Author Topic: Losses in inductors  (Read 563 times)

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Offline ZeroResistance

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Losses in inductors
« on: January 21, 2019, 11:08:58 am »
So lets say I have a electromechanical relay with a coil rated for 12V and 12ohm, which means its rated wattage is 12W.
If I give it 12VDC, a current of 1A flows thru the coil and it the magnetic energy attracts the contacts of the relay.
However when I calculate the power I^2 * R lost in the resistance in the relay it turns out to be same as the rated power 12W.
My question is if I give the relay 12w some of it is converted to mechanical energy and some is lost in the winding resistance.
So would it be correct to say that if I use I^2 * R that would the power lost in the resistance of the winding of the relay.
But then how does one deduce how much energy was converted to mechanical and how much was lost?

TIA

 

Online Zero999

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Re: Losses in inductors
« Reply #1 on: January 21, 2019, 11:34:59 am »
Electrical energy is only converted to mechanical energy when the power is initially applied to the coil. Once the relay is activated, all of the energy from that point onwards is converted to heat.

Relays also have hysteresis. 75% of the rated voltage is required to activate the relay, but after that, the voltage can be reduced to around 10% of the rating and the relay will remain energised.

If the force and distance of travel required to switch the relay contacts, is known, then the amount of energy required to activate the relay can be calculated, which can be used to work out the efficiency.
 
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Offline ogden

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Re: Losses in inductors
« Reply #2 on: January 21, 2019, 12:13:12 pm »
If the force and distance of travel required to switch the relay contacts, is known, then the amount of energy required to activate the relay can be calculated, which can be used to work out the efficiency.

Relay is one of the most inefficient "motors" known. Who cares - it is 0.00000001% or 0.000001% efficient? Number is irrelevant anyway.
 

Offline ZeroResistance

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Re: Losses in inductors
« Reply #3 on: January 21, 2019, 02:35:01 pm »
Electrical energy is only converted to mechanical energy when the power is initially applied to the coil. Once the relay is activated, all of the energy from that point onwards is converted to heat.

Relays also have hysteresis. 75% of the rated voltage is required to activate the relay, but after that, the voltage can be reduced to around 10% of the rating and the relay will remain energised.

If the force and distance of travel required to switch the relay contacts, is known, then the amount of energy required to activate the relay can be calculated, which can be used to work out the efficiency.

Isn't the current flowing into the coil and into its resistance while the relay is doing work in attracting the contact?
 

Offline schmitt trigger

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Re: Losses in inductors
« Reply #4 on: January 21, 2019, 02:42:22 pm »
This book, has an excellent explanation related to the energy conversion on a solenoid (which is what a relay is, a solenoid pushing some contacts):

https://www.amazon.com/Electric-Machinery-Fitzgerald/dp/0073660094/ref=sr_1_2?s=books&ie=UTF8&qid=1548081614&sr=1-2&keywords=electric+machinery+fitzgerald
 
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Offline soldar

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Re: Losses in inductors
« Reply #5 on: January 21, 2019, 08:35:45 pm »
Isn't the current flowing into the coil and into its resistance while the relay is doing work in attracting the contact?
No work is done. Work is force multiplied by distance. Distance is zero.

In other words, the relay holding the armature in place is doing as much work as a shelf holding a book in place.
All my posts are made with 100% recycled electrons and bare traces of grey matter.
 
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Online T3sl4co1l

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Re: Losses in inductors
« Reply #6 on: January 21, 2019, 10:55:35 pm »
You are perhaps confusing work (energy e.g. in joules) with power (e.g. in watts).

This is a necessary aspect of the example, as given: it contains energy and power sinks -- the inductance and resistance, respectively.

We must separate these two elements, if we are to determine where the energy goes, and where the work goes.

We can assume a ZeroResistance coil for arguments sake. (Convenient, eh? :D )  This must be supplied from a current limited source, so that when turned on, its inductance charges up to some current, and eventually has zero voltage drop (because dI/dt = 0 and nothing is changing magnetically or mechanically).  And when turned off, its current (and motion) goes to zero, and eventually its voltage as well.

In particular, the voltage drop is only the EMF (electromotive force, induced voltage) of the coil and magnetic circuit.

(In practice, we can separate these by knowing the coil resistance, sensing the coil current, and constructing the voltage drop due to resistance as Vr = I*R.  Subtracting this out, we are left with the EMF of the coil -- and residual errors, which are not to be underestimated in such a situation.  In a real coil, the coil temperature will be varying, and its resistance varies with frequency as well.  A fairly complicated equivalent circuit would be needed to null this; in short, better not to try, but to understand it from a theoretical basis instead.)

So: coil, some air gap, the relay armature, and the pole piece wrapping around the coil.  When off, the initial inductance is modest, due to the air gap.  We can calculate this, given some approximations, and typical geometry.  (For most relays, the geometry is close to a 'U' or 'E' core, with the armature being an 'I' piece, gapped away from the 'U'/'E' piece by the air gap.  The core is soft iron, mu_r large enough that we can ignore its length.)

As current is applied, voltage is developed, according to: V = L * dI/dt.  Gradually, the magnetic field increases, which puts a force across the air gap, according to: p = B^2 / (2*mu_0), the Maxwell strain.

This gives us a pressure.  Integrated over the cross-section of the air gap, it gives us the force pulling the relay closed!  A pressure also has units of energy density.  integrated over the air gap volume, this gives the total energy stored in the air gap.    Neat, huh?

Eventually, the force exceeds the spring force holding the armature back, and it begins to move.  As it closes, the gap shortens, which causes the inductance to increase.  We refer back to the inductor equation, V = L * dI/dt.  If dI/dt is constant, then V will tick up when the armature begins to move -- in fact, V will be nearly proportional to the velocity of the armature.  But V is already proportional to dI/dt, so how can we tell which is which?

Well, since we know we are performing real work here, there is an energy loss due to something other than the magnetic field itself (namely, something moving).

We can perform the same decomposition, splitting the voltage into the component due to inductance, and the component due to work performed.

Actually, I don't think it can be the same decomposition, because we are varying a parameter, not simply adding components together.  Well, welcome to the realities of nonlinear analysis. :)

Suffice it to say, the fact that V is higher than our initial level, while dI/dt is ramping steadily up, is necessarily performing work, and the integral V*I dt during that ramp, gives us the total energy stored.

When the relay fully closes, the air gap goes approximately to zero, and the voltage generated by a given dI/dt will be very large (i.e., the inductance is very high).  That means the energy stored in the magnetic field is very small.  (That's kind of a counterintuitive way to put it; maybe it helps to say that, for a given applied voltage, the dI/dt, or Irms at a given frequency, is small, therefore the reactive power is small, and so too, the energy stored per unit time is small.)  With little energy in the magnetic field, we know that the energy we did deliver, must've gone into mechanical work!

For a lossless armature, that work will have gone into the spring, so we expect to get it back when we reverse this process.  The waveform will be identical to what we started with, just played back in reverse.  (Well, assuming a zero mass armature, too.  It will accelerate differently this time, just because it's starting from a different position, and different spring force, and fringing around the air gap and all that.  To a first approximation, and various other ideal constructions, it can be the same.  In any case, the total area under the curve (as flux, integral V dt, and as energy, integral V I dt) will be the same.)

In practice, again there are losses -- the armature accelerates, then stops suddenly as it slams into the contacts and coil, bouncing several times in the process.  The bounces dissipate mechanical energy as friction, and maybe acoustic and viscous energy in the air too.  So the kinetic energy of the armature is lost, and more energy will be required to pull it in, than is returned (by the spring) on the way out (which again loses energy when it slams into the other stop).

In general, these are called reluctance machines.  The attractive force of magnetized, permeable cores is varied to affect mechanical work.  Stepper motors are a continuous (rotating) application of this; relays and solenoids are a single-acting linear application.

Now that this has been covered -- we can consider an interesting riddle.  Back in the mid 2000s, there was a heavily promoted overunity motor, Stoern Orbo I think it was?  It had a rotor of permanent magnets, a stator of saturable cores, and nothing else (see-through acrylic structure).  The claim was that the saturable cores attract the magnets, then are driven into saturation (no more attraction), and that the energy used to saturate the cores is zero, because they are simple inductors and that energy is returned when they are discharged before the next magnet approaches.  Can you see what's wrong with this claim? :)

Tim
Seven Transistor Labs, LLC
Electronic design, from concept to prototype.
Bringing a project to life?  Send me a message!
 
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Offline Nerull

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Re: Losses in inductors
« Reply #7 on: January 22, 2019, 03:25:55 am »
Electrical energy is only converted to mechanical energy when the power is initially applied to the coil. Once the relay is activated, all of the energy from that point onwards is converted to heat.

Relays also have hysteresis. 75% of the rated voltage is required to activate the relay, but after that, the voltage can be reduced to around 10% of the rating and the relay will remain energised.

If the force and distance of travel required to switch the relay contacts, is known, then the amount of energy required to activate the relay can be calculated, which can be used to work out the efficiency.

Isn't the current flowing into the coil and into its resistance while the relay is doing work in attracting the contact?

A rock sitting on the ground is attracted towards Earth, but no work is being done on it. Similarly, two magnets which are stuck together are mutually attracted to each other, but are doing no work.
 
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Online Zero999

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Re: Losses in inductors
« Reply #8 on: January 22, 2019, 09:35:50 am »
Electrical energy is only converted to mechanical energy when the power is initially applied to the coil. Once the relay is activated, all of the energy from that point onwards is converted to heat.

Relays also have hysteresis. 75% of the rated voltage is required to activate the relay, but after that, the voltage can be reduced to around 10% of the rating and the relay will remain energised.

If the force and distance of travel required to switch the relay contacts, is known, then the amount of energy required to activate the relay can be calculated, which can be used to work out the efficiency.

Isn't the current flowing into the coil and into its resistance while the relay is doing work in attracting the contact?

A rock sitting on the ground is attracted towards Earth, but no work is being done on it. Similarly, two magnets which are stuck together are mutually attracted to each other, but are doing no work.
That's not the case with a relay turning on. When a relay is being activated, work is being done. There will be losses in the coil, due to the resistance. T3sl4co1l has provided a very good explanation of this.
 

Offline ZeroResistance

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Re: Losses in inductors
« Reply #9 on: January 27, 2019, 07:08:36 am »
You are perhaps confusing work (energy e.g. in joules) with power (e.g. in watts).

This is a necessary aspect of the example, as given: it contains energy and power sinks -- the inductance and resistance, respectively.

We must separate these two elements, if we are to determine where the energy goes, and where the work goes.

We can assume a ZeroResistance coil for arguments sake. (Convenient, eh? :D )  This must be supplied from a current limited source, so that when turned on, its inductance charges up to some current, and eventually has zero voltage drop (because dI/dt = 0 and nothing is changing magnetically or mechanically).  And when turned off, its current (and motion) goes to zero, and eventually its voltage as well.

In particular, the voltage drop is only the EMF (electromotive force, induced voltage) of the coil and magnetic circuit.

(In practice, we can separate these by knowing the coil resistance, sensing the coil current, and constructing the voltage drop due to resistance as Vr = I*R.  Subtracting this out, we are left with the EMF of the coil -- and residual errors, which are not to be underestimated in such a situation.  In a real coil, the coil temperature will be varying, and its resistance varies with frequency as well.  A fairly complicated equivalent circuit would be needed to null this; in short, better not to try, but to understand it from a theoretical basis instead.)

So: coil, some air gap, the relay armature, and the pole piece wrapping around the coil.  When off, the initial inductance is modest, due to the air gap.  We can calculate this, given some approximations, and typical geometry.  (For most relays, the geometry is close to a 'U' or 'E' core, with the armature being an 'I' piece, gapped away from the 'U'/'E' piece by the air gap.  The core is soft iron, mu_r large enough that we can ignore its length.)

As current is applied, voltage is developed, according to: V = L * dI/dt.  Gradually, the magnetic field increases, which puts a force across the air gap, according to: p = B^2 / (2*mu_0), the Maxwell strain.

This gives us a pressure.  Integrated over the cross-section of the air gap, it gives us the force pulling the relay closed!  A pressure also has units of energy density.  integrated over the air gap volume, this gives the total energy stored in the air gap.    Neat, huh?

Eventually, the force exceeds the spring force holding the armature back, and it begins to move.  As it closes, the gap shortens, which causes the inductance to increase.  We refer back to the inductor equation, V = L * dI/dt.  If dI/dt is constant, then V will tick up when the armature begins to move -- in fact, V will be nearly proportional to the velocity of the armature.  But V is already proportional to dI/dt, so how can we tell which is which?

Well, since we know we are performing real work here, there is an energy loss due to something other than the magnetic field itself (namely, something moving).

We can perform the same decomposition, splitting the voltage into the component due to inductance, and the component due to work performed.

Actually, I don't think it can be the same decomposition, because we are varying a parameter, not simply adding components together.  Well, welcome to the realities of nonlinear analysis. :)

Suffice it to say, the fact that V is higher than our initial level, while dI/dt is ramping steadily up, is necessarily performing work, and the integral V*I dt during that ramp, gives us the total energy stored.

When the relay fully closes, the air gap goes approximately to zero, and the voltage generated by a given dI/dt will be very large (i.e., the inductance is very high).  That means the energy stored in the magnetic field is very small.  (That's kind of a counterintuitive way to put it; maybe it helps to say that, for a given applied voltage, the dI/dt, or Irms at a given frequency, is small, therefore the reactive power is small, and so too, the energy stored per unit time is small.)  With little energy in the magnetic field, we know that the energy we did deliver, must've gone into mechanical work!

For a lossless armature, that work will have gone into the spring, so we expect to get it back when we reverse this process.  The waveform will be identical to what we started with, just played back in reverse.  (Well, assuming a zero mass armature, too.  It will accelerate differently this time, just because it's starting from a different position, and different spring force, and fringing around the air gap and all that.  To a first approximation, and various other ideal constructions, it can be the same.  In any case, the total area under the curve (as flux, integral V dt, and as energy, integral V I dt) will be the same.)

In practice, again there are losses -- the armature accelerates, then stops suddenly as it slams into the contacts and coil, bouncing several times in the process.  The bounces dissipate mechanical energy as friction, and maybe acoustic and viscous energy in the air too.  So the kinetic energy of the armature is lost, and more energy will be required to pull it in, than is returned (by the spring) on the way out (which again loses energy when it slams into the other stop).

In general, these are called reluctance machines.  The attractive force of magnetized, permeable cores is varied to affect mechanical work.  Stepper motors are a continuous (rotating) application of this; relays and solenoids are a single-acting linear application.

Now that this has been covered -- we can consider an interesting riddle.  Back in the mid 2000s, there was a heavily promoted overunity motor, Stoern Orbo I think it was?  It had a rotor of permanent magnets, a stator of saturable cores, and nothing else (see-through acrylic structure).  The claim was that the saturable cores attract the magnets, then are driven into saturation (no more attraction), and that the energy used to saturate the cores is zero, because they are simple inductors and that energy is returned when they are discharged before the next magnet approaches.  Can you see what's wrong with this claim? :)

Tim


Apologies for the delay in replying.
It took me some time digesting the information provided.
Having said that, you have dished out yet another masterpiece. This is simply amazing!
Im still unable to comprehend quite a bit of the data provided so just let me regurgitate it a bit longer.
And I will get back with some comments!
Thanks again!!
 


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