Well, I need to justify it if I'm going to ignore it. As far as I can tell, the losses are pretty significant. First off, air is technically compressible, but at velocities slower than Mach 0.3, I don't think we need to consider that. Technically it could be travelling at Mach 0.37 through the 2" saddle, but I wouldn't expect that to throw off my number by anything significant.
Minor Headloss = K*V^2/(2g).
The volumetric flow of water leaving the pipe (and thus the volumetric flow of air that must enter) is 1300gpm (about 2.9cfs).
The air is therefore travelling through the 3" line at 59.0ft/s.
K for a sudden contraction equals 0.5(1-β^2)*sqrt(sin(θ/2))/β^4, where β = 3" diameter / 2" diameter; K = 1.41. (θ here is 180 degrees; see source)
Therefore the headlosses equal 1.41*(59ft/s)^2/(64.4ft/s^2) = 76 ft of head = 33 psi...
Since the air starts at ambient pressure of 14.7psi, this tells me we'd end up with a vacuum. This isn't obviously something I'm very familiar with, but any guidance would be very much appreciated; thanks!
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