Electric Energy by Direct Conversion from Gravitational Energy: a Gift from Superconductivity
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Energy Literacy Essential Principles and Fundamental Concepts for Energy Education
Energy Literacy Essential Principles and Fundamental Concepts for Energy Education A Framework for Energy Education for Learners of All Ages About This Guide Energy Literacy: Essential Principles and Intended use of this document as a guide includes, Fundamental Concepts for Energy Education but is not limited to, formal and informal energy presents energy concepts that, if understood and education, standards development, curriculum applied, will help individuals and communities design, assessment development, make informed energy decisions. and educator trainings. Energy is an inherently interdisciplinary topic. Development of this guide began at a workshop Concepts fundamental to understanding energy sponsored by the Department of Energy (DOE) arise in nearly all, if not all, academic disciplines. and the American Association for the Advancement This guide is intended to be used across of Science (AAAS) in the fall of 2010. Multiple disciplines. Both an integrated and systems-based federal agencies, non-governmental organizations, approach to understanding energy are strongly and numerous individuals contributed to the encouraged. development through an extensive review and comment process. Discussion and information Energy Literacy: Essential Principles and gathered at AAAS, WestEd, and DOE-sponsored Fundamental Concepts for Energy Education Energy Literacy workshops in the spring of 2011 identifies seven Essential Principles and a set of contributed substantially to the refinement of Fundamental Concepts to support each principle. the guide. This guide does not seek to identify all areas of energy understanding, but rather to focus on those To download this guide and related documents, that are essential for all citizens. The Fundamental visit www.globalchange.gov. Concepts have been drawn, in part, from existing education standards and benchmarks. -
DESIGN of a WATER TOWER ENERGY STORAGE SYSTEM a Thesis Presented to the Faculty of Graduate School University of Missouri
DESIGN OF A WATER TOWER ENERGY STORAGE SYSTEM A Thesis Presented to The Faculty of Graduate School University of Missouri - Columbia In Partial Fulfillment of the Requirements for the Degree Master of Science by SAGAR KISHOR GIRI Dr. Noah Manring, Thesis Supervisor MAY 2013 The undersigned, appointed by the Dean of the Graduate School, have examined he thesis entitled DESIGN OF A WATER TOWER ENERGY STORAGE SYSTEM presented by SAGAR KISHOR GIRI a candidate for the degree of MASTER OF SCIENCE and hereby certify that in their opinion it is worthy of acceptance. Dr. Noah Manring Dr. Roger Fales Dr. Robert O`Connell ACKNOWLEDGEMENT I would like to express my appreciation to my thesis advisor, Dr. Noah Manring, for his constant guidance, advice and motivation to overcome any and all obstacles faced while conducting this research and support throughout my degree program without which I could not have completed my master’s degree. Furthermore, I extend my appreciation to Dr. Roger Fales and Dr. Robert O`Connell for serving on my thesis committee. I also would like to express my gratitude to all the students, professors and staff of Mechanical and Aerospace Engineering department for all the support and helping me to complete my master’s degree successfully and creating an exceptional environment in which to work and study. Finally, last, but of course not the least, I would like to thank my parents, my sister and my friends for their continuous support and encouragement to complete my program, research and thesis. ii TABLE OF CONTENTS ACKNOWLEDGEMENTS ............................................................................................ ii ABSTRACT .................................................................................................................... v LIST OF FIGURES ....................................................................................................... -
STARS in HYDROSTATIC EQUILIBRIUM Gravitational Energy
STARS IN HYDROSTATIC EQUILIBRIUM Gravitational energy and hydrostatic equilibrium We shall consider stars in a hydrostatic equilibrium, but not necessarily in a thermal equilibrium. Let us define some terms: U = kinetic, or in general internal energy density [ erg cm −3], (eql.1a) U u ≡ erg g −1 , (eql.1b) ρ R M 2 Eth ≡ U4πr dr = u dMr = thermal energy of a star, [erg], (eql.1c) Z Z 0 0 M GM dM Ω= − r r = gravitational energy of a star, [erg], (eql.1d) Z r 0 Etot = Eth +Ω = total energy of a star , [erg] . (eql.1e) We shall use the equation of hydrostatic equilibrium dP GM = − r ρ, (eql.2) dr r and the relation between the mass and radius dM r =4πr2ρ, (eql.3) dr to find a relations between thermal and gravitational energy of a star. As we shall be changing variables many times we shall adopt a convention of using ”c” as a symbol of a stellar center and the lower limit of an integral, and ”s” as a symbol of a stellar surface and the upper limit of an integral. We shall be transforming an integral formula (eql.1d) so, as to relate it to (eql.1c) : s s s GM dM GM GM ρ Ω= − r r = − r 4πr2ρdr = − r 4πr3dr = (eql.4) Z r Z r Z r2 c c c s s s dP s 4πr3dr = 4πr3dP =4πr3P − 12πr2P dr = Z dr Z c Z c c c s −3 P 4πr2dr =Ω. Z c Our final result: gravitational energy of a star in a hydrostatic equilibrium is equal to three times the integral of pressure within the star over its entire volume. -
Energy Conservation in a Spring
Part II: Energy Conservation in a Spring Activity 4 Title: Forces and Energy in a Spring Summary: Students will use the “Masses and Springs” simulation from PhET to investigate the relationship between gravitational potential energy, spring potential energy, and mass. Standard: PS2 (Ext) - 5 Students demonstrate an understanding of energy by… 5aa Identifying, measuring, calculating and analyzing qualitative and quantitative relationships associated with energy transfer or energy transformation. Specific Learning Goals: 1. Understand how to calculate the force constant of a spring using the formula F = -k ∆x 2. Understand that even though energy is transformed during the oscillation of a spring among gravitational potential energy, elastic potential energy, and kinetic energy, the total energy in the system remains constant. 3. Understand how to calculate the gravitational potential energy for a mass which is lifted to a height above a table. 4. Understand how to calculate the elastic potential energy for a spring which has been displaced by a certain distance. Prior Knowledge: 1. The restoring force that a spring exerts on a mass is proportional to the displacement of the spring and the spring constant and is in a direction opposite the displacement: F = -k ∆x 2. Gravitational potential energy is calculated by the formula: PEg = mgΔh 2 3. Elastic potential energy is calculated by the formula: PEs = ½ k ∆x 4. In a mass and spring system at equilibrium, the restoring force is equal to the gravitational force on the suspended mass (Fg = mg) Schedule: 40-50 minutes Materials: “Masses and Springs 2.02 PhET” Engage: Three different masses are suspended from a spring. -
Potential Energy
Potential Energy • So far: Considered all forces equal, calculate work done by net force only => Change in kinetic energy. – Analogy: Pure Cash Economy • But: Some forces seem to be able to “store” the work for you (when they do negative work) and “give back” the same amount (when they do positive work). – Analogy: Bank Account. You pay money in (ending up with less cash) - the money is stored for you - you can withdraw it again (get cash back) • These forces are called “conservative” (they conserve your work/money for you) Potential Energy - Example • Car moving up ramp: Weight does negative work ΔW(grav) = -mgΔh • Depends only on initial and final position • Can be retrieved as positive work on the way back down • Two ways to describe it: 1) No net work done on car on way up F Pull 2) Pulling force does positive F Normal y work that is stored as gravitational x potential energy ΔU = -ΔW(grav) α F Weight Total Mechanical Energy • Dimension: Same as Work Unit: Nm = J (Joule) Symbol: E = K.E. + U 1) Specify all external *) forces acting on a system 2) Multiply displacement in the direction of the net external force with that force: ΔWext = F Δs cosφ 3) Set equal to change in total energy: m 2 m 2 ΔE = /2vf - /2vi + ΔU = ΔWext ΔU = -Wint *) We consider all non-conservative forces as external, plus all forces that we don’t want to include in the system. Example: Gravitational Potential Energy *) • I. Motion in vertical (y-) direction only: "U = #Wgrav = mg"y • External force: Lift mass m from height y i to height y f (without increasing velocity) => Work gets stored as gravitational potential energy ΔU = mg (yf -yi)= mg Δy ! • Free fall (no external force): Total energy conserved, change in kinetic energy compensated by change in m 2 potential energy ΔK.E. -
How Long Would the Sun Shine? Fuel = Gravitational Energy? Fuel
How long would the Sun shine? Fuel = Gravitational Energy? • The mass of the Sun is M = 2 x 1030 kg. • The Sun needs fuel to shine. – The amount of the fuel should be related to the amount of – The Sun shines by consuming the fuel -- it generates energy mass. from the fuel. • Gravity can generate energy. • The lifetime of the Sun is determined by – A falling body acquires velocity from gravity. – How fast the Sun consumes the fuel, and – Gravitational energy = (3/5)GM2/R – How much fuel the Sun contains. – The radius of the Sun: R = 700 million m • How fast does the Sun consume the fuel? – Gravitational energy of the Sun = 2.3 x 1041 Joules. – Energy radiated per second is called the “luminosity”, which • How long could the Sun shine on gravitational energy? is in units of watts. – Lifetime = (Amount of Fuel)/(How Fast the Fuel is – The solar luminosity is about 3.8 x 1026 Watts. Consumed) 41 26 • Watts = Joules per second – Lifetime = (2.3 x 10 Joules)/(3.8 x 10 Joules per second) = 0.6 x 1015 seconds. • Compare it with a light bulb! – Therefore, the Sun lasts for 20 million years (Helmholtz in • What is the fuel?? 1854; Kelvin in 1887), if gravity is the fuel. Fuel = Nuclear Energy Burning Hydrogen: p-p chain • Einstein’s Energy Formula: E=Mc2 1 1 2 + – The mass itself can be the source of energy. • H + H -> H + e + !e 2 1 3 • If the Sun could convert all of its mass into energy by • H + H -> He + " E=Mc2… • 3He + 3He -> 4He + 1H + 1H – Mass energy = 1.8 x 1047 Joules. -
Gravitational Potential Energy
An easy way for numerical calculations • How long does it take for the Sun to go around the galaxy? • The Sun is travelling at v=220 km/s in a mostly circular orbit, of radius r=8 kpc REVISION Use another system of Units: u Assume G=1 Somak Raychaudhury u Unit of distance = 1kpc www.sr.bham.ac.uk/~somak/Y3FEG/ u Unit of velocity= 1 km/s u Then Unit of time becomes 109 yr •Course resources 5 • Website u And Unit of Mass becomes 2.3 × 10 M¤ • Books: nd • Sparke and Gallagher, 2 Edition So the time taken is 2πr/v = 2π × 8 /220 time units • Carroll and Ostlie, 2nd Edition Gravitational potential energy 1 Measuring the mass of a galaxy cluster The Virial theorem 2 T + V = 0 Virial theorem Newton’s shell theorems 2 Potential-density pairs Potential-density pairs Effective potential Bertrand’s theorem 3 Spiral arms To establish the existence of SMBHs are caused by density waves • Stellar kinematics in the core of the galaxy that sweep • Optical spectra: the width of the spectral line from around the broad emission lines Galaxy. • X-ray spectra: The iron Kα line is seen is clearly seen in some AGN spectra • The bolometric luminosities of the central regions of The Winding some galaxies is much larger than the Eddington Paradox (dilemma) luminosity is that if galaxies • Variability in X-rays: Causality demands that the rotated like this, the spiral structure scale of variability corresponds to an upper limit to would be quickly the light-travel time erased. -
Gravitational Potential and Energy of Homogeneous Rectangular
View metadata, citationGravitational and similar papers potential at core.ac.uk and energy of homogeneous rectangular parallelepipedbrought to you by CORE provided by CERN Document Server Zakir F. Seidov, P.I. Skvirsky Department of Physics, Ben-Gurion University of the Negev, Beer-Sheva, 84105, Israel Gravitational potential and gravitational energy are presented in analytical form for homogeneous right parallelepiped. PACS numbers: 01.55.+b, 45.20.Dd, 96.35.Fs Keywords: Newtonian mechanics; mass, size, gravitational fields I. INTRODUCTION: BASIC FORMULAE As it is well known, Newtonian gravitational potential, of homogeneous body with constant density ρ,atpoint (X,Y,Z) is defined as triple integral over the body's volume: ZZZ U(X; Y; Z)=Gρ u(X; Y; Z; x; y; z) dxdydz (1) with −1=2 u(X; Y; Z; x; y; z)=[(x − X)2 +(y − Y )2 +(z − Z)2] ;(2) here G stands for Newtonian constant of gravitation. In spite of almost 400-year-long attempts since Isaac Newton' times, the integral (1) is known in closed form (not in the series!) in quite a few cases [1]: a) a piece of straight line, b) a sphere, c) an ellipsoid; note that both cases a) and b) may be considered as particular cases of c). As to serial solution, the integral (1) is expressed in terms of the various kinds of series for external points outside the minimal sphere containing the whole body (non-necessary homogeneous), as well as for inner points close to the origin of co-ordinates. However in this note we do not touch the problem of serial solution and are only interested in exact analytical solution of (1). -
Formula Sheet for LSU Physics 2113, Third Exam, Fall ’14
Formula Sheet for LSU Physics 2113, Third Exam, Fall '14 • Constants, definitions: m g = 9:8 R = 6:37 × 106 m M = 5:98 × 1024 kg s2 Earth Earth m3 G = 6:67 × 10−11 R = 1:74 × 106 m Earth-Sun distance = 1.50×1011 m kg · s2 Moon 30 22 8 MSun = 1:99 × 10 kg MMoon = 7:36 × 10 kg Earth-Moon distance = 3.82×10 m 2 2 −12 C 1 9 Nm −19 o = 8:85 × 10 2 k = = 8.99 ×10 2 e = 1:60 × 10 C Nm 4πo C 8 −19 c = 3:00 × 10 m/s mp = 1:67 × 10−27 kg 1 eV = e(1V) = 1.60×10 J −31 Q Q Q dipole moment: ~p = qd~ me = 9:11 × 10 kg charge densities: λ = ; σ = ; ρ = L A V 2 2 4 3 Area of a circle: A = πr Area of a sphere: A = 4πr Volume of a sphere: V = 3 πr Area of a cylinder: A = 2πr` Volume of a cylinder: V = πr2` • Units: Joule = J = N·m • Kinematics (constant acceleration): 1 1 2 2 2 v = vo + at x − xo = 2 (vo + v)t x − xo = vot + 2 at v = vo + 2a(x − xo) • Circular motion: mv2 2πr Fc = mac = , T = , v = !r r v • General (work, def. of potential energy, kinetic energy): 1 2 ~ K = 2 mv Fnet = m~a Emech = K + U W = −∆U (by field) Wext = ∆U = −W (if objects are initially and finally at rest) • Gravity: m1m2 GM Newton's law: jF~j = G Gravitational acceleration (planet of mass M): ag = r2 r2 M dVg GM Gravitational Field: ~g = −G r^ = − Gravitational potential: Vg = − r2 dr r 2 ! 4π m1m2 Law of periods: T 2 = r3 Potential Energy: U = −G GM r12 m1m2 m1m3 m2m3 Potential Energy of a System (more than 2 masses): U = − G + G + G + ::: r r r I 12 13 23 Gauss' law for gravity: ~g · dS~ = −4πGMins S • Electrostatics: j q1 jj q2 j Coulomb's law: jF~j = k Force on a charge in -
Mechanical Energy
Chapter 2 Mechanical Energy Mechanics is the branch of physics that deals with the motion of objects and the forces that affect that motion. Mechanical energy is similarly any form of energy that’s directly associated with motion or with a force. Kinetic energy is one form of mechanical energy. In this course we’ll also deal with two other types of mechanical energy: gravitational energy,associated with the force of gravity,and elastic energy, associated with the force exerted by a spring or some other object that is stretched or compressed. In this chapter I’ll introduce the formulas for all three types of mechanical energy,starting with gravitational energy. Gravitational Energy An object’s gravitational energy depends on how high it is,and also on its weight. Specifically,the gravitational energy is the product of weight times height: Gravitational energy = (weight) × (height). (2.1) For example,if you lift a brick two feet off the ground,you’ve given it twice as much gravitational energy as if you lift it only one foot,because of the greater height. On the other hand,a brick has more gravitational energy than a marble lifted to the same height,because of the brick’s greater weight. Weight,in the scientific sense of the word,is a measure of the force that gravity exerts on an object,pulling it downward. Equivalently,the weight of an object is the amount of force that you must exert to hold the object up,balancing the downward force of gravity. Weight is not the same thing as mass,which is a measure of the amount of “stuff” in an object. -
On the Magnetic Dipole Energy Expression of an Arbitrary Current
On the magnetic dipole energy expression of an arbitrary current distribution Keeyung Lee Department of Physics, Inha University, Incheon, 402-751, Korea (Dated: October 16, 2012) We show that the magnetic dipole energy term appearing in the expansion of the magnetic potential energy of a localized current distribution has the form U = +m · B which is wrong by a sign from the well known −m · B expression. Implication of this result in relation to the electric dipole energy −p·E, and the force and torque on the magnetic dipole based on this energy expression is also discussed. PACS numbers: 13.40 Em, 21.10. Ky I. Introduction netic potential energy and show that the expansion of Potential energy of a charge distribution in an external the magnetic energy of an arbitrary current distribution electric field or the potential energy of a current distribu- system results in the expression +m·B for the dipole en- tion in an external magnetic field is an important concept ergy term. Implication of this result in connection with in the electromagnetic theory. It is well known that in the force and torque on the dipole is also discussed. the energy expansion of a localized charge or of a local- ized current distribution, the dipole energy term becomes II. Magnetic Energy Expansion important. Before discussing the magnetic energy case, let us For an electrically neutral system, since the monopole consider first the expansion of electric potential energy term vanishes, electric dipole energy which has the form U = dV ρ(r)φ(r) of a static charge distribution with U = −p · E becomes the most important term in most chargeR density ρ(r) under the influence of an external cases. -
Magnetic Energy Dissipation and Emission from Magnetars
Magnetic energy dissipation and emission from magnetars Andrei Beloborodov Columbia University Magnetars • strong and evolving B • large variations in emission and spindown • internal + external heating • energy budget E ~ t L ~ 1047 −1048 erg Building up magnetic stresses Hall drift Goldreich Reisenegger (1992) ambipolar diffusion Observed quasi-thermal surface emission Kaminker et al. (2009) heating in the outer crust is required with E! ~ 1036 −1037 erg/s Crustal motions and internal heating • No cracks • No slippage except along magnetic flux surfaces • Collapse of ideal crystal? (Chugunov, Horowitz 2010) • Plastic flow (motion of dislocations) heating q! = σ s! heat is conducted toward the core and surface (Kaminker et al. 2009; Jose Pons) • QPOs (externally triggered) External (magnetospheric) dissipation Sun Recorded in extreme ultraviolet from NASA’s Transition Region and Coronal Explorer satellite. Sun: convective motions twist the magnetic field anchored to the surface Dissipated/radiated power: L = I Φ vacuum: I = 0 force-free: Φ = 0 Voltage regulated by e+- discharge Φ ~ 109−10V surface radiation: !ω ~ 3kT ~ 1 keV B 2 Landau energy: !ωB = mec BQ 3 4 resonant scattering: γω ≈ ωB when γ ~ 10 −10 (σ res ≈ πreλ) 2 + − scattered photon: E ~ γωB ~ γ ω → e + e Magnetosphere Twisted c j = ∇ × B ≠ 0, j || B force free 4π (cf. solar corona) Filled with plasma Dynamic -- Changing magnetic moment (spindown) -- Changing pulse profiles -- Bursts Flares δt ~ 0.1-0.3 s Starquake? Excitation of Alfven waves on field lines with length > cδt Reconnection in the magnetosphere? (Thomspon, Duncan 1996) Twisted magnetospheres and flares Parfrey et al. 2013 Twist energy W = W0 for untwisted dipole Loss of magnetic equilibrium and reconnection Parfrey et al.