# Dart Energy from Compressed Air

1. May 22, 2012

### Billwillett

Hi Everyone,

I am trying to calculate the exit velocity of a dart which is propelled by compressed air. Please see the attached jpeg for diagram and my work so far.

My final formula appears to be incorrect. I'd appreciate any help figuring out what is wrong.

A pressure tank of Volume Vt is has a tube with a valve at the end. It is pressurized. A dart slides over the tube. When the valve opens, pressure acts on the dart and it begins to move off the tube, measured as distance x. Once the dart exits the tube, there is no longer any force applied.

What is the energy of the dart at this point? From this I can calculate its velocity.

I chose to integrate the instantaneous energy as the dart moves from x = 0 to x = L when the dart leaves the tube. See attached. I basically integrate Fdx where F varies as a function of X. This gives me total energy expended to move the dart.

The result is E = initial pressure x initial volume x ln(1+Area of dart*length of dart/initial volume).

Peering into this formula I notice that initial pressure x initial volume equates to the initial energy stored in the tank. Also, (area of dart x length of dart)/initial volume = the ratio of dart volume to tank volume.

So E of the dart = stored energy * ln(1+ratio of the volumes). (see attachement for more readable form of formulas)

If the dart volume is extremely small and the tank large, then the ln equates to zero. There is no energy transferred to dart. That makes sense.

It seems the best case is when the dart volume is huge. If big enough nearly all the energy stored in the tank is applied to dart. Less is wasted as the tank vents to atmospheric when the dart exits.

But this is where I think the formula fails. If we chose a large dart volume relative to tank volume then the log evaluates to a number larger than 1! That would mean I am getting more energy out of the tank then the initial stored energy of PtVt.

What am I missing?

#### Attached Files:

• ###### dartphysics.jpg
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Last edited: May 22, 2012
2. May 23, 2012

### haruspex

You've taken pressure as inversely proportional to volume. That's only true at constant temperature (isothermal expansion). What you have here is adiabatic expansion. There's no time for heat to flow. The temperature will drop because of the work done in pushing the dart. Since you've not allowed for that work done, it isn't surprising you got too much energy out in some scenarios.