1. FLUIDS A fluid is any substance that flows and conforms to the boundaries of its container. A fluid could be a gas or a liquid. An ideal fluid is assumed to be incompressible (so that its density does not change), to flow at a steady rate, to be non-viscous (no friction between the fluid and the container through which it is flowing), and to flow without rotation (no swirls or eddies).

2. DENSITY AND SPECIFIC GRAVITY The density (ρ) of a substance is defined as the quantity of mass (m) per unit volume (V): For solids and liquids, the density is usually expressed in (g/cm 3 ) or (kg/m 3 ). The density of gases is usually expressed in (g/l). The specific gravity (SG) of a substance is the ratio of the density of the substance to the density of another substance that is taken as a standard. SG = density of substance/density of water The density of pure water at 4°C is usually taken as the standard (1 g/cm 3 or 1x10 3 kg/m 3 ).

3. PRESSURE Any fluid can exert a force perpendicular to its surface on the walls of its container. The force is described in terms of the pressure it exerts, or force per unit area: Units: N/m 2 or Pascal (Pa) One atmosphere (atm) is the average pressure exerted by the earth’s atmosphere at sea level 1.00 atm = 1.01 x10 5 N/m 2 = 101.3 kPa

4. PRESSURE IN FLUIDS A static (non-moving) fluid produces a pressure within itself due to its own weight. This pressure increases with depth below the surface of the fluid. Consider a container of water with the surface exposed to the earth’s atmosphere.

5. PRESSURE IN FLUIDS The pressure P 1 on the surface of the water is 1 atm. If we go down to a depth from the surface, the pressure becomes greater by the product of the density of the water ρ the acceleration due to gravity g, and the depth h. Thus the pressure P 2 at this depth is: P 2 = P 1 + ρ g h

6. Note that the pressure at any depth does not depend on the shape of the container, but rather only on the pressure at some reference level and the vertical distance below that level.

7. GAUGE PRESSURE AND ABSOLUTE PRESSURE Ordinary pressure gauges measure the difference in pressure between an unknown pressure and atmospheric pressure. The pressure measured is called the gauge pressure and the unknown pressure is referred to as the absolute pressure. P abs = P gauge + P atm ΔP = P abs - P atm

8. PRESSURE GAUGES All these gauges work by measuring the force exerted on a known area by comparing it to either a spring of some sort or a known pressure.

9. 10.1 A 5.0-kg solid cylinder has a 0.100-m radius and a height of 0.0500-m. a. Determine the density m = 5.0 kg r = 0.1 m h = 0.05 m V = Ah = πr 2 h = π(0.1) 2 (0.05) = 1.57x10 -3 m 3 = 3183 kg/m 3 b. Determine the specific gravity of the object SG = density of substance/density of water = 3183/1000 = 3.18

10. 10.2 A 30.0-kg piston holds compressed gas in a tank of volume 100 m 3. The radius of the piston is 0.050 m. a. Determine the absolute pressure m = 30 kg V = 100 m 3 r = 0.05 m = 3.7x10 4 Pa P abs = P PISTON + P atm = 3.7x10 4 + 1.01x10 5 = 1.38x10 5 Pa b. Determine the gauge pressure of the gas in the tank. P GAUGE = P PISTON = 3.7x10 4 Pa

11. 10.3 When a submarine dives to a depth of 120 m, to how large a total pressure is its exterior surface subjected? The density of seawater is about 1.03 g/cm 3. h = 120 m ρ = 1030 kg/m 3 P total = P atm + P water = P atm + ρgh = 1.01x10 5 + 1030(9.8)120 = 1.31x10 6 Pa

12. PASCAL'S PRINCIPLE Pascal’s Principle states that pressure applied to a confined fluid is transmitted throughout the fluid and acts in all directions.

13. The principle means that if the pressure on any part of a confined fluid is changed, then the pressure on every other part of the fluid must be changed by the same amount. This principle is basic to all hydraulic systems. P out = P in

14. 10.4 In a hydraulic system, one 200-kg cylinder has a sectional area of 100 cm 2. The other cylinder has an area of 10 cm 2. a. Determine the weight required to hold the system in equilibrium. m 1 = 200 kg A 1 = 100 cm 2 A 2 = 10 cm 2 P 1 = P 2 = 196 N

15. b. If the large cylinder is pushed down 5.0 cm, determine through what distance the small cylinder will move. h = 5 cm V = Ah V 1 = V 2 A 1 h 1 = A 2 h 2 = 50 cm

26. BUOYANCY AND ARCHIMEDES' PRINCIPLE Archimedes’ Principle states that a body wholly or partly immersed in a fluid is buoyed up by a force equal to the weight of the fluid it displaces. An object lowered into a fluid “appears” to lose weight. The force that causes this apparent loss of weight is referred to as the buoyant force. The buoyant force is considered to be acting upward through the center of gravity of the displaced fluid. F B = m F g = ρ F g V F

27. The pressure difference between the top and bottom of this object must support the weight of the fluid that would be there in equilibrium if the object were not there. In general, objects floats on a fluid if its density is less than that of the fluid. At equilibrium (when floating) the buoyant force on an object has a magnitude equal to the weight of the object.

28. Buoyant Force Due to a Fluid

29. The fraction of the floating object that is submerged is equal to the ratio of the density of the object to that of the fluid.

30. 10.5 An aluminum object has a mass of 27.0-kg and a density of 2.70x10 3 kg/m 3. The object is attached to a string and immersed in a tank of water. a. Determine the volume of the object m = 27 kg ρ Al = 2700 kg/m 3 ρ water = 1000 kg/m 3 = 0.01 m 3

31. b. Determine the tension in the string when it is completely immersed. V water = V Al Fg water = ρ w V w g = 1000(0.01)(9.8) = 98 N F B = Fg water ΣF = F T + F B - F g = 0 F T = Fg - F B = 27(9.8) - 98 = 166.6 N FTFT FBFB FgFg

32. 10.6 A piece of alloy has a mass of 86 g in air and 73 g when immersed in water. Find its volume and its density. m air = 0.086 kg m water = 0.073 kg ρ water = 1000 kg/m 3 F B = Fg - Fg apparent = 0.086(9.8) - (0.073)(9.8) = 0.1274 N F B = ρ w V w g = 1.3x10 -5 m 3 = 6.6x10 3 kg/m 3

33. 10.7 A solid aluminum cylinder with density of 2700 kg/m 3 has a measured mass of 67 g in air and 45 g when immersed in turpentine. Find the density of turpentine. ρ Al = 2700 kg/m 3 m air = 0.067 kg m water = 0.045 kg F B = Fg - Fg apparent = (0.067-0.045)(9.8) = 0.2156 N = 2.48x10 -5 m 3 = 887.1 kg/m 3

34. 10.8 The density of ice is 917 kg/m 3. What fraction of the volume of a piece of ice will be above water when floating in fresh water? ρ ice = 917 kg/m 3 ρ water = 1000 kg/m 3 When floating F B = Fg w = Fg i or m w g = m i g and m = ρ V ρ i V = ρ w V’ (V’ = volume of displaced water) fraction: = 0.917 V = 1 - 0.917 = 0.083

36. 10.9 A partly filled beaker of water sits on a scale, and its weight is 2.3 N. When a piece of metal suspended from a thread is totally immersed in the beaker (but not touching the bottom), the scale reads 2.75 N. What is the volume of the metal? Fg = 2.3 N Fg w = 2.75 N ρ water = 1000 kg/m 3 F B = Fg - Fg apparent = 2.75 - 2.3 = 0.45 N F B = ρ w V w g = 46x10 -6 m 3 = 46 cm 3

37. 10.10 A 60-kg rectangular box, open at the top, has base dimensions 1.0 m by 0.80 m and a depth of 0.50 m. a. How deep will it sink in fresh water? m = 60 kg A = 1 x 0.8 m 2 w = 0.5m F B = Fg box = ρgV = ρgAh’ FBFB Fg bo x = 0.075 m

38. b. What weight of ballast will cause it to sink to a depth of 30 cm? depth = 0.3 m ΣF = F B - Fg box - Fg ballast = 0 Fg ballast = F B - Fg box = ρgV - mg = ρgAh - mg = 1000(9.8) (1)(0.8)(0.3) - 60(9.8) = 1764 N FBFB Fg box Fg ballast

39. FLUIDS IN MOTION The equations that follow are applied when a moving fluid exhibits streamline flow. Streamline flow assumes that as each particle in the fluid passes a certain point it follows the same path as the particles that preceded it. There is no loss of energy due to internal friction (viscosity) in the fluid. In reality, particles in a fluid exhibit turbulent flow, which is the irregular movement of particles in a fluid and results in loss of energy due to internal friction in the fluid. Turbulent flow tends to increase as the velocity of a fluid increases.

41. FLOW RATE Consider a fluid flowing through a tapered pipe:

42. The flow rate is the mass of fluid that passes a point per unit time: m/t = ρ A vUnits: kg/s Where ρ is the density of the fluid, A is the cross- sectional area of the tube of fluid at the particular point in question, and v is the velocity of the fluid at the point in question.

43. Since fluid cannot accumulate at any point, the flow rate is constant. This is expressed as the equation of continuity. ρ A v = constant In streamline flow, the fluid is considered to be incompressible and the density is the same throughout. The equation of continuity can then be written in terms of the volume rate of flow (R) that is constant throughout the fluid: R = Av = constant Units: m 3 /s or: ρ 1 A 1 v 1 = ρ 2 A 2 v 2

44. 10.11 Oil flows through a pipe 8.0 cm in diameter, at an average speed of 4 m/s. What is the flow in m 3 /s and m 3 /h? r = 8/2 = 0.04 m v = 4 m/s A = πr 2 = π(0.04) 2 = 0.005 m 2 R = Av = 0.005(4) = 0.02 m 3 /s R = 0.02 m 3 /s (3600 s/h) = 72.4 m 3 /h

45. BERNOULLI’S EQUATION In the absence of friction or other non- conservative forces,the total mechanical energy of a system remains constant, that is, PE 1 + K 1 = PE 2 + K 2 There is a similar law in the study of fluid flow, called Bernoulli’s principle, which states that the total pressure of a fluid along any tube of flow remains constant.

46. Assume incompressible fluid in a laminar flow with no viscosity. Bernoulli’s equation is a form of the energy conservation law.

47. Bernoulli’s principle: Where the velocity of fluid is high the pressure is low. Where the velocity of fluid is low the pressure is high.

48. Find the velocity of a liquid flowing out of a spigot: The pressure is the same P 1 =P 2. The velocity v 2 » 0. Bernoulli’s equation is: This is called Torricelli’s Theorem.

50. Fluid Pressure in a Pipe of Varying Elevation

51. Another special case: Fluid flow but no change in height: y 1 = y 2. Bernoulli’s equation: When pressure is low (high), velocity is high (low).

52. 10.12 What volume of water will escape per minute from an open top tank through an opening 3.0 cm in diameter that is 5.0 m below the water level in the tank? r = 1.5x10 -2 m Top (1)Bottom (2) h 1 = 5 mh 2 = 0 If tank is large v 1 = 0 (Torricelli’s equation) R = v 2 A 2 = 9.9 (π(1.5x10 -2 ) 2 ) = 7x10-3 m 3 /s = 9.9 m/s

53. 10.13 Water flows at the rate of 30 mL/s through an opening at the bottom of a tank in which the water is 4.0 m deep. Calculate the rate of escape of the water if an added pressure of 50 kPa is applied to the top of the water. R before = 30 mL/s h = 4m P = 5x10 4 Pa v 1 = 0 Before the pressure was added: P 1 = P 2 After the pressure is added:

54. R before = 30 mL/s h = 4m P = 5x10 4 Pa v 1 = 0 dividing both equations: = 1.51

55. R = Av R after = R before (ratio) = 30(1.51) = 45 mL/s

56. 10.14 A water tank springs a leak at position 2 in the figure, where the water pressure is 500 kPa. What is the velocity of escape of the water through the hole? P 1 - P 2 = 5x10 5 N/m 2 h 1 = h 2 v 1 = 0 =31.6 m/s

57. 10.15 The pipe shown in the figure has a diameter of 16 cm at section 1 and 10 cm at section 2. At section 1 the pressure is 200 kPa. Point 2 is 6.0 m higher than point 1. When oil of density 800 kg/m 3 flows at a rate of 0.030 m 3 /s, find the pressure at point 2 if viscous effects are negligible. ρ = 800 kg/m3 r 1 = 0.08 m r 2 = 0.05 m P 1 = 2x105 Pa h 1 = 0 m h 2 = 6 m V 1 A 1 = V 2 A 2 = R = 1.49 m/s = 3.82 m/s

58. = 2x10 5 + ½ (800)(1.49 - 3.82) + 800 (9.8) (0-6) = 1.48x10 5 kPa