Is this equation for propeller thrust correct?
From the comments:what I want is a derivation of thrust as a function of power absorbed, prop diameter and prop axial velocity..


Power absorbed $P$ is engine torque  times prop angular velocity $\omega$: $P =  \cdot \omega \tag{1}$. 
Prop axial velocity at the tip $V_t =\omega \cdot D \tag{2}$


$P$ is the power applied in the axial direction, while thrust is generated in the direction perpendicular to the propeller plane - there must be some function describing the relationship between the forces in both directions. This function would be one describing efficiency and losses, since the primary force application is the engine torque and the thrust is the net resulting force.

If the goal is like is stated in OP ...propeller of diameter  moving at an axial speed  is 100% converted to the acceleration of a mass of air... this means that there are no losses, and we can use the simple impulse theory of the magic disk. This is comprehensively explained in this post, which results in the following equation between thrust and power:

$$P = \sqrt{\frac{F^3}{2\rho A}} \tag{3}$$

If we want $V_t$ in there we need to substitute (1) and (2):

$$\frac { \cdot V_t}{D} = \sqrt{\frac{F^3}{2\rho A}} \tag{4}$$

Since A = propeller disk area = $\frac {\pi}{4} \cdot D^2$, (4) further boils down to:

$$F^{3/2} = k \cdot  \cdot V_t \cdot \rho^{1/2} \tag{5}$$ 

Note that (3) contains $P$ but no $V_t$, while (5) contains $V_t$ but no P. We could lump the motor torque into the force $F$ dimension, but all my engineering genes scream to not do that because of the origin and effect relationship: no engine torque, no net thrust. 

Both (3) and (5) contain $\rho$, which is a must considering that the thrust is obtained by acceleration of a mass of air. The mass flow dimension can disappear into a force dimension as well when multiplied by a velocity, for instance a propeller tip velocity...

This answer shows the relationship between thrust and power of a helicopter rotor in the hover (and therefore of a propeller) using modified momentum theory with losses:

$$C_P = \frac{\kappa \cdot {C_T}^{3/2}}{\sqrt{2}} + \frac{\sigma \cdot {C_D}_0}{8}$$

with $\kappa$ = loss factor and $\sigma$ = blade solidity. So the exponent of the function (induced power) vs. (thrust) is 3/2. Also, there is an additional term that the dimensional analysis cannot catch, which is power required in the propeller plane.