# Elevation Head

## Problem 14 - Bernoulli's Energy Theorem

**Problem 14**

Water discharges through an orifice in the side of a large tank shown in Figure 4-06. The orifice is circular in cross section and 50 mm in diameter. The jet is the same diameter as the orifice. The liquid is water, and the surface elevation is maintained at a height h of 3.8 m above the center of the jet. Compute the discharge: (a) neglecting loss of head; (b) considering the loss of head to be 10 percent of h.

## Problem 13 - Bernoulli's Energy Theorem

**Problem 13**

The 150-mm pipe line shown in Figure 4-05 conducts water from the reservoir and discharge at a lower elevation through a nozzle which has a discharge diameter of 50 mm. The water surface in the reservoir 1 is at elevation 30 m, the pipe intake 2 and 3 at elevation 25 m and the nozzle 4 and 5 at elevation 0. The head losses are: from 1 to 2, 0; from 2 to 3, 0.6 m; from 3 to 4, 9 m; from 4 to 5, 3 m. Compute the discharge and make a table showing elevation head, pressure head, and total head at each of the five points.

## Energy and Head of Flow

Energy is defined as ability to do work. Both energy and work are measured in Newton-meter (or pounds-foot in English). Kinetic energy and potential energy are the two commonly recognized forms of energy. In a flowing fluid, potential energy may in turn be subdivided into energy due to position or elevation above a given datum, and energy due to pressure in the fluid. Head is the amount of energy per Newton (or per pound) of fluid.

Kinetic Energy and Velocity Head

Kinetic energy is the ability of a mass to do work by virtue of its velocity. The kinetic energy of a mass M having a velocity v is ½Mv^{2}. Since M = W/g,

$\text{Velocity head} = \dfrac{K.E.}{W} = \dfrac{v^2}{2g}$

Elevation Energy and Elevation Head

In connection to the action of gravity, elevation energy is manifested in a fluid by virtue of its position or elevation with respect to a horizontal datum plane.

$\text{Elevation head} = \dfrac{\text{Elevation energy}}{W} = z$

Pressure Energy and Pressure Head

A mass of fluid acquires pressure energy when it is in contact with other masses having some form of energy. Pressure energy therefore is an energy transmitted to the fluid by another mass that possesses some energy.

$\text{Pressure head} = \dfrac{\text{Pressure energy}}{W} = \dfrac{p}{W}$