Pressure and Depth. In a static, non-moving fluid

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1 Pressure and Depth In a static, non-moving fluid

2 Static Fluids Being on the surface of the earth, you can say that we dwell on the bottom of an ocean of air. The pressure we experience is primarily caused by the weight of the air, or the gravitational pull/force on that air. (Similarly, the force applied to you when someone stands on you that force is caused by the gravitational force acting on their mass.) The pressure is also caused by the collisions the air molecules make with your body which is determined by the molecule s Kinetic Energy! Since we do not experience an acceleration from the interactions between the air molecule s and us (in other words the air molecules colliding with you right now are not causing your body to accelerate), and these air molecules are, relatively speaking, at rest they re not rushing past you as in a waterfall situation a=0, ΣF = F net = 0 simply implies that the fluid is in equilibrium, and we can say that the fluid is static The following properties apply to static fluids.

3 If you were standing at the bottom of a swimming pool, where would the pressure be the greatest on your body? At your feet! Because your feet are lower (deeper) than your head there is more water weight pushing down on your feet than your head. P=F/A F = P A

4 F 1 =P 1 A F g =mg F 2 =P 2 A ΣF = 0 and F = P A ΣF =F 2 + (-F 1 ) + (-F g )=0 ΣF = F 2 - F 1 - F g = 0 F 2 = F 1 + F g P 2 A = P 1 A + mg Recall: =m/v m= V And V=l w h = A h m= A h P 2 A = P 1 A + mg P 2 A = P 1 A + A h g P 2 = P 1 + h g P 2 P 1 = h g

P 2 P 1 = ΔP = h g The volume of water does not affect this pressure change, only the height difference.

5 Fluid (or hydrostatic) pressure depends on depth, not volume Going from one depth to another, resulting in a difference of depth, h, causes a change in pressure, P. P 2 P 1 = ΔP = h g The volume of water does not affect this pressure change, only the height difference. P 2 > P 1 b/c fluid at P 2 is deeper than P 1 fluid h P 1 P 2 P 1 =pressure of that fluid at that place P 2 =pressure of that fluid at that place

6 For liquids, density,, does not vary with depth, but for gasses, like our atmosphere, can vary. P 2 = P 1 + h a g works for small h values with gasses where variation in is minimal. Who experiences more pressure, someone 6 feet deep in a swimming pool or someone 6 feet deep in Lake Shasta? Answer Same: 6 feet is 6 feet is 6 feet!! Pressure depends on depth! Show glass tube! Show bottles w/holes in side. Show siphoning.

7 Where is the dam the thickest? Why? Fluid pressure depends on depth, not volume.

8 Is there a difference between the pressure at point A and point B? NO!

9 When you are underwater, what is causing pressure on you? Both the water AND the air above you are causing pressure! P total = P water + P atmosphere P water = gh P atmosphere = X 10 5 Pa =1 atm At what depth would you experience 2atm of pressure? Assume air s pressure is X 10 5 Pa P water = gh = 1atm =1.013 X 10 5 Pa Height below surface of water=h h=p water /( water g) h=1.013 X 10 5 Pa/[(1000kg/m 3 )(9.80m/s 2 )] h=10.34m 34 ft below surface of water

10 Plumbing system ressure.asp

11 Why is your blood pressure taken on your upper arm? Because the upper arm is approximately level with your heart. No depth difference No h!

12 A water pump can be placed at the bottom of a well or at ground level. If a well is deep, which method works better?

13 Answer: pump at bottom of well. As long as the pump is powerful enough, when water is pumped or shoved into the pipe (by the pump), the water will move up the pipe. Pump at ground level does not push water up the pipe Instead, it works like a straw the pump removes air from w/in pipe, water is pushed into the pipe by the atmosphere. The pump at ground-level relies on air pressure to push water into pipe. This pump s maximum pressure will be that of the atmospheric pressure. The ground level pump will only work for shallow depths.

14 Siphoning Which side has a greater gh?

15 Pressure Gauges Mercury A mercury barometer is a type of pressure gauge that measures atmospheric pressure. P 2 = P 1 + gh = 0 + gh P atm = gh Weather pressure is often described in terms of the height of mercury compared to Standard Atmospheric Pressure = 760 mm Hg = 29.9 in Hg = 1 atm = X 10 5 Pa Standard Atmospheric Pressure

P 2 P atm Where P 2 P atm is called the Gauge Pressure (difference between

16 Open-Tube Manometer According to this situation P 1 =P atm and P 2 > P 1 P 2 > P atm P 2 is greater than P atm by an amount gh P 2 = P atm + gh, or gh = P 2 P atm Where P 2 P atm is called the Gauge Pressure (difference between atmospheric pressure and gas pressure). And P 2 is called the Absolute Pressure.

17 Gauge Pressure, Absolute Pressure Gauge pressure = the pressure over and beyond the atmospheric pressure = the amount of pressure exceeding the atmospheric pressure. Example: most tire gauges register the pressure over and above atmospheric pressure. Absolute pressure = the absolute pressure includes both the atmospheric and the gauge pressure. To determine the absolute pressure, one must add the atmospheric pressure to the gauge pressure. P absolute = P atm + P gauge Example: If a tire gauge registers 2.20X10 5 Pa, what s the absolute pressure? P absolute = P atm + P gauge = 1.013X10 5 Pa X10 5 Pa P absolute = 3.21X10 5 Pa

18 Gage Pressure

Blood begins to flow again when the pressure created by the heart at the peak of its beating cycle exceeds the cuff

19 The sphygmomano -meter Squeeze the bulb, inflate the cuff in order to cut off blood flow from artery below cuff. Open release valve, cuff pressure drops. Blood begins to flow again when the pressure created by the heart at the peak of its beating cycle exceeds the cuff pressure. Using a stethoscope, the operator listens for the initial flow and then reads the cuff gauge pressure with an open-tube manometer. Called the systolic pressure.

21 Pascal s Principle For a completely enclosed fluid, if you increase the pressure in one part, then it is transmitted undiminished to all parts of the fluid and the enclosing walls. Example: manual brakes on a car P foot = P brake F/A = F/A assuming h=0 Increasing area, A, causes force, F, to increase for same pressure, P!

22 Hydraulics! Movable piston The pressure (pressure, not force!) that the left piston exerts against the water (at rest) will be exactly equal to the pressure that the water exerts against the right piston if the levels are at the same height. P 1 = P 2 P in = P out Fluid is at same level, h=0 F/A = F/A

23 Let s say that 1N of force is exerted on the left piston which has an area of 1 cm 2. If the right piston s area is 50 cm 2, then what is the force exerted by the water on the right piston? F 2 = 50N When 1N of force is exerted on the left piston, the additional pressure of 1N per 1cm 2 is transmitted throughout the fluid and up against the larger piston (A= 50 cm 2 ). The additional pressure of 1N/cm 2 is exerted against every square centimeter of the larger piston. And, since there are 50 cm 2, the total extra force exerted on the larger piston is 50N. This is 50X greater than the force on the smaller piston! 1N input = 50N output!

24 Forces can be multiplied with such a device! 1N input = 50N output! This principle can be applied to multiply forces to any amount Thus the hydraulic press! Example: manual brakes in your car P foot-pedal = P brake-pad F f /A f = F b /A b Increase A and F increases for same P For Pascal s Principle problems, usually A is of a circle; A= r 2.

In a hydraulic car lift, the input piston has a radius of 0.0120 m and a negligible weight. The output plunger has a radius of 0.150 m. The combined weight of the car and plunger is 20,500 N.

26 In a hydraulic car lift, the input piston has a radius of m and a negligible weight. The output plunger has a radius of m. The combined weight of the car and plunger is 20,500 N. The lift uses hydraulic oil that has a density of 800 kg/m 3. What input force (F 1 ) is needed to support the car and the output plunger when the bottom surfaces of the piston and the plunger are at (a) the same level, and (b) the levels shown in the figure with h=1.10m?

27 What input force (F 1 ) is needed to support the car and the output plunger when the bottom surfaces of the piston and the plunger are at (a) the same level? P 1 = P 2 P in = P out Fluid is at same level, h=0 F 1 /A 1 = F 2 /A 2 F 1 = A 1 (F 2 /A 2 ) F 1 =[ (0.0120m) 2 ] (20,500N)/[ (0.150m) 2 ] F 1 = 131 N

28 What input force (F 1 ) is needed to support the car and the output plunger when the bottom surfaces of the piston and the plunger are at (b) the levels shown in the figure with h=1.10m? P 1 + h g = P 2 F 1 /A 1 + h g = F 2 /A 2 F 1 /A 1 = F 2 /A 2 h g F 1 = [F 2 /A 2 h g] A 1 F 1 ={(20,500N)/[ (0.150m) 2 ] [(800kg/m 3 ) (9.80m/s 2 ) (1.10m)]} [ (0.0120m) 2 ] F 1 = 127 N

29 A backhoe uses a hydraulic fluid to generate a large output force, starting with a small input force.

30 Hydraulic cylinders

31 Guess who??

Properties of Fluids SPH4C

Properties of Fluids SPH4C Fluids Liquids and gases are both fluids: a fluid is any substance that flows and takes the shape of its container. Fluids Liquids and gases are both fluids: a fluid is any substance