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Chapter 10 Fluids
Units of Chapter 10 ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Units of Chapter 10 ,[object Object],[object Object],[object Object]
10-1 Phases of Matter The three common phases of matter are  solid ,  liquid , and  gas . A solid has a definite  shape  and  size . A liquid has a fixed  volume  but can be any shape. A gas can be any shape and also can be easily  compressed . Liquids and gases both  flow , and are called  fluids .
10-2 Density and Specific Gravity The  density   ρ  ( lower case Greek  rho)   of an object is its  mass  per unit  volume : The SI unit for density is kg/m 3 . Density is also sometimes given in g/cm 3 ; to convert g/cm 3  to kg/m 3 , multiply by 1000. Water at 4 ° C has a density of 1 g/cm 3  = 1000 kg/m 3 . The  specific gravity  of a substance is the ratio of its density to that of water. SG = ( ρ / ρ water ) = 10 -3 ρ (10-1)
10-3 Pressure in Fluids Pressure  is defined as the  force  per unit  area . Pressure is a scalar; the units of pressure in the SI system are  pascals : 1 Pa = 1 N/m 2 Pressure is the same in every  direction  in a fluid at a given  depth ; if it were not, the fluid would flow.
10-3 Pressure in Fluids Also for a fluid at rest, there is no component of force  parallel  to any solid surface – once again, if there were the fluid would flow.
10-3 Pressure in Fluids The  pressure  at a depth  h  below the surface of the liquid is due to the  weight  of the liquid above it. We can quickly calculate: This relation is valid for any liquid whose density does not change with depth. ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Section 10-4: Atmospheric Pressure ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
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10-4 Atmospheric Pressure and Gauge Pressure Most  pressure gauges  measure the pressure  above  the atmospheric pressure – this is called the  gauge pressure . The  absolute pressure  is the sum of the atmospheric pressure and the gauge pressure.
Conceptual Example 10-4 ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
10-5 Pascal’s Principle If an  external  pressure is applied to a  confined  fluid, the pressure at  every  point within the fluid increases by that amount. This principle is used, for example, in hydraulic  lifts  and hydraulic  brakes .
10-6 Measurement of Pressure; Gauges and the Barometer There are a number of different types of  pressure gauges  (Most use  P - P A  =  ρ gh = P G   =  gauge pressure ) . This one is an open-tube  manometer . The pressure in the open end is  atmospheric  pressure; the pressure being measured will cause the fluid to  rise  until the pressures on both sides at the same height are  equal .
10-6 Measurement of Pressure; Gauges and the Barometer Here are two more devices for measuring pressure: the  aneroid gauge  and the  tire pressure gauge .
Various Pressure Units ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
10-6 Measurement of Pressure; Gauges and the Barometer This is a  mercury  barometer, developed by Torricelli to measure  atmospheric pressure . The  height  of the column of  mercury  is such that the pressure in the tube at the surface level is  1 atm . Therefore, pressure is often quoted in  millimeters  (or inches) of mercury.
10-6 Measurement of Pressure; Gauges and the Barometer Any liquid can serve in a Torricelli-style barometer, but the most  dense  ones are the most convenient. This barometer uses  water .
Prob. 17:   (A variation on Example 10-2)   Tank depth  =  5 m Pipe length  =  110 m Hill slope  =  58º Gauge Pressure  P G   = ? Height water  H  shoots  from broken pipe at  bottom? Height of water level in tank from house level:  h = (5 + 110 sin58º) = 98.3 m P G  =  ρ water gh = (1  10 3  kg/m 3 )(9.8 m/s 2 )(98.3 m) = 9.6  10 5  N/m 2   Conservation of energy:  H = h = 98.3 m (Neglects frictional effects, etc.)
10-7 Buoyancy and Archimedes’ Principle This is an object  submerged  in a fluid. There is a net  force  on the object because the pressures at the top and bottom of it are different. The  buoyant force  is found to be the  upward  force on the  same  volume of water:
Archimedes Principle ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
10-7 Buoyancy and Archimedes’ Principle The  net force on  the object is then the difference between the  buoyant force  and the  gravitational  force.
10-7 Buoyancy and Archimedes’ Principle If the object’s  density  is  less  than that of water, there will be an  upward  net force on it, and it will rise until it is partially  out  of the water.
10-7 Buoyancy and Archimedes’ Principle For a  floating  object, the fraction that is  submerged  is given by the ratio of the object’s  density  to that of the fluid.
10-7 Buoyancy and Archimedes’ Principle This principle also works in the air; this is why  hot-air  and  helium  balloons rise.
10-8 Fluids in Motion; Flow Rate and the Equation of Continuity If the flow of a fluid is smooth ( neighboring layers of the fluid slide smoothly ), it is called  streamline  or  laminar ( in layers )  flow (a). Above a certain speed, the flow becomes  turbulent  (b). Turbulent flow has  eddies (or  eddy currents : erratic, small, whirlpool-like circles) ; the  viscosity   ( internal friction )  of the fluid is much greater when eddies are present.
We will deal with  laminar  flow. The  mass flow rate  is the mass that passes a given point per unit time. The flow rates at any two points must be  equal , as long as no fluid is being added or taken away. 10-8 Fluids in Motion; Flow Rate and the Equation of Continuity This gives us the  equation of continuity :
10-8 Fluids in Motion; Flow Rate and the Equation of Continuity If the density doesn’t change – typical for liquids – this simplifies to   . Where the pipe is  wider , the flow is  slower .
Example 10-11: Estimate Blood Flow ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
10-9 Bernoulli’s Equation A fluid can also change its  height . By looking at the work done as it moves, we find: This is  Bernoulli’s  equation. One thing it tells us is that as the  speed  goes up, the  pressure  goes down.
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Sect. 10-9: Bernoulli’s Eqtn   ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
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Flow on the level   ,[object Object],[object Object],[object Object],[object Object],[object Object]
10-10 Applications of Bernoulli’s Principle: from Torricelli to Airplanes, Baseballs, and TIA Using Bernoulli’s principle, we find that the speed of fluid coming from a  spigot  on an  open  tank is:  This is called  Torricelli’s  theorem. (10-6) P 1  + (½) ρ (v 1 ) 2  +  ρ gy 1  = P 2  + (½) ρ (v 2 ) 2  +  ρ gy 2 P 1  = P 2 P 2 P 1
10-10 Applications of Bernoulli’s Principle: Lift  on an airplane  wing  is due to the different air  speeds  and  pressures  on the two surfaces of the wing. P TOP  < P BOT      LIFT force! A 1      Area of wing top,  A 2      Area of wing bottom F TOP   = P TOP  A 1  F BOT   = P BOT  A 2   Plane will fly if    ∑ F = F BOT  -  F TOP   - Mg  > 0 !
10-10 Applications of Bernoulli’s Principle: from Torricelli to Airplanes, Baseballs, and TIA A ball’s path will  curve  due to its  spin , which results in the air speeds on the two sides of the ball not being equal.
10-10 Applications of Bernoulli’s Principle: from Torricelli to Airplanes, Baseballs, and TIA A person with  constricted  arteries will find that they may experience a temporary lack of blood to the brain ( TIA; Transient Ischemic Attack ) as blood speeds up to get past the  constriction , thereby  reducing  the pressure.
10-10 Applications of Bernoulli’s Principle: from Torricelli to Airplanes, Baseballs, and TIA A  venturi meter  can be used to measure fluid  flow  by measuring  pressure  differences.
10-10 Applications of Bernoulli’s Principle: from Torricelli to Airplanes, Baseballs, and TIA Air flow  across the top helps smoke go up a  chimney , and air flow over multiple openings can provide the needed  circulation  in underground burrows.
Problem 46: Pumping water up   Street level:   y 1  = 0 v 1  = 0.6 m/s, P 1  = 3.8 atm  Diameter  d 1  = 5.0 cm  ( r 1  = 2.5 cm ).   A 1  =  π (r 1 ) 2 18 m up :  y 2  = 18 m,   d 2  = 2.6 cm   ( r 2  = 1.3 cm ).  A 2  =  π (r 2 ) 2   v 2  = ? P 2  = ? Continuity:   A 1 v 1  = A 2 v 2      v 2  = (A 1 v 1 )/(A 2 ) = 2.22 m/s  Bernoulli: P 1 + (½) ρ (v 1 ) 2  +  ρ gy 1   = P 2 + (½) ρ (v 2 ) 2  +  ρ gy 2      P 2  = 2.0 atm
Summary of Chapter 10  ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Summary of Chapter 10  ,[object Object],[object Object],[object Object],[object Object],[object Object]
Summary of Chapter 10  ,[object Object]

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Lecture Ch 10

  • 2.
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  • 4. 10-1 Phases of Matter The three common phases of matter are solid , liquid , and gas . A solid has a definite shape and size . A liquid has a fixed volume but can be any shape. A gas can be any shape and also can be easily compressed . Liquids and gases both flow , and are called fluids .
  • 5. 10-2 Density and Specific Gravity The density ρ ( lower case Greek rho) of an object is its mass per unit volume : The SI unit for density is kg/m 3 . Density is also sometimes given in g/cm 3 ; to convert g/cm 3 to kg/m 3 , multiply by 1000. Water at 4 ° C has a density of 1 g/cm 3 = 1000 kg/m 3 . The specific gravity of a substance is the ratio of its density to that of water. SG = ( ρ / ρ water ) = 10 -3 ρ (10-1)
  • 6. 10-3 Pressure in Fluids Pressure is defined as the force per unit area . Pressure is a scalar; the units of pressure in the SI system are pascals : 1 Pa = 1 N/m 2 Pressure is the same in every direction in a fluid at a given depth ; if it were not, the fluid would flow.
  • 7. 10-3 Pressure in Fluids Also for a fluid at rest, there is no component of force parallel to any solid surface – once again, if there were the fluid would flow.
  • 8.
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  • 11. 10-4 Atmospheric Pressure and Gauge Pressure Most pressure gauges measure the pressure above the atmospheric pressure – this is called the gauge pressure . The absolute pressure is the sum of the atmospheric pressure and the gauge pressure.
  • 12.
  • 13. 10-5 Pascal’s Principle If an external pressure is applied to a confined fluid, the pressure at every point within the fluid increases by that amount. This principle is used, for example, in hydraulic lifts and hydraulic brakes .
  • 14. 10-6 Measurement of Pressure; Gauges and the Barometer There are a number of different types of pressure gauges (Most use P - P A = ρ gh = P G = gauge pressure ) . This one is an open-tube manometer . The pressure in the open end is atmospheric pressure; the pressure being measured will cause the fluid to rise until the pressures on both sides at the same height are equal .
  • 15. 10-6 Measurement of Pressure; Gauges and the Barometer Here are two more devices for measuring pressure: the aneroid gauge and the tire pressure gauge .
  • 16.
  • 17. 10-6 Measurement of Pressure; Gauges and the Barometer This is a mercury barometer, developed by Torricelli to measure atmospheric pressure . The height of the column of mercury is such that the pressure in the tube at the surface level is 1 atm . Therefore, pressure is often quoted in millimeters (or inches) of mercury.
  • 18. 10-6 Measurement of Pressure; Gauges and the Barometer Any liquid can serve in a Torricelli-style barometer, but the most dense ones are the most convenient. This barometer uses water .
  • 19. Prob. 17: (A variation on Example 10-2) Tank depth = 5 m Pipe length = 110 m Hill slope = 58º Gauge Pressure P G = ? Height water H shoots from broken pipe at bottom? Height of water level in tank from house level: h = (5 + 110 sin58º) = 98.3 m P G = ρ water gh = (1  10 3 kg/m 3 )(9.8 m/s 2 )(98.3 m) = 9.6  10 5 N/m 2 Conservation of energy: H = h = 98.3 m (Neglects frictional effects, etc.)
  • 20. 10-7 Buoyancy and Archimedes’ Principle This is an object submerged in a fluid. There is a net force on the object because the pressures at the top and bottom of it are different. The buoyant force is found to be the upward force on the same volume of water:
  • 21.
  • 22. 10-7 Buoyancy and Archimedes’ Principle The net force on the object is then the difference between the buoyant force and the gravitational force.
  • 23. 10-7 Buoyancy and Archimedes’ Principle If the object’s density is less than that of water, there will be an upward net force on it, and it will rise until it is partially out of the water.
  • 24. 10-7 Buoyancy and Archimedes’ Principle For a floating object, the fraction that is submerged is given by the ratio of the object’s density to that of the fluid.
  • 25. 10-7 Buoyancy and Archimedes’ Principle This principle also works in the air; this is why hot-air and helium balloons rise.
  • 26. 10-8 Fluids in Motion; Flow Rate and the Equation of Continuity If the flow of a fluid is smooth ( neighboring layers of the fluid slide smoothly ), it is called streamline or laminar ( in layers ) flow (a). Above a certain speed, the flow becomes turbulent (b). Turbulent flow has eddies (or eddy currents : erratic, small, whirlpool-like circles) ; the viscosity ( internal friction ) of the fluid is much greater when eddies are present.
  • 27. We will deal with laminar flow. The mass flow rate is the mass that passes a given point per unit time. The flow rates at any two points must be equal , as long as no fluid is being added or taken away. 10-8 Fluids in Motion; Flow Rate and the Equation of Continuity This gives us the equation of continuity :
  • 28. 10-8 Fluids in Motion; Flow Rate and the Equation of Continuity If the density doesn’t change – typical for liquids – this simplifies to . Where the pipe is wider , the flow is slower .
  • 29.
  • 30. 10-9 Bernoulli’s Equation A fluid can also change its height . By looking at the work done as it moves, we find: This is Bernoulli’s equation. One thing it tells us is that as the speed goes up, the pressure goes down.
  • 31.
  • 32.
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  • 34.
  • 35. 10-10 Applications of Bernoulli’s Principle: from Torricelli to Airplanes, Baseballs, and TIA Using Bernoulli’s principle, we find that the speed of fluid coming from a spigot on an open tank is: This is called Torricelli’s theorem. (10-6) P 1 + (½) ρ (v 1 ) 2 + ρ gy 1 = P 2 + (½) ρ (v 2 ) 2 + ρ gy 2 P 1 = P 2 P 2 P 1
  • 36. 10-10 Applications of Bernoulli’s Principle: Lift on an airplane wing is due to the different air speeds and pressures on the two surfaces of the wing. P TOP < P BOT  LIFT force! A 1  Area of wing top, A 2  Area of wing bottom F TOP = P TOP A 1 F BOT = P BOT A 2 Plane will fly if ∑ F = F BOT - F TOP - Mg > 0 !
  • 37. 10-10 Applications of Bernoulli’s Principle: from Torricelli to Airplanes, Baseballs, and TIA A ball’s path will curve due to its spin , which results in the air speeds on the two sides of the ball not being equal.
  • 38. 10-10 Applications of Bernoulli’s Principle: from Torricelli to Airplanes, Baseballs, and TIA A person with constricted arteries will find that they may experience a temporary lack of blood to the brain ( TIA; Transient Ischemic Attack ) as blood speeds up to get past the constriction , thereby reducing the pressure.
  • 39. 10-10 Applications of Bernoulli’s Principle: from Torricelli to Airplanes, Baseballs, and TIA A venturi meter can be used to measure fluid flow by measuring pressure differences.
  • 40. 10-10 Applications of Bernoulli’s Principle: from Torricelli to Airplanes, Baseballs, and TIA Air flow across the top helps smoke go up a chimney , and air flow over multiple openings can provide the needed circulation in underground burrows.
  • 41. Problem 46: Pumping water up Street level: y 1 = 0 v 1 = 0.6 m/s, P 1 = 3.8 atm Diameter d 1 = 5.0 cm ( r 1 = 2.5 cm ). A 1 = π (r 1 ) 2 18 m up : y 2 = 18 m, d 2 = 2.6 cm ( r 2 = 1.3 cm ). A 2 = π (r 2 ) 2 v 2 = ? P 2 = ? Continuity: A 1 v 1 = A 2 v 2  v 2 = (A 1 v 1 )/(A 2 ) = 2.22 m/s Bernoulli: P 1 + (½) ρ (v 1 ) 2 + ρ gy 1 = P 2 + (½) ρ (v 2 ) 2 + ρ gy 2  P 2 = 2.0 atm
  • 42.
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  • 44.