Set 54 Problem number 8

Problem

A uniform cylinder is filled with water to a depth of 120 meters above a small hole through which water can escape.  We wish to determine the velocity of outflowing water by energy considerations, assuming that no dissipative forces are present. 

• We imagine that water is raised from the level of the hole to the top of the cylinder, raising the water level to 120.39 meters, and that water is then allowed to flow freely from the hole.
• Determine the PE gain as the water is raised, and the PE loss of the system as an equal quantity of water then escapes.
• Use these results to infer the velocity of the escaping water.

After solving the problem using energy considerations, solve using Bernoulli's Equation. 

Solution

The volume of water raised will be the volume of the added cylinder, which has cross-sectional area 55 m^2 and altitude .39 meters, and therefore volume

• volume of raised water = ( 55 m^2) * ( .39 meters) = 21.45 m^3.

This water will have mass

• mass of raised water = 1000 kg / m^2 * ( 21.45 m) = 21450 kg.

Its weight will be

• weight of raised water = (9.8 m/s^2) * 21450 kg = 210210 Newtons.

The water will be raised to an average altitude of ( 120 + 120.39 ) / 2 meters = 120.195 meters. The energy required to raise the water will be the product of the force 210210 Newtons required to raise the water against the influence of gravity and the average vertical distance 120.195 meters through which it is raised. The increase in potential energy will be equal to this work, so potential energy change is

• PE change  = 210210 Newtons * 120.195 meters = 2.526619E+07 Joules.

When the water is released, the 2.526619E+07 Joules of added potential energy will be lost; this lost PE is converted to 2.526619E+07 Joules of KE.  If we assume that the water in the cylinder has an insignificant velocity and hence an insignificant kinetic energy, then the KE will all be in the outflowing water and the 21450 kg of outflowing water will have a KE of 2.526619E+07 Joules.

For this mass we therefore have

• .5 m v^2 = KE,

so average outflow velocity is

• outflow velocity = v = `sqrt(2 KE / m) = `sqrt ( 2 * 2.526619E+07 Joules / 21450 kg) = 48.53681 m/s.

Bernoulli's Equation can be applied to this situation. 

• We assume that the container is open to the atmosphere, so that the pressure at altitude 120.39 meters is atmospheric pressure.  The water at the average altitude 120.195 meters is close enough to the top that its pressure does not vary significantly from atmospheric, and as we have previously assumed this water does not move with any significant velocity. 
• We thus have at this position a state where pressure is atmospheric, velocity is 0 and altitude is 120.195 meters.

• At the outflow position pressure is again atmospheric and altitude is 0 meters.  The outflow velocity is to be found.

• There is no change in pressure, so we have `rho g y + .5 `rho v^2 = constant.

• We easily calculate the change in `rho g y.   The change is 1000 kg / m^3 * 9.8 m/s^2 * (- 120.195 meters) = - 1177911 N / m^2.

• The change in .5 `rho v^2 is equal to - (- 1177911 N/m^2) = 1177911 N/m^2.
• Since at the first position v = 0, the change is simply equal to .5 `rho v^2, where v is the exit velocity.
• Setting .5 `rho v^2 = 1177911 N/m^2, we easily solve for v to obtain exit velocity v = 48.53681 m/s.

Generalized Solution

When a fluid of constant mass density (denoted by the Greek letter `rho) is added to a uniform cylinder of cross-sectional area A filled with fluid to a depth h above some reference level, the potential energy of the added fluid increases the total potential energy of the system. If the added depth is `dh, then the added volume is

added volume = A * `dh,

the added mass is the product

added mass = `rho * A * `dh

of the mass density and be added volume, and the weight of the added fluid is

added weight = `dw = mg = `rho * A * `dh * g, where g is the acceleration of gravity.

The average height to which the fluid is raised is `dh / 2 above the original depth h, or h + `dh/2, so the potential energy increase is

PE increase = work to raise = mg * (h + `dh/2) = (`rho * A * `dh) * g ( h + `dh / 2).

For values of `dh which are small compared to h, the PE increase is close to the approximate value

approximate PE increase = mg * h = (`rho * A * `dh) * g * h.

When this amount of fluid later exits from the reference level, an equal amount of fluid returns to the original level. At this point the system will have returned to its initial state and the added potential energy will have been converted to kinetic energy and dissipated energy. Since the dissipated energy is assumed to be negligible here, we have

.5 m v^2 = mg * (h + `dh / 2),

from which we easily obtain

v = `sqrt( 2 g (h + `dh / 2) ), or if `dh is small v = `sqrt(2 g h). 'ppThis is the desired velocity.

More generally, when water depth is h, in a short time interval water flowing out changes its altitude from h + `dh to zero, a change of approximately -h (the `dh is small in a short time interval; its contribution to the altitude change of the outflowing water can therefore be neglected). We therefore have

KE change = .5 m v^2 = .5 (`rho * A * `dh) * v^2

equal to

PE loss = - PE change = -(m g * -h) = m g h = (`rho * A * `dh) * g * h.

We can say that KE change + PE change = 0 (KE increases, PE decreases), so that .5 m v^2 + m g h remains constant, consistent with energy conservation. If we divide this expressioin by the volume of the exiting fluid we get .5 `rho v^2 + `rho g h for the constant expression.