Fluids and Thermodynamics

1. Fluids and Thermodynamics

2. Fluids • Fluids are substances that can flow, such as liquids and gases, and even a few solids. • In Physics B, we will limit our discussion of fluids to substances that can easily flow, such as liquids and gases. • D = m/V • D: density (kg/m3) • m: mass (kg) • V: volume (m3) • You should remember how to do density calculations from chemistry! Density can also be represented by - ρ ρ= m/V ρ: density (kg/m3)

3. Pressure • P = F/A • P : pressure (Pa) • F: force (N) • A: area (m2) • Pressure unit: • Pascal ( 1 Pa = 1 N/m2) • The force on a surface caused by pressure is always normal (or perpendicular) to the surface. This means that the pressure of a fluid is exerted in all directions, and is perpendicular to the surface at every location.

4. Atmospheric Pressure • Atmospheric pressure is normally about 100,000 Pascals. • Differences in atmospheric pressure cause winds to blow. Low atmospheric pressure inside a hurricane’s eye contributes to the severe winds and the evelopment of the storm surge..

5. Calculate the net force on an airplane window if cabin pressure is 90% of the pressure at sea level, and the external pressure is only 50% of that at sea level. Assume the window is 0.43 m tall and 0.30 m wide. Atmospheric pressure = 100000 Pa Pcabin = 90000 Pa Poutside = 50000 Pa Cabin Outside P = F/A P = 90000Pa – 50000Pa = 40000Pa A = lw A = 0.3m · 0.43m = 0.129m2 F = AP F = 0.129m2· 40000Pa = 5160N

6. The Pressure of a Liquid • P = ρgh • P: pressure (Pa) • ρ : density (kg/m3) • g: acceleration constant (9.8 m/s2) • h: height of liquid column (m) • This type of pressure is often called gauge pressure. Why? • The gauges we use to measure pressure do not take into account the atmospheric pressure which is very large! • If the liquid is water, this is referred to as hydrostatic pressure. Why? • Hydro – water + static - stationary

7. Absolute Pressure • Absolute pressure is obtained by adding the atmospheric pressure to the hydrostatic pressure. • Pabs = Patm + ρgh Hydrostatic Pressure in Dam Design The depth of Lake Mead at the Hoover Dam is 180 m. What is the hydrostatic pressure and what is the absolute pressure at the base of the dam?

8. Hydrostatic Pressure in Dam Design • The depth of Lake Mead at the Hoover Dam is 180 m. What is the hydrostatic pressure and what is the absolute pressure at the base of the dam? ρfresh water = 1000 kg / m3 P = ρgh P = 1000 kg / m3· 9.8m/s2 · 180 m = 1764000 Pa Pabs = Patm + ρgh Pabs = 100000Pa + 1764000Pa = 1864000 Pa

9. Hydrostatic Pressure in Levee Design Hurricane Katrina August 2005 A hurricane’s storm surge can overtop levees, but a bigger problem can be increasing the hydrostatic pressure at the base of the levee.

10. New Orleans Elevation Map • New Orleans is largely below sea level, and relies upon a system of levees to keep the lake and the river at bay Max ht of the levee 17.5 ft Surge elevation set at 11.5 ft Normal lake level at 1.0ft

11. Calculate the increase in hydrostatic pressure experienced by the levee base for an expected (SPH Design) storm surge. How does this compare to the increase that occurred during Hurricane Katrina, where the water rose to the top of the levee? • Not to mention there is atmospheric changes during hurricanes which effect the absolute pressure 16.5ft = 12in / 1ft · 2.54 cm / 1 in · 1 m / 100 cm = 5.03m hlevee = 16.5ft ρsalt water = 1025 kg /m3 P= ρgh P= 1025 kg /m3·9.8m/s2 · 5.03m = 50526.35 Pa

12. Pascal’s Principle • Applies when a pressure is applied to a container holding a fluid • Applies to closed systems • Pascal’s principle applies in hydraulic systems • F1 /A1 = F2 / A2

13. A person stands on a piston. The system is full of hydraulic fluid. The car with a mass of 1500 kg is sitting on a piston that has a radius of 3.1 m. A person is standing on the other piston that has a radius of 0.40 m. What must the weight of the person be to cause the car to rise? mc = 1500kg rc = 3.1m rp = 0.40m Fp = ? Ac =πr2 = π(3.1m)2 = 30.19m2 Ac =πr2 = π(0.40m)2 = 0.50m2 Fc / Ac = Fp / Ap Fp = FcAp / Ac Fp = (1500· 9.8m/s2) · 0.50m2 / 30.19m2 = 243N This is equal to 54 lbs, so a child could stand on the piston and make the car move!!!

14. Calculate the pressure exerted on the bottom of a glass that moves upward in elevator with constant acceleration, increasing from rest to 1.2 m/s in 2.7 s. The height of the water is 15 cm. The radius of the cup is 2 cm. The glass is filled with water which has a density of 1000 kg /m3.Do not include atmospheric contribution in your answer. V = Acircle · h = 0.0013 m2 · 0.15m = 0.000195 m3 ΣF = ma h FN - mg= ma m = Vρ = 0.000195 m3 · 1000 kg / m3 = 0.195 Kg a = (vf – vi) / t FN = ma + mg a = (1.2 m/s – 0m/si) / 2.7s = 0.44m/s2 A FN = 0.195 Kg(0.44m/s2 + 9.8m/s2) = 1.9968 N A = πr2 A = π(0.02m)2 = 0.0013 m2 FN = ma + mg

15. Calculate the pressure exerted on the bottom of a glass that moves upward in elevator with constant acceleration, increasing from rest to 1.2 m/s in 2.7 s. The height of the water is 15 cm. The radius of the cup is 2 cm. The glass is filled with water which has a density of 1000 kg /m3.Do not include atmospheric contribution in your answer. P1 = FN / A h P2 = ρgh P = ρgh + FN / A A P = 1000kg/m3 · 9.8m/s2 · 0.15m + 1.9968 N / 0.0013 m2 = 3006Pa

16. Buoyancy

17. An upward force counteracts the force of gravity for these objects. This upward force is called the buoyant force • Floating is a type of equilibrium Fbuoyancy mg

18. Archimedes Principle • Archimedes’ Principle: a body immersed in a fluid is buoyed up by a force that is equal to the weight of the fluid it displaces. • The Buoyant Force • Fbuoy = ρVg • Fbuoy: the buoyant force exerted on a submerged or partially submerged object. • V: the volume of displaced liquid. • ρ : the density of the displaced liquid. • When an object floats, the upward buoyant force equals the downward pull of gravity. • The buoyant force can float very heavy objects, and acts upon objects in the water whether they are floating, submerged, or even sitting on the bottom.

19. A sharks body is not neutrally buoyant, so a shark must swim continuously or he will sink deeper. • Buoyant force on submerged object Fbuoy = ρVg mg Fbouy≠ mg

20. SCUBA divers use a buoyancy control system to maintain neutral buoyancy (equilibrium!). • Buoyant force on submerged object Fbuoy = ρVg mg Fbouy= mg

21. If the diver wants to rise, he inflates his vest, which increases the amount of water he displaces, and he accelerates upward • Buoyant force on submerged object Fbuoy = ρVg mg Fbouy= mg

22. If the object floats on the surface, we know for a fact Fbuoy = mg! The volume of displaced water equals the volume of the submerged portion of the ship. • Buoyant force on floating object Fbuoy = ρVg mg Fbouy= mg

23. Assume a wooden raft has 80.0% of the density of water. The dimensions of the raft are 6.0 meters long by 3.0 meters wide by 0.10 meter tall. How much of the raft rises above the level of the water when it floats? Fbuoy = ρVg h y w l mg ΣF: Fbouy – mg = 0 We also know that the volume of displacement is equal to the volume of water displaced ρglwy= (0.8ρ)glwh ρgVsub– (0.8ρ)gVtotal = 0 y= (0.8) h y= 0.8 · 0.1m = 0.08m The height above the water is 0.02m

24. Buoyant Force • The buoyant force can be extremely strong. • Incredibly massive objects can float, even when they are not intended to…

25. Moving Fluids

26. When a Fluid Flows… • …mass is conserved. • Provided there are no inlets our outlets in a stream of flowing fluid, the same mass per unit time must flow everywhere in the stream. • http://library.thinkquest.org/27948/bernoulli.html • Fluid Flow Continuity • The volume per unit time of a liquid flowing in a pipe is constant throughout the pipe. • We can say this because liquids are not compressible, so mass conservation is also volume conservation for a liquid.

27. Fluid Flow Continuity (cont.) • V = Avt • V: volume of fluid (m3) • A: cross sectional areas at a point in the pipe (m2) • v: speed of fluid flow at a point in the pipe (m/s) • t: time (s) • A1v1 = A2v2 • A1, A2: cross sectional areas at points 1 and 2 • v1, v2: speed of fluid flow at points 1 and 2

28. A pipe of diameter 6.0 cm has fluid flowing through it at 1.6 m/s. How fast is the fluid flowing in an area of the pipe in which the diameter is 3.0 cm? How much water per second flows through the pipe? A1v1 = A2v2 πr21 v1 = πr22 v2 Apipe = πr2 v2 = πr21 v1 / πr22 v2 = π(0.06m)2· (1.6 m/s) / π(0.03m)2= 6.4 m/s

29. The water in a canal flows 0.10 m/s where the canal is 12 meters deep and 10 meters across. If the depth of the canal is reduced to 6.5 meters at an area where the canal narrows to 5.0 meters, how fast will the water be moving through this narrower region? • What will happen to the water in an open waterway if it cannot flow as fast as it wants to through a narrow region in a channel? • It will rise and flow out of the channel V2 = l1w1v1 / l2w2 V2 = A1v1 / A2 A1v1 = A2v2 V2 =(12m · 10m · 0.1 m/s) / (6.5m · 5m) = 0.37 m/s

30. Natural Waterways • Flash flooding can be explained by fluid flow continuity

31. Artificial Waterways Flooding from the Mississippi River Gulf Outlet was responsible for catastrophic flooding in eastern New Orleans and St. Bernard during Hurricane Katrina.

32. Mississippi River Gulf Outlet levees are overtopped by Katrina’s storm surge. • Fluid Flow Continuity in Waterways A hurricane’s storm surge can be “amplified” by waterways that become narrower or shallower as they move inland.

33. Moving Fluids and Bernoulli Effect

34. Bernoulli’s Theorem • The sum of the pressure, the potential energy per unit volume, and the kinetic energy per unit volume at any one location in the fluid is equal to the sum of the pressure, the potential energy per unit volume, and the kinetic energy per unit volume at any other location in the fluid for a non-viscous incompressible fluid in streamline flow. • All other considerations being equal, when fluid moves faster, the pressure drops.

35. Bernoulli’s Theorem • p + ρgh + ½ ρv2 = Constant • p : pressure (Pa) • ρ : density of fluid (kg/m3) • g: gravitational acceleration constant (9.8 m/s2) • h: height above lowest point (m) • v: speed of fluid flow at a point in the pipe (m/s)

36. Longer d, higher v, lower p • Knowing what you know about Bernoulli's principle, design an airplane wing that you think will keep an airplane aloft. Draw a cross section of the wing. Flift Fthurst Fdrag Flift = ΔPA mg

37. Problem: An above-ground swimming pool has a hole of radius 0.10 cm in the side 2.0 meters below the surface of the water. How fast is the water flowing out of the hole? How much water flows out each second? po + ρgh + ½ ρv2 = Constant h1 = 2m po + ρgh1 + ½ ρv12 = po + ρgh2 + ½ ρv22 ρgh1 + ½ ρv12 = ρgh2 + ½ ρv22 h2 = 0 m v1 = 0 m/s ρgh1 + = ½ ρv22 gh1 = ½v22 What about energy mgh = ½ mv2 v2 = √(2gh1) gh = ½ v2 v2 = √(2·9.8 m/s2 · 2.0m) = 6.26 m/s v2 = √(2gh1) Same equation!!! Bernoulli’s principle

38. Problem: An above-ground swimming pool has a hole of radius 0.10 cm in the side 2.0 meters below the surface of the water. How fast is the water flowing out of the hole? How much water flows out each second? v2 = 6.26 m/s h1 = 2m V = Avt V/t = Av V/t = πr2v V/t = π(0.001 m)2 · (6.26m/s) = 1.97 x 10-5 m3/s

39. Water travels through a 9.6 cm diameter fire hose with a speed of 1.3 m/s. At the end of the hose, the water flows out of a nozzle whose diameter is 2.5 cm. What is the speed of the water coming out of the nozzle? If the pressure in the hose is 350 kPa, what is the pressure in the nozzle? A1v1 = A2v2 v2 = A1v1 / A2 v2 = πr21v1 / πr22 p1 = 350000 Pa v1 = 1.3 m/s p2 = ? Pa v2 = ? m/s v2 = π(0.096m)21· 1.3 m/s / π(0.025m)22 =19.17 m/s

40. Water travels through a 9.6 cm diameter fire hose with a speed of 1.3 m/s. At the end of the hose, the water flows out of a nozzle whose diameter is 2.5 cm. What is the speed of the water coming out of the nozzle? If the pressure in the hose is 350 kPa, what is the pressure in the nozzle? p1 + ρgh1 + ½ ρv12 = p2 + ρgh2 + ½ ρv22 h1 = 0m h2 = 0m p1 = 350000 Pa v1 = 1.3 m/s p2 = ? Pa v2 = 19.17 m/s p1 + ½ ρv12 = p2 + ½ ρv22 p2 = p1 + ½ ρv12 - ½ ρv22 p2 = p1 + ½ ρ(v12 - v22) p2 =350000Pa + ½ · 1000 kg / m3((1.3 m/s)2 – (19.17m/s)2) = 167100.55 Pa

41. Bernoulli’s Principle and Hurricanes • In a hurricane or tornado, the high winds traveling across the roof of a building can actually lift the roof off the building. • http://video.google.com/videoplay?docid=6649024923387081294&q=Hurricane +Roof&hl=en

42. Make a device just like the above problem with a whole give them the information and they must predict where the water will hit the floor.

43. Thermodynamics

44. Thermodynamics • Thermodynamics is the study of heat and thermal energy. • Thermal properties (heat and temperature) are based on the motion of individual molecules, so thermodynamics is a lot like chemistry. • Total energy • E = U + K + Eint • U: potential energy • K: kinetic energy • Eint: internal or thermal energy • Potential and kinetic energies are specifically for “big” objects, and represent mechanical energy. • Thermal energy is related to the kinetic energy of the molecules of a substance.

45. Temperature and Heat • Temperature is a measure of the average kinetic energy of the molecules of a substance. Think of it as a measure of how fast the molecules are moving. The unit is ºC or K. • Temperature is NOT heat! • Heat is the internal energy that is transferred between bodies in contact. The unit is Joules (J), or sometimes calories (cal). • A difference in temperature will cause heat energy to be exchanged between bodies in contact. When two bodies are at the same temperature, no heat is transferred. This is called Thermal Equilibrium.

46. Thermal Expansion • Most substances expand when their temperature goes up. • ΔL = αLoΔT • Δ L is change in length • α is coefficient of linear expansion • Lo is original length of substance • ΔT is change in temperature

47. Sample problem: If the Eiffel tower is 301 m tall on a day when the temperature is 22º C, how much does it shrink when the temperature drops to 0º C? The coefficient of linear expansion is 11 x 10-6ºC-1 for the iron the tower is made from. L0 = 301m T1 =22º C T2 =0º C α =11x10-6 /ºC ΔL = αL0ΔT ΔL = 11x10-6 /ºC · 301m · (0 ºC - 22 ºC ) = 0.073 m

48. Ideal Gas Law • P1 V1 / T1 = P2 V2 / T2 • P1, P2: initial and final pressure (any unit) • V1, V2: initial and final volume (any unit) • T1, T2: initial and final temperature (in Kelvin!) • Temperature in K is obtained from temperature in oC by adding 273.

49. Suppose an ideal gas occupies 4.0 liters at 23oC and 2.3 atm. What will be the volume of the gas if the temperature is lowered to 0oC and the pressure is increased to 3.1 atm. P1 = 2.3 atm V1 = 4.0 L T1 = 23ºC + 273 = 296 K P2 = 3.1 atm V2 = ? L T2 = 0ºC + 273 = 273 K P1V1/T1 = P2V2/T2 V2 = (P1V1 T2)/(T1P2) V2 = (2.3 atm · 4.0 L · 273 K)/ (296 K · 3.1 atm) = 2.737 L

50. Ideal Gas Equation • P V = n R T • P: pressure (in Pa) • V: volume (in m3) • n: number of moles • R: gas law constant • 8.31 J/(mol K) • T: temperature (in K) PV = N/m2· m3 = Nm = J • nRT = mol · J/(mol · K) · K = J Both sides = J