This is the last of three articles devoted to fluids. This article covers two applications of Bernoulli's equation. The first application combines this principle with the condition that describes how pressure varies with height in a static column of fluid. The second solves a physics problem by combining Bernoulli's equation with the equation of continuity.
I do have to use some unusual notation: (1) Most variables are represented by capital letters with lower case letters used as subscripts. For example "initial (i) time (T) would be written Ti. (2) Powers are designated by ^. For example, initial time squared would be written Ti^2, and "ten to the fifth" would be written 10^5.
Problem. A boy blows across the left side of a tube shaped like a U. The tube is about 50% to 75% filled with water (Pw = 1000 kg/m^3). If the left column of the water is 1.5 cm above the right one, what is the speed of the air leaving the boy's lips? Assume the density of the air is Pa = 1.20 kg/m^3.
Analysis. First, we'll apply Bernoulli's equation to determine the pressure difference between the left (o) and right (r) tops of the tube. Since the tops of the tubes are open, this is the same as the pressure difference between the tops of the left and right water columns. Let's assume the speed of the air blowing across the top of the left column is Vo. The air at the top of the right column is stagnant so Vr = 0. The tops of the two sides of the tube are at the same height, so Yo = Yr. We now have with Bernoulli's equation
..............................................Bernoulli's Equation
......................Po + Pa(Vo^2)/2 + PaGYo = Pr + Pa(Vr^2)/2 + PaGYr.
Since Yo = Yr and Vr = 0, this equation reduces to
..........................................Pr - Po = Pa(Vo^2)/2.
The tube is open at both ends, so the pressure at the tops of both sides is the same as the pressure at the tops of both water columns. For the pressure variation of the static water column, we have
.....................................Pressure Variation with Height
.................................................Pr = Po + PwGH
so..............................................Pr - Po = PwGH.
Comparing the two expressions for the pressure difference, we have
........................................PwGH = Pa(Vo^2)/2,
and..............................Vo = SQRT(2PwGH/Pa).
With the known values for Pw and Pa along with H = 0.015 m, we find with this last equation that
...............................................Vo = 15.7 m/s.
Problem. This is a standard problem in basic physics. A small opening (o) develops in a water tower at a point a distance H below the surface (s) of the water. The cross-sectional areas of the opening and of the water column are Ao and As, respectively. What is the speed Vo of the stream of water that leaves the opening?
Analysis. We have with the equation of continuity
.....................................................Equation of Continuity
..............................................................AsVs = AoVo
so...........................................................Vs = AoVo/As.
Since Ao << As, Vs is very small; consequently, we'll assume that Vs = 0. Because the top of the water in the tank and the hole in the side are open to the atmosphere, Ps = Po = Patm. Using Pw for the mass density of water, we now have with Bernoulli's equation
..................................................Bernoulli's Equation
....................Ps + Pw(Vs^2)/2 + PwGYs = Po + Pw(Vo^2)/2 + PwGYo
..............................Patm + 0 + PwGYs = Patm + Pw(Vo^2)/2 + PwGYo
so...............................Vo = Sqrt(2G(Ys - Yo)) = SQRT(2GH),
because........................................H = Ys - Yo.
This is the same speed that a stone would acquire when falling a distance H.
I have just presented two more examples of what, I believe, all teachers should do when presenting physics problem solutions to their students. When you start every important part of a problem solution with a statement of the appropriate fundamental principle, the students are much more likely to understand that solution. This method also teaches students an approach they can use advantageously throughout all of their academic and professional careers.
Dr William Moebs is a retired physics professor, who taught at two Universities: Indiana-Purdue Fort Wayne and Loyola Marymount University. You can see hundreds of examples illustrating how he emphasizes fundamental principles by consulting PHYSICS HELP.
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