Happy Holidays from EngineeringExpert.net, LLC and the Engineering Expert Witness Blog.
Pulleys Make Santa’s Job Easier
Copyright 2016 – Philip J. O’Keefe, PE
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 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
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
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
As an engineering expert with 14 years’ electric utility experience, I’ve dealt with all types of electrical power generators, including many similar to the dynamo that James Prescott Joule used in his Experiment With Electricity. Today we’ll look inside Joule’s dynamo and see how it contributed to creating electricity as well as another of Joule’s discoveries, the Joule Heating Effect.
Dynamo-Circa Early 19th Century
In Joule’s Experiment With Electricity, the dynamo was powered by a steam engine, which enabled the dynamo’s shaft to spin. As it spun, the magnet located inside the dynamo also spun, thus creating a rotating magnetic field that surrounded the dynamo’s internal copper wire coils.
The interaction between the magnetic field and wire produced electric current which flowed inside the coils. The current ultimately made its way out of the dynamo by way of external wires, to which any number of devices could be powered when attached. The net result was the engine’s mechanical energy had been converted into electrical. To learn more about the process of producing electricity with magnets see my blog on, Coal Power Plant Fundamentals – The Generator.
As electrical energy flowed through the dynamo’s wiring, some of it was converted into heat energy. This was due to resistance posed by impurities present in the makeup of the wire, impurities which served to impede the overall flow of electric current. When electrons flowing through the wire collided with these impediments, they caused heat to build up inside the wire, a phenomenon which came to be known as the Joule Heating Effect. To read more on electrical resistance and Joule heating go to my blog, Wire Size and Electric Current.
The net result of Joule’s Experiment With Electricity was to further prove the link between chemical, heat, mechanical and electrical energies as set out in the Law of Conservation of Energy. And I suspect that knowledge was later put to use by Joule’s family for the betterment of their brewery business.
Next time we’ll use Joule’s experimental findings in conjunction with de Coriolis’ Kinetic Energy Formula to quantify the energy of the falling coffee mug we’ve been watching.
Copyright 2015 – Philip J. O’Keefe, PE
In my work as an engineering expert I’ve dealt with all forms of energy, just as we’ve watched James Prescott Joule do. He constructed his Joule Apparatus specifically to demonstrate the connection between different forms of energy. Today we’ll see how he furthered his discoveries by building a prototype power plant capable of producing electricity, a device which came to be known as Joule’s Experiment With Electricity.
Joule’s Experiment With Electricity
As the son of a wealthy brewer, Joule had been fascinated by electricity and the possibility of using it to power his family’s brewery and thereby boost production. To explore the possibilities, he went beyond the Apparatus he had built earlier and built a device which utilized electricity to power its components. The setup for Joule’s experiment with electricity is shown here.
Coal was used to bring water inside a boiler to boiling point, which produced steam. The steam’s heat energy then flowed to a steam engine, which in turn spun a dynamo, a type of electrical generator.
Tracing the device’s energy conversions back to their roots, we see that chemical energy contained within coal was converted into heat energy when the coal was burned. Heat energy from the burning coal caused the water inside the boiler to rise, producing steam. The steam, which contained abundant amounts of heat energy, was supplied to a steam engine, which then converted the steam’s heat energy into mechanical energy to set the engine’s parts into motion. The engine’s moving parts were coupled to a dynamo by a drive belt, which in turn caused the dynamo to spin.
Next time we’ll take a look inside the dynamo and see how it created electricity and led to another of Joule’s discoveries being named after him.
Copyright 2015 – Philip J. O’Keefe, PE
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 h1 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 h1 is,
h1 = 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 h2 will be,
h2 = 847 BTU/lb
and the amount of useful work that the turbine can perform with the condenser in place would therefore be,
W = h1 – h2 = 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 h1 and h2 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 water-to-steam 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.
Last time we learned that the amount of useful work, W, that a steam turbine performs is calculated by taking the difference between the enthalpy of the steam entering and then leaving the turbine. And in an earlier blog we learned that a vacuum is created in the condenser when condensate is formed. This vacuum acts to lower the pressure of turbine exhaust, and in so doing also lowers the enthalpy of the exhaust steam. Putting these facts together we are able to generate data which demonstrates how the condenser increases the amount of work produced by the turbine.
To better gauge the effects of a condenser, let’s look at the differences between its being present and not present. Let’s first take a look at how much work is produced by a steam turbine without a condenser.
The steam entering the turbine inlet has a pressure of 2000 pounds per square inch (PSI) and a temperature of 1000°F. Knowing these turbine inlet conditions, we can go to the steam tables in any thermodynamics book to find the enthalpy, h1. Titles such as Fundamentals of Classical Thermodynamics by Gordon J. Van Wylen and Richard E. Sonntag list enthalpy values over a wide range of temperatures and pressures. For our example this volume tells us that,
h1 = 1474 BTU/lb
where BTU stands for British Thermal Units, a unit of measurement used to quantify the energy contained within steam or water, in our case the water to steam cycle inside a power plant. Technically speaking, a BTU is the amount of heat energy required to raise the temperature of one pound of water by one degree Fahrenheit. The term lb should be a familiar one, it’s the standard abbreviation used for pound, so enthalpy is the measurement of the amount of energy per pound of steam flowing through, in this case, the turbine.
Since there is no condenser attached to the steam turbine’s exhaust in our illustration, the turbine discharges its spent steam into the surrounding atmosphere. The atmosphere in our scenario exists at 14.7 PSI because our power plant happens to be at sea level. Knowing these facts, the steam tables inform us that the value of the exhausted steam’s enthalpy, h2, is:
h2 = 1015 BTU/lb
Combining the two equations we are able to calculate the useful work the turbine is able to perform as:
W = h1 – h2 = 1474 BTU/lb – 1015 BTU/lb = 459 BTU/lb
This equation tells us that for every pound of steam flowing through it, the turbine converts 459 BTUs of the steam’s heat energy into mechanical energy to run the electrical generator.
Next week we’ll connect a condenser to the steam turbine to see how its efficiency can be improved.
Last time we learned how enthalpy is used to measure heat energy contained in the steam inside a power plant. The higher the steam pressure, the higher the enthalpy, and vice versa, and we touched upon the concept of work, or the potential for a useful outcome of a process. Today we’ll see how to get the maximum work out of a steam turbine by attaching a condenser at the point of its exhaust and making the most of the vacuum that exists within its condenser.
Let’s revisit the equation introduced last time, which allows us to determine the amount of useful work output:
W = h1 – h2
Applied to a power plant’s water-to-steam cycle, enthalpy h1 is solely dependent on the pressure and temperature of steam entering the turbine from the boiler and superheater, as contained within the purple dashed line in the diagram below.
As for enthalpy h2, it’s solely dependent on the pressure and temperature of steam within the condenser portion of the water-to-steam cycle, as shown by the blue dashed circle of the diagram.
Next week we’ll see how the condenser, and more specifically the vacuum inside of it, sets the platform for increased energy production, a/k/a work.
Last time we learned how the formation of condensate within a power plant’s turbine results in a vacuum being created. This vacuum plays a key role in increasing steam turbine efficiency because it affects a property known as enthalpy, a term used to denote total heat energy contained within a substance. For the purposes of our discussion, that would be the heat energy contained within steam which flows through the turbine in a power plant.
The term enthalpy was first introduced by scientists within the context of the science of thermodynamics sometime in the early 20th Century. As discussed in a previous blog article, thermodynamics is the science that deals with heat and work present within processes. Enthalpy is a key factor in thermodynamics, and is commonly represented in engineering calculations by the letter h and denoted as,
h = u + Pv
where u is the internal energy of a substance, let’s say steam; P is the pressure acting upon a specific volume, v, of the steam; and P and v are multiplied together. Pressure is force per unit area and is measured in psi, pounds per square inch. For the purposes of our discussion, it’s the amount of pressure that steam places on pipes containing it.
Looking at the equation above, simple math tells us that if we increase the pressure, P, the result will be an increase in enthalpy h. If we decrease P, the result will be a decrease in h. Now, let’s see why this property is important with regard to the operation of a steam turbine.
When it comes to steam turbines, thermodynamics tells us that the amount of work they perform is proportional to the difference between the enthalpy of the steam entering the turbine and the enthalpy of the steam at the turbine’s exhaust. What is meant by work is the act of driving the electrical generator, which in turn provides electric power. In other words, the work leads to a useful outcome. This relationship is represented by the following equation,
W = h1 – h2
In terms of the illustration below, W stands for work, or potential for useful outcome of the turbine/generator process in the form of electricity, h1 is the enthalpy of the steam entering the inlet of the turbine from the superheater, and h2 is the enthalpy of the steam leaving at the turbine exhaust.
We’ll discuss the importance of enthalpy in more detail next week, when we’ll apply the concept to the work output of a steam turbine.
Last time we discussed the key functions of the make-up valve in the power plant water-to-steam cycle. Today we’re going to talk about a vacuum. No, not the kind you use around the house, the kind that’s created by the condenser inside a power plant.
As discussed previously, the condenser is a piece of equipment that turns turbine exhaust steam back into water. The water that’s formed during this process is known as condensate, and its density is higher than that of the steam it shares space with inside the condenser. That difference in density is what creates the vacuum inside the condenser vessel. Put another way, the increase in density along with the condenser’s airtight design prevent air from rushing in from outside to occupy any of the space inside the condenser, a desirable condition from an efficiency standpoint.
But to understand how all this works we’ll first have to gain an understanding of what is meant by density. A textbook would define it as the mass of a substance divided by the amount of space that that substance occupies. Let’s take steam and water for example. One pound of steam at 212°F forms a vapor cloud that occupies 26.78 cubic feet of space. If we condensed that pound of steam back into water at the same temperature, it would just about fit into a 16 ounce glass and occupy a mere 0.017 cubic feet.
The huge difference in their volumes is due to the fact that steam contains more than five times the heat energy that unheated water does. That energy makes the molecules in a cloud of steam more active, causing them to collide against each other with great force, spread apart, and occupy a larger space.
If you’re wondering what change in density has to do with vacuum in the condenser, allow me to offer an analogy. Ever canned any produce, like tomatoes, in glass jars to over-winter? Not likely, as this once common survival tactic has nearly become a lost art. But the vacuum created inside the condenser is much like the vacuum created within a mason jar during canning.
Inside the glass mason jar, a small space is intentionally left between the tomatoes and lid. During the process of boiling, or heat sterilization, this space fills with steam. Then during cooling the trapped steam condenses into water. This condensation creates the vacuum that sucks down on the jar’s lid, giving it an airtight seal, a condition which won’t allow bacteria to grow on our canned foods. You see, like us bacteria need oxygen to live, but thanks to the vacuum inside our cooked mason jar no air containing oxygen will remain inside to harbor it.
Next time we’ll continue our discussion on vacuum to see how it’s used to increase a steam turbine’s efficiency.