How to Calculate Steam Turbine Isentropic Efficiency
Calculate steam turbine isentropic efficiency using actual and isentropic enthalpy drops. Covers the efficiency formula, how to find enthalpies from steam tables, and what affects efficiency.
Related Calculators
What Is Isentropic Efficiency?
Isentropic efficiency (η_s) compares the actual work output of a steam turbine to the theoretical maximum work output if the process were isentropic (reversible adiabatic). No real turbine is perfectly isentropic due to friction, turbulence, and heat losses.
For a turbine (work-producing device):
η_s = Actual Work Output / Isentropic Work Output = (h₁ − h₂_actual) / (h₁ − h₂_isentropic)
Where: - h₁ = specific enthalpy at turbine inlet (kJ/kg) - h₂_actual = specific enthalpy at actual turbine outlet - h₂_isentropic = specific enthalpy at outlet if process were isentropic
How to Find Enthalpies
Step 1: Inlet conditions (state 1) Given inlet pressure P₁ and temperature T₁ (or quality if wet steam), look up h₁ and s₁ from steam tables.
Step 2: Isentropic outlet conditions (state 2s) The isentropic process has: s₂s = s₁ (entropy is constant) Given outlet pressure P₂, find h₂s from steam tables at s₂s = s₁.
Step 3: Actual outlet enthalpy Apply the efficiency formula: h₂_actual = h₁ − η_s × (h₁ − h₂s)
Or if actual outlet state is measured: read h₂_actual directly from steam tables at actual P₂ and T₂.
Worked Example
Conditions: - Inlet: P₁ = 8 MPa, T₁ = 500°C - Outlet: P₂ = 20 kPa - Isentropic efficiency: η_s = 0.85
From steam tables (superheated steam at 8 MPa, 500°C): - h₁ = 3399.5 kJ/kg - s₁ = 6.727 kJ/(kg·K)
Isentropic outlet (P₂ = 20 kPa, s₂s = s₁ = 6.727): At 20 kPa: s_f = 0.832, s_fg = 7.907, s_g = 8.739 kJ/(kg·K) Since s₂s = 6.727 < s_g = 8.739 → wet steam at outlet Dryness fraction: x = (s₂s − s_f) / s_fg = (6.727 − 0.832) / 7.907 = 0.745
h₂s = h_f + x × h_fg = 251.4 + 0.745 × 2358.3 = 251.4 + 1757.0 = 2008.4 kJ/kg
Actual outlet enthalpy: h₂_actual = h₁ − η_s × (h₁ − h₂s) = 3399.5 − 0.85 × (3399.5 − 2008.4) = 3399.5 − 0.85 × 1391.1 = 3399.5 − 1182.4 = 2217.1 kJ/kg
Actual work output: w_actual = h₁ − h₂_actual = 3399.5 − 2217.1 = 1182.4 kJ/kg Isentropic work: w_s = 1391.1 kJ/kg
Isentropic efficiency confirmed: 1182.4 / 1391.1 = 85% ✓
Typical Isentropic Efficiency Ranges
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Frequently Asked Questions
- What is steam turbine isentropic efficiency?
- Isentropic efficiency = Actual work output ÷ Isentropic (ideal) work output = (h₁ − h₂_actual) / (h₁ − h₂s). It measures how close a real turbine comes to the perfect isentropic case. A 85% efficient turbine produces 85% of the theoretical maximum work for the given inlet and outlet pressures.
- What causes turbine isentropic efficiency to be less than 100%?
- Real turbines lose efficiency through: friction at blade surfaces and bearings, turbulence in the steam flow, heat loss through casing, leakage past seals, and deviation from ideal steam flow paths. Modern large utility turbines achieve 85–93% isentropic efficiency through optimized blade design and steam path.
- How do you find h₂s (isentropic outlet enthalpy)?
- Set s₂s = s₁ (entropy is constant for isentropic process). Then use steam tables at the outlet pressure P₂ with s₂s. If s₂s falls between saturated vapor and liquid values at P₂, the outlet is wet steam — calculate dryness fraction x = (s₂s − s_f) / s_fg, then h₂s = h_f + x × h_fg.
- What is typical isentropic efficiency for a steam turbine?
- Large modern utility steam turbines: 88–93% isentropic efficiency. Small back-pressure turbines: 70–80%. Multi-stage reheat turbines in combined cycle plants: 88–92% for individual turbines, with overall thermal efficiency of 45–55%. Single-stage small turbines are significantly less efficient due to blade geometry limitations.
Last updated 7/28/2026