# Neglecting Head Loss

## Problem 15 - Bernoulli's Energy Theorem

**Problem 15**

A pump (Figure 4-07) takes water from a 200-mm suction pipe and delivers it to a 150-mm discharge pipe in which the velocity is 2.5 m/s. At A in the suction pipe, the pressure is -40 kPa. At B in the discharge pipe, which is 2.5 m above A, the pressure is 410 kPa. What horsepower would have to be applied by the pump if there were no frictional losses?

## 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.

## 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.

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