1 / 16

180 likes | 982 Views

PHY 2048. Lecture 9. CHAPTER 12. Fluid Mechanics. Density. Density of a uniform fluid is the mass per unit volume of that fluid. Let m (kg) be the mass of a fluid that occupies a volume V (m 3 ). Density ρ = m/V (kg/m 3 ). A small area of fluid.

Download Presentation
## PHY 2048

**An Image/Link below is provided (as is) to download presentation**
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.
Content is provided to you AS IS for your information and personal use only.
Download presentation by click this link.
While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

**PHY 2048**• Lecture 9**CHAPTER 12**Fluid Mechanics**Density**Density of a uniform fluid is the mass per unit volume of that fluid Let m (kg) be the mass of a fluid that occupies a volume V (m3) Density ρ = m/V (kg/m3) A small area of fluid Force on the small area from the fluid to the right of it Force on the small area from the fluid to the left of it dA The net force has to be zero since the whole fluid is at rest The force is present at all points on the small area and the net force is F from the left and right Pressure Pressure is force per unit area: F = p dA Pressure, p, has units of N/m2 (also called Pascals and abbreviated as Pa)**(p+dp)A**dp + ρg dy = 0 d ρ(A dy)g pA Atmospheric pressure The air in the atmosphere can be thought of as a fluid The pressure due to air at sea-level under normal conditions is called atmospheric pressure Normal atmospheric pressure at sea-level = 1 atm = 101,325 Pa The force on the top surface of your hand (assuming an area of 5.0 cm2) due to the air at sea-level is 1.01325 x 105 X 5.0 x 10-4 = 51 N. This is like holding a 5.2 kg ball on your hand! But this is balanced by an equal force on the bottom surface of your hand. The element of fluid of area A and thickness dy is at rest The net force from the left cancels the net force from the right The net force from the front cancels the net force from the back The net force from the top along with the weight of the element of fluid has to be balanced by the net force from the bottom Pressure has to vary with height of the fluid and has to increase with depth p: pressure from the fluid on the bottom surface p+dp: pressure from the fluid on the top surface dp < 0 Mass of the fluid element is density (ρ) times volume (A dy) pA = (p+dp) A + ρ (A dy) g (dp/dy) = -ρg p = p0 + ρgd p0 is the pressure at the top**Absolute Pressure and Gauge Pressure**Absolute pressure is the net force per unit area Gauge pressure is the excess pressure over the atmospheric pressure**A U-tube is filled with water, and the two arms are capped.**The tube is cylindrical, and the right arm has twice the radius of the left arm. The caps have negligible mass, are watertight, and can freely slide up and down the tube. A one-inch depth of sand is poured onto the cap on each arm. The sand is removed and a 1 kg mass block is placed on each cap. The locations of the two caps at equilibrium are now as given in this figure. The dashed line represents the level of the water in the left arm. Since the weight is the same on both sides, the force is the same. Since the area on the right is four times bigger, the pressure on the right is four times smaller. Therefore, the height of the water column on the right will be higher than the left. If we want the heights on both sides to be the same, we need to place a weight on the right side that is four times larger than the one on the left for the two forces to exert the same pressure. The excess mass of water on the right side should have a mass of 3 kg so that the pressure on the right that is at the same level as shown by the dashed line is the same. The amount of sand on the right will be four times more and therefore has four times more weight. But the area on the right is four times more than the left. The pressure exerted by the sand is the same on both sides and the water level on both sides will be at the same height.**As the reservoir behind a dam is filled with water, the**pressure that the water exerts on the dam increases. Eventually, the force on the dam becomes substantial, and it could cause the dam to collapse. There are two significant issues to be considered: First, the base of the dam should be able to withstand the pressure ρgh, where ρ is the density of the water behind the dam, h is its depth, and g is the magnitude of the acceleration due to gravity. This means that the material of which the dam is made needs to be strong enough so that it doesn't crack (compressive strength). The second issue has to do with the strength of the foundation of the dam. The water pressure exerts a clockwise torque on the dam. The foundation of the dam should be strong enough so that the dam does not topple. The material has to be strong enough that the dam does not snap (shear strength). To study this phenomenon, consider the simple model of a dam depicted in the diagram. A reservoir of water (density ρ) behind the dam is filled to a height h. Assume that the width of the dam (the dimension pointing into the screen) is L. The torque about P due to this force is Consider a horizontal layer of the dam wall of thickness dx located a distance x above the reservoir floor. dF = ρg(h-x) (Ldx) dF dx x A problem that uses integration Integrate from 0 to h to get the total torque**A ball of mass mb and volume V is lowered on a string into a**fluid of density ρf. Assume that the object would sink to the bottom if it were not supported by the string. T B=ρfVg mbg Archimedes’ principle Apparent weight of a body immersed in a liquid = wbody - B = wbody - wfluid What is the tension T in the string when the ball is fully submerged but not touching the bottom, as shown in the figure? Net force on the ball is zero T = mbg - ρfVg**Work done =**Change in potential energy = Change in kinetic energy = remains a constant at all points within a fluid flow Incompressible fluid flow - Bernoulli’s equation At time t+dt Total mass has to remain the same at all times ρ A1ds1 = ρ A2ds2 = ρ dV Continuity equation for an incompressible fluid Work has to be done to make the fluid flow There is change in kinetic energy and potential energy At time t**A cylindrical open tank needs cleaning. The tank is filled**with water to a height H, so you decide to empty it by letting the water flow steadily from an opening at the side of the tank, located near the bottom. The cross-sectional area of the tank is A square meters, while that of the opening is a square meters. Plug in for v Set up to do the integral Final result A dh H h a Let h be the height of the water at time t Let dh (<0) be the reduction in height in a time dt The volume of water that went out in time dt is -A dh All of this water went out through the opening with cross-sectional area a Therefore, the water flowing out must have travelled a distance of (-A dh)/a in time dt The speed of the outgoing water at time t is v = -(A/a) (dh/dt) By energy conservation, kinetic energy of the outgoing water at the bottom should be the same as the loss in potential energy of the water at the top Since the mass of the water lost on the top is the same as the mass of the outgoing water in the bottom, (1/2) v2=gh**CHAPTER 15**Mechanical waves**Wave number =**Transverse waves Wave speed = v = λ/T = λf**We will accept without any additional mathematics that the**speed of a transverse wave in a stretched string is given by Speed of a transverse wave on a string Assume a string has been stretched so that it has a uniform tension T Let the mass per unit length of the string be μ Let us now pluck on some part of the string so that a wave travels through it If the string has a lot of tension, it is intuitive that the neighboring points in the string respond fast to any transverse motion. This would imply that the velocity of a wave in the string will increase if T increases. If the string is very heavy (larger mass per unit length), there is a larger inertia for a neighboring point to respond to a transverse motion. This would imply that the velocity of a wave in the string will decrease if μ increases. T/μ has units of (m/s)2**A large ant is standing on the middle of a circus tightrope**that is stretched with tension T. The rope has mass per unit length μ. Wanting to shake the ant off the rope, a tightrope walker moves her foot up and down near the end of the tightrope, generating a sinusoidal transverse wave of wavelength λ and amplitude A. Assume that the magnitude of the acceleration due to gravity is g. What is the minimum value for A that will shake the ant off. We will assume that the mass of the ant does not affect the wave motion. The transverse acceleration due to the wave motion is given by the acceleration in simple harmonic motion which is Aω2 where ω is the angular frequency. We want the transverse acceleration to be larger than g so that the ant will lose contact with the rope**Note that**is the negative of the speed of wave propagation Wave motion is characterized by two velocities: the velocity with which the wave moves in the medium (e.g., air or a string) and the velocity of the medium (the air or the string itself). Consider a transverse wave traveling in a string. The mathematical form of the wave is y(x,t) = A sin(kx-ωt). The string only moves in the y direction. Therefore, the x-component of the velocity of the string is zero. The y-component of the velocity of the string is (dy/dt)(x,t) = -Aω cos(kx-ωt) The speed of propagation of the wave is ω/k. This can be seen as moving with the wave such that (kx-ωt) is a constant. This would mean we will not see the wave move. (dy/dx)(x,t) is the slope of the string at the location x at time t and this is given by Ak cos(kx-ωt)

More Related