FLUIDS: Potential Energy, Kinetic Energy, Momentum, and Torricelli’s Theorem

Q: A 0.2 m container is full of a fluid of unknown density ( ρ ). A spigot at the bottom of the container is opened to allow fluid to flow at an unknown velocity ( v ) onto the ground. With what velocity will the fluid flow through the spigot?

A: In a separate exercise, this question was solved using the following equations:

KE₁ + PE₁ = KE₂ + PE₂

P₁ + ½ ρv₁² + ρgh₁ = P₂ + ½ ρv₂² + ρgh₂

Pressure here represents the energy ( E ) per unit volume ( V ) contained in a quantity of fluid within an open-air container. This container has an open spigot through which fluid falls to the earth. Our task is to use an appropriate term from the equation with the equation that relates kinetic energy ( KE ) to momentum ( p ):

p = mv

p² = m²v²

( p² / m ) = mv²

( p² / 2m ) = ½ mv²

If we choose the term for kinetic energy per unit volume from above, we’d have to have some insight into the velocity of the fluid leaving the container ( perhaps ); however, the velocity through which the fluid flows through the spigot is what we want to determine. Fortunately, the kinetic energy per unit volume as expressed in the equation has a potential energy ( PE ) per unit volume equivalent:

( KE / V ) = ( PE / V )

½ ρv² = ρgh

Note: In order for our substituted term to have the correct units, we must multiply it by ( V ) to convert it from energy per unit volume to energy:

( p² / 2m ) = ρghV

Mass ( m ) can be expressed as a product of density ( ρ ) and volume as follows:

m = ρV

We now substitute this value into the previous equation:

( p² / 2ρV ) = ρghV

p² = 2ρ²V²gh

p = √ ( ρ²V² )( 2gh )

mv = √ ( ρ²V² )( 2gh )

( ρV )( v ) = √ ( ρ²V² )( 2gh )

( ρV )( v ) = ( ρV )( √ 2gh )

v = √ 2gh

Wow!!!

This is the same result derived from an arguably more tedious process, but it shows that physics is physics regardless of what disguise it attempts to hide behind. Let’s now see what would have happened it we’d inserted the kinetic energy per unit volume term into our initial equivalency:

½ ρv²  = ( KE / V )

( ½ ρv² )( V ) = KE

( p² / 2m ) = ( ½ ρv² )( V )

( 2p² / 2m ) = ( ρv² )( V )

( p² / ρV ) = ( ρv² )( V )

p² = ( ρ²v² )( V² )

p = ( ρV )( v )

p = mv

Alas, our answer is correct, but its units are of no help in solving the problem. Remember, the ( KE ) term on the left was derived from momentum ( p ), so the algebra reduced both sides of the equation to momentum terms. Only by relating ( PE ) and ( KE ) did we arrive at a final term containing ( h ), which is the only useful bit of information given to us in the question at hand.