Quantitative demonstration -- Bernoulli's equation

Photo of my class by Frederic Lamontagne

A straightforward problem with Bernoulli's equation leads to a fun and dramatic demonstration, along with a fascinating discussion of the meaning of "negligible" and some use of the continuity equation.  Here we go...

I start with an empty tennis ball container.  With a pocket knife, I poke a small (<1 cm diameter) hole near the bottom of the container.  I show the class the container, and explain:  I'm going to cover the hole, and fill to the top with water.  The water clearly will shoot out of the hole.  What will the speed of this water jet be?

The class understands that a flowing fluid problem requires Bernoulli's equation.  So we write the equation:

P₁ + rgy₁ + ½rv₁² =  P₂ + rgy₂ + ½rv₂²

Consider point 1 to be the top of the water, and point 2 to be as the water exits the container.  We want to find v₂.  Let's see what terms can be cancelled or set to zero.

The water at the surface and at the exit is exposed to the atmosphere.  Any time a liquid, flowing or not, is exposed to the atmosphere, its pressure is atmospheric.  So the P terms cancel.

We can always take the lower y position to zero in Bernoulli or conservation of energy problems.  So get rid of the y₂ term. 

Now we have a class discussion / argument.  Is the speed of the water at the top zero?  Certainly not.  But the water at the top is moving very slowly relative to the jet at the bottom.  So let's try setting that v₂ to zero, and see what happens.

We measure the distance from the hole to the top of the container to be about 15 cm.  Solving Bernoulli's equation, we find the speed at the position of the hole to be 1.7 m/s.  Awesome!

But this is a quantitative demonstration.  We don't just solve the problem... we have to verify the answer.  How?  I can't just put a Vernier motion detector on the water jet.  The class brainstorms for a bit, and comes up with a great approach:  treat the water jet as a projectile.  Given the jet's initial horizontal speed of 1.7 m/s, we should be able to predict how far the jet will travel horizontally before hitting the ground.

What a great way to reinforce projectile motion problems, which we covered two months ago!  The horizontal acceleration is zero, the vertical acceleration is 10 m/s², the vertical initial velocity is zero (because the jet comes out of the tennis ball container horizontally).  I place a large beaker on the ground to catch the stream of water; we measure the vertical distance from the top of the beaker to the hole in the container for use in the projectile problem.

When all is said and done, we calculated a 65 cm horizontal distance for the jet to travel.  While I filled up the container (and kept my finger carefully over the hole!), the students measured 65 cm along the ground and placed the large beaker at the correct spot.  I released the water....

... and the jet landed right in the beaker.  Physics works.

FOLLOW UP:  We all saw obviously that the speed of the water at the top was NOT zero.  But our prediction of a 1.7 m/s speed was accurate!  What's up?

We go back through the problem, this time not assuming that the speed at the top v₁ is zero.  Instead, we use the continuity principle ( A₁v₁ = A₂v₂ ) to get another equation relating the speeds at top and bottom.  We can solve the two-variable system for the speed at the bottom.  We need to measure the cross-sectional area of the hole, and the area of the tennis ball can.  We find that the can has a diameter of about 7 cm, while the hole only is 1 cm wide.  Therefore the area of the can is 49 times that of the hole.  We resolve for the speed... and find that we still get 1.7 m/s.  Any difference would be in a later, very insignificant digit.  So, the speed of the water at the top is negligible compared to the speed of the water jet.