Bernoulli Equation - Fluid Flow - Handout, Exercises of Fluid Dynamics

Topics covered in this course include fluid properties, fluid statics, fluid kinematics, control volume analysis, dimensional analysis, internal flows, differential analysis, external flows CFD, compressible flow and turbomachinery. Key words for this lecture are: Bernoulli Equation, Water Draining from a Tank, Energy Grade Line and Hydraulic Grade Line, Head Form of the Energy Equation, Atmospheric Pressure, Conservation of Energy Equation, Grade Lines in a Fluid Flow, Venturi Tube, Airplane Wi

Typology: Exercises

2012/2013

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M E 320 Professor John M. Cimbala Lecture 14
Today, we will:
Do another example problem – head form of the energy equation
Discuss grade lines – energy grade line and hydraulic grade line
Derive and discuss the Bernoulli equation
5. Examples (continued)
Example – Water draining from a tank
Given: Water drains by gravity from a tank exposed to atmospheric pressure. The vertical distance
from the pipe outlet to the surface of the water in the tank is Δz = 0.500 m. The irreversible head
losses in the piping system (due to friction in the pipe, losses through the valve, elbow, etc.) are
estimated as h
L
= 0.400 m of equivalent water column height. Note: You will learn how to calculate
the irreversible head losses associated with piping systems on your own in Chapter 8.
Open valve
V
2
Δz
To do: Calculate the average velocity at the outlet, V
2
.
Solution:
From previous lecture…use the head form of the conservation of energy equation:
22
11 2 2
1 1 pump,u 2 2 turbine, e
12
22
L
PV P V
zh zh h
gg gg
αα
ρρ
⎛⎞
+++ =+ ++ +
⎜⎟
⎝⎠
∑∑
pf3
pf4
pf5

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M E 320 Professor John M. Cimbala Lecture 14

Today, we will :

  • Do another example problem – head form of the energy equation
  • Discuss grade lines – energy grade line and hydraulic grade line
  • Derive and discuss the Bernoulli equation
    1. Examples (continued)

Example – Water draining from a tank

Given : Water drains by gravity from a tank exposed to atmospheric pressure. The vertical distance

from the pipe outlet to the surface of the water in the tank is Δ z = 0.500 m. The irreversible head

losses in the piping system (due to friction in the pipe, losses through the valve, elbow, etc.) are

estimated as h L

= 0.400 m of equivalent water column height. Note : You will learn how to calculate

the irreversible head losses associated with piping systems on your own in Chapter 8.

Open valve

V 2

Δ z

To do : Calculate the average velocity at the outlet, V 2

.

Solution :

From previous lecture…use the head form of the conservation of energy equation:

2 2

1 1 2 2

1 1 pump,u 2 2 turbine, e

1 2

L

P V P V

z h z h h

g g g g

Example of Grade Lines in a Fluid Flow

D. The Bernoulli Equation

  1. Derivation

Begin with the head form of the conservation of energy equation along a streamline:

2 2

1 1 2 2

1 pump,u 2 turbine, e

1 2

L

P V P V

z h z h h

ρ g g ρ g g

At any location in the

duct, the difference

between EGL and HGL

is V

2

/(2 g ).

At point 0, HGL = EGL inside the

tank, since the fluid is at rest ( V = 0).

Neither EGL or HGL can rise above

this value unless work is added to the

flow (e.g., with a pump).

Since the jet exits at atmospheric

pressure at the outlet of the pipe,

P 3

= P atm

, and HGL is equal to

the height of the free surface of

the liquid.

EGL continually falls due to

irreversible losses, but HGL can

rise or fall. Overall, however,

HGL also must fall. In fact, HGL

can never rise above EGL.

FIGURE 5-