Problem 02 - Bernoulli's Energy Theorem

Problem 2
From Figure 4-01, the following head losses are known: From (1) to (2), 0 m; from (2) to (3), 0.60 m; from (3) to (4), 2.1 m; from (4) to (5), 0.3 m. Make a table showing elevation head, velocity head, pressure head, and total head at each of the five points. How high above the center of the pipe will water stands in the piezometer tubes (3) and (4)?
 

04-002-reservoir-to-pipe.gif

 

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 ½Mv2. Since M = W/g,

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

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

 

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.

$\text{Elevation energy} = Wz$

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

 

010-egl-hgl-diagrams.gif

 

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.

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

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

 

01 How to calculate the discharge and the velocity of flow

Problem 1
Compute the discharge of water through 75 mm pipe if the mean velocity is 2.5 m/sec.
 

Problem 2
The discharge of air through a 600-mm pipe is 4 m3/sec. Compute the mean velocity in m/sec.
 

Problem 3
A pipe line consists of successive lengths of 380-mm, 300-mm, and 250-mm pipe. With a continuous flow through the line of 250 Lit/sec of water, compute the mean velocity in each size of pipe.
 

Discharge or Flow Rate

Discharge (also called flow rate)
The amount of fluid passing a section of a stream in unit time is called the discharge. If v is the mean velocity and A is the cross sectional area, the discharge Q is defined by Q = Av which is known as volume flow rate. Discharge is also expressed as mass flow rate and weight flow rate.

Volume flow rate, $Q = Av$

Mass flow rate, $M = \rho Q$

Weight flow rate, $W = \gamma Q$

 

009-continuous-flow.gif

 

Rotation - Rotating Vessel

When at rest, the surface of mass of liquid is horizontal at PQ as shown in the figure. When this mass of liquid is rotated about a vertical axis at constant angular velocity ω radian per second, it will assume the surface ABC which is parabolic. Every particle is subjected to centripetal force or centrifugal force CF = mω2x which produces centripetal acceleration towards the center of rotation. Other forces that acts are gravity force W = mg and normal force N.
 

008-rotating-vessel.gif

 

Stability of Floating Bodies

Any floating body is subjected by two opposing vertical forces. One is the body's weight W which is downward, and the other is the buoyant force BF which is upward. The weight is acting at the center of gravity G and the buoyant force is acting at the center of buoyancy BO. W and BF are always equal and if these forces are collinear, the body will be in upright position as shown below.
 

005-floating-body-upright-position.gif

 

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