Chapter 4FLUID DYNAMIC

(Part 2)

2

1) Pitot Tube

It is a simple velocity measuring device. If a stream of uniform velocity flows into a

blunt body, the streamlines take a pattern similar to figure below:

Streamlines around a blunt body

APPLICATIONS OF THE BERNOULLI

EQUATION

3

Note: Some move to the left and some to the right. But one, in the centre, goes

to the tip of the blunt body and stops. It stops because at this point the velocity is

zero - the fluid does not move at this one point. This point is known as the

stagnation point.

From the Bernoulli equation, we can calculate the pressure at this point. Apply

Bernoulli along the central streamline from a point upstream where the velocity is

u1 and the pressure P1 to the stagnation point of the blunt body where the

velocity is zero, u2 = 0. Also z1 = z2.

4

This increase in pressure which bring the fluid to rest is called the dynamic pressure.

or converting this to head

The total pressure is known as the stagnation pressure (or total pressure)

or in terms of head

5

The blunt body stopping the fluid does not have to be a solid. It could be a static

column of fluid. When the static and stagnation of a flowing liquid are greater than

atmospheric pressure, two piezometers, one as normal and one as a Pitot tube

within the pipe can be used in an arrangement shown below to measure velocity

of flow.

A Piezometer and a Pitot tube

Using the above theory, we have the equation for P2

The expression for velocity obtained from two pressure measurements and

the application of the Bernoulli equation.

( )121

2

112

2

112 22

1

2

1hhgVVghghVPP −=+=+=

6

EXAMPLE 1

In water treatment plant, it is required to monitor the maximum velocity in

the water mains. A pitot-tube is inserted in the center of the pipe and

connected to two piezometer tubes. If the differential pressure head is 250

mm, estimate the velocity of flow.

Using Bernoulli’s Equation:

𝑉1 = 2𝑔(ℎ2 − ℎ1)

= 2 9.81 0.25= 2.21 m/s

7

The Pitot static tube combines the tubes and they can then be easily connected

to a manometer. A Pitot static tube is shown below. This device is applicable if the

pressures to be measured are below atmospheric, or if measuring pressure in

gases. The holes on the side of the tube connect to one side of a manometer and

register the static head (h1), while the central hole is connected to the other side

of the manometer to register, as before, the stagnation head (h2).

2) Pitot Static Tube

8

Consider the pressures on the level of the centre line of the Pitot tube and using

the theory of the manometer,

( )

( ) ghhXgPgXP

PP

ghhXgPP

gXPP

man

BA

manB

A

+−+=+

=

+−+=

+=

12

1

2

We know that , substituting this into the above gives2

1122

1uPPP stagnation +==

( )

( )

−=

+=−+

m

man

ghu

uPhgP

2

2

1

2

111

or

( )

121

2 PPu

−=

Density of the fluid in

the manometer

9

Application:

Fighter Jet

10

11

The basic technology of the pitometer log is similar to that of the pitot tube on an

aircraft. Typically, the pitometer has a long tube that penetrates the ship's hull

near the keel.

Application:

Pitometer in ship

12

EXAMPLE 2

A Pitot-static probe is used to measure the velocity of an aircraft flying at

3000 m (ρ = 0.909 kg/m3). If the differential pressure reading is 3 kPa,

determine the velocity of the aircraft.

From Pitot Static Tube Calculation:

𝑉1 =2(𝑃2 − 𝑃1)

𝜌

= 2(3000)

0.909

= 81.2 m/s

( )

( )

−=

+=−+

m

man

ghu

uPhgP

2

2

1

2

111

or

( )

121

2 PPu

−=

13

3) Venturi Meter (Confined Flows)

The Venturi meter is a device for measuring volume flow rate in a pipe. It consists

of a rapidly converging section which increases the velocity of flow and hence,

reduces the pressure. It then returns to the original dimensions of the pipe by a

gently diverging diffuser section. By measuring the pressure differences, the

discharge can be calculated. This is a particularly accurate method of flow

measurement as energy loss are very small.

A Venturi meter

14

Applying Bernoulli along the streamline from point 1 to point 2 in the narrow

throat of the Venturi meter, we have

By using the continuity equation,

we can eliminate the velocity u2,

Substituting this into and rearranging

the Bernoulli equation, we get

15

To get the theoretical volume flow rate, this is multiplied by the area. To get

the actual volume flow rate, taking into account the losses due to friction, we

include a coefficient of volume flow rate

16

This can also be expressed in terms of the manometer readings

Thus the discharge can be expressed in terms of the manometer reading:

17

Notice how this expression does not include any terms for the elevation or

orientation (z1 or z2) of the Venturimeter. This means that the meter can be at any

convenient angle to function.

The purpose of the diffuser in a Venturi meter is to assure gradual and steady

deceleration after the throat. This is designed to ensure that the pressure rises

again to something near to the original value before the Venturi meter. The angle

of the diffuser is usually between 6 and 8 degrees. Wider than this and the flow

might separate from the walls resulting in increased friction and energy and

pressure loss. If the angle is less than this the meter becomes very long and

pressure losses again become significant. The efficiency of the diffuser of

increasing pressure back to the original is rarely greater than 80%.

Venturi meter with digital readout

18

A vertical venturi meter measures the flow of oil of specific gravity 0.82 and

has an entrance of 125 mm diameter and throat of 50 mm diameter. There

are pressure gauges at the entrance and at the throat, which is 300 mm

above the entrance. If the coefficient for the meter is 0.97 and pressure

difference is 27.5 kN/m2, calculate the actual volume flow rate in m3/s.

EXAMPLE 3

z2z1

19

( ) 2

2

1 01226.04

125.0142.3mA ==

23

21 /105.27 mNpp =−

2/2.8044100081.982.0 mN==

mzz 3.021 −=−

Cd = 0.97

( ) 2

2

2 001964.04

050.0142.3mA ==

SOLUTION:

Area of Entrance

Area of Throat

20

( )

smQ

g

Q

A

A

zzg

ppg

ACQ

actual

actual

dactual

/01535.0

1001964.0

01226.0

3.02.8044

105.272

01226.097.0

1

2

3

2

3

2

2

1

2121

1

=

−

−

=

−

−+

−

=

21

4) Flow Through A Small Orifice (Free Jet)

We are to consider the flow from a tank through a hole in the side close to

the base. The general arrangement and a close up of the hole and

streamlines are shown in the figure below

Tank and streamlines of flow out of the sharp- edged orifice

22

The shape of the holes edges are as they are (sharp) to minimize

frictional losses by minimizing the contact between the hole and the liquid

- the only contact is the very edge.

Looking at the streamlines, you can see how they contract after the orifice

to a minimum value when they all become parallel, at this point, the

velocity and pressure are uniform across the jet. This convergence is

called the vena contracta. (From Latin: contracted vein.). It is necessary

to know the amount of contraction to allow us to calculate the flow.

We can predict the velocity at the orifice using the Bernoulli equation.

Apply it along the streamline joining point 1 on the surface to point 2 at

the centre of the orifice.

At the surface velocity is negligible (u1 = 0) and the pressure

atmospheric (P1 = 0). At the orifice, the jet is open to the air, so again the

pressure is atmospheric (P2 = 0). If we take the datum line through the

orifice, then z1 = h and z2 = 0, leaving

23

This is the theoretical value of velocity. Unfortunately it will be an overestimate

of the real velocity because friction losses have not been taken into account. To

incorporate friction, we use the coefficient of velocity to correct the theoretical

velocity,

So the volume flow rate through the orifice is given by

Each orifice has its own coefficient of velocity, they usually lie in the range (0.97

- 0.99). To calculate the volume flow rate through the orifice, we multiply the area

of the jet by the velocity. The actual area of the jet is the area of the vena

contracta not the area of the orifice. We obtain this area by using a coefficient

of contraction for the orifice

24

Tank emptying from level h1 to h2.

The tank has a cross sectional area of A. In a time dt, the level falls by dh or the

flow out of the tank is

Time for the tank to empty

(-ve sign as dh is falling)

Expression for the discharge from the tankghACQ orificedactual 2=

25

Rearranging and substituting the expression for Qactual through the orifice gives

This can be integrated between the initial level, h1, and final level, h2, to give

an expression for the time it takes to fall this distance

26

EXAMPLE 4

A pressurized tank of water has a 10 cm diameter orifice at

the bottom, where water discharges to the atmosphere. The

water level is 3 m above the outlet. The tank air pressure

above the water level is 300 kPa (absolute) while the

atmospheric pressure is 100 kPa. Neglecting frictional effects,

determine the initial discharge rate of water from the tank.

27

121

22

2

222

1

211

2

22z

g

PP

g

Vz

g

V

g

Pz

g

V

g

P+

−=→++=++

m/s 21.4

m) 3)(m/s 81.9(2N 1

m/s kg1

kPa1

N/m 1000

kg/m1000

kPa)100300(22

)(2 222

3121

2

=

+

−=+

−= gz

PPV

/sm 0.168 3==== m/s) 4.21(4

m) 10.0(

4

2

2

2

2orifice

V

DVAV

SOLUTION:

v1 = 0, z2 = 0

P1 ≠ 0 since pressure in the tank

is not atmospheric

28

EXAMPLE 5

The water level in a tank is 20 m above the ground. A hose is

connected to the bottom of the tank, and the nozzle at the end

of the hose is pointed straight up. The tank cover is airtight,

and the air pressure above the water surface is 2 atm gage.

The system is at sea level. Determine the maximum height to

which the water stream could rise.

29

1

gage1,

11

2211

2

222

1

211

22z

g

Pz

g

PPzz

g

Pz

g

Pz

g

V

g

Pz

g

V

g

P atmatm +=+−

=→+=+→++=++

m 40.7=+

= 20

N 1

m/skg 1

atm 1

N/m 325,101

)m/s 81.9)(kg/m 1000(

atm 2 22

232z

SOLUTION:

z2 is considered since

there is height change

V1 = V2 = 0 since velocity is in

same direction and negligible

30

EXAMPLE 6

Oil of specific gravity 0.82 discharges from an open tank through an orifice of

diameter 14 mm. The coefficient of velocity is 0.88 and the coefficient of

contraction is 0.62. The center of the orifice is at a depth of 0.9 m from the

surface of the oil. Determine the diameter of the Vena Contracta and the volume

flow rate of oil through the orifice.

0.9 mØ 14 mm

31

SOLUTION :

Data given:

Specific gravity = 0.82; d = 0.014 m; Cv = 0.88; Cc = 0.62; h = 0.9 m

2422

1054.14

014.0

4m

dAo

−=

=

=

o

cc

A

AC =

Substituting the values,

24

4

10955.0

1054.162.0

mA

A

c

c

−

−

=

=

or

mmmdc 02.1101102.0 ==

(a) Finding diameter of Vena Contracta

Aactual = CcAorifice

32

(b) Theoretical velocity

smghv /0202.49.081.922 ===

Theoretical volume flow rate

smvAQ o /1047.61054.102.4 344 −− ===

Coefficient of volume flow rate

546.062.088.0 === cvd CCC

Actual volume flow rate = Cd x Theoretical volume flow rate

sL

sm

/353.0

/1053.3

1047.6546.0

34

4

=

=

=

−

−

33

For the tank shown in above, find the time required to drain the tank from a level

of 3.0 m to 0.50 m. The tank has a diameter of 1.50 m and the nozzle has a

diameter of 50 mm.

EXAMPLE 7

34

SOLUTION:

35

End of Chapter 4

Any Question?