Posts Tagged ‘work’
Tuesday, March 15th, 2016
Last time we watched as the kinetic energy of our falling coffee mug was transformed into the work of creating a crater in a pan of soft kitty litter. Shock absorbing materials are often placed strategically to cushion valuable objects should they fall, and as an engineering expert I’ve sometimes had to implement breakitsfall solutions. Today we’ll place our mug into a less kind scenario, one in which it makes impact with the unforgiving hardness of a concrete floor. In so doing we’ll compare the mug’s ceramic to the floor’s concrete, and we’ll familiarize ourselves with the Mohs Scale of Hardness.
The Mohs Scale of Hardness, Ceramic vs. Concrete
Material hardness is commonly measured by the Mohs Scale of Hardness, which ranks the relative hardness of a material by observing how resistant it is to scratching by other materials harder than itself. This standard was developed by German mineralogist Friedrich Moh in 1812, and it rates objects’ hardness on a scale from 1.0, very soft, to 10.0, very hard. A fingernail, for example, ranks 2.5 on the scale, while a diamond ranks 10.0.
Now let’s take a look at the materials in our scenario, a ceramic mug and concrete floor, and see how they compare. The mug’s ceramic was created by mixing together clay, water, and other materials and then heating them in a kiln, a process known as firing. This firing causes a chemical reaction that bonds the individual materials tightly together, and when it cools it becomes the product we know as ceramic, a hard, brittle solid which registers at about 7.5 on the Mohs Scale.
The floor the mug falls to is pouredinplace cement, a compound consisting of primarily limestone, clay, pebbles and sand. When these materials are combined with water a chemical bonding takes place and forms the hard, stonelike matter we know as concrete, which comes in at about 8.0 on the Mohs Scale.
Although the mug’s ceramic is comparably hard to the floor’s concrete, its inherent brittleness, along with certain design features, most notably its handle, causes it to be fragile. Anyone broken a coffee mug lately?
As for the concrete floor the mug falls onto, it won’t yield to the mug’s freefall kinetic energy and form a crater like the litter did. So where does the mug’s energy go?
According to the WorkEnergy Theorem, most of the mug’s kinetic energy is still converted into work, just as it was when it met up with the litter, but because the concrete floor is harder and thicker than the mug’s thin ceramic, the mug’s kinetic energy at impact falls back on itself rather than transferring externally into the concrete. The result is a shattered mug and a mess to clean up.
But we haven’t yet accounted for all the mug’s energy. We’ll find out what happens to the rest of it next time.
Copyright 2016 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: brittleness, cement, ceramic, concrete, energy, engineering expert, force, freefall kinetic energy, hardness, Mohs Scale of Hardness, shatters, work, workenergy theorem
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Monday, February 8th, 2016
So far we’ve applied the WorkEnergy Theorem to a flying object, namely, Santa’s sleigh, and a rolling object, namely, a car braking to avoid hitting a deer. Today we’ll apply the Theorem to a falling object, that coffee mug we’ve been following through this blog series. We’ll use the Theorem to find the force generated on the mug when it falls into a pan of kitty litter. This falling object scenario is one I frequently encounter as an engineering expert, and it’s something I’ve got to consider when designing objects that must withstand impact forces if they are dropped.
Applying the WorkEnergy Theorem to Falling Objects
Here’s the WorkEnergy Theorem formula again,
F × d = ½ × m × [v_{2}^{2} – v_{1}^{2}]
where F is the force applied to a moving object of mass m to get it to change from a velocity of v_{1} to v_{2} over a distance, d.
As we follow our falling mug from its shelf, its mass, m, eventually comes into contact with an opposing force, F, which will alter its velocity when it hits the floor, or in this case a strategically placed pan of kitty litter. Upon hitting the litter, the force of the mug’s falling velocity, or speed, causes the mug to burrow into the litter to a depth of d. The mug’s speed the instant before it hits the ground is v_{1, }and its final velocity when it comes to a full stop inside the litter is v_{2}, or zero.
Inserting these values into the Theorem, we get,
F × d = ½ × m × [0 – v_{1}^{2}]
F × d = – ½ × m × v_{1}^{2}
The right side of the equation represents the kinetic energy that the mug acquired while in freefall. This energy will be transformed into Gaspard Gustave de Coriolis’ definition of work, which produces a depression in the litter due to the force of the plummeting mug. Work is represented on the left side of the equal sign.
Now a problem arises with using the equation if we’re unable to measure the mug’s initial velocity, v_{1}. But there’s a way around that, which we’ll discover next time when we put the Law of Conservation of Energy to work for us to do just that.
Copyright 2016 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: energy, engineering expert, falling objects, GaspardGustave de Coriolis, impact forces, kinetic energy, law of conservation of energy, work, workenergy theorem
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Wednesday, January 20th, 2016
In my work as an engineering expert I’ve never had to convert Joules of workenergy into calories, but that’s exactly what we’ll be doing together today. Last time we applied the WorkEnergy Theorem to the progress of Santa’s sleigh and found that an opposing wind force of 3848.7 Newtons –or 865.2 pounds for those of us who are American– slowed his team from an initial velocity of 90 meters per second to a final velocity of 40 meters per second and that it happened over a distance of 760 meters. Today we’ll see how many calories the reindeer need to expend to get them back up to full delivery speed.
Prancer Loves Oats
Now we know that Santa successfully made all his deliveries on time this past Christmas, so that means that at some point his reindeer team was able to get back up to full sleighflying speed. They did it by expending extra energy. We’ll use the WorkEnergy Theorem to find out how much energy that equates to. Here’s the Theorem again,
W = ½ × m × [v_{2}^{2} – v_{1}^{2}]
where W is the work/energy required to speed up the sleigh team’s mass, m, from an initial velocity v_{1} to a final velocity v_{2}. For a refresher on the special relationship between work and energy, see our past blog on the subject.
If Santa’s sleigh has a mass of 900 kilograms and its speed must increase from 40 to 90 meters per second, then the work required to do so is calculated as,
W = ½ × (900 kilograms) × [(90 meters/second)^{2} – (40 meters/second)^{2}]
W = ½ × (900 kilograms) × (6,500 meters^{2}/second^{2})
W = 2,925,000 kilogram^{2} · meters^{2} per second^{2} = 2,925,000 Joules
So Rudolph and his buddies had to expend 2,925,000 Joules of energy to perform 2,925,000 Joules of work. To understand where Rudolph and his team got that energy, we must state things in terms of nutritional value, that is, units of calories.
Did you know that 1 calorie is equal to 4,184.43 Joules? Applying that equivalency to our situation we get,
Nutritional Energy Required = (2,925,000 Joules) × (4,184.43 Joules/calorie)
= 699.02 calories
The net result is Santa’s team expended a total of 699.02 calories for all the reindeer to regain their full speed of 90 meters per second. That’s the nutritional energy found in slightly more than one cup of oats. Now everybody knows that Santa takes good care of his reindeer, so undoubtedly they were fed plenty of oats and hay before takeoff. This was stored in their body fat for future, ondemand use.
Sadly, Christmas is over, and it’s time to get back to the more mundane aspects of life. Next time we’ll apply the principles behind the WorkEnergy Theorem to calculate the braking force required to stop a car in motion.
Copyright 2015 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: calories, energy, engineering expert, Joules, mass, speed, velocity, work, workenergy theorem
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Friday, January 1st, 2016
As an engineering expert I’ve applied the WorkEnergy Theorem to diverse situations, but none as unique as its most recent application, the progress of Santa’s sleigh. Last week we saw how Santa and his reindeer team encountered a wind gust which generated enough force to slow them from an initial velocity of v_{1} to a final velocity, v_{2,} over a distance, d. Today we’ll begin using the WorkEnergy Theorem to see if Santa was able to keep to his Christmas delivery schedule and get all the good boys and girls their gifts in time.
Before we can work with the WorkEnergy Theorem, we must first revisit the formula it’s predicated upon, de Coriolis’ formula for kinetic energy,
KE = ½ × m × v^{2} (1)
where, KE is kinetic energy, m is the moving object’s mass, and v its velocity.
The equation behind the WorkEnergy Theorem is,
W = KE_{2 }– KE_{1} (2)
where W is the work performed, KE_{1} is the moving object’s initial kinetic energy and KE_{2} its final kinetic energy after it has slowed or stopped. In cases where the object has come to a complete stop KE_{2} is equal to zero, since the velocity of a stationary object is zero.
In order to work with equation (2) we must first expand it into a more useful format that quantifies an object’s mass and initial and final velocities. We’ll do that by substituting equation (1) into equation (2). The result of that term substitution is,
W = [½ × m × v_{2}^{2} ] – [½ × m × v_{1}^{2}] (3)
Factoring out like terms, equation (3) is simplified to,
W = ½ × m × [v_{2}^{2} – v_{1}^{2}] (4)
Now according to de Coriolis, work is equal to force, F, times distance, d. So substituting these terms for W in equation (4), the expanded version of the WorkEnergy Theorem becomes,
F × d = ½ × m × [v_{2}^{2} – v_{1}^{2}] (5)
Next time we’ll apply equation (5) to Santa’s delivery flight to calculate the strength of that gust of wind slowing him down.
Copyright 2015 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: de Coriolis, distance, energy, engineering expert, force, kinetic energy, mass, velocity, wind force, work, workenergy theorem
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Friday, December 18th, 2015
My work as an engineering expert sometimes involves computations of energy expended, as when I must determine how much energy is required to move something. But sometimes the opposite needs to be calculated, that is, how much energy is required to stop something already in motion. That’s the subject of today’s discussion, which we’ll approach by way of the WorkEnergy Theorem.
The WorkEnergy Theorem states that the work required to slow or stop a moving object is equal to the change in energy the object experiences while in motion, that is, how its kinetic energy is reduced or completely exhausted. Although we don’t know who to attribute the Theorem to specifically, we do know it’s based on the previous work of Gaspard Gustave de Coriolis and James Prescott Joule, whose work in turn built upon that of Isaac Newton’s Second Law of Motion.
Consider the example shown here. A ball of mass m moves unimpeded through space at a velocity of v_{1 }until it is met by an opposing force, F. This force acts upon the ball over a travel distance d, resulting in the ball’s slowing to a velocity of v_{2}.
The Work – Energy Theorem Illustrated
Does the illustration make clear the WorkEnergy Theorem dynamics at play? If not, return for the second part of this blog, where we’ll clarify things by getting into the math behind the action.
Copyright 2015 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: energy, engineering expert witness, GaspardGustave de Coriolis, Isaac Newton's Second Law of Motion, James Prescott Joule, kinetic energy, moving object, work, workenergy theorem
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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)
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 second^{2}.
Putting equation (1) in longhand notation terms, we arrive at,
Work = 178 kilogram • meters per second^{2} × 3 meters (2)
Work = 534 kilogram • meters^{2} per second^{2} (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 • meters^{2} per second^{2}, 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
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Tags: de Coriolis' formula to compute work, electric motors, energy, engineering expert, generate electricity, joule, Newton meters, Newtons, power plants, steam turbines, unit of energy, units of force, work, work energy relationship, work required
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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 Coriolis‘ Principle 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
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
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Tags: de Coriolis’ Principle of Work, dynamode, engineering expert, joule, lawsuit, work
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Tuesday, October 6th, 2015
As an engineering expert I’ve often witnessed energy change forms, something our example coffee mug has been experiencing as it moves from a shelf to the floor. The mug’s various energies were proven to be mathematically equivalent, expressed as, 4.9 kg•meter^{2 }/second^{2 }, which is read as, “kilogram meter squared per second squared.” This mouthful led to the renaming of the measurement to the Joule, in honor of James Prescott Joule, a scientist who successfully demonstrated the interrelationship between different forms of energy. We’ll focus on one of his experiments, the Joule Apparatus, today.
Back in the 1840s Joule built his Apparatus, a device which demonstrated the interrelationship between different forms of energy.
The Joule Apparatus
The Joule Apparatus consisted of a weight suspended by string over a pulley, which in turn was wound around a winding drum. As long as the drum remained stationary, the weight remained motionlessly suspended. While motionless, the weight’s potential energy lay latent within it, just as it had in our example coffee mug resting on a shelf.
But when the pressure keeping the winding drum stationary was released, the weight was set free to fall, and its potential energy began converting to kinetic. In the process, the string the weight was attached to unwound from the drum, which caused the drum to turn and along with it the paddle wheel it was attached to.
Joule’s Apparatus followed energy through many forms. From the quiet of potential energy to the kinetic energy demonstrated by the falling weight. The kinetic energy in turn was converted into mechanical energy, made manifest by the interaction between the moving drum and paddle wheel. The rotating paddles agitated the water, causing its temperature to rise. Observing this, Joule concluded that the mechanical energy of the spinning paddle wheel had been converted into heat energy, which temperature measurement proved was transferred into the water. Joule’s experiment thus proved the link between potential, kinetic, mechanical, and heat energies.
Joule’s work paved the way to make possible the later development of a host of modern mechanical devices that also converted heat energy into mechanical energy, or vice versa. These devices include a car’s engine and your kitchen’s refrigerator.
Next time we’ll see how Joule demonstrated a link between electrical and other forms of energy, including mechanical and heat. We’ll then use his discoveries to convert our falling coffee mug’s kinetic energy into yet another form.
Copyright 2015 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: electrical energy, energy engineer, energy of falling objects, engineering expert witness, forensic engineer, forms of energy, James Prescott Joule, Joule Apparatus, Joule energy unit, Joule's experiments, kinetic energy, mechanical energy, physics of falling objects, potential energy, work
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Wednesday, September 2nd, 2015
Last time we calculated the potential energy hidden within a coffee mug resting on a shelf. The concept of a passive object possessing energy may not be something all readers can identify with, but the secret behind that mug’s latent energy lies within The Law of Conservation of Energy, the topic we’ll be discussing today.
Julius Robert von Mayer, a German physicist of the mid 19th Century, is the man behind the Law. He posited that energy cannot be created or destroyed, it can only be transferred from one object to another or converted from one form of energy to another. Forms of energy include potential, kinetic, heat, chemical, mechanical, and electrical, all of which have the ability to become another form of energy.
Let’s take our coffee mug for example. Where did its potential energy come from? Ultimately, from the radiant energy emitted by the sun. The sun’s radiant energy was absorbed by plants and then converted to chemical energy through the process of photosynthesis, enabling them to grow. When they were later eaten by humans and other animals, the plants’ chemical energy became incorporated into their bodies’ cells, including the arm muscles used to lift the mug to the shelf.
In the act of lifting the cup, the arm’s muscle cells converted their own chemical energy into mechanical energy. And because lifting a mug to a shelf is work, for some of us greater than others, some of the arm’s chemical energy became heat energy which was lost to the environment.
Because of the elevated perch provided to the mug by the arm, which was in direct defiance of Earth’s gravitational pull, the arm muscles’ mechanical energy was transferred to the mug and converted to latent potential energy, because without that shelf to support it, the mug would fall to the ground. The mug’s potential energy would realize its full potential if it should be sent crashing to the floor, at which time it would become another form of energy. The mischievous orange kitty seems to have just that in mind.
We’ll talk more about the mug’s potential energy being converted to other forms next time.
Copyright 2015 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: chemical energy, engineering expert witness, falling objects, gravity, kinetic energy, law of conservation of energy, mechanical energy, potential energy, radiant energy, work
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Wednesday, December 4th, 2013
Last time we ran our basic power plant steam turbine without a condenser. In that configuration the steam from the turbine exhaust was simply discharged to the surrounding atmosphere. Today we’ll connect it to a condenser to see how it improves the turbine’s efficiency.
As discussed in a previous blog, enthalpy h_{1} is solely dependent on the pressure and temperature at the turbine inlet. For purposes of today’s discussion, turbine inlet steam pressure and temperature will remain as last time, with values of 2,000 lbs PSI and 1000°F respectively, and calculations today will be based upon those values. So to review, the inlet enthalpy h_{1 }is,
h_{1} = 1474 BTU/lb
If the condenser vacuum exists at a pressure of 0.6 PSI, a realistic value for a power plant condenser, then referring to the steam tables in the Van Wylen and Sonntag thermodynamics book, we find that the enthalpy h_{2 }will be,
h_{2} = 847 BTU/lb
and the amount of useful work that the turbine can perform with the condenser in place would therefore be,
W = h_{1} – h_{2} = 1474 BTU/lb – 847 BTU/lb = 627 BTU/lb
So essentially with the condenser present, the work of the turbine is increased by 168 BTU/lb (627 BTU/lb – 459 BTU/lb). To put this increase into terms we can relate to, consider this. Suppose there’s one million pounds of steam flowing through the turbine each hour. Knowing this, the turbine power increase, P, is calculated to be,
P = (168 BTU/lb) ´ (1,000,000 lb/hr) = 168,000,000 BTU/hr
Now according to Marks’ Standard Handbook for Mechanical Engineers, a popular general reference book in mechanical engineering circles, one BTU per hour is equivalent to 0.000393 horsepower, or HP. So converting turbine power, P, to horsepower, HP, we get,
P = (168,000,000 BTU/hr) ´ (0.000393 HP/BTU/hr) = 66,025 HP
A typical automobile has a 120 HP engine, so this equation tells us that the turbine horsepower output was increased a great deal simply by adding a condenser to the turbine exhaust. In fact, it was increased to the tune of the power behind approximately 550 cars!
What all this means is that the stronger the vacuum within the condenser, the greater the difference between h_{1} and h_{2} will be. This results in increased turbine efficiency and work output, as evidenced by the greater numeric value for W. Put another way, the turbine’s increased efficiency is a direct result of the condenser’s vacuum forming action and its recapturing of the steam that would otherwise escape from the turbine’s exhaust into the atmosphere.
This wraps up our series on the power plant watertosteam cycle. Next time we’ll use the power of 3D animation to turn a static 2D image of a centrifugal clutch into a moving portrayal to see how it works.
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Tags: automobile engine, BTU, BTU/lb, coal power plant, energy, engineering expert witness, enthalpy, forensic engineer, horsepower, mechanical engineer, power, power engineer, power industry expert, power plant efficiency, power plant expert, power plant training instructor, power plant training seminars, steam pressure, steam turbine, steam turbine expert witness, steam water cycle, thermodynamics, vacuum, work
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