Library / Thermodynamics / THE-05

Thermodynamics

Isentropic turbine efficiency

How close a real adiabatic turbine comes to the ideal isentropic work.

Ref: THE-05
Rev: v1
Field: Thermodynamics
Result: η_t

η_{t} = \frac{h_{1} - h_{2a}}{h_{1} - h_{2s}}

Adiabatic turbine with negligible kinetic and potential energy changes. Compare the actual outlet 2a with the isentropic outlet 2s at the same discharge pressure.

The efficiency is the work actually extracted divided by the best-case work for the same expansion.

Calculator

Solve for

η_tTurbine efficiencyComputed

h₁Inlet enthalpy

kJ/kg

5004,000

h_2aActual outlet enthalpy

kJ/kg

5003,500

h_2sIsentropic outlet enthalpy

kJ/kg

5003,500

Actual work510 kJ/kg
Isentropic (ideal) work600 kJ/kg

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

Turbine efficiency () current value 0.85

h_2sIsentropic outlet enthalpyKey driver

−5.15.1

h_2aActual outlet enthalpy

−0.54.5

h₁Inlet enthalpy

0.2291.77

Modern steam and gas turbines reach 85–92%.
For 3200 kJ/kg steam with h_2s = 2600 kJ/kg and η = 0.85, the actual work is 510 kJ/kg.

Chart data as tables

Ranked variables

Variable	Typical range	Lowest result	Highest result	Swing ()	Local step	Change in result
h_2s Isentropic outlet enthalpy	500 to 3,500 kJ/kg	−5.1	5.1	10.2	+10%	+0.65
h_2a Actual outlet enthalpy	500 to 3,500 kJ/kg	−0.5	4.5	5	+10%	−0.448
h₁ Inlet enthalpy	500 to 4,000 kJ/kg	0.2286	1.771	1.543	+10%	+0.0522

Sweep of Actual outlet enthalpy

Actual outlet enthalpy (kJ/kg)	Turbine efficiency
500	4.5
875	3.875
1,250	3.25
1,625	2.625
2,000	2
2,375	1.375
2,750	0.75
3,125	0.125
3,500	−0.5

Sweep one variable

Vary

Turbine efficiency ()Hover, touch, or use the arrow keys on the chart

Variables and how each one contributes

η_t

Turbine efficiency

Result ·

What the equation gives you. Everything else feeds it.

h₁

Inlet enthalpy

kJ/kg · typical 500 to 4,000

Inlet specific enthalpy. Raises the work in both numerator and denominator.

h_2a

Actual outlet enthalpy

kJ/kg · typical 500 to 3,500

Actual outlet enthalpy. Raises it and the efficiency falls.

h_2s

Isentropic outlet enthalpy

kJ/kg · typical 500 to 3,500

Outlet enthalpy for an ideal expansion. Raises the ideal work denominator downward and efficiency rises.

Worked example with the values above

Actual work: h₁ − h_2a = 3,200 − 2,690 = 510 kJ/kg.
Ideal work: h₁ − h_2s = 600 kJ/kg.
η_t = 510 / 600 = 0.85.

Assumptions and limits

Adiabatic with negligible ke and pe changes.
Same discharge pressure for 2a and 2s.

Common mistakes

Inverting the ratio (that is for compressors).
Comparing states at different pressures.

Statistics and uncertainty

Efficiency is a ratio of two enthalpy differences, so a 1% error in each can produce up to about 2% in η.

Where it is used

Turbine performance testing.
Gas-turbine and steam-cycle design.

Maths you need

Enthalpy: h
Isentropic process: constant entropy

Related formulas

Practise this formula