DISSIPATING HEAT from TRANSFORMER 3

[TheRichYS]

Can anybody help me on calculation of getting dissipating heat from transformer? I guess there is a chart or something that I can get the heat that comes out from transformer.

I need the quantity of heat dissipated from 3 transformers in kcal or in Joule

I need this information to calculate the size of the Exhaust fan and supplying louver for substation room.

Room Size: 85'-4"(L) x 30'(W) x 18'(H)
Location: side of the building

Sizes of 3 transformers in substation room:

TR1
2000KVA, 13857.5V, 45.31A, 1066KW, PF=1

TR2
2667KVA, 13840V, 58.13A, 1296KW, PF=-0.96

TR3
2667KVA, 13872.5V, 27.81A, 634KW, PF=1

Supplying Air Temperature via Supplying Louver = 70F
Current Room Temperature of Substation Room = 90F

Thank you very much for your kind replies in advance.

 

[DanDel]

You can calculate heat loss from a transformer directly by using the losses (in Watts) from the factory test sheets. Whatever power goes into the line side and doesn't come out the load side will be dissipated as heat loss.

Obviously, the losses are proportional to the load current, and you may have to estimate the load.

Watts measured over time become energy (Watt-hours), which can be converted to Joules (1 Watt-hour = 3600 Joules).

 

[TheRichYS]

I have found the load loss which was 19696 W in total.
Can anyone help me to find out how to calculate Heat Loss in Watt-Hour from the load loss in Watt?

 

[davidbeach]

Look at your units/dimensions. If you have Watts and you want Watt-Hours what do you need to do to make the units work?

 

[dpc]

Watts are power. Energy is power over time. If heat loss is 19.6 kW, in one hour this is 19.6 kW-h of energy. Just multiply this by how many hours you're interested in.

What DanDel is saying is that ALL losses in a transformer are heat loss. All losses (eddy current, resistance, etc) all end up as heat rejected into the oil or the surrounding air.

 

[davidbeach]

Actually, Watts are a much more useful value for calculating cooling requirements than Joules anyway. The cooling has to match the heat production at all times, so time cancels out and you are left with heat in vs. heat out.

 

[TheRichYS]

Thank you very much for all your input!!!
I have got handled this matter using all your help.
Thank you.!!