Isentropic Process

An isentropic process is a thermodynamic process, in which the entropy of the fluid or gas remains constant. It means the isentropic process is a special case of an adiabatic process in which there is no transfer of heat or matter. It is a reversible adiabatic process. An isentropic process can also be called a constant entropy process. In engineering such an idealized process is very useful for comparison with real processes.

Since there are changes in internal energy (dU) and changes in system volume (∆V), engineers often use the enthalpy of the system, which is defined as:

H = U + pV

In many thermodynamic analyses it is convenient to use the enthalpy instead of the internal energy. Especially in case of the first law of thermodynamics.

Isentropic Process and the First Law

The first law of thermodynamics in terms of enthalpy:

dH = dQ + Vdp

or

dH = TdS + Vdp

Isentropic Process - characteristics

Table of main characteristics

See also: First Law of Thermodynamics

See also: Ideal Gas Law

See also: What is Enthalpy

In this equation the term Vdp is a flow process work. This work,  Vdp, is used for open flow systems like a turbine or a pump in which there is a “dp”, i.e. change in pressure. As can be seen, this form of the law simplifies the description of energy transfer. At constant entropy, i.e. in isentropic process, the enthalpy change equals the flow process work done on or by the system:

Isentropic process (dQ = 0):

dH = Vdp     →     W = H2 – H1     →     H2 – H1 = Cp (T2 – T1)    (for ideal gas)

Isentropic Expansion – Isentropic Compression

See also: What is an Ideal Gas

In an ideal gas, molecules have no volume and do not interact. According to the ideal gas law, pressure varies linearly with temperature and quantity, and inversely with volume.

pV = nRT

where:

  • p is the absolute pressure of the gas
  • n is the amount of substance
  • T is the absolute temperature
  • V is the volume
  • R  is the ideal, or universal, gas constant, equal to the product of the Boltzmann constant and the Avogadro constant,

In this equation the symbol R is a constant called the universal gas constant that has the same value for all gases—namely, R =  8.31 J/mol K.

The isentropic process (a special case of adiabatic process) can be expressed with the ideal gas law as:

pVκ = constant

or

p1V1κ = p2V2κ

in which κ = cp/cv is the ratio of the specific heats (or heat capacities) for the gas. One for constant pressure (cp) and one for constant volume (cv). Note that, this ratio κ  = cp/cv is a factor in determining the speed of sound in a gas and other adiabatic processes.

Other p, V, T Relation

p,V,T relation - isentropic process

On a p-V diagram, the process occurs along a line (called an adiabat) that has the equation p = constant / VκFor an ideal gas and a polytropic process, the case n = κ  corresponds to an isentropic process.

Example: Isentropic Expansion in Gas Turbine

P-V diagram - isentropic process

P-V diagram of an isentropic expansion of helium (3 → 4) in a gas turbine.

Assume an isentropic expansion of helium (3 → 4) in a gas turbine. Since helium behaves almost as an ideal gas, use the ideal gas law to calculate outlet temperature of the gas (T4,is). 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).

Solution:

The outlet temperature of the gas, T4,is, can be calculated using p, V, T Relation for isentropic process (reversible adiabatic process):

p,V,T relation - isentropic process

In this equation the factor for helium is equal to κ=cp/cv=1.66. From the previous equation follows that the outlet temperature of the gas, T4,is, is:

isentropic process - example

Example: Isentropic Expansion in Gas Turbine

first law - example - brayton cycle

Ideal Brayton cycle consist of four thermodynamic processes. Two isentropic processes and two isobaric processes.

Let assume the ideal Brayton cycle that describes the workings of a constant pressure heat engineModern gas turbine engines and airbreathing jet engines also follow the Brayton cycle.

Ideal Brayton cycle consist of four thermodynamic processes. Two isentropic processes and two isobaric processes.

  1. isentropic compression – ambient air is drawn into the compressor, where it is pressurized (1 → 2). The work required for the compressor is given by WC = H2 – H1.
  2. isobaric heat addition – the compressed air then runs through a combustion chamber, where fuel is burned and air or another medium is heated (2 → 3). It is a constant-pressure process, since the chamber is open to flow in and out. The net heat added is given by Qadd = H- H2
  3. isentropic expansion – the heated, pressurized air then expands on turbine, gives up its energy. The work done by turbine is given by WT = H4 – H3
  4. isobaric heat rejection – the residual heat must be rejected in order to close the cycle. The net heat rejected is given by Qre = H- H1

As can be seen, we can describe and calculate (e.g. thermal efficiency) such cycles (similarly for Rankine cycle) using enthalpies.

See also: Thermal Efficiency of Brayton Cycle

Isentropic Processes in Thermodynamic Cycles

Ideal Carnot Cycle

  • Isentropic compression
  • Isentropic expansion

Ideal Rankine Cycle

  • Isentropic compression in a pump
  • Isentropic expansion in a turbine

Ideal Brayton Cycle

  • Isentropic compression in a compressor
  • Isentropic expansion in a turbine

Ideal Otto Cycle

  • Isentropic compression
  • Isentropic expansion

Ideal Diesel Cycle

  • Isentropic compression
  • Isentropic expansion

Isentropic Efficiency – Turbine, Compressor, Nozzle

In previous chapters we assumed that the gas expansion is isentropic and therefore we used T4,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 ηT,  ηC, ηN, 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 ηT is typically 0.7 to 0.9 (70–90%).

Isentropic Efficiency - equations

Isentropic vs. adiabatic compression

Isentropic vs. adiabatic expansion

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.

Example: Isentropic Turbine Efficiency

Isentropic vs. adiabatic expansion

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%).

Solution:

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

Nuclear and Reactor Physics:

  1. J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading, MA (1983).
  2. J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.
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  8. Robert Reed Burn, Introduction to Nuclear Reactor Operation, 1988.
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Advanced Reactor Physics:

  1. K. O. Ott, W. A. Bezella, Introductory Nuclear Reactor Statics, American Nuclear Society, Revised edition (1989), 1989, ISBN: 0-894-48033-2.
  2. K. O. Ott, R. J. Neuhold, Introductory Nuclear Reactor Dynamics, American Nuclear Society, 1985, ISBN: 0-894-48029-4.
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  4. E. E. Lewis, W. F. Miller, Computational Methods of Neutron Transport, American Nuclear Society, 1993, ISBN: 0-894-48452-4.