Incorrect answer for the drag power?

xuhq · Oct 26, 2010

A 150 lbm cartop carrier measures 51 in long by 35 in wide by 18 in high. If a 3000 lbm car is driven at 65 mph into a 10 mph headwind, the added net motor power required when the carrier is used is most nearly

a. 5.9 hp

b. 13 hp

c. 15 hp

d. 36 hp

Cd=1.18,

Fd=Cd*A*rou*v^2=1.18*(35/12)*(18/12)*0.075*((65+10)*5280/3600)^2/2/32.2=72.75lbf

Power=Fd*v=72.75*65*5280/3600/550=12.6 hp

Therefore, the answer should be b. However, the answer given in the book used the relative speed (65+10) for the v,

not the absolute speed, which I believe is incorrect.

If you redo and let the car stop, the headwind is still there that will give you nonzero power. That is a simple way to prove that the answer given in the book is incorrect.

Do I miss anything?

navyasw02 · Oct 26, 2010

What answer was given in the book? I worked the same problem and used the 65+10 to account for the headwind.

Capt Worley PE · Oct 26, 2010

The headwind does increase the drag.

Sschell · Oct 26, 2010

xuhq said:
If you redo and let the car stop, the headwind is still there that will give you nonzero power. ~~That is a simple way to prove that the answer given in the book is incorrect.~~
Do I miss anything?

If there is nothing to resist this force, the car will move backwards...

Relvinim · Oct 26, 2010

Why wouldn't you account for the headwind? Try riding a bike in a headwind...your heartrate will increase dramatically and so will your power output.

xuhq · Oct 26, 2010

We need to account for the headwind to calculate the drag force (as shown in my first post). However, I don't think you should account for it for the speed you use to calculate power. A stopped car will not need to do any work against the windhead to keep stopped.

Power=F*V, where V should be absolute velocity.

Relvinim said:
Why wouldn't you account for the headwind? Try riding a bike in a headwind...your heartrate will increase dramatically and so will your power output.