Here is the model I wish to entertain:
Isolate the control volume:
So we have the force of drag acting downward ##F_D##, Weight of air in control volume ##W_{cv}##, uniformly distributed (approximately atmospheric) ##P_{atm}## acting across the inlet and outlet cross sectional areas. The immediate goal is to find the drag force. Also velocities are assumed uniformly distributed across the inlet/outlet. Begin with (1):
$$ \sum \mathbf{F} = \frac{d}{dt} \int_{cv} \rho \mathbf{v} ~ dV\llap{-} + \int_{cs} \rho \mathbf{v} \left( \mathbf{V} \cdot d \mathbf{A} \right) \tag{1}$$
$$ -F_D - W_{cv} -P_{atm} A(y) + P_{atm}A_i = \frac{d}{dt} \int_{cv} \rho \mathbf{v} ~ dV\llap{-} + \int_{cs} \rho \mathbf{v} \left( \mathbf{V} \cdot d \mathbf{A} \right) $$
I neglect the difference in the pressure terms and the weight of the control volume. I am left with:
$$ -F_D = \frac{d}{dt} \int_{cv} \rho \mathbf{v} ~ dV\llap{-} + \int_{cs} \rho \mathbf{v} \left( \mathbf{V} \cdot d \mathbf{A} \right) $$
On to the momentum side of the equation:
Since the mass inside the control volume is taken to be very small, I expect the rate of momentum accumulation within the control volume to be small, hence:
$$ -F_D = \cancel{\frac{d}{dt} \int_{cv} \rho \mathbf{v} ~ dV\llap{-}}^{\approx 0} + \int_{cs} \rho \mathbf{v} \left( \mathbf{V} \cdot d \mathbf{A} \right) $$
$$ -F_D = \int_{cs} \rho \mathbf{v} \left( \mathbf{V} \cdot d \mathbf{A} \right) $$
Next, with the uniformly distributed velocities we evaluate the integrals over the control surfaces:
$$ -F_D = \int_{outlet} \rho \mathbf{v}^2(y) d A(y) - \int_{inlet} \rho \mathbf{v}^2 d A_i $$
Applying continuity ##Q = v_{inlet} A_{inlet} = v(y) A(y)##:
$$-F_D = \rho Q^2 \left( \frac{1}{A(y)} - \frac{1}{A_i} \right)$$
$$ \implies F_D = \rho Q^2 \left( \frac{1}{A_i} - \frac{1}{A(y)} \right) $$
The area of the annulus is given by:
$$ A(y) = \pi ( ky + r_i)^2 - A_p $$
Now apply this to drag force to the puck:
$$ F_D - W = m \ddot y $$
$$ \rho Q^2 \left( \frac{1}{A_i} - \frac{1}{A(y)} \right) - W = m\ddot y $$
$$ \rho Q^2 \left( \frac{1}{A_i} - \frac{1}{\pi ( ky + r_i)^2 - A_p} \right) - W = m\ddot y $$
From here you get the steady state solution for ##y## by setting ##\ddot y = 0 ## and solving.
Now, what seems to be sensible (to me) is this model for the expanding flow area as energy is robbed from it (flow work from drag) in the form of heat.
How can a sensible ##k## be found from first principles seems to be the remaining thorn.
Thoughts?
EDITS: corrected some of the mathematics pertaining to the evaluation of the control surface integrals as well as some careless subscript usage.
