As I make this control volume bigger, I might expect to capture all of the momentum imparted by the airplane. Once I'm satisfied that I have captured all of the momentum, I sum it up and find that the rate of momentum addition is equal in magnitude to the airplane lift.
This is not true in general. The wing is not the only force which imparts momentum to the air in the volume --- there are also the pressure distributions on the top and bottom planes of the volume. These can be almost anything, depending on the planes' fore/aft extent, left/right extent, and distance below/above the wing or 2-D airfoil. Below are some particular cases. The top and bottom planes are not quite flat, but are chosen to lie along streamlines, so that the inflow and outflow momentum change delta(M) occurs only through the vertical planes.
2-D airfoil, volume is infinitely long fore/aft, finite height.
Vertical velocities at inflow and outflow are zero, so we expect no vertical momentum change.
F2 = Lift -> delta(M) = 0 (verified)
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2-D airfoil, volume is finite fore/aft, height is infinite.
Pressures forces at top and bottom are equal and opposite.
F2 = 0 -> delta(M) = -Lift
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2-D airfoil, volume is a square, very large in extent. A detailed analysis reveals....
Half of the wing's downward lift force is canceled by forces on top and bottom planes.
The remaining force imbalance of -Lift/2 produces a nonzero delta(M)
F2 = Lift/2 -> delta(M) = -Lift/2
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One difference between the real world and one in which you can imagine arbitrarily large control volumes is that real world flows have viscosity. With viscosity included, I would expect to be able to truly capture all of the momentum imparted by an airplane inside a finite control volume. What is the rate of momentum transfer in this case?
No. Viscosity doesn't change the picture at all. The only effect it has is that F1 and F2 might also include viscous shear forces in addition to the pressure forces.
One other question about your reply. Your case a) with a cubic control volume would have a square Trefftz plane. You suggested that delta(M) would be zero in this case. I get something like half the lift.
Correct for the 2-D case (see 3rd case above). In 3-D, the momentum imabalance goes to zero.
An airplane cruising at 30000 ft has an overpressure "footprint" on the ground covering many square miles. This overpressure integrated over the whole footprint is equal to the lift on the airplane. So the ground pushes up on the atmosphere with a force equal and opposite to the airplane's downward force. The net force iz zero, and so the net momentum change in the atmosphere is also zero.
F1 + F2 = -Lift+Lift = 0 -> delta(M) = 0