Chapter 15: Fluid Motion

1. Chapter 15: Fluid Motion • Characteristics of fluids • Microscopically molecules of a fluid do not have long-range • order. But liquids do have short-range order unlike gases • Fluids can flow and conform to the boundaries of a container Fluids • Fluids cannot sustain a shearing stress

2. Some useful quantities • Density for mass ,volume for uniform solid or liquid unit kg/m3 (at STP for a gas) in SI unit Fluids (cont’d) • Pressure for area and normal force in SI unit unit pascal other useful units: height of a column of Hg corresponding to this pressure (mm)

3. Pressure of a liquid at rest (uniform density) fluid level A Pressure weight of the imaginary box imaginary box A

4. Pressure of a fluid at rest : gauge pressure atmospheric pressure gauge pressure: fluid level A Pressure (cont’d) imaginary box atmospheric pressure A

5. Pressure of a gas at rest ( ) When A simple model for atmospheric pressure

6. Pressure measurement using a liquid: (measures absolute pressure) From as atmospheric pressure Barometer If the liquid is mercury, for 1 atm :

7. Pressure measurement using a liquid : (measures gauge pressure) level 1 Manometer gauge pressure level 2 liquid tank manometer

8. incompressible • Pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and walls of the containing vessel weight piston (area A) pressure at P Pascal’s law pressure due to weight w: w/A atmospheric pressure w liquid pressure due to liquid above P P

9. Hydraulic lever Pascal’s law (cont’d)

10. Origin of buoyancy • Consider a submerged massless object filled with the same fluid • as the fluid that surrounds the object The object is at rest mass of fluid in the object buoyancy opposite dir. Buoyancy buoyant force • Now fill the object with another material mfg

11. Origin of buoyancy (cont’d) • Consider a portion of fluid at rest in a container surrounded by an • imaginary boundary represented in dashed line. • Since the portion of the fluid defined by the surface in dashed line • is at rest, the net force on this portion due to pressure must be • equal to that of the weight of the fluid inside the surface, and • opposite in direction. Buoyancy Fnet buoyant force X mg • The same argument can be applied when the • imaginary portion of the fluid is replaced by an • object that occupies the same space.

12. Archimedes’s principle • The buoyant force on a partly or completely submerged object • is equal to the weight of the displaced fluid: Buoyancy Object sinks • If apparent weight • If Object floats The object will rise until a part of it comes out above the fluid Surface when the average density increases to

13. Example What fraction of an iceberg is submerged in the sea water? Let’s assume that the total volume of the iceberg is Vi. Then the weight of the iceberg Fgi is Let’s then assume that the volume of the iceberg submerged in the sea water is Vw. The buoyant force B caused by the displaced water becomes Since the whole system is at its static equilibrium, we obtain Therefore the fraction of the volume of the iceberg submerged under the surface of the sea water is About 90% of the entire iceberg is submerged in the water!!!

14. Example • A fake or pure gold crown? Is the crown made of pure gold? Tair =7.84 N Twater =6.86 N rgold=19.3x103 kg/m3

15. Example • Floating down the river What depth h is the bottom of the raft submerged? A=5.70 m2 rwood=6.00x102 kg/m3

16. Example • Oil and water r=0.700 g/cm3 h1=8.00 m r=1025 kg/m3 h2=5.00 m Density and Pressure

17. Ideal fluids in motion • Incompressible ( density is constant at any position) • No internal friction (no viscosity) • Steady (non-turbulent) flow- the velocity at a point is • constant in time. flow tube flow line : The path of an individual particle in a moving fluid steady flow: A flow whose pattern does not change with time. Every element passing through a given point follows the same flow line streamline : A curve whose tangent at any point is in the direction of the fluid velocity at that point flow tube : The flow lines passing through the edge of an imaginary area such as A Ideal fluid flow A flow lines

18. Continuity equation I (incompressible fluid) The mass of a moving fluid does not change as it flows. The volume of the fluid that passes through area A during a small time interval dt : flow tube Continuity equation • In an ideal fluid the density is constant. • In a time interval dt the mass that flows • into Area 1 is the same as the mass that • flows out of Area 2.

19. Volume flow rate volume flow rate Continuity equation • Continuity equation II (compressible fluid)

20. Work done by pressure • Change in kinetic energy Bernoulli’s equation • Change in potential energy

21. Energy conservation Bernoulli’s equation

22. Venturi meter Example h

23. Torricelli’s theorem The velocity of the fluid coming out of a hole in a tank as shown in the figure can be calculated using Bernoulli’s equation. At the top surface the velocity of the fluid is zero. The pressures at the top surface and at the hole are the same, namely, the atmospheric pressure. Example

24. Siphon Suppose a U-shaped piece of pipe is completely submerged in water, filled with water, and then turned upside down under water. As you slowly pull the top of the U-shaped piece of pipe out of water, the water does not run out of the pipe. WHY? Example

25. Siphon Suppose a U-shaped piece of pipe is completely submerged in water, filled with water, and then turned upside down under water. As you slowly pull the top of the U-shaped piece of pipe out of water, the water does not run out of the pipe. WHY? Example Air cannot enter the pipe. As the water starts running out of the pipe, a near vacuum is created in the topmost region of the inverted U. The pressure here drops to near zero. The atmospheric pressure on the surface of the water in the bucket pushes the water into the U-shaped pipe.

26. Siphon If a U-shaped hose or pipe connects a liquid-filled container at a higher altitude to a container at a lower altitude over a barrier, the liquid can be siphoned into the container at the lower altitude. Atmospheric pressure helps to push the liquid over the barrier. Example

27. Siphon When P1>P2, the fluid can be siphoned from the left to the right bucket. Example

28. Water garden A water hose 2.50 cm in diameter is used by a gardener to fill a 30.0-liter bucket. The gardener notices that it takes 1.00 min to fill the bucket. A nozzle with an opening of cross-sectional area 0.500 cm2 is then attached to the hose. The nozzle is held so that water is projected horizontally from a point 1.00 m above the ground. Over what horizontal distance can the water be projected? Example

29. h =0.500 m y1 =3.00 m • Example : A water tank • Consider a water tank with a hole. • Find the speed of the water • leaving through the hole. y x (b) Find where the stream hits the ground.

30. Example : Fluid flow in a pipe A2=1.00 m2 A1=0.500 m2 h =5.00 m Find the speed at Point 1.

31. Viscosity Viscosity is internal friction in a fluid, and viscous forces oppose the motion of one portion of a fluid relative to another. Viscosity and turbulence

32. Drag If a fluid in laminar flow flows around an obstacle, it exerts a viscous drag on obstacle. Frictional forces accelerate the fluid backward against the direction of flow and the obstacle forward in the direction of flow. adjacent layers of fluid slide smoothly past each other and flow is steady laminar flow Viscosity and turbulence

33. Turbulence When the speed of a flowing fluid exceeds a certain critical value the flow is no longer laminar. The flow patter becomes extremely irregular and complex, and it changes continuously in time. There is no steady flow pattern. This chaotic flow Is called turbulence. Viscosity and turbulence

34. Problem 1 The upper edge of a gate in a dam runs the water surface. The gate is 2.00 m high and 4.00 m wide and is hinged along the horizontal line through its center. Calculate the torque about the hinge arising from the force due to the water. 2.00 m 4.00 m Solution Problems Denote the width and depth at the bottom of the gate by w and H. The force on a strip of vertical thickness dh at a depth h is: and the torque about the hinge is After integrating from h=0 to h=H, you get the torque:

35. Problem 2 • An object with height h, mass M, and a uniform cross-sectional area A • floats upright in a liquid with density r. • Calculate the vertical distance from the surface of the liquid to the • bottom of the floating object in equilibrium. • (b) A downward force with magnitude F is applied to the top of the object. • At the new equilibrium position, how much farther below the surface of • the liquid is the bottom of the object than it was in part (a)? • Calculate the period of the oscillation when the force F is suddenly • removed. Solution (a) From Archimedes’s principle so (b) The buoyant force is: With the result of part (a) solving for x gives: (c) The force is always in the direction toward the equilibrium, namely, a restoring force Therefore the “spring constant” is and the period of the oscillation is

36. Problem 3 You cast some metal of density rm in a mold, but you are worried that there might be cavities within the casting. You measure the weight of the casting to be w, and the buoyant force when it is completely surrounded by water to be B. (a) Show that is the total volume of any enclosed cavities. (b) If your metal is copper, the casting’s weight is 156 N, and the buoyant force is 20 N, what is the total volume of any enclosed cavities in your casting? What fraction is this of the total volume of the casting? Solution (a) Denote the total volume V. If the density of air is neglected, the buoyant force in terms of the weight is: Therefore The total volume of the casting is (b) The cavities are 12.4% of the total volume.

37. Problem 4 A A A U-shaped tube with a horizontal portion of length contains a liquid. What is the difference in height between the liquid columns in the vertical arms (a) if the tube has an acceleration toward the right? (b) if the tube is mounted on a horizontal turntable rotating with an angular speed with one of the vertical arms on the axis of rotation? Solutions • Consider the fluid in the horizontal part of the tube. This fluid with mass • , is subject to a net force due to the pressure difference between • the ends of the tube, which is the difference between the gauge pressures • at the bottoms of the ends of the tubes. Now this difference is • and the net force on the horizontal part of the fluid is • or • (b) Similarly to (a) consider the fluid in the horizontal part of the tube. As in (a) • the fluid is accelerating. The center of mass has a radial acceleration of • magnitude so the difference in heights between the columns • is