It’s easy to know that EXAIR’s Vortex Tubes can be used to cool down parts and other items, but did you know that our other engineered compressed air products can be used to cool down these same things? It’s the same process as cooling down hot food by blowing on it. And we can use the physical properties of any material – whether it’s the massive billets of steel in the photo up top, or the bowl of soup to the right, to calculate the amount of air flow required to change a certain mass of the material from one temperature to another.
For any material, there’s a certain amount of energy required to cause a certain temperature change of a certain mass of the material. This property is called Specific Heat (Cp), and it’s commonly expressed in Joules per gram per degree Celsius (J/g°C), or Btu’s per pound (mass) per degree Fahrenheit. (Btu/lbm°F). The Specific Heat of the material allows us to calculate the amount of heat that has to be removed to cool it from its starting to its desired temperature, using a standard heat transfer equation:
q = mCp ΔT, where:
- q is the amount of energy it’ll take to cause the temperature change.
- m is the mass of the material that you want to change the temperature of.
- Cp is the Specific Heat we talked about above.
- ΔT is the starting temperature, minus the desired temperature.
Once we know the amount of heat to be removed, we can then apply units of time, and calculate the rate of cooling you’ll need to achieve in order to get the material to the temperature you want, in the time that you want. Let’s work through an example, using a piece of steel weighing 50lbs that needs to be cooled from 300 °F to 200°F:
q = m * Cp * ΔT, where:
- m = 50lbm
- Cp = 0.117 Btu/lbm°F
- ΔT = 300°F – 200°F = 100°F
- q = 50lbm * 0.117 Btu/lbm°F * 100°F = 585 Btu of energy (heat) to be transferred
Now, let’s say we have two minutes to cool this piece of steel:
585 Btu/2 minutes X 60 minutes/hr = 17,550 Btu/hr
That’s the rate of cooling required for this application. Now, we can use another equation that’s commonly used in the HVAC industry to determine the amount of room temperature (70°F) air flow that’ll remove that amount of heat. It’s called the cooling power formula:
Q̇ = 1.0746 * ΔT * ṁ, where:
- Q̇ is the rate of heat transfer
- 1.0746 is a constant
- ΔT is the difference between the desired temperature and the air temperature
- ṁ is the flowrate of air in cubic feet per minute
Since “Q̇” is the unknown value, we have to get to use a little algebra and rearrange the equation:
ṁ = Q̇/(1.0746 * ΔT), where:
- Q̇ = 17,550 Btu/hr
- 1.0746 = 1.0746 (remember, it’s a constant)
- ΔT = 100°F – 70°F = 30°F
- 17,550 Btu/hr/(1.0746 * 30°F) = 544.4 cubic feet per minute
Now, this assumes that equilibrium will be reached (i.e. all of the heat than CAN be transferred to the air flowing past the steel WILL be transferred), but that’s not going to happen. Depending on the geometry of the material to be cooled, there are ways to maximize the contact time between the material and the cooling medium. For example, constructing a tunnel over a section of a conveyor so the airflow can blow in the opposite direction that the material is traveling. Even then, though, it’s unlikely you’ll reach equilibrium, so we’ll apply a service factor, and say our airflow is going to be 30% efficient in cooling the steel (which is really quite high) so we’ll need:
544.4 CFM/0.3 = 1,815 CFM
EXAIR Air Amplifiers are an excellent option for providing this kind of cooling flow. They’re compact, quiet, and efficient. Using the following table, we see that a 3″ Adjustable Air Amplifier supplied at 80psig has a total developed flow rate (Air Volume at Outlet) of 774 SCFM:
Another thing I like about the Adjustable Air Amplifiers for an application like this is that they’re, well, adjustable (it’s right there in the name). Turning the exhaust plug in or out will decrease or increase the air flow – this is how you can make gross adjustments to the air flow. A Pressure Regulator in the supply line then allows for precise ‘tweaks’ so you can dial in the performance to the level you need, without using any more compressed air than you have to.
If you have any questions about using compressed air for cooling, give me a call.
Russ Bowman, CCASS
Application Engineer
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