HVAC Calculations

It would be almost impossible for there to be a one-size-fits-all or rule-of-thumb anawer to this question. The control room may have various relays, contactors, lights, computers, and the like whose coils will add heat to the room. The room will need to be maintained at an acceptable temperature. If the exterior of the room is hotter, then you will need to remove heat from the room. On the other hand, if the exterior is colder, you may need to add heat.

You need to know the insulation value of the room's envelope (floor, ceiling, walls); heat infiltration/exfiltration through doors and other openings; heat sources in the room such as people and the aforementioned electrical loads, etc. Then you will need to select equipment to add or remove heat as needed. There isn't really any good shortcut.

The reason I even assumed there is one as there is a rule of the thumb for kw cosumption in watts per sq. ft for commercial buildings but, of course depending on type and application of structure.

Thanks for your input.

This is not correct for a control room, maybe for a house or small commercial venture where the "rule of thumb" has been proved out by repetitive calculations in similar situations.

I have performed this work on a 20'x20' (400sf) equipment/server room where the components were stacked in a controlled environment. The heat dissipation from the equipment required 5 tons of cooling with an additional 5 ton system (read redundancy, extra capacity and room for growth) on stand-by as a second stage of cooling as well back-up in case one system failed. This room was the centre for a 24/7/365 international corporation, critical for the operation of their business. Also, we required a water-less fire protection system. Very complex and detailed.

how every one ton require 1.25 where 1 ton = 3.6 KW please clarify on my mail

mohab.nashed@orascomci.com

mouhab_01234@yahoo.com

just i need to know, may be i'am wrong, so please reply.

thank you

as a rule of thumb I've always used 12 watts per sq ft

The rules of thumb given so far may be reasonable for generic residential structures, or even such occupancies as schools, hotels, prisons, and the like. But not for industrial occupancies or control rooms, which need to be analyzed more closely. For instance, if you want to put a control room next to a bank of blast furnaces, you had better ignore the simplistic solutions.

For comfort applications like residential a.c for a tropical ambient conditions thump rule is 150 to 200 Sq.Ft floor area per TOR. But again it depends upon the inside conditions. But when it is a control room you have mainly sensible heat load. You may have to check the capacity by doing heat load calculations. If it is a small room say 5X 4 M with normal height 1.5 T R window unit is enough to provide 24 to 25 C.

Hi jraubsr,

Hope information below will help you.

Heat Load - the amount of heat generated is know as the heat load. Heat is measured in either British Thermal Units (BTU) or Killowatts (kW). 1 kW is equivalent to 3412BTUs.

The heat load depends on a number of factors, by taking into account those that apply in your circumtances and adding them together a reasonably accurate measure of of the total heat can be calculated.

Factors include:

The floor area of the room

The size and position of the window

The number of room occupants (if any)

The heat generated by equipment

The heat generated by lighting

Floor area

The amount of cooling required depends on the area of the room. to calculate the are in square meter

To get Room Area BTU = Length (m) x Width (m) x 337

Window Size and Position

If your room has no window then you can ignore this part of calculation, how ever as rule of thumb there are window you need to take and orientation into account.

South window BTU = south facing window Lenght (m) x Width (m) x 870

North window BTU= North facing window Lenght (m) x Width x 165

If there no blinds on the window multiply the results by 1.5

Add together all the BTUs for the window

Window(s) BTU = South window(s) BTU + North window (s) BTU

Occupants

You will have to take into account people who normally working in the space. the heat output is around 400 BTU per person.

Total Occupants BTU = Niumber of Accupants x 400

Equipment

Clearly most heat in the roo is generated by the 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.

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

Kind regards

Roman

No, there is no rule of thumb for this calculation. Rules of thumb are not a replacement for proper calculations, they used for engineering checks of properly calculated and designed systems or for ball-parking costs and sizing in preliminary design (usually with generous margin of error).

For a control room you will need to perform heat loss and heat gain calculations. You are asking for heating requirements, yet I noticed most of the answers in this thread are addressing cooling requirements. A control room will require cooling as most control rooms will generate significant heat while heating is a secondary concern, and may not require any depending on the geographic location of the building and other design factors of the structure. Lack of cooling can be detrimental to equipment and occupants.

Hire a Mechanical Engineer specializing in HVAC and have this done properly. There is no short-cut, it can be quite an involved process. Also, you may be missing something that you do not have the training to handle, including such issues as fire protection, and air cleanliness. If something goes wrong here, it is you who failed to perform due diligence without the proper calculations to back up your design, and it will be you (or your company) named in the lawsuit, or otherwise held responsible. I bet the equipment in the control room is worth substantially more than your fee so why take the risk?

If someone is asking you for equipment loads to do their HVAC calculations, then it is up to the Electrical Engineer (or client) to provide a detailed equipment list including power consumption of all devices to be installed in the room. With this information the Mechanical engineer can incorporate this information into the calc's which will also include other environmental factors, such as solar fenestration, outdoor air temperatures, etc. Equipment to be installed is generally sized to a "design day" or other point. Also, future growth may have to be incorporated into the design.

It's just not that easy!

Duane Tilden, P.Eng

I am also facing same problem. I have a metal cage with punched hole. (http://en.wikipedia.org/wiki/Faraday_cage)which contain electric panel, transformer and ac-dc converter with generate around 935kW of heat load. now if i want to calculate heat dissipation by natural air and how i calculate requirement of HVAC?

935 kW seems to be an extraordinarily large heat load, unless perhaps the panel/transformer/converter are supplying 935 kW of power to some external heaters. Can you clarify this?

Its a simple in my case we have around 6 transformer inside cage. now from manuf. data we abel to find a loss. And in-the case of transformer all the losses consider as heat load (bcoz it is dissipated by surface of TR to environment)

Dear Mr. Jim,

The best way is to get the heat dissipation figures of each electric component. However, if such data is not available, you can take approx. 7-8 kW heat dissipation. This figure we take for estimating heat dissipation of Engine Contol Rooms on Ships. Again, this also depends on the size of the room and no. of electric components.

Regards.

Anant.