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Telephone Equipment Room Cooling

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NDEngineer

Mechanical
May 25, 2004
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Forgive me for my ingnorance, but I have what may be a elementary question regarding heat gain and DX cooling. I am cooling a telephone equipment room with the following loads:

Transmition and Solar Loads: 5500 BTU/hr
Equipment Loads: 20500 BTU/hr
~300 CFM F.A.: 9200 BTU/hr (sensible) 6000 BTU/hr (latent)

Total sensible cooling: 38000 BTU/hr (w/ supply fan load)
Total coil load: 44800 BTU/hr

OK, the problem is that the space is to be maintained at 65 deg F (O.A. design - 95 db, 74 wb). To keep the space at this temperature, the required CFM (~2500 assuming a leaving db temp of 55 db) is substantially higher than the rated CFM on the unit the owner INSISTS on using (1500 CFM for the 4 TON unit).

I'm not sure if the fan is direct or belt drive (I'm meaning to find out). IF it is a belt drive, can I just increase speed of the fan until it is able to give an output of 2500 CFM? Will the coil be able to keep up with this?

IF the fan is direct drive, would it be OK to oversize the unit (say a 5 TON w/ 1700 CFM)? Any other ideas?

Sorry for the long post, but I'm pretty lost as to what to do in this situation. Thanks in advance for any input.
 
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Is this a split system? What unit is the owner insisting on using -- the air handler, or condensing unit? How are you coming up with a need for 2500 CFM airflow?


Concerning the increasing of the fan speed:
If the air handler is designed for a nominal 4-tons, then increasing the airflow by that much will probably give you an excessive coil velocity. This becomes a big problem, because water condensing on the coil is carried downstream, instead of falling into the drain pan.

---KenRad

 
Thank you for the reply. Yes, it is a split system...the unit the owner wants to use is the air handler. Whatever size I choose, I plan to just match the condensing unit accordingly.

I can understand the problem of excess coil velocity, which is why I'm searching for an alternate solution for this problem.

Here's how I come up with 2500 CFM:

CFM = (zone sensible load)/(1.08*dT)

CFM = (20500+5500)/(1.08*(65-55))

CFM = 2407 (actual) = ~2500 (standard)

This number is confirmed by the Carrier Bloack Load Program. If I use 74 db at the inside design temp, everything would match up just fine...it is the 65 deg that is throwing me off.

 
it is not "uncommon" practice to use a 5 ton blower (around 2000cfm) and evaporator coil with a 4 ton condensing unit. If it is a heat pump application, i might stray from this approach due to volume mismatch during refrigerant charging. Check with the equipment manufacturer.

Something else to think about...your leaving air temp will not be 55 anymore.
 
So would this arrangement work?

5 Ton Air Handler (1700 CFM, 57,500 BTU/hr cooling cap.)
5 Ton condensing unit

Should I use a 4 ton condensing unit instead? Would I be able to use the cooling coil packaged with the 5 ton air handler or would I have to match it to the 4 ton CU? I'm very new to this but I'm willing to learn!
 
The air quantity is substantially high due to your requirement of 65 deg F. That must be very cold I suppose !!! High amount of sensible heat removal resulting in higher air quantity. This would also mean that the ADP is lower than normal resulting in lower Saturated Suction Temperature (SST). A standard unit would most likely not wotk in these conditions, unless you go for the specialised close control or precision control units. The other way would be to select an indoor unit which meets the air quantity and capacity. Use the rating charts and not rule of thumb. Select the condensing unit based on the SST obtained from the indoor unit selection. Most companies have computerised selection charts these days. They should be able to help you.

HVAC68
 
Thanks for the informative reply HVAC68. What exactly are you refering to when you suggest an "indoor unit"?

Apparently the units I was referring to early are a packaged deal...they contain a compressor and condenser coil, so there isn't a whole lot of options. I'm wondering what will happen if I use the 5 Ton unit (1700 CFM, 57,500 BTU/hr cooling cap) for this application with the given conditions. Is there a good chance of the coil freezing?
 
I don't know of anyone wanting 65°F in a telephone equipment room. Packaged AC systems are rated for 80°F indoor and 62°F to 67°F entering wet bulb temperature. Maybe 65°F is minimum, not design indoor. If room is critical consider backup AC from main house system.
 
If you have a standard packaged type of equipment, then chances are that they are rated for 80 deg F indoor temperature and around 62 to 67 deg F WBT as indicated by lilliput1.

By indoor unit, I meant an Air Handling Unit or Fan Coil Unit. With a standard packaged type of equipment, you may not be able to achieve the kind of capacity / air quantity requirements as would be required for a high sensible load application. Either try using separate indoor unit and outdoor units (mix-match) or try with close control unit manufacturers.

Last, but not the least, check the design criteria of 65 deg F. Seems tooooooooooo low for a telephone equipment room application.

HVAC68
 
Thanks for all of the help guys. You're right...the units are rated for 80/67 inside temp.

The design guidelines I recieved stated that the space was to be maintained at 65 F and 50% relative humidity. I tracked down where this info came from, and the source told me 65-68 F. After that, I contacted the person that maintains the same type of building with similar units, and he said they keep the temp around 68-70 F. Cooling to 65 and cooling to 70 is a big difference!

So I think I have it figured out...one last question out of curiosity if anyone is willing to answer: If the capacity is rated at 80/67, how do I convert that capacity to a different indoor condition (say 70/60)?
 
Better contact the equipment manufacturer. They should have rating charts. I don't know of any thumb rules or multipliers to convert from 80/67 to 70/60.

HVAC68
 
There are some flaws in your design IMHO.

1. ADP corresponding to 65F and 50% RH is 45F and part quantity of air should be cooled down to this temperature to get required moisture reduction.

2. Considering 2500 cfm and 10F rise in the control space, bypass quantity comes out to be as high as 43.4%(with 10% fresh air). You may have to go with separte ducting for bypass for this huge volume and control of the two streams becomes difficult.

3. 10F seems to be critical as you have not considered heat gain from ducting and other unaccounted losses.

4. 45F dew point with DX system is just at the lower limit.

I feel that the owner is on right track knowingly or unknowingly. I suggest you to design your system for 20F temperature and reducing the air flow rate to 1500(approximately). With a 5% bypass factor across the DX coil, the problem may not be severe.

If RH control is very critical then I suggest you to check desiccant dehumidifiers.

Regards,




 
Regarding the conversion of ratings from 80/67 to 70/60, I am assuming reference to wb/db temps in deg F.
If this assumption is correct, can anybody explain what is incorrect about plotting the 2 points (two mentioned above) on a Psychomnetric chart, finding the difference in enthalpy (kJ/kg in my language) and then one can easily rate the machine for the new conditions:

Knowing air =1.2kg/1000 liters
i.e. OA=95/74> 35C/23.3C > 111kJ/kg (out of interest)

Condition 1: 80/27> 27C/19.5C > 55kJ/kg
Condition 2: 70/60> 21C/15.5C > 45kJ/kg

I shall use 2000CFM as an example:
2000CFM = 944.6 liters/sec assume 1000 Liters/sec > 1.2kg/sec
Diff between Conditions 1 and 2, are 10kJ per sec which equates to 10kW extra cooling required. Condition 2 therefore requires extra cooling. Can anybody please comment on the above approac, I hope it is what you are looking for???

Sorry for the metric system, but this is 2004 :)

K-v3



 
What your calculation shows is the increase in required capacity of the system.

When refrigeration systems are run below the designed temperatures, their capacity gets reduced and this is called derating. This can be quantified (only)experimentally. Manufacturers can provide, upon request, derating data at various temperatures. Variables are mass flowrate of refrigerant, expansion device size, condenser and evaporator heat transfer areas etc.

Regards,


 
I took a moment and thought about this:

You want the ability to maintain 65°F in the space. At that low temperature, I chose a 60% (not 50%) humidity for the design condition. This is more reasonable using standard HVAC equipment.

First, use your known gain (44,800 BTU/hr) in the room and the difference in two known enthalpies (supply air at 52°F, 50.8°F dew pt. which is slightly below saturated given cooling coil bypass factor) and required room air of 65°F and 60% RH. This difference is 24.23 BTU/lbm - 21.03 BTU/lbm = 3.2 BTU/lbm.

From the equation: Q = 4.5 cfm dH and solving for cfm gives 3,111 (see * below).

Because of the use of the space, very little outside air but a lot of cooling is required. I assumed the recirculated air to outside air ratio to be 90:10. This mixture combines (based on your given OA conditions of 95 db, 74 wb combining with room conditions of 65 db, 60% RH) to 68°F and 57.6% RH. This is the total delivery air (3,111 cfm) that passes through the cooling coil.

We know the Q of the room; you gave that. Solving now for Q of the coil based on change in enthalpy between mixed conditions (68 db, 57.6% RH = 25.533 BTU/lbm) and delivery condition (52 db, 50.8 dew point), you get 63,080 BTU/hr or a 5.3 ton unit.

* Stated heat outputs are often significantly less than actual heat outputs. Design criteria often significantly exceed the actual equipment heat output. Basically, the calculated 3,111 cfm unit or 2,500 cfm unit might both be oversized. I say go with what the owner insists on, collect data, and let us know the result!

CB
 
CB,

If the supply air temperature is at 52[sup]0[/sup]F and room temperature is to be at 65[sup]0[sup]F, then with 3111cfm, the total heat flowrate will be 1.08x3111x(65-52) = 43678 btu/hr. But as the room load is purely sensible(from the data of OP) this makes the room much cooler if we don't provide reheat. I may be missing something. Can you please let me know why you are using total enthalpy difference across the coil to calculate air flow rate rather than room sensible load?

Regards,


 
Quark, you're right. I didn't read the givens closely enough. Room heat gain is 26,000 BTU/hr sensible, which gives 1,852 cfm given a 52°F/50.8°F dew pt supply and a 65°F/60% RH room (this calc's to about the same using 4.5*cfm*dH, as it should with no change in latent).

Now we know we need 1,852 cfm of supply and this will be a cooled and dehumidified mixture of 1,552 cfm 65°F/60% RH return air plus 300 cfm of 95°F/74°F wb outside air.

Solving for Q based on enthalpy difference at the coil, we get 44,212 BTU/hr or 3.7 tons...

Hopefully this makes more sense. Thanks for pointing that out, Quark. -CB
 
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