POWER TRANSFORMER EFFICIENCY

[JAMartinezB]

Hi Everyone, I have a double ended system with two transformer of 1 MVA 12.5kV/480V Y/Y transformer. Both are energize all the time but only one supply the power to the load.

The maximum demand of the system is 350 kW. My concern is about the efficiecy of this system since one trafo is just loaded to the 33% and the other is idle. Moreover, the utility is charging me for low PF, the average PF of the system is 0.82 lagging.

I would like to know is there is any standard curve of transformer efficiency vs load, and what is the contribution of the transformers to the low PF; since one is only 33% loaded and the other is idle.

Regards.

 

[waross]

I suspect that most of your power factor issues are due to the exciting current of the transformers. If you are able to measure the exciting current of the idle transformer, you will be able to compute the exciting kVA. Take about 90% of this value and connect that many kVARs of capacitors to each transformer. That should much improve your power factor enough to get you above the PF penalty zone.


Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[dpc]

Transformers are extremely efficient. The idle transformer has only core losses. If you have transformer test reports you can find the core loss in kW.

I really doubt that the magnetizing current of the idle transformer is having any significant impact on your overall power factor if you have 300 kW of load. The exciting current is very small and is actually not as reactive as generally imagined because the core losses show up as a resistive load.

If you have the transformer test data, you should find the no-load current. With that and the no-load losses, you can determine the vars being consumed by the transformer.

 

[waross]

Hi dpc;
I have seen the power factor on a large unloaded transformer, (only exciting current) as low as 0.1, or 10%. The power factor penalty was about 80%, but applied to a very small consumption charge. If you pick a realistic X/R ratio and compare the X to the Z you will find little difference. 80% or 90% is a safe figure for capacitors without over correcting.
I have used high voltage ammeters from A B Chance to measure exciting current. ce, in a jam, we de-energized the transformer and hung a clamp on ammeter on a primary lead. We then energized the transformer and read the current from a safe distance.
Don't forget that this is 300kW on 2 MW of transformer capacity.
With the old type of PF monitoring, you just need a given amount of KVARs a month added to escape the penalties. When the penalty cutoff is 0.9 and you correct the transformers to close to 1.0, you are adding KVARs 24/7. If the plant is operating an 8 hr. shift, those extra hours of excess KVARHrs add up.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[jghrist]

You would increase the efficiency by serving half of the load from each transformer. Load losses would be cut in half. Reactive losses would also be cut in half.

There is no standard curve of efficiency vs load. This depends on the particular load and no-load losses of the transformers which you can get from the test reports.

As dpc said, you can get the reactive losses from test data and the loading. How much this contributes to your pf penalty depends on the rate structure of the penalty.

 

[dpc]

I pulled out an old test report (admittedly for a larger transformer) and find that exciting current was 0.182% of Rated Current and Exciting current power factor was 0.47.

I'm still skeptical that the exciting current is having much impact on the overall system power factor. It certainly is contributing, but I think the OP will have to look to his loads to make much improvement in pf.

If I wasn't so lazy, I'd run some calcs to see, but I hate being confused by a lot of facts.

Cheers,

Dave

 

[wolf39]

jghrist is right: The best total efficiency for your application can be achieved by loading each transformer to 175 kVA. The total core losses remain the same whereas the total load (copper) losses can be cut in half. Seems that one transformer is in the stand-by mode, not connected to the load.

dpc gave us some information regarding the excitation current, being a mere 0.182% for a large transformer. It may well be that the exciting power factor was only 0.47 under no-load conditions. But we have to remember that under no-load conditions there is no MW throuput, its just the core losses which favourably increases the power factor. The reactive load can easily be calculated if the excitation current is known.

The utility shouldn't charge you for a poor power factor but for reactive kVAr power instead. The power factor can deteriorate if the system load decreases, i.e. in case some drives work under reduced load. The average power factor of your system (0.82 lagging) has its origin in the drives and other power consuming appliances. Your system transformers have very little impact on this.

Regards,

Wolf

http://www.hydropower-consult.com

 

[waross]

I will withdraw my statements and defer to dpc's imformation in regards to the small magnitude of the transformer VARs in comparison to the KVA and KVAR through put.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[JAMartinezB]

Thank you for your answers. Unfortunately I don´t have the trafo test data. But I have read that the trafos are less efficient if they are loaded less than 80% of the capacity.

Also, that the reactive power is around 5% of the kVA. How can I calculate the reactive power consumed and wouldn´t be more efficient to have two 500 kVA trafos rather than the two 1000 kVA.

Regards.

 

[dpc]

Transformer losses have two components: Core losses and winding (copper) losses.

Core losses are constant (for constant voltage) regardless of load.

Copper losses vary with the square of the load current.

Load losses at full load might be 4 to 5 times the core losses.

Reactive power doesn't directly have much to do with efficiency. That is a function of the losses. Actual losses are always purely resistive.

You are really looking at two unrelated questions: Power factor penalty and transformer losses.

Splitting the load between the two transformers will reduce the load losses quite a bit. It will not reduce the core losses at all and it will not change the var requirements of the transformer exciting currents.

De-energizing one transformer will reduce core losses and will improve power factor somewhat, but I'm doubtful it will have a big impact on your overall power factor penalty.

Cheers,

Dave

 

[davidbeach]

Of course transformer efficiency falls off as load goes down. There is a fixed amount of no-load loss that is there regardless of load, so as load goes down the no-load loss becomes a higher percentage of the load. Losses are going down, but just not as fast as load is going down, so don't get too hung up on lower efficiency.

 

[wolf39]

Gentlemen:

The transformer efficiency doesn't necessarily go down with decreasing load. It depends on the ratio of core losses to copper losses.

To give you one example. Some time ago I designed a power transformer as follows:

31,500 kVA
110/10.5 kV
50Hz

The resulting losses were as follows:

Core loss 18 kW
Copper loss 135 kW

The core loss figure turned out to be very low because the utility specified a transformer with an extreme low noise level. Therefore the core flux density had to be low as well.

Under the assumption that the power factor is 1.0, we get the following efficiencies:

100% load 99.52%
75% load 99.60%
50% load 99.67%
25% load 99.66%

For this example the efficiency goes up with decreasing output and only goes down at outputs of 25% and less.

Regards

Wolf

http://www.hydropower-consult.com

 

[davidbeach]

Peak efficiency is generally at the load level where core losses and copper losses are equal. The transformer in the example from wolf39 probably had its peak efficiency around 36% load. So, yes, you could have a transformer that went quite low in loading before the efficiency started to decline but even then the total losses are declining with load. If you compare losses to transformer full load rather than actual load you will find that lightly loaded transformers can look quite good.

 

[prc]

1) Maximum efficiency in a transformer will be when the load conditions are such that load loss =no-load loss.This will ocuur at a load =square root of no-load loss divided by load loss.(in pu)In modern transformers this will be at a load of 40-50 % depending on the ratio of no-load loss to load loss.
2) In modern transformers no-load power factor is quite high 0.5-0.9, thanks to better core material and better core processing and core joints provided.These measures brought down the no-load cuurent,for the same core loss.The other day I was seeing a 360 MVA 400 kV GSU with 0.05 % excitation current and 0.91 no-load power factor. PF will improve marginally for EHV units as the exciation current will be partially compensated by the not insignificant capacitive current.
3) Coming to the specific case- in a modern transformer of this size, the no-load losses and load losses will be approx 0.14 % & 1 % of the rating.The no-load current may be approx 0.2 %. Based on these,no-load excitaing KVA is only 2 KVA,negligible compared to load kVA and hence little effect on pf. .No-load pf will be 0.7.Max efficiency will be at a loading of 370-500 KVA depending on the vintage of the unit.In any way it is better to put the units in parallel so as to share the load.The efficiency difference will be negligible,but it will keep both your units warm avoiding moisture ingress in to insulation.

 

[jghrist]

The peak efficiency load level is not the same as the economic load level. For the economic load level, you have to consider the cost of the transformer.

 

[waross]

In support of your post, jghrist;
I was thinking, great. If we rerate our 1000KVA transformer down to 300KVA we can achieve higher efficiency. Is 1% or 2% more efficiency worth 300% more capital investment?
BTW. Both transformers paralleled has protection and arc flash implications.Merely energizing one transformer without load will generate sufficient heat to avoid moisture ingress.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[jghrist]

BTW. Both transformers paralleled has protection and arc flash implications.

Loading both transformers probably can be done only if the loads can be split without paralleling the transformers (half on one, half on the other, not connected together). This would avoid the protection and arc flash problems.

 

[JAMartinezB]

Thanks again for all your answers. Now I have a best picture of the pf and the efficiency situation. From your answers I could say that 5% of the KVA of a trafo is the not-load kVar is not a good rule of thumb.

Regards.

 

[prc]

1) Let us remember that there is only small difference in overall efficiency between maximum and minimum levels in a transformer.

2) Normally breaker will have the capacity for two paralled transformers.Otherwise load can be split.Warming up from a no-load ccondition and slighly loaded condition is different.Under no-load condition practically no winding heating. Overall losses also very small to create any worth heating.It is always better to keep windings warmto avoid water accumulation in winding insulation.

3)In modern transformers no load kvar is only 0.05-0.1% for power transformers and 0.5-2 % for distribution trfs. Excitation KVAR (%)is less with increased rating.