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 22 Monday: Read through Chapter 16 1

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 Now, let’s discuss another very important equation in fluid dynamics, which is Bernoulli’s equation, which is basically a restatement of energy conservation. In the previous example, the flow speed is increasing. Because the flow speed is increasing, it is not difficult to see that the kinetic energy is increasing also. If we were to track, for example, 1 kilogram of water going from this region to this region, in this region it is flowing 4 times faster, which means that its kinetic energy has increased by a factor of 16. The question is, where is that increase in the kinetic energy coming from? 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 1) As the speed increases, the pressure must decrease. P1+1/2 ρ v12+ρgy1=P2+1/2 ρ v22+ρgy2 P+1/2 ρ v2+ρgy= constant

Toricelli’s Law P + ½ r v2 = const A B P0 = 1 atm Toricelli’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 are there in a handfull of stuff? One billion 1015 1024 10100 None of the above Thermodynamics is basically a set of macroscopic laws that describe different states of matter and how these different states transform into each other. 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, it is a measure of how fast the atoms, molecules are moving. In a solid for example, temperature is a measure of how much the individual atoms that form the solid are vibrating. In a gas, temperature is a measure of how fast the atoms are flying around. When we say something is hot, we mean that It has more thermal energy when compared to something cold. Absolute zero means that 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. 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 1) Note that you can condense stuff by changing the pressure, without changing the temperature.

R=8.31 J/mol K: universal gas constant 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 We have discussed the ideal gas law, so here basically under certain assumptions, all gases whether it is O2, H2, neon, or air, satisfy a universal law, where PV=n R T. Here P is the pressure of the gas, V is the volume, n is the number of moles, and T is the temperature. 2) Now, why all gases satisfying these assumptions should have the same universal relationship with the same value of R is not obvious at all, it requires some thinking, and we will talk about this more when we discuss chapter 18. 3) However, there are also a lot of things about this relationship that intuitively makes sense. Consider a container at a fixed volume V. You would expect that the more gas you put in the container, the more pressure you would have, which is exactly what this equation tells you. With V fixed, as n gets larger, P gets larger as well. Similarly, the higher the temperature of the has, you would expect more pressure, which is exactly what this equation tells you. 4) Discuss microscopic theory of pressure!!!

P V = constant if n, T fixed Suppose that we have a sealed container at a fixed temperature. If the volume of the container is doubled, what would happen to pressure? Double Remain the same Halved None of the above 1) We will use the ideal gas law frequently over the next few weeks. Here is a quick problem. P V = n R T P V = constant if n, T fixed

kB: Boltzmann’s constant P V = n R T n: # of moles n=N/NA P V = n R T=(N/NA) R T =N (R/NA) T P V= N kB T kB: Boltzmann’s constant There is also a slightly different way to write the ideal gas law where instead of the number of moles we use the number of particles. So once again, we have PV=nRT, where n is the number of moles. Now, the number of moles is the number of particles N divided by the avagadros number, 6.02 x 10^23. So if we have avagadro’s number of particles, we have 1 mole, if we have twice the avagadro’s number of particles, we have two moles and so on. Now if we just plug-in this expression for the number of moles, in the ideal gas equation, we get….. Boltzmann’s constant can be thought as the gas constant per unit particle.

PV diagrams Pressure 3 atm 1 atm 3 Liters 1 Liter Volume Now, it is very useful to represent ideal gas processes on diagrams called PV diagrams. Here, we plot pressure versus volume. Each point on the graph represents a state of the gas. For example, here we have a gas that starts at this point, with a pressure of 1 atm, and the volume of the gas is 1 liter. Now although, we only show pressure and volume, other state variables of the gas can typically be found by using the ideal gas law. If we know the number of moles for example, or the number of particles, for each pressure and volume point, we can find the temperature by using the ideal gas law. In this example, the gas ends up at this point, and this line shows the particular trajectory that defines the process. Not only the end-points but also the trajectory is critical. As we will see in a moment, the work done on the gas is different for different trajectories even though the end points may be identical. 3 Liters 1 Liter Volume

PV diagrams: Important processes Isochoric process: V = const (aka isovolumetric) Isobaric process: P = const Isothermal process: T = const Now, a process that an ideal gas goes through can be quite complicated, so that trajectory can be quite complicated. There are, however, a number of special cases, that are very frequently encountered, and for these processes, the PV diagrams look particularly simple. Many important gas processes take place in a container of constant, unchanging volume. Such constant volume processes are called isochoric processes. Iso is a prefix that means constant or equal. We frequently encounter processes where the pressure of the gas does not change, these processes are called isobaric. So P is constant Now, processes where the temperature remains fixed are called isothermal processes.

Which one of the following PV diagrams describe an isobaric process (P=constant) Volume Pressure 1 2 Volume Pressure 1 2 Volume Pressure 1 2 Discuss how you could implement each process Discuss mixed processes V constant, isochoric PV constant, isothermal P constant, isobaric