Does Not Equate

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MiguelM

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Location
Angel Fire, NM
My father-in-law has a PV power back-up system installed. The voltage supplied by each inverter is rated for 105V (it is a stacked system) but metered, shows 99-101V. He wants to use it to run his well pump. My concern is that the if the well pump motor is not supplied with 120-110V, it may overheat and "burn up". I don't have the specs on the well pump.
Here's where my limited knowledge derails me:

My theory in electrical tells me that when the voltage is lower, the current being pulled to supply the power required by the motor should increase. However,I ran the well pump on the inverters and checked the voltage at the ac supply to the pump. The meter read 99 volts on each leg or 199 across both. I then metered the current through each leg with a loop meter and read 5.24A.
I did the same, this time with the supply to the pump coming from the residential supply. Each leg read 124V or 250V across both legs. The current read at 5.9A through each leg.
In other words, the current being pulled by the motor was higher with a lower voltage-opposite of what I expected.
I did the same experimenet with a halogen light and again with a shop vac. The same result. The a/c current drawn by each of these items was higher when the voltage supplied was lower.
Please explain this before I have to pull out my electrical theory books from the attic.
Thanks.
Miguel M.
 

K8MHZ

Senior Member
Location
Michigan. It's a beautiful peninsula, I've looked
Occupation
Electrician
My concern is that the if the well pump motor is not supplied with 120-110V, it may overheat and "burn up".

This is a likely possibility.

If the motor does not get sufficient voltage to operate, the rotor will spin slower than designed. Since the spinning of the rotor creates reverse electromagnetic force which acts as impedance, the result of less REMF is that more of the supplied voltage is being seen at the windings, which are very low impedance. The result of the low impedance of the windings not being counteracted by the impedance from the REMF is what creates excess heat.

My theory in electrical tells me that when the voltage is lower, the current being pulled to supply the power required by the motor should increase.

Your theory is incorrect. You are making the assumption that the motor will have the same output regardless of voltage, which is not true. If you drop voltage from a linear supply to a motor, the current will drop as well, the output will drop and the heat will rise because more of the power is being turned into heat than mechanical output.

In other words, the current being pulled by the motor was higher with a lower voltage-opposite of what I expected.
I did the same experimenet with a halogen light and again with a shop vac. The same result. The a/c current drawn by each of these items was higher when the voltage supplied was lower.
Please explain this before I have to pull out my electrical theory books from the attic.

What you are observing is the PV inverter making adjustments to seek the maximum power point. A PV inverter is not linear. It will change the voltage and current to seek the best combination, the one that provides the most amount of power. If you were using a wind turbine and an inverter designed for that purpose, you would not see the current rise as the voltage drops as wind inverters do not seek maximum power points.

Try the same experiment with a linear 99 volt power supply. In that case, you will see that the current and the voltage are proportional.
 
Last edited:

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
110622-0838 EDT

MiguelM:

Get a Variac and then run your experiment from power company power (likely a fairly good sine wave). Your inverter output may be only a crude approximation to a sine wave.

A tungsten filament incandescent bulb is essentially a pure resistance, but its resistance is a function of the ambient temperature, received radiation, and the current thru the filament.

For a typical bulb in a normal operating range as a load it is between an invariant resistance and a constant current load. You can calculate current vs voltage from my plot at
http://beta-a2.com/EE-photos.html P9 --- Photo 2769A1
The current ranges from 0.45 A at 60 V (133 ohms) to 0.625 A at 120 V (192 ohms).
Because a filament bulb has a long thermal time constant relative to 60 Hz you see what is approximately a constant resistance load at any particular applied RMS current.

RMS current and RMS voltage to a resistive load will be related by Ohm's law when the resistance is a constant throughout the AC cycle. This is independent of waveform. May not be so for a non-RMS meter.

If the load is not resistive, then RMS meters may also fail to be related by Ohm's law.

If you compare a load current from a source that is not a sine wave at one voltage with the load current from a sine wave source at a different voltage, then you may expect unexpected results. I believe this was your experiment.

.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
110622-0915 EDT

MiguelM:

K8MHZ has brought up some good points. He is assuming a PV inverter that is grid tied. In that case the inverter is designed as a current source with optimization to output maximum power available from the PV array.

I assumed your inverter was a constant voltage inverter.

K8MHZ I disagree with your motor description in a normal operating range.

First, consider a DC motor which is the simplest to describe. Also assume the motor is 100 % efficient and the load is a constant power load independent of speed. Then input electrical power has to equal output mechanical power. It is quite clear in this case that input current rises as input voltage decreases. Add reasonable losses and the results are not much different.

Second, change to an AC synchronous motor. Speed is absolutely constant until you overload the motor. Again the same result as the DC motor analysis. Input electrical power has to equal output mechanical power. Not much different when you consider a low rotor resistance induction motor. Speed is reasonably constant with changing applied voltage.

.
 

MiguelM

Member
Location
Angel Fire, NM
Thanks for the feedback-

Thanks for the feedback-

The inverter attempts to emulate what's termed in the manual as a "square sine wave". Your explanation and that of "sparky's" may be the reason for the behavior of current / voltage not being what I expected. I thank you.
Just for clarification, i am summarizing what I observed. I believe I confused the issue in my initial post by describing a scenario that contradicts my observation and for that, I apologize.

Summary:
On Battery power
Inverter output = 99-101V at plug-in, each leg
Measured amps at motor supply = 5.24A each leg
On residential supply
124V at plug-in, each leg
Measured amps at motor supply = 5.8A - 6A.
I expected lower voltage, higher amps
I received higher voltage, higher amps

Same behavior with the shop vac and halogen light.
I believe I understand Ohm's law, but it does not follow in my observation. The non-steady-state behavior of the inverter seems the probable explanation.

Just to dirty up the scenario more, what if my inverter output is at 50Hz versus 60Hz? Could the frequency difference account for less amps being pulled?
I appreciate all you who try to help.


110622-0838 EDT

MiguelM:

Get a Variac and then run your experiment from power company power (likely a fairly good sine wave). Your inverter output may be only a crude approximation to a sine wave.

A tungsten filament incandescent bulb is essentially a pure resistance, but its resistance is a function of the ambient temperature, received radiation, and the current thru the filament.

For a typical bulb in a normal operating range as a load it is between an invariant resistance and a constant current load. You can calculate current vs voltage from my plot at
http://beta-a2.com/EE-photos.html P9 --- Photo 2769A1
The current ranges from 0.45 A at 60 V (133 ohms) to 0.625 A at 120 V (192 ohms).
Because a filament bulb has a long thermal time constant relative to 60 Hz you see what is approximately a constant resistance load at any particular applied RMS current.

RMS current and RMS voltage to a resistive load will be related by Ohm's law when the resistance is a constant throughout the AC cycle. This is independent of waveform. May not be so for a non-RMS meter.

If the load is not resistive, then RMS meters may also fail to be related by Ohm's law.

If you compare a load current from a source that is not a sine wave at one voltage with the load current from a sine wave source at a different voltage, then you may expect unexpected results. I believe this was your experiment.

.
 

K8MHZ

Senior Member
Location
Michigan. It's a beautiful peninsula, I've looked
Occupation
Electrician
110622-0915 EDT

MiguelM:

K8MHZ has brought up some good points. He is assuming a PV inverter that is grid tied. In that case the inverter is designed as a current source with optimization to output maximum power available from the PV array.

I assumed your inverter was a constant voltage inverter.

K8MHZ I disagree with your motor description in a normal operating range.

First, consider a DC motor which is the simplest to describe. Also assume the motor is 100 % efficient and the load is a constant power load independent of speed. Then input electrical power has to equal output mechanical power. It is quite clear in this case that input current rises as input voltage decreases. Add reasonable losses and the results are not much different.

Second, change to an AC synchronous motor. Speed is absolutely constant until you overload the motor. Again the same result as the DC motor analysis. Input electrical power has to equal output mechanical power. Not much different when you consider a low rotor resistance induction motor. Speed is reasonably constant with changing applied voltage.

.

Then why do we need VFD's to slow motors down without burning them up?

If we lower the voltage, we must also change the frequency in order to reduce slip. The greater the slip, the less REMF and the lower the opposition to current flow. And, since the rotor is now producing less mechanical power, a greater proportion of power is being wasted as heat.

A DC motor is a totally different animal and it's speed is not dependent on supply frequency. A motor designed for DC most likely will operate on AC, whereas the opposite is not true. (Old portable drill motors had AC/DC brush motors in them).

This has come up before and I have said that someday I will try to locate my motor book from school. It has a much better way of explaining what goes on in an AC motor when the supply voltage is dropped without changing the frequency. The problem is, I have about 100 books in several piles from my move out here and haven't sorted them out yet.
 

K8MHZ

Senior Member
Location
Michigan. It's a beautiful peninsula, I've looked
Occupation
Electrician
I received higher voltage, higher amps

That is consistent with Ohm's Law.

The only way that volts and amps are not proportional through a fixed load is by manipulating the power supply, like a grid-tied inverter does.

A motor can't really be considered a fixed load, but unless there is some other form of voltage-current bias control, it still behaves as expected, just not linearly. That is, if you lower the voltage the current will also lower.

I thought what you were observing was changes being made by the inverter, but it appears not to be that way and what you were observing is normal, common and acceptable.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
110622-1228 EDT

K8MHZ:

I suggest you get a book like "Alternating-Machinery", by Bailey and Gault. It is apparently available at Western Michigan University in the Dwight B. Waldo Library. To completely study this book would take way more than one semester. It has the basic theory, it is 1951, and does not directly cover VFDs.

AC and DC machines are not totally different in some respects. One being that input power = losses + load power.

A DC motor can be made to have a very flat torque-speed curve or a very steep one. In either case the input-output power relationship still exists, but there maybe some small change in the loss part relative to the load part.

One important way to analyze many devices is from the conservation of energy perspective, instead of getting hung up in all sorts of other details..

Don't make comparisons of shop-vac motors with induction motors. Shop-vac motors are DC series motors, and happen to work on AC.

The reason VFDs are used with AC synchronous, and induction motors is that the rotating magnetic field or impulsed field is proportional to frequency and by changing frequency you can change the speed of rotation of the field, and thus of the motor.

Basically in a synchronous or induction motor the rotor follows the rotation of the spatial magnetic field. In the induction motor it slips a little bit and slip is proportional to load, and in the synchronous motor there is no slip, but there is an angular phase shift of the rotor relative to the rotating magnetic field that is proportional to torque.

Not all DC motors will work on AC.

.
 

Besoeker

Senior Member
Location
UK
I don't quite understand your explanation or your question?

This is simple OHM'S LAW. It always works.
For a passive resistance of constant value.
An electric motor is not a passive resistance of constant value.
You can't treat it that way.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
110622-1257 EDT

MiguelM:

I believe your measurement problem, and ability to correlate with what you expect, results from the characteristics of the instruments, the loads, and the different waveforms.

To determine whether you have a problem using the inverter probably would be best evaluated by operation of the load under full load conditions and evaluate temperature rise in the load and the inverter. Not easy, but maybe approximately doable.

.
 

K8MHZ

Senior Member
Location
Michigan. It's a beautiful peninsula, I've looked
Occupation
Electrician
110622-1228 EDT

K8MHZ:

I suggest you get a book like "Alternating-Machinery", by Bailey and Gault. It is apparently available at Western Michigan University in the Dwight B. Waldo Library. To completely study this book would take way more than one semester. It has the basic theory, it is 1951, and does not directly cover VFDs.

AC and DC machines are not totally different in some respects. One being that input power = losses + load power.

A DC motor can be made to have a very flat torque-speed curve or a very steep one. In either case the input-output power relationship still exists, but there maybe some small change in the loss part relative to the load part.

One important way to analyze many devices is from the conservation of energy perspective, instead of getting hung up in all sorts of other details..

Don't make comparisons of shop-vac motors with induction motors. Shop-vac motors are DC series motors, and happen to work on AC.

The reason VFDs are used with AC synchronous, and induction motors is that the rotating magnetic field or impulsed field is proportional to frequency and by changing frequency you can change the speed of rotation of the field, and thus of the motor.

Basically in a synchronous or induction motor the rotor follows the rotation of the spatial magnetic field. In the induction motor it slips a little bit and slip is proportional to load, and in the synchronous motor there is no slip, but there is an angular phase shift of the rotor relative to the rotating magnetic field that is proportional to torque.

Not all DC motors will work on AC.

.

I agree with everything you said, but I think you left out something.

The reason VFDs are used with AC synchronous, and induction motors is that the rotating magnetic field or impulsed field is proportional to frequency and by changing frequency you can change the speed of rotation of the field, and thus of the motor.

In a VFD, along with the changing of the frequency, there is also a change in voltage. As the frequency of the field slows and causes the rotor to slow, the applied voltage has to be dropped as the opposing REMF has been lowered. If only the frequency was lowered and not the voltage, the motor would eventually fail just like not changing the frequency and applying too much voltage.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
110622-1945 EDT

K8MHZ:

Sure the voltage has to be reduced because of magnetic saturation. Fundamentally there is a volt-time integral problem. But the real point is that speed is a function of frequency, and not of voltage. Saturation problems does not change the basic theory of speed vs frequency.

If you have a high resistance rotor induction motor then the speed torque curve is quite different than for a low resistance rotor. Thus, with certain types of load and fixed frequency excitation speed is adjustable with input voltage or adjustment of the rotor resistance by external means connected to the rotor thru slip-rings.

.
 
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