# What Is Heat Flux, and Why Is It Important?

by Nathaniel F

Why do some cartridges burn barrels more than others? What makes a smaller-caliber, higher velocity round toastier to your bore, and how can you compare one caliber to another in this respect? While it’s not the whole answer to this question, today we’re going to talk about a major component to this equation, and that’s heat flux.

Note: For all the engineers out there, the units we’ll be working in are kJ/cm^2, which are not the normal units for heat flux in engineering (those would be Watts/unit area). The reason for this is that for the purposes of a gunshot, the missing unit of time is not important. In other words, we are measuring the energy dumped per gunshot. This is very useful, as we can combine this unit with a rate of fire figure to approximate the actual heat flux (in W/unit area) of a gun over a given course of fire.

Oh, and for those following my math, I am using approximate unit conversion factors of 25.4 mm per inch, 15.43 grains per gram, and 3.28 feet per meter.

It may be easy to tell which of the pair on the left will erode the barrel faster, but it's much more ambiguous for the pair on the right. Having a way to compare cartridges in this respect is useful.

First, we need a couple of examples to compare. Let’s take the common 5.56x45mm M193 Ball loading, for starters. We know a few relevant things about this cartridge:

– It has a bore diameter of 5.56mm/0.219″, and a groove diameter of 5.69mm/0.224″. Either of these dimensions can be used for calculating heat flux, as long as which one is chosen is kept constant when comparing different calibers. Since groove diameter is the more commonly available figure – it is usually the same as the bullet diameter, we will use it.

– It has uses a charge of approximately 1.72g/26.5grs.

– It produces a muzzle velocity from a 50.8cm/20″ barrel of about 997 m/s or 3,270 ft/s.

– It has a bullet weight of 3.56 grams or 55 grains.

From these figures alone we can calculate heat flux. We should start by calculating the muzzle energy of the round, using the kinetic energy equation KE = 0.5 * (mass of the bullet) * (muzzle velocity)^2. I prefer doing my calculations in metric, so that would be:

MKE = 0.5 * 3.56 g * (997 m/s)^2 = 1,769 Joules

Next, we need to calculate the total energy produced by the cartridge. This is not the same as the muzzle energy, as guns are not perfectly efficient engines, so to approximate this we need to take the powder charge in grams and find out how much energy it produces when it burns. There’s not an easy way to do this exactly, so instead we’ll take a shortcut by just using a fixed figure of 3,870 Joules produced per gram, which is converted from a figure of 185 ft-lbs per grain estimate for IMR brand powders. This step is where most of the error will be introduced as that figure may not be correct for every kind of propellant (though it shouldn’t be wildly off for nitrocellulose-based propellants), and because for some barrel length-cartridge combinations a lot of unburnt powder may be ejected out the muzzle (and therefore its energy won’t go into heating the barrel). However, for a quick rule-of-thumb estimate, it’s not bad.

For M193, we find:

TKE = 3,870 J/g * 1.72 grams = 6,656 Joules

Giving us a wasted energy figure of:

6,656 Joules – 1,769 Joules = 4,887 Joules

Or 4.887 kilojoules (kJ). Now, find the area of the bore, or in this case the approximated area from the groove diameter:

5.69 mm / 2 = 2.845 mm

π * (2.845 mm)^2 = 25.4 mm^2 = 0.254 cm^2

Now, finally, we can find the heat flux per shot by dividing the approximate wasted heat in kilojoules by the approximate area of the bore in square centimeters!

4.887 kJ / 0.254 cm^2 = 19.2 kJ/cm^2

And that’s the approximate heat flux for a single shot of M193 Ball ammunition, through a 20″ barrel. By calculating values for different calibers, you can compare them to get a better idea of which calibers burn barrels out faster than others.

You can also use the excellent Powley computer hosted at kwk.us to do this calculation as well, though it’s not necessary. To do so, fill in all the values for case capacity, case length, cartridge overall length, bullet length, etc, and adjust the pressure to the correct value by clicking the checkbox next to the “Pressure” section. Then use it to find the total kinetic energy value by dividing the muzzle energy by the efficiency (convert the efficiency from percent to decimals first), and pick up the instructions above from there! A screenshot of what a filled in Powley sheet looks like is below:

Nathaniel F

Nathaniel is a history enthusiast and firearms hobbyist whose primary interest lies in military small arms technological developments beginning with the smokeless powder era. He can be reached via email at nathaniel.f@staff.thefirearmblog.com.

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##### Join the conversation
• Mazryonh on Mar 14, 2016

So will caseless rounds leave behind more heat because there's no heat-conducting casing to take some away? How about rounds like the VOG-25 grenade?

Plenty of kids think they want "blaster rifles." For the sake of argument, would a laser weapon putting out the same energy on target as a 5.56mm M193 round at, say, 100 meters from an M4A1 be more or less efficient (resulting in more heat buildup in the weapon) than the M4A1?

• Georgesteele on Mar 31, 2016

I keep staring at your calculations and cannot make sense of why you would use the cross-sectional area of the bore instead of the area of the internal barrel surface. You are dissipating the heat of combustion of the powder across the internal surface area of a 20" long tube - so the equation wouldn't be pi*r^2, which gives the area of the circle that is the bore, but rather pi*D*L - the first two giving the circumference and the latter being the barrel length.

That provides the surface area of the cylinder that is the barrel internal bore area, through which the heat is dissipated (that is, the component that is transferred from the high temperature gas minus that which vents from the barrel in the form of hot gas and unburned powder). Also, the heat generated within the first few moments, at the highest chamber temperature and pressure, would be (partially) transferred to the throat and leade area, but it would diminish as the bullet traversed the barrel, the pressure and temperature diminished, and the last of the powder contributing to thrust against the bullet burned, so it's not a linear condition - of course, barrels burn out from the chamber forward, that being one of the reasons.

By "not linear", I mean that once the bullet leaves the barrel, the temperature and pressure drop off very rapidly, so the muzzle is subjected to the lowest pressure and temperature, and for the shortest time. The area subjected to the highest combustion heat, for the longest time, is in the throat/leade area - both as the gases pass by the bullet as it jumps from the case neck across the leade into the first "engraving" portion of the rifling to create the barrel seal, and as the pressure and temperature skyrocket as the bulk of the powder ignites. As the bullet progresses down the barrel, the increasing volume behind the bullet drops the temperature and pressure by expansion, so heat transfer is reduced because the barrel gas temperature is reduced. But the leade area is still at the highest temperature, as the powder was burning in that area at the highest pressure level.

Back to the equation, what is happening within the center of the "tube" of burning gas behind the bullet shouldn't be as important as what is happening at the surface of that tube, which is in contact with the barrel grooves and rifling. Conductive (vs. radiative) heat transfer from the center of the cylinder of combustion gas isn't instantaneous through the surrounding gas to the outer edge of that expanding cylinder, nor is the heat transferred instantaneously from the gas in contact with the barrel surface - but as you say, the idea is to compare one round/barrel combination with another, rather than to measure exactly the thermal flux value.

So I'm just not seeing why the cross-sectional area equation pi*r^2 is used instead of the cylinder surface area equation, pi*D*L. Is the radiative component and combustion mass more predictive of the heat flux than the conductive/convective? Is there a source that might clarify this?