This problem deals with lift, drag, and how we can calculate power knowing both of those
quantities.
So we have a light air craft.
It has a wing area.
So our area equals 250 ft^2 and a lift coefficient of 0.40, a drag coefficient of 0.05 and giving
a velocity of 150 ft/s.
What we want to find is the lift and the drag, and the power required to maintain level flight.
The first thing we need to do is define our lift.
The lift is Cl times 1/2 times rho, u^2, over area.
So the only thing that we are missing there is rho, and it is the density.
Remember we have to use our English units.
So that is 2.38 times 10 to the minus 3*slugs/ft^2.
So now lets put them all together.
This is 0.4 times 0.5 times 2.38 times 10 to the minus 3rd Slugs/ft^2.
Multiplied by (150 ft/s)^2 and when we calculate all out what we end up ith is 2677.5 now lets
look at our units slugs*ft/s^2.
We want to change that to lbf, and the conversion is this.
So if you notice.
It is the exact same number, but we are butting it in lbf.
So this lift is also going to equal the weight of the aircraft.
Now lets figure out the drag.
So the drag is going to equal Cd times 1/2 times rho, times u^2, times A. So the only
thing that changes in this is our Cd.
So our drag is going to equal 0.05 times 0.5 times 2.38 times 10 to the minus 3rd.
I am not putting the units in here because there are exactly the same as before, and
when we figure out the drag, but what we come out with is 335 lbf, and we expect the drag
less then the lift.
So how do we get the power from that.
So the power is going to equal our 335 lbf times our velocity, which is 150 ft/s.
This equals 50200 lbf*m/s, and if we want to put this in a more recognizable form.
All we have to do is convert it to watts.
We do this by taking 50200 lb*m/s, and multiply it by 1.356 and we end up with 68100 W.