Question

Find the approximate value of velocity of best climb. Consider Thrust loading as 1.08, wing loading as 377 unit, CD0 as 0.025, K as 0.0075 and sea level density as 1.225 unit.

(a) 98.46

(b) 37.45

(c) 12.1

(d) 189.982

The question was asked in homework.

Question is taken from Flight Mechanics topic in division Performance and Flight Mechanics of Aircraft Design

Answer 1

Correct answer is (a) 98.46

For explanation: Given, Thrust loading T = 1.08, wing loading w=377, CD0=0.025, K = 0.0075, density d=1.225 unit

Velocity of best climb = \(\sqrt{\frac{w*[T+\sqrt{T^2+12CD0*K}]}{3*d*CD0}}\)

= \(\sqrt{\frac{377*[1.08+\sqrt{1.08^2+12*0.025*0.0075}]}{3*1.225*0.025}}\)

= \(\sqrt{4488.09*2.16}\) = 98.46 unit.