Calculations show that a certain airplane will produce 1332

Calculations show that a certain airplane will produce 1332 lb of drag at 450 mph at standard SL conditions. The airplane is equipped with an NACA 5868-9, variable pitch propeller of12-ft diameter with spinner. The engine is delivering 2000 horsepower (SHP) to the propeller at 1300 RPM and standard SL conditions. Calculate the advance ratio, J Determine the blade pitch angle (at 0.75R) & efficiency from the chart below. Is the airplane capable of attaining a speed of 450 mph? If so, what excess engine horsepower is available at this speed Efficiency curves for the NACA 5868-9 three-bladed propeller with spinner for various blade pitch angles at 0.75R.

Solution

a) Advance ratio = V / nD where V is the velocity of the aircraft, D - diameter of the propeller,

J = 450 x 1.46 x 60 / 1350 x 12 = 2.43

b) The blade pitch angle at 0.75 R is 50 deg with a max efficiency range 0.8 - 1

c) The propeller thrust eqn is given by,

F = 1/2 x rho x A x (Ve^2 - Vo^2)

considering the velocity of the flight to be 450 mph or 660 ft /s and rho as 1.9 slugs/ft^3

F = 0.5 x 660^2 x 1.9 x 3.14 x 6^2 = 413820 slugs- ft/s2 or 413 x 10^3 lbf

Clearly thrust tfrom the propeller is greater than the drag 1332 lbf produced by the aircraft at 450 mph.

d) Excess power available is the difference of shaft power and propeller power,

Propeller power = Thrust x velocity of flight = 413 x 10^3 x 660 = 272.58 x 10^6 lbf -ft / s

Excess power = 272.58 - 0.002 = 272.58 x 10^6 lbf -ft / s or 4542 x 10^3 lbf - ft / min