Pressure relief hole sizing

[CCR]

I am looking for some input on sizing the hole to bleed pressure for a parachute compartment to prevent premature seperation. I was double checking things for my level 3 attempt next weekend at Argonia and realized that I have no basis for the size of the pressure relief hole. This is not the sampling hole for the altimeter. The compartment is 8" in dia and has an effective length of 26". The cert flight sims to 7200ft and 780 fps but other motors that will fit, push this to 9000 ft and around 1000 fps. I am looking for some kind of formula if there is one as I also have another flight planned for a 54 mm md bird that could pass 16000 ft and hit around mach 2, so the pressure differential will rise much more quickly. Everthing I've read just indicates to drill a hole below the nosecone shoulder but no mention of size. I have a half inch hole in the big bird's upper bay and a 3/8" in the drouge area which is much smaller.

Any thoughts, experience or hard data would be welcomed.

Craig

 

[daveyfire]

I've usually drilled holes about that size, too. But it's always bugged me -- what size is right? -- so I did a little math to prove to myself it was right. The mass flow rate through an orifice is described as

.

I assumed the rocket was beginning at sea level and finishing at 16,000 feet (worst case for your 54mm bird), and so I assumed the pressure difference (deltaP above) was thaat between sea level pressure (14.7ish psi) and pressure at 16k (7.96ish psi, according to Atmosculator). Using the equation of state, it's then possible to calculate density at both altitudes, too. I assumed a Cd of 0.77 (standard for a clean-edged, drilled hole), and that you wanted pressure to equalize over the motor burn time -- 2 seconcds, roughly. The only matter left to solve hole area A was to calculate the difference in gas mass at the two altitudes; using the equation of state to solve density again, and the volume of the payload section for the L3 rocket, there's a mass difference of ~0.025 lbm between sea level pressure gas and 16,000 foot gas. So the flow rate is then 0.025/2 = 0.0125 lbm/sec. Then, just back out for area -- I came up with 0.0475 in^2, or one hole at 0.246" diameter.

Of course, this was making a lot of assumptions -- the two big ones that I can see were assuming the instantaneous appearance of a high pressure drop to drive the gas (which isn't quite the case, the pressure drops slowly as the vehicle ascends), and that you could live with a 2 second lag for the pressures to equalize. During the actual flight, the pressure will be equalizing constantly, so it won't take 2 seconds to happen. But at the same time, you don't have a high deltaP driving the flow, so the flow rate will be significantly less. The end result seemed reasonable, though, so I bet these assumptions were decent.

Bottom line, sounds like you've got plenty of vent to me, both from personal experience and from the numbers. Good luck with the L3!

 

[CCR]

Thank you for the info David.

 

[Handeman]

I don't quite follow why you want to equalize the pressure over the motor burn time. The rocket certainly won't go from sea level to 16K ft in 2 seconds. I see the 2 seconds as a safety margin in the calculation, but I would think a 10 second equalization time would work just as well.

Am I missing something?

 

[mparker59]

I don't quite follow why you want to equalize the pressure over the motor burn time. The rocket certainly won't go from sea level to 16K ft in 2 seconds. I see the 2 seconds as a safety margin in the calculation, but I would think a 10 second equalization time would work just as well.

Am I missing something?

I think he tried to make the point that the time was very short compared to the actual 'up' time of the flight. Heck 19,000 doing Mach 2 all the way is almost 20 seconds and since the rocket is decelerating most of the way it's closer to 20 seconds. But then he pointed out that the calc was also done as if the full pressure differential was driving the flow out the hole the whole 2 secs when, in fact, the pressure differential builds up slowly as well. The hope seems to be that the two assumptions cancel each other out. Otherwise we have to start doing calc and, for me at least, it has been too long.

Thanks to Daveyfire for a nice approximation. I always thought my 1/4" holes were ridiculously large - guess now I'll keep drilling them that way.

But, I drill two. I try not to pack my chutes so tight that air can't flow around them, but the chute must impede the flow in some way. There is launch level pressure below the chute - even if you push the chute all the way down in the bay there's air in the MMT unless you're flying on a max size motor. Between the vents and the shear pins and the baro ports what's one more hole amongst friends.

Mike

 

[daveyfire]

I don't quite follow why you want to equalize the pressure over the motor burn time. The rocket certainly won't go from sea level to 16K ft in 2 seconds. I see the 2 seconds as a safety margin in the calculation, but I would think a 10 second equalization time would work just as well.

The reasons were twofold -- one of which was the driving pressure approximation, as Mike explained. The second reason was, somehow in my mind, I reasoned that the air pressure from the vehicle's forward motion would hold the nose cone on during the motor burn time. After motor burnout, there's still dynamic pressure there, but there's also deceleration and parasite drag (Lovelace effect?!) happening and trying to pull the sections apart. I figured it' be best to be fully equalized to keep from adding a third force here. Shear pins help a lot, too.

The big question is: how fast does the air have to leak out to keep from building up enough pressure in the airframe to pop the nose section off? That's a function of ascent time, nose cone tightness, parachute packing-ness, and a whole bunch of other stuff I was too lazy to account for. So I figured the "worst case" sledgehammer approach was best, provided it gave a reasonable answer. Which it did, so I was happy

I did the math with the 8" x 26" hole volume, so a 1/4" hole seems like it will work for even a big rocket. I usually drill at least two in a vehicle that size, myself!

 

[UhClem]

Some information I put together on the subject. Inspired by the formula posted by Bob Krech.

 

[daveyfire]

Some information I put together on the subject. Inspired by the formula posted by Bob Krech.

What an elegant solution. These tools also appear suited to the analysis of avionics bay hole sizing -- something that's *always* bothered me. Thanks David and Bob!

 

[UhClem]

What an elegant solution. These tools also appear suited to the analysis of avionics bay hole sizing -- something that's *always* bothered me. Thanks David and Bob!

Avionics bays were the original context in which Bob posted the formula. That post on Rocketry Online is now lost.

 

[Handeman]

The second reason was, somehow in my mind, I reasoned that the air pressure from the vehicle's forward motion would hold the nose cone on during the motor burn time. After motor burnout, there's still dynamic pressure there, but there's also deceleration and parasite drag (Lovelace effect?!) happening and trying to pull the sections apart. I figured it' be best to be fully equalized to keep from adding a third force here. Shear pins help a lot, too.

That's the part I missed. As soon as the motor burns out you have the forces causing "drag separation", It certainly makes a lot of sense to equalize the tube pressures so they don't add to and possibly cause that very undesirable event.

Thanks guys...

 

[ErasmoBibiano]

Algumas informações eu juntei sobre o assunto. Inspirado na fórmula postada por Bob Krech.

Hello, do you still have this information?