I'd enjoy seeing that.
Ok, here goes...
We start with Newton's law:
Essentially, the propeller accelerates a mass of air, and in that equation m will be the mass of air and a will be the acceleration of the air and then F will be the resulting force, that is the thrust.
So, what we do is to consider what happens in one second. Firstly, we need to calculate how much air will pass through the prop, and then multiply that by the density of the air so that we have the mass. Then we multiply that by the acceleration of the air. We do that by assuming that the air will form a circular shaft, with the same diameter as the prop. We will use letters from the Greek alphabet and typically a change in any quantity is normally preceded by the Greek letter delta, so delta v will be the change in velocity. We use the Greek letter Rho for the density of the air.
So now we can substitute into #1
That is just the mass airflow through the prop per second, multiplied by the change in velocity during that second. The problem is that we have the term delta v in there and later we will want to eliminate it from the equation, so we need some more relationships.
Now we need to introduce the term for efficiency and we use the Greek letter eta, and efficiency is just output power divided by the input power and the output power is the thrust times the velocity and the input power is the power of the engine, and we express that like this:
rearranging that we get:
Also, the power absorbed by the propeller can be expressed as the thrust times the average velocity through it, like this:
If we equate #4 and #5 and then solve for delta v we get:
Now if we solve #5 for the thrust we get:
Now we can equate that to #2 to get:
Now we can substitute the value of delta v from #7 to get:
That looks complicated, but we can simplify it to this:
and we can rearrange that like this...
That is one of the standard forms of a cubic equation, and you notice that two of the terms in there appear to be the same. So we can simplify this even more by letting:
Then #16 is just:
That's an amazingly simple result for the efficiency of the prop, and it can be solved. All we need to do is evaluate a in #17 and substitute it into the standard solution shown below:
Remember, that's going to give us the maximum theoretically possible efficiency, but there will be losses, so one practical solution recommended by Perkins and Hage is to multiply the result by 0.8587. We will do that here for simplicity, but it is possible to estimate the losses and include them in the calculations.
So, let's do an example using the data from earlier in the thread and calculate the efficiency of the F6F prop at 96mph.
Then we have:
D = 13.08
Rho = 0.0024
P = 2000 x 550
v = 140.8
So now we calculate the value of a:
Now we substitute that into the solution given above, evaluate it and apply the reduction factor recommended by Perkins and Hage as follows:
So we have a prop with 64% efficiency, slightly more than my earlier guess.
Hope that helps...
Badboy
