- #1
Lyphta
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Homework Statement
A 3250 km aircraft takes 12.5 minutes to achieve its cruising altitude of 10 km and cruising speed of 850 km/h. If the plane's engines deliver, on the average, 1500 hp of power during this time, what is the efficiency of the engines?
Homework Equations
Efficiency = Power output / Power input x 100
P = W/t --> P= mgh/t, P=Fdcos(theta)/t
The Attempt at a Solution
m = 3250 kg
t = 12.5 min --> 750 sec
h = 10 km --> 10000 m
Pin = 1500 hp --> 1119000 Watts
v = 850 km/h --> 236.11 m/s
vf = at +vi (attempt to find the velocity components but don't add up right)
236.11 = a (750) + 0
a = .315 m/s^2
average v = vf/2
= 236.11/2
average v = 118 m/s
Poutx = F(average v) = ma(average v)
= 3250(.315)(118)
= 120802.5 W
Piny = Fd/t = mgh/t
= 3250(9.81)(10000)/750
= 425100 W
The real answer is 4.091 x 10^8 J ; e = 48.7% but I don't know how to get that answer...