FLUIDS A fluid is any substance that flows

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).

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 1 x 103 kg/m 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 x 105 N/m 2 = 101. 3 k. Pa

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.

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

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.

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. Pabs = Pgauge + Patm ΔP = Pabs - Patm

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.

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. 57 x 10 -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. 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. 7 x 104 Pa Pabs = PPISTON + Patm = 3. 7 x 104 + 1. 01 x 105 = 1. 38 x 105 Pa b. Determine the gauge pressure of the gas in the tank. PGAUGE = P PISTON = 3. 7 x 104 Pa

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 Ptotal = Patm + Pwater = Patm + ρgh = 1. 01 x 105 + 1030(9. 8)120 = 1. 31 x 106 Pa

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

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. Pout = Pin

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

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