If you guestimate that the prop is 80% efficient, then with 1590 hp input (full throttle), then you get .8 x 1590 hp = 1272 hp output.
Note, the engine is geared, 3000 rpm translates into 1437 rpm based on this article.
http://www.airnews.co.za/home/index.php?option=com_content&view=article&id=340:flying-the-p-51d-mustang&catid=83:aircrafteviews&Itemid=67
The math here is simply based on power and density of air.
Let Vp = velocity of air at the prop
Let Ve = velocity of air downstream of prop (when it's pressure returns to ambient)
Let Mf = mass flow of air
Area swept by prop is (pi x ((11 ft 2 in) / 2)^2) = 97.935 ft^2
1272 hp = thrust (lbs) x Vp (ft / sec) / 550
thrust = 1272 x 550 / (Vp) = Mf * (V
e- V
0)
Volume flow = Vp (ft/sec) x 97.935 (ft^2)
(Using Ve doesn't work because the cross section is smaller than the prop swept
area and unknown).
Ignoring compression effects for mass flow:
Mf = volume flow (ft^3 / sec) * .002330 slug / ft^3 (for air temp around 65 degrees)
Mf = Vp x 97.935 * .002330 slug / sec
thrust = 1272 x 550 / Vp = Mf * (Ve - V
0)
thrust = 1272 x 550 / Vp = Vp x 97.935 * .002330 * Ve
For static thrust, V
0 = 0
Ve = 2 Vp
thrust = 1272 x 550 / Vp = Vp x 97.935 * .002330 * 2 * Vp
1272 x 550 = Vp^3 x 97.935 * .002330 * 2
Vp^3 = 1272 x 550 / (97.935 * .002330 * 2)
Vp = 115.3 ft / sec
static thrust = 6067.5 lbs
Ve = 230.6 ft / sec
The pressure delta across the prop disk for a check
delta p = thrust / area = thrust / (97.935 ft^2) = 61.95 lbs/ft^2
delta p = .5 density (V
e^2 + V
0^2) = 61.95 lbs/ft^2
delta p = .43 psi
I'm not sure of this calculation. I don't know the efficiency in a static situation. One prop calculator shows 3200 lbs of static thrust for a 134 inch diameter, 4 bladed prop at 1435 rpm, which would indicate a much lower efficiency. Another calculator shows 3350 lbs of force.
For thrust = 3300 lbs:
delta p = thrust / area = 3300 / 97.935 = 33.696 lb/ft^2 = .5 x .002330 x Ve^2
Ve = 170 ft / sec
Vp = 85 ft / sec
Output power = 510 hp, effciency = 32.1%
Note this agrees with the math shown here for the static thrust situation:
prop_efficiency.htm
To use the units from above, for power, the unit is 1 ft lb / sec
1 hp = 550 ft lb / sec
Plugging this into the static thrust equation from the web site
FOM = 32.0884821%
Pavail = 1590 hp
Power output = FOM x Pavail = 510.2068654 hp
density = .002330 slug / ft^3
area = 97.935 ft^2
.320884821 x 1590 hp = 510.2068654 hp = 280613.776 lb ft / sec
Thrust = (280613.776 / ((1/2) x (1/(.002330x97.935))^(1/2)))^(2/3)
Thrust = (280613.776 / (((1/2) x (1/(.22818855)))^(1/2)))^(2/3)
Thrust = (280613.776 / (((1/2) x (4.382340832))^(1/2)))^(2/3)
Thrust = (280613.776 / ((2.191170416)^(1/2)))^(2/3)
Thrust = (280613.776 / (1.480260253))^(2/3)
Thrust = (189570.5674)^(2/3)
Thrust = 3300 lbs
Vp = 85.03447757 ft / sec
For a given airspeed, air density, and prop efficiency, you'd have to do similar math
output power = 1590 hp x efficiency
Ve = 2Vp - V
0
Mf = Vp (ft / sec) x density_of_air (slug / ft ^3) x 97.935 ft^2
Here is another propeller math oriented link:
http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node86.html
Another thing to note is that assuming density doesn't change, then the cross sectional area of the air affected by a propeller decreases inversely with speed. In a static thrust situation, the cross sectional area or the prop wash at the "exit" is 1/2 the cross sectional area at the prop. This creates a paradox in this simplified math, because at zero speed in front of the prop, the cross sectional area would be infinite. Look at figure 11.25 in section 11.7.2 and note what would happen if V
0 = 0.