Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gun is a kind of heat engine?

  1. Nov 15, 2008 #1
    Gun is a kind of heat engine? if so ,the efficiency of a gun is limited by carnot heat engine?
  2. jcsd
  3. Nov 15, 2008 #2
    Guns aren´t thermal engines because they don´t operate in a cyclic way.
    For example. you may consider a gun whose barrel is made of ceramic (or any adiabatic material). Then, by the first law you achive full conversion of heat into work (ignoring friction forces). This a well known example of a system that seems to defy the second law, but such a system is not an engine for the reason I gave above
  4. Nov 15, 2008 #3


    User Avatar

    Staff: Mentor

    Guns have nozzles, which are a component of a heat engine, but like gordianus said, guns themselves are not heat engines.
  5. Nov 15, 2008 #4
    Theoreticly a gun could have 100% efficiency: if the barrel were very long and frictionless, the bullet very heavy and we would fire it in space, then nearly all of the thermal energy of the gas would be transfered to kinetic energy of the bullet. The efficiency would approach 100% when the length of the barrel and mass of the bullet would approach infinity. You can check that with the formula for the work done by ideal gas at adiabatic change.
    Carnot law applies only to cyclic heat engines. Obviously it can not be applied in case of a gun, because a gun does not even need a thermal reservoir. If we wanted to change the firing of a gun into a cyclic process, we would have to compress the gass back to the original volume, which would require work. Also we would need a thermal reservoir to keep the gas cool during the compression, otherwise we could not get a positive work from this system.
  6. Nov 15, 2008 #5


    User Avatar

    How about water gun or pneumatic gun, can it be consider a heat engine?
    In both cases you have pressure reservoir (pressure can be consider as a heat reservoir). In such case we have to consider the idea of external combustion.

    There might be another consideration:
    In both cases , guns and heat engine can be categorized as a pressure to velocity converters.
  7. Nov 16, 2008 #6


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    While it is strictly true that a gun is not a heat engine, that doesn't mean that Carnot's law doesn't apply. In fact, what happens is that the chemical combustion of the powder generates heat (and also a chemical transformation of a solid into a gas), which then expands almost adiabatically and does work on the bullet, by accelerating it. As such a certain amount of the initial energy content of the powder is converted into mechanical energy, but there is also a certain amount of energy now in the final gas, which is higher than if we had expanded that gas from initial room temperature to its final volume adiabatically (it would be very cold then). In other words, there is also heat "dumped" in the final gas, at best at about room temperature. That will follow Carnot's law.
  8. Nov 16, 2008 #7
    I believe we´re disscussing a "theoretical" gun. No friction, reversible adiabatic expansion and such. Within this framework the powder´s energy can be totally transformed into kinetic energy. Don´t forget Carnot´s theorem also considers ideal conditions but it shows that even under such conditions, heat can´t be totally converted into work in a cyclic engine. The keyword is "cyclic" because any useful engine operates in such a way. It´s important to review Carnot´s theorem hypothesis.
  9. Nov 17, 2008 #8


    User Avatar
    Gold Member

    Doesn't this overlook the significant amount of heat absorbed by both the barrel and the bullet, or are you postulating for your 'ideal' gun that such doesn't occur?
  10. Nov 17, 2008 #9


    User Avatar
    Science Advisor
    Gold Member

    Adiabatic: Adj. Occurring without loss or gain of heat.

  11. Nov 17, 2008 #10


    User Avatar
    Gold Member

    Thanks, Brewski. One of the main disadvantages of not having a formal education is in being somewhat clueless about terminology. :redface:
  12. Nov 17, 2008 #11


    User Avatar

    Staff: Mentor

    Just to make sure here, everyone saw the other caveats in Lojzek's post, right? In order for the gun to use all of the energy in the gas, it must have an infinite length and be located in space (in a vacuum) otherwise it has a heat of rejection (as vanesch said).

    I say that, because in thermodynamics class, you often assume a process is adiabatic, but you virtually never disregard the heat of rejection. That's one of the more critical issues in thermodynamics.
  13. Nov 17, 2008 #12
    It is definitely a heat engine. While heat engine typically go through thermodynamic cycles for continuous work output, that is not a requirement. Even if it was required, a machine gun would be a heat engine.
  14. Nov 17, 2008 #13
    I don't think a water gun or pneumatic gun can be considered a heat engine, since they are
    not powered by heat: they are powered by work of the pump that maintains pressure. Water or air is just used to transfer force or temporary store energy. Converting heat to work is a much more difficult task, which can only be done if there is a way to put away the excess entropy gained from the heat reservoir.
  15. Nov 20, 2008 #14
    I would like to explain the theoretical "100 % efficiency gun" a little better:

    let's say that we have a very long (but finite) frictionless tube with length L and cross-section S, closed at one end. The tube contains ideal gas at a very high temperature T on a small volume S*dL near the closed end. The second end of that volume is closed by a heavy bullet, which is fixed by the trigger. The whole system is put into a low pressure environment (with any temperature, but very low density).

    What happens when we pull the trigger? The gas adiabaticly expands and since the final
    volume is much greater than the initial volume (L/dL aproaches infinity), the final temperature of the gas (when the bullet leaves the tube) approaches zero. So the final thermal energy of the gas is negligible when compared to the initial thermal energy. The kinetic energy of the gas is also negligible, since the bullet is heavy. The efficiency of this
    gun can come arbitrary close to 100% and it can work even if the outside temperature is
    higher than the maximum temperature of the propeling gas.

    This experiment does not obey the Carnot law because it is not Cyclic and because of the
    entropy of mixing. Carnot law can be derived only in case when the final state of the engine is the same as the initial state and both thermal reservoirs only exchange entropy due to reversible addition and extraction of heat: then the entropy of the engine itself hasn't changed and we know that the entropy removed from the high temperature reservoir must be smaller or equal to the entropy added to the low temperature reservoir
    In case of the above experiment we joined the gas in the gun and the surrounding gas, so
    me must consider the entropy of mixing and the formula dQ1/T1+dQ2/T2>=0 and Carnot
    law are not valid.
  16. Nov 22, 2008 #15
    Thank you Lojzek, i agree with you ! From this point of view, i think the "rocket engine "does not obey Carnot's theorem too.
  17. May 23, 2009 #16


    User Avatar

    I am going to try to formulate the questions concerning efficiency of guns in rather different way.

    1) Adiabatic – yes or no?

    In thermodynamic considerations we often assume the action to proceed in reversible – adiabatic mode. If we introduce model notions of real engine cycle course or of ballistic cycle into our considerations, we cannot avoid description of the phenomena difficult to mathematically analyze and experimentally verify. It will be necessary to make up and solve systems of differencial equations. It can happen that we will lose the view of influence of individual parameters on physical process we are interested in.

    In this view the concept of idealized reversible processes is useful.

    2) Heat engine – yes or no?

    Let us imagine, that the conrod journal of cyclic working steam engine breaks and the piston shoots through the wall of machine room like a gun shell. Does the steam engine stop being a heat engine?
    Internal combustion engine utilizes for its work similar fuel based on C, H, O and N as a gun. One of them is a heat engine and the other not?

    I think the mistake lies in the definition. The founding fathers simply forgot guns.

    In my opinion both internal combustion engine and gun are of course heat engines, one of them, however, working in cycles, the other not.

    Therefore there will be differences in efficiencies between these two types of heat engines.

    Gun as a non-cyclic machine has to have higher efficiency because it does not consume a part of energy to bring the system back to original state. It is so even in case of automatic weapons.

    Loading of a new cartridge cannot be compared to the return of the system to original state by compression and heating of working gas. So, machine gun is not a cyclic working heat engine. Gas offtake to power weapon mechanism is the same factor as barrel shortening – higher temperature, lower efficiency.

    While at cyclic engine the working fluid does not matter, at non-cyclic working gun damned does. At cyclic heat engine the efficiency depends on the initial – final temperature ratio only, whilst at non-cyclic engine the efficiency has to depend also on gas properties. On the specific heats ratio k at minimum. Assessment, however, is not an easy task, because in ballistics all thermodynamic functions go against each other.

    There is one more reason why neither the phenomena taking place in a gun nor in engine with internal combustion can be compared to Carnot cycle, namely the fact that injection of working fluid = combustion of propellant (air – fuel mixture) is not taken into consideration. Both technical devices work at similar expansion ratios, similar are burning temperatures, chemical composition of the fuel is similar, absolute pressures are different. As shown hereafter, by burning of the propellant the bullet gains some two thirds of its energy, the value which simply cannot be ignored.

    Even if I were not right and gun were not heat engine, the question concerning its efficiency is legitimate. We have a source of chemical (heat) energy to be transferred to bullet kinetic energy. With what efficiency?
    Here, I would like to modify the question asked by MAGNETAR and explanation provided by LOJZEK to a certain extent.

    Even an ideal heat engine has to have “reasonable dimensions“ so that it is possible to compare it to real engine and calculate efficiency. Steam engine must be comparable with locomotive and the gun should have the barrel as long only, that we are not forced to clear the forest we are hiding the gun in.

    For practical reasons the gun barrel cannot be infinitely or very long to gain 100% energy. It will be e.g. 5 m only (usual value) and then only we can ask what is the efficiency of this shortened gun.

    Really, in ballistics the efficiency is calculated as a portion of bullet kinetic energy (Ek) and propellant heat of combustion (Qv), but it is just total propellant enthalpy which could be utilized at infinite barrel at complete expansion of propellant gases on standard state conditions. This value, however, says nothing about what was the working pressure in gun chamber, what was the barrel length, what are physical properties of the gas etc. The differences in construction quality vanish. If there will be technical parameters of two guns of the same efficiency side by side in a table, it will attract our attention if the lengths of their barrels differ or not.

    Let us ask the following question: what is the efficiency of real gun, i.e. real firing from the weapon in which ideal firing could also take place under the same state conditions? This question is dramatically more interesting.

    What could actually be the course of ideal ballistic cycle?

    Carnot cycle consists of two expansions and two compressions. At firing, however, it is not possible to ignore injection of working fluid = burning of propellant, because it shifts the bullet up to the half of barrel.

    Let us imagine the model, where the propellant occupies almost all cartridge chamber. The chamber is closed on one end and on the other end sealed with the bullet so there is practically no free room left in it. This is so not to consume energy to pressurize the empty room. Really, modern days weapons have their cartridge chambers almost fully filled.

    The propellant is ignited.

    There are no heat or friction losses and the process is reversible.
    Propellant burns under constant pressure p (= permitted barrel strength pressure) and temperature (isobaric flame temperature) and combustion gases shift the bullet by the volume dV. The same effect would be achieved if the valve would be opened of big working hot gas reservoir and the gas would be discharged into the room behind the bullet. In this case neither temperature nor pressure drop in the reservoir occur – they remain constant. In this view the ballistic cycle resembles the first half of Diesel engine cycle. pdV = dnRT can be put down, which is (for simplicity’s sake) state equation of ideal gas. Initial number of moles is null, therefore we can put down: nRT. Product nRT is known in ballistics as force f or impetus of the propellant.

    To make it more interesting let me provide concrete values. If the heat of combustion of propellant is 4,000 kJ/kg, then propellant force is some 1,000 kJ/kg. So, it is approsimately 1/4 of propellant heat of combustion. Heat gun efficiency ξ=Ek/Qv*100, however, usually varies around 35 %.

    The rest of energy, i.e. 10 %, the bullet obtains from the internal energy of gas. We can put down:

    cvdT or cv(T1-T2)

    Sum of both energies is in fact a part of total gas enthalpy:

    dH = pdV + cvdT

    For the time being I call it the energy utilizable for ballistic cycle. I would appreciate somebody’s bringing a better term.

    One more think I want to point out is f/Ek ratio. It is 25/35, which is 0.714. It means that the bullet obtains more than 70 % energy from the impetus of propellant. The rest - 10/35 = 0.286 is cvdT/Ek.

    For 1 kg propellant the equation for energy can be re-written:


    where R – universal gas constant, Tv – isobaric flame temperature, Mr – mean molecular mass, κ – ratio of specific heats, V1 – volume of the barrel behind the shell at the end of combustion of propellant, Vb – total volume of the barrel, Wig – work done by ideal gun.

    Efficiency is a quotient of measured bullet energy Ek and work done by ideal gun Ek/Wig*100. This way calculated efficiency distinguishes weapon performances at various propellant charge weights, various propellant energies, various pressures and various velocities and weights of bullets. It is not an easy task to orientate oneself in measured shot parameters. Calculation of efficiency provides a very good technical estimation of propellant quality.

    The Table presents concrete values for medium calibre gun.

    Translation of projectile 32.0 %
    Rotation of projectile 0.14 %
    Frictional losses 2.17 %
    Total work done on projectile
    (area under pressure – travel curve 34.31 %
    Translation of recoil parts 0.12 %
    Translation of propelling gases 3.14 %
    Heat loss to gun and projectile 20.17 %
    Sensible and latent heat energy
    in the propelling gases 42.26
    Total propellant potential 100 %

    How is it possible to achieve ideal ballistic cycle in practice?

    Loading densities used during WW II were approximately 0.57 kg/dm3. Loading density is the propellant charge weight – combustion chamber (cartridge) volume ratio.

    At present, however, loading densities fired are ca 1.05 kg/dm3. It means that the propellant in the cartridge is slightly overpressed. Chamber is almost filled. The volume to be pressurized is small.

    This has been achieved by impregnation of surface layers of propellant grains with burning moderants or deterrents.

    At initial stages of ballistic cycle, when a free volume in the chamber is small, deterred layers of the propellant burn off slowly. When the volume behind the bullet increases, energetically richer cores of propellant grains start burning. The course of pressure p – V approaches the ideal one.

    Efficiencies of the modern days guns are very high. More than 80 % of theoretically utilizable energy is utilized. This value, which is some 85%, is similar to that of rocket motors nozzles efficiencies.

    During the time I have been dealing with these problems (30 years) the muzzle velocities of bullets have been increased by 200 m/s (let say from 1,000 to 1,200 m/s). It is highly probable that only a small increase in guns efficiency can be expected in future.

    It is obvious that guns have achieved the top of their technical perfectness.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?