## Isentropic Efficiency – Turbine, Compressor, Nozzle

In previous chapters we assumed that the gas expansion is **isentropic** and therefore we used **T _{4,is }** as the outlet temperature of the gas. These assumptions are only applicable with ideal cycles.

Most **steady-flow devices** (turbines, compressors, nozzles) operate under **adiabatic conditions**, but they are not truly isentropic but are rather idealized as isentropic for calculation purposes. We define parameters *η*_{T}*, **η*_{C}*, η*_{N}** , **as a

**ratio**of

**real work done**by device to

**work by device when operated under isentropic conditions**(in case of turbine). This ratio is known as the

**Isentropic Turbine/Compressor/Nozzle Efficiency**.

These parameters describe how efficiently a turbine, compressor or nozzle approximates a corresponding isentropic device. This parameter reduces the overall efficiency and work output. For turbines, the value of *η***_{T}** is typically 0.7 to 0.9 (70–90%).

## Example: Isentropic Turbine Efficiency

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; p_{3 }= **6.7 MPa**; T_{3} = 1190 K (917°C)) from a heat exchanger and exhaust it to another heat exchanger, where the outlet pressure is p_{4} = **2.78 MPa **(point 4)**.** The temperature (for isentropic process) of the gas at the exit of the turbine is T_{4s} = 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 **η**_{T}** = 0.91 (91%)**.

**Solution:**

From the first law of thermodynamics, the work done by turbine in an isentropic process can be calculated from:

**W**_{T}** = h**_{3}** – h**_{4s}** → W**_{Ts}** = c**

_{p}

*(T*

_{3}

*– T*

_{4s}

*)*From Ideal Gas Law we know, that the molar specific heat of a monatomic ideal gas is:

*C*_{v}** = 3/2R = 12.5 J/mol K** and

*C*

_{p}

*= C*

_{v}

*+ R = 5/2R = 20.8 J/mol K*We transfer the specific heat capacities into units of **J/kg K via:**

*c*_{p}* = C*_{p}* . 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:

*W*_{T,s}* = c*_{p}* (T*_{3}* – T*_{4s}*) = 5200 x (1190 – 839) = 1.825 MJ/kg*

The real work done by gas turbine in adiabatic process is then:

*W*_{T,real}* = c*_{p}* (T*_{3}* – T*_{4s}*) . **η*_{T}*= 5200 x (1190 – 839) x 0.91 = 1.661 MJ/kg*