Transformer Heat Loss

[dudikr2]

How do I calculate the power lost (heat loss) of a transformer? I have a 3750 kVA, 13.2kV-4.16kV, 1200 Amp, indoor transformer. I was trying to calculate the power factor, but i kept getting 4.224, and i know it's supposed to be less than 1. Also, i was curious how to calculate the power. I know it's voltage multiplied by current, but i wasn't sure which voltage value to use. Any info would be extremely helpful.

 

[dudikr2]

also, i was using this as a guide. https://www.mne.ksu.edu/people/personal/white/pdf-files/heat2.pdf

 

[charlie b]

Let me start by saying welcome to the forum.

Let me continue by saying you have a bit to learn about 3 phase calculations. Here is a bit of a start:

• The relationship between power, voltage, and current is as follows: VA = V x I x 1.732 (i.e., the square root of 3).
• In almost all calculations related to three phase systems, the value of the square root of 3 is going to come into the equation somewhere. The tricky part is knowing where (i.e., does it go into the numerator or the denominator).
• You always use the line-to-line voltage.
• So for the primary side, 3,750,000 VA = 13,200 V x I x 1.732, which tells us the primary rated current is 164 amps.
• For the secondary side, 3,750,000 VA = 4,160 V x I x 1.732, which tells us the secondary rated current is 520 amps.
• So I would say that I don't know where your value of 1,200 amps comes from. If the secondary were 480 volts, the rated current would be 4,512 amps.

As to power factor, you don't have enough information to calculate that. You would need to know either the KW value or the power angle. Power factor is equal to KW/KVA, and is also equal to the cosine of the power angle.

Finally, as to power loss in a transformer, my rule of thumb is 5% of the rated KVA. Someone else might have a better way of calculating that.

Just curious: how did you come up with the value of 4.224?

 

[dudikr2]

Thanks for the response Charlie. I am very new to this (I was just recently hired as an electrical engineer) so I have a lot to learn. I found the 1200 amps on the SKM drawing I'm using, but that may be the fuse size. It is a large schematic, and the only wattage i can find is 1750 kW. That would give me a power factor of 0.47 which seems low. Does that sound right?

 

[ron]

The typical efficiencies are listed here. https://www.siemens.com/download?BTLV_44103

Any inefficiency results in heat.

 

[dudikr2]

Ron, I did stumble upon tables like that, however I have a 3750 kVA transformer, and those tables usually only go up to 1000 kVA.

 

[dudikr2]

Charlie, I did respond to your comment, however it did not post for some reason. I found 1750 kW as a possible power for the transformer (i found it on my schematic but i'm not sure what it's referring to), however that would give me a 0.467 power factor which seems low.

 

[charlie b]

Ron, I did stumble upon tables like that, however I have a 3750 kVA transformer, and those tables usually only go up to 1000 kVA.

But the numbers are fairly constant throughout the range. As I said earlier, when the mechanical engineer asks me about heat loss (so that they can plan their HVAC system), I use a value of 5% of the tranformer's rated KVA. That may be a bit conservative, but it is a reasonable starting point.

 

[dudikr2]

Charlie, that is the situation I am in now. A mechanical guy asked me to calculate this for HVAC stuff he's working on, however, i am new and do not know how to do this. Basically he wants documentation on what the heat loss percentage is, and how I got that answer. He was saying something about it being around 2%.

 

[dudikr2]

Sorry for all of the questions guys, I'm new and just trying to learn. So how do I use that table that ron posted in order to find the heat loss? Meaning, how does efficiency relate to heat loss? Is it the same thing?

 

[dudikr2]

Here's what i found. "Full load losses for 150 [degrees] C transformers range from about 4% to 5% for 30 kVA and smaller to 2% for 500 kVA and larger." So if I have a 3750 kVA transformer, that means it would be 2% heat loss correct?

 

[Iron_Ben]

Sorry for all of the questions guys, I'm new and just trying to learn. So how do I use that table that ron posted in order to find the heat loss? Meaning, how does efficiency relate to heat loss? Is it the same thing?

A transformer that size went though final testing when manufactured. Efficiency at different load levels is a part of such testing afaik. There is a hard copy of a test report somewhere which would be a huge help to you. Of course if the unit is 20 years old, it may be long gone or tough to locate. Not to put to fine a point in it, but there are no-load losses (iron) and load losses (copper). Both are important and together they are the total losses. Transformer efficiencies are high, 98 to 99% commonly.

 

[Besoeker]

Charlie, that is the situation I am in now. A mechanical guy asked me to calculate this for HVAC stuff he's working on, however, i am new and do not know how to do this. Basically he wants documentation on what the heat loss percentage is, and how I got that answer. He was saying something about it being around 2%.

I wouldn't disagree with that figure.
A lot of what I have done has required guaranteed efficiencies for power electronics systems, and mainly variable speed drive systems.

These are actual figures from a 400kVA 11kV/560V transformer.

(s) Efficiency :
(1) losses at full load, 0.8 pf:
(2) losses at 75% full load 0.8 pf:
(3) losses at 50% full load 0.8 pf:

97.49 %
97.83 %
98.00 %

 

[dudikr2]

I wouldn't disagree with that figure.
A lot of what I have done has required guaranteed efficiencies for power electronics systems, and mainly variable speed drive systems.

These are actual figures from a 400kVA 11kV/560V transformer.

(s) Efficiency :
(1) losses at full load, 0.8 pf:
(2) losses at 75% full load 0.8 pf:
(3) losses at 50% full load 0.8 pf:

97.49 %
97.83 %
98.00 %

Okay thanks!

 

[Jraef]

Okay thanks!

I got these figures from ASHRAE once, from a document on estimating heat loss of electrical equipment for sizing of HVAC systems. Unfortunately I lost the original document so I only have this as data in a spreadsheet I use.



Heat gain from electrical equipment are indicated below:
Transformers

150 kVA and smaller : 50 Watts/kVA
150 - 500 kVA : 30 Watts/kVA
500 - 1000 kVA : 25 Watts/kVA
1000 - 2500 kVA : 20 Watts/kVA
larger than 2500 kVA : 15 Watts/kVA

Switchgear

Low voltage breaker 0-40 Amps : 10 Watts
Low voltage breaker 50-100 Amps : 20 Watts
Low voltage breaker 225 Amps : 60 Watts
Low voltage breaker 400 Amps : 100 Watts
Low voltage breaker 600 Amps : 130 Watts
Low voltage breaker 800 Amps : 170 Watts
Low voltage breaker 1600 Amps : 460 Watts
Low voltage breaker 2000 Amps : 600 Watts
Low voltage breaker 3000 Amps : 1100 Watts
Low voltage breaker 4000 Amps : 1500 Watts
Medium voltage breaker/switch 600 Amps : 1000 Watts
Medium voltage breaker/switch 1200 Amps : 1500 Watts
Medium voltage breaker/switch 2000 Amps : 2000 Watts
Medium voltage breaker/switch 2500 Amps : 2500 Watts

Motor Control Centers

Section : 500 Watts per section
Low voltage starters size 00 : 50 Watts
Low voltage starters size 0 : 50 Watts
Low voltage starters size 1 : 50 Watts
Low voltage starters size 2 : 100 Watts
Low voltage starters size 3 : 130 Watts
Low voltage starters size 4 : 200 Watts
Low voltage starters size 5 : 300 Watts
Low voltage starters size 6 : 650 Watts
Medium voltage starters size 200 Amp : 400 Watts
Medium voltage starters size 400 Amp : 1300 Watts
Medium voltage starters size 700 Amp : 1700 Watts

Variable Frequency Drive

Variable frequency drive : 2 - 6% of kVA rating
Bus duct : 0.015 Watts /ft Amp
Capacitors : 2 Watts / KVAR

I added the following:
SCR heating controllers and Reduced Voltage Solid State Soft Starters (without bypass): 4-1/2W / RLA
Soft Starters with bypass: same as motor starters of the same size.

 

[Julius Right]

In my opinion, the transformer manufacturer catalogue has to show 2 kinds of losses:
no-load losses and load losses. The no-load losses are approx. equal the iron laminate losses.
Subtracting no-load losses from load losses you’ll get copper losses.
Of course, it is an approximate solution. However it is practical for transformer resistance appreciation and even as HVAC data.
If you intend to calculate the losses at another load current you have to keep no-load constant and to multiply the copper losses by (Inew/Irated)^2.
If you have not the manufacturer guarantee report where it is noted the above losses it will be difficult to know the losses since the core losses and copper losses could be different-mainly according to price difference between copper and iron laminate at bid date.
See for instance :
http://www.coned.com/es/specs/EO-5015.pdf
https://www.mne.ksu.edu/people/personal/white/pdf-files/heat2.pdf
For a transformer of 3750 kVA the total losses have to be from 35 kW to 75 kW-in my opinion.

 

[kwired]

Sorry for all of the questions guys, I'm new and just trying to learn. So how do I use that table that ron posted in order to find the heat loss? Meaning, how does efficiency relate to heat loss? Is it the same thing?

If it were 100% efficient it would not give up any heat. Heat is caused by resistive losses, you will always have some of those.

Here's what i found. "Full load losses for 150 [degrees] C transformers range from about 4% to 5% for 30 kVA and smaller to 2% for 500 kVA and larger." So if I have a 3750 kVA transformer, that means it would be 2% heat loss correct?

A transformer at full load is usually more efficient then one at no load, but that is efficiency as it relates to the load conditions. Transformer efficiency may decrease if lightly loaded, but also is derived from a lower overall threshold.

 

[Besoeker]

If it were 100% efficient it would not give up any heat. Heat is caused by resistive losses, you will always have some of those.

And iron losses.