## Calculation of air temperature rise due to large,short term heat load

## Calculation of air temperature rise due to large,short term heat load

(OP)

I am an electrical engineer that needs a little HVAC help. The guys I know hadn't done this before. I have a medium voltage generator control switchboard which utilizes resistors to disipate heat generated in case of a ground fault. My system is 2400/4160/3ph. The resistor can sustain 400 amps for up to 10 seconds. The calculated energy I need to dissipate is 960,000 watts which can be generated for up to ten seconds. Thats about 3.3 million btuh for ten seconds. Can anyone tell me the associated rise in the adjacent area? The resistors consist of stainless steel bars with ceramic insulators in a stainless ventilated enclosure on top of a switchboard in a room that's 60'x40'x 24'high. Oh yeah, there's a couple of 1000kw Generators and their silencer's in there too. But I am just concerned with the area around the resistors. How hot is it going to get? Any clues? Any help is appreciated. Thanks

## RE: Calculation of air temperature rise due to large,short term heat load

Let me take a stab. Everyone get yer pencils out and shoot me down if I'm off base. You're going to release 9,100 BTU into the surrounding air in ten seconds. Even if half the volume of your room is filled with equipment, there are still 28,000 cubic feet of air in which the heat will eventually diffuse. At 80 degrees F, that's about 2060 lbm. The specific heat at those conditions is about 0.24 BTU/lbm/degF. So,

9100 BTU / (2060 lbm * 0.24 BTU/lbm/degF) = 18 degF temperature rise if there is zero ventilation.

It's linear with how much air is in the room, and if there is enough ventilation to equal one air change per hour the effect will be almost totally canceled out within one hour.

Of course, inside the enclosure with the resistors it will get much hotter in the short term. That would be a tough calculation...

Hope this helps a bit!

DRWEig

## RE: Calculation of air temperature rise due to large,short term heat load

when I read both of the posts ..I guess my brain must be torqued but the numbers don't jive(match) or am I not seeing something?

## RE: Calculation of air temperature rise due to large,short term heat load

So this is my view of EEJaime’s situation:

The resistors release 960,000 watts into the room in 10 secs.

That is equivalent to:

96kWatt/sec x 3412Btu/kWatt = 327,600 Btu/sec!

The room vol. is 60 x 40 x 24 = 57,600 ft^3

Therefore the average temp rise in the room is:

delta T = 327,600Btu per sec/(57,600ft^3 x 0.02 btu/ft^3/deg F)

= 284 degF/sec!!!

This is an average temp rise. Clearly the temp rise local to the resistors will be huge because this quantity of heat would not dissipate into the whole volume of the room very quickly.

Unfortunately my knowledge of mathematics does not allow me to calculate the temp at a given distance although I’m tempted to think that the temp decay wrt distance from the source would be proportional to the square of the distance.

The above result seems bizarre but looked at another way what we have is a MegaWatt(almost)of energy being released in 10 secs. into a room which is approx the volume of a two storey house!

Am I out to lunch here? Have I missed something? Anybody any other ideas; it’s a fascinating problem.

I do have one question Jaime. If 400 amps at 2400/4160/3ph go through a resistor does all of that energy get dissipated? I presume your calc is 400 amps x 2400 volts = 960,000. Is that correct?

## RE: Calculation of air temperature rise due to large,short term heat load

Is the resistor bank sitting naked in the room and convectively cooled? I think not. The total energy for the 10 seconds is 9.6 MJ, which would Joule-heat any material to melting point without some sort of mitigation.

Most such loads have to be massively cooled:

http://www.simplexdirect.com/LoadBank/specs09.html

For the Saturn model rated for 1800kW, the required airflow is 40,000 cfm. Based on that spec, the airflow is removing 56 MJ/min under steady-state conditions, with an 83°F air temp rise. Not clear about the specifics of the internal thermal conductivity.

But, given the air flow, after 5 minutes, you'd have 7 air-exchanges, which should be long enough to get to a new steady state.

TTFN

## RE: Calculation of air temperature rise due to large,short term heat load

The load would be there for up to ten seconds then the circuit breaker would be tripped, but the energy would heat up the resistor, which would in turn release the heat into the surrounding space. It would be released as radiated heat but the majority would be in the stainless steel bars which make up the resistor. It would be radiate into the space until the temparature returned to a steady state condition between the resistor and the room.

There should be some ventilation because the generators are feeding the emergency power system which in turn feeds the fans,(four fans exhausting 22,600cfm each). There are two generators in parallel, (eventually four), so if there were a fault in one, the other would probably still be available to provide power. The chances of a ground fault in multiple machines at the same time is remote unless something really disastrous occured and my question would probably be moot.

In any case, thank you all for your posts, it certainly gives me more information I can use.

## RE: Calculation of air temperature rise due to large,short term heat load

The resistors will get very hot, and then cool down much more gradually. The resistor manufacturer should be able to provide data on the heat transfer characteristics.

## RE: Calculation of air temperature rise due to large,short term heat load

In 10-seconds, the heat loss from the resistors to the air is not going to amount to more than a few BTU's unless these resistors are a whole lot larger physically than I would expect. The resistors are going to heat up quickly for that 10-second period; then the breakers will trip; then the resistors will cool down, yielding their heat to the air.

If the resistor weighs 200 pounds, most of which is the Stainless bars and the dielectric materials around it, along with some extent of the conductors on both sides...The heat rise is going to be something like 380 deg. F., ignoring the heat loss to the air.

If these are just simply exposed bars inside a box, the box is going to get warm as well...to know how warm requires a lot of insight into its geometry, ventilation characteristics and weight as well.

But as indicated by others, the heat will eventually dissipate and raise the local air temperature, such that if it takes 20-min for all to cool down, the temperature profile of the room interior will be locally high in the immediate area of the box.

How high has very little to do with air changes but has a whole lot more to do with the dimensions of the resistors and their containment. I'd be more worried about somebody trying to open the box then I would be about the local air temp.

## RE: Calculation of air temperature rise due to large,short term heat load

EEJAIME

## RE: Calculation of air temperature rise due to large,short term heat load

you may want to consider a dry sprinkler system with initiation from a sensor(heat/smoke) further away

Roger

## RE: Calculation of air temperature rise due to large,short term heat load

## RE: Calculation of air temperature rise due to large,short term heat load

960kWatt x 3412Btu/kWatt x 10 secs/(3600 secs/hr)

= +/- 9,100 Btu's

Hopefully I only confused myself.