Heat load of a Data room 1

[Hileg]

How can I calculate the heat load of a computer room. the power consumtion is 210,000 watts of variuos types of servers and mainframe. The room is 80' by 65' by 8'. the room temp is maintained @ 70 degrees. How would I calculate the tempature rise of this room if all the lieberts units were turned off. I need to know how quickly the temp. would come up in say 10 minutes. Please help. thanks

 

[friartuck]

Your heat load seems reasonable. Generally 650W/m2 is accepted giving 313000W load.

The heat rise would be very quick so I would suggest some sort of standby.

I assume you are using close control cabinet units (downflow into suspended floor)

So 4 x 75kW units would give the duty...but 5 units at 75 would give some redundancy.(Dx type Assumed) Otherwise use same priciple on your chiller selection with some redundancy buit in.

Some mainframe /server companies have there own spec for how much standby is needed...for instance if ICI or HSBC banking went down, the cost would be millions a minute loss...so cost can sometimes be no object.

The rise in temp would be related to a number of factors, but if we assume a typical modern well insulated building with low thermal mass, then the rise will be measured in minutes rather than hours.

if you are good at maths the simply equate heat in=heat out

300000W in={Area walls x U Value x Delta T}+
{Area Floor x U Value x Delta T}+
{Area ceiling or roof x U x DT) etc

where Delta T is Ti-To

Ti is inside temp
To is outside temp (Say 27 celcius in UK)


This is simplistic as it ignires solar gain etc but it will give you a good idea. I would not be surprised at a temp of 50-60 or do degrees in 5-10 mins of total AC failure.

The estimated internal space temp at equilibrium for average U values and 650W/m2 incidentally is around 300(yes 300) degrees C. The PC's would have melted long before.

So general advice..get a back up plan

Friar Tuck of Sherwood

 

[HVAC68]

These days close control units come with dual cooling coils - i.e. a chilled water coil as well as a DX cooling to act as a back-up. In case there is a problem with the central chilled water system, then the DX condensing unit starts and delivers cool air. This way space which is a premium in most places can be saved. You can also have standby motors for the evaporator fan.

HVAC68

 

[friartuck]

For high quality close control units, go to airedale.com

Friar Tuck of Sherwood

 

[imok2]

Hileg, Here's what i see
1. Room cuft = 41,800
2. 210,000 watts = 714,000 BTUH
3. 714,000 BTUH devided by 6 (10min)= ~120,000 btuh/10min
Assuming even distribution of heat: then
4. 120,000 btuh devided by 41,800 ~ 3*F

 

[friartuck]

Like your approach IMOK

But i get:

Q=mc DT

210kW=1177m3 x 1.205(density)x1.01kJ/kGk(spec ht)x DT/60sec

=8.78deg C per minute excluding heat sink effect and coincidental losses

so in 10 mins temp would be 87 degrees above room temp (celcius)

Friar Tuck of Sherwood

 

[imok2]

If I have a heat load of 210,000 watts that equals 60 tons of refrigeration/hour, now if I devide that hour by by 6 (10min) I get ~ 120,000btu. Now I have a room of 41,800 cu ft to absorb the heat.So if I take the total volume and devide it into the total heat generated in 10 min I get about 3*F. That seem quite reasonable to me I can't for the life of me ;see how you will get 87C in 10 min when the total heat generated is 714,000BTU in 1 hour. The time involved is 1/6 of that. Please explain in more detail. In addition,its watts not KW

 

[friartuck]

Hi Imok

The equation you use is Q=mc DT

Presumably you need somewhere to convert volume to mass and also you need the specific heat capacity. I'm not sure what the SHC of air is in BTU/Lb

or the density in lb/cu ft

However if you say Q=mc DT

density =m/v so m= density x vol (1197m3 x 1.20g kG/m3 for metric density)

C= 1.01 kJ/kG deg C

so for a rise of 8.78 deg in 1 min

Q= 1197x1.205x1.01x8.78deg/60sec=213kW (rounding up error)

likewise if you used the imperial equivalent you should get the same result

cheers

tucky

Friar Tuck of Sherwood

 

[imok2]

Frier tuck Thank you for your patience and your help in understanding what I was doing wrong I worked it out in Imperial and came up with the following:

Amount of Heat Gain/Loss
BTU: 120000/10 min.
Weight of the Substance: 3082 lbs
Specific Heat of the Substance: .24

Change in Temperature: 162.18*F
I am amazed at the temp rise! Never too old to learn

 

[quark]

Imok2!

Actually I spotted the error yesterday before your posting. You were calculating hourly volumes in your initial equation though your equation is implicitly expressed in hours causing double reduncancy (1.08 = 0.075x0.24x60).

But I was sure that you would find out this and you did it. Moreover, an engineer gets better satisfaction if he himself comes to know the error.

Keep it up.

PS: No tags attached

Regards,

 

[ChasBean1]

Imok, trust your instincts.

Friar, your calcs give a result in KJ, not KW and look to be meant for a fixed delta-T. Rate of change of temperature would rise in a square root relationship and be asymptotic to some more reasonable number, not the spontaneous combustion values shown.

No answers now, just criticism (hey, it's a lot easier to bust up someone else's good work than to create my own)...

Best,
CB

 

[friartuck]

Hi Chasbean

KJ/sec =kW

I couldn't show m as mass flow (M with a little dot above it) but if you look at the calc, it has secs in it

ciao

Friar Tuck of Sherwood

 

[McCormick93]

I recently did an analysis like this for my data center. Some things to be aware of:

1. If the 210,000 Watts are the total UPS load, don't forget to take out the losses incurred by the UPS (which can be 10-20%). Unless, of course the UPS systems are in the same space as the servers.
2. Not all of the power consumed by the servers is converted to heat. This is especially true for automated tape storage robots, big onboard cooling fans, and standalone air conditioning units which are converting electricity into mechanical work instead of heat.

My method for calculating temperature rise goes like this:
210,000 Watts = 12,000 Btu/min

Assuming we start at 70F and 50% RH, The mass of 41,600 ft3 of dry air is (check psychrometric chart):
41,600 ft3 / (13.5 ft3/lb) = 3081 lb

So every minute we are adding 12,000 Btu/3081 lb,
or 3.9 Btu/lb

Again referring to the psychrometric chart, at the starting point of 70F/50RH the heat content is about 25 Btu/lb. Add 3.9 Btu/lb (moving horizontally because we assume no humdification), we end up at 28.9 Btu/lb, where the dry bulb temperature is around 85F and RH=30%.

15 degrees per minute is pretty severe, but it also seems that an 8ft ceiling is kind of low for a 40W/ft2 data center.

Imok2 is vindicated in my opinion!

 

[friartuck]

Again, like your different approach McCormick, it just shows that theres more than one way to skin a cat (not that I dislike cats). Your figues give 15 F rise per min or 8.3C per min. (I tried it in kJ and C and the result is spot on)

I must admit, whenever I do the calcs for heats gains due to competers, I tend to get confused what figure to use. Heat gains from PC's and servers changes every time a new chip comes out. I have worked on any value from 90W to 300W gain for a PC. Some modern chips run at a very high temperature, though the processor doesn't run at full blast all the time.

ASHRAE I believe currently suggests using about 180W per PC which I assume allows for diversity since the power supply will be around 300W or more.

I once used a current transformer to measure the current passing through my home PC (pentium 4) and was surprised how low it was. I assume that when we estimate loads, we must generally over estimate by quite a bit.

The point of this garbled message is therfore, does anybody have any new test figures for realistic heat loads from PCs and associated equipment?



Friar Tuck of Sherwood

 

[eddie11]

Friar Tuck,

here is some empiracle data from Sun Microsystems:

http://docs.sun.com/source/816-1613-13/Chapter5.html#34676

http://docs.sun.com/source/816-1613-13/Chapter4.html