Black Powder Ejection Charge - Calculating "C"

I'm looking at a L3 document submitted that uses the following equation for determing correct charge size:

Wbp=C * D^2 * L

where C = PSI * 0.0004
D = diameter
L = length

My question is where or how to i derive the 0.0004 number. Obviously this comes from the PV=nRT equation in regards to black powder, but i don't want to blindly follow that number.

Any help appreciated. Thanks
Dan

Here, make it simple for youself:

If you have up to a 6" airframe, start with two grams and work your way up in one gram increments until you get what you want.

Larger than 6 and 4 grams is usually a good starting point.

On our Talon ground test videos, you see what 4 and 6 grams did to our 12" diameter rocket.

Thanks. I will be doing that, but first i want to at least come up with some calculated values for a comparison.

eastvolt said:

i want to at least come up with some calculated values for a comparison.

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Here's some information:

https://rocketryforum.com/showpost.php?p=445965&postcount=10

You are correct that the factor comes from PV=nRT. I went through the calculation a while back because of some interesting ground test results. Specifically, I had a piston that didn't come out of the rocket when the charge was fired. So, I wanted to go through the calculation to find out how much gas was actually being produced. Turns out that for a relatively gas-tight piston, it shouldn't have come out (learn something new every day).

The two key assumptions are what combustion products are actually being produced, and what is the temperature at which the gas pressure is determined. I found a number of references on the combustion products. When I factored this into the equation, the calculated temperature was something on the order of 2200F if I recall correctly.

I'm not sure if these assumptions have any practical value for determining the size of an ejection charge, but that's where the factor comes from.

SpartaChris said:

Here, make it simple for youself:

If you have up to a 6" airframe, start with two grams and work your way up in one gram increments until you get what you want.

Larger than 6 and 4 grams is usually a good starting point.

On our Talon ground test videos, you see what 4 and 6 grams did to our 12" diameter rocket.

Click to expand...

I'll agree with Chris here. I've seen plenty of folks use calculations and have airframes come in on flat spin as the nose on the main charge didn't make it all the way out.

On the other hand, doing the ground tests makes sure you know several times what's going to happen with the main.

In theory calculations should always be able to model what happens in practice. However in practice I've found theory was a nice aproximation.

Yes, but the calculations give you a good starting point at which to conduct your ground tests as opposed to just haphazardly loading up BP charges.

eastvolt said:

Yes, but the calculations give you a good starting point at which to conduct your ground tests as opposed to just haphazardly loading up BP charges.

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I'll disagree with that one too. A buddy of mine used a calculator that told him he needed to use an excessive amount of BP to seperate part of his rocket. I convinced him to cut it in half, an he begrudgingly did so. The end result was that the lesser amount he used in that test MORE than did the job, but had he followed the calculations, he quite possibly would have damaged his rocket. The thing ripped apart as it was, seperating both drogue and main bays with just the one charge in place.

Either way, just remember that 1 gram of BP, if contained properly, will produce a LOT more energy than most folks give it credit for.

As an aside, at LDRS, every time myself or someone in our group had a rocket RSO'ed, we were told that we didn't have enough of an ejection charge, that we needed to double it in most cases. I kid you not, EVERY time. It was a lot of fun proving the guy wrong each time.

OK, back to the thread.

I'll disagree with that one too. A buddy of mine used a calculator that told him he needed to use an excessive amount of BP to seperate part of his rocket. I convinced him to cut it in half, an he begrudgingly did so. The end result was that the lesser amount he used in that test MORE than did the job, but had he followed the calculations, he quite possibly would have damaged his rocket. The thing ripped apart as it was, seperating both drogue and main bays with just the one charge in place.

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Well no. Firstly, you said "excessive" amount of BP. That should trip a red flag right away. If the calculations seem a bit out of whack based on what your experience or general rules of thumb there are, then you should treat the results with caution. Secondly, you said, "buddy of mine used a calculator" sets off another red flag in that you are trusting that the calculator was correct. One thing i hate is using those automated calculators in which you have no idea how they are deriving the calculation etc... Just because its on a website, doesn't mean its correct.

Finally, its obvious the results of the calculator weren't correct. So my original point doesn't even apply in this case. I am assuming that the equations being used have already been validated through experimental results.

In general, I wouldn't trust a formula (especially one with magic numbers in it) that doesn't have the units specified...

But in any case, you can derive it by starting with the reaction associated with black powder 2(KNO3) + 3(C) + S => K2S + N2 + 3(CO2) and PV=nRT.

For now lets assume that we have an airframe that is 10cm in diameter, and 100cm long. The volume of that is V=pi*(d/2)^2*h, or 7700 cm^2, or 7.8L.

Now we want to increase the pressure of our airframe to be 15PSI, or one atmosphere, more than what was in there. So that means that we need to add enough BP to fill a 7.8L volume to 1atm.

The only variable left now is T, which needs to be the temperature of the gas produced by the BP... This is a tricky one, as the gas expands it will cool, but it is generally considered safe to assume it is 1500K


We rearrange PV=nRT to solve for n, giving us n=PV/RT, and use R = .082 L atm/K mol for the ideal gas constant, which gives us .056moles of gas necessary.

Then we can examine the equation for the combustion of black powder, and note that there are 5 moles of gas on the product side of the reaction. For simplicity's sake, we will only consider the N2 produced, as we can determine from the coefficients in the equation that N2 will consist of exactly 1/5th of the total gas produced. We cam then use ratios to solve for the amount of N2 needed, (1mol N2 / 5 mol total) = (X mol N2 / .056 mol total), which solves to .011mol N2.

We can then use this value to solve for the amounts of KNO3, C and S by using their molar masses, and their coefficient in the reaction. For S we simply multiply the molar mass of S (32g/mol) times the coefficient of S in the reaction (1) and then the number of moles of N2 disired (.011mol), which gives .36g. The same is true for C and KNO3, which gives .40g and 2.2g respectively. These are then added, to give a total of 3g necessary.

But since no one wants to sit down and go through that every time they need to fill their rocket with BP, we can wite an equation that relates whatever variables you want (airframe diameter, length, pressure, weight of BP) being careful to keep the powers correct.
ie, Wbp = L * D^2 * P * x
Plugging in the values used (3g, 100cm, 10cm, 15psi respectively) and solving for x would give .00002.
If we wanted to use inches instead, we could convert the measurements to inches at this point, and solve again (3g = 40in * 16in * 15psi *x ), which gives us x=0.0003, quite close to the value presented in the first post

Of course there are better ways to solve this, but sometime the long way helps people understand what is going on...

So what if I am only 16 years old

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. . . hmmmm . . . seems i've seen you somewhere else. Perhaps 4hV?

Yes, I am also on 4HV