# Energy Loss The final thread(Hopefully)

Discussion in 'General Electronics Discussion' started by Moha99, Apr 6, 2012.

1. ### Moha99

261
0
Nov 18, 2011
Hi everyone I know there is another post that consists of the same topic but that post helped me a lot in understanding the laws of Thermodynamics and Conservation of Energy a lot more. However, I needed to make a clear understanding and summing everything up for my own personal education of everything. Feel free to post you're replies I'm morst grateful and appreciated for them I understood a lot on this forum's in a very very short time I thank you all for that!

Now to the main topic!

I've been researching something lately and I wanted to make sure that I'm on the right picture here.

Now this is all related to energy loss because im trying to narrow it down and understand why to make up a a perfect understanding.

Now in electric generation lets say that 100J was applied in mechanical energy to transfer it to electrical energy. That 100J inputed in the output can never exceed that unless another energy is applied to it correct?

And we cant generate 100J in electricity due to loss of energy that is caused from friction and heat and any other resistance correct? But is it mainly friction and heat? So maybe if I'm lucky I can get 90J in electricity... Now if I had perfect wide,cold,short conductive wire I could get a chance of getting over 90J is that possibile? The heat is lessened so thus there isn't much loss of energy and in addition to the high frequencies a larger surface will be applied also.

This is all for my own good understanding of things and this also applies to another example like a rock thrown upward to the sky

If it was thrown upward with the energy of 100J it will dorp also with a loss due to air resistance and friction correct? Also transformers also have the same thing they have loss of energy as well right?

I would appreciate that you all would give me a detailed description if I misunderstood it all or what I'm missing.

I really hope that I'm in the same page here .

Thank you all!

2. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
Not originally a reply here, but the quotes should be ok...

well, yeah, you won't get more than 100J out (in fact you'll get less).

Also consider how you got the 100J of energy.

You may have burnt some fuel which may have a theoretical yield of 200J. That energy was converted to heat and used to heat water. some energy was lost in the exhaust gasses as unburnt fuel, heat, and even mechanical energy. Some heat was lost through conduction to other than the water.

The water is heated and converted to steam, the energy is transferred to a turbine. Some energy is lost in friction in the bearings, not all of the energy in the steam is converted to mechanical energy (some heats the turbine, some remains in the steam exiting the turbine).

And so you end up with 100J of mechanical energy.

True. The actual amount of energy you can get is debateable, but this step is pretty efficient. Your losses will include mechanical (friction), electrical (resistance), and magnetic. you can safely assume that most of this will end up as heat. A vanishingly small amount will be radiated as very low frequency EM radiation.

But your losses don't stop there. The losses in transmission are huge, and are mostly I^2R losses and are consequently why transmission lines operate at high voltages.

That may reduce resistive losses, but you want wire that has a large cross-sectional area to reduce resistance, and (even at 50Hz) to take advantage of the skin effect (you might even use tubular conductors).

Yeah, it's kind of a similar process.

When you throw the rock up, you're converting the energy from kinetic to potential energy. There are no losses in the conversion, but kinetic energy is leaky. The air resistance (for a thrown rock) is the major loss. In addition, you may have put a spin on the rock as you threw it. This won't be converted to potential energy and will either be lost due to air resistance or when the rock falls on the ground.

As the rock stops and starts to fall, the stored potential energy gets converted back to potential energy (losslessly), but the kinetic energy is again subject to losses as air resistance again slows the rock (and if this heats the rock, the heat gets radiated away)

So in a transformer the same thing happens. The electrical energy is converted to a magnetic field, and the change in magnetic field is converted back to electrical energy. There are electrical losses in the coils of wire, and some of the magnetic energy creates eddy currents which lose more energy. However, high power transformers can easily exceed 99% efficiency.

3. ### Moha99

261
0
Nov 18, 2011

In electric generation specifically. There are 3 main problems that eventually turn to heat:

mechanical (friction), electrical (resistance), and magnetic all will turn into heat.
Now the heat were can it be found? in the generator's interior? Where the coils and magnets are found, or is it found throughout the output copper cables? And what degree does the heat reach to? hundreds or maybe thousands of degrees of heat?

That heat is caused from those there i can imagine it more when im rubbing my two hands to gather very quickly because of the friction heat is generated I can image all of those three applied together will increase heat drastically.

4. ### Moha99

261
0
Nov 18, 2011
So Heat is the main and final cause of energy loss...

Since all those 3 factors add up and eventually turn up to heat.

Question is... Where that heat is specifically found? On the output wire or inside the generator...?

5. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
Well, I^2R heating is in conductors, and heat cause by friction in mechanical devices builds up in the bearings, surfaces, and/or lubricants.

From there the heat is conducted, convected, and radiated to other components and eventually to the atmosphere, coolant, etc.

6. ### Moha99

261
0
Nov 18, 2011
So that heat is eventually cooled up by the atmosphere after its done the damage of energy loss?

So if the conductors were cold from before the heat will be absorbed right?

Can friction be solved by those mechanical devies?

7. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
Cold is relative. Heat does not equal temperature. And heat flows through thermally conductive materials.

So the temperature may not change much, even though heat is flowing through some part -- and almost certainly it will reach an equilibrium where the temperature stops changing (there will be a temperature gradient that causes heat to flow just as a potential gradient causes current to flow).

The heat may all end up in the atmosphere. Some may end up in the oceans, the ground, or in space. Some small amount may remain in the system because the rate of flow of the heat out of the system depends on a number of things, but a temperature gradient is required. If you heat up the environment, you therefore reduce the temperature differential, and therefore slow the rate at which you transfer heat to it.

Pretty simply, the change in energy of a system is equal to the energy going in less the energy coming out. If either one exceeds the other instantaneously, it means that there is some store of energy in the system.

Looking at a generator, if we assume the incoming energy is mechanical, energy comes out in the form of electrical energy, heat, sound, and EM radiation (and possibly others). Energy is stored in the system by means of mc^2, heat, magnetic fields, the potential energy of any mass, and the kinetic energy of moving masses (think a spinning turbine). Most of those stored forms ae useful, however some of the heat isn't. (oh, and the mc^2 energy of mass is typically not going to play a part)

8. ### Moha99

261
0
Nov 18, 2011

That is really interesting never thought of heat that way...

I know this may sound stupid but for example if i had a big 500W generator and connected a thermometer to the output copper cables without rotating the shaft so there is no electricity begin generated at all. When the generation process begins I connected the thermometer again to the output I'll find a MAJOR difference in the temperature or a slight one?

I think the rotation rate of the shaft is also important in that measurements so lets say I rotated the generators shaft to MAXIMUM rpm as required lets say for example 3000RPMS I think there will be a major difference in temperature.

And that unuseful heat is the mainly the highest cause of energy loss right? If that is solver the energy loss rate will significantly drop?

I mean products that are rated at 90% or 99% efficiency they tried to solve energy loss... In transformers, Generators and all that.

Last edited: Apr 8, 2012
9. ### Moha99

261
0
Nov 18, 2011
Also. One of the main reasons why generators and motor make very noisy sounds from their rotation is due to friction.

Now is the fiction cause from the motor/generators shaft touching another objects and under very very high rpm that causes a LOT of heat.

10. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
You should. It flows like water and electricity and things like that.

Presumably you read the ambient temperature.

Probably a slight increase. Depending on the load on the generator and how long it's been running, it may be immeasurable.

Remember heat is not temperature. A small amount of heat (energy) distributed through a gauge device will yield a very low temperature (average thermal energy) change.

Maybe not. The bearings are likely to be very good and will not contribute to a rapid temperature rise. Over a period of time they will have generated enough heat (when I say generated, I mean caused enough "other" energy to be converted to heat) to cause an appreciable temperature rise.

You've ignored the I^2R losses which I would imagine would be a significant cause of heating.

In pretty much all high energy applications, loss of energy to heat is very carefully managed. It's just that some processes (typically those involving heat) do not lend themselves to extremely high efficiency.

I've been through an operating turbine hall several times (and several turbine halls -- one of which was actually outside!).

There are many sources of sound. I suspect that the largest is due to the turbulence of the flow of steam through the turbines themselves with the consequent vibration being transmitted to the shaft and then to the bearings. Another source of sound is that caused by the air movement around the spinning rotor. In fact, when the door to the huts over the generators were opened (in the outdoor turbine hall) the noise level increased significantly).

I would imagine that the bearings are relatively quiet. If they had significant friction I'd be very worried. Any noise from them is more likely to be due to noise transmitted through them.

Think of any normal (electric) motor. The heat comes from the windings, not the bearings.

11. ### Moha99

261
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Nov 18, 2011
Thanks Steve.

So far as I said before I admire you for that help!

Im reading about Copper Loss right now and understanding more of what you said above.

I'll read more about energy loss and friction I believe it could be solved... Both of them I mean.(Not completely but a high majority of it)

When we look at that example of that rock thrown up and falling done if it was in a vacuum where the is no air I guess it would fall back 100J or at least close to that.
I'll read and study and come back with more questions if needed thanks!

Last edited: Apr 8, 2012
12. ### Moha99

261
0
Nov 18, 2011

Looking at it more "Resistance" of a certain material is the main cause of energy loss in a conductive circuit is that.
I read if we were able to cool the material at a very low temperature its possibile to reach a 0 resistance.
So we can avoid energy loss in a circuit.
What do you all think about that?

And in some Generators they have improved the bearing's by lessing the friction.
And the windings of the material inside the generators "probably copper" is designed differently to decrease the amount of energy loss in total.

So they factored out every energy loss and reduced the effect wich is pretty cool.

But I guess what's left is the copper wires.

Last edited: Apr 9, 2012
13. ### timothy48342

218
1
Nov 28, 2011
Superconductors.

It was first discovered that when some materials are cooled to near absolute zero degrees, the resistance drops to near zero. (It is a non-linear drop, not a gentle slope.)

Many materials exibit this phenomenon and each at various temperatures. In most cases this occurs, as above, near absolute zero.

The usefulness of that discovery alone is limited. You can't expect to keep a motor or generator at near absolute zero temperatures without spending some extra energy for cooling. (Although, once you get them there, there is a lot less heat produced, and so a lot less energy spent on cooling.)

Later it was discovered that some materials exibit this phenomenon at tempertures that are not so extreme. Copper is not one of them, but some alloys of Copper are.

Work is being done to find materials that exibit this phenomenon at higher and higher temperatures. According to wikipedia 133Kelvin is the current record, which is still pretty darn cold. Some of these materials may be quite expensive. When finding applications for them, the cost is going to have an affect. If a room temperature or higher superconductor could be discovered, that was easy to produce...(cheap to produce)... I think.. That would change the world a bit.

(I use the term "phenomenon" because it is a phenomenon to me. Scientists probably know what is going on and why.)
--tim
wikipedia on "High-temperature superconductivity":
http://en.wikipedia.org/wiki/High-temperature_superconductivity#Examples

14. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
There are several problems with superconducting generators.

Firstly those high temperature superconductors tend to be brittle, not soft like copper. So it's harder to make wires, even harder to wind them into coils and then really hard to stop them breaking.

Secondly, superconductors tend to be sensitive to magnetic fields, becoming non-superconducting in a strong magnetic field. Unfortunate because inside the generator you have these strong magnetic fields.

There may be ways around some of these, but it adds a certain difficulty factor.