75 kVA transformer load with no load

[gar]

081118-1148 EST

iwire:

The KWH meter is reading the cumulative energy consumed. For each small increment of time the energy is measured (watts times seconds) and that is added to the previously accumulated value. The adding is accomplished by the mechanical counter (the gear box and pointers). On a mechanical watt-hour meter the rate of rotation (speed - rpm) of the disk is a measure of the instantaneous power being used.

The instantaneous power is V*I*PF. The power measured by the rate of rotation of the KWH meter will differ from the calculation of V times A by whatever the PF (power factor) is.

I can not directly measure power factor, but I can measure VA and watts and from this determine PF.

There is a relationship between VA and watts but it depends upon the waveforms and their phase relationship.

.

 

[winnie]

When we talk about AC circuits, we generally speak in terms of RMS voltage and RMS current. The RMS value is a _single_ number that represents the continuously _changing_ voltage or current flowing through the circuit. It is analogous to using the 'average' to represent a large set of numbers.

When we talk about DC circuits, we know that power is simply volts times amps. But in an AC circuit, if you multiply RMS volts * RMS amps, you don't correctly calculate the power delivered to the load.

There are different causes of power factor; I'll just talk about 'displacement power factor' since that gets the basic idea across:

The reason is that in an AC circuit you need to know the phase relationship between the current and the voltage, and the RMS numbers tell you nothing about this phase relationship. If the phase relationship between the current and the voltage is poor, then lots of current can flow, with very little power being delivered to the load.

For things like transformers and motors, energy is actually stored in the machine magnetic field. This magnetic field is changing with the line frequency...which means that energy needs to be stored and released at the line frequency. If you look closely at the power flow to something like a transformer, what you find with a 'poor' power factor is that lots of power _is_ being delivered to the load, but only for part of the AC cycle, and that lots of power is actually flowing from the load back to the source in the other part of the AC cycle. In the transformer magnetizing case, what we find is that for 1/4 of the cycle, lots of power is going to store energy in the magnetic field, then for 1/4 cycle lots of power is being released from the magnetic field and going back to the source, and then this repeats.

See http://www.allaboutcircuits.com/vol_2/chpt_11/1.html for both more math, and for some good pictures of the relationship between voltage, current, and power delivered to the load.

-Jon

 

[jim dungar]

Bob, maybe you can explain it to me.

The OP is reading H1 = 2.4A, H2 = 2.6A, H3 = 4.4A at 480 volts, that sounds to me to be a lot more power then 400 watts

An unloaded transformer is a very strange beast. It consists of a whole lot of inductive current and relatively little resistance. This combination results in a very low Power Factor. And, it will have a current waveform that is very distorted which will cause a non "true-rms" meter to report inaccurate values. Both of these factors mean a simple formula with standard assumptions may be misleading.

I would not be surprised for a current generation transformer to have about 1-1.5% core losses.

 

[zog]

Bob, maybe you can explain it to me.

The OP is reading H1 = 2.4A, H2 = 2.6A, H3 = 4.4A at 480 volts, that sounds to me to be a lot more power then 400 watts

Bob,

The majority of that measured current is from reactive load. The total current measured is a compoent of the apparant power. There are three ways of expressing power in an AC circuit:

?Real power (P) expressed in watts (W) is the power that is required to perform work.

?Apparent power (S) expressed in volt-amperes (VA) is the total power flowing is the circuit. Transformers are rated in their apparent power expression using units in kVA (kilovolt-amperes), as this value is proportional to the total heating in the windings

?Reactive power (Q) expressed in volt-amperes reactive (VAR) is the power which does no useful work but is required to assist in performing work, such as setting up magnetic fields in motors and transformers. It is often called ?imaginary power.?

The reactive compenent of that current basically goes back and forth between the transformers expanding and collapsing field and the transformer secondary that is feeding its expanding and collapsing field. Lets say for arguements sake the Pf on this circuit is 0.5 lagging (Assuming no other loads on the transformer feeding this one), that means half of the current he is measuring on the H1/H2/H3 is reactive load, and the other half is the losses in the transformer, which is being converted to heat, and will be seen as Watts on the W/Hr meter.

Make sense?

 

[iwire]

Sorry guys, I know your trying but I give up.

We have no idea if his amp clamp was true RMS or not, so while the EEs fall back to formulas I fall back to test results.

If I see 2, 2 & 4 amps on my True RMS clamp meter I assume the KWH meter is spinning as well.

 

[iwire]

Make sense?

NO! gosh Darn it, it is not making a bit of sense to me. (Mad at me not you all)

Not saying your wrong (or any of the others) only that I don't understand it.

It sounds like you are all saying the amp clamp and KWH meter are entirely different animals and what one measures the other does not.

There may be truth to that with a cheap clamp meter but I am still left confused.

 

[zog]

Ok lets try this, look at the attached picture. Forget about voltage, it dosent matter for this discussion, just think about the current.

The energy being used (Watts, turning into heat in this case) makes up the horitontial part of the triangle, this is some of the current measured by the clamp on meter.

The reactive load is the vertical part of the triangle, that current is being absorbed by the field of this transformer to make it expand and 1/120 second later as the field collapses it "pushes" the current back the other direction, no work was performed, no real energy was used. However, there is current flow, and it is measured by the clamp on meter.

The hypotonuse (SP?) of the triangle is the total power (Current) being measured by the clamp on ammeter.

A sqaured + B squared = C squared. A is watts, B is VARS, C is VA.

I used to explain this to my students using a superball bouncing on the floor, it work great if I am standing in front of you bouncing my superball explaing this but hard to do in this forum.

 

[winnie]

It sounds like you are all saying the amp clamp and KWH meter are entirely different animals and what one measures the other does not.

Exactly!

The amp clamp might be a fancy schmancy true rms device, or it might be a cheapo device, but its job is to measure the AC current flowing in the wire. It doesn't know about voltage, power factor, reactive power, etc. It just knows about the current flowing in the wire. The readout is smoothed over time, so that you don't see the 60Hz constantly changing current value, but instead some sort of average that represents that constantly changing value.

The KWH meter is measuring something very different. It is measuring both voltage and current, and reporting the 'instantaneous' product of the two. It doesn't actually know the voltage or the current as a separate measurement, it only knows the product. The power readout is smoothed over time, and represented as the rate of the spinning disk. On top of this, the meter counts the number of disk revolutions to get the total energy consumption. (Note: electronic meters do measure voltage and current separately, and then calculate the product, but the same point holds: the meter is reporting the instantaneous product of voltage and current, smoothed over time.)

To get to the next step, you need to understand how to find the 'instantaneous' product of voltage and current. This simply means that at any moment in time, you measure the current at that moment, the voltage at that moment, and multiply the two. These are _not_ the RMS values smoothed over time, but instead the voltage or current at that given moment. The next moment, both values, and their product will be different. The instantaneous product of voltage and current is a constantly changing AC value.

This is where power factor comes in. If the voltage and current are 'in phase', then when the voltage is positive, the current will be positive; when the voltage is negative the current will be negative. In both cases, the instantaneous product, and thus the power, will be positive. But when the voltage and current are not 'in phase', then you will have one positive when the other is negative, resulting in negative power. Negative power sounds strange until you realize that it simply means power flowing from the 'load' to the 'source'.

When voltage and current are not in phase, then for part of the AC cycle, the load is actually returning energy to the source. This doesn't come out of thin air; this energy was simply 'borrowed' from the source during the other part of the AC cycle. The load is not consuming the energy (not consuming KWH) but current is flowing to shuffle this energy back and forth.

-Jon

 

[Luketrician]

Isn't this why some factories (with poor pf, .85 or less)and such are billed in KVA instead of KW?

It takes more power to over come the pf in these situations to perform actuall work correct? Power, that if only billed in KW would not actually catch all of the Power consumed...true and reactive power correct?

 

[zog]

Isn't this why some factories (with poor pf, .85 or less)and such are billed in KVA instead of KW?

It takes more power to over come the pf in these situations to perform actuall work correct? Power, that if only billed in KW would not actually catch all of the Power consumed...true and reactive power correct?

I dont know of any factoories that are billed in kVA, but, they have a Pf range they have to maintain and are fined large penalties if they have to low of a Pf, thats why they use Pf correction capacitors.

 

[Greg Swartz]

Good Job!

Good Job!

eddy current and hysteresis.

Man, I haven't heard those words since school...
Went over quite a bit of hysteresis.

The majority of that measured current is from reactive load. The total current measured is a compoent of the apparant power. There are three ways of expressing power in an AC circuit:

?Real power (P) expressed in watts (W) is the power that is required to perform work.

?Apparent power (S) expressed in volt-amperes (VA) is the total power flowing is the circuit. Transformers are rated in their apparent power expression using units in kVA (kilovolt-amperes), as this value is proportional to the total heating in the windings

?Reactive power (Q) expressed in volt-amperes reactive (VAR) is the power which does no useful work but is required to assist in performing work, such as setting up magnetic fields in motors and transformers. It is often called ?imaginary power.?

Straight out of the text book.

This is a very good thread. Guys, I think you answered the questions quite well.
It'd be great if the other side of the forum came here and looked at this one.
Greg

 

[MCBC]

Wow Guys,

I guess I opened a real can of worms with this one, but its great info for readers. The clamp meter I use is a Greenlee CM-410 (not true RMS) however the readings are accurate as I did a comparison with a Fluke true RMS meter from a electrical contractor friend of mine.

Thanks so much for the discussion, I does not appear I can do any kind of cost savings on this one as I have to run my refrigderation units 24/7 and cannot shut down the transformer.


Regards

MCBC

 

[MCBC]

Also,

The tranformer has common X0 to the LV panel and a bare ground wire from X0to the buildings plumbing (Doubled Up under the same lug). The transformer cabinet is grounded from the HV curcuit. I know in shielded curcuits that grounding both ends will cause a ground loop. Could this be an issue? or is it normal.

MCBC

 

[zog]

Wow Guys,

I guess I opened a real can of worms with this one, but its great info for readers. The clamp meter I use is a Greenlee CM-410 (not true RMS) however the readings are accurate as I did a comparison with a Fluke true RMS meter from a electrical contractor friend of mine.

Thanks so much for the discussion, I does not appear I can do any kind of cost savings on this one as I have to run my refrigderation units 24/7 and cannot shut down the transformer.


Regards

MCBC

What was your question? We love worms here,welcome to the forum.

 

[zog]

Iwire, please tell me you get it now. If not I will have to come to MA with my superball.

 

[gar]

090213-1659 EST

iwire:

See if you can find a good quality capacitor and a wattmeter suitable for the current. 25 mfd is about 106 ohms at 60 Hz, This capacitor across 120 V will have about 1.1 A current flow. The VA reading will be about 135.

Also connect a wattmeter to read the power input to the capacitor. It will be way below 135 W.

Practical capacitors are generally are more efficient than inductors. Meaning less heat loss for a given current.

I have just measured the current of a Sprague AB-1201, 12-25 mfd 125 VAC 60 Hz. Across 120 V the current was 0.92 A. Tonight I will measure the power input with a 75 W meter. If you do the experiment I suggest that you put a resistor in series with the capacitor. Maybe 50 ohms 50 W. This will allow two tests, one with the resistor and the other with it shorted.

Inductors and capacitors have the same general characteristic, that of a phase shifted current. I am suggesting this experiment with a capacitor because the ratio of VA to watts will be very great.

.

 

[gar]

090213-2001 EST

To continue with my expewriments I am using a Fluke 27 for current measurement, a Superior Electric Powerstat for variable voltage, a Simpson 75 W wattmeter, Sprague AB-1201 21-25 mfd capacitor, GE 45F277 12.5 mfd 400 V capacitor, an Ohmite 10 ohm resistor, and a 46 ohm resistor. You will see a big difference in the internal power loss of the two capacitors.

Two voltages were used, 60 and 120.

AB-1201 no external series resistance 21-25 mfd
060 V 0.52 A 3.5 W 31 VA
120 V 1.20 A 22 W 144 VA equivalent internal series resistance 22/1.44=15.2 ohms

AB-1201 plus 10 ohms external series resistance
060 V 0.52 A 7.0 W 31 VA
120 V 1.11 A 30 W 144 VA equivalent total series resistance 30/1.23=24.3 ohms or 14.3 ohms internal


GE 45F277 no external series resistance 12.5 mfd
060 V 0.28 A 0.5 W 16.8 VA
120 V 0.61 A 1.5 W 73.2 VA equivalent internal series resistance 1.5/.372=4.0 ohms

GE 45F277 plus 10 ohms external series resistance
060 V 0.26 A 1.0 W 15.6 VA
120 V 0.56 A 4.0 W 67.2 VA equivalent total series resistance 4/0.314=12.8 ohms or 2.8 ohms internal

GE 45F277 plus 56 ohms external series resistance
060 V 0.25 A 4.0 W 15.0 VA
120 V 0.54 A 17.5 W 64.8 VA equivalent total series resistance 17.5/0.292=60.0 ohms or 4.0 ohms internal

Note: the very low power readings on a 75 W scale are open to considerable error. This is not a well controlled experiment, but there is such a stark contrast that it is easy to see the point that the wattmeter can read a very low power dissipation with substantial current flowing thru a reactive component.

.