# Brayton Cycle: 15 Facts You Should Know

## Brayton cycle diagram

• Process 1-2 : Reversible Adiabatic compression or Isentropic compression ambient air is drawn into the compressor
• Process 2-3 : Constant Pressure heat addition, heat is added to the compressed air when it runs through a combustion chamber
• Process 3-4 : Reversible Adiabatic expansion or Isentropic expansion; Heated compressed air is passed through the turbine
• Process 4-1 : Constant Pressure heat rejection, heat is rejected to ambient air

## Brayton cycle P-V diagram | Brayton cycle T-S diagram

• Process 1-2 : Reversible Adiabatic compression or Isentropic compression ambient air is drawn into the compressor
• Process 2-3 : Constant Pressure heat addition, heat is added to the compressed air when it runs through a combustion chamber
• Process 3-4 : Reversible Adiabatic expansion or Isentropic expansion;  Heated compressed air is passed through the turbine
• Process 4-1 : Constant Pressure heat rejection, heat is rejected to ambient air

## Ideal Brayton cycle | Thermal efficiency of Brayton cycle

The thermal efficiency of Ideal Brayton cycle is given by

$\eta= \frac{T_2-T_1}{T_2}=1-[\frac{P_1}{P_2}]^\frac{\gamma-1}{\gamma}$

Where ϒ = adiabatic index = 1.4 for air

## Thermal efficiency of Brayton cycle | Brayton cycle derivation | Closed Brayton cycle | Brayton cycle analysis

• Process 1-2 : Reversible Adiabatic compression or Isentropic compression
• Process 2-3 : Constant Pressure heat addition
• Process 3-4 : Reversible Adiabatic expansion or Isentropic expansion
• Process 4-1 : Constant Pressure heat rejection

$\frac{T_2}{T_1}=[\frac{V_1}{V_2 }]^{\gamma-1}=r^{\gamma-1}=r_p^\frac{\gamma-1}{\gamma}$

Where,

r is the compression ratio = V1/V2

rp is Pressure ratio = P2/P1

re = Expansion ratio = V4/V3

Process 2 -3: Heat addition at constant pressure is calculated as,

Qin = m Cp[T­3-T2].

Process 3-4: Reversible adiabatic expansion is calculated as

$\frac{T_3}{T_4}=r_p^\frac{\gamma-1}{\gamma}$

Process 4 -1: Heat-rejection at constant pressure will be

QR = m Cp[T­4-T1]

Work done = Qin – QR.

Efficiency of the Brayton cycle is represented as.

$\\\eta=1-\frac{Q_R}{Q_{in}}\\\\ \eta=1-\frac{mC_p (T_4-T_1)}{mC_p (T_3-T_2 )}\\ \\\eta=1-\frac{(T_4-T_1)}{(T_3-T_2 )}$

Since,

$\frac{T_3}{T_4}=\frac{T_2}{T_1}=r_p^\frac{\gamma-1}{\gamma}$

$\eta=1-\frac{1}{r^{\gamma-1}}=1-\frac{1}{r_p^\frac{\gamma-1}{\gamma}}$

## Brayton Refrigeration Cycle | Inverted Brayton Cycle | Joule Brayton Cycle | Reverse Brayton Cycle

It is also known as Bell-Coleman Cycle. It finds its use in Aircraft applications. It is also an inversion of reversed Carnot cycle.

• Process 1-2 : Reversible Adiabatic compression or Isentropic compression
• Process 2-3 : Constant Pressure heat rejection
• Process 3-4 : Reversible Adiabatic expansion or Isentropic expansion
• Process 4-1 : Constant Pressure heat addition

Process 1-2: Reversible Adiabatic compression or isentropic compression

$\frac{T_2}{T_1}=[\frac{V_1}{V_2 }]^{\gamma-1}=r^{\gamma-1}=r_p^\frac{\gamma-1}{\gamma}$

Where,

r is the compression ratio = V1/V2

rp is Pressure ratio = P2/P1

re = Expansion ratio = V4/V3

Process 2-3: Heat-rejection at constant pressure will be

QR = m Cp [T1 – T­4]

Process 3-4: Reversible Adiabatic expansion or isentropic expansion

$\frac{T_3}{T_4}=r_p^\frac{\gamma-1}{\gamma}$

Process 4-1: Constant Pressure heat addition

Qin = m Cp [T2 – T­3].

We know that,

$\frac{T_3}{T_4}=\frac{T_2}{T_1}=r_p^\frac{\gamma-1}{\gamma}$

Work consumed = Qin – QR.

COP of the cycle is represented as.

$\\COP=\frac{Q_R}{W} \\\\ \\COP=\frac{m C_p [T_1-T_4]}{m C_p [T_2-T_3 ]-m C_p [T_1-T_4]}\\\\ \\COP=\frac{1}{\frac{T_2-T_3}{T_1-T_4}-1}\\\\ \\COP=\frac{1}{r_p^\frac{\gamma-1}{\gamma}-1}$

## A Brayton cycle with regeneration using air

In this process, heat from exhaust gases is utilized to preheat the compressed air. Thus air enters at higher temperature in combustion chamber. No additional heat is added in combustion chamber and no excess turbine work is produced and no excess compressor work is done. The efficiency of cycle increases.

Energy balance in regenerator

Heat lost by hot combustion gases = heat gained by compressed air

Cp [T5 – T­6] =Cp [T3 – T­2]

[T5 – T­6] =[T3 – T­2]………. (1)

Effectiveness of regenerator

$e=\frac{T_3-T_2}{T_5-T_2}$

For ideal case e = 1

$1=\frac{T_3-T_2}{T_5-T_2}$

T3 = T­5

From (1)

T6 = T­2

Efficiency of the Brayton cycle is represented as.

$\\\eta_{IR}=1-\frac{Q_R}{Q_{in}}\\\\ \eta_{IR}=1-\frac{mC_p (T_6-T_1)}{mC_p (T_4-T_3)}$

For ideal regenerator

T3 = T­5

T6 = T­2

$\eta_{IR}=1-\frac{mC_p (T_2-T_1)}{mC_p (T_4-T_5)}$

$\eta_{IR}=1-\frac{T_1 (\frac{T_2}{T_1}-1)}{T_5 (\frac{T_4}{T_5}-1)}$

$\frac{T_2}{T_1}=[\frac{V_1}{V_2 }]^{\gamma-1}=r^{\gamma-1}=r_p^\frac{\gamma-1}{\gamma}$

Process 4-5: Constant Pressure heat addition

$\\\frac{T_4}{T_5}=r_p^\frac{\gamma-1}{\gamma}\\\\ \frac{T_4}{T_5}=\frac{T_2}{T_1}$

For eq. (A)

$\\\eta_{IR}=1-\frac{T_1}{T_5 }\\\\ \eta_{IR}=1-\frac{T_1}{T_4 }*\frac{T_4}{T_5 }\\\\ \eta_{IR}=1-\frac{T_{min}}{T_{max} }r_p^\frac{\gamma-1}{\gamma}$

## Actual Brayton cycle

• Process 1-2a :Non- Isentropic compression
• Process 2a-3 :Constant Pressure heat rejection
• Process 3-4a :Non-Isentropic expansion
• Process 4a-1 : Constant Pressure heat addition

The variation due to consideration of efficiencies of turbine and compressor can be evaluated as

Ratio of ideal work to actual work for turbine

$\eta_T=\frac{h_3-h_{4a}}{h_3-h_{4s}}$

Ratio of actual work to ideal work for compressor

$\eta_c=\frac{h_{2s}-h_1}{h_{2a}-h_1}$

## Brayton cycle process

• Process 1-2 : Reversible Adiabatic compression or Isentropic compression
• Process 2-3 : Constant Pressure heat addition
• Process 3-4 : Reversible Adiabatic expansion or Isentropic expansion
• Process 4-1 : Constant Pressure heat rejection

$\frac{T_2}{T_1}=[\frac{V_1}{V_2 }]^{\gamma-1}=r^{\gamma-1}=r_p^\frac{\gamma-1}{\gamma}$

Where,

r is the compression ratio = V1/V2

rp is Pressure ratio = P2/P1

re = Expansion ratio = V4/V3

Process 2 -3: Heat addition at constant pressure is calculated as,

Qin = m Cp[T­3-T2].

Process 3-4: Reversible adiabatic expansion is calculated as

$\frac{T_3}{T_4}=r_p^\frac{\gamma-1}{\gamma}$

Process 4 -1: Heat-rejection at constant pressure will be

QR = m Cp[T­4-T1]

Work done = Qin – QR.

## Brayton cycle gas turbine

• Process 1-2 : Reversible Adiabatic compression or Isentropic compression
• Process 2-3 : Constant Pressure heat addition
• Process 3-4 : Reversible Adiabatic expansion or Isentropic expansion
• Process 4-1 : Constant Pressure heat rejection

$\frac{T_2}{T_1}=[\frac{V_1}{V_2 }]^{\gamma-1}=r^{\gamma-1}=r_p^\frac{\gamma-1}{\gamma}$

Where,

r is the compression ratio = V1/V2

rp is Pressure ratio = P2/P1

re = Expansion ratio = V4/V3

Process 2 -3: Heat addition at constant pressure is calculated as,

Qin = m Cp[T­3-T2].

Process 3-4: Reversible adiabatic expansion is calculated as

$\frac{T_3}{T_4}=r_p^\frac{\gamma-1}{\gamma}$

Process 4 -1: Heat-rejection at constant pressure will be

QR = m Cp[T­4-T1]

Work done = Qin – QR.

Efficiency of the Brayton cycle is represented as.

$\\\eta=1-\frac{Q_R}{Q_{in}}\\\\ \eta=1-\frac{mC_p (T_4-T_1)}{mC_p (T_3-T_2 )}\\ \\\eta=1-\frac{(T_4-T_1)}{(T_3-T_2 )}$

Since,

$\frac{T_3}{T_4}=\frac{T_2}{T_1}=r_p^\frac{\gamma-1}{\gamma}$

$\eta=1-\frac{1}{r^{\gamma-1}}=1-\frac{1}{r_p^\frac{\gamma-1}{\gamma}}$

## Application of Brayton cycle

Brayton cycle is applied in jet propulsion engine. They are durable light weight and can produce high power output. It depends upon high temperature and pressure at the exhaust of turbine to provide thrust force. They are also used in helicopters and military vehicles.

## How to increase efficiency of Brayton cycle

1. Increase Turbine inlet temperatures: According to ideal gas law, rise in temperature is directly related to rise in pressure ratio. By Ideal Brayton efficiency equation, It is directly related to pressure ratio. Thus efficiency is increased.
2. Increase Turbine and Compressor Efficiency: increasing efficiency of turbine and compressor will lead to less mechanical loss thus increasing the overall efficiency.

3. Modifications to the Simple Brayton Cycle:  Using regeneration, intercooling, reheating or everything combined will improve the overall efficiency.

## Open Brayton cycle

• Process 1-2 : Reversible Adiabatic compression or Isentropic compression ambient air is drawn into the compressor
• Process 2-3 : Constant Pressure heat addition, heat is added to the compressed air when it runs through a combustion chamber
• Process 3-4 : Reversible Adiabatic expansion or Isentropic expansion; Heated compressed air is passed through the turbine
• Process 4-1 : Constant Pressure heat rejection, heat is rejected to ambient air

$\frac{T_2}{T_1}=[\frac{V_1}{V_2 }]^{\gamma-1}=r^{\gamma-1}=r_p^\frac{\gamma-1}{\gamma}$

Where,

r is the compression ratio = V1/V2

rp is Pressure ratio = P2/P1

re = Expansion ratio = V4/V3

Process 2 -3: Heat addition at constant pressure is calculated as,

Qin = m Cp[T­3-T2].

Process 3-4: Reversible adiabatic expansion is calculated as

$\frac{T_3}{T_4}=r_p^\frac{\gamma-1}{\gamma}$

Process 4 -1: Heat-rejection at constant pressure will be

QR = m Cp[T­4-T1]

Work done = Qin – QR.

Efficiency of the Brayton cycle is represented as.

$\\\eta=1-\frac{Q_R}{Q_{in}}\\\\ \eta=1-\frac{mC_p (T_4-T_1)}{mC_p (T_3-T_2 )}\\ \\\eta=1-\frac{(T_4-T_1)}{(T_3-T_2 )}$

Since,

$\frac{T_3}{T_4}=\frac{T_2}{T_1}=r_p^\frac{\gamma-1}{\gamma}$

$\eta=1-\frac{1}{r^{\gamma-1}}=1-\frac{1}{r_p^\frac{\gamma-1}{\gamma}}$

## Brayton cycle problems and solutions | Brayton cycle with regeneration example

Q.1) Find power of the turbine operating on Brayton cycle operating with minimum temperature 300 K and maximum temperature 1600 K. It has compression ratio 14:1. Minimum cycle pressure is 100 kPa.  Output power of the cycle is 10 MW. Evaluate the fraction of output power of turbinne required to drive the compressor and the thermal efficiency of the cycle?

Solution: T1= 300 K, T3= 1600 K, P2/P1=14,

$\\a)\;Assume \;s_2= s_1\\\\ T_2=T_1(\frac{P_2}{P_1})^{\frac{k-1}{k}}= 300(14)^{\frac{1.4-1}{1.4}} = 638.1 K\\\\ -w_C = -w_{12 }= h_2-h_1= C_{P0}(T_2-T_1) = 1.004*(638.1-300)= 339.5 kJ/kg\\\\ Also, s_4 = s_3\\ \\T_4 = T_3(\frac{P_4}{P_3})^{\frac{k-1}{k}}= 1600*[1/14]*0.286 = 752.2 K\\\\ w_T = w_{34} = h_3-h_4 = C_{P0}(T_3-T_4) = 1.004*(1600-752.2) = 851.2 kJ/kg\\\\$

$\\w_{NET} = 851.2-339.5=511.7 kJ/kg\\\\ \dot{m}= \frac{P}{w_{NET}} = \frac{100000}{511.7}=195.4 kg/s\\\\ W_T = \dot{m}*w_T=195.4*851.2=166.32 MW\\\\ \frac{-w_C}{w_T} = \frac{339.5}{851.2} = 0.399\\\\ b) q_H = C_{P0}(T_3- T_2) = 1.004*(1600 – 638.1) = 965.7 kJ/kg\\\\ \eta_{TH} = w_{NET}/q_H = 511.7/965.7 = 0.530$

Q.2) Consider asimple Brayton cycle having pressure ratio as 8 and maximum cycle temperature as 1400 K. inlet compressor temperature is 300K. compressor efficiency is 80%. Assume air as ideal gas neglecting changes in kinetic and potential energies. Evaluate power required by the compressor in kW/kg.

T1=300 K, T3 = 1400 K, η (isentropic) = 0.8, cp = 1.004 kJ/kg.k, ϒ = 1.4, P2 /P1 = 8

$\\T_2=T_1*(\frac{P_2}{P_1})^{\frac{\gamma-1}{\gamma}} = 300*(8)^{\frac{0.4}{1.4}} = 543.41 K\\\\ -w_C = -w_{12}= h_2-h_1= C_P_0(T_2- T_1) = 1.004*(543.41-300)=244.39 kJ/kg$

$\\\eta_{isen}=\frac{w_i}{w_{ac}}\\ \\\0.8=\frac{244.39}{w_{ac}}\\ w_{ac}=304.3 kJ/kg$

Q.3) Calculate ratio of the thermal efficiency of the simple cycle to that of the regenerative cycle for a simple Brayton cycle and a gas turbine cycle with perfect regeneration. Pressure ratio is constant at 6 for both cycles.  In regenerative cycle, the ratio of minimum cycle temperature to maximum cycle temperatures is 0.3. Assume ϒ = 1.4

For simple Brayton cycle

$\eta=1-\frac{1}{r_p^\frac{\gamma-1}{\gamma}}$

$\eta=1-\frac{1}{6^\frac{1.4-1}{1.4}}=0.40=40\%$

For ideal regeneration

$\eta_{IR}=1-\frac{T_{min}}{T_{max} }r_p^\frac{\gamma-1}{\gamma}$

$\eta_{IR}=1-0.3*6^\frac{1.4-1}{1.4}=49.94\%$

Ratio of efficiency of cycles

$R=\frac{0.40}{0.4994}=0.80$