Table of Contents
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.
ηII = 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
T0: 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 = h1 -h2 … (1)
We know
af = h – (T * s)
hence
af2 = h2 – (T0 * s2)
af1 = h1 – (T0 * s1)
Difference in specific flowability at inlet and outlet
af1 – af2 = h1 – h2 + T0 (s2 – s1) … (2)
from equation (1) and (2)
af1 – af2 = (∂W/∂m) + T0 (s2 – s1) … (3) (Maximum exergy intake)
For an ideal or reversible process
s2 – s1 = 0
Hence for a reversible process equation (3) becomes
af1 – af2 = ∂W/∂m (Minimum exergy intake)
Now, second law efficiency is
ηII = Minimum exergy intake to perform given task/ Actual exergy intake to perform the same task
ηII = (∂W/∂m)/ [(∂W/∂m) + T0 (s2 – s1)]
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 = h2 -h1 … (4)
af2 = h2 – (T0 * s2)
af1 = h1 – (T0 * s1)
Difference in specific flowability at inlet and outlet
af2 – af1 = h2 – h1 – T0 (s2 – s1) … (5)
From equation (4) and (5)
af2 – af1 = (∂W/∂m) – T0 (s2 – s1) … (6) (Maximum exergy intake)
For an ideal or reversible process
s2 – s1 = 0
Hence for a reversible process equation (6) becomes
af2 – af1 = ∂W/∂m (Minimum exergy intake)
Now, second law efficiency is
ηII = Minimum exergy intake to perform given task/ Actual exergy intake to perform the same task
ηII = (∂W/∂m)/ [(∂W/∂m) – T0 (s2 – s1)]