B Air pressure generated by driving at 60mph

[david66ad]

If I place a 5' L x 12" W open box at front and a 1/8" hole at the back of the box and drove my car at 60 mph how much psi would I generate at the 1/8" hole?

 

[erobz]

It's probably difficult to say without determining the drag coefficient of the box. A small hole like that it isn't going to alter the drag much.

I think you would basically have:

$$ P = \frac 1 2 C_D \rho_{air} v^2 $$

 

[Baluncore]

Welcome to PF.

The size of the box is not critical, so long as the hole is small by comparison.
https://en.wikipedia.org/wiki/Ram_pressure#Imperial_units
60 mph = 0.064 psi = 9.20 lbs/sqft
You will get about 1/16 of one psi.

 

[Lnewqban]

Welcome @david66ad !

If your car does not over-heat, it should be about this value:
https://en.wikipedia.org/wiki/Stagnation_pressure

 

[erobz]

Its interesting that it isn't dependent on the drag coefficient? The drag force varies with it. The drag coefficient encapsulates form drag and pressure drag. I guess the pressure drag portion ( which is what is of interest here) must be independent of the form drag.

 

[david66ad]

It's probably difficult to say without determining the drag coefficient of the box. A small hole like that it isn't going to alter the drag much.

I think you would basically have:

$$ P = \frac 1 2 C_D \rho_{air} v^2 $$

If I replaced the box with 5' diameter funnel down to a 1/8" spout, what would the outgoing PSI be from the spout?

 

[Baluncore]

If I replaced the box with 5' diameter funnel down to a 1/8" spout, what would the outgoing PSI be from the spout?

It would be the same. A pressure is independent of area.

 

[erobz]

If I replaced the box with 5' diameter funnel down to a 1/8" spout, what would the outgoing PSI be from the spout?

I don't know. Others are saying the pressure is independent of the shape, size, so the drag coefficient is irrelevant. So, it would seem the same as the box. I'm still a bit suspicious to what degree of an approximation that is.

 

[russ_watters]

Its interesting that it isn't dependent on the drag coefficient? The drag force varies with it. The drag coefficient encapsulates form drag and pressure drag. I guess the pressure drag portion ( which is what is of interest here) must be independent of the form drag.

Stagnation pressure is essentially the maximum drag, or a 1.0 drag coefficient. Can't do better than that, and it isn't hard to achieve. That's the point of a pito-static tube for airspeed measurement.

The OP appears to be under the common but false belief that you can do better with a funnel.

 

[erobz]

Stagnation pressure is essentially the maximum drag, or a 1.0 drag coefficient. Can't do better than that, and it isn't hard to achieve. That's the point of a pito-static tube for airspeed measurement.

The OP appears to be under the common but false belief that you can do better with a funnel.

The only problem I'm seeing that optimization is apparently automatic independent of shape?

 

[russ_watters]

The only problem I'm seeing that optimization is apparently automatic?

I don't understand what you mean.

 

[erobz]

I don't understand what you mean.

You can't alter the shape to reduce the pressure. Its apparently automatically maximized.

The cone and the box will have very different coefficients of drag, but the pressure acting on them is the same; independent of the drag coefficient.

 

[russ_watters]

You can't alter the shape to reduce the pressure. Its apparently automatically maximized.

The cone and the box will have very different coefficients of drag, but the pressure acting on them is the same; independent of the drag coefficient.

Yes. The key feature is just a featureless hole. There's not much that can be done to make it better or worse.

Drag is basically stagnation pressure (force) minus regain. For the OP's question we don't care about regain.

 

[cjl]

You can't alter the shape to reduce the pressure. Its apparently automatically maximized.

The cone and the box will have very different coefficients of drag, but the pressure acting on them is the same; independent of the drag coefficient.

The pressure acting at the stagnation point is the same, yes. The overall pressure distribution (and of course the overall drag) is going to highly depend on the shape of course.

Also, interestingly, it is possible for drag coefficient to exceed 1, but you can't exceed stagnation pressure on the pressure side of an object. To create a greater drag coefficient, you need a shape that also creates a low pressure region behind it, creating a greater net force than simply the dynamic pressure applied across the frontal area.

 

[russ_watters]

Also, interestingly, it is possible for drag coefficient to exceed 1...

[Google] I stand corrected!