FWB Air Cylinder volume

[Mr. Wonderful]

Hello Friends:

I am amazed at the knowledge of everyone here. Simply amazing. Don't post much as I am not at a level worth posting.

I have 2 questions I thought to ask:

What is the cylinder volume for a Feinwerkbau 800 Field Target gun and...
similarly, what is the cylinder volume for a Feinwerkbau P44 pistol (not short barrel).

I cannot find the answer anywhere.

Appreciate it.

[David M]

Volume should be engraved on tank.
Similar tank size (same shot numbers) is the Morini 162 at 0.08 litre

[spektr]

My P44 tanks do not have a volume engraved on them.

[ojh]

Presuming that the tank is cylindrical, you may calculate a rough estimate.

Measure the length of the tank (L).
Measure the diameter of the tank (d).
The outside volume of the tank is then Vo = L*3.14*(d/2)^2.
If we guestimate that the volume for air is 10% less than the outside volume, then the air volume V = 0.9*Vo.
If your measurements were in millimeters, the volume is now in cubic millimeters.
In liters the final volume is Vl = V/1000000, in cubic centimeters it's Vc = V/1000.

An example: let's say you get L = 150 mm, d = 30 mm.
Vo = (150 mm)*3.14*((30 mm)/2)^2 = 105975 mm^3
V = 0.9*105975 mm^3 = 95377.5 mm^3
Vl = 0.095 L
Vc = 95 cm^3

[Mr. Wonderful]

Well I want to know the volume when the cylinder is filled to 2900 psi = 200 bar.
Since air is being pressurized, it will contain more volume than at 14.7 psi = 1 atmosphere.

Maybe I should be using Boyle's Law.

[Mr. Wonderful]

Volume should be engraved on tank.
Similar tank size (same shot numbers) is the Morini 162 at 0.08 litre

I'll look.

[Rover]

I'm a little confused by your question. Your concern should be the number of shots per fill, which will drop after every fill.

[Green_Canoe]

Well I want to know the volume when the cylinder is filled to 2900 psi = 200 bar.
Since air is being pressurized, it will contain more volume than at 14.7 psi = 1 atmosphere.

Maybe I should be using Boyle's Law.

It's as simple as 200 * original volume since you are correct in using Boyle's Law P1V1=P2V2. Solved for V1 (the volume you have to compress to get 200 BAR) you get (200 BAR * 0.08l)/1BAR = 16l

(I borrowed the cylinder size of 0.08l from another thread and it seems to closely match the measured dimensions of my Steyr LP10 cylinder.)

[David M]

Well I want to know the volume when the cylinder is filled to 2900 psi = 200 bar.
Since air is being pressurized, it will contain more volume than at 14.7 psi = 1 atmosphere.

Volume of Morini 162 tank at 15 C and 1013.2mb
Sea level 0.08 litre
80 bar (empty non firing) 6.31 litre
200 bar Full 15.78 litre

[Mr. Wonderful]

I'm a little confused by your question. Your concern should be the number of shots per fill, which will drop after every fill.

I am a scuba diver with multiple scuba tanks of varying volume. Example: I have a 100 cu ft tank at 3400 psi when full. Also have a 80 cu ft tank with a top pressure of 3500 psi. Each tank psi will decrease with filling my rifles but instead of calculating pressures, its better to do so with volume.

[Mr. Wonderful]

Well I want to know the volume when the cylinder is filled to 2900 psi = 200 bar.
Since air is being pressurized, it will contain more volume than at 14.7 psi = 1 atmosphere.

Maybe I should be using Boyle's Law.

It's as simple as 200 * original volume since you are correct in using Boyle's Law P1V1=P2V2. Solved for V1 (the volume you have to compress to get 200 BAR) you get (200 BAR * 0.08l)/1BAR = 16l

(I borrowed the cylinder size of 0.08l from another thread and it seems to closely match the measured dimensions of my Steyr LP10 cylinder.)

Thank you.

[Mr. Wonderful]

And in any case someone wondered, I have written to FWB and came back with no answers.

[Dewol49172]

To determine the volume of a cylinder filled with air at different pressures, you can use Boyle's Law, which states that the pressure of a given amount of gas is inversely proportional to its volume at constant temperature. The formula for Boyle's Law is: � 1 ⋅ � 1 = � 2 ⋅ � 2 P 1 ​ ⋅V 1 ​ =P 2 ​ ⋅V 2 ​

where: � 1 P 1 ​ is the initial pressure, � 1 V 1 ​ is the initial volume, � 2 P 2 ​ is the final pressure, � 2 V 2 ​ is the final volume. Given that the initial pressure is 14.7 psi (1 atmosphere) and the final pressure is 2900 psi (200 bar), we can set up the equation: 14.7 ⋅ � 1 = 2900 ⋅ � 2 14.7⋅V 1 ​ =2900⋅V 2 ​ Now, to find the relationship between � 1 V 1 ​ and � 2 V 2 ​ (the volume at 1 atmosphere and 200 bar, respectively), we can rearrange the equation: � 1 = 2900 14.7 ⋅ � 2 V 1 ​ = 14.7 2900 ​ ⋅V 2 ​ This tells us that the initial volume ( � 1 V 1 ​ ) is equal to the ratio of the final pressure to the initial pressure multiplied by the final volume ( � 2 V 2 ​ ).