Skip to main content

Steam turbine 10 MPa, 550°C: isentropic efficiency

Short answer

The isentropic efficiency is 83.5%. Actual work per kg of steam is 940.9 kJ/kg, giving a power output of 9.41 MW for 10 kg/s.

Steam enters an adiabatic turbine at 10 MPa and 550°C. Exit conditions are 0.06 MPa and a quality of 96%. Determine the isentropic efficiency and actual work output for a mass flow rate of 10 kg/s.

Steam table values (read from tables)

Inlet, 10 MPa, 550 °C: h₁ = 3502.0 kJ/kg, s₁ = 6.7584 kJ/kg·K.

At 0.06 MPa (T_sat = 85.9 °C): h_f = 359.84, h_fg = 2293.0, h_g = 2652.9 kJ/kg; s_f = 1.1452, s_fg = 6.3859, s_g = 7.5311 kJ/kg·K.

Actual exit (x₂ = 0.96)

h₂ = h_f + x₂ h_fg = 359.84 + 0.96 × 2293.0 = 2561.1 kJ/kg

Isentropic exit (s₂s = s₁)

x₂s = (6.7584 − 1.1452)/6.3859 = 0.8790 h₂s = 359.84 + 0.8790 × 2293.0 = 2375.4 kJ/kg

Isentropic efficiency

η_T = (h₁ − h₂)/(h₁ − h₂s) = (3502.0 − 2561.1)/(3502.0 − 2375.4) = 940.9/1126.6

η_T = 0.835 (83.5 %)

Actual power

w_actual = h₁ − h₂ = 940.9 kJ/kg (adiabatic, kinetic energy neglected)

Ẇ = ṁ w = 10 × 940.9 = 9409 kW

Answer: η_T ≈ 83.5 %, actual work = 940.9 kJ/kg, power ≈ 9.41 MW.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Fundamental of Thermodynamics & Heat Transfer chapter 5 questions →