^ drag force on an object on an automobile

Hi,

Suppose there is a block 1 m by 1m by 1m sitting on top of an automobile while driving. Is there a way to estimate the additional drag force and hence the increased fuel consumption due to this block. I will appreciate any help.

Thanks.

Reply to
aguy
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A lot of the answer depends on how fast you plan on driving, and on what sort of driving you'll be doing. Here's a shot at an estimate though:

Most cars have a drag coefficient (Cd) of around .3 to .5. Most cars also have a frontal area of between 2 and 3 m^2.

For this type of problem, drag can be approximated by the equation: F=.5*Cd*A*rho*v^2

where Cd is the drag coefficient, A is the frontal area, rho is the atmospheric density (=1.28 kg/m^3), and v is velocity (m/s). For the purpose of the rest of the analysis, we'll assume rho and velocity are constant.

So say your car has a drag coefficient of .4 and a frontal area of

2.5m^2. That's a combined Cd*A value of 1.0. The block has a frontal area of 1m^2 and a Cd of about .8 (typical drag for cubes)... so your cube has a combined Cd*A value of .8. So you're basically increasing the drag on your car by about 80%. Leaving out rolling resistance, which usually is a low single digit percentage of the total energy expended, you're basically reducing the gas efficiency of the car by about 44%. In other words, you're only getting 56% as many miles per gallon than if you didn't have the cube. So if you normally get 25 miles per gallon, with the cube you'll probably get around 14 (that's about an extra 8 cents per mile at $2.59/gallon).

That's just a really rough estimate, and is other variables can influence the gas efficiency also.

Hope that helps, Dave

Reply to
dave.harper

Brilliant. Best regards. Ignacio Simón Yarza. Mech&electrical-electronic engineer.

"dave.harper" escribió en el mensaje news: snipped-for-privacy@g47g2000cwa.googlegroups.com...

Reply to
Ignacio Simón Yarza

My copy of Fluid Mechanics by Frank White lists the 3-D drag coefficient of a cube as 1.07 face on and 0.81 corner on for Re ~10^5.

Corner on the projected frontal area is (2) ^1/2 = 1.414 so :

(Cd*A )_faceon = 1.07*1=1.07

(Cd*A)_cornereon = 0.81*1.414 = 1.15

Which would seem to tell you to put the cube face on. There will be a change for the interaction with the vehicle, as the flow isn't the same as that around a cube in free space.

If one looks at 2-D coefficients White gives 2.1 and 1.6, leading to 2-D Cd*As of 2.1 vs. 2.26, respectively. So again this implies you want to face the cube face forward.

---------- Ed Ruf Lifetime AMA# 344007 ( snipped-for-privacy@EdwardG.Ruf.com)

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Reply to
Ed Ruf

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