2.4 Analyze the ideal Brayton gas-turbine cycle | University of Alberta MEC E 340: Applied Thermodynamics

University of Alberta MEC E 340: Applied Thermodynamics

Gas Power Cycles

State-by-state energy and performance analysis

We analyze a closed ideal Brayton cycle containing air. State 1 is the compressor inlet, state 2 the compressor exit, state 3 the turbine inlet, and state 4 the turbine exit. The model assumes an ideal gas with constant specific heats, isentropic compression and expansion, constant-pressure heat addition and rejection, and no pressure losses. These are analysis assumptions, not measured property data. Use property values supplied in a problem or explicitly stated assumptions; do not infer numerical properties from a schematic plot.

What you will learn

Identify the four Brayton-cycle states and ideal processes.
Use the ideal-gas isentropic relation with stated constant-specific-heat assumptions to find temperatures.
Calculate compressor work, turbine work, heat transfers, net work, and thermal efficiency.
Check energy closure, signs, units, and physical plausibility.

1. Define the cycle and review the needed balance

The cycle proceeds in this order:

1\rightarrow2

is isentropic compression,

2\rightarrow3

is constant-pressure heat addition,

3\rightarrow4

is isentropic expansion, and

4\rightarrow1

is constant-pressure heat rejection. States 1 and 4 share the low pressure; states 2 and 3 share the high pressure. The pressure ratio is the high pressure divided by the low pressure.

For a complete cycle, the working fluid returns to its initial state, so its net energy change is zero. Take heat input, heat rejection, compressor work input, and turbine work output as positive magnitudes. Net work output is turbine work output minus compressor work input. It must also equal heat input minus heat rejected.

With no pressure losses, the compressor and turbine have equal pressure-ratio magnitudes. Keep device work directions distinct: the compressor requires work input, while the turbine delivers work output.

w_{net}=w_t-w_c=q_{in}-q_{out}

State order: compressor inlet, compressor exit, turbine inlet, turbine exit.
The ideal cycle has two isentropic processes and two constant-pressure processes.
Net work output is turbine work output less compressor work input.

2. Find the state temperatures

For an ideal gas with constant specific heats, the isentropic pressure–temperature relation connects a pressure ratio to an absolute-temperature ratio. Apply it separately to the compressor and turbine. The exponent uses the heat-capacity ratio, defined as specific heat at constant pressure divided by specific heat at constant volume.

Use kelvins in this relation. It is appropriate only when the stated model assumptions support it. If a problem supplies variable-specific-heat data or requests property-table use, follow that method and use only the provided data.

Once temperatures are known, constant-specific-heat enthalpy changes follow from temperature differences. Specific work and heat are energy transfers per unit mass, commonly reported in kilojoules per kilogram.

\frac{T_2}{T_1}=\left(\frac{p_2}{p_1}\right)^{(k-1)/k},\quad \frac{T_3}{T_4}=\left(\frac{p_3}{p_4}\right)^{(k-1)/k}

Use compression to find state 2 and expansion to find state 4.
Use absolute temperatures in the isentropic relation.
Use only stated or supplied gas-property assumptions.

3. Apply energy balances and calculate performance

Treat the compressor and turbine as steady-flow devices. Neglect changes in kinetic and potential energy. The first-law balance then relates shaft work to enthalpy change: compressor work input equals the enthalpy rise, and turbine work output equals the enthalpy drop. Heat input and rejection follow from the enthalpy changes on the constant-pressure legs.

For constant specific heat, an enthalpy change is the specific heat multiplied by the corresponding temperature difference. Calculate compressor work input, turbine work output, heat input, and heat rejection with their physical directions in mind. Then find net work and thermal efficiency. Thermal efficiency is net work output divided by heat input.

A pressure–specific-volume or temperature–entropy plot can show the state order and the two pressure levels. Treat it as schematic unless the problem provides enough property information for exact coordinates. If mass flow rate is given, multiply a specific transfer by mass flow rate to obtain a transfer rate. Kilograms per second multiplied by kilojoules per kilogram gives kilowatts.

Check that net work agrees with heat input minus heat rejected, allowing for rounding. A power-producing cycle must have positive net work. When using a mass flow rate, check that power and heat-transfer rates have units of kilowatts.

The thermal efficiency can also be expressed in terms of heat rejected. \eta_{th} = w_{net}/q_{in} = 1 - q_{out}/q_{in}

Use enthalpy changes for steady-flow work and heat under the stated assumptions.
Thermal efficiency is net work output divided by heat input.
Check energy closure, signs, units, and positive net work.

4. A reliable solution sequence

Record the working fluid, known temperatures, pressure ratio, and supplied properties. State the ideal assumptions: isentropic compressor and turbine processes, constant-pressure heat-transfer processes, and no pressure losses. Label all four states and identify the high and low pressures.

Use the pressure ratio and isentropic relation to find exit temperatures. Then calculate compressor work, turbine work, heat input, and heat rejection from temperature differences. Find net work and efficiency after the individual transfers are clear. Convert specific results to rates only when mass flow rate is known.

Finish by checking units, energy balance, and signs. Negative net work means the stated conditions do not produce net power in this model. Recheck state labels, pressure ratio, assumptions, and arithmetic rather than changing a sign to force a positive result.

w_{c,in} = c_p(T_2-T_1), w_{t,out} = c_p(T_3-T_4)

State assumptions before applying the isentropic relation.
Calculate specific quantities before converting them to rates.
Check signs, SI units, energy closure, and positive net work.

Worked example

Find temperatures and efficiency

An ideal Brayton cycle uses air with supplied constant properties

c_p=1.004\ \mathrm{kJ/(kg\cdot K)}

and

k=1.4

. The compressor inlet temperature is

300\ \mathrm{K}

, the turbine inlet temperature is

1200\ \mathrm{K}

, and the pressure ratio is 6. Find the state temperatures, specific work transfers, heat transfers, and thermal efficiency.

Ideal Brayton cycle on schematic pv axes

The plot is schematic and not to scale.

Set the pressures
States 1 and 4 are at the low pressure, while states 2 and 3 are at the high pressure. With no pressure losses, both devices have the given pressure-ratio magnitude.
$\frac{p_2}{p_1}=\frac{p_3}{p_4}=6$
Find the exit temperatures
Apply the constant-specific-heat isentropic relation to compression and expansion. Temperatures are rounded to one decimal place.
$T_2=T_1(6)^{(1.4-1)/1.4}\approx500.5\ \mathrm{K},\quad T_4=T_3/(6)^{(1.4-1)/1.4}\approx719.3\ \mathrm{K}$
Calculate specific transfers
Use the supplied specific heat and the temperature differences to calculate compressor work input $w_c=c_p(T_2-T_1)$ , turbine work output $w_t=c_p(T_3-T_4)$ , heat input q_{in}=c_p(T_3-T_2), and heat rejection $q_{out}=c_p(T_4-T_1)$ .
Calculate and check efficiency
Divide net work by heat input. The heat difference agrees with net work within rounding. The efficiency is \eta_{th} = (w_t-w_c)/q_{in} \approx 40.1%.

Answer: The state temperatures are

300\ \mathrm{K}

, approximately

500.5\ \mathrm{K}

,

1200\ \mathrm{K}

, and approximately

719.3\ \mathrm{K}

. Net specific work is approximately

281.4\ \mathrm{kJ/kg}

and thermal efficiency is approximately 40.1%.

Check: The heat difference is approximately

702.3-420.9=281.4\ \mathrm{kJ/kg}

. Turbine work exceeds compressor input, so net work is positive.

Worked example

Analyze a cycle with a higher pressure ratio

An ideal Brayton cycle has

T_1=290\ \mathrm{K}

,

T_3=1100\ \mathrm{K}

, pressure ratio 8,

c_p=1.004\ \mathrm{kJ/(kg\cdot K)}

, and

k=1.4

. These air properties are supplied assumptions. Find the state temperatures, net specific work, heat input, heat rejection, and thermal efficiency.

Ideal Brayton cycle on schematic pv axes

The plot is schematic and not to scale.

Find the exit temperatures
The ideal compressor and turbine share the same pressure-ratio magnitude. Apply the isentropic relation.
$T_2=T_1(8)^{(0.4/1.4)}\approx525.3\ \mathrm{K},\quad T_4=T_3/(8)^{(0.4/1.4)}\approx607.3\ \mathrm{K}$
Calculate work and heat
Use the supplied constant specific heat and the temperature differences. w_c=c_p(T_2-T_1), w_t=c_p(T_3-T_4), q_{in}=c_p(T_3-T_2), q_{out}=c_p(T_4-T_1)
Calculate efficiency and check
Net work is positive because turbine output exceeds compressor input. \eta_{th}=(w_t-w_c)/q_{in}\approx44.8%

Answer: The state temperatures are approximately

290\ \mathrm{K}

,

525.3\ \mathrm{K}

,

1100\ \mathrm{K}

, and

607.3\ \mathrm{K}

. Net specific work is approximately

258.5\ \mathrm{kJ/kg}

; heat input is approximately

577.0\ \mathrm{kJ/kg}

, heat rejection approximately

318.5\ \mathrm{kJ/kg}

, and efficiency approximately 44.8%.

Check: The heat difference is approximately

577.0-318.5=258.5\ \mathrm{kJ/kg}

, consistent with the net work. Turbine output exceeds compressor input.

Worked example

Convert specific performance to rates

An ideal Brayton cycle uses supplied air properties

c_p=1.005\ \mathrm{kJ/(kg\cdot K)}

and

k=1.4

. Given

T_1=300\ \mathrm{K}

,

T_3=1000\ \mathrm{K}

, pressure ratio 4, and mass flow rate

5.00\ \mathrm{kg/s}

, find net power and heat-input rate.

Ideal Brayton cycle on schematic pv axes

The plot is schematic and not to scale.

Find the exit temperatures
Use the same pressure-ratio magnitude for compression and expansion.
$T_2=T_1(4)^{(0.4/1.4)}\approx445.8\ \mathrm{K},\quad T_4=T_3/(4)^{(0.4/1.4)}\approx673.0\ \mathrm{K}$
Calculate specific net work and heat input
Use constant-specific-heat enthalpy differences. w_{net}=(c_p(T_3-T_4))-(c_p(T_2-T_1)), q_{in}=c_p(T_3-T_2)
Convert to rates
Multiply each specific transfer by mass flow rate to obtain powers in kilowatts. \dot W_{net}=\dot m w_{net}, \dot Q_{in}=\dot m q_{in}

Answer: The cycle delivers approximately

911\ \mathrm{kW}

of net power and requires approximately

2785\ \mathrm{kW}

of heat input.

Check: Turbine work per kilogram exceeds compressor input, giving positive net power.

Common mistakes and how to avoid them

Treating compressor work as work output.

Correction: The compressor consumes work. Subtract compressor input from turbine output to find net work.

Using a different pressure ratio for the turbine without a stated pressure loss.

Correction: In the ideal cycle, both constant-pressure legs connect the same high and low pressures, so compressor and turbine pressure-ratio magnitudes match.

Dividing heat input by net work to calculate thermal efficiency.

Correction: Thermal efficiency is net work output divided by heat input.

Reading exact properties from a schematic cycle plot.

Correction: Use stated conditions, supplied property data, and governing relations for numerical values.

Lesson summary

The ideal Brayton cycle has two isentropic device processes and two constant-pressure heat-transfer processes.
With ideal-gas, constant-specific-heat assumptions, use the isentropic relation to find exit temperatures.
Use enthalpy differences for work and heat; calculate net work and thermal efficiency from their definitions.
Check energy closure, signs, SI units, and positive net work.

Check your understanding

Question 1

What ideal process occurs from state 3 to state 4?

Isentropic expansion through the turbine
Constant-pressure heat addition
Isentropic compression through the compressor
Constant-pressure heat rejection

Show answer and explanation

Isentropic expansion through the turbine

State 3 is the turbine inlet and state 4 its exit, so the ideal process is isentropic expansion.

Question 2

A cycle receives

800\ \mathrm{kJ/kg}

of heat and rejects

500\ \mathrm{kJ/kg}

. What are its net work output and thermal efficiency?

$300\ \mathrm{kJ/kg}$ and 37.5%
$1300\ \mathrm{kJ/kg}$ and 62.5%
$300\ \mathrm{kJ/kg}$ and 60.0%
$500\ \mathrm{kJ/kg}$ and 37.5%

Show answer and explanation

300\ \mathrm{kJ/kg}

and 37.5%

Net work is

800-500=300\ \mathrm{kJ/kg}

. Efficiency is

300/800=0.375

, or 37.5%.

Key terms

Pressure ratio: The high cycle pressure divided by the low cycle pressure.
Isentropic process: An ideal process with constant entropy; it models compression or expansion in the ideal Brayton-cycle devices.
Specific work: Work transfer per unit mass, commonly reported in kilojoules per kilogram.
Thermal efficiency: Net work output divided by heat supplied to the cycle.

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2.4 · Analyze the ideal Brayton gas-turbine cycle

University of Alberta MEC E 340: Applied Thermodynamics

State-by-state energy and performance analysis

What you will learn

1. Define the cycle and review the needed balance

2. Find the state temperatures

3. Apply energy balances and calculate performance

4. A reliable solution sequence

Find temperatures and efficiency

Analyze a cycle with a higher pressure ratio

Convert specific performance to rates

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Key terms

Continue through MEC E 340

About this lesson and its review