Is this equation for propeller thrust correct?

Question

From dimensional considerations, and assuming that the power $P$ applied to a propeller of diameter $D$ moving at an axial speed $V$ is 100% converted to the acceleration of a mass of air with density $rho$, I have arrived to the following expression for the thrust $F$:

$F=k(P·rho·V)^{1/2}·D$

where $k$ is a non-dimensional constant.

Is that expression basically correct? I'm now editing to include below the derivation, a bit long...

We have a propeller with a diameter $D$, absorbing a power $P$ when moving at an axial velocity $V$ in air of density $\rho$. Let’s assume that there exists a function f such that the thrust $F$ of the propeller is:

$F = f(P,D,V,\rho)$

The variables are thrust $F$, dimensions $MLT^{–2}$; Power $P$, dimensions $ML^2T^{–3}$; prop diameter $D$, dimensions $L$ and air density $\rho$, dimensions $ML^{–3}$

Five variables are too much. The system can’t be solved unless we have four, one dependent and tree independent… So, in place of $V$ and $D$, we take the volume $W$ swept by the spinning propeller in the unit of time, dimensions $L^{3}T^{–1}$

$W = π/4 · D^{2} · V$

There is an non-dimensional constant $k$ such that:

$k = F^a\cdot P^b\cdot \rho^c \cdot W^d$ where $a,b,c,d$ are numbers to be determined.

Switching to the dimensions:

$M^0 L^0 T^0 = (MLT^{–2})^a (ML^2T^{–3})^b (ML^{–3})^c (L^3 T^{–1})^d$

Then, the system is:

$0 = a + b + c\\ 0 = a + 2b –3c + 3d \\ 0 = –2a –3b –d$

$F$ is the dependent variable, so we make a=1

Solving the system:

$b = –1/2 \\ d = –1/2 \\ c = –1/2$

Inserting in $k = F^a\cdot P^b\cdot \rho^c \cdot W^d$ where $a,b,c,d$ the values of the exponents, and solving for $F$,

$F = k\cdot P^{1/2}\cdot \rho^{1/2} \cdot W^{1/2}$

But $W = π/4 · D^{2} · V$

Now $k$ may absorb the constant π/4, and we get:

$F = k\cdot P^{1/2}\cdot \rho^{1/2} \cdot V^{1/2}\cdot D$

Answer 1

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.