CVE 240 – Fluid Mechanics

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Transcript CVE 240 – Fluid Mechanics

Fluid Mechanics

CHAPTER 4

EULER’S EQUATION

Dr . Ercan Kahya

Engineering Fluid Mechanics 8/E

by Crowe, Elger, and Roberson Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.

Review of Definitions

Steady flow:

Unsteady flow: velocity is constant with respect to time velocity changes with respect to time

Uniform flow: velocity is constant with respect to position

Non-uniform flow: velocity changes with respect to position

Local acceleration:

– change of flow velocity with respect to

time

– occurs when flow is unsteady •

Convective acceleration:

– change of flow velocity with respect to

position

– occurs when flow is non‐uniform

EULER’S EQUATION

• To predict

pressure variation in moving fluid

• Euler’s Equation is an extension of the hydrostatic equation for accelerations other than gravitational • RESULTED FROM APPLYING NEWTON SECOND LAW TO A FLUID ELEMENT IN

THE FLOW OF INCOMPRESSIBLE, INVISCID FLUID

Assume that the viscous forces are zero

EULER’S EQUATION

F l

ma l

ΔP Δl

γsinα

ρa

l Taking the limit of the two terms at left side at a given time as Δl → 0   

l

(

p

 γz)  ρa l

ACCELERATION IS IN THE DIRECTION OF DECREASING PIEZOMETRIC PRESSURE!!! When “a = 0” → Euler equation reduces to hydrostatic equation!

In the x direction, for example:   

x

(

p

 γz)  ρa x (

p

 γz) 2  (

p

 γz) 1   

x

ρa x “2” and “1” refer to the location with respect to the direction l (When l = x direction, then “2” is the right-most point. When l = z direction, “2” is the highest point.)

EULER’S EQUATION

An example of Euler Equation is to the uniform acceleration of in a tank: Open tank is accelerated to the right at a rate a x For this to occur ; a net force must act on the liquid in the x-direction To accomplish this ; the liquid redistributes itself in the tank (A’B’CD) – The rise in fluid causes a greater hydrostatic force on the left than the right side → this is consistent with the requirement of “ F = ma ” – Along the bottom of tank, pressure variation is hydrostatic in the vertical direction

EULER’S EQUATION

  

l

(

p

γz)

ρa

l •

The component of acceleration in the l direction : a

x cosα

Apply the above equation along A’B’ 

d dl

( γz) 

ρa x Cos

 Apply the above equation along DC tan   -

a x g dz dl

 -

a x Cos

g

 sin 

Example 4.3

: Euler’s equation

A • The truck carrying gasoline (

γ = 6.60 kN/m3) and is slowing down

at a rate of 3.05 m/s 2 .

• 1) What is the pressure at point A?

• 2) Where is the greatest pressure & at what value in that point?

  

l

(

p

 γz)  ρa l

Solution:

• Apply Euler’s equation along the top of the tank; so z is constant • Assume that deceleration is constant • Pressure does not change with time 

d dl

(

p

γz)

ρa

l Along the top the tank 

dp dl

 ρa l

p

  ρa l

l

C

Euler’s equation in vertical direction: ( Note that a z =0 ) 

d dz

(

p

 γz)  ρa z (

p bottom

 γz

bottom

)  (

p top

 γz

top

) Pressure variation is hydrostatic in the vertical direction

Centripetal (Radial) Acceleration

a r

V t

2

r

  2

r

• For a liquid rotating as a rigid body:

V = ω r

• a r = centripetal (radial) acceleration, m/s 2 • V t = tangential velocity, m/s • r = radius of rotation, m • ω = angular velocity, rad/s

Pressure Distribution in Rotating Flow

• A common type of rotating flow is the flow in which the fluid rotates as a rigid body. • Applying Euler Equation in the direction normal to streamlines and outward from the center of rotation (

OR

INTEGRATING EULER EQUATION IN THE RADIAL DIRECTION FOR A ROTATING FLOW ) results in 

p

z

 

2

2

r g

2 

C Pressure variation in rotating flow

Note that this is not the Bernoulli equation • When flow is rotating, fluid level will rise away from the direction of net acceleration

Example 4.4:

Find the elevation difference between point 1 and 2

p

1  

z

1   2 2

r

1 2

g

p

2  

z

2   2 2

r

2 2

g

p 1 = p 2 = 0 and r 1 = 0 , r 2 = 0.25m then →

z

1 

z

2   2 2

r

2 2

g

z 2 – z 1 = 0.051m & Note that the surface profile is parabolic

Pressure Distribution in Rotating Flow

p

1  

z

1   2 2

r

1 2

g

p

2  

z

2   2 2

r

2 2

g Another independent equation;

The sum of water heights in left and right arms should remain unchanged

p = pressure, Pa γ = specific weight, N/m3 z = elevation, m ω = rotational rate, radians/second r = distance from the axis of rotation

Bernoulli Equation

Integrating Euler’s equation along a streamline in a steady flow of an

incompressible, inviscid

fluid yields

t

he Bernoulli equation:

V

2 2

g

P

 

z

C

z: Position p/ γ : Pressure head V 2 /2g: Velocity head C: Integral constant

Application of Bernoulli Equation Bernoulli Equation: – Piezometric pressure :

p + γz

– Kinetic pressure :

ρV 2 /2

For the steady flow of incompressible fluid inviscid fluid the sum of these is constant along a streamline

Application of Bernoulli Equation: Stagnation Tube

V 1 2 2 g  p 1   V 2 2 2 g  p  2

V

1 2  2  (

P

2 

P

1 )

P

1  

d

P 2   (

l

d

)

V

1 2  2  (  (

l

d

)  

d

)

V

1  2

gl

Stagnation Tube

p 1 /  p 2 /  1 2  h=V 2 /2g V 1 2 2 g  p 1   V 2 2 2 g  p  2 V 2 =0 & z 1 = z 2 V 1  2 g   p 2  p 1     2 g  h

Application of Bernoulli Equation: Pitot Tube

Bernoulli equation btw static pressure pt 1 and stagnation pt 2; V 1 2 2 g  p 1   z 1  V 2 2 2 g  p  2  z 2  H V 2 = 0 then

Pitot tube equation

; h 1  1 2 1 

V

1  2

g

 

p

2  

p

1    2 

p

 Stagnation point p 1   h 1   1  h    h 1   h   p 2 p 2  p 1    h     1  1     h   s 1 s  1  

VENTURI METER

The Venturi meter device measures the flow rate or velocity of a fluid through a pipe. The equation is based on the Bernoulli equation, conservation of energy, and the continuity equation.

Solve for flow rate Solve for pressure differential

Class Exercises:

(Problem 4.42)

Class Exercises:

(Problem 4.59)