Say what????

 

What's the point of this post?

 

I noticed the last formula in the QuickLoad manual and was interested in how it had been derived from fundamental principles. I am probably the only one!

"Of what use is a newborn baby." James Clerk Maxwell, on being questioned about the use of his new electromagnetic theory.

 

Just how many people on this board actually read and understand these mathematic/chemical equations? Unless one has a doctorate in the related sciences, they mean nothing to the "average Joe" shooter, hunter, reloader. Why take up band-width with something that means little to the average shooter?
Just my personal observations.

Allen

 

Why take up bandwidth commenting on them? Don't click!

This is high school algebra and a few definitions. I am 'doing it for the children'.

 

F2G1D, I enjoy reading about your knowledge and experience concerning gunsmithing, reloading and shooting, but when you lay out stuff like this, it is just babble to the majority of people that utilize this site. Sorry if I rubbed a sore spot, but you are speaking of things that belong on a chemical engineers forum. Sorta of like talking politics on this site, too many of us don't care for it! I take a step back and will not include any others in my statement, that is not my place. I am not trying to make enemies with you, it is just that you speak too far above the meaning of this board.

Allen

 

No offense taken. I am used to it. I was actually holding back.


Heat of Explosion (Combustion):

Consider a volume of ideal gas V0, subject to a free adiabatic expansion.

Consider the work done by the gas

W = ∫ PdV

For an adiabatic process

PVγ = P0V0γ
P = P0(V0/V)γ

Substituting

W = ∫ P0(V0/V)γdV, evaluated between V0 and V

Integrating

W = [1/(1 - γ)](P0V0γ)V(1-γ), evaluated between V0 and V

W = [1/(1 - γ)](P0V0)γ[V(1-γ) - V0(1-γ)]

Simplifying

W = [1/(1 - γ)](P0)[V0γV(1-γ) - V0γV0(1-γ)]

Simplifying

W = [1/(1 - γ)](P0)[V0γV(1-γ) - V0]

Factoring V0

W = [1/(1 - γ)](P0)(V0)[V0(γ-1)V(1-γ) - 1]

V(1-γ) = 1/V0(γ-1)

Substituting

W = [1/(1 - γ)](P0)(V0){V0(γ-1)[1/V(γ-1)] - 1}

Rearranging

W = [P0V0/(1 - γ)][(V0/V)(γ-1) - 1]

Let V → ∞

W = [P0V0/(1 - γ)](- 1)

Simplifying

W = P0V0/(γ − 1)

U = Q - W

Since U = 0 for an infinite expansion

Q = W

Substituting

QEX = P0V0/(γ - 1) Joules, the heat of explosion

For an ideal gas

P0V0 = nRT0, where T0 is the adiabatic flame temperature

Substituting

QEX = nRT0/(γ - 1) Joules

All γ terms are exponents except /(...)

 

That's it, rub it in!

Allen

 

If even one of the two and one-half billion people connected to the internet find this interesting, I have served my purpose. I am sitting on thousands of pages like this on a variety of subjects. Don't tempt me! ,

 

F2G1D

I noticed the last formula in the QuickLoad manual and was interested in how it had been derived from fundamental principles. I am probably the only one!


May I ask from what fundamentals principles of What? I also noticed that you do not take PM messages


I am not in any way trying to be a wise guy? i think if you play at my lower math level and explain the pertinence level of the last level of the quickload formulae and the importance of the understanding the formula you have shown would be greatly appreciated

 

F2G1D


May I ask from what fundamentals principles of What? I also noticed that you do not take PM messages

The fundamental principles of thermodynamics.

PM on.

 

Interesting F2 stuff. But the ideal gas law only works for an ideal gas, and no real compound is an ideal gas. The law is a good tool to teach the relationship of Pressure, Volume, and Temperature, but of no practical use in ballistic calculations...just shows a relationship.

I wasn't aware that Quickload was based upon it. If so it explains why QuickLoad is just a rough approximation and best not used. My 1960s Powerly Calculator was an interesting tool/device but get seldom use with the much better manuals available today.

I believe your definition of " δ (gm/gm), the density of water" is incorrect. it should be " (gm/ml)". .

It is important to expand ones mind, keep going.

 

It is definitely an approximation. There is no reason that we couldn't use the Nobel-Abel equation. In fact that is the use of the covolume. The density is actually expressed as gm/cm³ but water is one gm = 1 cm³. The advantage is that you can express it in dimensionless terms. That is the origin of expressing case capacity in grains of water , for example, speaking of Powley.

Maximum Charge the case will hold is: dBV0, where dB is the bulk density of the powder. For

example (for IMR 4350), C = .882V0 ‘gr. For IMR 4350, dB = 13.6 gr/cm3 = .882 gm/cm3 = .882

(gm powder/gm water), since 1 cm3 water is equivalent to 1 gm, any common unit is sufficient, such

as .882 (gr/gr of water), or simply .882. A case with a powder capacity of 60 gr of water has a

volume of .237 in3. If the case is filled with the maximum charge of (.882)(60) gr = 52.9 gr of

powder with a dB of 13.6 gr/cm3, the powder occupies a volume of 3.89 cm3 = .237 in3. The case is

full. LDMax = (dBV0)/V0 = dB.

 

You caught my interest so i did some research to find my own answers thru campfire 24 and other research it appears this may be true "The covolume is a rough approximation. More detailed calculations are done in internal ballistics simulations."

 

Coppock's Model

Search for "covolume".

Nobel-Abel Equation of State

Search for 'co-volume'.

 

How close an approximation to a small arms firing event do you find adiabatic expansion equations to be? Given the large amount of heat apparently absorbed by the cartridge case, barrel, and the bullet itself, I should have though it not terribly useful, but then I've been away from any need of any significant math for long enough that I'd probably have to go back and take a class or two to be able to run through many formerly-simple calcs (to my very great chagrin, of late!).

 

According to Ballistics Theory and Design of Guns and Ammunition, Carlucci,"...expansion of the gas after charge burnout is neither adiabatic nor isentropic, however the result is usually within about 5% with respect to pressure."


Referring to an adiabatic expansion of an ideal gas.

 

gas expansion ratios I think this is where covolume can come into play

 

It does. The volume occupied by the gas molecules themselves may fill the case - not as a gas (with free space), which is normal, but like ping pong balls. If the expansion ratio is 6, the 'covolume' is 1/6 of the volume of the case and barrel.

 

Does that reference (or another) give an estimate prior to burnout?

 

It gives a thorough analysis starting with the burning properties of the propellant.