4
$\begingroup$

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:

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

$\endgroup$
  • $\begingroup$ this seems related, and Peter's answer seem to contain what you need: aviation.stackexchange.com/questions/8819/… $\endgroup$ – Federico Mar 7 '18 at 19:19
  • $\begingroup$ How did you calculate the exponents? I can identify four unknown exponents, but only three dimesional equations. $\endgroup$ – Gypaets Mar 7 '18 at 20:12
  • $\begingroup$ @Federico In the –quite likely– case that Kämpf's answer is right, the expression I've derived is probably wrong... $\endgroup$ – xxavier Mar 7 '18 at 20:15
  • 1
    $\begingroup$ @xxavier. Thrust is caused by a change in momentum, or/and an increase in static pressure, acting over an area. So, for a gas turbine, Net thrust (FN) = M x (Ve - Va) / g + Ae x (Pse - Pamb). Va is the aircraft velocity, Ve is the exhaust velocity. The physics is the same for a prop, but when the flow is subsonic, Pse = Pamb. So, you are left with the first term, and your velocity term should be a change in velocity, of the mass flow going through the prop. This confers with the general form of Peter's answer. $\endgroup$ – Penguin Mar 8 '18 at 10:38
  • 1
    $\begingroup$ @xxavier. Fully understand the dimensional analysis approach. But you can use DA and an understanding of the physics to arrive an equation which is dimensional correct and physically correct. The two are not mutually exclusive. And... your question is asking if your equation is correct, so I am answering that by showing one of the errors. $\endgroup$ – Penguin Mar 9 '18 at 12:08
0
$\begingroup$

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 𝕋 times prop angular velocity $\omega$: $P = 𝕋 \cdot \omega \tag{1}$.
  • 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 𝐷 moving at an axial speed 𝑉 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):

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

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

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

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.

$\endgroup$
  • $\begingroup$ I see a problem: Equation (3) is derived from momentum considerations for the case of static thrust. that's an special case, since, in that condition, the angles of attack 'seen' by the blades are much greater than the angle 'seen' by the blades when there is a significant axial velocity of the prop disk... $\endgroup$ – xxavier Sep 1 at 11:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.