Analysis of depth vs y including energy and continuity equation

Relevant Ideas are outlined below.

As water flows out of the container the PE of the water in the container decreases. 

• During a short time interval water depth y1 decreases to y1 - | `dy |, so that a 'slice' of water at the y1 position is replaced by an equal amount of water at y = 0.
• Intuitively we know that when a mass at one altitude is replaced by the same mass at a lower altitude the PE of the system decreases.

We will express our results in terms of density rho, vertical displacement y1, acceleration of gravity g and cross-sectional area A.  We will think of the change occuring between clock times t = t1 and t = t1 + `dt, for some small `dt.

Formally recall that the change in the PE of a system is equal to the work done BY the system against the total conservative force.

In this case the conservative force is that of gravity.

We will consider what happens to the PE of the system when a slice of water is vertically displaced thru `ds = -y1.  We note that this displacement is in the direction of the gravitational force.

It is easier to think of the work done by gravity ON the system.  The work done BY the system against gravity will be the negative of this quantity.

To find the work done by gravity on the slice we find the force exerted by gravity on the slice, and multiply by the vertical displacement, assuming a downward vertical displacement.

• To find the force exerted by gravity we must find the mass m of the slice; the force exerted by gravity on the slice will then be m g.
• To find the mass m we can multiply the density of the liquid by the volume of the slice.
• The volume of the slice is equal to the product of its cross-sectional area and its thickness.

Our final conclusion is the FgravON, the force exerted by gravity on the slice, is

• FgravON = rho A `dy * g, where density is rho, cs area is A and thickness is `dy.

This force acts parallel to the displacement, so the work done by gravity on the system is positive and we have

• `dWgravON = | FgravON | * | y1 |.

It follows that the change in potential energy, which is equal to the work done by the system against gravity, is

• `dPE = `dWgravBY = - `dWgravON = - | FgravON | * | y1 |.

We see that, as we saw intuitively, the PE of the system decreases.

The law of conservation of energy tells us that `dKE + `dPE + `dWnoncons = 0, where `dWnoncons is the work done by the system against nonconservative forces.

• It follows that if `dWnoncons = 0 the change in KE is `dKE = - `dPE.  Since the system in this situation loses PE, we see that KE will be gained.
• If the system does positive work against a nonconservative force (e.g., friction) then `dKE = -`dPE - `dWnoncons will be less than in the case where `dWnoncons = 0.  In the presence of significant friction, for example, the system will gain less KE than if friction were negligible.

In the system we are considering the velocity of the water in the container is presumably small so that the KE of the water in the system is negligible.  In this case, if we begin with a stationary system (or a system moving with constant velocity) and open the tube we will after a short time have a system with reduced PE, therefore with increased KE.

• Almost all of the increased KE is in the outflowing water.
• In the absence of friction the KE of the outflowing water will be equal to the PE lost by the system (`dKE = - `dPE).
• If friction is significant the outflowing water will have less KE than for the ideal case in which friction is negligible.

The exit velocity of the water is related to the velocity with which the water surface moves.  We can see this relationship by comparing the amount of water lost from the container to the amount of water leaving the tube in some short time interval `dt.  These amount will of course be equal. 

• The change in the water level will be `dy = vSurface * `dt.  So the change in volume will be `dV = Acontainer * `dy = Acontainer * vSurface * `dt, where Acontainer is the cs area of the container.
• The water exiting the tube in a short time interval `dt will form a cylinder whose volume is the product of its length and the cs area of the tube.  The length of the stream is equal to the product  vExit * `dt so that `dV = Atube * vExit * `dt.

Since what comes out of the tube is equal to what left the container, the two expressions for the volume must be equal. 

Setting the two expressions equal we easily solve to find that vExit = vSurface * Acontainer / Atube.  That is, the exit velocity is equal to the velocity of the surface multiplied by the ratio of the cs areas.

Expressing this result using subscripts 1 and 2 instead of 'surface' and 'tube' we obtain the equivalent general equations

• v2 = v1 * A1 / A2
• v2 / v1 = A1 / A2
• v1 A1 = v2 A2.

These equations apply to any incompressible fluid in a closed system.  The third of these is the usual expression of this result, and is the usual form of what we call the Continuity Equation.  All three forms are, of course, equivalent.