# Wind past a flat roof, without hitting the sidewall

1. Dec 2, 2014

### leviterande

Hi,
I have became confused about a "certain situation" belonging to Bernoulli Principle :a "wind on flat roof situation" more precisely. The big misconceptions on the net didn't make it easier. There are very contradicting explanations from experts on the web. However, I hope to get a clarification and help from this forum.

Consider fig. 1 and fig. 2 (sorry for the inaccurate drawings)

-Fig.1- Wind past a FLAT roof example
See figure 1. In books, plenty of examples and questions are given on calculation of uplift force on FLAT ROOFS of houses. WIndspeed and area of the FLAT roof is given to calculate the upward net uplift on the roof. Now if I understand this correctly, the only REAL reason for the lower pressure above roof/uplift force on a FLAT roof is because the constriction of the incoming flowing wind hitting the side wall. After this wind stream hits the sidewall, it now bends/flows up towards and around the roof where now the streamlines gets therefore closer together and reaches now a higher velocity and thus the pressure here gets lower too around and above the flat roof (all because of conservation of momentum of the initial wind velocity and Bernoulli). The flat Roof feels upward lift due to the higher indoor pressure compared to the upper outdoor air. So, the airflow-constriction i.e. the sidewall of the house is the original CAUSE for the lower pressure on the roof and thus the uplift force on the roof(?) correct? I believe so

-Fig. 2- WInd past a FLAT ROOF but wind doesn't hit the sidewall i.e. no airflow constriction
To demonstrate my point, see now the hypothetical fig. 2. It is the same example as fig. 1 except now, the wind stream portion is flowing higher from the ground and is just flowing right above the FLAT roof without meeting the "sidewall constriction" or at least minimally touching the sidewall. Now in this hypothetical situation, my question is very simple: would there be a lower pressure on the roof and thus an uplift force on the roof in fig 2.? I explained this to several people and got mixed answers....
I believe that there would be no force on the flat roof in figure 2 because there is no constriction or change/deflection of airflow(?), right?

Your input is highly appreciated as I am really wanting to know. I tried some experiments but here, even slight crudeness unfortunately destroy results.
Thanks

2. Dec 3, 2014

### Staff: Mentor

So you'd also be wondering whether the wind of a hurricane can lift man-hole covers off utility accesses in the street?

3. Dec 3, 2014

### leviterande

Yes. I didnt read about any man-holes covers being lifted off. But yes I would be wondering that too.

Last edited: Dec 3, 2014
4. Dec 3, 2014

### leviterande

So, is there anyone who can give an answer/clarify my wondering?

5. Dec 3, 2014

### klimatos

The Bernoulli effect has nothing to do with the constriction or compression of the air flowing over the wall. You are confusing the Bernoulli effect with the Venturi effect. Look at a diagram of a perfume atomizer and you will see what I mean. The flow of air down the bulb pipe flows evenly across the opening of the reservoir pipe and creates a drop in pressure on that feed pipe. The flow of any fluid across a surface (real or imaginary) will result in a drop in pressure (compared to static conditions) upon that surface. This is the drop in pressure that causes sheets of plywood to be sucked out of the beds of speeding pickup trucks, that causes the fabric convertible top on a moving car to bow outward instead of inward, that causes large boulders to be plucked from the beds of flooding streams, that causes your shower curtain to move toward the flow of water and not away from it.

Picture a standard tri-axial orthogonal reference system. Set one of these three axes as the wind axis. When the wind blows, only some of the air molecules will have a component of motion in the windward direction--albeit a majority. The others will have a component of motion in the leeward direction. The increase in total molecular kinetic energy of translation in the windward direction is greater than the decrease in the leeward direction. The difference must be made up by decreases in the total kinetic energy of translation of the remaining four axial arms. Since the total kinetic energy of translation along any axial arm and the pressure normal to that arm are rigorously related, each arm registers a drop in pressure normal to the axial arm and parallel to the wind axis. Or, as commonly stated, the flow of a fluid across a surface creates a drop in pressure upon that surface.

6. Dec 4, 2014

### leviterande

Hi Klimatos. Thanks for your reply. Your explanation is what I typically also find as an explanation on the web, and one would perhaps think this is the picture of how it works. However there are other many sources that obviously indicate that the Bernouilli effect is widely misunderstood/explained wrong even in textbooks and that the fact is that speed alone would never cause a drop of pressure! Some kind of deflection is needed for any change of pressure(which kind of makes sense to me ). Look for example at this reply from rcgldr in a short thread on physics forum where a guy asked about force on a car-roof:

The interesting reply from rcgldr: "There would have to be something deflecting air away from a flat roof in order to produce lift. Speed alone isn't going to create lift on a flat plate parallel to the relative wind. Bernoulli principle doesn't apply here"

Furthermore, airplane altimeters work on the same principle, a flush mounted static port is mounted on the aircraft side where the airflow of course always goes past it(as the plane flies)So if I understand correctly and as rcgldr also mentions: the airplane can be stationary or flying at very high speed, yet the pressure measured is the same, unless the altitude is of course higher. Otherwise(and if moving air is all that is needed to create lower pressure), the altimeter in airplanes would show constantly and forever more decreasing pressure/higher altitude the faster the plane flies, but that is obviously not the case.
rcgldr sais "You could place a flush mounted static port in a flat roof to measure pressure of the moving air outside and it would indicate the same ambient pressure if the car was stopped or moving ."

Now, as I understand also, all cars of course have a windshield, all flat roofed houses have a sidewall, this is what deflects the airflow and thus lowers the pressure and thus produces uplift on the car -roof or on fabric convertible top or on a flat roof.

rcgldr sais "For a real car, the windshield deflects the air away from the roof, and the overall shape of most cars is similar to a common wing and tends to produce some lift.." just like a sidewall on my flat-roofed house example."

It seems there are conflicting conclusions between your and his posts. Do you see the dilemma? I would really need to understand what is correct here as this is why I posted the question. Perhaps, thanks to NascentOxygen, a much better determining example and issue-defining question would thus now be:

Would a horizontal parallel wind cause an uplift force on a street-manhole cover?
According to you Kilmatos, the answer is yes. According to rcgldr, the answer is no and the web equally apparently shows conflicting conclusions. What is the truth?
Thanks

7. Dec 4, 2014

Sure it would. You are making a mistake here in separating a change of speed from a deflection as if they are two disjoint phenomenon. An isentropic change of speed for a parcel of fluid will always result in a lower static pressure. It just so happens that the change of speed is generally brought about by some sort of deflection of constriction. Bernoulli's equation is a statement of the conservation of energy in a fluid. Neglecting gravity, it is a statement that along a streamline (or everywhere if the flow is isentropic), total pressure is constant, or in symbols, $p_0 = p +\rho v^2/2 = \mathrm{constant}$. For any change in $v$, there is a change in $p$ such that $p_0$ stays the same along a given streamline, regardless of how the change in $v$ came about. Consider that $p_0$, the total (or stagnation) pressure, is essentially the measure of the total energy density in the fluid. $\rho v^2/2$ is the kinetic energy density in the fluid as a result of the bulk fluid motion, and $p$ is like a spring potential energy density in the fluid. In order to conserve the total energy, if you increase the kinetic energy of the flow by increasing the speed, the potential energy goes down.

The problem here is that this explanation in that thread is incorrect. Bernoulli's principle certainly does apply on the roof of a car. The problem with that thread is that it has nowhere near enough information to answer the question.

Well, that is sort of correct. The static port will always measure the static pressure, so as long as that is in a location where the static pressure is equal to ambient, then yes, the pressure measured will always be the same for a given altitude. The problem is that not all points on a plane meet that criteria. For example, if you placed a flush-mounted hole on the top of the wing, that would not be at ambient pressure and would generally read lower than ambient pressure. In the case of the car, unless the care was shaped like an upside-down trapezoid or something, the flow over the top of the car has very likely been accelerated, so a pressure port on the top of the car is not likely to be measuring ambient static pressure, but the static pressure as a result of an accelerated flow.

The problem here is that the air in the house and the wind don't necessarily originate in the same reservoir condition. You might be sitting in your nice, air-conditioned house, while the wind outside is being generated by this gigantic heat engine that is a hurricane. Houses are not air-tight, so over a long time the pressure will tend to reach equilibrium with that outside, so if you have a very gradual decrease in external pressure over the course of a lengthy storm, then there are probably no issues as the inside air and outside air would be at (or nearly at) the same static pressure. On the other hand, that isn't likely the case for a gust of wind that comes on rather suddenly.

The problem then goes back to that of the uncommon reservoir. There is nothing to suggest that total pressure should be conserved inside a house and outside of it, so you can't just apply Bernoulli's equation using the pressure inside the house and the velocity over the roof to find the pressure over the roof. If you think about it, the pressure inside your house is already standing essentially still and is at a total pressure that equals the ambient pressure inside your house. On the other hand, if you go stand outside in the hurricane and put your hand out so that it stops the wind, the pressure you feel on that windward side of your hand is the total pressure and is very likely higher than the ambient pressure (and total pressure since the air is stagnant) in your house.

A similar situation is the bogus Bernoulli's principle explanation of why a sheet of paper lifts when you blow over it. You can't use Bernoulli's principle there because the air under the paper is still and has a total pressure that is different from the air going over the top since that moving air originated in your lungs and its reservoir was therefore pressurized to a higher total pressure than ambient when you started blowing. The paper lifts because the added pressure from your lungs is still rather small, so if you apply Bernoulli's equation from the point the air starts in your lungs to the point it moves over the paper, the static pressure is still going to be a bit lower than ambient so you get lift. Once again, though, you can't combine the wind velocity there with the pressure under the paper that didn't originate at the same reservoir condition as the moving air. It doesn't work that way.

So then the question is, given the pressure inside your house, what is the pressure over the flat roof for a given outside total pressure and wind velocity. The end result may well be that the flat roof (or a manhole cover) gets sucked off, but the common explanation using Bernoulli's equation with the inside and outside pressure is wrong. The same concept applies to the manhole cover.

8. Dec 4, 2014

### leviterande

Hi, and thank you for the answer. I didnt actually separate a change of speed from a deflection, perhaps my point was not clear or I didnt express it well, but yes I am aware that change of speed and deflection happen together, that is basically what I have said. Therefore I said in earlier posts that a constant speed alone without any change should therefore not create any change in pressure. If the speed was changed, there would be a change in pressure. and as you said, it just happens that the change of speed happens by some sort of deflection. Perfectly agreed.

Anyway
Now first lets take a look at the manhole cover as it seems to be the best example to clarify the issue better for me:
You are saying that a parallel horizontal wind flowing past a flat street manhole cover,(without any bumps, constrictions etc that would induce speed changes) could induce
an uplift force on the cover? Correct? Am I getting you correctly? or not?

Thank you a lot for your time

9. Dec 4, 2014

It could but that doesn't mean it necessarily will. It all depends on what the pressure on each side of the manhole cover is at a particular point in time, and the two are not necessarily easily related through a relationship like Bernoulli's equation since they don't originate from the same total conditions. Moving air can produce a force over a manhole cover, though. I hate to cite Mythbusters, but they did an episode where they looked at a race care driving fast over a manhole cover, and while they certainly didn't lift the cover off the road, they were able to measure a force on the cover. It isn't perfect science, but it should serve to illustrate that moving air over a flat surface can result in a lift force, but it isn't necessarily due to Bernoulli's principle since the moving air has had energy added to it beyond the air beneath the flat surface in these sorts of cases.

10. Dec 4, 2014

### CWatters

Interesting. Is this because the KE is proportional to V2 rather than V? eg...

In one direction you have (Vt + Vw)2 and in the other (Vt - Vw)2 where

Vt is the "temperature velocity" (I'm sure there is a better term)
Vw is the wind velocity

11. Dec 4, 2014

### klimatos

In essence, yes. If you measure both pressure and total kinetic energy of translation along a single axial arm, you will find that the pressure in Pascals is always exactly four times the total kinetic energy of translation measured in joules per axial arm. I have trouble with Latex, but let me see if I can construct an equation that illustrates this:

p = 2np2=4(½np2=KET)

p is the pressure in pascals, np is the mean number of molecules per cubic meter having a component of motion in the axial arm direction, m is the mean molecular impulse mass (not quite the same as the mean molecular mass because faster molecules are more likely to impact on a surface than slower ones and lighter molecules are faster than heavier ones at the same temperature) and σ2 is the variance of the molecular speed distribution along a single axis. σ itself is the standard deviation of that speed distribution and is usually referred to as the root-mean-square molecular velocity. The pressure sensor is presumed to be normal to the axial arm. In still air, np is exactly one-half of the total number density (n). In moving air, its proportion becomes an error function of the wind velocity expressed in standard deviations.

This is true both under conditions of equilibrium (still air) and when the wind is blowing, and is valid for all six axial arms.

The change in molecular velocity along the wind axis is not linear with the wind velocity as your equations show. It is a bit more complex than that because the molecules that "change allegiance" when the wind blows all have molecular speeds along the wind axis that are equal to or less than the wind velocity.

Last edited: Dec 4, 2014
12. Dec 4, 2014

### klimatos

Leviterande, I assume that you are talking about graduate-level college textbooks and not grade-school level. When you find a web source that contradicts such current textbooks, I suggest you go with the textbook explanation until such time as the textbook is modified to accept the new hypothesis. Bernoulli's principles have been tested in both the laboratory and in the real world tens of thousands of times over the centuries. They work.

However, don't lose your skepticism. Science progresses by proving the experts wrong--usually slightly wrong, but sometimes grossly wrong.

13. Dec 4, 2014

### leviterande

About that race car test, Interesting indeed. However, perhaps it should be anticipated that there would be a lift force measured from the mythbuster racecar episoede afterall. I mean, couldnt the reason be simply because there is a moment where the wind speed actually changed/accelerated relatively to the manhole cover? The moment before the racecar came close and over the cover, the air was stationary above and near the cover, the moment when the car now was very close and right above the cover the air STARTED to speed up, and finally moment when the racecar passed. The final picture is that the air relative to the cover was moving/ accelerating during the "car passing" and thus naturally should induce a change in the pressure over the plate. Isnt then this race car example at least different from the earlier street manhole cover example where the already established, steady, paralell to ground, and constant speed wind was flowing over manhole cover? I am just wondering if this view might have some merit?

- Racecar example: Existence of air acceleration relative to and above the cover, where all points in the air space along the path-line above the cover simultaneously move at different speeds.

-Oridinary wind example: One steady air speed relative to and above the cover, where all points in the air space along the path- line above the cover simultaneously move at the same speed.
Thanks

14. Dec 4, 2014

### rcgldr

In the case of high downforce race cars like Indy Racing League and Formula 1, when they run on street courses, the manhole covers are welded shut to prevent them from getting popped loose due to the low pressure under these type of race cars. Formula 1 race cars have front and rear wings, and a rear diffuser section that reduce pressure under the car. Indy Racing League have wings, diffuser, and underbody channeling to reduce pressure under the car (Formula 1 cars have "skidboards", no underbody channeling).

The pressure of the air coming out of an atomizer is slightly higher than ambient. Some atomizers divert some of the pressure into the container, while others like the "Flit Gun" pesticide sprayer rely on vortices being generated on the open end of a tube exposed to a flow perpendicular to the tube. The vortices reduce the pressure from above ambient to below ambient. Carburetors combine venturi effect and the open tube vortice effect to atomize fuel. If the end of the tube was flush mounted onto a flat plate parallel to the flow, then it would act as a static port and sense the slightly higher than ambient pressure from the atomizer pump.

Last edited: Dec 4, 2014
15. Dec 5, 2014

### leviterande

Exactly rcgldr. The manhole cover lifts here due to the lower pressure created above it from the passing car. But what/how is the car passing really mainly doing this?: Answer: Constriction/deflection of airflow through the wings/under body shape. The car with its wings is deflecting the moving air(airspeed change which creates a pressure differential) and naturally there is a force to be felt on the manhole cover. So pardon me if I am wrong but just for the above reason, I think this race car example may be inadequate for the clarifying of the issue here and is different from say an already established parallel natural straight wind that is flowing above a manhole cover(?)

Thanks

16. Dec 5, 2014

### rcgldr

The race cars wings and the angled diffuser section scoop or "suck" air away from the pavement by accelerating the air upwards. Underbody channeling (and the diffuser exit path for underbody air flow) uses a venturi like method to increase air speed and decrease pressure through the channel.

With an established horizontal air flow, the friction from the flat surface and viscosity in the air result in a boundary layer, and the pressure that the boundary layer exerts on the flat surface will be the same as the static pressure of the flow, as observed from a frame of reference moving at the same velocity as the flow. For a manhole cover, the gap in the seam around the manhole cover should be small enough to "hide" under the boundary layer (othewise the gap could generate vortice flow).

17. Dec 5, 2014

### leviterande

Agreed and understood.

Pardon me that I am not sure if I got you. If the gap is small enough as its effects to be neglected, did you mean that the cover would generally not feel a uplift force?
Thanks

18. Dec 5, 2014

### rcgldr

Within reason, no lift. The assumption is that the static pressure within the flow is the same as the pressure below the manhole cover.

19. Dec 5, 2014

### leviterande

Excellent! Understood and makes sense as I originally thought. Now however, it seems (unless I hopefully misinterpreted boneh3ad) in post #9, that his conclusions to be opposite to yours? ha? I interpreted from post 9 that lift would occur on a manhole plate just because there is a parallel wind and not because any possible variables like irregularities in the terrain. (But, perhaps boneh3ad was really simply talking about the possible pressure differences from the small irregularities like gaps, bumps etc?) You rcgldr however are saying that there is no lift and I am confident this view is right.

Now, keep in mind I am ignoring any small ring gap distances and assuming that there are no bumps anywhere near or on the plate and I assume that the wind is already established flowing at a constant velocity(not like a race car scenario). Disregarding any small effects/pressure differences from gaps, bumps etc, there shouldn't be any force on the plate, right boneh3ad? why? well, near the manhole-plate, the wind airflow would be straight and parallel (at least nearly) and hasn't encountered a constriction, thus hasn't speed up, and thus no pressure difference resulted and thus no force on the manhole cover, correct?

Perhaps the whole main original question was a very obvious one which therefore led to a plethora of branches and interpretations.

Regards

20. Dec 5, 2014

### rcgldr

The only measured force was with race cars, most of which reduce pressure under the car to some extent, just not as much at Indy or Formula cars. The other issue pointed out in post #9, is that the static pressure of the wind could be different than the static pressure under the manhole since they come from different sources. For example, street drains feed into the area under manholes, and a street drain facing the wind could experience increased pressure due to the wind being slowed down as it enters the chamber with the street drain, and if there was a manhole cover just above that street drain, it would experience the difference in pressure.