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4.1 · Trace energy through a combined gas–vapour cycle

Learn to trace energy through a combined gas–vapour cycle through clear examples and targeted practice.

University of Alberta MEC E 340: Applied Thermodynamics

Combined Power and Cogeneration

Following energy through two power loops linked by an HRSG

A combined gas–vapour plant uses two working fluids in separate power loops. Hot gas transfers energy to water or steam in a heat-recovery steam generator (HRSG); the fluids do not mix. The vapour loop can then produce additional work. We will trace these energy flows with steady-operation balances and check them at component, loop, and plant boundaries. Unless a problem says otherwise, take heat into a control volume and work out of it as positive, and neglect kinetic- and potential-energy changes. Use property data only when supplied by the problem or its named source.

What you will learn

  • Identify the gas and vapour loops, their working fluids, and the heat-recovery link between them.
  • Apply component, cycle, and whole-plant energy balances using consistent signs.
  • Distinguish internal heat transfer from external energy input.
  • Calculate combined net power and plant thermal efficiency from supplied data.

1. Define the loops, states, and boundary

Begin by identifying the two working fluids and the equipment in each loop. A gas loop may include a compressor, heater, and gas turbine. A vapour loop may include a pump, HRSG, steam turbine, and condenser. The HRSG links the loops by transferring energy while keeping the streams separate.
Give each fluid its own state labels. For example, g1g_1 and g2g_2 can mark the gas entering and leaving the HRSG, while v1v_1 and v2v_2 mark the vapour entering and leaving it. Labels organize the bookkeeping; they do not specify temperatures, pressures, phases, or enthalpies. Obtain those properties only from the problem or its named source.
Choose the boundary that matches the question. An HRSG balance concerns energy crossing that component boundary. A vapour-loop balance includes its heat input and rejection and its turbine and pump work. A whole-plant boundary encloses both loops: energy transferred inside the plant is not an external input. One loop diagram cannot represent the full combined plant faithfully, so describe the two loops and their connection.
  • Keep gas and vapour state labels distinct.
  • State whether a balance applies to a component, a loop, or the whole plant.
  • The HRSG transfer is internal when the boundary encloses both loops.

2. Balance the HRSG and the separate loops

For a steady-flow component with one inlet and one outlet, mass flow in equals mass flow out when there is no accumulation. With negligible kinetic- and potential-energy changes, the energy rate depends on heat transfer, work, and the enthalpy carried by the stream. Specific enthalpy hh is energy per unit mass; multiplying its change by mass flow rate gives an energy rate.
For an HRSG with negligible heat loss to the surroundings and no shaft work, the gas-side energy decrease equals the vapour-side energy increase. The two mass flow rates need not be equal, so multiply each stream's enthalpy change by that stream's own mass flow rate. If heat escapes from the HRSG, the vapour receives less energy than the gas loses.
A complete cycle returns its working fluid to its initial state, so its net energy change over the cycle is zero. A turbine delivers work; a pump or compressor consumes work. Calculate a loop's net work by subtracting work input from work output. If the loop receives heat and rejects heat, both belong in its cycle energy balance.
With heat into a control volume and work out positive, the steady-flow energy balance for one inlet and one outlet, when kinetic- and potential-energy changes are negligible, is given below.
\dot Q - \dot W = \dot m(h_{out} - h_{in})
  • Use energy rates for steady-flow component balances.
  • Declare whether heat loss and shaft work are neglected.
  • Subtract pump or compressor work input from turbine work output.

3. Combine the loop balances at the plant boundary

First calculate each loop's net work output. Add the gas-loop and vapour-loop net work to obtain total plant net work. The HRSG transfer leaves the gas loop and enters the vapour loop; when the two loop balances are added, this internal transfer appears with opposite signs and cancels.
For plant thermal efficiency, divide total net work output by the external energy supplied on the stated basis. If fuel-energy rate is given, use that as the denominator. HRSG heat supplied to the vapour loop is not another external plant input: it came from the gas stream.
Check the signs and the balance. Under the stated assumptions, the vapour-side HRSG energy gain cannot exceed the gas-side energy decrease unless another energy source is specified. Keep all rates in consistent units, such as megawatts. If energy also leaves as rejected heat or in other energy-carrying streams, net plant work must be less than external energy supplied.
Plant thermal efficiency is defined here as total net plant work divided by the external energy input on the stated basis.
The plant thermal efficiency is given by the following equation: ηplant=W˙net,totalQ˙external,in\eta_{\mathrm{plant}} = \frac{\dot W_{\mathrm{net,total}}}{\dot Q_{\mathrm{external,in}}}.
  • Add loop net work outputs, not just turbine outputs.
  • Count external energy input once at the whole-plant boundary.
  • The internal HRSG transfer cancels when the loop balances are combined.

4. A practical tracing sequence

Describe both loops and mark the HRSG connection. Record the working fluids, state labels, chosen boundary, sign convention, and idealizations. Write the relevant balances before inserting numbers; this makes clear which transfers cross the boundary.
Match every enthalpy to the correct fluid and state. Multiply a specific enthalpy change by its corresponding mass flow rate before comparing it with an energy rate. Do not guess missing property data or assume a particular HRSG recovery unless the problem supplies it.
Compare component, loop, and whole-plant balances. The HRSG term has opposite signs in the two loop balances and disappears from the whole-plant balance. That cancellation is the central bookkeeping idea in tracing energy through a combined gas–vapour cycle.
W˙net,total=W˙net,gas+W˙net,vapour\dot W_{net,total} = \dot W_{net,gas} + \dot W_{net,vapour}
  • Write balances for clearly defined boundaries.
  • Keep each property tied to its fluid and state.
  • Check the combined balance after checking the individual loops.

Worked example

1. Energy transferred in an HRSG

An HRSG operates steadily with negligible heat loss to the surroundings and no shaft work. The gas stream gives up energy at a rate of 26 MW as it cools through the HRSG. Find the energy rate received by the vapour stream and the fraction of the gas-side energy decrease transferred to it.
HRSG energy transfer
HRSG energy transferHRSGHot gas and feedwater/vapourCooled gas and heated vapourControl-volume schematic

Schematic control-volume representation, not to scale. The gas and vapour streams remain separate; operation is steady, heat loss is negligible, and there is no shaft work.

  1. Choose the boundary
    Take both HRSG streams as the control volume. The gas and vapour remain separate. With no external heat loss or shaft work, the energy lost by the gas is transferred to the vapour.
  2. Apply the HRSG balance
    The gas-side energy decrease is 26 MW. Under the stated assumptions, the vapour receives the same energy rate. \dot E_{vapour,in}=26\ MW\mathrm{MW}
  3. Find the transfer fraction
    Divide the energy rate received by the vapour by the gas-side energy decrease.
    26 MW26 MW=1.00\frac{26\ \mathrm{MW}}{26\ \mathrm{MW}}=1.00
Answer: The vapour receives 26 MW. The transferred fraction is 1.00, or 100%, under the stated no-heat-loss assumption.
Check: The result follows from the idealized HRSG balance. If heat loss to the surroundings were specified, the vapour would receive less than the gas-side energy decrease.

Worked example

2. Balance a vapour power cycle

For a steady vapour power cycle, supplied data are: HRSG heat input, 20 MW; turbine work output, 8.0 MW; and pump work input, 0.40 MW. Neglect other heat transfers during the cycle. Find the condenser heat-rejection rate and the cycle net work output.
Vapour power loop
Vapour power loopSpecific volume, vPressure, PPumpHRSGTurbineCondenser11223344Schematic · not to scale

Schematic P–v cycle, not to scale. State positions are illustrative only; no property values or exact coordinates are implied.

  1. Set the cycle boundary
    Take the complete vapour loop as the system. Over a cycle, the working fluid returns to its starting state.
  2. Write the cycle balance
    There is no net energy accumulation over a cycle. Heat entering from the HRSG plus pump work input equals turbine work output plus condenser heat rejection.
    20+0.40=8.0+Q˙cond,out20 + 0.40 = 8.0 + \dot Q_{cond,out}
  3. Solve for rejection and net work
    Rearrange for condenser rejection, then subtract pump work input from turbine work output for net work.
    Q˙cond,out=12.4 MW,W˙net=7.60 MW\dot Q_{cond,out} = 12.4\ \mathrm{MW}, \quad \dot W_{net} = 7.60\ \mathrm{MW}
Answer: The condenser rejects heat at 12.4 MW, and the vapour cycle produces 7.60 MW of net work.
Check: The balance closes: 20 MW plus 0.40 MW equals 8.0 MW plus 12.4 MW. Net work is less than turbine output because the pump consumes power.

Worked example

3. Find combined output and plant efficiency

A combined plant has a gas-cycle net work output of 18 MW. Its vapour cycle receives 20 MW from the HRSG and produces 8.0 MW of turbine work while consuming 0.40 MW of pump work. The gas turbine receives 50 MW of external fuel energy. Assume steady operation and no other work-producing equipment. Find total net work and plant thermal efficiency.
  1. Find vapour-loop net work
    Subtract pump work input from turbine work output.
    W˙net,vapour=8.0−0.40=7.60 MW\dot W_{net,vapour} = 8.0 - 0.40 = 7.60\ \mathrm{MW}
  2. Add the loop outputs
    Add gas-loop net work and vapour-loop net work. HRSG transfer is internal.
    W˙net,total=18+7.60=25.60 MW\dot W_{net,total} = 18 + 7.60 = 25.60\ \mathrm{MW}
  3. Calculate plant efficiency
    Use the external fuel-energy rate for the denominator.
    ηplant=25.60 MW50 MW=0.512\eta_{plant} = \frac{25.60\ \mathrm{MW}}{50\ \mathrm{MW}} = 0.512
Answer: The combined plant produces 25.60 MW of net work. Its thermal efficiency is 51.2%.
Check: Total net work is below the 50 MW external energy input. The internal HRSG transfer is excluded.

Common mistakes and how to avoid them

Counting HRSG heat as both an external plant input and an input to the vapour cycle.
Correction: Include it in the vapour-loop balance, but treat it as an internal transfer in the whole-plant balance.
Adding turbine work outputs without subtracting pump or compressor inputs.
Correction: Find each loop's net work by subtracting work consumed from work produced, then add the loop net outputs.
Comparing a stream's specific enthalpy change directly with an energy rate.
Correction: Multiply the specific enthalpy change by its mass flow rate to obtain an energy rate.

Lesson summary

  • A combined gas–vapour plant has separate working-fluid loops linked by HRSG energy transfer.
  • Apply steady-flow component and cycle energy balances with a clear sign convention.
  • Treat HRSG transfer as internal to the whole plant and add net work from both loops.
  • Calculate efficiency using the stated external energy input and check the overall balance.

Check your understanding

Question 1

A vapour loop receives 12 MW from an HRSG, produces 4 MW of turbine work, and uses 0.5 MW of pump work. If there are no other heat transfers, what heat rate must it reject?
  1. 7.5 MW
  2. 8.5 MW
  3. 11.5 MW
  4. 16.5 MW
Show answer and explanation
8.5 MW
The cycle balance is 12 MW plus 0.5 MW pump input equals 4 MW turbine output plus heat rejected. Therefore, rejection is 8.5 MW.

Question 2

When calculating whole-plant efficiency, how should heat transferred from gas exhaust to the vapour loop be treated?
  1. Count it as additional external input on top of fuel energy.
  2. Count it as plant work output.
  3. Treat it as an internal transfer, not an additional external input.
  4. Subtract it from the gas turbine's net work.
Show answer and explanation
Treat it as an internal transfer, not an additional external input.
The HRSG transfers energy between plant loops. Include it in the loop balances, but not as an additional external input to the whole plant.

Key terms

Heat-recovery steam generator (HRSG)
A component that transfers energy from hot gas to water or steam without mixing the fluids.
Net work output
Work delivered by a cycle after subtracting the work required by its pump or compressor.
Thermal efficiency
For a power plant, net work output divided by external energy supplied on the stated basis.

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