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## Adiabatic Process

An **adiabatic process** is a thermodynamic process, in which there is **no heat transfer** into or out of the system (Q = 0). The system can be considered to be **perfectly insulated**. In an adiabatic process, energy is transferred only as work. The assumption of no heat transfer is very important, since we can use the adiabatic approximation only in **very rapid processes**. In these rapid processes, there is not enough time for the transfer of energy as heat to take place to or from the system.

In real devices (such as turbines, pumps, and compressors) **heat losses** and losses in the combustion process occur, but these losses are usually low in comparison to overall energy flow and we can approximate some thermodynamic processes by the adiabatic process.

## Adiabatic Process and the First Law

For a closed system, we can write the **first law of thermodynamics in terms of enthalpy**:

**dH = dQ + Vdp**

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**. In adiabatic process, the **enthalpy change** equals the **flow process work** done on or by the system:

**Adiabatic process (dQ = 0):**

**dH = Vdp → W = H**_{2}** – H**_{1}** → H**_{2}** – H**_{1}** = C**

_{p}

*(T*

_{2}

*– T*

_{1}

*)**(for an ideal gas)*

## Adiabatic Expansion – Adiabatic 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 **adiabatic process** can be expressed with the **ideal gas law** as:

*pV ^{κ} = constant*

or

**p _{1}V_{1}^{κ} = p_{2}V_{2}^{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**is a factor in determining the speed of sound in a gas and other adiabatic processes.

_{p}/c_{v}**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 / V ^{κ}**.

**For an ideal gas and a polytropic process, the case**

*n = κ*

**corresponds to an adiabatic process.**

## Example of Adiabatic Expansion

Assume an **adiabatic 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 (**T _{4,real}**). In this turbines the high-pressure stage receives gas (point 3 at the figure; p

_{3 }=

**6.7 MPa**;

**T**(917°C)) from a heat exchanger and exhaust it to another heat exchanger, where the outlet pressure is p

_{3}= 1190 K_{4}=

**2.78 MPa**(point 4)

**.**

**Solution:**

The outlet temperature of the gas, T_{4,real}, can be calculated using **p, V, T Relation** for adiabatic process. Note that, it is the same relation as for the isentropic process, therefore results must be identical. It this case, we calculate the expansion for different gas turbine (less efficient) as in case of **Isentropic Expansion in Gas Turbine.**

In this equation the factor for helium is equal to **κ****=c _{p}/c_{v}=1.66**. From the previous equation follows that the outlet temperature of the gas,

**T**, is:

_{4,real}See also: **Mayer’s relation**

## Adiabatic Process in Gas Turbine

Let assume the **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 Brayton cycle consist of four thermodynamic processes. Two adiabatic processes and two isobaric processes.

**adiabatic compression**– ambient air is drawn into the compressor, where it is pressurized (1 → 2). The work required for the compressor is given by**W**_{C}= H_{2}– H_{1}.**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**Q**_{add}= H_{3 }– H_{2}**adiabatic expansion**– the heated, pressurized air then expands on turbine, gives up its energy. The work done by turbine is given by**W**_{T}= H_{4}– H_{3}**isobaric heat rejection**– the residual heat must be rejected in order to close the cycle. The net heat rejected is given by**Q**_{re}= H_{4 }– H_{1}

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 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*

## Free Expansion – Joule Expansion

These are** adiabatic processes** in which **no transfer of heat** occurs between the system and its environment and **no work is done** on or by the system. These types of adiabatic processes are called** free expansion**. It is an** irreversible process** in which a gas expands into an insulated evacuated chamber. It is also called **Joule expansion**. For an ideal gas, the temperature doesn’t change (see: Joule’s Second Law), however, real gases experience a temperature change during free expansion. In free expansion Q = W = 0, and the first law requires that:

**dE**_{int}** = 0**

A free expansion can not be plotted on a P-V diagram, because the process is rapid, not quasistatic. The intermediate states are not equilibrium states, and hence the pressure is not clearly defined.