Presentation on theme: " PROGRAM OF “PHYSICS” Lecturer: Dr. DO Xuan Hoi Room 413"— Presentation transcript:
1 PROGRAM OF “PHYSICS” Lecturer: Dr. DO Xuan Hoi Room 413
2PHYSICS 2 (FLUID MECHANICS AND THERMAL PHYSICS) 02 credits (30 periods)Chapter 1 Fluid MechanicsChapter 2 Heat, Temperature and the Zeroth Law of ThermodynamicsChapter 3 Heat, Work and the First Law of ThermodynamicsChapter 4 The Kinetic Theory of GasesChapter 5 Entropy and the Second Law ofThermodynamics
3References :Halliday D., Resnick R. and Walker, J. (2005), Fundamentals of Physics, Extended seventh edition. John Willey and Sons, Inc.Alonso M. and Finn E.J. (1992). Physics, Addison-Wesley Publishing CompanyHecht, E. (2000). Physics. Calculus, Second Edition. Brooks/Cole.Faughn/Serway (2006), Serway’s College Physics, Brooks/Cole.Roger Muncaster (1994), A-Level Physics, Stanley Thornes.
9States of Matter Solid Liquid Gas Plasma Has definite volume Has definite shapeMolecules are held in specific location by electrical forces and vibrate about equilibrium positionsCan be modeled as springs connecting molecules
10States of Matter Solid Liquid Gas Plasma Crystalline solid Atoms have an ordered structureExample is salt (red spheres are Na+ ions, blue spheres represent Cl- ions)Amorphous SolidAtoms are arranged randomlyExamples include glass
11States of Matter Solid Liquid Gas Plasma Has a definite volume No definite shapeExist at a higher temperature than solidsThe molecules “wander” through the liquid in a random fashionThe intermolecular forces are not strong enough to keep the molecules in a fixed positionRandom motion
12States of Matter Solid Liquid Gas Plasma Has no definite volume Has no definite shapeMolecules are in constant random motionThe molecules exert only weak forces on each otherAverage distance between molecules is large compared to the size of the molecules
13States of Matter Solid Liquid Gas Plasma Matter heated to a very high temperatureMany of the electrons are freed from the nucleusResult is a collection of free, electrically charged ionsPlasmas exist inside stars or experimental reactors or fluorescent light bulbs!For more information:
14Is there a concept that helps to distinguish between those states of matter?
15DensityThe density of a substance of uniform composition is defined as its mass per unit volume:some examples:Object is denser Density is greaterThe densities of most liquids and solids vary slightly with changes in temperature and pressureDensities of gases vary greatly with changes in temperature and pressure (and generally 1000 smaller)UnitsSIkg/m3CGSg/cm3 (1 g/cm3=1000 kg/m3 )
16PressurePressure of fluid is the ratio of the force exerted by a fluid on a submerged object to areaUnitsSIPascal (Pa=N/m2)Example: 100 N over 1 m2 is P=(100 N)/(1 m2)=100 N/m2=100 Pa.
171. Variation of Pressure with Depth 1.1 Pressure and Depth If a fluid is at rest in a container, all portions of the fluid must be in static equilibriumAll points at the same depth must be at the same pressure (otherwise, the fluid would not be in equilibrium)Three external forces act on the region of a cross-sectional area AExternal forces: atmospheric, weight, normal
18Test 1You are measuring the pressure at the depth of 10 cm in three different containers. Rank the values of pressure from the greatest to the smallest:4. It’s the same in all three10 cm
19Pressure and Depth equation Po is normal atmospheric pressure1.013 x 105 Pa = 14.7 lb/in2The pressure does not depend upon the shape of the containerOther units of pressure:76.0 cm of mercuryOne atmosphere 1 atm = x 105 Pa14.7 lb/in2
20Example 1:Find pressure at 100 m below ocean surface.
211.2 Absolute Pressure and Gauge Pressure The excess pressure above atmospheric pressure is usually called gauge pressure (gh), and the total pressure is called absolute pressure.
22PROBLEM 1A storage tank 12.0 m deep is filled with water. The top ofthe tank is open to the air. What is the absolute pressure atthe bottom of the tank? The gauge pressure?SOLUTIONThe absolute pressure :The gauge pressure :
23PROBLEM 2The U-tube in Fig. 1 contains two liquids in staticequilibrium: Water of density pw = 998 kg/m3 is in the rightarm, and oil of unknown density px is in the left.Measurement gives l = 135 mm and d = 12.3 mm.What is the density of the oil?SOLUTIONIn the right arm:In the left arm:
24The hydraulic press is an important application of Pascal’s Principle A change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and to the walls of the container.The hydraulic press is an important application of Pascal’s PrincipleAlso used in hydraulic brakes, forklifts, car lifts, etc.Since A2 > A1, then F2 > F1 !!!
251.4 Measuring PressureOne end of the U-shaped tube is open to the atmosphereThe other end is connected to the pressure to be measuredPressure at B is Po+ρghA long closed tube is filled with mercury and inverted in a dish of mercuryMeasures atmospheric pressure as ρghThe spring is calibrated by a known forceThe force the fluid exerts on the piston is then measured
26QuestionSuppose that you placed an extended object in the water. How does the pressure at the top of this object relate to the pressure at the bottom?1. It’s the same.2. The pressure is greater at the top.3. The pressure is greater at the bottom.4. Whatever…
271.5 Buoyant ForceThis force is called the buoyant force.What is the magnitude of that force?P1AP2A= mg
28Buoyant ForceThe magnitude of the buoyant force always equals the weight of the displaced fluidThe buoyant force is the same for a totally submerged object of any size, shape, or densityThe buoyant force is exerted by the fluidWhether an object sinks or floats depends on the relationship between the buoyant force and the weight
29Archimedes' PrincipleAny object completely or partially submerged in a fluid is buoyed up by a force whose magnitude is equal to the weight of the fluid displaced by the object.This force is buoyant force.Physical cause: pressure difference between the top andthe bottom of the object
30Archimedes’ Principle: Totally Submerged Object The upward buoyant force is B = ρfluid gVobjThe downward gravitational force is w = mg = ρobj g VobjThe net force is B – w = (ρfluid - ρobj) g VobjDepending on the direction of the net force, the object will either float up or sink!
31The net force is B - w=(ρfluid - ρobj) g Vobj The object is less dense than the fluid ρfluid < ρobjThe object experiences a net upward forceThe object is more dense than the fluid ρfluid > ρobjThe net force is downward, so the object accelerates downward
32Test 2Two identical glasses are filled to the same level with water. One of the two glasses has ice cubes floating in it.Which weighs more?1. The glass without ice cubes.2. The glass with ice cubes.3. The two weigh the same.NOTE : Ice cubes displace exactly their own weight in water.
33Weight of the whole iceberg : PROBLEM 3An iceberg floating in seawater, as shown in figure, is extremelydangerous because much of the ice is below the surface. Thishidden ice can damage a ship that is still a considerable distancefrom the visible ice. What fraction of the iceberg lies below thewater level ? The densities of seawater and of iceberg areW = 1030 kg/m3 and I = 917 kg/m3SOLUTIONWeight of the whole iceberg :Buoyant force :(VW : volume of the displaced water = volume of the ice beneath the water)The fraction of ice beneath the water’s surface:
34Chapter 8 Fluid Mechanics 1. Variation of Pressure with Depth2. Fluid Dynamics
352.1 Fluids in Motion: Streamline Flow Streamline flow (also called laminar flow)every particle that passes a particular point moves exactly along the smooth path followed by particles that passed the point earlierStreamline is the pathdifferent streamlines cannot cross each otherthe streamline at any point coincides with the direction of fluid velocity at that pointLaminar flow around an automobile in a test wind tunnel.
362.1 Fluids in Motion: Turbulent Flow The flow becomes irregularexceeds a certain velocityany condition that causes abrupt changes in velocityEddy currents are a characteristic of turbulent flowHot gases from acigarette made visible by smokeparticles. The smoke first moves inlaminar flow at the bottom andthen in turbulent flow above
37Fluid Flow: ViscosityViscosity is the degree of internal friction in the fluidThe internal friction is associated with the resistance between two adjacent layers of the fluid moving relative to each other
382.2 Characteristics of an Ideal Fluid The fluid is nonviscousThere is no internal friction between adjacent layersThe fluid is incompressibleIts density is constantThe fluid is steadyIts velocity, density and pressure do not change in timeThe fluid moves without turbulenceNo eddy currents are present
392.3 Equation of Continuity The product of the cross-sectional area of a pipe and the fluid speed is a constantSpeed is high where the pipe is narrow and speed is low where the pipe has a large diameterAv is called the volume flow rateThe mass is conserved :
40(a) The speed of the oil: PROBLEM 4As part of a lubricating system for heavy machinery, oil of density850 kg/m3 is pumped through a cylindrical pipe of diameter 8.0 cmat a rate of 9.5 liters per second. The oil is incompressible.What is the speed of the oil? What is the mass flow rate?If the pipe diameter is reduced to 4.0 cm, what are the newvalues of the speed and volume flow rate?SOLUTION(a) The speed of the oil:The mass flow rate:(b)Oil incompressible: volume flow rate has the same value:
413. Bernoulli’s Equation Magnitude of the force exerted by the fluid in section 1: P1A1The work done by thisforceW1 = F1x1 = P1A1x1 = P1V( V: volume of section 1)The work done by by the fluid in section 2:W2 = - F2x2 = - P2A2x1 = - P2V(W2 < 0 : the fluid force opposes the displacement)The net work done by two forces: W = (P1 - P2)V
42Theorem of the variation of kinetic energy : Bernoulli’s equation applied to an ideal fluid :
43Bernoulli’s Equation Relates pressure to fluid speed and elevation Bernoulli’s equation is a consequence of Conservation of Energy applied to an ideal fluidAssumes the fluid is incompressible and nonviscous, and flows in a nonturbulent, steady-state mannerStates that the sum of the pressure, kinetic energy per unit volume, and the potential energy per unit volume has the same value at all points along a streamline
44Measure the speed of the fluid flow: Venturi Meter EXAMPLEApplication of Bernoulli’s EquationMeasure the speed of the fluid flow: Venturi MeterShows fluid flowing through a horizontal constricted pipeSpeed changes as diameter changesSwiftly moving fluids exert less pressure than do slowly moving fluidsHow to measure the speed v2 ?
45Measure the speed of the fluid flow: Venturi Meter Application of Bernoulli’s EquationMeasure the speed of the fluid flow: Venturi MeterEquation of Continuity :
464. Poiseuille’s law Rate of flow : the volume of fluid which passes through a given surface per unit time (m3/s) Poiseuille's equation :RP1P2vL : viscosity of the fluid
47PROBLEM 5A horizontal pipe of 25-cm2 cross-section carries water at a velocity of 3.0 m/s. The pipe feeds into a smaller pipe with cross section of only 15 cm2. W=103kg/m3(a) What is the velocity of water in the smaller pipe ?(b) Determine the pressure change that occurs from the larger-diameter pipe to the smaller pipe.(a)A1v1A2v2SOLUTION(b)
48PROBLEM 6A large pipe with a cross-sectional area of 1.00 m2 descends 5.00 m and narrows to m2, where it terminates in a valve. If the pressure at point 2 is atmospheric pressure, and the valve is opened wide and water allowed to flow freely, find the speed of the water leaving the pipe.SOLUTIONhv2v12P2=P0P1=P0
50PROBLEM 7There is a leak in a water tank. The hole is very small compared to the tank’s cross-sectional area.(a) If the top of the tank is open to the atmosphere, determine the speed at which the water leaves the hole when the water level is above the hole.A2P2 =P0(a)SOLUTIONhP0v1y2A1y1
51PROBLEM 7There is a leak in a water tank. The hole is very small compared to the tank’s cross-sectional area.(b) Where does the stream hit the ground if the hole is 3.00 m above the ground ?y(b)A2P2 =P0SOLUTIONhP0v1y2A1y1x
52PROBLEM 8An airplane has wing, each wing area 4.00 m2, designed so that air flows over the top of the wing at 245 m/s and under the wing at 222 m/s. Find the mass of the airplane such that the lift on the plane will support its weight, assuming the force from the pressure difference across the wings is directed straight upwards.SOLUTION
53The lift on the plane supports the plane’s weight :
54PROBLEM 9BLOOD PRESSURE WITH DEPTH:Human blood has a density of approximately1.05 x 103 kg/m3.(a) Use this information to estimate the difference in blood pressure between the brain and the feet in a person who is approximately 1.6 m tall.SOLUTION(a)The difference in pressure is given by:
55PROBLEM 9BLOOD PRESSURE WITH DEPTH:Human blood has a density of approximately1.05 x 103 kg/m3.(b) Estimate the volume flow rate of blood from the head to the feet of this person. Assume an effective radius of 24 cm.The viscosity of blood is N.s/m2.SOLUTION(b)Poiseuille's equation :