when a transformer there is a inrush and the xfmr is energized, if you have a transformer and the secondary load is lower than the rating of the transformer does the primary still see the fla of the xfmr or the load connected to it? if you powered a load on the secondary of lets say 30kw on a 75 kva xfmr, and put a meter on the line side would it see the xfmr full load or the secondary of the load 30kw at the primary voltage once up and operating?
transformer loading??
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mdshunk
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Cool question. What I can say for sure is that I've seen unloaded transformers trip the primary side breaker during restart after a power outage.Mike01 said:
LarryFine
Master Electrician Electric Contractor Richmond VA
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The primary load would be the secondary load, in this case 30kw, plus the magnetizing current, plus energy lost as heat.Mike01 said:
ok
ok
Ok then so your actual load draw would be the magnetizing of the core, the load losses (usually around 3%) plus the load, interisting thanks again.
how do you know what the load is for the magnetizing of the core?? Marc, how would an unloaded xfmr trip the ocpd I know the inrus is typically 10-15 times on newer ee xfmrs it can reach 25-30, but if it energized originally why not on re-enerigize? any thoughts??
ok
Ok then so your actual load draw would be the magnetizing of the core, the load losses (usually around 3%) plus the load, interisting thanks again.
how do you know what the load is for the magnetizing of the core?? Marc, how would an unloaded xfmr trip the ocpd I know the inrus is typically 10-15 times on newer ee xfmrs it can reach 25-30, but if it energized originally why not on re-enerigize? any thoughts??
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mdshunk
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Long story short... improper OCPD type selected in the first place.Mike01 said:
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Immediately at energization a transfromer effectively only sees the inrush current, the load is not considered.
After the intial inrush period the transfromer only supplies the load current +copper losses+winding losses.
After the intial inrush period the transfromer only supplies the load current +copper losses+winding losses.
gar
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Mike01:
Core losses which are hystersis and eddie current can be measured with a wattmeter with no load on the transformer. Then measure input RMS current with noload. Measure primary resistance. I^2*R will give you the primary resistive loss due to core losses. So core losses alone are the no load power input minus the resistive loss from the core losses current.
Core losses remain approximately constant with loading of the transformer.
Since the core loss current is non-sinusoidal and phase shifted you can not directly add the core loss current to the load currnet. But you know that input current is a result of the core loss current and the reflected current from the secondary load. The secondary current reflected to the primary is the inverse of the turns ratio. For a secondary current of 1000 A and a primary to secondary ratio of 2 to 1 the primary current due to the secondary load is 500 A. In addition to the core losses you have the I^2*R losses due to primary and secondary resistances.
Another perspective is that there has to be additional current in the primary over that of the reflected current from the secondary load to provide all the power lost in the transformer. With high efficiency transformers you are not talking about a lot of power loss.
On inrush current. The last magnetization state of the transformer core at turn off and the instantaneous voltage and polarity at turn on determines the inrush current. If turn off occurs near saturation in a positive flux direction, and if at turn on the direction of flux change is in the same direction, then you severely saturate very quickly and high inrush occurs.
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Mike01:
Core losses which are hystersis and eddie current can be measured with a wattmeter with no load on the transformer. Then measure input RMS current with noload. Measure primary resistance. I^2*R will give you the primary resistive loss due to core losses. So core losses alone are the no load power input minus the resistive loss from the core losses current.
Core losses remain approximately constant with loading of the transformer.
Since the core loss current is non-sinusoidal and phase shifted you can not directly add the core loss current to the load currnet. But you know that input current is a result of the core loss current and the reflected current from the secondary load. The secondary current reflected to the primary is the inverse of the turns ratio. For a secondary current of 1000 A and a primary to secondary ratio of 2 to 1 the primary current due to the secondary load is 500 A. In addition to the core losses you have the I^2*R losses due to primary and secondary resistances.
Another perspective is that there has to be additional current in the primary over that of the reflected current from the secondary load to provide all the power lost in the transformer. With high efficiency transformers you are not talking about a lot of power loss.
On inrush current. The last magnetization state of the transformer core at turn off and the instantaneous voltage and polarity at turn on determines the inrush current. If turn off occurs near saturation in a positive flux direction, and if at turn on the direction of flux change is in the same direction, then you severely saturate very quickly and high inrush occurs.
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gar
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steveng:
First, you need to define the primary voltage.
Inrush current is a random value from near zero to some maximum value and varies over this range depending on the point in the AC cycle that turn on occurs and the last magnetic residual state of the transformer core.
If there is a load on the transformer, then this adds to the inrush current. If you had an all incandescent load you might have 10 times the initial current from these lamps that you have for their steady state conditions.
Worst turn on time point for incandescent lamps is near the peak of the voltage waveform.
Worst turn on time point for the magnetic core is a voltage point such that the volt-time integral causes the magnitude of the core magnetic flux to increase from the residual flux state. This quickly puts the core into saturation. You need to be familar with hysteresis curves of magnetic materials to understand this problem.
The source impedance in the supply to the transformer will have a significant effect on peak inrush current.
You might expect a peak inrush current from 5 to 20 times the full rated transformer primary current. Very dependent on the various factors mentioned. In a worst case with no secondary load the maximum possible current in the primary is determined by the source voltage, source impedance, and the DC resistance of the primary.
For the most part this is an 8 to 16 MS transient.
Jim will probably give you a simpler answer, but if you can follow the theory, then you can reason a similar problem for yourself in the future.
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steveng:
First, you need to define the primary voltage.
Inrush current is a random value from near zero to some maximum value and varies over this range depending on the point in the AC cycle that turn on occurs and the last magnetic residual state of the transformer core.
If there is a load on the transformer, then this adds to the inrush current. If you had an all incandescent load you might have 10 times the initial current from these lamps that you have for their steady state conditions.
Worst turn on time point for incandescent lamps is near the peak of the voltage waveform.
Worst turn on time point for the magnetic core is a voltage point such that the volt-time integral causes the magnitude of the core magnetic flux to increase from the residual flux state. This quickly puts the core into saturation. You need to be familar with hysteresis curves of magnetic materials to understand this problem.
The source impedance in the supply to the transformer will have a significant effect on peak inrush current.
You might expect a peak inrush current from 5 to 20 times the full rated transformer primary current. Very dependent on the various factors mentioned. In a worst case with no secondary load the maximum possible current in the primary is determined by the source voltage, source impedance, and the DC resistance of the primary.
For the most part this is an 8 to 16 MS transient.
Jim will probably give you a simpler answer, but if you can follow the theory, then you can reason a similar problem for yourself in the future.
.
brian john
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Not an engineer but that is what I thought.
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steveng said:
Predicting the amount of current inrush is very difficult if not outright impossible, so estimates are used instead. The amount of inrush is not dependent on the size of the transformer. Inrush is almost entirely related to the transformer design (i.e. magnetic properties of the core steel and the inductance of the windings). While not all new energy efficient (TP-1) transformers have higher inrush than older units, most do.
From one manufacturer's new EE 480-208Y/120 units:
15kVA = 10.18x
30kVA = 7.5x
45kVA = 5.52x
75kVA = 10.04x
112.5kVA = 12x
The industry standard is that inrush current values are given at 0.1 Sec, so you need to set the instantaneous trip of your OCPD to be higher than the inrush current.
For estimating "rules of thumb" many people will use 12X for new transformer inrush and 10X for the magnetic trip point of a non-adjustable circuit breaker
gar
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brian john:
All external loads on a transformer (meaning load on the secondary) and the magnetic characteristics of the transformer determine the primary current at any instant of time.
In many cases the normal load on the transformer output may be a small percentage of the worst case transformer inrush current due to saturation, and thus the load current can be ignored. This is reasonable in many cases.
But probably not true for a large number of CNC machines with capacitor input filters, and incandescent lamps.
(edit) But note because of the control circuitry in a CNC machine, or most other industrial machines, if power is lost CRM (the main machine contactor) will probably be dropped and thus the machine is not a load when power is reapplied to the main supply transformer.
(end edit)
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brian john:
All external loads on a transformer (meaning load on the secondary) and the magnetic characteristics of the transformer determine the primary current at any instant of time.
In many cases the normal load on the transformer output may be a small percentage of the worst case transformer inrush current due to saturation, and thus the load current can be ignored. This is reasonable in many cases.
But probably not true for a large number of CNC machines with capacitor input filters, and incandescent lamps.
(edit) But note because of the control circuitry in a CNC machine, or most other industrial machines, if power is lost CRM (the main machine contactor) will probably be dropped and thus the machine is not a load when power is reapplied to the main supply transformer.
(end edit)
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kingpb
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The problem with assuming too high of an inrush for setting the OCPD, is that you could overshoot the amount of fault current and then not trip on a fault as quickly as you will want too. You need to look at the TCC curve.
magdumshirish
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Loading A Trnsformer
Loading A Trnsformer
Respected Sir,
Thank You Very Much For Reply.
Sir, My Problem Is As Follows
Transformer Specification Are:
Cpacity: 315kva
Voltage Ratio: 11kv/433v
Fl Pry Current: 16a
Fl Sec Current :420a
Continious Sc Current: 170a Const Non Motorised Load
Now I Want To Switch On Motor Say 6ohp
Suppose The Startting Current Is 6*fl(90)=360a
Or 360/1.732 =207a.
So Mometoraly Sec Current Is 170+360=53o For Some
Milliseconds
Or 170+207=377a
I Am Worried About First Case Of 530a
Please Guide Me About Secondary Susrrenttainable
Current Insuch Cases.
I Am Worried About Fluction Or Flicuring In Secondary
Votage.
Loading A Trnsformer
Respected Sir,
Thank You Very Much For Reply.
Sir, My Problem Is As Follows
Transformer Specification Are:
Cpacity: 315kva
Voltage Ratio: 11kv/433v
Fl Pry Current: 16a
Fl Sec Current :420a
Continious Sc Current: 170a Const Non Motorised Load
Now I Want To Switch On Motor Say 6ohp
Suppose The Startting Current Is 6*fl(90)=360a
Or 360/1.732 =207a.
So Mometoraly Sec Current Is 170+360=53o For Some
Milliseconds
Or 170+207=377a
I Am Worried About First Case Of 530a
Please Guide Me About Secondary Susrrenttainable
Current Insuch Cases.
I Am Worried About Fluction Or Flicuring In Secondary
Votage.
brian john
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- Retired after 52 years in the trade.
In my experience typical inrush (captured with high speed disturbance analyzers) is been in the 4X-6X range. No engineering data just field experience.
The worse case scenarios are with transformers connected to the load of an ATS with no in phase monitor and transfer occurs with both sources energized (usually returning to utility) after a power outage. Even then if both sources are available and the transfer is to generator the inrush is significantly lower than transferring to utility. I would assume this is due to the available current of the source being transferred too.
With the transformers inrush currents I have had looked while energizing the transformer, (mostly 30 kva-500 kva) the secondary load appeared to be insignificant to the inrush current when we did multiple primary input CB operations, typically these transformers were about 15%-50% loaded, which is typical in my experience for 480-208/120 VAC dry type transformers in this part of the country.
The worse case scenarios are with transformers connected to the load of an ATS with no in phase monitor and transfer occurs with both sources energized (usually returning to utility) after a power outage. Even then if both sources are available and the transfer is to generator the inrush is significantly lower than transferring to utility. I would assume this is due to the available current of the source being transferred too.
With the transformers inrush currents I have had looked while energizing the transformer, (mostly 30 kva-500 kva) the secondary load appeared to be insignificant to the inrush current when we did multiple primary input CB operations, typically these transformers were about 15%-50% loaded, which is typical in my experience for 480-208/120 VAC dry type transformers in this part of the country.
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