**Brayton cycle**describes the workings of a

**constant-pressure heat engine**. The Brayton cycle is one of the most common

**thermodynamic cycles**found in gas turbine power plants or airplanes.

The** thermal efficiency** in terms of the compressor **pressure ratio** (PR = p_{2}/p_{1}), which is the parameter commonly used:

In general, **increasing the pressure ratio** is the most direct way to increase the overall thermal efficiency of a Brayton cycle.

In 1872, an American engineer, **George Bailey Brayton,** advanced the study of heat engines by patenting a constant pressure internal combustion engine, initially using vaporized gas but later using liquid fuels such as kerosene. This heat engine is known as “Brayton’s* Ready Motor”*. The

**original Brayton engine**used a

**piston compressor**and

**piston expander**instead of a gas turbine and gas compressor.

Today,** modern gas turbine engines** and **airbreathing jet engines** are also constant-pressure heat engines. Therefore we describe their thermodynamics by the **Brayton cycle**. In general, the **Brayton cycle** describes the workings of a **constant-pressure heat engine**.

It is one of the most common **thermodynamic cycles** found in gas turbine power plants or airplanes. In contrast to the Carnot cycle, the **Brayton cycle** does not execute isothermal processes because these must be performed very slowly. In an **ideal Brayton cycle**, the system executing the cycle undergoes a series of four processes: two isentropic (reversible adiabatic) processes alternated with two isobaric processes.

Since **Carnot’s principle** states that no engine can be more efficient than a reversible engine (**a Carnot heat engine**) operating between the same high temperature and low-temperature reservoirs, a gas turbine based on the Brayton cycle must have lower efficiency than the Carnot efficiency.

A large single-cycle gas turbine typically produces 300 megawatts of electric power and has 35–40% thermal efficiency. Modern Combined Cycle Gas Turbine (CCGT) plants, in which the thermodynamic cycle consists of two power plant cycles (e.g.,, the Brayton cycle and the Rankine cycle), can achieve a thermal efficiency of around 55%.

## Types of Gas Turbines

In general, heat engines and also gas turbines are categorized according to a combustion location as:

**Turbines with internal combustion**. Most gas turbines are internal combustion engines. In these turbines, the high temperature is achieved by burning the fuel-air mixture in the combustion chamber.**Turbines with external combustion**. A**heat exchanger**is usually used in these turbines, and only a clean medium with no combustion products travels through the power turbine. Since the turbine blades are not subjected to combustion products, much lower quality (and therefore cheaper) fuels can be used. These turbines usually have lower thermal efficiency than turbines with internal combustion.

## Types of Brayton Cycle

**Open Brayton Cycle (keywords)**

Since most gas turbines are based on the **Brayton cycle** with **internal combustion** (e.g.,, jet engines), they are based on the **open Brayton cycle**. In this cycle, air from the ambient atmosphere is compressed to the compressor’s higher pressure and temperature. In the combustion chamber, the air is heated further by burning the fuel-air mixture in the airflow. Combustion products and gases expand in the turbine either to near atmospheric pressure (engines producing mechanical energy or electrical energy) or to a pressure required by the jet engines. The open Brayton cycle means that the gases are discharged **directly into the atmosphere**.

**Closed Brayton Cycle**

In a **closed Brayton cycle,** the working medium (e.g.,, helium) **recirculates in the loop,** and the gas expelled from the turbine is reintroduced into the compressor. A **heat exchanger** (external combustion) is usually used in these turbines, and only a clean medium with no combustion products travels through the power turbine. The **closed Brayton cycle** is used, for example, in closed-cycle gas turbines and high-temperature gas-cooled reactors.

**Reverse Brayton Cycle – Brayton Refrigeration Cycle**

A Brayton cycle that is driven in the reverse direction is known as the reverse Brayton cycle. Its purpose is to move heat from the colder to the hotter body rather than produce work. In compliance with the second law of thermodynamics, **Heat cannot spontaneously flow** from cold system to hot system without external work being performed on the system. Heat can flow from colder to the hotter body, but **only when forced by external work**. This is exactly what refrigerators and heat pumps accomplish. These are driven by electric motors requiring work from their surroundings to operate. One of the possible cycles is a reverse Brayton cycle, which is similar to the ordinary Brayton cycle, but it is driven in reverse via a network input. This cycle is also known as the gas refrigeration cycle or the Bell Coleman cycle. This cycle is widely used in jet aircraft for air conditioning systems using air from the engine compressors. It is also widely used in the LNG industry. The largest reverse Brayton cycle is for subcooling LNG using 86 MW of power from a gas turbine-driven compressor and nitrogen refrigerant.

## Brayton Cycle – Processes

In a **closed ideal Brayton cycle**, the system executing the cycle undergoes a series of four processes: two isentropic (reversible adiabatic) processes alternated with two isobaric processes:

**Isentropic compression**(compression in a compressor) – The working gas (e.g.,, helium) is compressed adiabatically from state 1 to state 2 by the compressor (usually an axial-flow compressor). The surroundings work on the gas, increasing its internal energy (temperature) and compressing it (increasing its pressure). On the other hand, the entropy remains unchanged. The work required for the compressor is given by**W**_{C}**= H**_{2}**– H**_{1}**.****Isobaric heat addition**(in a heat exchanger) – In this phase (between state 2 and state 3), there is a constant-pressure heat transfer to the gas from an external source since the chamber is open to flow in and out. In an open ideal Brayton cycle, the compressed air 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}**= H**_{3 }**– H**_{2}**Isentropic expansion**(expansion in a turbine) – The compressed and heated gas expands adiabatically from state 3 to state 4 in a turbine. The gas works on the surroundings (blades of the turbine) and loses an amount of internal energy equal to the work that leaves the system. The work done by the turbine is given by**W**_{T}**= H**_{4}**– H**_{3}**.**Again the entropy remains unchanged.**Isobaric heat rejection (in a heat exchanger)**– In this phase, the cycle completes by a constant-pressure process in which heat is rejected from the gas. The working gas temperature drops from point 4 to point 1. The net heat rejected is given by**Q**_{re}**= H**_{4 }**– H**_{1}

During a Brayton cycle, work is done on the gas by the compressor between states 1 and 2 (**i****sentropic compression**). Work is done by the gas in the turbine between stages 3 and 4 (**i****sentropic expansion**). The difference between the work done by the gas and the work done on the gas is the network produced by the cycle, and it corresponds to the area enclosed by the cycle curve (in the pV diagram).

As can be seen, it is convenient to use **enthalpy** or **specific enthalpy** and express the first law in terms of enthalpy in the analysis of this thermodynamic cycle. This form of the law **simplifies the description of energy transfer**. **At constant pressure**, the **enthalpy change** equals the **energy** transferred from the environment through heating:

**Isobaric process (Vdp = 0):**

**dH = dQ → Q = H**_{2}** – H**_{1}

**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 = H**_{2}** – H**_{1}

See also: **Why power engineers use enthalpy? Answer: dH = dQ + Vdp**

## 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**. The assumption of no heat transfer is very important since we can use the adiabatic approximation only in **very rapid processes**.

**Isentropic Process and the First Law**

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

**dH = dQ + Vdp**

**or**

**dH = TdS + Vdp**

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

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

_{p}

*(T*

_{2}

*– T*

_{1}

*)**(for ideal gas)*

**Isentropic Process of the Ideal Gas**

The **isentropic process** (a special case of 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}## Isobaric Process

An** isobaric process** is a thermodynamic process in which the **system’s pressure **

**remains constant**(p = const). The heat transfer into or out of the system does work and changes the system’s internal energy.

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**

**Isobaric Process and the First Law**

The classical form of the first law of thermodynamics is the following equation:

**dU = dQ – dW**

In this equation, dW is equal to **dW = pdV** and is known as the boundary work. In an isobaric process and the ideal gas, **part of the heat added** to the system will be used to **do work,** and **part of the heat** added will increase the **internal energy** (increase the temperature). Therefore it is convenient to use **enthalpy** instead of internal energy.

**Isobaric process (Vdp = 0):**

**dH = dQ → Q = H**_{2}**– H**_{1}

**At constant entropy**, i.e., in the isentropic process, the **enthalpy change** equals the **flow process work** done on or by the system.

**Isobaric Process of the Ideal Gas**

The **isobaric process** can be expressed with the **ideal gas law** as:

or

On a **p-V diagram**, the process occurs along a horizontal line (called an isobar) with the equation p = constant.

See also: Charles’s Law

## Brayton Cycle – pV, Ts diagram

The **Brayton cycle** is often plotted on a pressure-volume diagram (**pV diagram**) and a temperature-entropy diagram (**Ts diagram**).

When plotted on a **pressure-volume diagram**, the isobaric processes follow the isobaric lines for the gas (the horizontal lines), adiabatic processes move between these horizontal lines, and the area bounded by the complete cycle path represents the **total work** that can be done during one cycle.

The **temperature-entropy diagram** (Ts diagram) in which the thermodynamic state is specified by a point on a graph with specific entropy (s) as the horizontal axis and absolute temperature (T) as the vertical axis. Ts diagrams are a useful and common tool, particularly because it helps to visualize the **heat transfer** during a process. For reversible (ideal) processes, the area under the T-s curve of a process is the** heat transferred** to the system during that process.

## Thermal Efficiency of Brayton Cycle

In general, the **thermal efficiency**, *η***_{th}**, of any heat engine is defined as the ratio of the work it does,

**W**, to the heat input at the high temperature, Q

_{H}.

The **thermal efficiency**, *η***_{th}**, represents the fraction of

**heat**,

**Q**

**, converted**

_{H}**to work**. Since energy is conserved according to the

**first law of thermodynamics**and energy cannot be converted to work completely, the heat input, Q

_{H}, must equal the work done, W, plus the heat that must be dissipated as

**waste heat Q**

**into the environment. Therefore we can rewrite the formula for thermal efficiency as:**

_{C}This is a very useful formula, but we express the thermal efficiency using the first law in terms of enthalpy.

To calculate the thermal efficiency of the **Brayton cycle** (single compressor and single turbine), engineers use the first law of thermodynamics in terms of enthalpy rather than internal energy.

The first law in terms of enthalpy is:

**dH = dQ + Vdp**

In this equation, the term ** Vdp** is a

**flow process work.**This work,

**, is used for**

*Vdp***open flow systems**like a

**turbine**or a

**pump**in which there is a

**“dp”**, i.e., change in pressure. There are no changes in the control volume. As can be seen, this form of the law

**simplifies the description of energy transfer**.

There are expressions in terms of more familiar variables such as temperature and pressure:

*dH = C*_{p}*dT + V(1-αT)dp*

Where **C**** _{p}** is the

**heat capacity at constant pressure**and

**is the (cubic) thermal expansion coefficient. For**

*α***ideal gas**αT = 1 and therefore:

*dH = C*_{p}*dT*

**At constant pressure**, the **enthalpy change** equals the **energy** transferred from the environment through heating:

**Isobaric process (Vdp = 0):**

**dH = dQ → Q = H**_{3}** – H**_{2}** → H**_{3}** – H**_{2}** = C**

_{p}

*(T*

_{3}

*– T*

_{2}

*)***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 = H**_{4}** – H**_{3}** → H**_{4}** – H**_{3}** = C**

_{p}

*(T*

_{4}

*– T*

_{3}

*)*The **enthalpy** can be made into an **intensive** or **specific** variable by dividing by the mass. **Engineers use the** **specific enthalpy** in thermodynamic analysis more than the enthalpy itself.

Now, 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. This cycle consist of four thermodynamic processes:

Isentropic compression – ambient air is drawn into the compressor, 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, 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}**= H**_{3 }**– H**_{2} - 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}**= H**_{4}**– H**_{3} - Isobaric heat rejection – the residual heat must be rejected to close the cycle. The net heat rejected is given by
**Q**_{re}**= H**_{4 }**– H**_{1}

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

The thermal efficiency of such a simple Brayton cycle for ideal gas and in terms of specific enthalpies can now be expressed in terms of the temperatures:

where

- W
_{T }the work done by the gas in the turbine - W
_{C }the work done on the gas in the compressor - c
_{p }is the**heat capacity ratio**

### Pressure Ratio – Brayton Cycle – Gas Turbine

The** thermal efficiency** in terms of the compressor **pressure ratio** (PR = p_{2}/p_{1}), which is the parameter commonly used:

In general, **increasing the pressure ratio** is the most direct way to increase the overall thermal efficiency of a Brayton cycle because the cycle approaches the Carnot cycle.

According to Carnot’s principle, higher efficiencies can be attained by increasing the temperature of the gas.

But there are also **limits on the pressure ratios** that can be used in the cycle. The highest temperature in the cycle occurs at the end of the combustion process, and it is limited by the **maximum temperature** that the **turbine blades** can withstand. As usual, metallurgical considerations (about 1700 K) place upper limits on thermal efficiency.

Consider the effect of compressor pressure ratio on thermal efficiency when the turbine inlet temperature is restricted to the maximum allowable temperature. There are two Ts diagrams of Brayton cycles having the same turbine inlet temperature but different compressor pressure ratios on the picture. As can be seen for a fixed-turbine inlet temperature, the network output per cycle (W_{net} = W_{T} – W_{C}) decreases with the pressure ratio (**Cycle A**). But the Cycle A has greater efficiency.

On the other hand, **Cycle B** has a larger network output per cycle (enclosed area in the diagram) and thus the greater network developed per unit of mass flow. The work produced by the cycle times a mass flow rate through the cycle is equal to the power output produced by the gas turbine.

Therefore with less work output per cycle (Cycle A), a larger mass flow rate (thus a **larger system**) is needed to maintain the same power output, which may not be economical. This is the key consideration in the gas turbine design since here. Engineers must balance thermal efficiency and compactness. In most common designs, the pressure ratio of a gas turbine ranges from about 11 to 16.

### Thermal Efficiency Improvement – Brayton Cycle

There are several methods, how can be the thermal efficiency of the Brayton cycle improved. Assuming that the maximum temperature is limited by metallurgical consideration, these methods are:

**Reheat, Intercooling and Regeneration in Brayton Cycle**

As was discussed,** reheat and inter-cooling** are complementary to** heat regeneration**. By themselves, they would not necessarily increase the thermal efficiency, however, when inter-cooling or reheat is used in conjunction with heat regeneration, a significant increase in thermal efficiency can be achieved, and the network output is also increased. This requires a gas turbine with two stages of compression and two turbine stages.

## Isentropic Efficiency – Turbine, Compressor

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 the device to

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

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

See also: Irreversibility of Natural Processes

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%).

## Brayton Cycle – Problem with Solution

Let assume the closed **Brayton cycle**, one of the most common **thermodynamic cycles** found in modern gas turbine engines. In this case, assume a** helium gas turbine** with a single compressor and single turbine arrangement. One of the key parameters of such engines is the maximum turbine inlet temperature and the compressor pressure ratio (PR = p_{2}/p_{1}), which determines the thermal efficiency of such engines.

In this turbine, the high-pressure stage receives gas (point 3 at the figure) from a heat exchanger:

- p
_{3 }= 6.7 MPa; - T
_{3}= 1190 K (917°C)) - the isentropic turbine efficiency is
*η*= 0.91 (91%)_{T}

and exhaust it to another heat exchanger, where the outlet pressure is (point 4):

- p
_{4}= 2.78 MPa - T
_{4,is}= ?

Thus the compressor pressure ratio is equal to PR = 2.41. Moreover, we know that the compressor receives gas (point 1) at the figure:

- p
_{1 }= 2.78 MPa; - T
_{1}= 299 K (26°C) - the isentropic compressor efficiency η
_{K}= 0.87 (87%).

The heat capacity ratio for helium is equal to** =c _{p}/c_{v}=1.66**

- the heat added by the heat exchanger (between 2 → 3)
- the compressor outlet temperature of the gas (T
_{2,is}) - the real work done on this compressor when the isentropic compressor efficiency is η
_{K}= 0.87 (87%) - the turbine outlet temperature of the gas (T
_{4,is}) - the real work done by this turbine, when the isentropic turbine efficiency is η
_{T}= 0.91 (91%) - the thermal efficiency of this cycle

**Solution:**

1) + 2)

From the first law of thermodynamics, the net heat added is given by **Q _{add,ex} = H_{3 }– H_{2}** [kJ] or Q

_{add}= C

_{p}.(T

_{3}-T

_{2s})

**,**but in this case we do not know the temperature (T

_{2s}) at the outlet of the compressor. We will solve this problem in intensive variables. We have to rewrite the previous equation (to include

**η**

**) using the term (**

_{K}**+h**) to:

_{1}– h_{1}Q_{add} = h_{3} – h_{2} = h_{3} – h_{1} – (h_{2s} – h_{1})/η_{K} ** **[kJ/kg]

Q_{add }**= **c_{p}(T_{3}-T_{1}) – (c_{p}(T_{2s}-T_{1})**/**η_{K})

Then we will calculate the temperature, **T _{2s}**, using p, V, T Relation (from Ideal Gas Law) for the adiabatic process between (1 → 2).

In this equation the factor for helium is equal to =c_{p}/c_{v}=1.66. From the previous equation follows that the compressor outlet temperature, T_{2s}, is:

Using this temperature and the isentropic compressor efficiency we can calculate the heat added by the heat exchanger:

**Q _{add }= **c

_{p}(T

_{3}-T

_{1}) – (c

_{p}(T

_{2s}-T

_{1})

**/**η

_{K}) = 5200.(1190 – 299) – 5200.(424-299)/0.87 = 4.633 MJ/kg – 0.747 MJ/kg =

**3.886 MJ/kg**

3)

The work done on the gas by the compressor in the isentropic compression process is:

*W*_{C,s}* = c*_{p}* (T*_{2s}* – T*_{1}*) = 5200 x (424 – 299) = 0.650 MJ/kg*

The real work done on the gas by the compressor in the adiabatic compression is then:

*W*_{C,real}* = c*_{p}* (T*_{2s}* – T*_{1}*). η*_{C}*= 5200 x (424 – 299) / 0.87 = 0.747 MJ/kg*

4)

The turbine outlet temperature of the gas,** T _{4,is}**, can be calculated using the same

**p, V, T Relation**as in 2) but between states 3 and 4:

The previous equation follows that the outlet temperature of the gas, T_{4,is}, is:

5)

The work done by gas turbine in the isentropic expansion 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 the adiabatic expansion is then:

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

6)

As was derived in the previous section, the **thermal efficiency** of an ideal Brayton cycle is a function of **pressure ratio** and **κ**:

therefore

*η*_{th }*=* 0.295 = **29.5%**

The thermal efficiency can be also calculated using the work and the heat (without η_{K}):

*η*_{th,s }*= (**W*_{T,s}* – W*_{C,s}*)*** / **Q

_{add,s}= (1.825 – 0.650) / 3.983

**0.295 =**

*=***29.5%**

Finally, the thermal efficiency including isentropic turbine/compressor efficiency is:

*η*_{th,real }*= (**W*_{T,real}* – W*_{C,real}*)* ** / **Q

_{add}= (1.661 – 0.747) / 3.886

**0.235 =**

*=***23.5%**