GenPhys (Finals) (1).pdf

Week 10: Rotational Equilibrium & 3. Angular Acceleration – rate of change of Rotational Dynamics angular velocity Why are door knobs installed at the opposite side of the hinge? To make it as easy as possible to open the Units: deg/s2, rev/s2, or ra...

Week 10: Rotational Equilibrium & 3. Angular Acceleration – rate of change of Rotational Dynamics angular velocity Why are door knobs installed at the opposite side of the hinge? To make it as easy as possible to open the Units: deg/s2, rev/s2, or rad/s2 door, the knobs are placed as far from the Linear acceleration: hinge as is feasible. This maximizes the lever arm, reducing the amount of force needed to open the door. Rotation – motion of a body turning about an axis, where each particle of the body moves along a circular path. Kinematics of Rotation 1. Angular Displacement - angle in radians Kinematic Equations for Uniformly through which a point revolves around a Accelerated Motion center or line has been rotated in a specified sense about a specified axis. Units: degrees, radians, or revolutions 1 revolution = 360° = 2π radians Linear distance: Examples: Ex 1: A rider notices that the wheels of his bicycle make 12 rev in 15 s. (a) What is the average angular speed of the wheel in 2. Angular Velocity – rate at which angular radians/s? (b)What distance in meters does displacement changes with time the wheel travel if its radius is 33 cm? Units: deg/s, rev/s, or rad/s Linear velocity: Moment of Inertia (rotational inertia) – rotational analog for mass; property of a rotating body to resist change in its state of rotation 1 For a system made up of several particles: Radius of gyration (k): Moments of Inertia of Composite Bodies of Simple Geometric Shapes: Torque (Moment of force) – effectiveness of a force in rotating a body on which it acts; vector quantity Lever arm or Moment arm (r) – perpendicular displacement of the line of action of the force from the pivot point Examples: Ex 1: Selena is a majorette during a parade. Her baton consists of a wood of mass 0.25 kg and diameter of 0.12 m at each end of a slim SI unit: meter-newton (m·N) rod. The rod is 0.50 m long. Find the moment of inertia and radius of gyration of the baton about an axis perpendicular to the rod and intersecting it at (a) the middle and (b) 0.4 m *positive – counterclockwise from one end of the rod. Assume that the rod *negative – clockwise has negligible mass. Which applied force will open the door most effectively? 2 It is easiest to rotate a door when the force is Dynamics of Rotation: applied farthest from the hinge. Rotation is Newton’s Laws of Motion in Rotating greatest when force is applied perpendicularly Bodies to the door. 1. Newton’s 1st Law – Law of Inertia Examples: “A body at rest will not rotate and a rotating Ex 1: A particle located at m is acted body will not change its angular velocity upon by a constant fo,rce of 2 N. Calculate unless acted upon by an external unbalanced the torque about the origin that the particle torque.” experiences as a result of this force. 2. Newton’s 2nd Law – Law of Acceleration “An unbalanced torque acting on a rigid body produces angular acceleration. This angular acceleration is directly proportional to and in thesame direction as the unbalanced torque and inversely proportional to the body’s moment of inertia about the axis of rotation.” 3. Newton’s 3rd Law – Law of Interaction Ex 2: Find the resultant torque about A and C “For every action torque, there is an equal but when the forces shown in the figure act on a opposite reaction torque.” 5.0 N rod. The rod is 6 m long, and its weight is acting at its center. Examples: Ex 1: What torque is needed to accelerate a Ferris wheel from rest to 3.25 rad/s in 15 s. Approximate the Ferris wheel to be a disk of radius 12.5 m and of mass 825 kg. Examples: Ex 2: A 1.20 kg object is suspended by means of string of negligible mass. The string passes over a pulley or radius 0.32 m and moment of inertia of 1.45 kg·𝑚 2. Find the (a) torque produced by the hanging object on the pulley, (b) angular acceleration of the pulley, and (c) acceleration of the object. 3  1.50 rev/s. What is the ratio of her moment of inertia with outstretched arms to the moment of inertia with arms tucked in close to her chest? Rotational Work and Kinetic Energy Work done in rotation: Power: Angular Momentum (L) – product of moment Kinetic Energy of a rotating rigid body: of inertia about an axis and the angular velocity of a body L=Iw Total Kinetic Energy of a rolling body: 𝟐 SI unit: kg·𝒎 /s Conservation of Angular Momentum “When the net external torque acting on a Examples: system is zero, the total angular momentum Ex 1: When you open the lid of your spin dryer, of the system is conserved.” a retarding torque of 12.5 m·N stops the dryer after 8 rev. (a) How much work is done by the torque to stop the dryer? (b) If the dryer stops after 4.0 s, how much power is done by the stopping device that provided the torque? Examples: Ex 1: A physics teacher sits on a rotating stool to demonstrate the conservation of angular momentum. With her arms outstretched, she rotates as 0.75 rev/s. when her arms are tucked in close to her chest, she rotates at 4 Equilibrium Examples: Statics – concerned with calculation of Ex 1: Eight boys are equally spaced about a forces acting on and within structures that basketball and each exerts a force upon it are in equilibrium. toward the center. Each boy in position 1, 3, Center of Gravity – point where the entire and 7 exerts a force of 70 N, and the rest of weight of a body may be assumed to be the boys exert 50 N each. What force is concentrated. needed to keep the ball at rest? Given: Conditions of Equilibrium 1. First Condition of Equilibrium: Translational Equilibrium “The sum of forces acting on an object in equilibrium is zero.” Week 11: Gravity Motion of Heavenly Bodies Kepler’s Laws of Planetary Motion 2. Second Condition of Equilibrium: 1. Law of Ellipses Rotational Equilibrium “Orbit of a planet around the sun is an ellipse “The sum of torques about any pivot must be with the sun at one point.” zero.” Equilibrant – additional force needed to 2. Law of Equal Areas balance the resultant to produce translation “Planet moves around the sun in such a way equilibrium if the resultant of the forces acting on a body is not zero; equal in magnitude to that a line drawn from the sun to the planet the resultant force but oppositely directed. sweeps out equal areas in equal intervals of time.” 3. Law of Periods “The ratio of the squares of the periods of any two planets revolving around the sun is equal to the ratio of the cubes of their mean distance from the sun.” 5 Examples: Ex 1: The mean solar distance of Mercury is 0.387 AU. What is its period? Ex 1: What is the acceleration due to gravity of an object falling at the surface of the Earth? (mass of Earth = 6 x 1024 kg; radius of Earth = 6.4 x 106 m) Newton’s Law of Universal Gravitation “Every object in the universe attracts every other object with a force proportional to the product of their masses and inversely proportional to the square of the distance between them.” Satellite Motion Gravitation – happens between all bodies by Satellite – any object that orbits a planet or a virtue of their masses. star – may be natural or artificial Syncom – synchronous communication Examples: satellite – satellite having the same period of Ex 1: Compare the gravitational force of revolution to the period of rotation of the attraction on a 1.0 kg rice when it is at the planet it orbits surface of the moon to the gravitational force on the same object when it is at the surface of the Earth. (mass of the moon = 7.35 x 2022 kg; radius of the moon =1.738 x 106 m; mass of Earth = 6 x 1024 kg; radius of Earth = 6.4 x Satellite dishes – stationary antennas that 106 m) are placed in our homes that receive signals without changing orientations from a revolving satellite. Examples: Ex 1: What should be the altitude of a satellite synchronous with Earth? Acceleration due to Gravity – rate at which an object changes its velocity due to the force of gravity. 6 Gravitational Potential Energy and Escape Characteristics of a Wave: Velocity 1. Amplitude (A) – maximum displacement of a body from its equilibrium position Gravitational Potential Energy – work 2. Period (T) – time required to make a needed to move an object from an initial complete to-and-fro motion position to a final position. Cycle – one complete to-and-fro motion 3. Frequency (f) – number of cycles per unit time; unit: hertz (Hz) 𝟏 𝒇= Escape Velocity – minimum velocity an 𝑻 object must have to escape its gravitational field without ever falling back. Characteristics of a Wave: Angular Frequency (ω) – a scalar measure of rotation rate; it refers to the angular displacement per unit time or the rate of change of the phase of a sinusoidal waveform, or as the rate of change of the argument of the sine function. Simple Harmonic Motion (SHM) – type of Examples: periodic motion where the restoring force is Ex 1: What is the escape speed from planet proportional to the displacement of the body Earth given that the mass of Each is 6 x 1024 from its equilibrium position. This restoring kg and its radius is 6.4 x 106 m? force acts in a direction opposite that of the displacement. Conditions for SHM: 1. The restoring force must be proportional to the displacement and act opposite to the direction of motion with no drag forces or friction. 2. The frequency of oscillation does not depend on the amplitude. Week 12: Periodic Motion Simple Harmonic Oscillators – systems Periodic Motion – motion that is repeated at exhibiting simple harmonic motion regular intervals of time Mass-spring system – simplest form of Ex: movement of the hands of a clock, harmonic oscillator; a body of mass m pendulum, rocking chair, heartbeat, rotation of oscillating on one end of an elastic spring the blades of an electric fan, movement of 𝑭 = 𝒌𝒙 Earth about its axis and about the sun. Equilibrium Position or Resting Position – position assumed by the body when it is not vibrating Restoring Force – force that tends to restore a body from its displacement to its equilibrium position 7 Examples: Ex 1: An oscillating body takes 0.8 s to complete four cycles. What is the (a) period, (b) frequency, and (c) angular frequency of the body? Pendulum Simple pendulum – consists of a concentrated mass called the bob suspended by a light thread and attached to a fixed Ex 2: A force of 3.0 N elongates a spring by support 6.0 cm. (a) What is the force constant of the spring? (b) How much force is needed to Restoring Force: 𝑭 = 𝒎𝒈 𝒔𝒊𝒏 𝜽 elongate the spring by an additional 6.0 cm? 𝑳 Period: 𝑻 = 𝟐𝝅√ 𝒈 Based on the equation, the period of a simple pendulum is governed by the following laws: 1. The period of a simple pendulum is directly proportional to the square root of its length. 2. The period is inversely proportional to the square root of the acceleration due to gravity. 3. The period is independent of the mass of Period, Frequency, and Angular the bob. Frequency of a Simple Harmonic Motion 4. The period is independent of the angular amplitude if the angular displacement is small, Maximum speed: say less than or equal to 10O. Period: Examples: Ex 1: A simple pendulum of length 50.0 cm 𝟏 𝒌 takes 5 s to make 10 complete back-and-forth Frequency: 𝒇 = √ 𝟐𝝅 𝒎 motion. (a) Find its period. (b) What will be its 𝒌 period when its length is increased to 200.0 Angular Frequency: 𝛚 = √𝒎 cm? Examples: Ex 1: When a 2.5 kg object is suspended from a spring, the spring stretches by 0.05 m. (a) What is its force constant? (b) If the suspended mass is set into vibration, what will be its frequency? 8 Physical Pendulum – hanging object is a rigid body Restoring gravitational torque: 3. Critically damped oscillation Period: – the system returns to equilibrium without oscillating but faster than overdamped oscillation; examples, door closing mechanisms and shock absorbers in cars Production and Properties of Waves Classification of Waves: based on medium Examples: 1. Mechanical Waves – require a medium to Ex 1: A 1.5 kg uniform meterstick at one end propagate; cannot travel in vacuum; ex: oscillates as a physical pendulum with a sound waves, ripples period of 1.25 s. Find its moment of inertia 2. Electromagnetic Waves – can travel in a with respect to the pivot point. vacuum; ex: light, radio waves, microwaves Classification of Waves: based on wave propagation 1. Transverse waves – particles of the medium are vibrating perpendicular to the direction of wave propagation; top of the wave is called crest, while the bottom of the wave is called trough. 2. Longitudinal waves – particles of the Damped Oscillation – when motion of an medium are vibrating parallel to the direction oscillator reduces due to an external force; of the wave of wave propagation; region periodic motions of gradually decreasing where particles of the medium are closer amplitude; object will eventually stop vibrating together is called compressions, while region where particles are farther apart is called Three types of damped oscillation: rarefaction. 1. Underdamped oscillation – the system oscillates with decreasing Basic Wave Relation amplitude until it becomes zero; example, a Basic relation: swing pushed once where: 𝒗 is the speed of the wave is m/s; 𝒇 is the frequency in hertz (Hz); is wavelength in meters 2. Overdamped oscillation – the system returns to equilibrium without oscillating; example, a spring mechanism that closes the door gradually 9 Mechanical Description of a Wave Power (P) – energy transported per unit time Wave Function – detailed description of the 𝑷 = 𝑰𝑺 = 𝑰(𝟒𝝅𝒓𝟐) position of any particle in the medium where Intensity (I) – energy transported per unit the wave is propagating area and per unit time Angular frequency (ω) – measure of rotation 𝑷 𝑬 𝑰= = rate: 𝝎 = 𝟐𝝅𝒇 𝑺 𝑺𝒕 𝑰 = 𝟐𝝅𝟐 𝒑𝒗𝒇𝟐 𝑨𝟐 Wave number (K) – number of wavelength per unit distance: where: S is the surface area of a sphere 𝑺 = 𝟒𝝅𝒓𝟐; r is the distance of the point from the source; E is energy of the wave; I is intensity Wave Function in W/m2 Wave function (for a wave traveling from left to right): Inverse Square Law for Wave Intensity “The intensity of the wave is inversely proportional to the square of the distance of the point under consideration from the source Wave function (for a wave traveling from right of wave.” to left): If power out of the source is constant, 𝑰 𝟏 𝒓𝟏 𝟐 = 𝑰 𝟐 𝒓𝟐 𝟐 Ex 1: A transverse wave has the following where: 𝑰𝟏 is intensity at distance 1; 𝑰𝟐 is properties: amplitude = 0.05 m, frequency = intensity at distance 2; 𝒓𝟏 is distance 1 from 2.5 Hz, wave speed = 15 m/s. The wave is source of wave; 𝒓𝟐 is distance 2 from source propagating in the +x-direction at t = 0 and x = 0, y = 0. Find the (a) angular frequency and (b) of wave wave number. (c) Write the wave function. Inverse Square Law for Wave Intensity Examples: Ex 1: You whisper something to your friend who is 25.0 cm away from you. The sound reaches your friend’s ear with an intensity of 2.0 x 10−10 W/m2. (a) How much power and energy does the source of sound emit in 1 Week 13: Mechanical Waves and Sound hour assuming its power output remains Energy of a Mechanical Wave (E) – amount constant? (b) What is the intensity of the of energy in a wave; related to its amplitude sound heard by another friend who is 50 cm and its frequency away from you? 𝑬 = 𝟐𝝅𝟐𝒎𝒇𝟐 𝑨𝟐 10  3. Partial Destructive Interference – happens when two waves of different amplitudes but are 𝟏𝟖𝟎° out of phase with each other meet; the resulting wave has an amplitude equal to the difference of the amplitudes of the component waves. Examples: Ex1: Show that when two identical sine waves that are 𝟏𝟖𝟎° out of phase with each other interfere, the resultant will be a total destructive interference. Interference – occurs when two or more waves meet while passing through the same medium at the same time Law of Superposition “When two or more waves of the same type travel simultaneously in the same direction, the resultant displacement at that point is equal to the sum of the displacements due to each individual wave” Standing Waves in a String Standing Wave – also known as stationary wave, is a wave which oscillates in time but whose peak amplitude profile does not 1. Constructive Interference – happens move in space when two waves of the same frequency, in Nodes – points where the particles of the phase, and traveling in the same direction medium are not displaced from their meet; the resulting wave is of the same equilibrium position frequency but with an amplitude equal to the Antinodes – points where particles sum of the amplitudes of the two component experience maximum displacement waves. Half-wavelength – distance from one node to the adjacent node or antinode to the adjacent antinode One-fourth of a wavelength – distance from one node to an antinode 2. Total Destructive Interference – happens when two waves of the same amplitude and frequency but are 𝟏𝟖𝟎° out of phase with each other meet. 11 Fundamental Frequency – lowest possible Sound Basics frequency Sound – longitudinal wave created by 𝒗 vibrating objects and capable of producing 𝒇𝟏 = 𝟐𝑳 a sensation in our auditory system where: L is the length of the string in m; v is Audible range – 20 Hz to 20 00 Hz speed of wave in m/s Infrasounds – sounds with frequencies below 20 Hz Ultrasounds – sounds with frequencies above 20 000 Hz Harmonic Frequencies – frequencies that are integral multiples of the fundamental Doppler Effect – apparent change in sound frequency frequency caused by motion of sound 𝒗 source and/or observer 𝒇𝑵 = 𝑵 𝟐𝒍 Doppler Shift – difference between the where: N = 1, 2, 3, … frequency from the source f and the If N = 1, the frequency emitted is the apparent frequency f’ fundamental frequency a.k.a first harmonic. The following symbols are used: If N = 2, the frequency is called second harmonic, and so on. Observer in Motion; Source at Rest Observer moving toward the source: Overtones – all frequencies higher than the fundamental frequency; thus, the second harmonic is the first overtone, the third Observer moving away the source: harmonic is the second overtone, and so on. Ex 1: The fundamental frequency of a string is 200 Hz. Find the first three overtones. Observer at Rest; Source in Motion Source moving toward the observer: Source moving away from the observer: 12 Observer in Motion; Source in Motion 𝒗 Both observer and source in motion: Case 1: 𝒇′ = ( )𝒇 = 𝒗 +𝒗𝒔 𝟑𝟒𝟔. 𝟏 𝒎/𝒔 − 𝟐𝟓𝒎/𝒔 = ( )2550𝐻𝒛 𝟑𝟒𝟔. 𝟏 𝒎/𝒔 𝑓′ = 2365.80468 Hz or 2365.80 Hz Week 14: Fluid Mechanics Density Fluids – liquid, gas and plasma Fluid Mechanics – study of the properties of Examples: fluids Ex 1: The frequency of a fire engine siren is 1 Hydrostatics or fluid statics – deals with 750 Hz. (a) What frequency will a stationary fluids at rest person hear if the fire truck is moving toward Hydrodynamics or fluid dynamics – him at 18.0 m/s? (b) What frequency will a deals with fluids in motion person hear if he moves away from the fire truck at 8.0 m/s? Assume the speed of sound Density or mass density – property of is 340 m/s. matter defined as the ratio of its mass to its volume; SI Unit: kg/m3 𝒎 𝛒= 𝑽 Density of an ideal gas: 𝑷𝑴 𝛒= 𝑹𝑻 R = 0.0821 liter·atm/mol·K. R = 8.3145 J/mol·K. R = 8.2057 m3·atm/mol·K. R = 62.3637 L·Torr/mol·K or L·mmHg/mol·K. Ex 2: Compare the apparent frequencies of a 2550 Hz source moving away from you at 25 m/s and when you are moving away from the source at 25 m/s. Assume a temperature of Which one is the densest? least dense? 30℃. Why? Given: f = 2550 𝑣𝑠 = 25𝑚/𝑠 𝑣 = 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑜𝑢𝑛𝑑 𝑎𝑡 30°𝐶 𝑖𝑠 𝑎𝑡 346.1 m/s 𝒗 𝒗 + 𝒗𝒔 Equation: 𝒇′ = (𝒗 +𝒗𝒔 )𝒇 𝒇′ = ( 𝒗 )𝒇 A is the least dense for the molecules inside the Unkown: 𝒇′1 =? 𝒇′ 2 =? cube are not more than C and is not compact. Solution: While C is the densest for its compactability. 𝒗 Case 1: 𝒇′ = (𝒗 +𝒗𝒔 )𝒇 = 𝟑𝟒𝟔. 𝟏 𝒎/𝒔 Which one is the densest? least dense? = ( )2550𝐻𝒛 Why? 𝟑𝟒𝟔. 𝟏 𝒎/𝒔 + 𝟐𝟓𝒎/𝒔 𝑓′ = 2378.21342 Hz or 2378.21 Hz 13  completely to all parts of the fluid. 𝑭𝟏 𝑭𝟐 = 𝑨𝟏 𝑨𝟐 A is dense because of its compact nature If you push the plunger of a syringe, why does because more moleculs make up the Cube, the liquid inside come out at the other end of while compared to the Cube C for its less the barrel? compact nature. Pressure – magnitude of the force acting When the plunger is pushed back in, the perpendicular per unit area of the surface; volume decreases and the pressure SI Unit: Pa increases. Once the pressure is greater than 𝑭 𝐏= that outside the syringe, the fluid inside the 𝑨 barrel will flow out. The operation of your 1 atm = 1.013 x 105 Pa lungs also can be explained using Boyle's = 1.013 bar Law. = 14.7 lb/in2 = 760 mmHg Practice Exercise 9.7 = 760 torr The plunger and the needle of a hypodermic syringe have areas of 0.5 cm2 and 0.0006 cm2, Which one is more difficult to use, a stiletto or respectively. To inject a vaccine, a nurse wedge? Why? pushes the plunger with a force of 5.5 N. What pressure is transmitted to the vaccine? Given: Here, Ap denotes the plunger’s Stilleto is difficult to use because your are area, An denotes the putting all of your weight on one point (the needle’s area, heel). While wedges are more comfortable and Fp denotes the and stable than traditional high heels because applied force on the of even weight distribution. plunger. Why are water dams built with a thicker base? Equation: The pressure exerted by liquids increases with depth. So, as the depth increases, more pressure is applied to the walls. That is why Solution: walls are made thicker at the bottom, so that they can handle the pressure exerted by Answer: P = 110 000 N/m2 water. Archimedes’ Principle and Buoyancy Pascal’s Buoyant force – upward force acting on an Principle object caused by fluid pressure “Any change in Archimedes’ Principle pressure in an “The magnitude of buoyant force FB on a enclosed fluid at submerged object is equal to the weight of the rest is transmitted fluid displaced by the object.” 14  (a) 𝐹𝐵 = 𝜌𝑤𝑎𝑡𝑒𝑟 𝑉𝑔 = 1000 𝑘𝑔/𝑚 3 × (2.3 × 10−3 𝑚 3 ) × 9.8 𝑚/𝑠 2 𝐹𝐵 = 22.54𝑁 Flotation Case 1: If the object is denser than the liquid, (b) thebuoyant force exerted by 26 𝑘𝑔 𝜌𝑜 = = 11304.34783 𝑘𝑔/𝑚3 the liquid is less than its (2.3 × 10−3 𝑚 3 ) weight. Hence, the object 𝜌𝑜 = 11304.35 𝑘𝑔/𝑚 3 will sink. 𝑽𝒅𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒅 = 𝑽𝒐𝒃𝒋𝒆𝒄𝒕 Hydrodynamics – study of fluids in motion Streamline – path followed by succeeding Flotation particles Case 2: If the object is Tube of flow – bundle of streamlines less dense than the liquid, Laminar flow (steady) then its weight in the – when the velocity of liquid is less than the flow is relatively slow weight of the displaced Turbulent flow – if the liquid. Buoyant force is velocity is gradually greater than the weight increased until above a of the object while it is immersed in the liquid. critical value; there are The object will rise until the buoyant force no streamlines but equals its weight. Thus, the object will be whirlpools or eddy partially submerged. currents 𝑽𝒔𝒖𝒃𝒎𝒆𝒓𝒈𝒆𝒅 𝝆𝒐𝒃𝒋𝒆𝒄𝒕 = 𝑽𝒐𝒃𝒋𝒆𝒄𝒕 𝝆𝒍𝒊𝒒𝒖𝒊𝒅 Continuity Equation – expression of the conservation of mass Flotation “The same volume of fluid Case 3: If the density that enters the pipe in a of the object is equal given time interval exits the to the density of the pipe in the same time liquid, buoyant force interval.” is equal to the weight 𝑨𝟏 𝑽𝟏 = 𝑨𝟐 𝑽𝟐 of the object. The where: A1 and A2 are object will float. cross-sectional areas of the pipe at sections A and B, Practice Exercise 9.8 respectively; v1 and v2 are the speeds of the An object displaces 2.3x10-3 m3 of water when flow of the fluid at sections A and B, totally immersed into it. (a) What is the respectively buoyant force exerted by the water? (b) What is the density of the object as its mass is 26 Volumetric flow rate kg? - product of area and Given: V = 2.3 × 10−3 m3 velocity m = 26kg 𝐐 = 𝐀𝐯 g = 9.8 m/s2 𝜌𝑤𝑎𝑡𝑒𝑟 = 1000𝑘𝑔/𝑚3 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐟𝐥𝐨𝐰 𝐫𝐚𝐭𝐞 = 𝐐𝛒 Equation: 𝐹𝐵 = 𝜌𝑤𝑎𝑡𝑒𝑟 𝑉𝑔 Bernoulli’s Equation 𝑚 𝜌𝑜 = 𝑉 Solution: 15 “An increase in speed of a fluid occurs 1000kg/m3 ×(0.42m2/s2 - 0.26672m2/s2) = 44.44 Pa 2 simultaneously with a decrease in pressure or a decrease in the fluid’s potential energy.” Week 15: Temperature and Heat Heat and Temperature pressure energy + potential energy + kinetic Temperature – measure of the average energy kinetic energy of molecules making up an object Bernoulli’s Equation Heat – energy in transit from one body to pressure head + elevation head + velocity another because of the difference in head temperature; it is the measure of the total kinetic energy of the particles of a body; SI Unit: joule (J) and calorie (cal) Heat Calorie – amount of heat needed to change note: h = y the temperature of one gram of water from (height/ vertical displacement) 14.5oC to 15.5oC at a pressure of 1 atm 1 calorie = 4.186 joules Mechanical equivalent of heat – the ratio 4.186 J/cal Temperature Scales Ice point – temperature at which ice melts under pressure of 1 atm Steam point – temperature at which water boils under a pressure of 1 atm Temperature Scales 1. Celsius and Fahrenheit Scales Practice Exercises 9.11 Suppose that a person afflicted with aneurysm has enlarged portion of aorta with a 2. Kelvin Scale cross-sectional area 50 percent bigger than the normal. (a) What is the speed of the blood through this enlarged portion? (b) What is the 3. Ranking Scale pressure difference between the normal and the enlarged aorta? Assume that the speed of blood in the normal aorta is 40 cm/s and that the person is lying down. Given: A2 = 1.5A1 V1=1.5v2 v2= 40 V1 = 40 cm/s Unkown: V2 = ? p2 = ? Equation: 𝑣1 𝑣2 = 𝐴1 𝐴2 𝑝2 − 𝑝1 = 𝜌/2(𝑣12 − 𝑣22 ) Solution: Examples: (A) The normal body temperature of a person is 40 cm/s 𝑣2 = = 26.67 𝑐𝑚/𝑠 98.6oF. Convert this temperature to (a) 1.5 (B) degree Celsius, (b) Kelvin, and (c) degree Rankine. 16 Given: TF = 98.6oF Unknown: TC = ? TK = ? TR = ? 5 Equation: TC = ( oF - 32) 9 TK = TC TR = TF 5 Solution: TC = ( 98.6oF - 32) = 37oC 9 TK = 37oC + 273.15K = 310.15K 3. dependent on the material of which the rod TR = 98.6oF +459.67 = 558.27oR is made. An aluminium rod will expand The temperature of an object is increased by differently from a copper rod of the same 50oF. What is the corresponding change in (a) length even if subjected to the same Rankine degree, (b) Celsius degree, and (c) temperature increase. Kelvin? ∆L = L − L0 = aL0∆T = aL0 (T − T0) Given: TF = 50oF Unknown: TC = ? TK = ? TR = ? where: α is the coefficient of linear expansion 5 o Equation: TC = ( F - 32) + TK = TC TR = TF of the material of which the rod is made 9 5 Solution: TC = ( 50oF - 32) = 10oC 9 Coefficient of linear expansion (α) – TK = 10oC + 273.15K = 283.15K change in length divided by the original length TR = 50oF +459.67 = 509.67oR and change in temperature Thermal Equilibrium and Zeroth Law of Thermodynamics Zeroth Law of Thermodynamics Coefficient of area expansion (γ) – change “Two bodies are said to be in thermal in area divided by the original area and equilibrium if they have the same change in temperature temperature.” Thermal Expansion Thermal Expansion of Solids For linear expansion of solids, the change in length (ΔL) of a solid, say a rod, is: Coefficient of volume expansion (β) – 1. proportional to the change in temperature change in volume divided by the original (ΔT). A rod subjected to a temperature volume and change in temperature increase of 40oC will expand twice than when subjected to a temperature increase of 20 oC. 2. proportional to its original length (ΔL o). When subjected to the same temperature change, long rods will have greater change in length than short rods of the same material. 17 Examples: 3. Gas particles are in a constant state of Ex 2: A machinist was tasked to insert an random motion and move in straight lines until aluminium piston into a steel cylinder. The they collide with another body. inside diameters of the cylinder and of the 4. The collisions exhibited by gas particles are piston are 0.500 m and 0.4998 m, completely elastic; when two molecules respectively, at 25oC. To what temperature collide, total kinetic energy is conserved. must they be heated to obtain a perfect fit? 5. The average kinetic energy of gas molecules is directly proportional to absolute temperatureonly; this implies that all molecular motion ceases if the temperature is reduced to absolute zero. Gas Laws: 1. Boyle’s Law “At constant temperature, pressure is inversely proportional to volume.” P1V1 = P2V2 2. Charles’s Law “At constant pressure, the volume of a gas is directly proportional to absolute temperature.” 𝑽𝟏 𝑽𝟐 = 𝑻𝟏 𝑻𝟐 3. Gay-Lussac’s Law “At constant volume, pressure and temperatureare directly proportional to each other.” 𝑷𝟏 𝑷𝟐 = 𝑻𝟏 𝑻𝟐 4. Combined Gas Law – combines the three gas laws: Boyle's Law, Charles' Law, and GayLussac's Law “The ratio of the product of pressure and volume, and the absolute temperature of a gas equals a constant.” 𝑷 𝟏 𝑽𝟏 𝑷 𝟐 𝑽𝟐 = 𝑻𝟏 𝑻𝟐 5. Ideal Gas Law: PV = nRT PV = NkBT Week 16: Ideal Gases and Laws of where: P is pressure in pascals; V is volume Thermodynamics (Part 1) in cubic meters; n is the number of moles; R is Ideal Gas Law the universal gas constant (R = 8.314 J/mol Ideal Gas – a hypothetical gas whose K); T is temperature in kelvin; N is the total molecules occupy negligible space and have number of molecules; and kB is the Boltzmann no interactions, and which consequently constant (kB = 1.38 x 10-23 J/K); N = nNA, obeys the gas laws exactly. where NA is the Avogadro’s number equal to 6.02 x 1023 particles per mole. Properties of Ideal Gases: 1. The volume occupied by the individual Examples: particles of a gas is negligible compared to Ex 1: What volume will 1 mol of an ideal gas the volume of the gas itself. occupy at STP? STP means standard 2. The particles of an ideal gas exert no temperature and pressure, which are at 0OC attractive forces on each other or on their and at 1.013 x 105 Pa or 1 atm, respectively. surroundings. 18 Given: Closed system – no mass enters or leaves 5 P = 1.013 x 10 Pa the system; however, energy may be n = 1mol exchanged with the surroundings J Isolated system – neither mass nor energy R = 8.31 mol × K enters or leaves the system T = 273.15K Unknown: V =? nRT Equation: PV = nRT V= P Solution: nRT Reversible process – system and its V= P surroundings can be returned to their initial J state before undergoing a process. (1 mol)(8.31 mol × K)273.15K V= 1.013 x 105 Pa V = 2.24 × 10−1 m3 Examples: If 1.5 m3 of an ideal gas initially at STP is compressed to 0.75 m3 and heated to 30OC, what would be its final pressure? Irreversible process – system and its Given: V1 = 1.5m3 V2 = 0.75m3 surroundings cannot return to their initial state P1 = 1.013 × 105 Pa T1 = 273.15K T2 = 30℃ + 273.15K = 303.15K Unknown: 𝑃2 =? Equation: 𝑃1 𝑉1 𝑃2 𝑉2 𝑃1 𝑉1 𝑇2 = 𝑃2 = 𝑇1 𝑇2 𝑇1 𝑉2 Solution: Cycle – series of processes that starts and (1.013 × 105 Pa)(1.5m3 )(303.15K) ends atthe same condition 𝑃2 = (273.15K)(0.75m3 ) 𝑃2 = 224 851.5102 Pa or 224 851.51 Pa First Law of Thermodynamics Thermodynamics – concerned with heat and its transformation to mechanical energy; literally means “moving or evolving heat” Sadi Carnot – father of Thermodynamics First Law of Thermodynamics System – object or collection of objects under “When heat is added to a system, some of it study remains in the system, increasing its internal Surroundings – everything around the energy, while the rest leaves the system as system the system does work.” Universe – constituted by the system and the Q = ∆U + W surroundings where: Q is the heat added to the system; ΔU is the change in internal energy; and W is the work done by the system Internal Energy – sum of the kinetic and Open system – mass and energy may be potential energies possessed by the added or removed from the system molecules of an object due to their motions and positions relative to each other 19  ∆𝐔 = 𝐧𝐂𝐯 ∆𝐓 1. Isochoric process where: n is the – constant volume process number of moles; (ΔV = V2 – V1 = 0); no work done (ΔW = 0) CV is the molar 𝐐 = 𝚫𝐔 specific heat at 2. Isobaric process constant volume; – constant pressure process and ΔT is the (ΔP = 0) change in 𝐖 = 𝐏(𝐕𝟐 − 𝐕𝟏 ) temperature Ideas about Internal Energy: 3. Isothermal process 1. If heat is added to the system and no work – constant temperature process (ΔT = 0); is done by the system, the internal energy in change in internal energy is zero (ΔU = 0) 𝑾 the system will be increased by the same 𝐕𝟐 𝐖 = 𝐧𝐑𝐓 𝐥𝐧 𝐖 amount of heat added. 𝐕𝟏 2. When a system does work on the 𝐏𝟐 = 𝐧𝐑𝐓 𝐥𝐧 surroundings and no heat is transferred into 𝐏𝟏 the system during the process, the internal energy decreases by the same amount of work done. 𝐕 Work: 𝐖 = ∫𝐕 𝟐 𝐏𝐝𝐕 𝟏 PV Diagram – graph of pressure versus volume Path – series of intermediate states 𝐂𝐩 where: γ = 𝐂 is the ratio of the specific heat 𝐯 between the initial capacity of an ideal gas at constant pressure and final states to the specific heat capacity of an ideal gas at when a system constant volume changes from its initial to final state State variables – variables that depend only on the initial and final states and not on the path taken; example, change in internal energy Process variables – variables that depend on the path; example, heat and work Thermodynamic Processes and PV Diagrams Examples: Thermodynamic Process Ex 1: Two moles of an ideal gas are taken around the cycle as shown in the PV diagram – change from an initial (γ = 1.4). Calculate the (a) temperature states state to a final state of a A, B, C, (b) change in internal energy of the system that usually gas, (c) total heat, (d) total work done by the involves a change in its gas for the entire cycle. pressure, volume, or temperature 20 Ex 2: Two moles of a hypothetical ideal gas are taken around the cycle ABCA. Process AB is isochoric, BC is isobaric, and CA is isothermal. Given that CV is 20.8 J/mol K, (a) fill in the table below, (b) draw the PV diagram for the cycle, and (c) compute the heat, change in internal energy, and work for the cycle. 21  2. Clausius Statement “Heat flows naturally from hot to cold objects.” 3. Entropy Statement “When a reversible process occurs, the total entropy of the universe remains the same. When an irreversible process occurs, the total entropy of the universe increases. Equivalently, the entropy of an isolated system remains the same or increases.” Heat Engines – devices that convert thermal energy to mechanical energy Working substance – substance inside the heat engine that undergoes cooling and/or heating, compression and/or expansion, and sometimes phase change Pattern of Operation of Heat Engines 1. Heat (QH) is supplied to the engine by an external source called hot reservoir or heat source; 2. Part of the heat is used to do work on an object; 3. The rest of the heat (QC) is released at a temperature lower than the input temperature to an external place called the cold reservoir or the heat sink. Week 17: Ideal Gases and Laws of Two major types of Heat Engines: Thermodynamics (Part 2) 1. Internal Combustion Engine – burns the The Second Law of Thermodynamics fuel inside the engine; examples, Otto and Second Law of Thermodynamics – limits diesel engines 2. External Combustion Engine – burns the the amount of work a heat engine can do for a certain amount of heat fuel outside the engine; example, steam engines 1. Kelvin-Planck Statement “No heat engine can completely convert heat energy to work. In other words, there is no 100 percent heat engine.” 22  Ex 3: An engine does 3250 J of work and Efficiency of Heat Engine discards 5750 J of heat. Find the efficiency of Efficiency (ε) – how much of the input is this engine. converted to work Given: W = 3250J QC = 5750J QC Unkown: ε = ? ε = (1 − ) × 100% QH Equation: Q ε = (1 − Q C ) × 100% W = QH − H where: ε is the efficiency of an heat engine; QC QC is the heat released to the cold reservoir; Solution: QH is heat supplied by the heat engine QC = QH + W work QC = 5750J + 3250J = 9000J ε= × 100% heat input W ε = ( ) × 100% QC Efficiency of Carnot engine: TC 3250J ε = (1 − ) × 100% ε=( ) × 100% = 36.11% TH 9000J Note: TH and TC must be expressed in kelvin Examples: Ex 1: An internal combustion engine having an efficiency of 40 percent produces 3000J of Ex 4: A heat engine takes in 1500 J of heat at heat as exhaust. How much work does the 2000K and rejects 950 J of heat at 1000 K. engine do? Find (a) the maximum efficiency of this engine and (b) its actual efficiency. Entropy – thermodynamic measure of Ex 2: An engineer designs a heat engine disorder operating between 70OC and 120OC. What is - Grk. “transformation”; proposed by Rudolf the maximum efficiency of this heat engine? Clausius Examples of increasing entropy: (a) when heat is added to an object, molecules tend to move faster; (b) when gas flows from a container under high pressure to a space under low pressure, just like spraying air freshener in a wide room; (c) when ice melts Change in Entropy (ΔS) – heat (QH) added or released during the process divided by the temperature (T) 𝐐 ∆𝐒 = 𝐓 23 Entropy when heat is added or removed from a solid or liquid: ∆S = Q/T 𝐓𝟐 2825 kJ ∆𝐒 = 𝐦𝐜 𝐥𝐧 ∆S = 𝐓𝟏 323.15 K SI Unit: J/K ∆S = 8.742070246 kJ/K ∆S = 8.74 kJ/K The entropy statement of the second law may be written as: ΔS of universe = 0 for reversible processes; and ΔS of universe > 0 for irreversible processes Alternative Entropy Statement of Second Law: “All natural or spontaneous processes tend toward a state of greater disorder.” ∆𝐒 ≥ 𝟎 Examples: Ex 1: Find the change in entropy when 3.5 kg of ice melts at 0OC. (Lf of water is 3.35 x 105 J/kg) Ex 2: Find the change in entropy when 1.25 kg of steam at 100OC condenses and cools to 50OC. Given: m = 1.25kg T1 = 100℃ + 273.15K = 373.15K T1 = 50℃ + 273.15K = 323.15K latent heat of vaporization (for steam), L= 2,260 kJ/kg Unkown: ∆S =? Equation: latent heat of vaporization, Q= mL Solution: Q = mL Q= 1.25 kg x 2260 kJ/kg = 2825 kJ 24 25