The Venturi tube has two large-diameter sections joined by one of significantly smaller diameter, each of which has a tap in the middle. A U-shaped piece of Tygon® tubing connects the middle section and the large section at right. (The tap on the remaining section is plugged.) When you flow air through the tube, its increase in speed as it goes through the narrow section causes the static pressure there to be lower than that under the large section, and the fluid (colored water) in the tubing rises on the side connected there.As the short description above states, the Venturi tube shown in the photograph has two large-diameter sections joined by one of much smaller diameter. Each section has a tap. The one on the large section at left is plugged. A piece of Tygon® tubing connects the tap on the middle section to the one on the large section at right. When you flow air through the tube at a particular rate – so much volume per time, e.g., liters/minute, m3/s, etc. – it exits the tube at the far end at the same flow rate. For a particular volume flow rate, V, the linear speed of the air as it flows through the tube at a particular point equals V divided by the cross-sectional area of the tube there, or V/A, or v = V/A. As the air goes from the large-diameter section into the narrower section, A decreases considerably, so v increases by the same factor. Another way of putting this is that for a constant volume flow rate, Av = constant, or for flow through two sections of a pipe that have different diameters, A1v1 = A2v2. The flowing air stream exerts both static pressure, the pressure that would be present if there were no flow, and dynamic pressure, the pressure that arises from the flow. Bernoulli’s theorem gives the relationship between them at a particular point in the flow:
p + (1/2)ρv2 + ρgy = constant.
The static pressure consists of the two terms p and ρgy, and the dynamic pressure is (1/2)ρv2. ρ is the density of the air, g is the acceleration of gravity, and y is the height at that point. (Strictly speaking, this equation applies to a nonviscous, incompressible fluid. Under the conditions that obtain in this demonstration, however, the compressibility of the air is not important.) For the two sections of the Venturi tube (that is, those connected by the section of Tygon® tubing), we have:
p1 + (1/2)ρv12 + ρgy1 = p2 + (1/2)ρv22 + ρgy2
Since the tube is level, and all points along it are thus at the same height, y1 and y2 are equal, and we have:
p1 + (1/2)ρv12 = p2 + (1/2)ρv22
and:
p1 - p2 = (1/2)ρ(v22 - v12) = ρmgh - ρgh,
where I have used the subscript m, for “manometer,” for the density of the colored water in the Tygon® tube, and h is the difference in height between the surface of the liquid on the left side of the Tygon® tubing and the surface of the liquid on the right side. We see that the difference in static pressure above each side of the Tygon® tubing goes as the difference of the squares of the velocities of the air flowing in the two sections of the Venturi tube. Thus, this pressure difference increases as you increase the air flow. As you increase the air flow, the fluid rises higher and higher in the left side of the Tygon® tubing.
The Venturi effect finds practical use in carburetors, which are still used in some engines to provide the fuel-air mixture, aspirator pumps, through which one flows water to draw gases in through a side arm, and in some flow meters.
From the relations above, we can write v2 = v1(A1/A2). We then have (1/2)ρ([v1(A1/A2)]2 - v12) = ρmgh - ρgh, or (1/2)ρv12[(A1/A2)2 - 1] = (ρm - ρ)gh. This can be rearranged to v12 = [2(ρm - ρ)ghA22]/[ρ(A12 - A22)], which gives
v1 = A2√([2(ρm - ρ)gh]/[ρ(A12 - A22)]).
Here, v1 is the speed of the air at the entrance of the tube (in the large-diameter section). The volume flow rate at the entrance then, as noted above, is V = v1A1.
One can arrive at the Bernoulli equation in one of at least two ways. It arises from a formal mathematical treatment of the equations of motion for a fluid, which are similar to those for electrodynamics. Since it is essentially a statement of the conservation of energy, one can also derive it by means of the work-energy theorem. We can envision fluid flowing through a tube with an entrance having a particular cross-sectional area, A1, and an exit at a different height, with an area A2. If we consider a particular volume of fluid at the entrance and at the exit, these correspond to a length l1 at the entrance and l2 at the exit. If we call the entrance pressure p1, the exit pressure p2, and the entrance and exit heights y1 and y2, respectively, we can calculate the total work done on the fluid (by the sum of pressure and gravitational forces) as p1A1l1 - p2A2l2 - mg(y2 - y1). The product of area and length is volume, which we can also write as m/ρ, so this becomes (p1 - p2)(m/ρ) - mg(y2 - y1). By the work-energy theorem, this work must equal the kinetic energy change for the volume element, which we can express as (1/2)mv22 - (1/2)mv12. Equating these gives:
(p1 - p2)(m/ρ) - mg(y2 - y1) = (1/2)mv22 - (1/2)mv12
which we can rearrange as p1 + (1/2)ρv12 + ρgy1 = p2 + (1/2)ρv22 + ρgy2. Since the subscripts can refer to any two points along the tube, we can drop them and replace the right side of this equation with a constant to give the Bernoulli equation as stated above.
References:
1) Sears, Francis Weston and Zemansky, Mark W. College Physics, Third Edition (Reading, Massachusetts: Addison-Wesley Publishing Company, 1960) p. 271.
2) Feynman, Richard P., Leighton, Robert B. and Sands, Matthew. The Feynman Lectures on Physics, Volume II, Mainly Electromagnetism and Matter (Reading, Massachusetts: Addison-Wesley Publishing Company, 1963) 40-2 to 40-9.
3) Resnick, Robert and Halliday, David. Physics, Part One, Third Edition (New York: John Wiley and Sons, 1977), pp. 386-392.