A propeller driven airplane is flying at constant speed in a straight line at a

Question

A propeller driven airplane is flying at constant speed in a straight line at a constant altitude. The air-propeller interaction produces a thrust that moves the plane forward. The air-fuselage interaction produces a drag force that acts to slow the plane. The air-wing interaction produces a lift force and the airplane-earth interaction produces a gravitational force Derive an expression for the magnitude of the thrust required for straight and level flight at constant speed vS. Answer in terms of CD, CL, M, g. Discuss your solution as CD > CL. and as CD

Explanation / Answer

a) For a straight and level flight with constant speed vs, FT = FD and FL = Fg

FL = Fg implies CLvs2 = Mg which gives vs2 = Mg/CL

FT = FD implies FT = CDvs2 = CD(Mg/CL)

Hence, FT = MgCD/CL

For CD > CL, FT > Mg

For CD < CL, FT < Mg

b) Along a straight line, v = (Mg/CL)1/2

So, tL = L/v = L(CL/Mg)1/2

We see that tL is inversely proportional to the square root of the weight of the plane. So, if plane weighs less, it will take more time to travel the same distance.

c) For banked turn,

FLcos = Mg

FLsin = MvB2/R

which gives banking speed, vB = (gRtan)1/2 = (gtanL/2)1/2

Now, since we have constant banking speed, |FTbanked| = |FD|

So, |FTbanked| = CDvB2 = CD[(gtanL/2)1/2]2 = FTLtan/2MCL

d) tB = 2R/vB = 2(L/2)/(gtanL/2)1/2 = (2L/gtan)1/2

Now, L = tL(Mg/CL)1/2

Substituting L, we get,

tB = [2tL/tan(M/gCL)1/2]1/2

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