Problem 06 - Bernoulli's Energy Theorem

Problem 6
As shown in Figure 4-03, the smaller pipe is cut off a short distance past the reducer so that the jet springs free into the air. Compute the pressure at 1 if Q = 5 cfs of water. D1 = 12 inches and D2 = 4 inches. Assume that the jet has the diameter D2, that the pressure in the jet is atmospheric and that the loss of head from point 1 to point 2 is 5 ft of water.
 

04-004-water-jet-at-reducer-end.gif

 

Problem 03 - Bernoulli's Energy Theorem

Problem 3
A 300-mm pipe is connected by a reducer to a 100-mm pipe. See Figure 4-02. Points 1 and 2 are at the same elevation, the pressure at 1 is 200 kPa. The discharge Q is 30 liters per second flowing from 1 to 2 and the energy lost from 1 to 2 is equivalent to 20 kPa.

  1. Compute the pressure at 2 if the liquid is water.
  2. Compute the pressure at 2 if the liquid is oil (sp gr = 0.80).
  3. Compute the pressure at 2 if the liquid is molasses (sp gr = 1.5).

 

04-003-pipes-connected-by-reducer.gif

 

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.
 

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