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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/kgI 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 |
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Answer» 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. |
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