View Single Post
voko
voko is offline
#1
Feb13-13, 05:31 AM
Thanks
P: 5,535
Video: http://www.youtube.com/watch?v=I9zBGgpzl0I

Description: http://arxiv.org/ftp/arxiv/papers/1301/1301.5188.pdf

So I though I would do a back-of-envelope check if that was even possible. I decided to model that as adiabatic expansion from the pressure plenum into the barrel, ignoring any effect the nozzle could make.

So ##p_i V_i^{\gamma} = p_f V_f^{\gamma}##, yielding ## p_f = p_i (\frac {V_i} {V_f})^{\gamma} ##. Then, the internal energy is ## U = \hat{c}_V n R T = \hat{c}_V p V ##, so ## U_i - U_f = \hat{c}_V (p_i V_i - p_f V_f) = \hat{c}_V (p_i V_i - p_i (\frac {V_i} {V_f})^{\gamma} V_f) = \hat{c}_V p_i V_i (1 - (\frac {V_i} {V_f})^{\gamma - 1}) ##.

From the dimensions given I computed ## V_i = 0.0042 m^3, \ V_f = (0.0042 + 0.0028) = 0.007 m^3 ##; two pressure values were mentioned, so I took ## p_i = 500 kPa ##. I assumed air is a diatomic ideal gas, so ## \gamma = 7/5, \hat{c}_V = 5/2 ##, so the final result is ## U_i - U_f = 970 J ##.

Then the mass of the ping pong ball is 2.7 g, the reported speed was 406.4 m/s, thus its kinetic energy was 223 J, which is quite a bit less than the diff in the air's internal energy, which makes the whole thing plausible - at least in this model.

Now, my question is, is there any obvious problem with the model and, perhaps, the computation? How would one model the effect of the nozzle?
Phys.Org News Partner Physics news on Phys.org
A 'quantum leap' in encryption technology
Using antineutrinos to monitor nuclear reactors
Bake your own droplet lens