# Confusing Units after using Bernoulli's Equation

1. Jun 3, 2013

### Mezentio

So in a spacecraft with regular air at 1 atm pressure, a hole gets punched in it (depressurizing it). I'm looking for the Exit Velocity of the air at t=0.

Using Units:
Internal Pressure: Pi = 1 atm = 101300 [kg / m*s2]
Air Density: $\rho$ = 1.2 [kg / m3]

And Equation:
Pi = 0.5 * $\rho$ * (Ve2)
...turns into
Ve = √( 2*Pi / $\rho$ )

The answer I get is Ve = 410 [ m2 / s2 ]

How is Distance2 over Time2 a Velocity in this context?
I'm hoping to compare another object's relative wind-speed against this, and I'm not sure how.
Would a stationary object next to the escaping air feel like it's going 410 [m/s]?

P.S. Equation and Units were from: http://www.spaceacademy.net.au/flight/emg/spcdp.htm

Last edited: Jun 3, 2013
2. Jun 3, 2013

### rcgldr

So the square root in the equation didn't reduce m^2/s^2 to m/s?

3. Jun 3, 2013

### Mezentio

Oh wow I'm really that dumb. Thanks.

4. Jun 3, 2013

Your equation doesn't work, however. You are using an equation meant for incompressible flows for a flow that is clearly compressible.

5. Jun 3, 2013

### Mezentio

Hmm you're right - Wikipedia says the simple equation only works for gas at low speeds.

Although maybe I can get away with not having to go further into it - The big question I'm trying to answer is:

I'm assuming an object traveling Mach 0.5 towards the spacecraft, suddenly entering the spacecraft (pressurized at 1 atm, etc), shouldn't create a Sonic Boom.

But say the spacecraft opens the window right before the object gets there. Since the object's relative air-speed would be larger (because of the decompressing air rushing towards it), could the resulting escaping air cause a Sonic Boom as the object passes through the window?

Or would the lowered pressure cancel out the gained air-speed, making a Sonic Boom equally unlikely?

I can post this as a new thread if you like. Thanks again.

(I'm talking about Star Trek: Into the Darkness btw, where Kirk and
fly from ship-to-ship.)

6. Jun 3, 2013

Well considering that you are out in space, no, there would be no sonic boom out in space. Sound can't propagate in a vacuum (or near vacuum, as it were). In relation to the movie, the two were flying through space where Mach number has no meaning since there is no speed of sound. Once they entered the atmosphere in the ship there certainly could be provided that their speed of movement was faster than the speed of sound relative to the moving air.

The answer to your question originally is that, given the pressures involved (1 atm inside, 0 atm outside), the air will be moving at precisely Mach 1 at the exit of the hatch, meaning that the two would pretty much have to be moving supersonically right as they enter that air and there would almost certainly be a sonic boom of some kind. The only real question is how long they would be traveling that fast once the air resistance starts. A sonic boom can only be heard behind the advancing wavefront from the shocks coming off the moving object, so if they slowed fast enough the boom may not be heard by anyone inside the hold.

7. Jun 3, 2013

### Mezentio

Ok interesting, thanks.

I got interested in this because in the movie, the android calls out their distance twice - "1800 meters", (waits 1.3 seconds), "1600 meters." Which means that according to the movie, they're going exactly 154 m/s (344 mph), and decelerate to 0 m/s in 16 seconds with nothing but air and the seat of their pants to slow them down.

Ever seen a Motorcycle wreck? Ever see one going 344 mph? Not pretty