Not for License Exam Fuel Consumption

Ahh, sure. With the understanding that this is purely academicif the Exams ask you use the equation KC listed where Con1=Cons2 then use that.

But for the realistic version using the first formula con2/con1= (v2/v1)^3 :

  1. You already had the information that 17.5kt burns 274bbl/d. Using that and the first formula you can solve for the consumption at 17kt. This comes to 251bbl/d
  2. You said you’re going 775nm at 17kt. So 17kt = 17nm/hr. 775nm divided by 17nm/hr = 45.6 hr. 45.6 hr divided by 24hr/day = 1.89 days
  3. 251bbl/day multiplied by 1.89 days = 477 bbl