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2.5 · Compare regeneration, intercooling, and reheat in gas turbines

Learn to compare regeneration, intercooling, and reheat in gas turbines through clear examples and targeted practice.

University of Alberta MEC E 340: Applied Thermodynamics

Gas Power Cycles

How heat recovery and staged compression or expansion affect work and heat input

Begin with an idealized closed gas-turbine cycle using air as the working fluid. The basic states are 1 at the compressor inlet, 2 at its outlet, 3 at the turbine inlet, and 4 at the turbine outlet. The processes are compression from 1 to 2, external heating from 2 to 3, turbine expansion from 3 to 4, and heat rejection from 4 to 1. A real gas turbine is an open-flow plant; this closed-cycle model helps compare component changes. Assume steady operation and negligible changes in kinetic and potential energy unless stated otherwise. Use ideal-gas, constant-specific-heat relations only when the problem gives or permits those assumptions. Regeneration recovers exhaust heat, intercooling removes heat during compression, and reheat adds heat during expansion. Compare their effects on both work and external heat input.

What you will learn

  • Describe where regeneration, intercooling, and reheat occur in a gas-turbine cycle.
  • Use simple energy balances to estimate effects on work or external heat transfer.
  • Compare the modifications without assuming that more turbine work means higher cycle efficiency.

1. Set up the comparison: states, signs, and balances

A useful comparison needs a clear basis. Identify the working fluid, state numbers, pressure levels, and assumptions. For a steady-flow component with one inlet and one outlet, mass flow in equals mass flow out. With negligible kinetic and potential energy changes, the energy balance links heat transfer, work transfer, and enthalpy change. Here heat into the working fluid and work out of it are positive; a compressor therefore has work input, while a turbine produces work output.
For an ideal gas with constant specific heat, enthalpy change per unit mass can be estimated from specific heat times temperature change. If a problem supplies enthalpies or other property data, use those data instead. Over a complete cycle, net work equals net heat transfer. Thermal efficiency compares net work output with total external heat input. Heat transferred internally in a regenerator is not additional external heat.
q-w=h_{out}-h_{in}
  • Compare net cycle work and external heat input, not just turbine work.
  • State what is held constant before comparing cases.

2. Regeneration: reuse turbine-exhaust heat

A regenerator transfers heat between two separate air streams: compressed air leaving the compressor and hot gas leaving the turbine. The streams do not mix. Call the compressed-air outlet from the regenerator state 2R. Turbine exhaust at state 4 enters the hot side and leaves cooler; compressed air at state 2 enters the cold side and leaves warmer at state 2R.
If the turbine exhaust is hotter than the compressed air entering the regenerator, heat flows to the compressed air. It then needs less external heating to reach the same turbine-inlet temperature. If compressor and turbine states are held fixed, regeneration does not directly change their work; it reduces external heat input. For equal mass flow rates and negligible heat loss to the surroundings, heat lost by the hot stream equals heat gained by the cold stream. Do not assume equal outlet temperatures unless an appropriate ideal limit is stated.
qcold=qhotq_{cold}=q_{hot}
  • Main effect: lower external heat input for the same turbine-inlet state.
  • Check the inlet temperatures before claiming that heat can flow in the desired direction.

3. Intercooling and reheat: change compression and expansion

Intercooling removes heat between compressor stages. In a two-stage arrangement, air is compressed, cooled, then compressed again. Cooling lowers the temperature entering the second compressor. For the same overall pressure ratio, idealized stages, and effective intercooling, compressor work is generally lower than for one compression stage. The intercooler rejects heat, and the lower final compression temperature can also affect the external heat needed later.
Reheat adds heat between turbine stages. In a two-stage arrangement, air expands, is reheated, then expands again. Reheating raises the temperature entering the second turbine stage, so that stage can produce more work than a comparable expansion without reheat. Reheat also requires extra external heat. More turbine work alone does not prove that cycle efficiency has increased.
The overall comparison depends on the states and what is held constant. Intercooling tends to reduce compressor work; reheat tends to increase turbine work; regeneration tends to reduce external heat input. To calculate efficiency, find net work and total external heat input on the same basis.
\eta_{th}=w_{net}/q_{in}
  • Intercooling: heat leaves the working fluid between compressor stages.
  • Reheat: heat enters the working fluid between turbine stages.
  • Regeneration: exhaust heat is transferred to compressed air.

4. A practical comparison method

First state the comparison conditions, such as overall pressure ratio and turbine-inlet temperature. Identify each component and its states, then show heat and work directions. A process sketch can clarify order, but it is not a source of exact property values.
Apply the steady-flow energy balance to the affected components. Use supplied property data, or the stated ideal-gas model when allowed. Calculate compressor or turbine enthalpy changes for work comparisons. For heat comparisons, distinguish external heating or cooling from internal recovery. Carry consistent units, such as kJ/kg.
Check physical direction and signs: an intercooler rejects heat, reheat adds heat, and a regenerator preheats air only when its hot-side inlet is hotter. A component-level work change is not automatically a cycle-efficiency change; efficiency requires both net work and total external heat input.
  • Use consistent assumptions and comparison conditions.
  • Check energy-transfer directions and distinguish component work from cycle performance.

Worked example

Estimate heat saved by regeneration

Compressed air enters a regenerator at 500 K, and turbine exhaust enters it at 700 K. The regenerator effectiveness is 0.75. Assume equal mass flow rates, negligible heat loss, and constant specific heat. Find the compressed-air outlet temperature and the reduction in external heat input per kilogram if the turbine-inlet temperature is 1100 K. Use cp=1.005 kJ/(kg⋅K)c_p=1.005\ \mathrm{kJ/(kg\cdot K)}.
Basic gas-turbine cycle
Basic gas-turbine cycleSpecific volume, vPressure, PCompressionExternal heatingExpansionHeat rejection1Compressor inlet2Compressor outlet3Turbine inlet4Turbine outletSchematic · not to scale

Air-standard closed-cycle model; schematic P–v plot, not to scale. Regenerator streams are not shown.

  1. Find the compressed-air outlet temperature
    Effectiveness is the actual cold-stream temperature rise divided by its maximum possible rise based on the inlet temperatures. Use the supplied effectiveness and inlet temperatures.
    T2R=500+0.75(700−500)=650 KT_{2R}=500+0.75(700-500)=650\ \mathrm{K}
  2. Compare external heat inputs
    With regeneration, external heating raises the air from 650 K to 1100 K. Without regeneration, it raises the air from 500 K to 1100 K. Their difference is the heat saved per kilogram under the stated constant-specific-heat assumption. \Delta q_{in}=1.005[(1100-500)-(1100-650)]=150.75\ kJ/kg\mathrm{kJ/kg}
Answer: The compressed-air outlet temperature is 650 K. Regeneration reduces external heat input by about 151 kJ/kg under the stated assumptions.
Check: The exhaust inlet is hotter than the compressed-air inlet, so heat transfer is in the stated direction. The positive heat saving corresponds to a 150 K smaller temperature rise during external heating.

Worked example

Estimate compressor-work reduction with intercooling

Air enters a compressor at 300 K. Compare one ideal compression stage with two ideal stages and perfect intercooling back to 300 K between stages. The total pressure ratio is 16, split equally between stages. Assume an ideal gas with constant γ=1.4\gamma=1.4 and cp=1.005 kJ/(kg⋅K)c_p=1.005\ \mathrm{kJ/(kg\cdot K)}. Estimate the compressor-work reduction per kilogram.
  1. Estimate the one-stage outlet temperature
    For the stated ideal-gas, isentropic compression model, the temperature ratio is the pressure ratio raised to (γ−1)/γ(\gamma-1)/\gamma. Apply the full pressure ratio.
    Tout=300(16)2/7≈662.4 KT_{out}=300(16)^{2/7}\approx662.4\ \mathrm{K}
  2. Estimate work with two stages
    Each stage has pressure ratio 4. Perfect intercooling returns the air to 300 K before the second stage, so both stages have the same temperature rise. Sum their enthalpy increases.
    w2=2(1.005)[300(42/7−1)]≈293.1 kJ/kgw_2=2(1.005)[300(4^{2/7}-1)]\approx293.1\ \mathrm{kJ/kg}
  3. Compare the work estimates
    The one-stage work is the enthalpy increase from 300 K to the one-stage outlet temperature. Subtract the two-stage work to find the reduction.
    w1=1.005(662.4−300)≈364.2 kJ/kg;w1−w2≈71.1 kJ/kgw_1=1.005(662.4-300)\approx364.2\ \mathrm{kJ/kg};\quad w_1-w_2\approx71.1\ \mathrm{kJ/kg}
Answer: Two-stage compression with perfect intercooling reduces compressor work by about 71 kJ/kg for these assumptions.
Check: The reduction is positive, as expected when cooling lowers the temperature entering the second compression stage. This compares compressor work only, not complete cycle efficiency.

Worked example

Estimate the effect of reheat on turbine work

Air enters a turbine at 1200 K and expands through two ideal stages. The overall pressure ratio is 16, split equally between stages. In the reheat case, the air is reheated to 1200 K between stages. Compare turbine work with a single ideal expansion through the same overall pressure ratio. Assume an ideal gas with constant γ=1.4\gamma=1.4 and cp=1.005 kJ/(kg⋅K)c_p=1.005\ \mathrm{kJ/(kg\cdot K)}.
  1. Calculate the single-expansion outlet temperature and work
    Use the ideal-gas isentropic temperature relation with the full pressure ratio. Turbine work per kilogram is the enthalpy decrease under these assumptions.
    Tout,1=1200(16−2/7)≈543.6 K;wt,1=1.005(1200−543.6)≈659.9 kJ/kgT_{out,1}=1200(16^{-2/7})\approx543.6\ \mathrm{K};\quad w_{t,1}=1.005(1200-543.6)\approx659.9\ \mathrm{kJ/kg}
  2. Calculate work from both reheated stages
    Each stage has pressure ratio 4. The first stage begins at 1200 K, and reheating returns the second-stage inlet to 1200 K. Both ideal stages therefore have the same temperature drop.
    Tout,stage=1200(4−2/7)≈807.6 K;wt,2=2(1.005)(1200−807.6)≈789.0 kJ/kgT_{out,stage}=1200(4^{-2/7})\approx807.6\ \mathrm{K};\quad w_{t,2}=2(1.005)(1200-807.6)\approx789.0\ \mathrm{kJ/kg}
  3. Compare turbine work
    Subtract single-expansion work from work with reheat. This measures the turbine-work increase only; reheat also requires external heat, so this calculation does not determine cycle efficiency.
    wt,2−wt,1≈129.1 kJ/kgw_{t,2}-w_{t,1}\approx129.1\ \mathrm{kJ/kg}
Answer: Reheat increases estimated turbine work by about 129 kJ/kg under these assumptions.
Check: The reheated second stage begins at a higher temperature than it would without reheat, allowing additional expansion work. The added external heat must also be counted in a cycle-performance comparison.

Common mistakes and how to avoid them

Assuming that more turbine work automatically means higher thermal efficiency.
Correction: Reheat adds external heat as well as turbine work. Compare net work with total external heat input before concluding that efficiency rises.
Treating a regenerator as if it adds energy to the cycle.
Correction: It transfers heat from turbine exhaust to compressed air. It can reduce external heat input but does not create energy.
Claiming intercooling reduces compressor work without stating comparison conditions.
Correction: Specify the same overall pressure ratio and the stage and intercooling assumptions before comparing work.
Assuming turbine exhaust can preheat compressed air in every case.
Correction: Check the two regenerator inlet temperatures. The exhaust must be hotter than the compressed air for heat transfer in that direction.

Lesson summary

  • Regeneration recovers turbine-exhaust heat to reduce external heat input.
  • Intercooling removes heat between compressor stages and can reduce compressor work under stated conditions.
  • Reheat adds heat between turbine stages and can increase turbine work, but also increases external heat supplied.
  • A cycle-efficiency comparison requires net work and total external heat input on a consistent basis.

Check your understanding

Question 1

Which modification transfers heat from turbine exhaust to compressed air?
  1. Intercooling
  2. Reheat
  3. Regeneration
  4. Compression
Show answer and explanation
Regeneration
Regeneration uses hot turbine exhaust to preheat compressed air.

Question 2

Why does increased turbine work from reheat not by itself prove that cycle thermal efficiency increases?
  1. Reheat also requires external heat input.
  2. Reheat always reduces compressor work to zero.
  3. Turbine work is not part of net cycle work.
  4. Reheat prevents the working fluid from changing temperature.
Show answer and explanation
Reheat also requires external heat input.
Thermal efficiency compares net work with external heat input, and reheat increases the heat supplied.

Question 3

What does intercooling do in a two-stage compressor?
  1. Adds heat before the second compressor stage.
  2. Removes heat between compressor stages.
  3. Removes heat between turbine stages.
  4. Transfers heat from compressed air only after the turbine.
Show answer and explanation
Removes heat between compressor stages.
Intercooling removes heat from the air between compression stages.

Key terms

Regeneration
Heat recovery from turbine exhaust to preheat compressed working fluid before external heating.
Intercooling
Heat removal from the working fluid between compressor stages.
Reheat
Heat addition to the working fluid between turbine expansion stages.
Thermal efficiency
The ratio of cycle net work output to external heat input.

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