#### dudikr2

##### Member

- Location
- New York, NY

- Thread starter dudikr2
- Start date

- Location
- New York, NY

- Location
- New York, NY

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

- Location
- Seattle, WA

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?

- Location
- New York, NY

- Location
- New York, 40.7514,-73.9925

Any inefficiency results in heat.

- Location
- New York, NY

- Location
- New York, NY

- Location
- Seattle, WA

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.

- Location
- New York, NY

- Location
- New York, NY

- Location
- New York, NY

- Location
- Lancaster, PA

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.

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 %

- Location
- New York, NY

Okay thanks!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 %

- Location
- San Francisco Bay Area, CA, USA

- Occupation
- Electrical Engineer

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.Okay thanks!

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.

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.

- Location
- NE Nebraska

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.

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.

And iron losses.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.