Physics 1501: Lecture 31 Today ’ s Agenda

1. Physics 1501: Lecture 31Today’s Agenda • Homework #10 (due Friday Nov. 18) • Midterm 2: Nov. 16 • Students with schedule issues … • Honor’s student … • Topics: Fluids • Pascal’s Principle (hydraulic lifts etc.) • Archimedes’ Principle • Fluid dynamics • Bernouilli’s equation • Example of applications

2. n A Fluids • What parameters do we use to describe fluids? • Density • Pressure units : kg/m3 = 10-3 g/cm3 units : 1 N/m2 = 1 Pa (Pascal) • Any force exerted by a fluid is perpendicular to a surface of contact, and is proportional to the area of that surface. • Force (a vector) in a fluid can be expressed in terms of pressure (a scalar) as:

3. Pb Gas (STP) H2O Steel         Bulk modulus (Pa=N/m2) Fluids • Bulk Modulus • LIQUID: incompressible (density almost constant) • GAS: compressible (density depends a lot on pressure)

4. Pressure vs. DepthIncompressible Fluids (liquids) • When the pressure is much less than the bulk modulus of the fluid, we treat the density as constant independent of pressure: incompressible fluid • For an incompressible fluid, the density is the same everywhere, but the pressure is NOT!

5. Any change in the pressure applied to an enclosed fluid is transmitted to every portion of the fluid and to the walls of the containing vessel. Pascal’s Principle • So far we have discovered (using Newton’s Laws): • Pressure depends on depth: Dp = rgDy • Pascal’s Principle addresses how a change in pressure is transmitted through a fluid. • Pascal’s Principle explains the working of hydraulic lifts • i.e. the application of a small force at one place can result in the creation of a large force in another. • Does this “hydraulic lever” violate conservation of energy? • Certainly hope not.. Let’s calculate.

6. Pascal’s Principle • Consider the system shown: • A downward force F1 is applied to the piston of area A1. • This force is transmitted through the liquid to create an upward force F2. • Pascal’s Principle says that increased pressure from F1 (F1/A1) is transmitted throughout the liquid. • F2 > F1 : Have we violated conservation of energy??

7. Pascal’s Principle • Consider F1 moving through a distance d1. • How large is the volume of the liquid displaced? • This volume determines the displacement of the large piston. • Therefore the work done by F1 equals the work done by F2 We have NOT obtained “something for nothing”.

8. M dA A1 A10 A) dA = (1/2)dB B) dA = dB C) dA = 2dB M dB A) dA = (1/2)dC B) dA = dC C) dA = 2dC A2 A10 • If A10 = 2´A20, compare dA and dC. dC M A1 A20 Lecture 31, ACT 1Hydraulics • Consider the systems shown to the right. • In each case, a block of mass M is placed on the piston of the large cylinder, resulting in a difference dI in the liquid levels. • If A2 = 2´A1, compare dA and dB.

9. Example Problems (1) • At what depth is the water pressure two atmospheres? It is one atmosphere at the surface. What is the pressure at the bottom of the deepest oceanic trench (about 104 meters)? Solution: d is the depth. The pressure increases one atmosphere for every 10m. This assumes that water is incompressible. P2 = P1 + rgd 2.02105 Pa = 1.01105 Pa + 103 kg/m3*9.8m/s2*d d = 10.3 m P2 = 1.01105 Pa + 103 kg/m3*9.8m/s2*104 m = 9.81107 Pa = 971 Atm For d = 104 m:

10. M1 h M2 Example Problems (2) • Two masses rest on two pistons in a U-shaped tube of water, as shown. If M1 = 3 kg and M2 = 4.5 kg, what is the height difference, h? The area of piston 1 is A1 = 200 cm2, and the area of piston 2 is A2 = 100 cm2. Solution: M1g/A1 + rgh = M2g/A2 h = (M2/A2-M1/A1)/r = 0.3 m The pressure difference due to the two masses must be balanced by the water pressure due to h. Note that g does not appear in the answer.

11. Barometer Using Fluids to Measure Pressure • Use Barometer to measure Absolute Pressure • Top of tube evacuated (p=0) • Bottom of tube submerged into pool of mercury open to atmosphere (p=p0) • Pressure dependence on depth:

12. Manometer p0 p1 Dh Using Fluids to Measure Pressure • Use Manometer to measure Gauge Pressure • Measure pressure of volume (p1) relative to atmospheric pressure (º gauge pressure) • The height difference (h) measures the gauge pressure 1 atm = 760 mm (29.9 in) Hg = 10.3 m (33.8 ft) H20

13. W2? W1 W1 > W2 W1 < W2 W1 = W2 Archimedes’ Principle • Suppose we weigh an object in air (1) and in water (2). • How do these weights compare? • Why? • Since the pressure at the bottom of the object is greater than that at the top of the object, the water exerts a net upward force, the buoyant force, on the object.

14. W2? W1 Archimedes: The buoyant force is equal to the weight of the liquid displaced. Archimedes’ Principle • The buoyant force is equal to the difference in the pressures times the area. • The buoyant force determines whether an object will sink or float. How does this work?

15. y F mg B Sink or Float? • The buoyant force is equal to the weight of the liquid that is displaced. • If the buoyant force is larger than the weight of the object, it will float; otherwise it will sink. • We can calculate how much of a floating object will be submerged in the liquid: • Object is in equilibrium

16. y F mg B The Tip of the Iceberg • What fraction of an iceberg is submerged?

17. Pb styrofoam A) It sinks B) C) D) styrofoam Pb Lecture 31, ACT 2Buoyancy • A lead weight is fastened to a large styrofoam block and the combination floats on water with the water level with the top of the styrofoam block as shown. • If you turn the styrofoam+Pb upside down, what happens?

18. Cup II Cup I (D) can’t tell (C) the same (A) Cup I (B) Cup II ACT 2-AMore Fun With Buoyancy • Two cups are filled to the same level with water. One of the two cups has plastic balls floating in it. • Which cup weighs more?

19. oil water (B) move down (C) stay in same place (A) move up ACT 2-BEven More Fun With Buoyancy • A plastic ball floats in a cup of water with half of its volume submerged. Next some oil (roil < rball < rwater) is slowly added to the container until it just covers the ball. • Relative to the water level, the ball will:

20. Fluids in Motion • Up to now we have described fluids in terms of their static properties: • density r • pressure p • To describe fluid motion, we need something that can describe flow: • velocity v • There are different kinds of fluid flow of varying complexity • non-steady/ steady • compressible / incompressible • rotational / irrotational • viscous / ideal

21. Simplest situation: consider ideal fluid moving with steady flow - velocity at each point in the flow is constant in time • In this case, fluid moves on streamlines streamline Ideal Fluids • Fluid dynamics is very complicated in general (turbulence, vortices, etc.) • Consider the simplest case first: the Ideal Fluid • no “viscosity” - no flow resistance (no internal friction) • incompressible - density constant in space and time

22. Ideal Fluids • streamlines do not meet or cross • velocity vector is tangent to streamline • volume of fluid follows a tube of flow bounded by streamlines streamline • Flow obeys continuity equation • volume flow rate Q = A·v is constant along flow tube. • follows from mass conservation if flow is incompressible. A1v1 = A2v2

23. Steady Flow of Ideal Fluids(actually laminar flow of real fluid)

24. a) 2 v1 b) 4 v1 c) 1/2 v1 c) 1/4 v1 Lecture 31 Act 3Continuity • A housing contractor saves some money by reducing the size of a pipe from 1” diameter to 1/2” diameter at some point in your house. v1 v1/2 1) Assuming the water moving in the pipe is an ideal fluid, relative to its speed in the 1” diameter pipe, how fast is the water going in the 1/2” pipe?

25. dV Bernoulli Equation Conservation of Energy for Ideal Fluid • Recall the standard work-energy relation • Apply the principle to a section of flowing fluid with volume dV and mass dm = r dV (here W is work done on fluid)

26. a) smaller b) same c) larger Lecture 31 Act 4Bernoulli’s Principle • A housing contractor saves some money by reducing the size of a pipe from 1” diameter to 1/2” diameter at some point in your house. v1 v1/2 2) What is the pressure in the 1/2” pipe relative to the 1” pipe?

27. Some applications • Lift for airplane wing • Enhance sport performance • More complex phenomena: ex. turbulence

28. More applications • Vortices: ex. Hurricanes • And much more …

29. Ideal Fluid: Bernoulli Applications • Bernoulli says: high velocities go with low pressure • Airplane wing • shape leads to lower pressure on top of wing • faster flow  lower pressure  lift • air moves downward at downstream edge  wing moves up

30. Ideal Fluid: Bernoulli Applications • Warning: the explanations in text books are generally over-simplified! • Curve ball (baseball), slice or topspin (golf) • ball drags air around (viscosity) • air speed near ball higher on bottom • lower pressure force  sideways acceleration or lift

31. Ideal Fluid: Bernoulli Applications • Bernoulli says: high velocities go with low pressure • “Atomizer” • moving air ‘sweeps’ air away from top of tube • pressure is lowered inside the tube • air pressure inside the jar drives liquid up into tube

32. area A Real Fluids: Viscosity • In ideal fluids mechanical energy is conserved (Bernoulli) • In real fluids, there is dissipation (or conversion to heat) of mechanical energy due to viscosity (internal friction of fluid) Viscosity measures the force required to shear the fluid: where F is the force required to move a fluid lamina (thin layer)of area A at the speed vwhen the fluid is in contact with a stationary surface a perpendicular distance y away.

33. area A air H2O oil glycerin       Viscosity (Pa-s) Real Fluids: Viscosity • Viscosity arises from particle collisions in the fluid • as particles in the top layer diffuse downward they transfer some of their momentum to lower layers • lower layers get pulled along (F = Dp/Dt)

34. L r p+Dp Q R p Real Fluids: Viscous Flow • How fast can viscous fluid flow through a pipe? • Poiseuille’s Law • Because friction is involved, we know that mechanical energy is not being conserved - work is being done by the fluid. • Power is dissipated when viscous fluid flows: P = v·F = Q ·Dp the velocity of the fluid remains constant power goes into heating the fluid: increasing its entropy

35. L/2 L/2 a) 3/2 b) 2 c) 4 Lecture 31 Act 5Viscous flow • Consider again the 1 inch diameter pipe and the 1/2 inch diameter pipe. 1) Given that water is viscous, what is the ratio of the flow rates, Q1/Q1/2, in pipes of these sizes if the pressure drop per meter of pipe is the same in the two cases?