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Heat rise 3

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PeacockBlue

Mechanical
Oct 29, 2009
1
A switchroom has 2 air conditioning units, each with a sensible capacity of 52kW.

The heat emmitted by the switchgear is 75 kW.

If one of the a/c units fail, how do I calculate what the heat rise will be and how fast the heat rise will occur?
 
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75 - 52 = 23 kW

So, in effect you are having a heating capacity of 23kW inside the room when one unit fails.

You want to know what temperature will occur in the room and how fast temperature will rise.

The temperature that will be reached is given by the equilibrium that is reached when the temperature in the room gets higher than the temperature of the surrounding area (outside?).

At which temperature(-difference) does 23kW "leak" from the room to the surrounding. If this is not outside then the 23kW leaking will also raise the temperature of the adjacent rooms, thus complicating the problem...

How fast the temperature will rise depends on accumulation of heat in various parts making up the room (walls, floors, ceiling, equipment not producing heat itself).

If your equipment room is not very large then 23kW will heat up the volume of air inside the room pretty quickly.

Depending on what temperature is the absolute maximum for the equipment in the room and depending on your local climate it might be possible to keep roomtemperature at a still acceptable level by venting with an appropriate amount of outside air.

Example: max. room temperature 40°C
max. outside temp. 30°C
difference = 10°C

7500 m3/h at 10°C = 25.2 kW cooling
 
Let’s assume your room is perfectly insulated (no heat loss)and your switchgear is the only heat source.

The temperature of your room will be increased of the amount dT per each hour

dT = (23*3600)/(V*Cp)

where:

Cp = 1.18 kJ/(m3*°C) specific heat of air
V = m3 room volume

In a real world you have to evaluate the heat loss through the walls, the ceiling, the floor, the doors, the windows, etc that must be algebraically added to heat generated by your source.

As general formula for each surface you have to take into account the amount:

Qloss = Ki*Si*dT’

Where:

Ki = [W/(m2*°C)] is the thermal transmittance for the surface i

Si = [m2] is the area of the surface i

dT’ = [°C] is the temperature difference between the inside and the outside your room
 
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