Chapter 14: Fluid MechanicsSlide 2Fluid MechanicsSlide 4Slide 5Density & Specific Gravity Not discussed much in your text! Plays the role for fluids that mass plays for solid objectsSlide 7Slide 8Slide 9Forces in FluidsSect. 14.1: Pressure Plays the role for fluids that force plays for solid objectsSlide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18ExampleExample: (A variation on the previous example)Atmospheric PressureGauge PressureConceptual ExamplePascal’s LawSlide 25Slide 26Slide 27Pascal’s Law, Other ApplicationsSlide 29Various Pressure UnitsSlide 31Mercury BarometerChapter 14: Fluid MechanicsCOURSE THEME: NEWTON’S LAWS OF MOTION!•Chs. 5 - 13: Methods to analyze dynamics of objects in Translational & Rotational Motion using Newton’s Laws! Chs. 5 & 6: Newton’s Laws using Forces (translational motion)Chs. 7 & 8: Newton’s Laws using Energy & Work (translational motion)Ch. 9: Newton’s Laws using Momentum (translational motion)Chs. 10 & 11: Newton’s Laws (rotational language; rotating objects).NOW•Ch. 14: Methods to analyze the dynamics of fluids in motion.First, we need to discuss FLUID LANGUAGE. Then, Newton’s Laws in Fluid Language!•The three (common) states or phases of matter are:1. Solid: Has a definite volume & shape. Maintains it’s shape & size (approximately), even under large forces.2. Liquid: Has a definite volume, but not a definite shape. It takes the shape of it’s container.3. Gas: Has neither a definite volume nor a definite shape. It expands to fill it’s container.NOTE! These definitions are somewhat artificial–The time it takes a substance to change its shape in response to an external force determines whether the substance is a solid, liquid or gasChapter 14 lumps 2. & 3. into the category ofFLUIDSFluid Mechanics•Fluids: Have the ability to flow. •A fluid is a collection of molecules that are randomly arranged & held together by weak cohesive forces & by forces exerted by the walls of a container.Both liquids & gases are fluidsFluids•Two basic categories of fluid mechanics: •Fluid Statics –Obviously, describes fluids at rest•Fluid Dynamics–Obviously, describes fluids in motion•The same physical principles (Newton’s Laws) that have applied in our studies up to now will also apply to fluids. But, first, we need to introduce Fluid Language. Fluid MechanicsDensity & Specific GravityNot discussed much in your text! Plays the role for fluids that mass plays for solid objects•Density, ρ (lower case Greek rho, NOT p!) of object, mass M & volume V: ρ  (M/V) (kg/m3 = 10-3 g/cm3)•Specific Gravity (SG): Ratio of density of a substance to density of water. ρwater = 1 g/cm3 = 1000 kg/m3See table!!ρ = (M/V) SG = (ρ/ρwater) = 10-3ρ (ρ water = 103 kg/m3)NOTE: 1. The density for a substance varies slightly with temperature, since volume is temperature dependent 2. The values of densities for various substances are an indication of the average molecular spacing in the substance. They show that this spacing is much greater than it is in a solid or liquid•Note: ρ = (M/V)  Mass of body, density ρ, volume V is M = ρV Weight of body, density ρ, volume V isMg = ρVgForces in Fluids•To do fluid dynamics using Newton’s Laws, we obviously need to talk about forces in fluids. •Unlike solids: –Static Fluids do not sustain shearing forces (stresses). Shearing forces are exerted parallel to fluid surfaces. –Static Fluids do not sustain tensile forces (stresses). Tensile forces are exerted perpendicular to the fluid surface.•The only force that can be exerted on an object submerged in a Static Fluid is one that tends to compress the object from all sides•The force exerted by a Static Fluid on an object is always perpendicular to the surfaces of the objectSect. 14.1: PressurePlays the role for fluids that force plays for solid objects•Consider a cross sectional area A oriented horizontally inside a fluid. The force on it due to fluid above it is F.•Definition: Pressure = Force/AreaF is perpendicular to A SI units: N/m21 N/m2 = 1 Pa (Pascal) FPA�•Consider a solid object submerged in a STATIC fluid as in the figure. •The pressure P of the fluid at the level to which the object has been submerged is the ratio of the force (due to the fluid surrounding it in all directions) to the area•At a particular point, P has the following properties: 1. It is same in all directions. 2. It is  to any surface of the object.FPA�If 1. & 2. weren’t true, the fluid would be in motion, violating the statement that it is static!•P is  any fluid solid surface: P = (F /A)•Note that pressure is a scalar, in contrast with force, which is a vector. It is proportional to the magnitude of the force•Suppose the pressure varies over an area. Consider a differential area dA. That area has a force dF on it and dF = P dA•The direction of the force producing a pressure is perpendicular to the area of interest.•A possible means of measuring the pressure in a fluid is to submerge a measuring device in the fluid. •A common device is shown in the lower figure. It is an evacuated cylinder with a piston connected to an ideal spring. It is first calibrated with a known force. •After it is submerged, the force due to the fluid presses on the top of the piston & compresses the spring.•The force the fluid exerts on the piston is then measured. Knowing the area A, the pressure can then be found. Pressure MeasurementsFPA�•Experimental Fact: Pressure depends on depth.•See figure. If a static fluid is in a container, all portions of the fluid must be in static equilibrium.•All points at the same depth must be at the same pressure–Otherwise, the fluid would not be static.•Consider the darker region, which is a sample of liquid with a cylindrical shape–It has a cross-sectional area A–Extends from depth d to d + h below the surface•The liquid has a density –Assume the density is the same throughout the fluid–This means it is an incompressible liquidSect. 14.2: Variation of Pressure with Depth•There are three external forces acting on the darker region. These are:–The downward force on the top, P0A–Upward force on the bottom, PA–Gravity acting downward, Mg•The mass M can be found from the density:•The net force on the dark region must be zero: ∑Fy = PA – P0A – Mg = 0•Solving for the pressure gives P = P0 + gh•So, the