SMK PERGURUAN CIKINI PHYSICS FLUID DYNAMICS. Created by Abdul Rohman Hal.: 2 FLUID DINAMICS LAMINER...

SMK PERGURUAN CIKINIPHYSICSPHYSICS

FLUID DYNAMICS

Hal.: 2 FLUID DINAMICS Created by Abdul Rohman

LAMINER AND TURBULENT FLOWS

The flows lines of moving fluid consist of two kinds, they are:

The flow of fluids that follow a straight or curved line that its end and starting point are clear and there are no crossing lines of flow.

The flow of fluid that is marked the existence of rotasi flow and its particlesdirection is different, or opposite with direction of the whole fluid motion..

Source: http://www.math.ucsb.edu/~hdc/res/rhomesh.gif

Aliran laminer dan aliran turbulen

1.

2.


THE EQUATION OF CONTINUITY

If a fluid flows in a pipe with section area A and the velocity of fluid v, then the amount of fluid (volume) that flows passing that section per unit of time is called debit.

In the form of equation, debit is formulated as follow:

vAQ t

VQ and

Note:Q = debit of fluid flow (m3/s)V = volume of fluid in motion (m3) t = time (s)v = the velocity of fluid flow (m/s)



If a fluid flows in a steady flow through a pipe with different section areas, then the volume of fluid that passes at each section is equally large in the same time interval.

The continuinity equation expresses that at ideal fluid flow, the cross product of fluid flow speed with its section area is constan.

Note:Q1 = debit of fluid flow part first (m3/s)Q2 = debit of fluid flow part second (m3/s)A1 = section area part first (m2)A2 = section area part second (m2)v1 = the velocity of fluid flow 1 (m/s)v2 = kecepatan cairan bagian 2 (m/s)

2211

21

vAvA

QQ



Example

1. The average velocity of water flow in a pipe with diameter 4 cm is 4 m/s. Calculate the amount of fluid (water) flowing per second (Q)!

Solution

d = 4 cm r = 2 cm = 2 x 10-2 m

v = 4 m/s

Q = …?

Q = A v = r2 v

= 3.14 (2 x 10-2 m) x 4 m/s

= 5.024 m3/s



2. A pipe with diameter 12 cm and the narrowing end point with diameter 8 cm. If the velocity of flow in the large diameter is 10 cm/s, calculate the velocity of flow at the narrowing end.

d1 = 12 cm r = 6 cm = 6 x 10-2 m

d2 = 8 cm r = 4 cm = 2 x 10-2 m

A1 = r12 = 3,14 x (6 cm)2 = 113, 04 cm2

A1 = r12 = 3,14 x (4 cm)2 = 50,24 cm2

V1 = 10 cm/s and v2 = ……?

A1 v1 = A2 v2

113.04 cm2 x 10 cm/s = 50.24 cm2

Solution

scmv

v

5.22

24.50

4.1130

2

2


BERNOULLI’S PRINCSIPLE

The pressure of fluid in a place with large velocity is smaller than that of in the place with small velocity.

constant 221 vhgp

Note:p = pressure (N/m2) = density of fluid (kg/m3)g = gravitasional acceleration (m/s2)h = height of fluid from reference point (m)v = velocity of fluid (m/s)

Bernoulli’s equation



There are two special cases related to Bernoulli’s equation.

1. The fluid at a rest or not flowing (v1 = v2 = 0)

)( 1221 hhgpp

This equation expresses the hydrostatic pressure in the liquid at certain height.

Note:p1 and p2 = pressure at point 1 and 2 (N/m2)h1 and h2 = the height of place 1 and 2 (m) = density of fluid (kg/m3) g = gravitasional acceleration (m/s2)



2. The fluid in motion in a horizontal pipe (h1 = h2 = h)

)(2

1 21

2221 vvpp

This equation expresses if v2 > v1, then p1 > p2, it means if the the velocity of fluid flow in a place is big then its pressure is small and holds the contrary.

Note:p1 and p2 = pressure at point 1 and 2 (N/m2)v1 and v2 = the velocity at point 1 and 2 (m/s) = density of fluid (kg/m3)


THE APPLICATION OF BERNOULLI’S PRINCIPLE

Determining the velocity and debit of water spray in perforated tank

ghv 2

ghAQ 2

Note:Q = the flow debit m3/sv = the velocity of water spray at the lack m/sh = the height of water above the hole mg = gravitation acceleration m/s2

A = section area of lack hole m2

hQ = A.v

water



Example

A tank contains water of 1.25 m high. There is a leak hole in the tank 45 cm from the bottom of tank. How far the place of water falls measured from the tank (g =10 m/s2)?

Solution

h1 = 1.25 m

h2 = 45 cm = 0.25 m

v = …? smsm

msm

mmsm

hhgv

/4/16

)80.0(/20

)45.0125(/102

)(2

22

2

2

21

The velocity of water flow from leak hole

1.25 cm

1.25 m water



The water path is partially a parabolic motion with elevation angle a = 0o (v0 horizontal direction)

st

st

t

tsmm

tsmm

tgtvy

sm

m

3.0

9.0

/545.0

)/10(045.0

sin

2

/5

45.0

22

2221

221

0

2

m

ssm

tvx

2.1

)3.0)(1)(/4(

)(cos0

Thus, the water falls at 1.2 m from the tank.



Venturimeter

]1)/[(

)(22

21

211

AA

PPv

flow velocity v2

flow velocity v1

Note:p1 = pressure of point 1st N/m2

p2 = pressure of point 2nd N/m2

= density of fluid kg/m3

v1 = velocity of fluid at point 1 m/s A1 = section area of point 1 m2

A2 = section area of point 2 m2

demonstrationSource:www.google.com



A venturimeter with the big section area 10 cm2 and small section area 5 cm3 is used to measure the velocity of water flow. If the height difference of water surface is 15 cm.

Calculate the velocity of water flow in the big and small section (g = 10 m/s2)?

Example

15 cm

A2

A1

v1 v2



Solution A1 = 10 cm2 = 10 x 10-4 m2

A2 = 5 cm2 = 5 x 10-4 m2

h = 15 cm = 15 x 102 m

g = 10 m/s2 , v2 = …?

1105

1010

1015/102

1

2

2

24

24

22

2

2

1

m

m

msm

A

A

hgv

To determine the velocity v2 use the equation of continuity.

sm

smm

m

vA

Av

vAvA

/2

/1105

101024

24

12

12

2211

Thus, the velocityof flow in the big and small section are 1 m/s and 2 m/s



Mosquito sprayer

If the piston is pumped, the air from the cylinder tube is forced out through the narrow hole. The air coming out from this narrow hole has height velocity so that decreasing the air pressure above the nozzle.

Because the air pressure above the nozzle is smaller than that of on liquid surface in the tube, then the liquid will spray out through the nozzle.

hole

low pressure

atmospheric pressure



Pitot tubeA pilot tube is an instrument used to measure the speed of a gas or air flow.

gh

v'2

Note:h = height difference of liquid columm surface in manometer (m)g = gravitation acceleration (m/s2) = density of gas (kg/m3)’ = density of liquid in manometer (kg/m3)v = speed of air or gas flow (m/s)



A pitot tube is used to measure the speed of oxigen gas flow with density 1.43 kg/m3 in a pipe. If the height difference of liquid at both feet of manometer is 5 cm and liquid density is 13600 kg/m3.

Calculate the speed of gas flow at that pipe! (g = 10 m/s2)

Example

Solution

= 1,43 kg/m3

’= 13600 kg/m3

h = 5 cm = 0,05 m

g = 10 m/s2

v =...?

sm

mkg

msmmkg

ghv

/52.97

/43.1

05.0/10/136002

'2

3

23

Thus, the speed of oxygen flow in the pipe is 97,52 m/s



Lift force of airplane’s wingsF2 = p2 A

F1 = p1 A

v2

v1

According to Bernoulli’s principle, if the speed of air flow at the up side of wing is larger than that of at the downside, then the air pressure at upside of wing smaller than that of at the downside.

AppFF )( 2121 Note:F1 = pushing force of plane to the upward (N)F2 = pushing force of plane to the downward (N)F1 – F2 = lift force of plane (N)p1 = pressure at the downside (N/m2)p2 = pressure at the upside (N/m2)A = section area of wing (m2)



The equation of lift force above can also be expressed as follow:

AvvFF )(2

1 21

2221

Note:F1 = pushing force of plane to the upward (N)F2 = pushing force of plane to the downward (N)F1 – F2 = lift force of plane (N)v1 = the velocity of air below the wing (m/s)v2 = the velocity of air above the wing (m/s) = the density of air (kg/m3)



If the velocity of air flow at the downside of a plane’s wings is 60 m/s, what is the velocity at the upside of the plane’s wings if the upward pressure obtained is 10 N/m2?

( = 1.29 kg/m3)

Example



sm

smv

mNsm

ppvv

ppvv

ppvv

hgvphgvp

/13.60

/5.3615

29.1

/)10(2)/60(

)(2

)(2

)(

221

22

1222

21

1222

21

1222

212

1

2222

121

212

11

Solution

p2 – p1 = 10 N/m

v2 = 60 m/s

h1 = h2

v1 = …?

Thus, the velocity of air flow at the upside of the plane’s wings is 60.13 m/s


FLUID DINAMICS

Exercise!

1. The density of ball weighing 0.5 kg and diameter 10 cm is …

2. The hidrostatic pressure on a vessel surface 30 cm under water surface with density 100 kg/m3 and g = 9.8 m/s2, is ….

3. Debit of fluid have dimention of ….

4. A tank that is 4 m high from land is filled up with water. A valve (tap) placed 3 meters under water level in the tank. If the valve is opened, what is the velocity of water spray?

SMK PERGURUAN CIKINI PHYSICS FLUID DYNAMICS. Created by Abdul Rohman Hal.: 2 FLUID DINAMICS LAMINER...

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