Finding transformer magnetizing kVARs? 5

[rhatcher]

Can anyone tell me how to determine the magnetizing kVARs for a transformer? Is there a general value that can be used for estimating purposes based on kVA capacity? The known values of the transformers in question are primary and secondary voltage, rated kVA, and percent impedance.

In order to help understand the nature of my question, my facility has about 9.5MVA of distribution in the form of eight transformers with 13.2kV primaries and various low voltage secondaries. I am in the process of determining our actual power usage, but I am estimating that it is in the range of 350-500kVA continuous with a peak of 1000kVA and a power factor of 80-90%. The difference in usage versus capacity is due to the fact that the building's original purpose required much more power than we do.

The problem is that our power factor as seen by the utility is very low and it is costing us a lot of money in penalties. Since the utility metering is on the primary side, I suspect that the reason for this is that we have so much idle transformer capacity. However, I need to put a number on it. If I can determine the magnetizing kVARS for a given transformer or for a given amount of kVA capacity, I can estimate how much all of that idle transformer capacity is costing and decide if it is worth it to do something about it. Right now, I could eliminate 3 transformers (4250kVA) and de-energize 2 more (2250kVA) if I transfered some loads to the remaining 3 transformers.

 

[electricuwe]

For transformers of the size involved in your case a magnetising current of 1,5..2% might be expected.

 

[jghrist]

Tranformer no-load losses may be a bigger cost low pf.

 

[edison123]

Typically the trafo pf is defined by the load to which it is connected. (which is why they are rated in KVA and not in KW). While I agree with electricuwe about the no-load amps range of your trafos, it does not contribute much to the low pf. I would start looking at the load end for any pf improvement.

 

[rhatcher]

Thanks for your answer electricuwe. I now have detailed info on my utility bill, electric rate plan, and power consumption. The results derived from your answer are later in the post.

For jghrist, you are correct. When asking my question, I was disregarding the real power no-load losses. My beginning focus was on the kVARs since my power factor is lower than expected. However, since my original post I have learned that my "penalties" are based on kVA demand instead of kVAR demand, so the real power no-load losses are worth considering as well. Can you or anyone else provide info on what the no-load kW losses for a transformer are based on kVA rating?

For Edison123, you are correct in stating that a transformer's power factor is determined by the load to which it is connected. However, as stated in my previous post, I am trying to determine the costs of having idle transformers in my facility so the no-load kVAR and kW (thanks again jghrist)losses are my concern.

As I stated in my previous post, I have 9MVA of transformer capacity and my utility metering is on the primary side. I now have detailed info on my power consumption. A typical reading is 249.6kW, 279.1kVAR, 374.4kVA, and 0.6667pf. Using electricuwe's figures (and ignoring kW losses for now), if I eliminated 6500kVA of idle transformer capacity I would also eliminate about 98 kVARs of consumption. This would change the above data to 249.6kW, 181.9kVAR, 308.8kVA, and 0.808pf. If I could waive a magic wand and replace all of the idle transformers with a single 500kVA unit, the results would be 249.6kW, 151.6kVAR, 292.0kVA, and 0.8547 pf.

So...although the actual power consumption is lower than the estimate given in my previous post, the "corrected" power factor of 0.85 is more in line with what I expected. If anyone can provide info on the no-load kW losses, it would allow me to come up with a more accurate estimate of the costs of these idle transformers.

Thanks to all and Merry Christmas

 

[jghrist]

The best thing to do is to get actual test data on the transformers. You will need percent exciting (magnetizing) current, no-load (iron) losses, and load (copper) losses. Otherwise, use typical values, but realize that there is a very wide range of losses, so "typical" may not be very accurate. Electricuwe has provided typical exciting currents.

http://oge.apogee.net/pd/aptit.htm

is a table of typical impedances including % resistance. Load loss % is basically the same as % resistance. Typically no-load loss will be 30% - 60% of load loss, per Westinghouse T&D Manual for 46 kV and below.

If you redistribute loads to eliminate some transformers, you will increase load losses, which are proportional to the square of the load. The load loss is given at rated load.

 

[edison123]

rhatcher,

On seeing your load service factors, I can understand why you would want to eliminate idling giants. I agree at the service fcators you have mentioned, the total system pf will be poor. But, if you trafos had different secondary voltages, how do you propose to integrate your loads to a single trafo ? Probably further step down trafos downstream so that all are in series instead the present parallel set up ?

jghrist,

I think the total load losses will remain the same before and after load integration.

 

[rhatcher]

Thanks jghrist. If I understand you, then the no load real power losses in percent of rated kVA are equal to 30-60% of the percent impedance. So, if most of my transformers are around 6% impedance and I assume that they are inefficient because of the older design and the fact that they are dry type, then my no load real power losses would be around 3% of rated kVA?

For edison123, I agree that I can eliminate no-load losses but that my load losses will remain constant. About the question of consolidating the transformers of different secondary voltages..that is the question. Other than the 8 transformers I have mentioned, there are only a few other step down transformers in my facility with a total capacity of less than 150kVA and they are obviously add-ons since the original tenants. This is because the design specification for the original tenants apparently did not allow stepdowns other than the primary substations. For primary substations, there are 4-440V delta, 2-240V delta, and 2-208Y/120V units in service for a total of 9MVA. There is also 1-2400V 1500kVA delta unit that has I haven't mentioned before now since it has already been disconnected at the primary switch.

I am thinking that I can completely remove 2-440V and one 240V units that were each dedicated to primary loads that no longer exist as long as I switch the few remaining circuits to other sources. As an example, one of these is a 750kVA 440V sub dedicated to an large oven that used to be electrically heated. The oven has since been converted to gas, so when you follow the circuit from the 750kVA transformer it leads through the unused oven electric power panel, portions of which are still energized (??), and ends up at the 150VA control power transformer for the new gas controller (LOL).

Anyway, of the remaining five, 2-440V, 1-240V, and 2-208Y/120V, one each of the 440V and the 208V transformers feed sections of the building that are mostly not used, so with the exception of a few circuits that need to be transferred, they can be deenergized at the primary switch until whenever they are truly needed again (if ever). That would leave me with 1-440V at 1000kVA, 1-240V at 750 kVA, and 1-208Y/120V at 1000kVA. This is still way more than I need, but I think at this point the cost to consolidate further would far outway any savings.

 

[Elecme]

There are two kvars that should be considered for any transformer:
Fixed kvar and this is related to the magnitizing current
Step kvar and this is dependant on the loading
Such figures are almost known and available at manufactureres.
Try

http://www.trafounion.com

Elecme

 

[jghrist]

The 30% - 60% factor I gave is times % load loss (=%R) not time % impedance.

Load loss will increase if the load is consolidated into fewer transformers because it is proportional to the square of the load. For example:

With 2-750 kVA transformer each 1/2 loaded

%R = 1.94% %Z = 5.75% No-load/Load Loss = 40%
No-load loss = 40% · 1.94% = 0.582%
No-load loss for each xfmr = 750 · 0.582% = 4.365 kW
Total no-load loss = 2 · 4.365 = 8.73 kW
Load loss for each xfmr = 750 · 1.94% · (1/2)² = 3.6375 kW
Total load loss = 2 · 3.6375 = 7.275 kW

With 1-750 kVA transformer fully loaded

No-load loss = 750 · 0.582% = 4.365 kW
Load loss = 750 · 1.94% = 14.55 kW

In addition to the demand savings, you will save energy (kWh). The no-load losses reduction will provide 24 hrs/day energy savings, so multiply the kW savings by the number of hours in a month to find the kWh reduction. Load loss energy increases are more complicated because the transformers are not continuously loaded and the loss is proportional to the square of the load. This means that you can't use the average load but have to look at the load profile.

 

[jbartos]

Suggestion: A little modeling by using software of:
e.g. SKM, Inc. (DAPPER, A-FAULT, etc.), OTI, Inc. (ETAP), EDSA, etc. may help you to find the right solution.
These businesses have their web sites.

http://www.skm.com

http://www.etap.com

http://www.edsa.com

Also, check the facility computers. They may already have some software for the electrical power system modeling installed.

 

[jbartos]

Suggestion: Sometimes, transformers are relatively idle since there is a redundancy design for the power distribution system. This is usually the case, if the power distribution system outage or downtime becomes costly. In that case, the owner has to pay some price for the redundancy. That however does not mean that the power factor has to be high. It may be compensated by using power factor compensating capacitor banks that automatically switch to adjust or optimize the site power factor value to a non-penalty level.

 

[busbar]

rhatcher, is there any chance you have access to CTs and PTs in the switchgear that serves the transformers? Excitation characteristics could be measured to compare data for actual costs. It looks like your calculations would allow scaling of a portable power analyzer or simply, voltmeter, ammeter and phase-angle meter connected to existing instrument-transformer secondaries.

 

[busbar]

Another long shot towards billing reduction with respect to no-load losses may be to offset var demand with a single small capacitor bank on the bus between transformer primaries and the utility meter.

Conceivably, small capacitor sets on the >600V secondaries could also offset transformer primary-side var losses.

 

[rhatcher]

Thanks to all for the excellent posts and stars for jghrist and busbar. I have enough info to go on for now. Like I described in my previous post, I know that 3 x-formers can be removed and 2 more can be deenergized at the primary until they are needed again if I can justify the cost to swap a few feeders around. As it is, you have confirmed my idea that this is the case. Although we will probably never completely use the remaining capacity, reducing it further would not be practical because it would be very expensive and it would never achieve cost payback. This is because there are 3 different secondary voltages distributed throughout the facility which is a 4 story light-industrial building with an adjacent high bay shop.

 

[jbartos]

Suggestion: The idea of transformer de-energization for would be needed energization in the future does not address the transformer maintenance in the de-energized condition to have it ready on short notice. There might be some expense involved. The same holds true for associated power distribution, e.g. buses, cables, protective devices, etc.
To compensate the kVAr by capacitors may be a better option. It would keep the transformers normally functioning, on standby if the power consumption increases. The facility of that size should not have any problems to maintain such power distribution equipment.
It is not clear from the original posting, in what kind of business is the facility in.