Energy cost of air displacement? 1

[ChrisGreaves]

New here. Aeronautics because I am curious about the cost of airflow around an object.
In my case a Toyota Rav4 driving down the highway; calm weather, dry etc. Just the Rav 4.
I first proposed this discussion on Eileen's Lounge, but a user suggested I check out at this site.

The Rav4 is five feet wide and five feet high in cross-section. Assume that there is no space between the road and the underside of the SUV. It moves down the road like a block of wood, 25 square feet in cross-section, pushing air out of the way.
If I assume that it pushes air five feet vertically, I calculate 4,500 foot pounds per second of work; that is lifting 60 square inches of air vertically by five feet, assuming 15 psi. At 60mph=88 fps= about 1,000 inches/second, this has left me contemplating that the car engine must be doing 4,500,000 foot-pounds of work each second.

But one horsepower is defined as 550 ft/lbs/second, which suggests that the car in my simple model is an 8,182 HP Toyota Rav4.

The Toyota RAV4 is 203 HP.
Please and Thank you, where is my error in thinking?

I can see that air might be pushed to the side rather than vertically, but then that air must do work to push other air to the side.. For me the bottom line is that where there was air occupying a 5'x5' cross section, there is now no air in the cross section - just steel, plastic, rubber etc.

I am trying to determine how much work is done just to make a car occupy air-space on a 3.5 hour trip along the highway.
Thanks for any advice on the calculation, also any tips about a more appropriate forum.
Cheers, Chris

 

[MintJulep]

The force needed to move your car through the air is given by the drag equation.

Work is force x distance.

 

[btrueblood]

"basic physics classes in high school said that while the kinetic energy of a particle is given by 1/2 m v^2, the kinetic energy of a fluid is 1/2 m v^3."

The part in red is not correct.

As Mint and others have said, force is proportional to v², Work is force x distance, power is (work/time) and thus power is proportional to v³, which is probably what your physics teacher was saying, and which you misheard or misinterpreted.

 

[rb1957]

"my basic physics classes in high school said that while the kinetic energy of a particle is given by 1/2 m v^2, the kinetic energy of a fluid is 1/2 m v^3."

1) I think you misunderstood the lesson. Energy is energy, it has units of Joules or lbf*ft or whatever. if you "suddenly" add a v (going from v^2 to v^3) you've changed the units and changed whatever it is you've calculated. Now it is just possible (though I don't know how or why) that the mass of a fluid was changed from kg or lbs to something else which off-sets the added velocity. This is possible 'cause we say the mass of the particle is this ... we can see the particle (or cannonball or bag of sugar); but "fluid" particles are harder to determine, to see and feel (though mass of fluid travelling along a pipe is easier to visualize). I don't think you can find any textbook to show this equation, unless possibly a typo.

2) If you were indeed taught this, then wow ! I wonder what else you were taught ??

3) Recognise that your thought experiment is flawed, and I'm not surprised that the folks at "Eileen's Lounge" couldn't see that