I am performing a number of simple heat transfer experiments with the application of a magnetic field around the fuel line to evaluate the effects. To attempt to quantify some of the heat losses in the system I have worked out how many moles of water will be produced by combustion of a butane-propane gas mixture in air. I am hoping to find a relationship between that mass of water and the energy lost from the system by producing that water.

I know the relationship between the HHV and LHV but that seems to only deal with water vapour that has been condensed into a liquid so I'm unsure if that is any use.

  • $\begingroup$ Your question is a little unclear to me. Are you trying to calculate how much total heat is given off in the combustion process or how much heat is carried away by the water. If you are asking about the first case, have you looked up the heat of combustion for butane and propane? $\endgroup$ Commented Apr 21, 2015 at 14:18
  • $\begingroup$ Apologies, I am concerned with the second case, how much heat is lost due to the water. $\endgroup$ Commented Apr 21, 2015 at 14:29
  • $\begingroup$ Separate from the question at hand. I should caution you that you should expect no changes from putting a magnet on the fuel line. There is no physical mechanism (that i'm aware of) by which such a device would have even the slightest effect. It will have do nearly nothing to the fuel as it passes (slight changes of nuclear spins), and none of those changes will be persistent. Do you have working theory as to why this will change anything? $\endgroup$
    – Dan
    Commented Apr 21, 2015 at 14:47
  • $\begingroup$ I agree, the majority of tests performed showed that no noticeable effects were observed. However, a few flow rate tests seemed to show an increase in flow upon the application of the neodymium magnets. I except the fact that this is more likely due to slight variations in the gas weight etc. but I did then find a research paper discussing the effects of magnetic fields on laminar fluid flow in pipes. $\endgroup$ Commented Apr 21, 2015 at 14:55

2 Answers 2


If you get complete combustion of any hydrocarbon fuel, your products will be primarily $CO_2$ and $H_2O$. Other products will be present in quantities too small to be relevant to the heat losses that you're looking at.

These two gases will change the heat capacity, conductivity and density of the flame and your exhaust gases. So heat heat transfer computations that you perform regarding convection and conduction should change accordingly. Note that these properties vary with temperature over the range that you're interested in (300–2,000 K, roughly), so it's best not to just pick one value.

The total energy from the fuel should be computed from the heat of combustion ($\Delta H_C$) of the gas being burnt. From balancing the chemical equations you can determine the amount of water produced for every mole of fuel. The heat stored in the water depends on it's temperature (look this up in a table or compute the integral of $c_p$ plus the heat of vaporization). The ratio $H_{H_2O}/\Delta H_c$ tells you how much energy went into the water.

These gases do thermally radiate as well and you may want to consider their emissivity, although the flame sheet itself (where combustion is still in process) is likely to be a more significant radiator.

The HHV-LHV difference is meaningful if you are trying to extract as much heat as possible from your exhaust gases. For example, after a turbine engine has extracted work from the combustion products, they'll be too cold to produce mechanical work, but could be used to heat a building. It just has to be cold enough for the water to condense so that the heat of vaporization can be released ( <100 °C).

You should be able to find most of the relevant thermal properties online (in the NIST Chem Web Book in the entries for 'fluid properties.' Any standard text of combustion (Law, Turns, Glass etc.) will usually have tabulated info (based on your project you should already have consulted at least one of those heavily).

  • $\begingroup$ Thank you very much for your input. I have taken a lot of info from a combustion engineering book recommended by my supervisor. $\endgroup$ Commented Apr 21, 2015 at 14:42
  • $\begingroup$ I will use the mass of water produced from the reaction in comparison to the heat of combustion for my mix and hopefully get a useful value for energy loss. Thank you again. $\endgroup$ Commented Apr 21, 2015 at 14:43
  • $\begingroup$ I'm not quite clear on why $H_2O$ constitutes a loss. It's a hot gas just like the $CO_2$ and $N_2$. What makes it different for your purposes? $\endgroup$
    – Dan
    Commented Apr 21, 2015 at 15:16
  • $\begingroup$ I am just trying to compare the theoretical amount of energy that should be produced, to the amount transferred via heating by accounting for where some of the energy is lost/transferred. Doing this for the water produced in the reaction seemed like a good starting point. $\endgroup$ Commented Apr 21, 2015 at 15:26

You need to consider the latent heat of vaporisation of the water produced. As from this excerpt from this source:

Due to high combustion temperatures, this water takes the form of steam which stores a small fraction of the energy released during combustion as the latent heat of vaporisation; in simple terms, as heat energy stored in the vapourised ‘state’ of water.

The total amount of heat liberated during the combustion of a unit of fuel, the HHV or HCV, includes the latent heat stored in the vapourised water.

The latent heat of water vaporisation is 2260 kJ/kg.

  • $\begingroup$ Does that mean that using the mass of water obtained from balancing the combustion equation, (0.279kg for 1mol of the gas mix) multiplied by 2260kJ/kg, will give the amount of energy used to produce the water from combusting 1mol of the gas mix? So (0.279 x 2260) = 630.54kJ $\endgroup$ Commented Apr 21, 2015 at 14:48

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