Example: Isentropic Turbine Efficiency
Isentropic process is a special case of adiabatic processes. It is a reversible adiabatic process. An isentropic process can also be called a constant entropy process.
Assume an isentropic expansion of helium (3 → 4) in a gas turbine. In this turbines the high-pressure stage receives gas (point 3 at the figure; p3 = 6.7 MPa; T3 = 1190 K (917°C)) from a heat exchanger and exhaust it to another heat exchanger, where the outlet pressure is p4 = 2.78 MPa (point 4). The temperature (for isentropic process) of the gas at the exit of the turbine is T4s = 839 K (566°C).
Calculate the work done by this turbine and calculate the real temperature at the exit of the turbine, when the isentropic turbine efficiency is ηT = 0.91 (91%).
From the first law of thermodynamics, the work done by turbine in an isentropic process can be calculated from:
WT = h3 – h4s → WTs = cp (T3 – T4s)
From Ideal Gas Law we know, that the molar specific heat of a monatomic ideal gas is:
Cv = 3/2R = 12.5 J/mol K and Cp = Cv + R = 5/2R = 20.8 J/mol K
We transfer the specific heat capacities into units of J/kg K via:
cp = Cp . 1/M (molar weight of helium) = 20.8 x 4.10-3 = 5200 J/kg K
The work done by gas turbine in isentropic process is then:
WT,s = cp (T3 – T4s) = 5200 x (1190 – 839) = 1.825 MJ/kg
The real work done by gas turbine in adiabatic process is then:
WT,real = cp (T3 – T4s) . ηT = 5200 x (1190 – 839) x 0.91 = 1.661 MJ/kg