## How does a diffuser/ride-height create drag?

Here are our CFD links and discussions about aerodynamics, suspension, driver safety and tyres. Please stick to F1 on this forum.
autogyro, you'd be surprised how much downforce you can put on an F1 car with no increase in drag...
newbie

Joined: 29 Sep 2009

No, no, no, what on earth is being said here.

"Taping up the inlets reduces the amount of air going under the car.
That increase DF from the under floor but it also increases drag from the under floor. This drag increase is larger than the reduction in drag on the body from the taped up and smoother nose.
Arguing the reverse is simply aero merchants continuing to con people by over complicating the issue."

The statement is a load of rubbish, how does decreasing the volumetric flow under the car, produce more downforce?

Do you understand trailing edge vortices and how they serve to alter the effective AoA of a wing? Also the energy required in the formation of a vortex?

Autogyro, you seem to think that going against someone's view and being brash about things makes you smarter? There doesn't seem to be any reasoning behind the posts and your intention always seems to talk down Formula 1 teams and their engineers. Things look awfully green to me.

For the original poster, the most important fact to remember about aerodynamics and the related forces is, there are only 2 mechanisms to impart the forces and moments generated by a body moving through a fluid, on to the body; 1. Pressure distribution over the body surface 2. Shear stress distribution on the body surface

It's quite easy to get yourself wrapped up in complex modes, geometries and fluid mechanics but the forces can only be imparted either by the pressures on the surface or through shear stress. In a lot of cases, the shear stress distribution is not something that is easily and conveniently altered, particularly on race cars.

If you really want to understand, you need to study aerodynamics properly. The common theory of how lift and drag forces are generated is in most parts wrong. People are taught that the flow on one side has to travel further to meet up with the flow at the trailing edge thus altering static pressure distribution, there is no evidence to show the flow has to meet up here. A symmerical airfoil section inclined in free stream can generate lift. Also the taught Newtonian approach where by the forces are generated by particles hitting the surfaces, well it's not wrong but for all flows except very very high speed flows, it's meaningless.

In regards to the diffuser, you must ask yourself, is the drag a cause of the pressure distribution on the diffuser or the shear stress? And as you can immagine, the ride height changes also serve to change the velocity of the air under the car which again, change the pressure distribution and the shear stresses. If you look up the variables that contribute to a simplifed shear stress calculation, you can see how the velocity comes in to it, then you can conclude how changing the airspeed would change the shear stress.

You could do a nice simple model with a flat plate above the ground at different heights.
F1_eng

Joined: 5 Aug 2009

Thank you F1_eng. I was totaly wrong and it was obvious as soon as you posted.
I am not an aerodynamacist as should be obvious. However I do know a little more than a 'silly child' kilco and a fair bit about motor racing and F1.
I still maintain that increasing DF will result in increased drag over all and that DF is the main factor preventing F1 development.
autogyro

Joined: 4 Oct 2009

autogyro wrote:
Add them to an inefficient wing with high drag and they increase DF which also increases drag if compared to the original wing.

From an example study in McBeath's Competition Car Aerodynamics:
No endplate: downforce = 769.2N, drag = 194.8N
large endplate: downforce = 900.1N, drag = 178.1N

Ooh look! A major increase in downforce with a significant reduction in drag.
However there is actualy no comparison, because without the end plates the wing would be an incomplete design and not subject to comparison in any meaningful way.

Wow, someone had better tell birds that their wings are incomplete...
Just_a_fan

Joined: 31 Jan 2010

Just_a_fan wrote:
autogyro wrote:
Add them to an inefficient wing with high drag and they increase DF which also increases drag if compared to the original wing.

From an example study in McBeath's Competition Car Aerodynamics:
No endplate: downforce = 769.2N, drag = 194.8N
large endplate: downforce = 900.1N, drag = 178.1N

Yes for the wing but what about drag for the complete car?

Ooh look! A major increase in downforce with a significant reduction in drag.

However there is actualy no comparison, because without the end plates the wing would be an incomplete design and not subject to comparison in any meaningful way.

Wow, someone had better tell birds that their wings are incomplete...

I was thinking of the rear wing not the front wing.
autogyro

Joined: 4 Oct 2009

Just for reference, I would be a little dubious of facts/data in Simon McBeath's book, doesn't have the greatest reputation.
F1_eng

Joined: 5 Aug 2009

F1_eng wrote:I am not going to talk anyone down on here without them being here to defend themselves, it is just a warning.

It's a bit late for that...

I suppose the problem is that doing a 'for dummies' style book on an extremely complex subject is always going to make the experts think "eek, that's not right".

For example, I've dipped in to some texts on CFD as part of the MSc I'm doing (not related to vehicle aero at all by the way) and to say that they're dense, complex and extremely specialist is understatement. A lot of the 'for dummies' explanations are only a very approximate view of the subject, therefore, but that is the only way that it can be explained to the casually interested.

Now, I wouldn't use McBeath's book to design a race car, but if someone (such as an F1 fan with some spare time) wants a taste of the subject it's got to be better than trying to read something full of PDEs
Just_a_fan

Joined: 31 Jan 2010

PDEs are where the fun starts
F1_eng

Joined: 5 Aug 2009

xpensive wrote:
ringo wrote:The other part concerning the expansion of the air in the diffuser. A diffuser is a steady state device so it does not consume work. So it does not consume power.

Speeding up the air under the car does not consume power? I think not.

Power is Volumetric flow times Pressure differential, that rarely comes for free. If air-speed under the car is 20 m/s higher than above, there's a price for it, one way or the other,

Where is the power being supplied from?
If you do that calculation you will get wrong results for what you think the diffuser is taking up.
example: say a diffuser that's .75m x .2m at the back and .75m x .05m at the throat.
it's understood, one pressure will be 4 times the other looking at the area ratios.

if the car is moving at 60m/s, under it as you say would be 20 more, so 80m/s.
volumetric flow = A x V = 0.0375 x 80 = 0.3 m3/s end pressure is atmospheric, so throat presure is Patm/4, so differential = 101325 Pascal x (1 -1/4) = 25331.25 Pa

power taken = 25331.25Pa x .3m^3/s = 7599.375W = 7.6kW, which is 10hp, which is not true for a diffuser. This is a wrong application.
If i put a diffuser on my road car and i was driving at that speed, i don't think it will rob my car of 10 horsepower.

A diffuser is adiabatic, there are no external forces imparted on the fluid, so there is no work hence no power use. This is why diffusers are put on industrial gas turbines as well.

Maybe i misunderstood what you were referring to, maybe what happens after that air leaves the diffuser, but within the diffuser itself no work is done on the fluid.
For work to be done energy has to cross the boundaries of the system. In the case of the diffuser system, no energy crosses it. Air simply enters and leaves, with viscous effects and other things taking place within the fluid, but not across the system boundary, to be strict in a thermodynamic sense.

I agree that It's not free, not because of it's function, but only because of what is required of the shape of the car to make it effective.
For Sure!!
ringo

Joined: 29 Mar 2009

End plates increase lift distribution but do produce some drag, especially if they are poorly designed.
Most aircraft avoid using end plates. They have to be well developed before they are considered beneficial, since fuel consumption is most important.

I think what autogyro is saying is that if there is a vehicles that is optimally designed. There is a point where any further attempt to increase down-force would result in an increase of drag. A good example is the RB6
We can't look at poorly designed systems that can pick up improvements in drag and lift,only because they were crappily designed to begin with. A good example the HRT, which can make improvements of lift and drag with no penalty because it wasn't optimal at all.
For Sure!!
ringo

Joined: 29 Mar 2009

ringo wrote:
xpensive wrote:
ringo wrote:The other part concerning the expansion of the air in the diffuser. A diffuser is a steady state device so it does not consume work. So it does not consume power.

Speeding up the air under the car does not consume power? I think not.

Power is Volumetric flow times Pressure differential, that rarely comes for free. If air-speed under the car is 20 m/s higher than above, there's a price for it, one way or the other,

Where is the power being supplied from?
If you do that calculation you will get wrong results for what you think the diffuser is taking up.
example: say a diffuser that's .75m x .2m at the back and .75m x .05m at the throat.
it's understood, one pressure will be 4 times the other looking at the area ratios.

if the car is moving at 60m/s, under it as you say would be 20 more, so 80m/s.
volumetric flow = A x V = 0.0375 x 80 = 0.3 m3/s end pressure is atmospheric, so throat presure is Patm/4, so differential = 101325 Pascal x (1 -1/4) = 25331.25 Pa

power taken = 25331.25Pa x .3m^3/s = 7599.375W = 7.6kW, which is 10hp, which is not true for a diffuser. This is a wrong application.
If i put a diffuser on my road car and i was driving at that speed, i don't think it will rob my car of 10 horsepower.

A diffuser is adiabatic, there are no external forces imparted on the fluid, so there is no work hence no power use. This is why diffusers are put on industrial gas turbines as well.

Maybe i misunderstood what you were referring to, maybe what happens after that air leaves the diffuser, but within the diffuser itself no work is done on the fluid.
For work to be done energy has to cross the boundaries of the system. In the case of the diffuser system, no energy crosses it. Air simply enters and leaves, with viscous effects and other things taking place within the fluid, but not across the system boundary, to be strict in a thermodynamic sense.

I agree that It's not free, not because of it's function, but only because of what is required of the shape of the car to make it effective.

Oh I thank you, you just proved my point to perfection, a 7.6 kW loss at a speed of 216 km/h makes perfect sense!
"Bernoulli is a nine-letter name"
xpensive

Joined: 22 Nov 2008

F1_eng wrote:PDEs are where the fun starts

Just_a_fan

Joined: 31 Jan 2010

ringo wrote:End plates increase lift distribution but do produce some drag, especially if they are poorly designed.
Most aircraft avoid using end plates. They have to be well developed before they are considered beneficial, since fuel consumption is most important.

I think what autogyro is saying is that if there is a vehicles that is optimally designed. There is a point where any further attempt to increase down-force would result in an increase of drag. A good example is the RB6
We can't look at poorly designed systems that can pick up improvements in drag and lift,only because they were crappily designed to begin with. A good example the HRT, which can make improvements of lift and drag with no penalty because it wasn't optimal at all.

I was responding to a sweeping generalisation with a similar, although slightly more focussed generalisation.

Anyway, modern airliners almost all use endplates of some type or other because they are so beneficial to the efficiency of the wing.
Just_a_fan

Joined: 31 Jan 2010

xpensive wrote:
ringo wrote:
Where is the power being supplied from?
If you do that calculation you will get wrong results for what you think the diffuser is taking up.
example: say a diffuser that's .75m x .2m at the back and .75m x .05m at the throat.
it's understood, one pressure will be 4 times the other looking at the area ratios.

if the car is moving at 60m/s, under it as you say would be 20 more, so 80m/s.
volumetric flow = A x V = 0.0375 x 80 = 0.3 m3/s end pressure is atmospheric, so throat presure is Patm/4, so differential = 101325 Pascal x (1 -1/4) = 25331.25 Pa

power taken = 25331.25Pa x .3m^3/s = 7599.375W = 7.6kW, which is 10hp, which is not true for a diffuser. This is a wrong application.
If i put a diffuser on my road car and i was driving at that speed, i don't think it will rob my car of 10 horsepower.

A diffuser is adiabatic, there are no external forces imparted on the fluid, so there is no work hence no power use. This is why diffusers are put on industrial gas turbines as well.

Maybe i misunderstood what you were referring to, maybe what happens after that air leaves the diffuser, but within the diffuser itself no work is done on the fluid.
For work to be done energy has to cross the boundaries of the system. In the case of the diffuser system, no energy crosses it. Air simply enters and leaves, with viscous effects and other things taking place within the fluid, but not across the system boundary, to be strict in a thermodynamic sense.

I agree that It's not free, not because of it's function, but only because of what is required of the shape of the car to make it effective.

Oh I thank you, you just proved my point to perfection, a 7.6 kW loss at a speed of 216 km/h makes perfect sense!

Ok, so it means if a diffuser takes power, then conversely ,a nozzle would give the car 10hp then!
the right equation to use is the word done by shear forces; which is still internal to the fluid.
For Sure!!
ringo

Joined: 29 Mar 2009

Ringo, think about it, 10 Hp to achieve a 20 m/s overspeed under you car is peanuts, when you will probably use at least 200 to get to 60 m/s anyway.
Play a little Bernoulli and see how much donforce those 80 m/s under your car would give you, probably thousands of Newtons.

F = A * Rho * (v1^2 - v2^2)/2
"Bernoulli is a nine-letter name"
xpensive

Joined: 22 Nov 2008

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