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Supersonic ping pong ball cannon

by voko
Tags: ball, cannon, ping, pong, supersonic
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Feb13-13, 05:31 AM
P: 5,794


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?
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