0
$\begingroup$

I am reposting this question here, as the Physics section closed my question as not in the correct area, and suggested Engineering.....

I have a very basic knowledge of fluid dynamics. I am interested in the gas system of a firearm. How it functions with specific ammunition & specific conditions. The system can be manually adjusted/modified. How these two pieces of the gas system function together is confusing me. Therefore the questions.

The gas system is adjustable. One setting, the smaller gas piston hole, is to be used when normal high velocity ammunition is being used. The other larger gas piston hole is to be used when subsonic, lower velocity ammunition is being used.

The gas port hole in the barrel is .050" in diameter. Fitting over this hole is the gas system piston, twisted to either the plus (.062") or minus (.041") setting. Here is what I do not understand.

Question: If the hole in the barrel is only .050" in size, how can the large .062" hole in the piston above, allow more gas through than a .050" hole as in the barrel?

I assumed that the .050" hole determines the amount of gas entering the system, not a second hole on top of the first, that is larger. Hope this makes sense.... Thanks!

Addition Correct. It is part of the MCX gas piston assembly, but the plug, not the piston. From the drawing you can see the gas port/hole in the top of the barrel. The gas block sits on top of this hole. The gas block itself has a much larger hole (.150”??) through it. Then the adjustment plug, which slips into the gas block has two holes (large & small) of its own in it to switch between.

The back of the plug is open to the piston. When the weapon is fired and passes the port in the barrel, gas goes up through these holes & into the plug, & back against the piston to force it rearward, functioning the action. Me question is very specific about fluid dynamics, that I don’t understand.

To me the smallest opening in this series would make the difference as to how much or how fast the gas applies pressure on the face of the piston to drive it back. But, if that were the case, then the largest hole needed would be the same size as the gas port hole in the barrel itself. A larger hole than that would be a waste. I am told that is not the case, & in fact larger holes in the block or plug can direct more or faster gas against the piston.

I need a simple explanation as to how is this possible? Thanks! enter image description here

Improve this question
$\endgroup$
1
  • $\begingroup$ Pistons dont have holes in them, I suspect you mean a port to let gas onto the piston. Give us a drawing. $\endgroup$ Commented Feb 23 at 8:05

1 Answer 1

Reset to default
0
$\begingroup$

So the answer is nothing to do with gas dynamics and everything to do with manufacturing tolerances. If you select the small port that maximum tolerance in positioning the port is (0.050-0.041)/2 =4.5 thou, and will give you 0.041 flow rate. If you select the large port the tolerance is (0.062-0.050)/2=6 thou, and you will get the flow rate through the 0.050" orifice.

The governing equation is stolen from wiki and looks like this

orifice equation

where beta is 0 , e is 1 , and Cd is 0.62.

Improve this answer
$\endgroup$
6
  • $\begingroup$ Thanks, could you give me the answer in a way that does not assume I am a scientist or mathematician? Are you saying that the large port/hole produces flow strictly based upon the .050" barrel port size? If so, then the size of the large port/hole would make no difference so long as it was sized at .050" or larger? Would that be correct or not? Thanks! $\endgroup$ Commented Feb 26 at 4:07
  • $\begingroup$ I know that I am trying to simplify something that is more complex. I am simply trying to understand why the large port/hole is .062", and not .050" the same size as the barrel port? $\endgroup$ Commented Feb 26 at 4:14
  • $\begingroup$ Because they are worried about the accuracy with which the two holes are aligned. If the large port was 0.050 you would have to centre it perfectly over the barrel port, otherwise it would occlude part of the barrel port. $\endgroup$ Commented Feb 26 at 5:42
  • $\begingroup$ Please simplify the response like I am in 6th grade. The barrel port is .050”. The gas block hole is .150”, mounted over barrel port. The inserted gas plug large hole is .062”. All holes are in alignment. Only a specific amount of gas pressure/speed can travel through these 3 holes. Now change the gas block & gas plug hole sizes to also be .050”. All holes are now the same .050” size and in alignment. Would the amount of gas able to travel through the system be the same or different before & after the 2 changes? I need a Yes or No before I can try to understand the why. $\endgroup$ Commented Feb 27 at 16:27
  • $\begingroup$ That would have a slight effect but the most important size is the smallest in the series of holes. There will be slight effects from the others, but fairly trivial. The system you are discussing is similar to an orifice plate - en.wikipedia.org/wiki/Orifice_plate. The precise shape of the hole in the plate affects Cd, but as you can see from the article it is not a huge effect. $\endgroup$ Commented Feb 27 at 21:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.