# What does Isentropic Efficiency actually mean?

I know that Isentropic Efficiency is a comparison between the actual performance of something(turbine for example) and the performance that would be achieved under idealized circumstances for the same inlet and outlet. Let's say a turbine has an isentropic efficiency of 0.75. What does 0.75 mean and the rest (0.25) mean? Thank you

It means that, assuming adiabatic operation, for the same pressures in and out, the shaft work done by the turbine is 25% less and the enthalpy change of the flow stream is 25% less. This, in turn, means that the enthalpy of the exit stream is higher by 25% of the isentropic work. The decrease in the work is equivalent to the increase in the outlet enthalpy. So the temperature of the exit stream is higher.

• Sorry Chet. Posted my answer before seeing yours. Commented Nov 13, 2020 at 16:02
• @BobD Oops. Missed that. Oh well, maybe our answers compliment one another. Commented Nov 13, 2020 at 16:05
• Yes, because yours pointed out the higher exit temperature and mine referred to a higher exit enthalpy. Commented Nov 13, 2020 at 16:07

The isentropic turbine efficiency is the ratio of the actual work output of the turbine to the work output of the turbine if the turbine undergoes an isentropic (constant entropy) process between the same inlet and exit turbine pressures. So if a turbine has an isentropic efficiency of 75% it means it does 25% less work than an ideal (isentropic) turbine due to entropy production in the process.

The work output of an adiabatic turbine is the difference between the inlet and exit enthalpies of the steam. An ideal turbine is both adiabatic and reversible, which makes it isentropic, making the exiting enthalpy a minimum. If it is adiabatic but not reversible, the exiting enthalpy will be greater, meaning there is less energy available for doing work. So we can define the isentropic efficiency, $$E_s$$ as follows

$$E_{s}=\frac{h_{i}-h_{e}}{h_{i}-h_{es}}$$

Where

$$h_{es}$$ = the exiting enthalpy for an isentropic process

$$h_{e}$$ = the actual exiting enthalpy

$$h_{e}>h_{es}$$

Hope this helps.