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Adiabatic steam turbine 50 MW: mass flow, isentropic efficiency

Short answer

(a) Mass flow rate from steady‑flow energy: w = (h₁−h₂)+(V₁²−V₂²)/2000 = 742.1 kJ/kg; ṁ = 50 MW / 742.1 kJ/kg = 67.4 kg/s. (b) Isentropic exit quality 0.9054, h₂s = 2427.2 kJ/kg; η_T = (h₁−h₂)/(h₁−h₂s) = 74.7%.

Steam enters an adiabatic turbine at 5 MPa, 500°C and with a velocity of 50 m/s and exits at 50 kPa, 100°C and with a velocity of 150 m/s. If the power output of the turbine is 50 MW, determine (a) the mass flow rate of steam flowing through the turbine, and (b) the isentropic efficiency of the turbine. [Refer attached table for the properties of steam]

Steam table values (read from tables)

State 1: 5 MPa, 500 °C: h₁ = 3434.5 kJ/kg, s₁ = 6.9778 kJ/kg·K. V₁ = 50 m/s. State 2 (actual): 50 kPa, 100 °C, superheated: h₂ = 2682.4 kJ/kg. V₂ = 150 m/s. At 50 kPa (saturated): h_f = 340.5, h_fg = 2304.7 (h_g = 2645.2); s_f = 1.0910, s_fg = 6.5019 (s_g = 7.5930).

(a) Mass flow rate

Steady flow, adiabatic (q = 0):

w = (h₁ − h₂) + (V₁² − V₂²)/2 w = (3434.5 − 2682.4) + (50² − 150²)/(2 × 1000) w = 752.1 + (−10) = 742.1 kJ/kg

ṁ = Ẇ/w = 50 000/742.1

ṁ = 67.4 kg/s

(b) Isentropic efficiency

Isentropic exit at 50 kPa has s₂s = s₁ = 6.9778 kJ/kg·K, which is less than s_g = 7.5930, so it is wet:

x₂s = (6.9778 − 1.0910)/6.5019 = 0.9054 h₂s = 340.5 + 0.9054 × 2304.7 = 2427.2 kJ/kg

η_T = (h₁ − h₂)/(h₁ − h₂s) = (3434.5 − 2682.4)/(3434.5 − 2427.2) = 752.1/1007.3

η_T = 0.747 (about 74.7 %)

(Including the kinetic energy change in both the actual and isentropic work gives 74.4 %, so the result is about 74–75 %.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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