smk perguruan cikini physics fluid dynamics. created by abdul rohman hal.: 2 fluid dinamics laminer...
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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.
Hal.: 3 FLUID DINAMICS Created by Abdul Rohman
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)
Hal.: 4 FLUID DINAMICS Created by Abdul Rohman
THE EQUATION OF CONTINUITY
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
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THE EQUATION OF CONTINUITY
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
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THE EQUATION OF CONTINUITY
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
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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
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BERNOULLI’S PRINCSIPLE
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)
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BERNOULLI’S PRINCSIPLE
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)
Hal.: 10 FLUID DINAMICS Created by Abdul Rohman
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
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THE APPLICATION OF BERNOULLI’S PRINCIPLE
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
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THE APPLICATION OF BERNOULLI’S PRINCIPLE
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.
Hal.: 13 FLUID DINAMICS Created by Abdul Rohman
THE APPLICATION OF BERNOULLI’S PRINCIPLE
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
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THE APPLICATION OF BERNOULLI’S PRINCIPLE
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
Hal.: 15 FLUID DINAMICS Created by Abdul Rohman
THE APPLICATION OF BERNOULLI’S PRINCIPLE
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
Hal.: 16 FLUID DINAMICS Created by Abdul Rohman
THE APPLICATION OF BERNOULLI’S PRINCIPLE
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
Hal.: 17 FLUID DINAMICS Created by Abdul Rohman
THE APPLICATION OF BERNOULLI’S PRINCIPLE
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)
Hal.: 18 FLUID DINAMICS Created by Abdul Rohman
THE APPLICATION OF BERNOULLI’S PRINCIPLE
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
Hal.: 19 FLUID DINAMICS Created by Abdul Rohman
THE APPLICATION OF BERNOULLI’S PRINCIPLE
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)
Hal.: 20 FLUID DINAMICS Created by Abdul Rohman
THE APPLICATION OF BERNOULLI’S PRINCIPLE
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)
Hal.: 21 FLUID DINAMICS Created by Abdul Rohman
THE APPLICATION OF BERNOULLI’S PRINCIPLE
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
Hal.: 22 FLUID DINAMICS Created by Abdul Rohman
THE APPLICATION OF BERNOULLI’S PRINCIPLE
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
Hal.: 23 FLUID DINAMICS Created by Abdul Rohman
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?
Hal.: 24 FLUID DINAMICS Created by Abdul Rohman
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