Electrical Energy

[Theredbarron]

Is it still impossible to multiply electrical energy?

[StringJunky]

55 minutes ago, Theredbarron said:

Is it still impossible to multiply electrical energy?

If you mean can you get more out than you put in then no.

[Theredbarron]

11 hours ago, StringJunky said:

If you mean can you get more out than you put in then no.

So I have a theory but I dont know if its a good idea for me to post something like this as in due to the possible value. I do have a question however. Are the motors of today the most efficient design out for armature's as in that's the only way they are made? I have found 2 different versions of todays but I'm sure there is more. Some are more specific to the applications like power generators for the hz and skew and alternators have no need for clean ac so they are much less specifically designed. Not just for power generating armatures but also in regular motors. I should say rotor mainly because what I'm referring to is the coil and the material that is being used to conduct the magnetism from it. I know its one in and one out for energy. What I want to know is that would making a lighter armature be like the difference between driving a semi up a hill vs a fiat 500 when creating electricity? What does reducing the weight of an armature do for power requirements?

[studiot]

1 hour ago, Theredbarron said:

So I have a theory but I dont know if its a good idea for me to post something like this as in due to the possible value. I do have a question however. Are the motors of today the most efficient design out for armature's as in that's the only way they are made? I have found 2 different versions of todays but I'm sure there is more. Some are more specific to the applications like power generators for the hz and skew and alternators have no need for clean ac so they are much less specifically designed. Not just for power generating armatures but also in regular motors. I should say rotor mainly because what I'm referring to is the coil and the material that is being used to conduct the magnetism from it. I know its one in and one out for energy. What I want to know is that would making a lighter armature be like the difference between driving a semi up a hill vs a fiat 500 when creating electricity? What does reducing the weight of an armature do for power requirements?

It takes more energy to run a heavy generator up to speed, but once it is running at speed there is only friction to overcome (mechanically) and the difference is insignificant.

Furthermore the heavier armature has a better flywheel effect.

Edited September 2, 2018 by studiot
spelling

[Theredbarron]

14 minutes ago, studiot said:

It takes more energy to run a heavy gererator up to speed, but once it is running at speed there is only friction to overcome (mechanically) and the difference is insignificant.

Furthermore the heavier armature has a better flywheel effect.

So the momentum is what your losing when the weight is reduced then? What about if it was a motor would the momentum be needed?

[studiot]

Yes but it is more usual to use torque and angular velocity when considering shaft work.

Torque is (time) rate of change of angular velocity.

just as

Force is (time) rate of change of linear velocity.

[swansont]

14 hours ago, Theredbarron said:

Is it still impossible to multiply electrical energy?

Define "multiply"

Energy is a conserved quantity.

[studiot]

40 minutes ago, swansont said:

Define "multiply"

Energy is a conserved quantity.

I don't think the OP actually means 'mathematically multiply', I think he wants to improve the generation efficiency.

However he is probably unaware that electrical generators are pretty damn efficient already 90% or better so any multiplication factor is going to be less than 1.2.

[Theredbarron]

I mean 800 watts in and 1600 watts out by multiply. 

And yes armature effecientcy 

9 hours ago, studiot said:

I don't think the OP actually means 'mathematically multiply', I think he wants to improve the generation efficiency.

However he is probably unaware that electrical generators are pretty damn efficient already 90% or better so any multiplication factor is going to be less than 1.2.

Yes this too. Im actually looking at cutting the rotating mass by half. I'm trying to find information on what the effects would be by that much weight? Either power generators or motors effects. 

[Strange]

6 hours ago, Theredbarron said:

I mean 800 watts in and 1600 watts out by multiply

That is power not energy. You could, in principle, get 1600W out for half the time. 

But, you can’t get out more energy than you put in. 

[studiot]

Yes you need to know the difference between power and energy.

This is very important.

As regards your quest for efficiency

You are a practical chap so here is a short description of testing a small electrical machine, along with a sample calculation from actual measurements.

 

There are many factors to consider when designing an armature.

Have you heard of eddy currents?

Materials play an important part so that magnetic saturation does not occur or the device will run out of steam prematurely.

Do you know why the windings are buried in slots in the armature ?

 

Here is a chart shown losses in a/c machines.

 

 

This should provide a start and some food for thought.

 

[Theredbarron]

4 hours ago, studiot said:

Have you heard of eddy currents?

Materials play an important part so that magnetic saturation does not occur or the device will run out of steam prematurely.

Do you know why the windings are buried in slots in the armature 

 

 

From what I'm getting is eddy currents are bad. Sounds like They act as to balance the magnetic effects in a conductor or feed back magnetism. Trying to put the electrons where they originally are maybe. Maybe power factor effects as well.

The one I want to redesign is more alternator rotor. The coil itself is parallel to the shaft and it uses the conductor to shape the magnetic field. I wanted to separate the movement of the coil to the conductor that is uses to direct the magnetic field. So a 2 piece armature.  I believe that this is where I would run into eddy currents from the movement between the coil of the armature and the conductor that's wrapped around it. 

I kind of always thought that the slots were there to accurately place the coil windings for the most amount of magnetic effect in the conductor that its using. It make me think this would help in creating clean ac voltage. 

6 hours ago, Strange said:

That is power not energy. You could, in principle, get 1600W out for half the time. 

But, you can’t get out more energy than you put in. 

So the energy required to turn a shaft and create 745 watts would have to be equal to that not including losses correct?

Edited September 3, 2018 by Theredbarron
added words

[studiot]

12 minutes ago, Theredbarron said:

I kind of always thought that the slots were there to accurately place the coil windings for the most amount of magnetic effect in the conductor that its using. It make me think this would help in creating clean ac voltage. 

By placing the windings in slots they are actually moved out of the position of maximum field.

However if the armature was smooth and the windings were on the surface, there would be considerable sideways force generated on them (it's still there in the slots) so they would tend to slip sideways.

It doesn't matter that the windings are out of the most intense part of the field because they still cut the same number of field lines.

The slots are there to stop the sideways slip and hold the windings in place.

 

The solution to eddy currents is sheet or segmental construction since the eddy conduction path is broken at every interface.

However there is a cost-benefit assessment to be made becasue the thinner the laminations the more expensive they are.

 

The principle of brake testing applies to all rotary machinery, I had thought you would be interested.

 

 

[Theredbarron]

1 minute ago, studiot said:

The principle of brake testing applies to all rotary machinery, I had thought you would be interested.

 

 

I am very interested. I am going to use this on what I have now so I can get numbers. My testing right now is showing 336 watts just at rated speed no extra load. I'm only starting with 800 so that's quite the loss. This is why I want to look at armatures. 

10 minutes ago, studiot said:

 The solution to eddy currents is sheet or segmental construction since the eddy conduction path is broken at every interface.

However there is a cost-benefit assessment to be made becasue the thinner the laminations the more expensive they are.

 

 

I am definitely considering this.

[studiot]

 

12 minutes ago, Theredbarron said:

My testing right now is showing 336 watts just at rated speed no extra load. I'm only starting with 800 so that's quite the loss.

 

What do you mean no extra load?

How are you measuring the 800 and 336 watts?

[Theredbarron]

 

4 minutes ago, studiot said:

 

 

What do you mean no extra load?

How are you measuring the 800 and 336 watts?

The motors capability is 800. The losses just to run the setup is 336. No extra electrical mechanical load being applied on the output of what I'm working on.

That only leave 464 left over. I pretty much cut my capability almost in half just due to rotating mass. 

Its like trying to race while driving a tractor trailer that's fully loaded. 

[studiot]

So 800 is the rating.

It will not draw that amount under no load.

 

This is a very very important electrical principle:_

 

The power or total energy is determined by the load, not by the supply.
The rating on a device is the maximum  that might be drawn. It does not imply that amount will (ever) be drawn from the supply.

 

So how did you measure the 336?

Edited September 3, 2018 by studiot

[Theredbarron]

1 minute ago, studiot said:

So 800 is the rating.

It will not draw that amount under no load.

 

This is a very very important electrical principle:_

 

The power or total energy is determined by the load, not by the supply.
The rating on a device is the maximum  that might be drawn. It does not imply that amount will (ever) be drawn from the supply.

 

So how did you measure the 336?

I a load I have placed on it. I used an amp clamp and came up with 1.4 amp at 240v single phase. The motor is moving something its just how much it cost in power to move it is what I want to change. 

It actually dropped a little lower then that once I changed the motor speed to its most efficient point which was 94% rated speed. 

[studiot]

So it's not on bo load?

[Theredbarron]

1 minute ago, studiot said:

So it's not on bo load?

Correct just no extra load

[studiot]

You should always compare zero load to some load to full load.

Not part load to full load.

 

Now I need to introduce you to power factor.

The actual power your motor is consuming is volts x amps x power factor.

 

The PF is typically 0.85 (85%) for a well designed motor

So you are actually using 1.4*240*0.85 = 286 real world watts.

I won't go into the technical details of why, just say this is due to the way volts and amps are measured in inductive machinery.

Here is an american site with some practical information.

https://www.ecmweb.com/motors/motor-efficiency-power-factor-and-load

[Theredbarron]

7 minutes ago, studiot said:

You should always compare zero load to some load to full load.

Not part load to full load.

 

Now I need to introduce you to power factor.

The actual power your motor is consuming is volts x amps x power factor.

 

The PF is typically 0.85 (85%) for a well designed motor

So you are actually using 1.4*240*0.85 = 286 real world watts.

I won't go into the technical details of why, just say this is due to the way volts and amps are measured in inductive machinery.

Here is an american site with some practical information.

https://www.ecmweb.com/motors/motor-efficiency-power-factor-and-load

I thought power factor was for 3 phase? By real world you mean what its delivering at its shaft in power?

[Theredbarron]

I remember now. Power factor is for inductive load and not resistive. So if the rotating mass is decreased what would this do to the power factor? 

[Theredbarron]

I know this might sound dumb but just to make sure are motors rated at the speed that it states? I'm pretty sure that if it states so many watts and whatever rpm means that it creates that force at the rpm it shows even though it may draw more then the rated power correct? 

[Theredbarron]

I'm not seeing why its impossible to multiply power output. I get that it takes as much in or more to get the same out but that applies to the power that's rotating that shaft. Cars have been using this type of physics for years now. Its too easy to just say what I'm talking about. The other thing is looping the electricity. I know that you cant just plug power back to the start. It takes energy to create energy so it would just stall itself out if you can even get it to run itself at all. But no one ever said that the energy had to go back to the same spot that it came from. So now all I'm waiting on is some parts and some tests so see what I can do with it. If motor ratings are correct and cars have been using this same logic that I'm about to use then what could go wrong!