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 (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
On a p-V diagram, the process occurs along a line (called an adiabat) that has the equation p = constant / Vκ. For an ideal gas and a polytropic process, the case n = κ corresponds to an isentropic process.
Example: Isentropic Expansion in 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):
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:
Example: Isentropic Expansion in Gas Turbine
Let assume the ideal Brayton cycle that describes the workings of a constant pressure heat engine. Modern 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.
- 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.
- 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 = H3 – H2
- 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
- isobaric heat rejection – the residual heat must be rejected in order to close the cycle. The net heat rejected is given by Qre = H4 – 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%).
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; 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
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