Cooling Products Using Fans

We learn as small children that you can cool things using an airstream.  From our earliest memories, we are taught to blow on our hot food, blow on our finger if it gets burned, etc.  We can use that experience to also cool products in industrial applications.

There are two definitions that you need to understand as we begin.  The intensity of the heat is commonly measured with a thermometer, and has units of degrees Fahrenheit (oF) or degrees Celcius (oC).  The amount of heat that needs to be removed is given the units of British Thermal Units (BTU).  One BTU is 1/180th of the amount of heat needed to raise or lower one pound of water by one degree Fahrenheit from 32 oF to 212 oF.  There is an equivalent value in the SI system, but since it is so much fun to use the IP system, we’ll continue using it in our examples.

An additional item you need to know is the specific heat of the material you are cooling.  This is the amount of heat required to change one pound of the material by 1 oF. 

These values are available from various sources, but here are some typical ones:

          Air                0.24             Aluminum      0.23

          Copper          0.09             Iron/Steel      0.12

          Water           1.00             Zinc              0.09


Knowing this value allows us to write an equation that describes the heat removal as:

          BTU = M * s * (t2-t1)           where:  M = mass in lbs.                               (Eq. 1)

                                                             S = specific heat in BTU/lb./oF

                                                             T2-t1 = temperature change needed in oF


Since we are using air to accomplish the cooling, we can replace M with CFM * density.  The specific heat of air in the range of temperatures and humidity typically used for cooling is 0.24, the equation can be simplified to :


          BTU/hr. = CFM * density * 60 * s * (t2-t1)                                              (Eq. 2)


Using standard air at sea level of 0.075 lb./ft3 density, the equation can be simplified even further, and rearranged to:


          CFM = BTU/hr. / (1.08 * (t2-t1))                                                            (Eq. 3)


You should make adjustments to the 1.08 factor to account for temperature and altitude.  These factors are available on any number of conversion factor websites.


You can use this concept to not only cool a product, but also to cool a space where heat is being generated, cool a large electric motor or generator, and similar applications.  But, the focus of this paper is to cool product, so let’s review how that is done.


Using equation 1 above, you need to predict the amount of heat you need to remove from the product.  As an example, let’s cool 1,000 lbs. of aluminum castings from 500 oF to 100 oF.  The BTU’s removed will be:


          BTU = 1,000 * 0.23 * (500-100) = 92,000


Other factors that changes the effectiveness of cooling with air is the velocity of the air, the amount of surface area exposed, and the temperature different between the hot product and the airstream.  Again, we know from experience that blowing hot food very lightly will eventually reduce the temperature.  Likewise, if we blast it with air, it won’t cool as fast as we might think.  We also know that you can’t cool a large mass by only applying cool air to a small fraction of the surface, unless we are willing to wait a long time. 


We also have to consider the cost of cooling with air.  If we try to speed up the process too much, then we consume huge amounts of power driving the fan.  A good compromise velocity is 1700 to 2000 fpm.  That velocity range allows the heat to be drawn out of the material at a good rate, will result in reasonable cooling times, and conserves energy.


In still air, the normal turbulence will cool at approximately 15 degrees per minute, but of course, the rate of cooling slows dramatically as the temperature difference between the air and the material if reduced.  At 1000 oF,  the rate is more like 45 degrees per minute, but once you get down to around 200 oF, you might be looking at only 6 degrees per minute.


At cooling velocities of 1700 FPM, typical castings or extrusions will release heat at the rate of 5 to 10 BTU/hr./sq. ft./oF temperature difference between the air and the product.  That gives us at least an estimate of how long it will take to cool the product.  Let’s use the average value of 7.5 BTU/hr./sq. ft./ oF.  We’ll call this value K.


Using the same 1,000 lbs. of castings with 200 sq. ft. of area exposed, with 90 oF cooling air at 1700 FPM to cool it from 500 oF to 100 oF, we get the following results.

The average temperature of the product would be (500 + 100) / 2, or 300 oF.  The air will heat up as it pulls the heat from the product, so let’s assume it averages 95 oF.  The average difference between the cooling air and the product would then become (300 – 95) = 205 oF.

          BTU/hr. = K * surface area * delta T

                     = 7.5 * 200 * 205

  = 307,500 BTU/hr.


Since we have 92,000 BTU to remove, and the capacity to remove 307,500 BTU/hr., we get an estimate of 18 minutes for the process.


You can also select the fan based on the airflow required.  That process will take a whole different paper, because the spread and throw of the fan will be important.  That result will determine how close you can set the fan, or conversely, if you have to place the fan a certain distance away, you can then estimate the performance required to get the airflow you need on the product.  The Hartzell design uses the tip-seal feature that we originated to greatly improve the throw of the fan.


All of these factors are estimates, but they will give you an indication of how to plan your process.