Question

What is the value of specific climb when the value of \(\frac{dH}{dt}\)=1 and the fuel flow is 833.33 kg?

(a) 0.0013ft/kg

(b) 0.0012ft/kg

(c) 0.0019ft/kg

(d) 0.002ft/kg

I have been asked this question in a job interview.

I would like to ask this question from Minimum Fuel Climb topic in division Climb and Descent Performance of Aircraft Performance

Answer 1

Correct answer is (b) 0.0012ft/kg

To elaborate: The answer is 0.0012ft/kg. The formula for specific climb is given by SC=\(\frac{dH/dt}{Q_f}\). Given, \(\frac{dH}{dt}\)=1 and Qf=833.33 kg. Substituting the values we get SC=\(\frac{1}{833.33}\).

We get SC=0.0012ft/kg.