(12) Pressure in Liquid - Pascal & U-Tube.pptx

pressure in liquid

1. 2. Pascal’s Principle When a fluid is confined in a container , the fluid exerts force on every part of the area of the walls and bottom of the container which fluid touches.

2. Application of Pascal’s principle in Hydraulic Machines - A Hydraulic Jack ( Dongkrak Hidrolik ) - hydraulic car brakes

3. Hydraulic machines  are  machinery  and tools that use liquid fluid power to do simple work. Heavy equipment  is a common example. In this type of machine ,  hydraulic  fluid is transmitted throughout the  machine  to various  hydraulic  motors and hydraulic  cylinders and becomes pressurised according to the resistance present.

4. Pascal’s equation: Where: P 1 = pressure in piston 1 (Pa) P 2 = pressure in piston 2 (Pa) F 1 = Force applied in piston 1 (N) F 2 = Force applied in piston 2 (N) A 1 = surface area in piston 1 (m 2 ) A 2 = surface area in piston 2 (m 2 ) D 1 = diameter in piston 1 (m) D 2 = diameter in piston 2 (m) P 1 = P 2 P 1 P 2

5. Example 1. An hydraulic jack used to lift up the car that has weight 1.500 N. the small area 50 cm 2 and large area 2.000 cm 2 . Calculate the force (F) needed to lift up the car? Answer: known : W = F 2 = 1.500 N A 1 = 50 cm2 A 2 = 2.000 cm2 unknown:F 1 =...? solution: A 1 = 50 cm 2 A 2 = 2000 cm 2

6. 2. Look at the picture below! Find the weight of the car (F 2 )? Answer known : D 1 = 50 cm D 2 = 200 cm F 1 = 150 N unknown: F 2 =...? solution:

7. Exercises Mr. Mark is raising a 2000 kg car on his hydraulic lift. If the area of the input piston is 9cm 2 , while the area of the output piston is 630 cm 2 , what force must be exerted on the input piston to lift the car? Answer: Known: m = 2000 kg F2 = w = m x g w = 2000 kg x 10 m/s2 = 20000 N A1 = 9 cm2 A2 = 630 cm2 Unknown: F1 = …? Solution P1 = P2 F1/ 9 cm2 = 20000 N/630 cm2 F1 = 2000 N/7 F1 = 285,71 N F2 = w = m x g = 2000 kg x 9,8 m/s2 w = 19600N So, F1 = 280 N

8. P 1 = P 2 ρ 1 x g x h 1 = ρ 2 x g x h 2 ρ 1 x h 1 = ρ 2 x h 2 3. U- t ube (Pipa U) ρ 1 = density for the 1 st liquid ρ 2 = density for the 2 nd liquid h 1 = height for the 1 st liquid h 2 = height for the 2 nd liquid

9. EXAMPLE U-tube filled by mercury (ρ 1 = 13,6 g/cm 3 ) and in the other side filled by water with height 34 cm, calculate the difference of height between mercury and water? Answer known : ρ 1 = 13,6 g/cm 3 ρ 2 = 1 g/cm 3 h 2 = 34 cm unknown: The difference height (h 2 -h 1 ) = … ? solution: ρ 1 x h 1 = ρ 2 x h 2 13,6 g/cm 3 x h 1 = 1 g/cm 3 x 34 cm h 1 = cm h 1 = 2,5 cm h 2 – h 1 = 34 cm – 2,5 cm = 31,5 cm

10. A column of an oil floats on water in a U-tube of uniform cross-section area as shown in figure. If the density of water is 1000 kg/m 3 , find the density of oil! (h 1 = 10 cm, h 2 = 9.2 cm) Exercise

11. Answer Known: h1 = 10 cm h2 = 9,2 cm p2 = 1000 kg/m3 Unknown: p1 = ….? Solution: p1 x h1 = p2 x h2 p1 x 10 cm =1000 kg/m3 x 9,2 cm p1= 920 kg/m3