1. SMK PERGURUAN CIKINI PHYSICS FLUID DYNAMICS

2. Created by Abdul Rohman Hal.: 2 FLUID DINAMICS 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.

3. Created by Abdul Rohman Hal.: 3 FLUID DINAMICS 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: and Note: Q = debit of fluid flow (m 3 /s) V = volume of fluid in motion (m 3 ) t = time (s) v = the velocity of fluid flow (m/s)

4. Created by Abdul Rohman Hal.: 4 FLUID DINAMICS 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: Q 1 = debit of fluid flow part first (m 3 /s) Q 2 = debit of fluid flow part second (m 3 /s) A 1 = section area part first (m 2 ) A 2 = section area part second (m 2 ) v 1 = the velocity of fluid flow 1 (m/s) v 2 = kecepatan cairan bagian 2 (m/s)

5. Created by Abdul Rohman Hal.: 5 FLUID DINAMICS 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 =  r 2 v = 3.14 (2 x 10 -2 m) x 4 m/s = 5.024 m 3 /s

6. Created by Abdul Rohman Hal.: 6 FLUID DINAMICS 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. d 1 = 12 cm  r = 6 cm = 6 x 10 -2 m d 2 = 8 cm  r = 4 cm = 2 x 10 -2 m A 1 =  r 1 2 = 3,14 x (6 cm) 2 = 113, 04 cm 2 A 1 =  r 1 2 = 3,14 x (4 cm) 2 = 50,24 cm 2 V 1 = 10 cm/s and v 2 = ……? A 1 v 1 = A 2 v 2 113.04 cm 2 x 10 cm/s = 50.24 cm 2 Solution

7. Created by Abdul Rohman Hal.: 7 FLUID DINAMICS BERNOULLI’S PRINCSIPLE The pressure of fluid in a place with large velocity is smaller than that of in the place with small velocity. Note: p = pressure (N/m 2 )  = density of fluid (kg/m 3 ) g = gravitasional acceleration (m/s 2 ) h = height of fluid from reference point (m) v = velocity of fluid (m/s) Bernoulli’s equation

8. Created by Abdul Rohman Hal.: 8 FLUID DINAMICS BERNOULLI’S PRINCSIPLE There are two special cases related to Bernoulli’s equation. 1. The fluid at a rest or not flowing (v 1 = v 2 = 0) This equation expresses the hydrostatic pressure in the liquid at certain height. Note: p 1 and p 2 = pressure at point 1 and 2 (N/m 2 ) h 1 and h 2 = the height of place 1 and 2 (m)  = density of fluid (kg/m 3 ) g = gravitasional acceleration (m/s 2 )

9. Created by Abdul Rohman Hal.: 9 FLUID DINAMICS BERNOULLI’S PRINCSIPLE 2. The fluid in motion in a horizontal pipe (h 1 = h 2 = h) This equation expresses if v 2 > v 1, then p 1 > p 2, it means if the the velocity of fluid flow in a place is big then its pressure is small and holds the contrary. Note: p 1 and p 2 = pressure at point 1 and 2 (N/m 2 ) v 1 and v 2 = the velocity at point 1 and 2 (m/s)  = density of fluid (kg/m 3 )

10. Created by Abdul Rohman Hal.: 10 FLUID DINAMICS THE APPLICATION OF BERNOULLI’S PRINCIPLE Determining the velocity and debit of water spray in perforated tank Note: Q = the flow debit m 3 /s v = the velocity of water spray at the lack m/s h = the height of water above the hole m g = gravitation acceleration m/s 2 A = section area of lack hole m 2 h Q = A.v water

11. Created by Abdul Rohman Hal.: 11 FLUID DINAMICS 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/s 2 )? Solution h 1 = 1.25 m h 2 = 45 cm = 0.25 m v = …? The velocity of water flow from leak hole 1.25 cm 1.25 m water

12. Created by Abdul Rohman Hal.: 12 FLUID DINAMICS THE APPLICATION OF BERNOULLI’S PRINCIPLE The water path is partially a parabolic motion with elevation angle a = 0 o (v 0 horizontal direction) Thus, the water falls at 1.2 m from the tank.

13. Created by Abdul Rohman Hal.: 13 FLUID DINAMICS THE APPLICATION OF BERNOULLI’S PRINCIPLE Venturimeter flow velocity v 2 flow velocity v 1 Note: p 1 = pressure of point 1 st N/m 2 p 2 = pressure of point 2 nd N/m 2  = density of fluid kg/m 3 v 1 = velocity of fluid at point 1 m/s A 1 = section area of point 1 m 2 A 2 = section area of point 2 m 2 demonstration Source:www.google.com

14. Created by Abdul Rohman Hal.: 14 FLUID DINAMICS THE APPLICATION OF BERNOULLI’S PRINCIPLE A venturimeter with the big section area 10 cm 2 and small section area 5 cm 3 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/s 2 )? Example 15 cm A2A2 A1A1 v1v1 v2v2

15. Created by Abdul Rohman Hal.: 15 FLUID DINAMICS THE APPLICATION OF BERNOULLI’S PRINCIPLE Solution A 1 = 10 cm 2 = 10 x 10 -4 m 2 A 2 = 5 cm2 = 5 x 10 -4 m 2 h = 15 cm = 15 x 10 2 m g = 10 m/s 2, v 2 = …? To determine the velocity v 2 use the equation of continuity. Thus, the velocityof flow in the big and small section are 1 m/s and 2 m/s

16. Created by Abdul Rohman Hal.: 16 FLUID DINAMICS 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

17. Created by Abdul Rohman Hal.: 17 FLUID DINAMICS THE APPLICATION OF BERNOULLI’S PRINCIPLE Pitot tube A pilot tube is an instrument used to measure the speed of a gas or air flow. Note: h = height difference of liquid columm surface in manometer (m) g = gravitation acceleration (m/s 2 )  = density of gas (kg/m 3 )  ’ = density of liquid in manometer (kg/m 3 ) v = speed of air or gas flow (m/s)

18. Created by Abdul Rohman Hal.: 18 FLUID DINAMICS THE APPLICATION OF BERNOULLI’S PRINCIPLE A pitot tube is used to measure the speed of oxigen gas flow with density 1.43 kg/m 3 in a pipe. If the height difference of liquid at both feet of manometer is 5 cm and liquid density is 13600 kg/m 3. Calculate the speed of gas flow at that pipe! (g = 10 m/s 2 ) Example Solution  = 1,43 kg/m 3  ’= 13600 kg/m 3 h = 5 cm = 0,05 m g = 10 m/s 2 v =...? Thus, the speed of oxygen flow in the pipe is 97,52 m/s

19. Created by Abdul Rohman Hal.: 19 FLUID DINAMICS THE APPLICATION OF BERNOULLI’S PRINCIPLE Lift force of airplane’s wings F 2 = p 2 A F 1 = p 1 A v2v2 v1v1 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. Note: F 1 = pushing force of plane to the upward (N) F 2 = pushing force of plane to the downward (N) F 1 – F 2 = lift force of plane (N) p 1 = pressure at the downside (N/m 2 ) p 2 = pressure at the upside (N/m 2 ) A = section area of wing (m 2 )

20. Created by Abdul Rohman Hal.: 20 FLUID DINAMICS THE APPLICATION OF BERNOULLI’S PRINCIPLE The equation of lift force above can also be expressed as follow: Note: F 1 = pushing force of plane to the upward (N) F 2 = pushing force of plane to the downward (N) F 1 – F 2 = lift force of plane (N) v 1 = the velocity of air below the wing (m/s) v 2 = the velocity of air above the wing (m/s)  = the density of air (kg/m 3 )

21. Created by Abdul Rohman Hal.: 21 FLUID DINAMICS 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/m 2 ? (  = 1.29 kg/m 3 ) Example

22. Created by Abdul Rohman Hal.: 22 FLUID DINAMICS THE APPLICATION OF BERNOULLI’S PRINCIPLE Solution p 2 – p 1 = 10 N/m v 2 = 60 m/s h 1 = h 2 v 1 = …? Thus, the velocity of air flow at the upside of the plane’s wings is 60.13 m/s

23. Created by Abdul Rohman Hal.: 23 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/m 3 and g = 9.8 m/s 2, 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?

24. Created by Abdul Rohman Hal.: 24 FLUID DINAMICS