Compressible flow and Destin Sandlin's 1050-MPH baseball cannon 2

[JoeFrickinFriday]

You've probably seen this by now, but just in case, last fall Destin Sandlin uploaded a video about a pneumatic baseball cannon he and some friends built that can launch baseballs at well over 1,000 MPH. There's a 23-minute video in which they describe the design, constructions, and early tests of it:

https://www.youtube.com/watch?v=cqidD7kVnxY

Destin uploaded a companion video a few months later, in which he and his fellows fire baseballs through a variety of items through which you should not be able to fire a baseball:

https://www.youtube.com/watch?v=4I-p8vjQ95s

My question concerns the first video. at 2:48, we see a 2-D CAD printout showing a critical flow venturi in the barrel just upstream of where the baseball begins its journey. This made sense to me, as it seems like a CFV would be needed to accelerate the driving gas past sonic velocity. Without a CFV, ISTM the tank and barrel together would form a CFV with an exit-to-throat ratio of 1 (so I guess something akin to a critical flow orifice), making it impossible to go faster than the local sonic velocity. But a little while later at 3:30, we see a SolidWorks model that shows no CFV in the barrel. So what gives? I asked this question in the comment section, but got no response (no hurt feelings on my parts, he's got a lot of videos with a lot of comments and can't respond to all of them).

So I'm wondering what the collective here has to say. Is a CFV necessary for achieving supersonic muzzle velocity in a pneumatic cannon like this, or am I missing something about compressible flow?

Thanks...

 

[SwinnyGG]

People keep mentioning choked flows....

This is real world, friends. This isn't on paper.

The process under scrutiny here is not adiabatic. The pressure and gas temperature are not constant.

The only variables that actually matter are the upstream pressure, and the pressure ratio. More pressure or a higher pressure ratio are going to get you more velocity, ad nauseum, until things start to break. It's very simple.

 

[rb1957]

The, well my, point is why "fuss" with a delaval throat, which is intended AFAIR to minimise pressure loss over the shock, when the major pressure loss is happening upstream ?

Yes, more pressure = more speed, but we are (well, I am) thinking about what's the minimum pressure needed to make the ball supersonic ?

another day in paradise, or is paradise one day closer ?

 

[SwinnyGG]

Well the original question that started the thread was 'is a critical flow venturi necessary to generate supersonic muzzle velocity from an air cannon' to which the answer is no.

 

[rb1957]

agreed ... I was wondering where the delaval nozzle talk started, since I didn't see one in the test.

another day in paradise, or is paradise one day closer ?

 

[SwinnyGG]

In the youtube video the original concept drawing shows a delaval nozzle at the root of the barrel (nozzle OD is barrel ID, nozzle inlet is directly at the tank); later on in the video the solidworks model is shown, without a nozzle at all. I believe this was the source of the question 'is a nozzle required'

 

[3DDave]

The pressure is transmitted via collisions between molecules in the gas - the temperature is indicative of the speed the molecules are traveling through the vacuum between the particles. The pressure wave/expansion cannot exceed the speed of the molecules, no matter how high the pressure is.

That's why I suspect the compression of gas into the chamber just before firing has significantly heated that gas, which allows it to sustain a sonic flow higher than in the air outside the gun.

 

[rb1957]

compressing gas heats it ... expanding gas cools it, no?

so the tank (at 500psi) is filled with gas from a 3000psi cylinder.
the original air in the tank has been compressed (from 15psi to 500psi)

another day in paradise, or is paradise one day closer ?

 

[3DDave]

Yeah - that's what I'm thinking. It was a factor that wasn't mentioned by Smarter Every Day. They instrumented for pressure but it's just as important to measure the internal temperature.

 

[rb1957]

depends what you want to know. You need pressure to ensure you don't explode the bomb you're loading ! You need temperature if you want to calc speed of sound.
But who cares about that ... unless you're looking into the performance of the flow around the valve ? I wonder if, for a really high pressure differential, you'd find a low Cd (coefficient of discharge) through the valve (cause of the shock). Then maybe gas would slow build behind the ball and either ...
1) the falling pressure differential would reduce the shock and increase the flow and ..., or
2) the low pressure building behind the ball is just enough to move it down the pipe and maybe it'd just fall out the end, followed by a flood of gas !??

another day in paradise, or is paradise one day closer ?

 

[JoeFrickinFriday]

3D Dave said:
>That's why I suspect the compression of gas into the chamber just before firing has significantly heated that gas, which allows it to sustain a sonic flow higher than in the air outside the gun.

rb1957 said:
>compressing gas heats it ... expanding gas cools it, no?
>
>so the tank (at 500psi) is filled with gas from a 3000psi cylinder.
>the original air in the tank has been compressed (from 15psi to 500psi)

The starting mass of air in the accumulator at 15 psi (before charging with high pressure gas) is pretty small. Then they admit air from a rack of bottles at 2500 psi, which is expanded to 500 psi upon entering the accumulator. In other words, just before firing, a small amount of the accumulator's total air mass has been compressed and heated, but the vast majority of it has been expanded and cooled. In the aggregate, it'll be extremely cold (upthread I estimated -65F, disregarding heat transfer through accumulator walls) when they push The Button.

I think SwinnyGG hit on the key upthread. Accumulator air can flow subsonically, and even sonically, across the choke point at the transition from accumulator to barrel and get the ball moving at subsonic speed. Once you've got the barrel behind the ball loaded with high-pressure air, that mass of air is already past the choke point (the transition from reservoir to barrel); it should have no problem expanding and driving the ball to supersonic speeds without an expansion nozzle. I think the inertia/mass of the ball is the thing that makes the difference. If it weren't there, then I'm pretty sure you'd only have sonic flow along the entire length of the barrel.

 

[3DDave]

Expanding the gas cools the volume it expands from. How does it also cool what it expands to when that is a region that is seeing increasing pressure? After all, the pressure in the original tank did work on the gas that is expelled and bringing that to a stagnation volume should recover that work in the form of pressure and temperature, basically taking sonic flow out of the 2000 psi tank and stopping it in the 500 psi tank.

 

[SwinnyGG]

should recover that work in the form of pressure and temperature

It does - but if the only medium performing any work is the pressurized gas in the bottles, than energy balance dictates that the gas can't recompress itself to a temperature that is any higher than the original temperature of the gas in the bottles. Which were at whatever temperature they were sitting outside in that field. Probably 80 degrees or something- not some extreme temperature that would result in a huge increase in the speed of sound through that gas.

The pressure wave/expansion cannot exceed the speed of the molecules, no matter how high the pressure is.

That's correct- locally. If the mass of air is moving at any non-zero velocity, the velocity of the pressure wave moving through the mass, in the same direction as the movement of the mass, is moving at a velocity faster than the speed of sound through the mass. This is true of any mass with a sound wave moving through it, whether the mass is a gas or not.

 

[3DDave]

Where did the work in the 1500 PSI drop go? It cannot recompress itself to a higher temperature at the initial pressure, but I see no reason why it cannot result in a far higher temperature at a much lower stagnation pressure.

If no work is done on the pipe or the restriction, then the energy in the gas at the accumulator should still be very high; since that's not in the pressure of the gas, it has to be in the temperature. The exact temp depends on the ability of the tanks to exchange energy with the surroundings, but if the pressurization event is short then that exchange is negligible.

 

[SwinnyGG]

Where did the work in the 1500 PSI drop go?

That work is negative, and it went to reducing the temperature of the gas and plumbing.

 

[JoeFrickinFriday]

Expanding the gas cools the volume it expands from. How does it also cool what it expands to when that is a region that is seeing increasing pressure? After all, the pressure in the original tank did work on the gas that is expelled and bringing that to a stagnation volume should recover that work in the form of pressure and temperature, basically taking sonic flow out of the 2000 psi tank and stopping it in the 500 psi tank.

If it helps, think of it this way. Start with a 1-m^3 accumulator at 101 kPa, 20C, and a rack of high-pressure bottles at 13786 kPa, 20C. Imagine a frictionless piston at one end of the accumulator. Now admit gas from the bottle rack on one side of the piston. The piston moves, compressing the gas that was already in the accumulator, heating it. But the gas you're admitting from the bottle racks has undergone expansion and is extremely cold. When you're done, the accumulator has two bodies of gas in it, both at 3446 kPa, but each at vastly different temperatures. I ran through this with an adiabatic calculator (assuming N2), and here's what I got:

Initial accumulator pressure: 101.325 kPa
Initial accumulator temp: 20C
Initial volume of primary gas in accumulator: 1 m^3
mass of primary gas in accumulator: 1.164 kg

Final pressure in accumulator: 3446 kPa
Final temperature of primary gas in accumulator: 529.8C
Final volume of primary gas in accumulator: 0.08054 m^3

Bottle rack pressure: 13,786 kPa
Bottle rack temperature: 20C
initial volume of gas taken from bottles: 0.34154 m^3

Final volume of gas taken from bottle rack: 0.91946 m^3 (add final volume of primary gas, you get 1 m^3)
Final temperature of gas taken from bottle rack: -75.85C
mass of gas taken from bottle rack: 54.12 kg

So you've got about 54 kg of gas at -76C, and 1.2 kg of gas at 529.8C, at the same pressure. Now mentally dissolve the piston, and let the two bodies of gas mix; they'll eventually arrive at an aggregate temperature of -62.8C in the accumulator prior to firing.

 

[danschwind]

Yeap, initially the gas inside the vessel would heat up due to compression but after it has some time (few seconds IMO) to heat exchange with the incoming cold N2 the overall mixture would have a temperature lower than atmospheric.

The initial mass of gas inside the vessel is very very low compared to the incoming N2 in the final pressurized vessel also.

Daniel
Rio de Janeiro - Brazil

 

[SwinnyGG]

The temperature of the gas inside the vessel is also heavily dependent on the rate at which the pressure changes, and it changes relatively slowly. They fill the tank over a period of many minutes; all that time the tank wall is working hard to bring the contained gas to ambient temperature.

 

[3DDave]

It's extremely cold due to being expanded. And the gas in the original bottle is cold due to being expanded.

Where did the energy go?

It's a bit frustrating to know that that energy has just disappeared from existence.

Lots of results. No equations. I think that's the problem.

When 2000 PSI hydraulic fluid is dropped to 500 PSI through an orifice or restriction it gets incredibly hot. But not when it's another fluid?

I think the calculated drop in Nitrogen temperature in the low pressure side assumes that work is done on the control volume in expanding the gas. Instead, the Mach 1 gas is blasting out of the bottle through a shock wave and that kinetic energy is partly converted to heat and partly captured as an increasing pressure.

 

[danschwind]

3DDave, it is just a simple isenthalpic expansion of the N2. Enthalpy before and after the restriction is equal. After a while, the vessel itself and its surroundings will heat transfer with the gas so enthalpy will eventually change, but the drop in temperature can be computed easily at first through an insenthalpic flash.

Daniel
Rio de Janeiro - Brazil

 

[3DDave]

I think the problem is this: it is only isenthalpic if no work is done. But the first bit of gas allowed into the accumulator compresses (does work) onto the existing gas. Then the next bit compresses both the first bit and the existing gas. It's not throttled to where the gas escaping from the pressure bottles is simply allowed to leave at some pressure; it is constantly compressing the gas in the accumulator.

From

https://en.wikipedia.org/wiki/Joule–Thomson_effect#/media/File:Throttling_in_Ts_diagram_01.jpg

it looks like the largest drop is from 300K (80F) at 135 bar (2000psia) to 270K (26F) at 1 bar (14.5 psia), an average of 285K

So 1 volume at 26 F (273K) + 1 volume at 80 F (300K) compressed to fit into 1 volume ->

I think that's T2 = T1(P2/P1)^((k-1)/k)

which will be 285K*(2)^((1.4-1)/1.4) =347K.

Each additional 1 volume adds less compression, but the pressure and temperature should still rise.

And it makes sense - if the equilibrium pressure was 2000 psi, then there would be no additional compression and no additional temperature rise.