One Dimensional Steady Flow One Dimensional Steady Flow

1- Euler’s Equation (Equation of Motion)

Fluid Dynamics

θ

P

VP+dP

W= Ads

V+dV

Z

dZ

1

2

ds

Applying Newton’s law:

∑ F = mass x acceleration

PA – (P+dP)A - Ads Cos = Ads V

ds

dV

ds

dzcosθ

dsdV

PA – (P+dP)A - Ads = AdsV dsdz

Dividing by Ads we obtain:

– – =dsdz

dsdP

γ1

g

1ds

22Vd

0dzg2

2Vd

γdP

Euler's Equation

2 -Bernoulli’s Equation 2 -Bernoulli’s Equation

From Euler’s Equation: for incompressible, one-dimensional by integration and take and g as constants.

Constantdzg2

2Vd

γdP

Hzg2

V

γP 2

Where: H is constant and termed as the total head Where: H is constant and termed as the total head

• Steady flow: The Bernoulli equation can also be written between any two points on the same streamline as

DATUM

z2z1

p2/g

v22/2g

p1/g

v12/2g

21

TOTAL HEAD

2

222

1

211 z

g2

V

γ

Pz

g2

V

γ

P

Hydraulic Grade Line (HGL) and Energy Grade Line (EGL)

• Each term in this equation has the dimension of length and represents some kind of “head” of a flowing fluid as follows:

• P/ρg is the pressure head; it represents the height of a fluid column that produces the static pressure P.

• v2/2g is the velocity head; it represents the elevation needed for a fluid to reach the velocity v during frictionless free fall.

• z is the elevation head; it represents the potential energy of the fluid.

2

.2

P vz H const

g g

In an idealized Bernoulli-type flow, EGL is horizontal and its height remains constant. But this is not the case for HGL when the flow velocity varies along the flow.

Static, Dynamic, and Stagnation Pressures

• The sum of the static, dynamic, and hydrostatic pressures is called the total pressure. Therefore, the Bernoulli equation states that the total pressure along a streamline is constant.

• The sum of the static and dynamic pressures is called the stagnation pressure, and it is expressed as

2

( )2stag

vP P kPa

Measurement of static and dynamic pressure

• When static and stagnation pressures are measured at a specified location, the fluid velocity at that location can be calculated from:

2

211

2

pvp

)pp(2

v 121

Pitot- Static Tube

Bernoulli’s equation is assumed to hold along the center streamlineBernoulli’s equation is assumed to hold along the center streamline

If the tube is horizontal, z1 = z2 and we can solve for V2:

We relate the velocities from the incompressible continuity relation

Example(8-1)

Water is flowing from a hose attached to a water main at 400 kPa gage. A child places his thumb to cover most of the hose outlet, increasing the pressure upstream of his thumb, causing a thin jet of high-speed water to emerge. If the hose is held upward, what is the maximum height that the jet could achieve?

Solution

2 21

2 3 2

400 1000 / 1 . /

(1000 / )(9.81 / ) 1 1

40.8

atmP P kPa N m kg m sz

g kg m m s kPa N

m

2

222

1

211

22gz

vpmgz

vpm

oo

Z1 = 0.0, v1 = 0.0, v2 = 0.0, p2 = patm

A piezometer and a pitot tube are tapped into a horizontal water pipe, to measure static and stagnation (static + dynamic) pressures. For the indicated water column heights, determine the velocity at the center of the pipe.

Example(8-2)

Solution• P1 = ρg(h1+h2)

• P2 = ρg(h1+h2 +h3)2 2

1 1 2 21 22 2

P v P vz z

g g g g

21 2 1

2

v P P

g g

Where z1 =0.0, v2 = 0.0 and z2 =0.0

21 2 3 1 21 2 1

3

( )

2

g h h h h hv P Ph

g g g

21 32 2(9,81 / )(0.12 ) 1.53 /v gh m s m m s

Applications of Bernoulli’s Equation

1 -Flow through Orifice:

Fluid

2

H

21

1

loss2

222

1

2

1 hzg2

V

γ

Pz

g2

V

γ

P

E1 = E2 + Losses1-2 H.G.L.

1hloss

A1V1 = A2V 2

A1<< A2 V1 = 0

From Continuity Equation Neglect hloss

2

222

11 z

g2

V

γ

Pz

γ

P

H g2V2

g2

V)z(z

γ

PP 22

2121

g2

VH

22

or

or

For ideal case without losses

Q = A2V2 H g2AVAQ 222 or

H g2ACQ d actual

H g2AQ ltheoretica

Where, Cd (Coefficient of discharge can be determined from calibration

Cd about 0.6

Cd =1 Hloss= 0or

2 -Venturi meter:

Is used to measure the flow rate for Liquid and gases.

1

2

3

throat

z1z2

z3Datum

section

section

ltheoretica

actuald Q

QC

1 2 3

P

P VV

Flow

Q = A1V1 = A2V 2= A3V 3

dA

21

1

loss2

222

1

2

1 hzg2

V

γ

Pz

g2

V

γ

P

)z(h)z(hg2

VV2211

21

22

Applying Bernoulli’s equation between sections (1) and (2)

E1 = E2 + Losses1-2

L1L2

Hg2

VV 21

22

A1V1 = A2V 2 21

21 V

A

AV

Neglect hloss

Hg2A

A1V

21

222

2

Hg2AA

AV

22

21

12

Hg2AA

AAQ

22

21

12

gH2AA

AACQ

22

21

12

d

Cd can be estimated experimentally by calibration and its value is about 0.96.

Hg2VA

AV 2

2

2

1

222

Then

To measure the total head H experimentally:

1 2 3

P

P VV

Flow dA

RL

R’L’

y y

h h

By using the U tube manometer.

PL = P1 + 1h + 1y

1γ

2γ

PR = P2 + 1h + 2y

PL = PR

P1 + 1h + 1y = P2 + 1h + 2y

P1- P2 = y (2 - 1)

1 H = y (2 - 1)

1

γ

γy

γ

)γ(γyH

1

2

1

12 gH2AA

AACQ

22

21

12

d

Example:

A nozzle as shown in figure has the following data:Q = 60 liter/sec. of water, d1 = 25 Cm., d2 = 15 Cm. and P1 = 1 bar. Find P2. Neglect lossesSolution: 1

2

Applying Bernoulli’s equation between sections (1) and (2)

2

222

1

2

1 zg2

V

γ

Pz

g2

V

γ

P 1

E1 = E2

Assuming no losses(1)

Q = A1V1 = A2V 2

2

22

1

21 V

4

πdV

4

πd60 1.222

(0.25)π

10604V

2

3

1

m/sec.

398.3)15(0.π

10604V

2

3

2

m/sec.

Substituting in (1) P2 = 0.9486 bar

Example:

A nozzle as shown in figure has the following data: For water, d1 = 20 Cm., d2 = 5 Cm., z1 = 5 m, z2 = 3 m, P1 = 5 bar, V1 = 1 m/sec. Find P2 and V2.

Solution: 1

2

Applying Bernoulli’s equation between sections (1) and (2)

2

222

1

2

1 zg2

V

γ

Pz

g2

V

γ

P 1

)1(

z1

z2

P2 = 3.9 bar

and

V2 = 1600 Cm./sec.

Q = A1V1 = A2V 2

3 -Orifice meter: Is used to measure the flow rate for Liquid and gases in a pipe.

2

d d/2

d1 do

Hg2AA

ACV

22

21

12

Hg2AA

AACQ

22

21

12

1

H

Applying Bernoulli’s and Continuity equations:

Flow

γ

)Pg(P2

AA

C1

CCAVAQ 21

2

1

22c

vc222

Orifice meter

Cc : is the area coefficient.

.theo

.actc A

AC

Cv : is the velocity coefficient.

.theo

.actv V

VC

< 1

<1

vc.theo

.actd CC

Q

QC 4

1

o

d

dd

1

CC

d1 do

Vena Contracta: vcd CCC

Where:

Z

yZ1

Z2

V1If V1 = 3 m/sec.

V2 = 10 m/sec.

z1 = ?? m.

z = 2 m.

z2 = 1 m.

y = ?? m.

Find:

2

222

1

2

1 zg2

V

γ

Pz

g2

V

γ

P 1

P1 = P2 = 0 z1 =( z + y ) m.

4 -Open Channel Flow

y = 3.64 m. and

V2

x

y

H

1

1

2

2

5 -Notches and Weir:

g2

Vh)y(x

g2

Vyx

2221

g2

Vh2V

2

2

1

g

b

hδh

Area of strip = b. δh

Velocity through the strip = gh2

Discharge through the strip = δh gh2

Integrating from h = 0 to h = H

H

0

th dhbhg2Q2

1

B

b2

3

Hg2B3

2Q th

If b = B = constant

2

θ

hδh

If V- Notch

25

H2

θtang2

15

8Q th