How does bernoulli's principle work?

Question

I have a science fair coming up soon in which you have to provide a presentation of a topic and a working model. I have deceided to make my project on bernoulli's principle, but I am kind of confused how it works. Also, would a hovercraft using air to push off follow bernoulli's principle? the project is on this website: http://www.all-science-fair-projects.com/science_fair_projects/89/401/932af3e19be52c06d1305dbb7bc3c39b.html

Answer 1

Bernoulli's principle works like this:



Objects (liquid, gas) under high pressure, tend to flow to an area of low pressure until the forces equal out. Okay, deep breath... let's explain:



I have a glass of coke. I put a straw in it. The downward atmospheric pressure on the glass of coke is 15 psi, the same as on EVERYTHING at roughly sea level. Okay, now i want to take a drink. I breath out to collapse my lungs. I put my mouth on the straw and expand my lungs. Since i'm not breathing air in, expanding my lungs like this creates a lower pressure area. The pressure OUTSIDE the straw on the liquid in the glass is still 15 psi. Because i've removed pressure from the straw, the 15 psi outside the straw will now push liquid up and into my mouth. Bernoulli's principle is, in fact, the entire reason straws work.



Vacuum cleaners work the same way, they create low pressure regions, which causes air to move through a carpet or over a surface and into the low pressure region (up inside the machine). This moving air often picks up the dust, etc, with it and is filtered through the vacuum.



HOWEVER, bernoulli's principle is NOT responsible for the flight of aircraft (this would be a good science fair topic... what really DOES cause a wing to lift a plane up). The wing is canted at a slight angle and air bouncing down off of it causes an equal and opposite reaction (newtons laws). It has nothing to do with high and low pressure areas. to experiment, in a fast moving car put your hand out the window and angle it... you will see the effects of lift as the angle of attack of your hand changes

Answer 2

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'Alexander,' your basic error is that Bernoulli's Principle applies along STREAMLINES, and only throughout a medium under various assumptions about conditions far upstream, or other restrictive conditions such as potential flow, or yet others too tedious to give here. (Think about how it is in fact derived --- by FOLLOWING the motion along a streamline!) Your supposed application of it simply violates the conditions under which it applies. That is the key point. Proceeding further, however: 'Bekki B' is correct in saying that "The water column is supported by pressure from below which differs because of Bernoulli's principle," but no-one seems to have seen whether the numerical details given in the picture are reasonable, or what they together imply. So let's look at that. For most practical purposes, water may be considered incompressible. Call its density 'ρ.' Although there will in practice be velocity differences across the pipe's cross-sections, let's assume that the velocity profiles are similar. (This could be the poorest assumption, but we need something like this, to proceed.) Then characterizing the flows by the values of the velocities along the central axis of (assumed) pipes of similar cross-sections A1, A2, as the figure invites us to do, we will have A1*v1 = A2*v2, or d1^2*v1 = d2^2*v2, that is v2 = (d1/d2)^2*v1. But d1/d2 = (90mm/40mm) = (3/2)^2, thus v2 = (3/2)^4 * v1, and v2^2 = (3/2)^8 * v1^2. Along the central streamline, with height h constant, we'll have v1^2 / 2 + P1/ρ = v2^2 / 2 + P2/ρ. Rearranging and inserting the above v1, v2 relationships we obtain: P1 - P2 = ρ * (v1^2 - v2^2) / 2, or numerically P1 - P2 = ρv1^2 [(3/2)^8 - 1]/2 = 12.31... ρv1^2. In cgs units, with ρ = 1gm/cm^3, this gives simply P1 - P2 = 12.31... v1^2. ......(A) VERTICALLY, the usual pressure support equation will hold. It does NOT have any reference to the horizontal velocity in it. That implies P1 - P2 = ρg (h1 - h2), where h1 - h2 = 37mm = 3.7cm. Then in cgs units, we have numerically P1 - P2 = 1*981*3.7 = 3630... cgs units. ......(B) Equations (A) and (B) imply: 12.31... v1^2 = 3630..., so that v1^2 = 294.8, or v1 = 17.2 cm/s. Then v2 = (3/2)^4 * v1 = 86.9 cm/s. These speeds, together with the pipe diameters given, seem reasonable for a laboratory experiment. There's a good chance that the flow would be laminar, which is of course implicitly built into the analysis. Live long and prosper. [Later edit: The algebraic result, before insertion of numerical values, follows from equating the two expressions for P1 and P2. That gives, simply: v2^2 - v1^2 = 2 g (h1 - h2). That is generally true, and shows very clearly how the speed-up from v1 to v2 is associated with a decrease in h2. It also shows how SMALL velocity differences at HIGH speeds are as effective as larger velocity differences at low speeds. To go further and derive the separate values of v1 and v2 requires an appeal to the ratios of the actual cross-sectional areas within which the flows are confined, such as that given by assuming d1^2*v1 = d2^2*v2, as I did above.] [ FURTHER EDIT: 'Alexander,' your second "Additional details" item misses the point, once again. You are in effect attempting to apply Bernoulli's Principle ACROSS rather than ALONG streamlines! There is NO streamline connecting the tops of those water columns, and the "Po's" in each of your three supposed applications of it will NOT be the same. THAT is what you have effectively assumed. ]

Answer 3

Bernoulli's principle is really nothing more than a re-statement of the conservation of total mechanical energy in a system. If you look at the kinetic energy available in a moving gas, it's pretty obvious that the faster the gas moves, the more kinetic energy it has. But since pressure is a measure of -potential- enrgy, for the velocity to decrease, the pressure must increase so that the total energy remains the same. Conversly, for the velocity to increase, the pressure must decrease. Bernoulli is kinda 'counter intuitive' until you see it that way.



HTH



Doug

Answer 4

Bernoulli Vacuum

Answer 5

increase velocity decrease pressure. so you decrease pressure from the top so it can float. so have the air push out in such a way that pressure up is high and pressure down is low. or something to that effect