10/9/2017 LET S PERFORM 4 EXPERIMENTS: UNIT 1 FLUID STATICS AND DYNAMICS CHAPTER 11 FLUIDS IN MOTION SNORING BERNOULLI'S PRINCIPLE

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1 1/9/17 AP PHYSICS LET S PERFORM 4 EXPERIMENTS: 1. Cans on a string. UNIT 1 FLUID STATICS AND DYNAMICS CHAPTER 11 FLUIDS IN MOTION. Blowing a piece of paper. 3. Index card & straw. 4. Ping Pong ball and hair dryer. BERNOULLI'S PRINCIPLE SNORING Bernoulli s Principle: The pressure that a fluid exerts on a surface decreases as the speed with which the fluid moves across the surface increases. Bernoulli's principle has many important applications, including fluid-flow implications in biological systems such as blood through blood vessels. FLOW RATE AND FLUID SPEED FLOW RATE AND FLUID SPEED Flow rate is defined as the volume V of fluid that moves through a cross section of a pipe divided by the time interval Δt during which it moved: The SI unit of flow rate is m 3 /s. 1

2 1/9/17 FLOW RATE AND FLUID SPEED FLOW RATE Q = A v [ m3 s ] = m [ m s ] Q = V t Q = A l t Q = A v careful with symbols v = velocity V = volume The heart pumps blood at an average flow rate of 8 cm 3 /s into the aorta, which has a diameter of 1.5 cm. Determine the average speed of the blood flow in the aorta. Write an expression for velocity before solving. Water runs through a water main with a velocity of 3 m/s. Calculate the radius of the cross sectional area, if the flow rate of the water in the pipe is 37.7 m 3 /s. Write an expression for radius before solving. v = Q π r v = 45.7 cm/s v =.45 m/s r = Q π v r = m r = cm PIPES Pipe with the same cross sectional area throughout Frequently the radius of a single pipe varies from one part of the pipe to another. Pipe with changing cross sectional area How does this affect the flow of rate and the speed of the moving fluid in the different parts of the fluid?

3 1/9/17 ASSUMPTIONS: The vessel does not expand. The liquid is incompressible. CONTINUITY EQUATION Fluid does not back up, therefore Q is constant. To keep Q constant, the speed of the fluid must change. CONTINUITY EQUATION The flow rate past cross section 1 will equal that past cross section. Q 1 = Q Water runs through a water main of cross-sectional area.4 m with a velocity of 6 m/s. Calculate the velocity of the water in the pipe when the pipe tapers down to a cross-sectional area of.3 m. Write an expression for velocity before solving. v 1 is the average speed of the fluid passing cross section A 1 and v is the average speed of the fluid passing cross section A. v = v 1 A 1 A v = 8 m/s Blood normally flows at an average speed of about 1 cm/s in a large artery with a radius of about.3 cm. Assume that the radius of a small section of the artery is reduced by half because of atherosclerosis, a thickening of the arterial walls. Determine the speed of the blood as it passes through the constriction. CAUSES AND TYPES OF FLUID FLOW Fluid flow is caused by differences in pressure. When the pressure in one region of the fluid is lower than the pressure in another region, the fluid tends to flow from the higher-pressure region toward the lower-pressure region. For example, large masses of air in Earth's atmosphere move from regions of high pressure into regions of low pressure. v = 4 v 1 v = 4 cm/s 3

4 1/9/17 STREAMLINE FLOW AND TURBULENT FLOW Streamline flow: Every particle of fluid that passes a particular point follows the same path as particles that preceded it. Turbulent flow: Characterized by agitated, disorderly motion. THE BERNOULLIS: A FAMILY OF GENIUSES Uncle: Jakob Bernoulli Mathematician Artist (who "was the first to discover the theory of probability") Father: Johann Bernoulli Mathematician Chemist (one of the "early developers" of calculus) Daniel Bernoulli Physicist Mathematician DANIEL BERNOULLI (17-178) Became very interested in determining a way to measure blood pressure by using physical principles. CONSERVATION OF MECHANICAL ENERGY SYSTEM: SHADED VOLUME OF WATER AND EARTH His quest led him to discover the most fundamental principal of Bernoulli's... the fluid behind does positive work.. the fluid in front does negative work F 1 V V F Dx 1 Dx W NET = W 1 + W W NET = KE + U G 4

5 1/9/17 W 1 + W = KE + U G F 1 X 1 F X = m v P = F A ρ = m V P 1 A 1 X 1 P A X = ρ V v V = A X + g m y + g ρ V y P 1 V P V = ρ V v + g ρ V y divide everything by V P 1 P = ρ v P 1 P = ρ v P 1 + ρ v 1 ρ v 1 + g ρ y distribute (un-delta-it) g ρ y 1 reorganize!!! 1 = P + ρ v P 1 + ρ v 1 BERNOULLI'S EQUATION 1 = P + ρ v USING BERNOULLI BAR CHARTS TO UNDERSTAND FLUID FLOW Bernoulli s Equation: relates the pressures, speeds, and elevations of two points along a single streamline in a fluid. The sum of the kinetic and gravitational potential energy densities and the pressure at position 1 equals the sum of the same three quantities at position. y TORRICELLI S LAW What is the speed with which water flows from a hole punched in the side of an open plastic bottle? ASSUMPTIONS y 1 P 1 P 1 = P atm (open to atmosphere) v 1 = m/s (water is not going down at any appreciable amount) P P = P atm (open to atmosphere) y = m (physicists set their anywhere they want) TORRICELLI S LAW P 1 + ρ v 1 1 = P + ρ v v 1 = m/s y = m P 1 = P = P 1tmospheric P atm + 1 = P atm + ρ v + g ρ y 1 = ρ v v = g y 1 g y 1 = v Water leaves at free fall speed! 5

6 1/9/17 TORRICELLI S LAW v = g y GENERAL STRATEGY FOR BERNOULLI S EQUATION If the container is modified: The water would shoot out the same height as the original height of the water (conservation of energy) Choose two points in the path of the fluid, and stick with those two points If one of the points is open to the atmosphere, use P atm as the external pressure. Feel free to set the zero height level to be where it is most convenient. Density of the fluid will not divide out of the equation don t forget the P external terms! A hole is located 1 cm below the water surface. A) What is the speed with which water flows from a hole punched in the side of an open plastic bottle? B) Construct a Bernoulli bar chart. B) Construct a Bernoulli bar chart. P 1, 1 = bottom 1, 8, P = top v 1 = 1.4 m s 6, 4,, P1 KE1 UG1 P KE UG At a desalinization plant, a large tank of saltwater ( = 1,5 kg/m 3 ) is 5 m in height, and open at the top. A small drain plug with a cross-sectional area of m is 5. m from the floor. A) Calculate the speed of the saltwater as it leaves the hole on the side of the tank when the hole is unplugged. B) If this were freshwater instead of saltwater, how would it affect your answer? C) Construct a Bernoulli bar chart. P 1 + ρ v 1 A) v P 1 = P atm v 1 m/s h 1 = 5 m SOLUTION 1 = P + ρ v m s P = P atm v =? h = 5 m B) Speed does not depend on the density of the fluid 6

7 1/9/17 C) Construct a Bernoulli bar chart. P 3, 1 = top P = bottom 5, 45,, 15, 1, 5, 196, 49, P1 KE1 UG1 P KE UG An underground irrigation system uses a subterranean pump to provide water to a field of pickle plants. The pump provides an underground pressure of 3 x 1 5 Pa, and the flow speed of the underground water is.3 m/s. A) What is the speed of the water as it exits the pipe at ground level? B) Construct a Bernoulli bar chart. P 1 + ρ v 1 P 1 + ρ v 1 1 = P + ρ v y 1 = m P = P atmospheric + = P atm + ρ v B) Construct a Bernoulli bar chart. 35, 3, 3, 5,, 199,685 v = ρ P 1 + ρ v 1 P atm g ρ y v m s 15, 1, 5,,645 4,9 P1 KE1 UG1 P KE UG BLOWING THE ROOF OFF A HOUSE Roofs can be blown from houses during tornadoes or hurricanes. How does that happen? On a windy day, the air inside the house is not moving, whereas the air outside the house is moving very rapidly. P NET = P - P 1 P NET = F NET A The air pressure inside the house is greater than the air pressure outside, creating a net pressure against the roof and windows that pushes outward. (Bernoulli s principle: high v, low P) 7

8 1/9/17 During a storm, air moves at a speed of 45 m/s (1 mi/h) across the top of a -m flat roof of a house. Estimate the net force exerted by the air pushing up on the inside of the roof and the air pushing down on the outside of the roof. air = 1.3 kg/m 3. P 1 + ρ v 1 P 1 + ρ v 1 1 = P + ρ v y 1 = y v = m/s = P atm P NET = P atm - P 1 P NET = F NET A F NET A = ρ v 1 F NET = A ρ v 1 F NET = 63,5 N EXPLAIN! Tip DISLODGING PLAQUE The physical principles of a roof being lifted from a house also explain how plaque can become dislodged from the inner wall of an artery. The plaque may block a considerable portion of the area where blood normally flows. The kinetic energy density is much greater in the constricted area. This pressure differential could cause the plaque to be pulled off the wall and tumble downstream, causing a blood clot. Blood flows through the unobstructed part of a blood vessel at a speed of.5 m/s. The blood then flows past a plaque that constricts the cross-sectional area to one-ninth the normal value. The surface area of the plaque parallel to the direction of blood flow is about.6 cm = 6. x 1 5 m. blood = 15 kg/m 3. P NET = P - P 1 P NET = F NET A y - y 1 = 8

9 1/9/17 (SOLUTION) a) Determine the speed of blood through point. v = 9 v 1 = 4.5 m s b) Find the pressure difference between points 1 and N P P 1 = m c) Estimate the net force that the fluid exerts on the plaque. F NET = N MEASURING BLOOD PRESSURE The air is slowly released from the cuff, decreasing the pressure. When the pressure in the cuff is equal to the systolic pressure, blood starts to squeeze through the artery past the cuff. The flow is intermittent and turbulent and causes a sound heard with the stethoscope. When the pressure decreases to less than the diastolic pressure, the artery is continually open and blood flow is laminar and makes no sound. VENTURI EFFECT VENTURI EFFECT The Venturi effect is the reduction in fluid pressure that results when a fluid flows through a constricted section (or choke) of a pipe. P Fluid:Air ASSUMPTIONS The Venturi effect is named after Giovanni Battista Venturi ( ), an Italian physicist P 1 v 1 = m/s v > m/s P air = ρ v VENTURI EFFECT VENTURI EFFECT P Fluid:water ASSUMPTIONS P air = P water ρ air v = ρ water g y 1.5 kg ρ air = m 3 P 1 y 1 = m y > m P water = ρ g y v = ρ water g y ρ air v = 3.5 m s 1 kg ρ water = m 3 y = 6.5 cm 9

10 1/9/17 After a rainstorm, your basement is filled with water to a depth of.1 m. The surface area of the basement floor is 15 m. A water pump with a short 1.-cm-radius outlet pipe connects to a.9-cm-radius hose, which in turn goes out the basement window to the ground outside,.4 m above the basement floor. The pump can remove water from the basement at a rate of 6.8 m 3 /h (18 gal/h). Assume P atm = Pa (SOLUTION) a) Determine the time interval (in seconds) needed to remove the water. t = V = s Q b) Determine the speed of the water at the pump outlet pipe. v 1 = Q 4.18 m = A 1 s c) Determine the speed of the water at the end of the hose. v = Q 7.4 m = A s d) Determine the water pressure produced at the pump outlet pipe. 143,35.6 N P 1 = m e) Construct a Bernoulli bar chart. e) Construct a Bernoulli bar chart. 16, 143,353 14, 1, 1, 8, 6, 4, 7,549 3,5, 8,717 P1 KE1 UG1 P KE UG AP EQUATION TABLES Continuity Bernoulli s 1