Heat Exchanger Question

[JasonVanHalle]

I'm working on a heat exchanger which has the purpose of cooling a semi-enclosed area. My question concerns how much the electronic devices in the enclosed area will increase the temperature. From what I remember of Thermodynamics, whatever the rated power consumption of the device is should be how much energy is in turn increasing the temperature of the area.

For example: If my electronic device is rated at 1kW for power consumption, is 100% of that 1kW going to be increasing the temperature of the surrounding area, or is there a measurable amount that goes towards making the device function?

If I have 10kW worth of devices in the area, do I need a heat exchanger capable of removing 10kW of heat in order for the temperature of the area around the devices to remain constant?

I'm almost sure of my answer on this, but I'm looking for some backup as well, because one of the higher-ups is challenging this idea.

Thanks!

[Kelly_Bramble]

Here's how one would handle an enclosed room when sizing HVAC.

Factors to include:
· The floor area of the room
· The size and position of windows, and whether they have blinds or shades
· The number of room occupants (if any)
· The heat generated by equipment
· The heat generated by lighting

Window Size and Position
If windows exist
South Window BTU = South Facing window Length (m) x Width (m) x 870
North Window BTU = North Facing windows Length (m) x Width (m) x 165
If there are no blinds on the windows multiply the result(s) by 1.5.
Obviously if you are in the Southern Hemisphere you would swap the conversion factors as the heat on North facing windows is then greatest.
Add together all the BTUs for the windows.
Windows BTU = South Window(s) BTU + North Window(s) BTU

Occupants
Purpose built Server Rooms don’t normally have people working in them, but if people do regularly work in your Server Room you will have to take that into account. The heat output is around 400 BTU per person.
Total Occupant BTU = Number of occupants x 400

Equipment
This is trickier to calculate that you might think. The wattage on equipment is the maximum power consumption rating, the actual power consumed may be less. However it is probably safer to overestimate the wattage than underestimate it.
Add together all the wattages for equipment and multiply by 3.5.
Equipment BTU = Total wattage for all equipment x 3.5
Lighting
Take the total wattage of the lighting and multiply by 4.25.
Lighting BTU = Total wattage for all lighting x 4.25

Total Cooling Required
Add all the BTUs together.
Total Heat Load = Room Area BTU + Windows BTU + Total Occupant BTU + Equipment BTU + Lighting BTU
This is the amount of cooling or heat removal required.

[JasonVanHalle]

Thanks Kelley - I appreciate your very thorough and very detailed answer. I was mainly looking for the 1 watt of equipment to 3.5 BTUs of cooling capacity.

I'm a machine design guy, so I haven't touched Thermo anything since college. I'm just trying to temporarily step into another division within my company to help troubleshoot issues they're having with the product.