\"A contributing factor to an airplane\'s duel consumption is the bypass ratio o
ID: 3307809 • Letter: #
Question
"A contributing factor to an airplane's duel consumption is the bypass ratio of the engine system. The bypass ratio is the amount of air passing around the engine core relative to the amount of air passing through the core. An airplane manufacturer is designing a new airplane and wants to determine the bypass ratio for the airplane's engine system. The airplane will fly 3,200 hours per year and will average 450 miles per hour. The amount of fuel that the airplane consumes can be expressed as:
z = 0.0521 - (7.64*10^-4) * y
for 4 < y < 12
where y is the bypass ratio (a unitless number) and z is the number of gallons of fuel consumed per mile flown by the airplane. The cost of fuel remains constant at $4.21 per gallon.
The initial cost of the engine system as a function of the bypass ratio is $257,000 + $2,800y^2.
The engine system will be used for 12 years. At the end of 12 years, the salvage value of the engine system as a function of bypass ratio is $12,000y. The airplane manufacturer wants to minimize the annual equivalent cost (AEC) of the engine system (which includes the initial cost, the annual cost of fuel, and the salvage value). The manufacturer's MARR is 14.4%. What is the optimal bypass ratio rounded to the nearest tenth that minimizes the AEC of the engine system?
(The optimal answer for the bypass ratio is between 4 to 12, but it should not be necessary to consider that constraint in your calculations.)"
Explanation / Answer
The total cost (AEC) = Initial cost + Annual cost of fuel +Salvage value
AEC= $257,000 + $2,800y^2.+ $12,000y+3200*450*z*$4.21
where z = 0.0521 - (7.64*10^-4) * y and y is the bypass ratio Airplave covers.
We can use calculus to find out the optimal ratio y from the above equation
for the AEC to be minimum the first derivative is taken and equated to zero
The first and second derivatives can also be used to look for maximum and minimum points of a function.
Hence we get
dAEC/dy =2*2800*y+12000-3200*450*7.64*10^-4*4.21 *12= 0
Therefore, Y= 43580.083/5600 = 7.78
By subsitituting this value in the second derivative we find whether the cost is less or more.
Hence the optimimal bypass ratio is 7.78 for which the AEC is minimum.
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