2853.0 kJ/kg is greater than h_g = 2700.1 kJ/kg at 0.175 MPa. So the actual exit state is superheated steam (about 190 °C), not wet steam. The quality is effectively x = 1 (moisture content = 0 %).

Answer: ṁ = 4.43 kg/s. The moisture content is 0 % (exit steam is superheated), so it is below the 2 % limit and excessive blade wear does not occur. (Only the ideal isentropic exit would be slightly wet, x = 0.9925; irreversibility raises the exit enthalpy and dries the steam.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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