so i figured out the swept volume of my 392 carbine

so i figured out the swept volume of my 392 carbine

Joined: December 14th, 2010, 5:08 am

June 22nd, 2012, 5:55 am #1

to simplify the first calculation i treated it as if i had a flat piston and valve and i came out to 22cc. but just to be on the safe side we'll just say 20cc or 1.22ci. now how do i figure out how to calculate that into valve pressure per stroke? ive searched the google and this forum but cant seem to find the right conversion chart or formula.

i'll sleep on it for tonight and could probably think of the right key words by then

DNADNNOID
Last edited by Dnadnnoid on May 3rd, 2013, 7:30 pm, edited 1 time in total.
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Joined: December 14th, 2010, 5:08 am

June 22nd, 2012, 6:23 am #2

i totally reverse engineered this little tidbit from an earlier thread of mine to figure it out:


quote from phil:

""Total of 8 pumps with a Benji is: roughly 28 cu.in. of swept air.
The factory valve volume is: .25 cu.in.

In a perfect world, at sea level, the above is a ratio of 112:1 or, 1,646psi

The Benji has a lot of losses going on, so the above figure is really too kind.

The piston seal is one big area of loss. ""


the formula is: air pressure x swept volume x number of pumps / valve volume = valve pressure

right?
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Joined: December 14th, 2010, 5:08 am

June 22nd, 2012, 7:43 am #3

desired valve pressure x valve volume / swept volume / air pressure = # of pumps needed to reach desired valve pressure
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Joined: June 7th, 2010, 12:14 am

June 22nd, 2012, 2:19 pm #4

The "dead space", or trapped volume. It does no good, but must be included in any compression ratio calculation.

Unswept volume includes both valve volume and the trapped regions.

Swept volume + unswept volume / unswept volume= compression ratio.
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Joined: February 17th, 2010, 3:30 am

June 22nd, 2012, 4:14 pm #5

If the compression ratio was just as you described, for a 392 using simplified numbers, it would be roughly:

(50 + (4.5+0.5) / 5 = 55 / 5 = 11:1 for a 39X.... That is roughly 11 x 14.7 = 162 psi.... and would mean that you couldn't raise the pressure inside the valve to more than that....

The check valve alters this, of course, isolating the headspace from the air in the valve, and effectively allowing a much higher compression ratio for subsequent pumps.... The maximum compression ratio is:

(swept volume + headspace) / headspace .... using the above numbers:

(50 + 0.5) / 0.5 = 50.5 / 0.5 = 101:1 for 0.5cc headspace.... That is roughly 101 x 14.7 = 1485 psi as a pressure limit.... That is why reducing the headspace is VERY important, as it limits the maximum pressure that can be reached.... Reducing the headspace by half would allow twice the maximum pressure.... assuming the linkage didn't fail first....

This pressure is never reached, as the check valve opens (at a progressively higher pressure each stroke) and allows the air to enter the valve....

Fortunately, none of this really matters.... You can approximate the pressure inside the valve by taking the total swept volume (per stroke times the number of strokes) and dividing by the total volume of the valve plus headspaces.... times an efficiency factor....

Press. = (14.7 x Eff. x (Swept Vol. x No. Strokes)) / (Valve Vol. + Dead Vol.)

At 100% efficiency, 10 strokes....

P = (14.7 x 1.00 x (50 x 10)) / (4.5 + 0.5) = (14.7 x 500) / 5.0 = 1470 psi

At 70% efficiency, however....

P = (14.7 x 0.70 x (50 x 10)) / (4.5 + 0.5) = (10.3 x 500) / 5.0 = 1029 psi

In actual fact, the efficiency probably decreases as the number of strokes (and the heat) increases.... so the resulting pressure inside the valve grows more slowly as you pump more.... The 70% number I used above is just an example, I have no idea what the actual efficiency is in a 39X....

Bob


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Joined: December 14th, 2010, 5:08 am

June 22nd, 2012, 8:07 pm #6

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Joined: February 17th, 2010, 3:30 am

June 22nd, 2012, 8:29 pm #7

that I use to calculate pressure in the valve.... It's much more complicated than the simplified version I gave above.... The pressure actually builds more quickly initally, then crosses over, and then builds more slowly than the simple formula I gave above.... However, since we don't know the efficiency (fudge factor) anyway.... there seems little point in complicating things.... Here are a couple of graphs from my spreadsheet using a swept volume of 58cc, and a valve volume of 4.2 cc....

[/IMG]

As I said, I don't know the efficiency, but you can see what happens on the first graph at various values.... On the second graph, I set the efficiency to be 70% and then showed the effect of headspace.... You will note the big gains to be made by reducing it....

Bob
Last edited by rsterne on June 22nd, 2012, 9:48 pm, edited 1 time in total.
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Joined: April 1st, 2009, 3:18 am

June 23rd, 2012, 1:11 am #8

to simplify the first calculation i treated it as if i had a flat piston and valve and i came out to 22cc. but just to be on the safe side we'll just say 20cc or 1.22ci. now how do i figure out how to calculate that into valve pressure per stroke? ive searched the google and this forum but cant seem to find the right conversion chart or formula.

i'll sleep on it for tonight and could probably think of the right key words by then

DNADNNOID
Now, keep in mind this is at 100% efficiency. 1.22 cubic inches of air with a .25 cubic inch valve:
1.22ci (Swept volume) /.25ci (valve volume) = 4.88:1 ratio. Now you have to multiply the value of 4.88ci by 14.7 (atmosphere pressure). So, you now have 4.88ci * 14.7 = 71psi per pump stroke.

Take in to account, for reality sake, 90% efficiency. 90 percent of 71psi is 63psi per stroke. These numbers are not dead on perfect, but will most certainly be within a digit to give you an idea of what is happening.

"The majority of things in our lives are created by folks no smarter than the rest. Afterall, the world is comprised, and operated by C average people intellctually, academically, and morally. These people are often the great pioneers that set the precedent for what excellence should be."
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Joined: June 7th, 2010, 12:14 am

June 23rd, 2012, 4:16 am #9

But since sealing and related efficiency is a looming uncertainty... close enough....

even if the math is wrong. it's not swept vs valve volume!
Last edited by CalG on June 23rd, 2012, 4:17 am, edited 1 time in total.
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Joined: April 1st, 2009, 3:18 am

June 23rd, 2012, 2:31 pm #10

The math is intended for each pump stroke with Trevor's given carbine design - not the total end result. The simple answer is in fact how much volume is being compressed into a given storage volume. In a perfect world (no loss) we would have 100% efficiency, sadly we do not not. So, for experiment sake, we can toss around the efficiecny number as a variable to see a 'what if' scenario. If we knew the exact efficiency, that given number would no longer be a variable.

Trevor's carbine, on ten pumps, isn't breaking the 800psi figure. So, for him the get the 500fps level on less than 800psi is actually pretty decent in the end.

Edit: Also, we assume there is loss (head space), but there may be no loss with his shortened piston.

"The majority of things in our lives are created by folks no smarter than the rest. Afterall, the world is comprised, and operated by C average people intellctually, academically, and morally. These people are often the great pioneers that set the precedent for what excellence should be."
Last edited by Duane30 on June 23rd, 2012, 2:36 pm, edited 1 time in total.
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