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4.5 · Calculate useful-energy utilization in cogeneration

Learn to calculate useful-energy utilization in cogeneration through clear examples and targeted practice.

University of Alberta MEC E 340: Applied Thermodynamics

Combined Power and Cogeneration

A control-volume method for counting useful power and useful heat

Cogeneration produces electrical or mechanical power and useful heat from the same energy supply. The key calculation is not a new property-table problem: it is a careful accounting problem. Draw a boundary around the plant, identify the fuel-energy input, and count only the power and heat that serve a useful purpose. In this lesson, the working fluid within the plant may include combustion products, steam, or another cycle fluid, but no property values are needed unless a problem supplies them. We assume steady operation and use rates in SI units. The study-guide label 4.5 identifies this topic; it is not an official curriculum expectation code.

What you will learn

  • Define a cogeneration plant boundary and identify its energy inputs and useful outputs.
  • Calculate useful-energy utilization from supplied fuel, power, and heat rates.
  • Distinguish useful process heat from heat rejected to the surroundings.
  • Check results against the steady-flow energy balance without double counting energy.

1. Define the plant boundary and the performance measure

Start with a control volume around the cogeneration plant or the plant components named in the problem. Mark the fuel-energy rate entering, the net power leaving, and any heat delivered to a process or building. A useful boundary can include a power cycle and a heat-recovery section together. If the question gives rates at the plant boundary, use them directly; do not invent individual states or equipment efficiencies.
Useful-energy utilization, also called the utilization factor in this calculation, is the sum of useful power and useful heat divided by the energy supplied. Power and heat rate are both measured in kilowatts, so the ratio is dimensionless. Fuel energy rate must use the same stated heating-value basis throughout a comparison, such as lower heating value (LHV) or higher heating value (HHV). Do not mix the two.
The word useful matters. Heat delivered at the temperature and time required by a process can count. Heat discarded to cooling water or the environment does not count as useful output merely because it crosses the plant boundary. Net power means power delivered after any internal power consumption specified in the problem.
EUF=W˙net+Q˙usefulE˙fuel\mathrm{EUF}=\frac{\dot W_{\mathrm{net}}+\dot Q_{\mathrm{useful}}}{\dot E_{\mathrm{fuel}}}
  • Count only outputs that perform a stated useful service.
  • Use consistent energy-rate units and fuel heating-value basis.
  • A boundary around the whole plant avoids counting internal transfers as external outputs.

2. Apply the steady energy balance

For a steady plant, energy does not build up inside the chosen boundary. In a simplified plant accounting, the fuel-energy rate is apportioned among net power, useful heat, and energy that leaves in other forms, including rejected heat and exhaust losses. The first law supports this accounting, but it does not say that every energy output is useful.
When the plant boundary has one fuel-energy input and the listed outputs account for all energy leaving, the balance is fuel input equals power output plus all heat and other energy losses. If a problem gives only fuel input, useful power, and useful heat, those three values are enough to calculate the utilization factor, but they may not be enough to determine every loss separately.
Do not add the useful heat to the fuel input, or subtract it from the denominator. It is an output that shares the same fuel input with power production. Also avoid adding gross turbine power when the given plant output is net power; internal loads have already reduced the net value.
E˙fuel=W˙net+Q˙useful+E˙other losses\dot E_{\mathrm{fuel}}=\dot W_{\mathrm{net}}+\dot Q_{\mathrm{useful}}+\dot E_{\mathrm{other\ losses}}
  • At steady state, there is no energy accumulation within the plant boundary.
  • Rejected heat belongs in a complete energy balance but not in useful output.
  • Use the power value specified at the chosen boundary: gross or net, not both.

3. Calculate, compare, and check

List the supplied rates before substituting. Convert megawatts to kilowatts if other rates are in kilowatts, and keep the numerator and denominator in the same units. The utilization factor is not the same as thermal efficiency: thermal efficiency counts power output relative to energy input, whereas useful-energy utilization also credits useful heat.
The factor should be between zero and one for the stated simple accounting, provided the fuel input is the only energy source and the categories do not overlap. If a result exceeds one, check for mixed units, double counting, an omitted energy input, or inconsistent boundaries. A high utilization factor indicates that a large share of the input is put to useful service; it does not by itself state how much power is produced or whether the heat is at a suitable temperature.
A fair comparison between cogeneration arrangements requires the same basis and clear output definitions. If a problem supplies property data for a cycle, those data may help determine power or heat rates, but they are not part of the utilization calculation once the rates are known. This lesson uses only the simple plant-level energy accounting needed for topic 4.5.
0≤EUF≤10\leq \mathrm{EUF}\leq 1
  • Check units, boundary, input basis, and output categories before accepting the ratio.
  • Useful-energy utilization credits both useful heat and net power.
  • The factor alone does not measure output quality or establish that two plants have equivalent services.

Worked example

Plant with power and process heat

A steady cogeneration plant receives fuel energy at 1.00 MW on an LHV basis. It delivers 0.32 MW net electrical power and 0.48 MW of useful process heat. Calculate its useful-energy utilization.
  1. Set the boundary and known rates
    Use the whole plant as the control volume. The supplied fuel input and outputs are all rates in megawatts, and the power is already net.
    E˙fuel=1.00 MW,W˙net=0.32 MW,Q˙useful=0.48 MW\dot E_{\mathrm{fuel}}=1.00\ \mathrm{MW},\quad \dot W_{\mathrm{net}}=0.32\ \mathrm{MW},\quad \dot Q_{\mathrm{useful}}=0.48\ \mathrm{MW}
  2. Count useful outputs
    Both the delivered power and the stated process heat provide useful services, so include each once in the numerator.
    EUF=0.32+0.481.00=0.80\mathrm{EUF}=\frac{0.32+0.48}{1.00}=0.80
Answer: The useful-energy utilization is 0.80, or 80%.
Check: The useful outputs total 0.80 MW, below the 1.00 MW fuel input. The remaining 0.20 MW is available for other energy losses under the simplified balance.

Worked example

Convert rates before calculating

A plant receives fuel energy at 2.4 MW. It exports 650 kW of net power and supplies 900 kW of useful heat. Find the useful-energy utilization.
  1. Put every rate in the same units
    Convert fuel input from megawatts to kilowatts so it matches the output rates. The fuel basis is taken as consistent with the supplied plant data.
    2.4 MW=2400 kW2.4\ \mathrm{MW}=2400\ \mathrm{kW}
  2. Calculate the fraction
    Add net power and useful heat, then divide their total by the fuel-energy input.
    EUF=650+9002400=0.646\mathrm{EUF}=\frac{650+900}{2400}=0.646
Answer: The useful-energy utilization is approximately 0.646, or 64.6%.
Check: The useful output is 1550 kW, less than the 2400 kW input. The difference, 850 kW, represents other energy leaving or losses in the simplified plant accounting.

Worked example

Exclude heat that is rejected

A cogeneration unit receives 5.0 MW of fuel energy. It delivers 1.2 MW net power and 2.1 MW of useful heating. It also rejects 0.9 MW to cooling water that is not used by any process. Calculate useful-energy utilization and state the unaccounted energy rate.
  1. Separate useful heat from rejected heat
    Only the 2.1 MW supplied for useful heating is credited in utilization. The cooling-water rejection is an energy output, but the problem states that it serves no useful purpose.
    Q˙useful=2.1 MW\dot Q_{\mathrm{useful}}=2.1\ \mathrm{MW}
  2. Calculate utilization and remaining energy
    Use the stated net power and useful heat in the utilization numerator. For the energy balance, subtract net power, useful heat, and rejected heat from the fuel input.
    EUF=1.2+2.15.0=0.66,E˙other=5.0−1.2−2.1−0.9=0.8 MW\mathrm{EUF}=\frac{1.2+2.1}{5.0}=0.66,\quad \dot E_{\mathrm{other}}=5.0-1.2-2.1-0.9=0.8\ \mathrm{MW}
Answer: The useful-energy utilization is 0.66, or 66%. The remaining energy rate not listed among the three specified outputs is 0.8 MW.
Check: The balance closes: 1.2 MW of power, 2.1 MW of useful heat, 0.9 MW rejected, and 0.8 MW other energy total 5.0 MW. The rejected heat is not counted as useful.

Common mistakes and how to avoid them

Counting all heat leaving the plant as useful heat.
Correction: Credit heat only when the problem identifies it as serving a process, building, or other useful demand.
Using gross power as well as net power in the numerator.
Correction: Use the power rate defined at the selected boundary. If net power is given, do not add internal power consumption back in.
Mixing kilowatts with megawatts or changing between LHV and HHV.
Correction: Convert all rates to one unit and keep the fuel heating-value basis consistent.
Adding the useful heat to the denominator.
Correction: The denominator is the energy input. Useful heat is an output and belongs with net power in the numerator.

Lesson summary

  • Choose a plant boundary and identify fuel-energy input, net power, and useful heat.
  • Calculate useful-energy utilization as useful power plus useful heat divided by fuel-energy input.
  • Exclude rejected heat from useful output while retaining it in a complete energy balance.
  • Use consistent units and heating-value basis, and check that the outputs do not exceed the input without another stated energy source.

Check your understanding

Question 1

A plant receives 800 kW of fuel energy, exports 240 kW net power, and supplies 360 kW useful heat. What is its useful-energy utilization?
  1. 0.30
  2. 0.45
  3. 0.75
  4. 1.33
Show answer and explanation
0.75
The useful outputs total 600 kW. Dividing by the 800 kW fuel input gives 0.75.

Question 2

A plant rejects 150 kW of heat to the surroundings, with no useful service from that heat. How should it be treated?
  1. Include it as useful heat in the numerator.
  2. Add it to the fuel-energy input.
  3. Exclude it from useful output, but include it in a complete energy balance.
  4. Subtract it from the net power.
Show answer and explanation
Exclude it from useful output, but include it in a complete energy balance.
Rejected heat is energy leaving the control volume, but it is not a useful output under the stated condition.

Key terms

Cogeneration
Producing useful power and useful heat from a shared energy supply.
Useful-energy utilization
The fraction of supplied fuel energy represented by the sum of useful power and useful heat.
Net power
Power delivered after internal power consumption included within the stated plant boundary.
Control volume
A chosen region in space across whose boundary mass and energy may flow.
Fuel-energy rate
The rate of energy supplied by fuel, evaluated using the heating-value basis stated for the data.

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Published by DoAssignment. This AI-assisted lesson follows University of Alberta MEC E 340: Applied Thermodynamics, study topic 4.5. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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