Ventilation Fan Calculation

What is the temperature of the incoming air?

You mean the external temperature? Is it about 90degF.

The specific heat of air at that temperature is 1.005 kJ/kg.degC. So for a 5 degree cooling the air can take away just over 5kJ/kg of air. divide that number into the total power dissipated as heat within the room and the answer is the number of kg per second that needs to be shifted through the room. The fan size follows.

http://www.engineeringtoolbox.com/air-properties-d_156.html

Then you are trying to limit your internal temperature rise to only 5°F. That's a very tight specification. As you may remember I do not do HVAC myself so my analysis maybe off. However, from having to keep electronics cool in a chassis I do have some related experience.

The fan specification will be in how many cubic feet per minute it can transfer with the fan opening being the significantly smallest restriction. In other words, if the fan opening itself is one square foot of area then the next largest flow restriction must be ten square feet for you to get the rated air flow. If you are drawing air through an otherwise air tight box that has only one other opening with the same area as your fan opening then the net flow will be reduced only by the square root of two. Smaller openings will cause a significant drop in flow rate so assuring that you have at least the same inlet area as outlet area will give you predictable flow rates. (Opening the door to the room will not cause a dramatic change in airflow.) Now since you wish to exhaust hot air and inlet cool air you should put the exhaust in the ceiling and the air inlets close to the ground.

Now the rest requires knowing how many watts are produced inside that room. Don't confuse this with the number of watts that pass through the wires to your loads. I'm talking about the the thermal watts produced by the wires, resistors, relays and connections in this room. Don't forget the lighting heat and any other heat producing devices (people) regularly occupying this room.

Maybe this guy can help Sizing Of Fan And Filter For Control Panel. He seems to have the same problem as you. Oh, and he seems to have a cooling problem, too.

Maybe a fan supplier can lead you.

Maybe get one of those infrared thermometers and find the hot spot and aim a fan at it.

This is a ventilation for a room not a panel. A room contains many panels.

A sure way to do it is to get one of these. The added benefit is, that it will take moisture out of the air also...............another thing that electrical stuff doesn't like. They're relatively cheap too.

I don't know that much about HVAC, but what I would do is: install an exhaust fan in the ceiling and control the air flow through windows in at least two of the walls. The windows would be on the sunny side and on the shady side. Open windows on the shady side for ventilation and close them when the sun shifts to that side. Then open the windows on the opposite side. You want to draw in air cool not hot air.

You have approximately 5000 cu ft of air volume in the room. A fan is rated in CFM (cubic feet per minute). A 5000CFM fan will change the air in one minute. A smaller fan, say 2500CFM will do it in two minutes, or a two minute air change. The higher the CFM, the higher will be the air velocity coming through windows and doors. If air velocity is too high, papers and other items will be flying around the room. A fan with a lower CFM will take longer to change the air, but will keep down dust and turbulance. You may need air filters over the windows to filter out dust. A 4 or 5 minute air change would be about right.

Cooling of air a needs refrigeration system. By ventilation of air will reduce the temperature if the fresh air temperature is less than the room temperature. Other economical method is evaporative cooling but it will be effective only when outside humidity is low. You need to refer ASHRAE standards for number of air changes per hour for electrical panel room and design the ventilation system accordingly.

I know the answer but you ask CR4 to do all your work for you...so I'll pass.

i agree with you.

Before coming here, home work is reqd.

The Institute of Electrical and Electronics Engineers (IEEE) publishes tables that show the watts of thermal loss in various electrical equipment. Watts can be converted to BTU. Contact them. An Internet search will provide you with their current address and phone number. Good luck with your project.

By the way, I am an electrical engineer. If I were designing the electrical installation in your room, I would want my equipment cooled lower than 95 degrees F, especially if there are any electronics involved. I think an air conditioner would be required. However, not knowing your project limitations and design parameters, my suggestion of an air conditioner may not be feasible.

You should calculate rate of increase of temp in fixed room volume of approx 5000 cu ft. After ventilating the room, close all doors and windows, and record temp at different parts of room at every 15 min. till it reaches to maximum. From the average rate of increase in temp, rate of heat dissipation from panels can be calculated ( approx weight of air in room in kg x increase in temp in deg C) in kJ. Neglect the heat dissipation from outer room walls & windows to atmosphere. Now to mitigate the heat added in room, calculate rate of fresh air volume at 90 deg F to be pushed in. It can be either induced or forced.