Posts Tagged ‘joule’

Work and Energy Share an Interesting Relationship

Thursday, December 10th, 2015

      My work as an engineering expert has often required that I perform calculations to quantify the energy consumed by electric motors and steam turbines, such as when they work together at power plants to generate electricity.   Today we’ll see how work and energy share an interesting relationship that is brought out by examining the units by which they are measured.

     Last time we used de Coriolis’ formula to compute work to calculate the amount of work performed while pushing a loaded wheelbarrow a distance of 3 meters.   We found that in order to move the wheelbarrow that distance, a gardener must exert a force equal to 534 Newton • meters of work.   That relationship is shown here,

Work = 178 Newtons × 3 meters = 534 Newton • meters           (1)


Work is force times distance

de Coriolis’ Formula to Compute Work


     The Newton, as discussed previously in this blog series, is shorthand notation for metric units of force, and we’ll use those units today to demonstrate the special relationship between work and energy.

We’ll start by supposing that you’re unfamiliar with the Newton as a unit of measurement.   In that case you’d have to employ longhand notation to quantify things, which means you’d be measuring units of force in terms of kilogram • meters per second2.

     Putting equation (1) in longhand notation terms, we arrive at,

Work = 178 kilogram • meters per second2 × 3 meters       (2)

Work = 534 kilogram • meters2 per second2                    (3)

     If you’ve been following along in this blog series, you’ll recognize that the unit of measurement used to compute work, namely, kilogram • meters2 per second2, is the same as was used previously to measure energy.  That unit is the Joule, which is considerably less wordy.

     Equations (2) and (3) bear out the interesting relationship between work and energy — they share the same unit of measure.   This relationship would not be apparent if we only considered the units for work presented in equation (1).

     So following standard engineering convention where work and energy are expressed in the same units, the work required to push the wheelbarrow is expressed as,

Work = 534 Joules

     Yes, work and energy are measured by the same unit, the Joule.   But, energy isn’t the same as work.   Energy is distinguished from work in that it’s the measure of the ability to perform work.    Stated another way, work cannot be performed unless there is energy available to do it, just as when you eat it provides more than mere pleasure, it provides your body with the energy required to perform the work of pushing a wheelbarrow through the garden.

     Next time we’ll see how work factors into the Work Energy Theorem, which mathematically relates work to energy.

Copyright 2015 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog



de Coriolis’ Principle of Work and Dynamode

Sunday, November 15th, 2015

      I was recently retained as an engineering expert in a lawsuit in which I had to determine the force acting upon an object.   In order to quantify that force, I employed Gaspard Gustave de CoriolisPrinciple of Work, an engineering concept he introduced in his 1829 textbook, Calculation of the Effects of Machines.   We’ll take our first look at de Coriolis’ Principle of Work today, along with the term he used to quantify work, the dynamode.

      As a scientist living during the time of the great Industrial Revolution, de Coriolis was interested in lots of things, and he was particularly interested in quantifying the effort involved to accomplish tasks, like how many men, horses, or steam engines were required to move a stationary object.   He defined this activity as work, and he hoped its study would lead to a broadly accepted engineering principle which could be applied across industrial functions.

      According to de Coriolis, work is the force acting upon a stationary object which causes it to move, multiplied by the distance moved.   Work could also be defined as the force acting upon an object already in motion, multiplied by the distance traveled before it comes to a stop.

     To quantify work, de Coriolis proposed the dynamode as its unit of measure, a term which derives from the Greek words dynamis, meaning power, and odos, meaning path.   He went on to define one dynamode as the amount of work required to lift an object with a mass of 1,000 Kg, or kilograms, one meter above the ground.


de Coriolis' Dynamode Quantity

de Coriolis’ Dynamode Quantity


      Catchy as it may sound, the word dynamode is all but forgotten today.   But de Coriolis’ Principle of Work and his formula to calculate work remain to the present day as fundamental concept in engineering.   We’ll present that formula next time.

Copyright 2015 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog


Converting Kinetic Energy to Electrical Energy

Tuesday, November 3rd, 2015

     When acting as an engineering expert I’m often called upon to investigate incidents where energy converts from one form to another, a phenomenon that James Prescott Joule observed when he built his apparatus and performed his experiments with electricity.   Today we’ll apply Joule’s findings to our own experiment with a coffee mug when we convert its kinetic energy into electrical energy and see how the units used to express that energy also change. 

     We had previously calculated the kinetic energy contained within our falling coffee mug to be 4.9 kg • meter2/second2, also known as 4.9 Joules of energy, by using de Coriolis’ Kinetic Energy Formula.   Now most of us don’t speak in terms of Joules of energy, but that’s easily addressed.   As we learned in a previous blog on The Law of Conservation of Energy, all forms of energy are equivalent and energy can be converted from one form to another, and when it does, the unit of energy used to express it also changes.

     Let’s say we want to put our mug’s 4.9 Joules of kinetic energy to good use and power an electric light bulb.   First we must first find a way of converting the mug’s kinetic energy into electrical energy.   To do so, we’ll combine Joule’s apparatus with his dynamo, and connect the mug to this hybrid device with a string.

Converting Kinetic Energy to Electrical Energy

                     Converting Kinetic Energy to Electrical Energy

     As the mug falls its weight tugs on the string, causing the winding drum to rotate.   When the drum rotates, the dynamo’s magnet spins, creating electrical energy.   That’s right, all that’s required to produce electricity is a spinning magnet and coils of wire, as explained in my previous blog, Coal Power Plant Fundamentals – The Generator.

     Now we’ll connect a 5 Watt bulb to the dynamo’s external wires.   The Watt is a unit of electrical energy named in honor of James Watt, a pioneer in the development of steam engines in the late 18th Century.

     Now it just so happens that 1 Watt of electricity is equal to 1 Joule of energy per a specified period of time, say a second.   This relationship is expressed as Watt • second.   Stated another way, 4.9 Joules converts to 4.9 Watt • seconds of electrical energy.   Let’s see how long we can keep that 5 Watt bulb lit with this amount of energy.    Mathematically this is expressed as,

Lighting Time = (4.9 Watt • seconds) ÷ (5 Watts) = 0.98 seconds

     This means that if the mug’s kinetic energy was totally converted into electrical energy, it would provide enough power to light a 5 Watt bulb for almost 1 second.

     Next time we’ll see what happens to the 4.9 Joules of kinetic energy in our coffee mug when it hits the floor and becomes yet another form of energy.

Copyright 2015 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog


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.

mechanical engineering expert witness falling objects

      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


Computing Potential Energy

Tuesday, August 25th, 2015

      Last week we discovered that objects acquire potential energy as it relates to gravity based on the height those objects are elevated above the ground.   We also introduced an equation to calculate the potential energy of a coffee mug perched on a shelf.   We’ll work with that equation today and compute the latent energy that’s hidden within that mug.

      Here again is the equation to determine potential energy, put in terms relating to gravity,

PEgravitational = m × g × h

where m is the mass of the mug, h is the height it’s been elevated above the floor, and g is the Earth’s acceleration of gravity factor, as explained in my previous blog entitled, Sir Isaac Newton and the Acceleration of Gravity.

      The equation above can be solved using either English or metric units.   In the US it’s generally standard practice to perform calculations using English units, such as feet and pounds.   But when measuring mass a less familiar English unit, the slug, comes into play.   If you’re interested in learning more about this unit, go to a previous blog article entitled, The Force of Gravity.

      The kilogram is the metric equivalent of a slug.   Since it’s the unit of mass most commonly used throughout the world, we’ll use it to perform our calculation.

mechanical engineering expert witness

      Let’s say our mug has a mass of 0.25 kilograms, the shelf it’s resting on is 2 meters above the floor, and g is 9.8 meters/second2.   The mug’s gravitational potential energy would then be expressed as,

PEgravitational = (0.25 kg) × (9.8 meters/second2) × (2 meters)

PEgravitational = 4.9 kg • meter2/second2

      Next time we’ll expand on our discussion of potential energy and discuss the Law behind the phenomenon and the fact that energy can only be converted from one form to another.

Copyright 2015 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog