Heat load calculation in Electrical rooms

[iceemani]

Calculatiing heat load for an electrical building, to give my HVAC counterpart for the design on air conditioning.

Reffering to standard practices, i notice the heat loss on demand load of the equipments are factored as directly proportional to the full load heat loss.

Where the heat loss is Square of (I) * R, so for a 50% demand, which is I/2 , the heat loss is a quarter(or 25%) of the full load heat loss, not 50% of full load heat loss.

i am trying to find technical standards. papers on this subject but whatever i read considering a direct proportional reduction on heat.

have anyone have experiance on this, where the "Square of (I) * R" function applied to arrive the heat load for the demand load. Highly apprecaite if one can share their experiance or thought.

 

[templdl]

I know I have information at one time in regard to the BTUs for various size breakers. So such information is available if you get to the right people. There should be similar data for motor starters. With transformers there are no load and full load losses data we hich is commonly available.
There may be data for panerlboards also.which most likely would be directly related to the breasker's installed in the panel. But there also are losses in bus and cable.
I know that I was able to provide the heating watts iin the past.

 

[Julius Right]

In my opinion, you could consult "Heat Loss from Electrical and Control Equipment in Industrial Plants"
www.mne.ksu.edu/people/personal/white/pdf-files/heat2.pdf

 

[Julius Right]

There are some devices in which the heat losses are not exactly direct proportional with current as
transformers and motors where part of the losses come from steel core [laminates] or enclosure and
the ventilation and friction losses in a motor. In a control relay-or contactor- for instance, the operating coil losses
do not depend on the current flowing through contacts-which losses usually are negligible.

 

[Haji]

Calculatiing heat load for an electrical building, to give my HVAC counterpart for the design on air conditioning.

Reffering to standard practices, i notice the heat loss on demand load of the equipments are factored as directly proportional to the full load heat loss.

Where the heat loss is Square of (I) * R, so for a 50% demand, which is I/2 , the heat loss is a quarter(or 25%) of the full load heat loss, not 50% of full load heat loss.

i am trying to find technical standards. papers on this subject but whatever i read considering a direct proportional reduction on heat.

have anyone have experiance on this, where the "Square of (I) * R" function applied to arrive the heat load for the demand load. Highly apprecaite if one can share their experiance or thought.

That is not the way to find heat load due to electrical equipment because the 'R' is not readily calculable/measurable unless the equipment is a pure resistor. Easier way to do so is through efficiency or rather, inefficiency of an electrical equipment as readily given by manufacturers.

 

[GoldDigger]

There are two components to the heat load in the electrical room.
One is the sum of all of the motors, contactors, etc. which are located in the room.
The other, and perhaps what the OP is referring to, is the I^2R loss in the wiring itself which is feeding loads everywhere else in the building. You could calculate that based on the estimated resistance if each wire and the calculated load on that wire.
The second contribution will be small and possibly not worth calculating exactly.
For a 200A 120/240 service an upper limit would be the load current times two times the expected voltage drop. The voltage drop inside the room will be less than 1 volt conservatively.
For this example an upper limit would be 400 watts.
Add in idling and running losses for any transformers in the room. That may well be the biggest number.