**Second law efficiency**

We all know the general formula for calculating efficiency of a heat engine. It is a ratio of output work to input heat.

This formula for calculating efficiency is in line with first law of thermodynamics. Which deals with the conservation of quantity of energy (not quality).

But when it comes to the second law of thermodynamics which deals with the quality of energy, a different formula for calculating efficiency is required.

This formula takes available energy into account. This efficiency is known as *Second Law Efficiency*.

Second law efficiency is the ratio of *minimum exergy intake to perform given task* to the *actual exergy intake to perform the same task.*

η_{II} = Minimum exergy intake to perform given task/ Actual exergy intake to perform the same task

Now let’s discuss the second law efficiency for turbine and compressor.

*Where:*

*h: Enthalpy*

*af: Availability (exergy) of a flowing stream (or specific flowability)*

*s: Entropy*

*T _{0}: Operating temperature of turbine*

*Note: In this calculation we are neglecting changes in kinetic and potential energies.*

**Second law efficiency of an adiabatic turbine**

Let us consider an adiabatic turbine (as shown in picture).

From steady flow energy equation (SFEE).

∂W/∂m = h_{1} -h_{2} … (1)

We know

af = h – (T * s)

hence

af_{2} = h_{2} – (T_{0} * s_{2})

af_{1} = h_{1} – (T_{0} * s_{1})

Difference in specific flowability at inlet and outlet

af_{1} – af_{2} = h_{1} – h_{2} + T_{0} (s_{2} – s_{1}) … (2)

from equation (1) and (2)

af_{1} – af_{2} = (∂W/∂m) + T_{0} (s_{2} – s_{1}) … (3) (Maximum exergy intake)

For an ideal or reversible process

s_{2} – s_{1} = 0

Hence for a reversible process equation (3) becomes

af_{1} – af_{2} = ∂W/∂m (Minimum exergy intake)

Now, second law efficiency is

η_{II} = Minimum exergy intake to perform given task/ Actual exergy intake to perform the same task

η_{II} = (∂W/∂m)/ [(∂W/∂m) + T_{0} (s_{2} – s_{1})]

**Second law efficiency of an adiabatic compressor**

Let us consider an adiabatic compressor (as shown in picture).

From steady flow energy equation (SFEE).

∂W/∂m = h_{2} -h_{1 }… (4)

af_{2} = h_{2} – (T_{0} * s_{2})

af_{1} = h_{1} – (T_{0} * s_{1})

Difference in specific flowability at inlet and outlet

af_{2} – af_{1} = h_{2} – h_{1} – T_{0} (s_{2} – s_{1}) … (5)

From equation (4) and (5)

af_{2} – af_{1} = (∂W/∂m) – T_{0} (s_{2} – s_{1}) … (6) (Maximum exergy intake)

For an ideal or reversible process

s_{2} – s_{1} = 0

Hence for a reversible process equation (6) becomes

af_{2} – af_{1} = ∂W/∂m (Minimum exergy intake)

Now, second law efficiency is

η_{II} = Minimum exergy intake to perform given task/ Actual exergy intake to perform the same task

η_{II} = (∂W/∂m)/ [(∂W/∂m) – T_{0} (s_{2} – s_{1})]