Wrap-up chapter 15 (fluids) Start chapter 16 (thermodynamics)

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Presentation transcript:

Wrap-up chapter 15 (fluids) Start chapter 16 (thermodynamics) Lecture 22 Goals: Wrap-up chapter 15 (fluids) Start chapter 16 (thermodynamics) Assignment HW-9 due Tuesday, Nov 23 Monday: Read through Chapter 16 1

To describe fluid motion, we need something that describes flow: Fluid dynamics To describe fluid motion, we need something that describes flow: Velocity v Ideal fluid model: Incompressible fluid. No viscosity (no friction). Steady flow So far, we have discussed fluids at static equilibrium, now we will start our discussion of fluid dynamics, where there will be fluid-flow, fluid-motion. As we discussed many times during this course, whenever we have motion, one of the most important kinematic quantities is the velocity of the motion. So here we will associate a velocity to fluid flow, keeping in mind that different portions of the fluid might be flowing with different velocities. Fluid flow is a very complicated subject. When the flow is turbulant in particular, there are almost no analytical techniques, and the problems are almost always analyzer by large-scale numerical calculations. So to make the problem more tractable, we will make simplifying assumptions, which are called the ideal fluid model: These conditions are called laminar flow.

Types of Fluid Flow When the flow is turbulent, you typically observe very erratic flow pattern, a lot of circular motion, and also the flow will not be steady that is at some point in time, you might see a flow pattern shown here, and at a little later time, the pattern may be quite different. So this is not what we will be considering in this course. So the flow that we will be considering, laminar flow, will have smooth velocity vectors, that don’t change with time as shown here.

Keep track of a small portion of the fluid: Streamlines Keep track of a small portion of the fluid: When we discuss fluid-flow, we frequently draw lines that are called streamlines, and in this slide I would like to introduce them. So basically, we concentrate on a small volume of the fluid and we follow it through as it flows. So these are the velocity vectors, and in this example, this small volume of the fluid is moving here, and then from here, to here If we connect the trajectory of this small volume, we call that a streamline. If we go though the same procedure at other positions of the fluid, we get the streamlines that characterize the flow. For laminar flow, the streamlines do not cross The velocity vector at each point is tangent to the streamlines

A1v1=A2v2 A1v1 : units of m2 m/s = volume/s Continuity equation A 1 2 v A1v1 : units of m2 m/s = volume/s A2v2 : units of m2 m/s = volume/s Now, one of the most important equations in fluid flow is the continuity equation, which you can view as mass conservation. 2) Considers cross sectional area A1 and assume that the flow has a uniform speed v1 at this first point 3) Similarly, at a later point consider cross sectional area A2 with velocity v2 4) Now if we multiply the area with the velocity, we get the volume of fluid per second that is flowing through the first point 5) Similarly, if we do that at the second point, we would get the volume of fluid per second that is flowing through the second point. 6) Now, whatever is coming in must be coming out. 7) To find the mass flow rate just multiply by the density, this is really a statement of mass conservation. A1v1=A2v2

Exercise Continuity 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 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? (A) 2 v1 (B) 4 v1 (C) 1/2 v1 (D) 1/4 v1

Energy conservation: Bernoulli’s eqn Δx F2=P2A2 F1=P1A1 v2 v1 In the previous example, the flow speed is increasing. Well, there is a mass that is flowing at a greater speed, so the kinetic energy is increasing. For that to happen, we need to do some work on the system. Now, a little bit of thinking shows that, there is no force in the system other than the forces due to the pressure of the liquid. So this increase in the kinetic energy must come from the work done by the pressure. Consider a volume of deltaV moving from point 1 to point 2. At places where the velocity is going up, the pressure must decrease. This argument is valid for a frictionless system. work=F1Δx=P1A1Δx=P1ΔV net work=(P1 ΔV-P2 ΔV)=(P1-P2 ) ΔV

P1+1/2 ρ v12=P2+1/2 ρ v22 P1+1/2 ρ v12+ρgy1=P2+1/2 ρ v22+ρgy2 net work=(P1-P2 ) ΔV =1/2 Δm v22-1/2 Δm v12 P1+1/2 ρ v12=P2+1/2 ρ v22 With height, the change in the potential energy must also be taken into account P1+1/2 ρ v12+ρgy1=P2+1/2 ρ v22+ρgy2 1) P+1/2 ρ v2+ρgy= constant

Torcelli’s Law P + ½ r v2 = const A B P0 + r g h + 0 = P0 + 0 + ½ r v2 P0 = 1 atm Torcelli’s Law P + ½ r v2 = const A B P0 + r g h + 0 = P0 + 0 + ½ r v2 2g h = v2 A B

F/A0 = Y (ΔL/L0) F/A0 = -B (ΔV/V0) Elastic properties Young’s modulus: measures the resistance of a solid to a change in its length. L0 L F F/A0 = Y (ΔL/L0) Bulk modulus: measures the resistance of solids or liquids to changes in volume. V0 V0 - V F If we clamp one end of a solid rod and apply a force to the other end, the rod will stretch. Depending on the material, the rod may stretch a lot if the material is soft for example, or very little, if the material is very hard. Young’s modulus, that we typically denote with capital Y quantifies the resistance of a solid to change its length either stretch or compress. Ofcourse, if the force is very large, then you will break the material, but well below that limit, the relationship between force per unit area and the fractional change in length is roughly linear, and that proportionality constant is Young’s modulus. Tensile stress, strain Young’s modulus has identical units to pressure The harder the material, the larger the young’s modulus F/A0 = -B (ΔV/V0)

1) Space elevator Carbon nanotube 100 x 1010

Thermodynamics: A macroscopic description of matter Roughly, how many atoms/molecules are there in a handfull of stuff? One billion 1015 1024 10100 None of the above 1) Order of magnitude answer!!!

Microscopic to macroscopic connection One mole of substance = 6.02 x 1023 basic particles O2 has atomic mass of 32 1 mole of O2 will have 32 grams of mass mass of one O2 molecule=32 grams/6.02 x 1023 Avagadro’s number

Temperature Three main scales Farenheit Celcius Kelvin 212 100 373.15 Water boils 32 273.15 Water freezes -273.15 -459.67 A measure of thermal energy When we say something is hot, we mean that It has more thermal energy when compared to something cold. Absolute zero means, when the thermal energy is zero. If we have a gas for example, absolute zero would be the point where the pressure drops to zero. Or where the motion of the atoms/molecules that form the gas come to a stop. At room temperature, the average velocity of the molecules around us are about 500 m/s. When we go to absolute zero temperature, the average velocity goes to zero. Nowadays, we can cool atoms to a temperature of pK using laser cooling techniques, we can basically create the coldest places in the universe. Absolute Zero

Phase-diagrams Recall “3” Phases of matter: Solid, liquid & gas All 3 phases exist at different p,T conditions Triple point of water: p = 0.06 atm T = 0.01°C Triple point of CO2: p = 5 atm T = -56°C 1) Note that you can condense stuff by changing the pressure, without changing the temperature.

R=8.31 J/mol K: universal gas constant Fini Ideal Gas Law Assumptions that we will make: “hard sphere” model for the atoms density is low temperature not too high P V = n R T n: # of moles R=8.31 J/mol K: universal gas constant Fini At STP, one mole of gas has a volume of 22.4 liters.

Next time: PV diagrams…important processes Isochoric process: V = const (aka isovolumetric) Isobaric process: p = const Isothermal process: T = const Volume Pressure 1 2 Isochoric Volume Pressure 1 2 Isothermal Volume Pressure 1 2 Isobaric