1 CTC 450 ► Bernoulli’s Equation ► EGL/HGL. Bernoulli’s Equation 2

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Transcript presentasi:

1 CTC 450 ► Bernoulli’s Equation ► EGL/HGL

Bernoulli’s Equation 2

3 Preview ► Bernoulli’s Equation  Kinetic Energy-velocity head  Pressure energy-pressure head  Potential Energy ► EGL/HGL graphs  Energy grade line  Hydraulic grade line

4 Objectives ► Know how to apply the Bernoulli’s equation ► Know how to create EGL/HGL graphs

5 Assumptions ► Steady flow (no change w/ respect to time) ► Incompressible flow ► Constant density ► Frictionless flow ► Irrotational flow

6 3 Forms of Energy ► Kinetic energy (velocity) ► Potential energy (gravity) ► Pressure Energy (pump/tank)

7 Units ► Energy (ft or meters)

8 Kinetic Energy (velocity head) ► V 2 /2g

9 Pressure Energy (pressure head) ► Pressure / Specific weight

10 Potential Energy ► Height above some datum ► Hint: It is best to set the datum=0 at the centerline of the lowest pipe

11 Bernoulli’s equation ► section 1 = section 2

Hints: ► If there is a reservoir (or large tank) pick a point at the surface of the water. The kinetic energy and pressure energy are zero, leaving only the potential energy. ► If water is discharging to the atmosphere pick a point just outside the pipe. The pressure energy=0. 12

13 Energy Grade Line ► Graphical representation of the total energy of flow of a mass of fluid at each point along a pipe. For Bernoulli’s equation the slope is zero (flat) because no friction loss is assumed

14 Hydraulic Grade Line ► Graphical representation of the elevation to which water will rise in a manometer attached to a pipe. It lies below the EGL by a distance equal to the velocity head. ► EGL/HGL are parallel if the pipe has a uniform cross-section (velocity stays the same if Q & A stay the same).

15 Hints for drawing EGL/HGL graphs ► EGL=HGL+Velocity Head ► EGL=Potential+Pressure+Kinetic Energies ► HGL=Potential+Pressure Energies ► Bernoulli’s equation assumes no friction; therefore the lines should not be sloped.

18 Reservoir Example ► Water exits a reservoir through a pipe. The WSE (water surface elevation) is 125’ above the datum (pt A) The water exits the pipe at 25’ above the datum (pt B). ► What is the velocity at the pipe outlet?

19 Reservoir Example ► Point A:  KE=0  Pressure Energy=0  Potential Energy=125’ ► Point B:  KE=v 2 /2g  Pressure Energy=0  Potential Energy=25’ (note: h=100’)

20 Reservoir Example ► Bernoulli’s: Set Pt A energy=Pt B energy  v 2 /2g=h  v=(2gh).5 Have you seen this equation before?  Velocity=80.2 ft/sec