## Posts Tagged ‘power’

### Another Specialized Application of the Euler-Eyelewein Formula

Tuesday, June 13th, 2017
 Last week we saw how friction coefficients as used in the Euler-Eyelewein Formula, can be highly specific to a specialized application, U.S. Navy ship capstans.   In fact, many diverse industries benefit from aspects of the Euler-Eytelwein Formula.   Today we’ll introduce another engineering application of the Formula, exploring its use within the irrigation system of a hydroponics plant. Another Specialized Application of the Euler-Eyelewein Formula         Pumps conveying water are an indispensable part of a hydroponics plant.   In the schematic shown here they are portrayed by the symbol ⊗.     In our simplified scenario to be presented next week, these pumps are powered by a mechanical power transmission system, each consisting of two pulleys and a belt.   One pulley is connected to a water pump, the other pulley to a gasoline engine.   A belts runs between the pulleys to deliver mechanical power from the engine to the pump.     The width of the belts is a key component in an efficiently running hydroponics plant.   We’ll see how and why that’s so next time.   Copyright 2017 – 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      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 ____________________________________

### How Condensers Increase Efficiency Inside Power Plants

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 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. ________________________________________

### Desuperheating in the Steam Turbine

Monday, September 2nd, 2013

Last time we learned that the addition of a superheater to the electric utility power plant steam cycle provides a ready supply of high temperature steam, laden with heat energy, to the turbine, which in turn powers the generator.   But this isn’t its only job.   One of the superheater’s most important functions is to regulate the ongoing process of desuperheating that takes place as the turbine consumes heat energy.   To understand this, let’s see what takes place if the superheater were to be removed from its position between the boiler and turbine.

## Figure 1

Without the superheater, the only available remaining source of sensible heat energy to the turbine would come from the meager amount present in phase C steam as shown in Figure 1.   If you’ll recall from a past blog, the sensible heat energy contained in superheated steam is the best source of energy for a steam turbine, because it’s able to keep it operating most efficiently.

As the turbine consumes the heat energy in phase C, starting at point 3 and continuing to point 2, the steam it’s consuming is in the process of desuperheating, as evidenced by the downward slope between the two points.   Desuperheating is an engineering term which means that as sensible heat energy is removed from the steam due to its use by the turbine, there will be a resulting drop in steam temperature.   And if this process were to continue without the compensatory function provided by the addition of a superheater to the steam cycle, the steam’s temperature would eventually return to mere boiling point, at point 2.   This is an undesirable thing.

With the steam’s temperature at boiling point, the only remaining source of heat energy to the turbine is the latent heat energy of phase B.   This heat energy will lead to an undesirable circumstance for the operation of our power hungry turbine as we will see next week.

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### Heat Energy Within the Power Plant – Water and Steam Cycle, Part 1

Monday, August 5th, 2013
 Last time we learned that electric utility power plant boilers are vessels that are reinforced with thick steel and are closed off from the surrounding atmosphere so as to facilitate the building up of highly pressurized steam.   This steam is laden with sensible heat energy, meaning it’s a useful energy, and it’s used to run steam turbines, which in turn drive electrical generators.   The end result is power to consumers.       Let’s now revisit our basic electric utility boiler diagram to see how water and steam flow.       Water is fed into the boiler, heat is applied externally, and steam exits through a pipe leading to the steam turbine.   You’ll notice that after the steam passes through the turbine, some of it is expelled into the surrounding atmosphere.       Since water is being continuously boiled off to produce steam, the boiler must be continuously replenished with a fresh supply.   This is typically supplied by a nearby body of water, hence one reason that power plants are often situated on a lake or river.       Since water contains both minerals and organic matter, including algae, a treatment system to remove these contaminants must be added to the water’s inlet area before it can be used.   This will keep operating parts such as the boiler and turbine free of damaging deposits.       The treatment system operates much like the water softener in your home, but on a larger scale.   Lake water is drawn into the system by a make-up pump, so named because it makes up, or replenishes spent water with a fresh supply.   The result is clean, mineral-free water that’s delivered to the boiler by a boiler feed pump, so named because its specific function is to feed water to the boiler.       Feeding water to the boiler on a continuous basis is no easy task because of the steam straining to break free, and boiler feed pumps are massively powerful devices built to accomplish this.   They effectively force water into the boiler even as high internal pressures try to force the water out.   This pressure is often greater than 1,500 pounds per square inch (PSI) in modern power plants.       So at this point we’ve discussed the fact that the boiler requires a continuous supply of fresh water, which is converted into high pressure steam, which is then sent on to spin a steam turbine.   The turbine powers an electrical generator, resulting in usable energy.       If you’ve been reading along closely, you will have identified that as things stand now it’s a rather inefficient and wasteful system, a point which we’ll address in next week’s blog. ___________________________________________

### Transistors – Voltage Regulation Part XVII

Monday, November 12th, 2012

Last time we learned about a new type of transistor called a bipolar transistor and how it controls the flow of electric current traveling from the collector to the emitter within our circuit.  We also saw how the bipolar transistor is integrated within a Zener diode voltage regulator circuit to make a new type of circuit called a transistor series voltage regulator.

Now let’s see how this all works by hooking our circuit up to both an unregulated power supply and an external supply circuit as shown in Figure 1.

## Figure 1

When voltage VUnregulated is applied to our transistor series voltage regulator circuit by way of an unregulated power supply, electric current flows through RLimiting into the base, B, of the transistor.  The transistor senses this current and responds by opening a path for current to flow from its collector, C, to its emitter, E.  With this path established, current flows freely from the unregulated power supply, through the transistor’s collector and emitter, on to the output terminal, and finally to the external supply circuit.  Total resistance of this circuit is said to be RTotal

At this point you’re probably wondering why the bipolar transistor base and Zener diode are connected to RLimiting.  Next time we’ll conclude our series by seeing how this connection is crucial to the functionality of our transistor series voltage regulator.

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### Industrial Control Basics – Introduction to Electric Relays

Tuesday, January 3rd, 2012
I’ve always considered science to be cool.  Back in the 5th grade I remember fondly leafing through my science textbook, eagerly anticipating our class performing the experiments, but we never did.  For some reason my teacher never took the time to demonstrate any.  Undeterred, I proceeded on my own.

I remember one experiment particularly well where I took a big steel nail and coiled wire around it.  When I hooked a battery up to the wires, as shown in Figure 1 below, electric current flowed from the battery through the wire coil.  This set up a magnetic field in the steel nail, thereby creating an electromagnet.  My electromagnet was strong enough to pick up paper clips, and I took great pleasure in repeatedly picking them up, then watching them unattach and fall quickly away when the wires were disconnected from the battery.

## Figure 1

Little did I know then that the electromagnet I had created was similar to an important part found within electrical relays used in many industrial control systems.  An example of one of these relays is shown in Figure 2.

## Figure 2

So, what’s in the little plastic cube?  Well, a relay is basically an electric switch, similar to the ones we’ve discussed in the past few weeks, the major difference being that it is not operated directly by human hands.  Rather, it’s operated by an electromagnet.  Let’s see how this works by examining a basic electrical relay, as shown in Figure 3.

## Figure 3

The diagram in Figure 3 shows a basic electric relay constructed of a steel core with a wire coil wrapped around it, similar to the electromagnet I constructed in my 5th grade experiment.  If the coil’s wires are not hooked up to a power source, a battery for example, no electric current will flow through it.  When there is no current the coil and steel core are not magnetic.  For purposes of our illustration and in accordance with industrial control parlance, this is said to be this relay’s “normal state.”

Next to the steel core there is a movable steel armature, a kind of lever, which is attached to a spring.  On one end of the armature is a pivot point, on the other end is a set of electrical switch contacts.  When the relay is in its normal state, the spring’s tension holds the armature against the “normally closed,” or N.C., contact.  If electric current is applied to the wire leading to the pivot point on the armature while in this state, it will be caused to flow on a continuous path through the armature and the N.C. contact, then out through the wire leading from the N.C. contact.  In our illustration, since the armature does not touch the N.O. contact, an air gap is created that prevents electric current from traveling through the contact from the armature.

Next week we’ll see how these parts come into play within a relay when electric current flows through the coil, turning it into an electromagnet.

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### Wire Size and Electric Current – Joule Heating

Sunday, March 20th, 2011
 Ever take a peek inside the toaster while you’re waiting for the toast to pop up?  If so, you would have noticed a bright orange glow.  That glow is produced when the toasting wires heat up, which in turn creates a nice crusty surface on your bread or waffle.  It’s the same phenomenon as when the filament inside an incandescent bulb glows.  The light and heat produced in both these cases are the result of the Joule, pronounced “jewel,” effect at work.      To understand Joule heating, let’s first refresh our memories as to electrical current resistance.  We learned previously that wire is not a perfect conductor, and as such resistance to flow is encountered.  This resistance causes power to be lost along the length of wire, in accordance with this equation: Power Loss = I2 × R Where I is the electric current flowing through a wire, and R is the total electrical resistance of the wire.  The power loss is measured in units of Joules per second, otherwise known as watts, “watt” denoting a metric unit of power.  It is named after the famed Scottish mechanical engineer, James Watt, who is responsible for inventing the modern steam engine.  A Joule is a metric unit of heat energy, named after the English scientist James Prescott Joule.  He was a pioneer in the field of thermodynamics, a branch of physics concerned with the relationships between different forms of energy.      Anyway, to see how the equation works, let’s look at an example.  Suppose we have 12 feet of 12 AWG copper wire.  We are using it to feed power to an appliance that draws 10 amperes of electric current.  Going to our handy engineering reference book, we find that the 12 AWG wire has an electrical resistance of 0.001588 ohms per foot, “ohm” being a unit of electrical resistance.  Plugging in the numbers, our equation for total electrical resistance becomes: R = (0.001588 ohms per foot) × 12 feet = 0.01905 ohms And we can now calculate power loss as follows: Power = I2 × R = (10 amperes)2 × (0.01905 ohms) = 1.905 watts      Instead of using a 12 AWG wire, let’s use a smaller diameter wire, say, 26 AWG.  Our engineering reference book says that 26 AWG wire has an electrical resistance of 0.0418 ohms per foot.  So let’s see how this changes the power loss: R = (0.0418 ohms per foot) × 12 feet = 0.5016 ohms Power = I2 × R = (10 amperes)2 × (0.5016 ohms) = 50.16 watts      This explains why appliances like space heaters and window unit air conditioners have short, thick power cords.  They draw a lot of current when they operate, and a short power cord, precisely because it is short, poses less electrical resistance than a long cord.  A thicker cord also helps reduce resistance to power flow.  The result is a large amount of current flowing through a superhighway of wire, the wide berth reducing both the amount of power loss and the probability of dangerous Joule heating effect from taking place.       Our example shows that the electric current flowing through the 12 AWG wire loses 1.905 watts of power due to the inconsistencies within the wire, and this in turn causes the wire to heat up.  This is Joule heating at work.  Joule heating of 50.16 watts in the thinner 26 AWG wire can lead to serious trouble.      When using a power cord, heat moves from the copper wire within it, whose job it is to conduct electricity, and beyond, on to the electrical insulation that surrounds it.  There the heat is not trapped, but escapes into the environment surrounding the cord.  If the wire has low internal resistance and the amount of current flowing through it is within limits which are deemed to be acceptable, then Joule heating can be safely dissipated and the wire remains cool.  But if the current goes beyond the safe limit, as specified in the American Wire Gauge (AWG) table for that type of wire, then overheating can be the result.  The electrical insulation may start to melt and burn, and the local fire department may then become involved.          That’s it for wire sizing and electric current.  Next time we’ll slip back into the mechanical world and explore a new topic: the principles of ventilation. _____________________________________________

### Transformers – The Voltage/Current Trade-Off

Sunday, December 26th, 2010

 As a child I considered the reindeer Rudolph, with his nose so bright, to be a marvel of engineering.  Now an adult, I remain perplexed as to the mystery behind the self-generating power source behind his nose.  Did it ever overheat? I wondered.  Perhaps today’s discussion can shed some light on the matter.      During the course of our discussion of electricity certain terms have been tossed about, like voltage and current.  For some the distinction between the two may be unclear, and that is what we’ll be addressing today.      Electricity is a rather abstract phenomenon, but you may consider the flow of electrical current through a wire to be much like water flowing through a garden hose.  The water won’t flow unless there’s sufficient pressure behind it, and that pressure is supplied by pumps, either at your city water works or your personal well.  Take away the pressure, and the water stops flowing through the hose.       Electricity flows in much the same manner.  It requires a pushing pressure to get it on its journey from power plant to home, and that pressure is voltage.  Take away voltage, and the current stops flowing through the wire.  Voltage is, of course, produced by an electrical generator at the power plant.       Last time we saw how an electrical transformer can reduce high voltage to low voltage and how this process also works in reverse.  But how can that be?  How can low voltage be turned into high?  Is it really possible to get “something from nothing”?  Let’s take a closer look.      When a light bulb burns out in your home, you routinely look at the bulb to see how many watts it is so you can replace it with the same type.  But what exactly is a “watt”?  It’s a unit of power, and the markings on the bulb tell you how much electrical power it consumes when you use it.  Generally speaking, this electrical power is related to voltage and current by this formula: Power = Volts × Electrical Current      Knowing this, if I have a 60 watt bulb in a table lamp, and I plug it into a 120 volt wall outlet, then how much electrical current is the lamp going to draw from the outlet?  Using the formula above and a little algebra, we get: Electrical Current = Power ÷ Volts Electrical Current = 60 watts ÷ 120 volts = 0.5 amperes      And believe it or not, this same formula that’s used to assess power  of a light bulb also applies to electrical transformers.  Basically, the power going into the transformer is equal to the power coming out.      To see how this works, consider the example step-up transformer shown in Figure 1, which converts a low voltage to a higher one.  By the way, “step up” transformers have all sorts of applications.  For example, they are used by electric utilities to raise the voltage produced by a power plant to make it more economical to transmit to far away customers.  We’ll get into that in another article. Figure 1 – A Step-Up Transformer      In this example the input voltage on the primary coil is stepped up from 120 volts to 480 volts on the secondary coil, and this works according to the formula we learned about in last week’s blog: NP ÷ NS = VP ÷ VS where NP and NS are the number of turns of wire in the primary and secondary coils respectively, and VP and VS are the voltages of the primary and secondary coils respectively.  Plugging in the numbers we get: 50 turns ÷ 200 turns = 120 volts ÷ VS [(200 turns ÷ 50 turns) × 120 volts] = VS = 480 volts      Okay, for the sake of our example let’s say that an electric motor is connected to the 480 volt secondary coil.  We have an electric meter hooked up to the primary coil and we measure a 2 ampere (a.k.a. “amps”) electrical current flowing through it.  Without having the benefit of another electric meter positioned at the secondary coil, how can we measure how much electrical current is flowing through it?  The current flowing through the secondary coil is found by equalizing the power in the primary and secondary coils: PowerP = PowerS      Another way of stating this is to say that electrical power is equal to volts times current, so the equation becomes: VP × IP = VS × IS where IP and IS are the primary coil and secondary coil currents, respectively.  Plugging in the numbers and working a little algebra we get the electrical current in the secondary coil: 120 volts × 2 amps = 480 volts × IS IS = (120 volts × 2 amps) ÷ 480 volts = 0.5 amps      This shows us that the current flowing in the secondary coil is lower than that of the primary coil.  It is therefore obvious that the voltage increase in the secondary coil comes at the expense of electrical current that can flow through the secondary coil.  Squeeze down on current, voltage goes up.  Squeeze down on voltage, current goes up.  The power flowing through the transformer stays the same.      Conversely, step-down transformers reduce the voltage coming in, and thereby produce the reverse effect.  There is an actual increase in current that can flow through the secondary coil.  This principle exemplifies the tradeoff process which is often present in science and engineering.      Next time we’ll explore how both step-up and step-down transformers are used by electric utilities to transmit power from power plants to its customers tied into the utility grid.  As for Rudolph and his power source, that’s still under investigation. _____________________________________________

### Transformers and The Magic of Electricity

Sunday, December 5th, 2010