Calculations for CFM for Forced Convection

No se si te ayude mucho, yo calculo la necesidad de cuantas veces requiero cambiar el aire de un cuarto multiplicando el largo, ancho y alto de la habitación, ejemplo una sala esteril HEPA requiere 20 recambios por hora si la habitación mide 15 pies (largo) x 12 pies (ancho) x 9 pies (alto) / 60 minutos, esto da = 459 cfm, con un equipo de 500 cfm cumplo con las especificacioones, (cfm = Pies cubicos por minuto).

There are too many unknowns to provide a definitative answer, however, let me suggest a guide for you to follow;

Do you know the capacity or output of the source of heat? If not, is the source large enough to heat the space? Assuming you have a source of heat..........

Let's assume you do have and it is enough to heat the space, and the output is XXXX BTU/hr. Now, the hotter the discharge air temperature off the source of heat, the less amount of air you need to heat the space, and visa versa. Also, circulating air at a rate of 0.75 to 2.00 cfm / sq ft (this applies to spaces with 8 - 10 ft ceiling heights) would provide a reasonable rate of circulation without undue drafts. And you will find air temperatures in excess of about 140 dF would produce some uncomfortable conditions and percieved large swings in temperature.

So, with these numbers in mind (space temperature=75dF, heater discharge=140 dF)you can use the following equation to arrive at a suitable CFM.

CFM = heat output BTU/hr / 1.08 / (140-75). (1.08 is a conversion factor)

Now, divide the space sq ft by the CFM, and if the answer falls between 0.75 and 2.00 you will be OK. If the number is on the low side, say below 1.00, you may reduce the discharge air temperaure and recalculate. This would provide an increased circulation rate with slightly cooler air, and probably a greater degree of comfort.

If you do not know any of the basic numbers, you can still make it work.

Measure the space you intend to heat, length and width to obtain the sq footage. If your winter low temperature is in the 0 dF range and the construction of the facility is average, you would probably be safe in using 50 BTU/hr/sq ft to estimate the heating load. This would give you a starting point in determining the heat output required. All of this is called a "SWAG" (scientific wild a** guess). If you're worried, "kick it up a notch" to 60 BTU/hr/sq ft. Multiplying the sq ft x BTU/hr/sq ft will start you on your merry way.

The more accurate you are with the actual heat output, the better your results.

Now, get out of here and go to work.........

You obtain a froced convection when the convection is not any more given by the density diffrences generated by heating(cooling) of environmental air. As soon as you use a fan to move the air in contact with you transfer surfaces you find yourself in forced convection situation. Now depending on the velocity you can give to the air flox you abtain diffrent convection coefficients in w/°c/m². The higher the velocity the higher the possibility to transfer heat from the surface to the stream. Attention the relationship is not linear. Further the higher the sped the more turbulence you reach and this has 2 effects: higher transfer but also higher energy losses.

So that you question is too general to get a true answer. The best is to look depending on the transfer proçblem you are confronted with in a specific book since although heat transfer is the same which ever application you have there specific recommandations to rspect in the different fields.

Your local Building Code should have a section that describes minimum air replenishments for various room uses on an hourly basis. Calculate from the volume of the target heating space and multiply by the code refreshment rate divided into minutes.

Mark

Our Friend from Peru said it succinctly--post#1

You also need to DIY: Look up tables On MEAN Heat Capacities c_m,kJ/(m³.°C)--:

Dry Air at 0°C=1.298 and at 100DegC=1.302 - VARIES LINEARLY

Water Vapour(for Relative Humidity?) has c_m =1.495 &1.1.507 AT 0&100°C

Rest is school Arithmetic!!