Isentropic Process

Isentropic Process and the First Law

The first law of thermodynamics in terms of enthalpy:

dH = dQ + Vdp

or

dH = TdS + Vdp

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 = C_p (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 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 the adiabatic process) can be expressed with the ideal gas law as:

pV^κ = constant

or

p₁V₁^κ = p₂V₂^κ

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

Other p, V, T Relation

On a p-V diagram, the process occurs along a line (called an adiabat) with the equation p = constant / V^κ. 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 the outlet temperature of the gas (T_4,is). In these 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 exhausts it to another heat exchanger, where the outlet pressure is p₄ = 2.78 MPa (point 4).

Solution:

The outlet temperature of the gas, T_{4 is}, can be calculated using p, V, T Relation for the isentropic process (reversible adiabatic process):

In this equation, the factor for helium is equal to κ=c_p/c_v=1.66. The previous equation follows that the outlet temperature of the gas, T_4,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.

The ideal Brayton cycle consists of four thermodynamic processes. Two isentropic processes and two isobaric processes.

1. Isentropic compression – ambient air is drawn into the compressor, pressurized (1 → 2). The work required for the compressor is given by W_C = H₂ – H₁.
2. Isobaric heat addition – the compressed air then runs through a combustion chamber, burning fuel, 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 Q_add = H₃ – H₂
3. Isentropic expansion – the heated, pressurized air then expands on a turbine, gives up its energy. The work done by the turbine is given by W_T = H₄ – H₃
4. Isobaric heat rejection – the residual heat must be rejected to close the cycle. The net heat rejected is given by Q_re = 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 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 a device to work by a device when operated under isentropic conditions (in the case of the 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 these 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 the 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 the turbine in an isentropic process can be calculated from:

W_T = h₃ – h_4s     →     W_Ts = c_p (T₃ – 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 (T₃ – 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 (T₃ – T_4s) . η_T = 5200 x (1190 – 839) x 0.91 = 1.661 MJ/kg