Question

An aircraft has cruise range of 1500NM and SFC C = 0.5 1/hr. Cruise velocity of jet aircraft is 570ft/s with L/D as 13.2. Aircraft is supposed to do loiter of 2 hrs then what should be the gross weight of aircraft? Given, empty weight fraction of 0.55, crew and payload OF 6500kg with 0.98 as mission segment fuel fraction for climb, take-off and landing.

(a) 37000 kg

(b) 37.8 ton

(c) 37880lb

(d) 3780 kg

I had been asked this question during a job interview.

This interesting question is from Conceptual Sketch Sizing in portion Conceptual Sketch Sizing of Aircraft Design

Answer 1

The correct answer is (b) 37.8 ton

To explain: Given, a jet aircraft, crew and payload = 6500 kg.

Since, mission profiles are given we need to find mission segment fuel weight fraction for each.

For take-off, climb and landing it is given as 0.98.

For cruise,

Range R 1500 nm= 9114000ft, C =0.5 1/hr. =0.0001389 1/s, L/D =13.2, velocity V=570 ft/s

Now cruise weight fraction is given by Range formula. For, prop-driven aircraft it is given by,

Wcruise / W0 = \(e^{\left(-\frac{R*C}{V*\frac{L}{D}}\right)}\) = \(e^{(-\frac{9114000*0.0001389}{570*13.2})}\) = 0.8451

Loiter: loiter of 2 hour is given hence,

Endurance E = 2*3600 = 7200 s

Now, loiter fuel fraction

Wloiter / W0 = \(e^{\left(-\frac{E*C}{\frac{L}{D}}\right)}\) = \(e^{(-\frac{7200*0.0001389}{13.2})}\) = 0.9270

Now from fuel fraction method,

Wx / W0 = multiplication of each phase fraction

= 0.98*0.98*0.8451*0.9270*0.98

= 0.7373

Now,

Wf / W0 = 1.06 * [1 – (Wx / W0)] = 1.06 * [1-0.7373] = 0.2784

Empty weight fraction We / W0 = 0.55

Gross weight of aircraft is,

W0 = Wcrew + Wpayload + Wfuel + Wempty

By re-arranging,

W0 = Wcrew + Wpayload / [1 – (Wf / W0) – (We / W0)]

= 6500 / [1-0.2784-0.55] = 37880 kg = 37.8