Set 54 Problem number 3

Problem

At a certain depth a `plug' of exiting water, .05 meters long, initially at rest and with cross-sectional area .00082 m^2, as it is forced out of a container by gauge pressure 52000 N/m^2.  Use work and energy considerations to find the velocity of the exiting water.

Solution

The water implicitly exits to gauge pressure zero, so the gauge pressure is the pressure difference between inside and outside the container.

The work done by the pressure is the product of the force exerted by the pressure and the distance through which it is exerted.

• The force is equal to the product of the pressure and the cross-sectional area, or F = 52000 N/m^2 * .00082 m^2 = 42.64 Newtons.
• The work is done by the force as the 'plug' forms, and is therefore done over a distance of .05 meters.  We thus have

• work = force * distance = 42.64 Newtons * .05 meters = 2.132 Joules.

The mass of the water is the product of its density 1000 kg/m^3 by its volume in m^3.  Volume is

• volume of 'plug' = cross-sectional area * length = .05 m * .00082 m^2 = .000041 m^3

so its mass is

• mass of 'plug' = volume * density = .000041 m^3 * 1000 kg/m^3 = .041 kg.

The KE change of the 'plug' is equal to the work done on it.  Since original KE is zero, the final KE will therefore be

• KE = .5 m v^2 = 2.132 Joules,

so

• v = `sqrt(2 * 2.132 Joules / m).

Substituting .041 kg for the mass we find that

• v = `sqrt(2 * 2.132 Joules / .041 kg) = 10.19 m/s.

Generalized Solution

When a 'plug' of length L and cross-sectional area A is forced from a container by pressure difference `dP we see that the work done on the 'plug' is

• work = force * distance =  ( `dP * A) * L

The mass of the plug is

• m = `rho * V = `rho * L * A.

If the water in the 'plug' is initially at rest and if no dissipative forces act, we have

• KE = .5 m v^2 = work done on plug,

where v is the exit velocity of the water.  We thus have

• .5 m v^2 = `dP  * A * L, or
• .5 `rho * L * A * v^2 = `dP * A * L.

Solving for v we obtain

• v = `sqrt( 2 * dP * A * L / (`rho * L * A) ) = `sqrt ( 2 `dP / `rho).