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8.4 · Distinguish higher and lower heating values

Learn to distinguish higher and lower heating values through clear examples and targeted practice.

University of Alberta MEC E 340: Applied Thermodynamics

Chemical Reactions and Combustion

MEC E 340 Applied Thermodynamics — Study topic 8.4

Consider a steady-flow combustion control volume: fuel and the required air enter, and combustion products leave. For this topic, focus on the water in those products and on the reference condition used to report the fuel’s heating value. Heating value is the energy released per amount of fuel when it undergoes complete combustion and the products are brought to the stated reference condition. The key distinction is whether product water is treated as liquid or as vapour. This is a definition and energy-accounting distinction; it does not require a cycle analysis or a detailed combustion model.

What you will learn

  • Define higher heating value (HHV) and lower heating value (LHV) using the state of the combustion-product water.
  • Explain why HHV is greater than LHV for a fuel that forms water.
  • Calculate the difference between HHV and LHV when the required water amount and latent heat are given.
  • Use a heating value consistently with its stated basis and avoid counting water-condensation energy twice.

1. Define the reference condition before comparing values

HHV and LHV report energy released per unit amount of fuel, commonly in kilojoules per kilogram or megajoules per kilogram. They describe the same fuel and combustion process, but use different reference conditions for the water in the products.
For the higher heating value, the product water is liquid in the final reference condition. The reported energy therefore includes the energy that could be recovered if water vapour in the products condensed to liquid. For the lower heating value, the product water remains vapour in the final reference condition. That condensation energy is not included.
These are heating-value conventions, not two different amounts of chemical energy in the fuel. A device that cools exhaust enough to condense water may recover some of the energy associated with condensation; a device that leaves the water as vapour does not recover that part as heat. The actual recovered heat depends on the device and its operating conditions, not just on the name of the heating value.
HHV−LHV=mw,formedhfg\mathrm{HHV}-\mathrm{LHV}=m_{w,\mathrm{formed}}h_{fg}
  • HHV: product water is liquid at the reference condition.
  • LHV: product water is vapour at the reference condition.
  • For a fuel that forms water, HHV is greater than LHV.

2. Use the water amount and energy basis consistently

In the comparison equation, mw,formedm_{w,\mathrm{formed}} is the mass of water formed per chosen amount of fuel, and hfgh_{fg} is the latent heat for the specified reference condition. The equation gives an energy difference for that same fuel basis. If the heating values are reported per kilogram of fuel, use kilograms of water formed per kilogram of fuel. If they are reported per kilomole of fuel, keep the water amount on a per-kilomole-of-fuel basis.
Use only data supplied by a problem or an identified property source. The latent heat depends on the stated reference condition; do not silently substitute a value at a different condition. If a problem directly gives HHV and LHV, compare them on their stated common basis rather than reconstructing the difference with an unstated water amount or property.
A simple idealized control-volume energy balance helps explain the accounting. With negligible changes in kinetic and potential energy and no shaft work, energy released by combustion leaves as heat and/or remains in the products. In an HHV reference, product water has been condensed; in an LHV reference it has not. The latent-heat term is the difference between these product reference states. It is not an additional energy source to add on top of HHV.
qdifference=mw,formedhfgq_{\mathrm{difference}}=m_{w,\mathrm{formed}}h_{fg}
  • Keep the fuel basis identical on both sides of a comparison.
  • Use latent heat only at the specified reference condition.
  • Do not add the condensation term twice.

3. Read tables and reported data without changing the definition

A heating-value table or equipment specification should identify the fuel, the amount basis, and whether the reported value is HHV or LHV. A value per kilogram is not directly interchangeable with one per kilomole or per unit volume unless the needed conversion data and conditions are also provided.
For a quick consistency check, ask what happens to the product water in the stated reference condition. Liquid water means HHV; water vapour means LHV. If the fuel forms no water, this particular condensation difference is absent. For water-forming fuels, a reported LHV exceeding the HHV on the same basis signals a likely mismatch, transcription error, or different reference assumptions.
When using a heating value in a later energy calculation, state the convention alongside the number. This makes clear whether the condensation contribution is included and prevents comparing fuel-energy figures that use different product-water states.
  • Check fuel identity, amount basis, and HHV/LHV label.
  • The product-water phase identifies the convention.
  • Compare values only after confirming matching reference assumptions.

Worked example

Find the difference per kilogram of fuel

A supplied combustion calculation states that 1.50 kg of water is formed per kilogram of fuel. At the stated reference condition, use hfg=2,400 kJ/kgh_{fg}=2{,}400\ \mathrm{kJ/kg}. Find HHV minus LHV per kilogram of fuel.
  1. Match the basis
    The water amount is given per kilogram of fuel, so the resulting energy difference will also be per kilogram of fuel. The supplied latent heat is already on a per-kilogram-of-water basis.
  2. Apply the definition
    Multiply the water formed by the latent heat at the stated reference condition. The units cancel to energy per kilogram of fuel.
    HHV−LHVmf=(1.50 kg water/kg fuel)(2,400 kJ/kg water)\frac{\mathrm{HHV}-\mathrm{LHV}}{m_f}=(1.50\ \mathrm{kg\ water/kg\ fuel})(2{,}400\ \mathrm{kJ/kg\ water})
Answer: HHV − LHV = 3,600 kJ/kg fuel = 3.60 MJ/kg fuel.
Check: The difference is positive, as expected when combustion forms water. The units reduce to kilojoules per kilogram of fuel.

Worked example

Convert an HHV to an LHV

For a specified fuel and reference condition, the supplied data are HHV = 46.0 MJ/kg fuel, water formed = 1.10 kg/kg fuel, and hfg=2,300 kJ/kgh_{fg}=2{,}300\ \mathrm{kJ/kg}. Calculate the corresponding LHV.
  1. Calculate the convention difference
    The supplied water amount and latent heat are on a consistent fuel basis. Their product gives the energy included in HHV but not in LHV.
    (1.10)(2,300)=2,530 kJ/kg fuel=2.53 MJ/kg fuel(1.10)(2{,}300)=2{,}530\ \mathrm{kJ/kg\ fuel}=2.53\ \mathrm{MJ/kg\ fuel}
  2. Subtract from HHV
    Because HHV includes the condensation contribution and LHV excludes it, subtract the difference from the given HHV.
    LHV=46.0−2.53 MJ/kg fuel\mathrm{LHV}=46.0-2.53\ \mathrm{MJ/kg\ fuel}
Answer: LHV = 43.47 MJ/kg fuel, or 43.5 MJ/kg fuel to three significant figures.
Check: The LHV is lower than the HHV by 2.53 MJ/kg fuel, matching the calculated condensation contribution.

Worked example

Compare conventions on a kilomole-of-fuel basis

A supplied idealized reaction basis states that burning 1 kmol of a fuel produces 2 kmol of water. At the specified reference condition, use hfg=40,000 kJ/kmol waterh_{fg}=40{,}000\ \mathrm{kJ/kmol\ water}. The supplied HHV is 800,000 kJ/kmol fuel. Find the LHV on the same basis.
  1. Keep the amount basis
    The data are given per kilomole of fuel, so retain that basis throughout. The supplied water amount is per kilomole of fuel.
  2. Find the convention difference
    Multiply the kilomoles of water formed by latent heat per kilomole of water. This yields kilojoules per kilomole of fuel.
    (2 kmol water/kmol fuel)(40,000 kJ/kmol water)=80,000 kJ/kmol fuel(2\ \mathrm{kmol\ water/kmol\ fuel})(40{,}000\ \mathrm{kJ/kmol\ water})=80{,}000\ \mathrm{kJ/kmol\ fuel}
  3. Obtain LHV
    Subtract the condensation contribution from HHV because the LHV reference leaves product water as vapour.
    LHV=800,000−80,000 kJ/kmol fuel\mathrm{LHV}=800{,}000-80{,}000\ \mathrm{kJ/kmol\ fuel}
Answer: LHV = 720,000 kJ/kmol fuel.
Check: The result is below the supplied HHV by exactly 80,000 kJ/kmol fuel, and both heating values use the same fuel basis.

Common mistakes and how to avoid them

Treating HHV and LHV as different chemical energy releases from the same fuel.
Correction: They differ by the reference condition for product water; HHV includes the water-condensation contribution and LHV does not.
Subtracting a latent heat value without checking its reference condition or amount basis.
Correction: Confirm that the water amount, latent heat, and heating values use compatible reference conditions and matching fuel bases.
Adding the HHV–LHV difference to HHV when calculating LHV.
Correction: Subtract the difference from HHV, since HHV is the larger value for a fuel that forms water.
Assuming a device always recovers the full HHV value as useful heat.
Correction: HHV is a reference heating value. Actual heat recovery depends on the device and whether product water condenses.

Lesson summary

  • HHV assumes product water is liquid at the reference condition; LHV assumes it remains vapour.
  • For water-forming combustion, HHV exceeds LHV by the latent heat associated with the water formed.
  • Use the supplied water amount and latent heat on consistent bases and at the stated reference condition.
  • State whether a heating value is HHV or LHV whenever using it in an energy calculation.

Check your understanding

Question 1

A fuel forms water during complete combustion. Which statement is correct on the same reference and fuel basis?
  1. LHV is greater because water vapour carries more energy than liquid water.
  2. HHV is greater because its reference condition includes product water condensed to liquid.
  3. HHV and LHV must be equal for a steady-flow control volume.
  4. The values cannot be compared unless the fuel is burned in a cycle.
Show answer and explanation
HHV is greater because its reference condition includes product water condensed to liquid.
HHV includes the energy associated with condensing product water; LHV leaves that water as vapour. Thus HHV is greater on a consistent basis.

Question 2

A supplied basis gives 0.80 kg water formed per kilogram of fuel and hfg=2,500 kJ/kgh_{fg}=2{,}500\ \mathrm{kJ/kg}. What is HHV minus LHV?
  1. 2,000 kJ/kg fuel
  2. 3,125 kJ/kg fuel
  3. 2,500 kJ/kg fuel
  4. 0.00032 kJ/kg fuel
Show answer and explanation
2,000 kJ/kg fuel
The difference is (0.80)(2,500)=2,000(0.80)(2{,}500)=2{,}000 kJ per kilogram of fuel. The water-mass units cancel.

Question 3

A table reports 42 MJ/kg but does not identify HHV or LHV. What is the soundest next step?
  1. Assume it is HHV because that value is always tabulated.
  2. Use it as LHV if the application is a combustion device.
  3. Check the table’s definition, reference condition, and fuel basis before using it.
  4. Add a latent heat value to convert it to either convention.
Show answer and explanation
Check the table’s definition, reference condition, and fuel basis before using it.
A numerical value alone does not identify the product-water reference state or guarantee a matching basis. Check the source definition rather than guessing.

Key terms

Higher heating value (HHV)
Heating value referenced to combustion products in which formed water is liquid.
Lower heating value (LHV)
Heating value referenced to combustion products in which formed water remains vapour.
Latent heat of vaporization
Energy per unit mass or amount associated with changing liquid water to vapour at a specified condition; condensation releases the same magnitude at that condition.
Fuel basis
The stated amount of fuel used to report a heating value, such as one kilogram or one kilomole.

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