Transformer Standby Losses

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delectric123

Senior Member
Location
South Dakota
I have a 25 KVA 480 to 240 1 ph. step-down transformer. under no load the input current is 3.8 amps. any idea what the power factor would be so I can calculate the standby losses?
 

texie

Senior Member
Location
Fort Collins, Colorado
Occupation
Electrician, Contractor, Inspector
I have a 25 KVA 480 to 240 1 ph. step-down transformer. under no load the input current is 3.8 amps. any idea what the power factor would be so I can calculate the standby losses?

Gee, this could get into quite an engineering discussion due to the variables involved such as transformer design. I might add that it would be over my head as I'm not an engineer. My inclination would be to simply use a true watt meter to check for the value and call it a day.
I do think it is a fair question as unloaded transformers can have considerable energy losses. I'll let the experts chime in.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
140211-2119 EST

I don't know for your size, but my guess is below 0.1 and probably lower.

On a 175 W transformer I measured 0.18 . A larger transformer should have a lower PF. Also the core material, and transformer design will have a considerable influence. Excitation voltage vs design voltage is a major factor.

See http://www.csemag.com/single-articl...oncerns/478c06eebfe849e4db9cbec2f3969fe7.html

If we assume 3% loss at 35% load and that 50% of the loss at 35% load is no load loss, then for your 25 kVA unit we get 0.03 * 0.35 * 0.5 * 25,000 = 131 W. This would be a PF of 0.005 .

Using older design criteria the result is PF = 0.03 * 0.5 = 0.015 . Both seem low. But I don't have a 25 kVA to test. A 2 kVA at the shop, no load, runs too hot to touch. Is 3% loss a valid criteria at 100% load in an older design?

Check the specification for your particular transformer.

.
 

broadgage

Senior Member
Location
London, England
If the transformer is reasonably new, then loss data should be available from the supplier.

In the case of an old unit, the only reliable way to determine losses is by measurement, either use a true watt meter, or a utility type KWH meter for a known time.

Note that the off load losses are virtualy the same no matter which way round the transformer is connected.
It might be easier to energise the lower voltage side for test purposes, for example if you have a 240 volt KWH meter to hand but do not have a 480 volt unit.
 

delectric123

Senior Member
Location
South Dakota
140211-2119 EST

I don't know for your size, but my guess is below 0.1 and probably lower.

On a 175 W transformer I measured 0.18 . A larger transformer should have a lower PF. Also the core material, and transformer design will have a considerable influence. Excitation voltage vs design voltage is a major factor.

See http://www.csemag.com/single-articl...oncerns/478c06eebfe849e4db9cbec2f3969fe7.html

If we assume 3% loss at 35% load and that 50% of the loss at 35% load is no load loss, then for your 25 kVA unit we get 0.03 * 0.35 * 0.5 * 25,000 = 131 W. This would be a PF of 0.005 .

Using older design criteria the result is PF = 0.03 * 0.5 = 0.015 . Both seem low. But I don't have a 25 kVA to test. A 2 kVA at the shop, no load, runs too hot to touch. Is 3% loss a valid criteria at 100% load in an older design?

Check the specification for your particular transformer.

.

with a PF of 0.005 * 3.8 * 480 = about 0.009 KW/H which is nothing to sneeze at. but if the PF would be something like 0.8 then that would be another story.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
140212-1359 EST

delectric123:

I think the power factors that I calculated based on the assumptions made are likely way to low.

Note: for your transformer that is only 9 watts. For the physical size of a 25 kVA transformer you would find it hard to feel the temperature rise of the exterior surface from 9 W internal dissipation. Actually I believe the transformer would feel quite warm. That probably means a 100 or more watts.

If the primary resistive impedance of the transformer was 1%, then the approximate resistive loss at no load would be about 18 W. Double this for total no load loss and it is about 40 W. You really need to find manufacturer data on the transformer.

If you can not find a data sheet, then the easiest way to measure it is with a wattmeter as mentioned by broadgage.

You can also do temperature rise of the transformer surface or the resistance change of the wire in the excited coil as a means of measuring power loss. But still a power meter is the easiest.

.
 

rcwilson

Senior Member
Location
Redmond, WA
FWIW, my old Cutler-Hammer spec book lists no-load losses for a 25 kVA, type DS-3, 150C rise transformer as 200 Watts. Total losses at 100% load = 790W
 

texie

Senior Member
Location
Fort Collins, Colorado
Occupation
Electrician, Contractor, Inspector
If the transformer is reasonably new, then loss data should be available from the supplier.

In the case of an old unit, the only reliable way to determine losses is by measurement, either use a true watt meter, or a utility type KWH meter for a known time.

Note that the off load losses are virtualy the same no matter which way round the transformer is connected.
It might be easier to energise the lower voltage side for test purposes, for example if you have a 240 volt KWH meter to hand but do not have a 480 volt unit.

I stated previously that it might be simplest to use a true watt meter and just measure it. After thinking about it, your suggestion of using manufacturer data might be more reliable. I'm not sure most watt meters (even a revenue grade) would be accurate at such low levels and likely funky wave form of a small unloaded transformer.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
140212-2054 EST

GoldDigger:

I like to label my graphs from DTE as kWh/h to mean power average over a one hour period.

DTE provides data in graphical form labeled kWh for the Y-axis, but the plot is not a monotonic function, but rather the average power over a 1 hour period in discrete increments. This can be described as a step plot. However, some plots are vertical bar plots, and possibly one could argue that kWh only applies to the amount of a single bar.

Further they tend to label the time as the start of the hour over which the averaging is performed, rather than at the end of the averaging time. Also they are presently on EDT rather than EST adding to the confusion.

.
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
I have no problem at all with kWh/h. It is just kW/h (rate of increase of energy usage?) that bugs me.
:)
Your usage makes perfect sense. But it could also be simplified as average kW.

Tapatalk!
 

iwire

Moderator
Staff member
Location
Massachusetts
I have no problem at all with kWh/h. It is just kW/h (rate of increase of energy usage?) that bugs me.
:)
Your usage makes perfect sense. But it could also be simplified as average kW.

This is a major problem and the world should change so you are not bugged. :angel:
 

delectric123

Senior Member
Location
South Dakota
I have no problem at all with kWh/h. It is just kW/h (rate of increase of energy usage?) that bugs me.
:)
Your usage makes perfect sense. But it could also be simplified as average kW.

Tapatalk!

Do I dare ask, what does Tapatalk mean?:dunce:

i'll see if I can find a data sheet for that transformer.
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
But of course. It is a tag inserted by the App I use for the forum from mobile. Since it is free and useful I figured I would leave the plug in place for awhile.

Tapatalk!
 

Besoeker

Senior Member
Location
UK
I have a 25 KVA 480 to 240 1 ph. step-down transformer. under no load the input current is 3.8 amps. any idea what the power factor would be so I can calculate the standby losses?

Transformer losses are mostly two main components.
Iron loss and copper loss.
Fe loss is approximately constant throughout the load range.
Cu loss is proportional to I^2.

For the systems I design/work on there is usually a transformer but often in the several hundred kVA range.
I have to calculate system efficiency. I use a constant 0.7% for Fe and 1% for Cu at rated load.

I do realise that this doesn't answer your specific question but the principle has served me well for decades.
 
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