# horsepower

## Problem 20 - Bernoulli's Energy Theorem

**Problem 20**

The 600-mm pipe shown in Figure 4-11 conducts water from reservoir A to a pressure turbine, which discharges through another 600-mm pipe into tailrace B. The loss of head from A to 1 is 5 times the velocity head in the pipe and the loss of head from 2 to B is 0.2 times the velocity head in the pipe. If the discharge is 700 L/s, what power is being given up by the water to the turbine and what are the pressure heads at 1 and 2?

- Read more about Problem 20 - Bernoulli's Energy Theorem
- Log in to post comments
- 57976 reads

## Problem 19 - Bernoulli's Energy Theorem

**Problem 19**

A pump draws water from reservoir A and lifts it to reservoir B as shown in Figure 4-10. The loss of head from A to 1 is 3 times the velocity head in the 150-mm pipe and the loss of head from 2 to B is 20 times the velocity head in the 100-mm pipe. Compute the horsepower output of the pump and the pressure heads at 1 and 2 when the discharge is: (a) 12 L/s; (b) 36 L/s.

- Read more about Problem 19 - Bernoulli's Energy Theorem
- Log in to post comments
- 57289 reads

## Problem 16 - Bernoulli's Energy Theorem

**Problem 16**

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 3.6 m/s. The pressure is -35 kPa at A in the suction pipe. The 150-mm pipe discharges horizontally into air at C. To what height h above B can the water be raised if B is 1.8 m above A and 20 hp is delivered to the pump? Assume that the pump operates at 70 percent efficiency and that the frictional loss in the pipe between A and C is 3 m.

- Read more about Problem 16 - Bernoulli's Energy Theorem
- Log in to post comments
- 38193 reads

## 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?

- Read more about Problem 15 - Bernoulli's Energy Theorem
- Log in to post comments
- 37423 reads

## 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.

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

- Read more about Energy and Head of Flow
- Log in to post comments
- 81878 reads