## Posts Tagged ‘energy conversion’

### Calculating Velocity — de Coriolis’ Kinetic Energy Formula

Monday, September 28th, 2015
 Last time we introduced Gaspard-Gustave de Coriolis’ formula to compute kinetic energy.  Today we’ll use it to determine the speed of descent, or velocity, of the coffee mug we’ve been watching closely in the last few blogs.  To calculate the mug’s velocity, we must bear in mind physicist Julius Robert von Mayer’s assertion that all forms of energy are interrelated, and in fact interchangeable, because energy can neither be created nor destroyed, it can only change forms.  For a refresher, see The Law of Conservation of Energy.       Let’s now put a practical spin on this concept and apply it to our coffee mug’s free fall to the floor.  Once again, de Coriolis’ formula, KE = ½ × m × v2                                               (1) where m is the mass of our falling object and v its velocity.  The ½ is an unchanging, constant term that’s present due to the mathematical Rules of Integration governing integral calculus.  Calculus and its derivations are beyond the scope of this blog, but if you’re interested in pursuing this, follow this link to, The Physics Hypertextbook – Kinetic Energy.       According to von Mayer’s Law, at the precise instant before the mug hits the floor its kinetic energy, KE, is equal to the potential energy, PE, it possessed when it rested passively on the shelf.  Stated another way, the instant before the mug makes contact with the floor, all its potential energy will have been converted into kinetic.       The mug’s PE was calculated previously to be equal to 4.9 kg • meter2/second2.  See Computing Potential Energy for a review.  Knowing this, the mathematical relationship between the mug’s potential and kinetic energies is expressed as, PE = KE = 4.9 kg • meter2/second2                         (2)       By substituting this mathematical representation for KE into equation (1) we arrive at, 4.9 kg • meter2/second2 = ½ × m × v2                   (3) We also know the mug’s mass, m, to be equal to 2.6 kilograms, so integrating that into the right side of equation (3) it becomes, 4.9 kg • meter2/second2 = ½ × ( 0.25kg) × v2         (4) That leaves the mug’s velocity, v2, as the only remaining unknown term.  We’ll use algebra to isolate this variable by dividing both sides of equation (4) by ½ × ( 0.25kg). (4.9 kg • meter2/second2) ÷ [½ × ( 0.25kg)] = v2 39.20 meter2/second2 = v2 Finally, we’ll take the square root of the equation to place it in terms of v. 6.26 meters/second = v The mug’s velocity an instant before impact equates to 6.26 meters/second, or almost 21 feet per second.       Next time we’ll discuss a metric unit used to measure energy known as the Joule and discover the man behind it. Copyright 2015 – Philip J. O’Keefe, PE Engineering Expert Witness Blog ____________________________________

### Coal Power Plant Fundamentals

Sunday, January 23rd, 2011

 Several years ago I was asked by power producers within the electric utility industry to write and then present a training course on the subject of coal power plant fundamentals.  The finished product was a two day introductory course on the energy transformation process within a coal fired plant.      Since that time my seminar, entitled Coal Power Plant Fundamentals, has been presented to a variety of audiences, including Mirant Corporation, Platte River Power Authority, and Integrys Energy Group, Inc.  Audience makeup has been diverse and has included equipment manufacturers, mining companies, power industry consultants, and regulatory agencies.      This seminar, which I continue to present today in meeting rooms across the country, covers all major systems in a typical power plant, from coal handling when the coal first enters the plant, to its eventual end destination, the electrical switch yard which facilitates power transmission to customers.  My Power Point presentation is embellished with ample illustrations, including photographs that I have taken during the course of my career and diagrams which I created using CAD, or Computer Aided Drawing software, one of which is featured below.  In addition to the overhead slides, I provide a 150-page bound book which is distributed to seminar attendees.  They use it to both follow along with my lecture and have a source of refresher material to take home with them.  I’ve been told that having my illustrations in front of them makes a world of difference towards their understanding of the subject matter.      The unique thing about my course is that it focuses on the simplified presentation of complex engineering concepts, much like my blogs do.  Of course it always helps to have an engineering background or scientific background of sorts, but I wrote the course to accommodate understanding of the subject matter by individuals without any technical background.  Accountants, salespersons, administrative staff, plant operating and maintenance workers, and journalists have all found the course to be easy to follow, interesting, and informative.      So how do you get electricity from coal?  To answer this question and give you a sampling of my seminar material let’s take a look at Figure 1.  Figure 1 – The Coal Power Plant Energy Transformation Process      Following along from left to right, the coal is first burned in order to transform the chemical energy which it contains into heat energy.  That heat energy is then absorbed by water inside a nearby boiler, where it is converted into steam.  The heat energy in the steam flows through a pipe into a steam turbine where it is again transformed, this time into mechanical energy that enables the turbine shaft to spin.  The mechanical energy in the turbine is then transmitted by its shaft, enabling it to turn an electrical generator.  And, finally, the mechanical energy is transformed by the generator into electrical energy for our usage.      Simple process, right?  Well, maybe, maybe not.  My illustration certainly helped to simplify things, but there are a lot of details that were purposely omitted so as not to “muddy the waters.”  It’s those details which have the potential to make things a lot more complicated, and next week we’ll begin to take a closer look at some of them.   _____________________________________________

### Dynamic Brakes

Monday, May 31st, 2010

 Last week we looked at how a mechanical brake stopped a rotating wheel by converting its mechanical energy, namely kinetic energy, into heat energy.  This week, we’ll see how a dynamic brake works.      Chances are you have directly benefited by a dynamic braking system the last time you rode in an elevator.  But, to understand the basic principle behind an elevator’s dynamic brake system, let’s first take a look at the electric braking system in Figure 1 below.  Figure 1 – A Simple Electric Braking System      Here the brake consists of an electric generator wired via an open switch to an electrical component called a resistor.  The weight is attached to a cable that is wound around a pulley on the generator’s shaft.   As the weight freefalls, the cable unwinds on the pulley, causing the pulley to turn the generator’s shaft.      Unlike last week’s mechanical brake which required a good deal of effort to employ, a dynamic braking system requires very little.  All that needs to be done is to close a switch as shown in Figure 2 below.  When the switch is closed, an electrical circuit is created where the resistor gets connected to the generator.  The resistor does as its name implies: it resists (but doesn’t stop) the electrical current flowing through it from the generator.  As the electrical current fights its way through the resistor to get back to the generator, the resistor gets hot like an electric heater.  This heat is dissipated to the cooler surrounding air.  At the same time, the weight begins to slow down in its descent.  But how is this happening?      The electric braking system can be thought of as an energy conversion process.  We start out with the kinetic, or motion energy, of the freefalling weight.  This kinetic energy is transmitted to the electrical generator by the cable, which spins the generator’s shaft as the cable unwinds.  Electrical generators are machines that convert kinetic energy into electrical energy.  This energy travels from the electric generator through wires and a closed switch to the resistor.  In the process the resistor converts the electrical energy into heat energy.  So, kinetic energy is drawn from the falling weight through the conversion process and leaves the process in the form of heat.  As the falling weight is drained of kinetic energy, it slows down.   Figure 2 – Applying the Electric Brake         Okay, now let’s get back to dynamic brakes on elevators.  An elevator is attached by a cable to a hoist that is powered by an electric motor.  When it’s time to stop at the desired floor, the automatic control system disconnects the elevator’s electric motor from its power source and turns the motor into a generator.  The generator is then automatically connected to a resistor like the one shown in the electric brake above.  The kinetic energy of the moving elevator is converted by the generator into electrical energy.  The resistor converts the electrical energy into heat energy which is then dissipated into the surrounding environment.  The elevator slows down in the process because it’s being robbed of kinetic energy.  When the dynamic brake slows the elevator down enough, a mechanical brake is introduced, taking over to bring the elevator to a complete stop.  This two-fold process serves to reduce wear and tear on the mechanical brake’s parts, lengthening the operational lifespan of the system as a whole.      Next time, we’ll tie everything together and show how mechanical and dynamic brakes work together in a diesel locomotive. _____________________________________________