Let’s Cool Things Down With Heat Transfer Equations

When it comes to cooling products, we get many questions on what would be the best method. Generally with larger parts with heavy mass and large surface areas, we would recommend the Super Air Amplifiers, Super Air Knives or Super Air Nozzles. We have to look at many factors to determine the correct method, but if look at the mass of the part, the ambient conditions, the speed of the conveyor, and the change in temperature, we can get a good start in setting up an application.

In determining a good estimate, we use a couple of heat equations to help. As with any customer, you want to make sure you have as much information to get a good platform to start. We use two equations to begin. The first equation is used for the product that needs to be cooled.

Equation 1

q = m * Cp * (T2 – T1)

Where:
q – heat (BTU) or (Kcal)
m – Mass (Lb) or (Kg)
Cp – specific heat (BTU/Lbm oF) or (Kcal/Kg oC)
ΔT – Temperature (oF) or (oC)

Once we have the amount of heat that we need to remove, then we can look at the product to cool it. With Super Air Amplifiers, we use an equation that is used in fan cooling. This is the second equation.

Equation 2

h = 1.08Qs(T2 – T1)       OR                h = 0.33Qs(T2 – T1)

Where:
h – heat rate (BTU/hr)                                   h – heat rate (Watts)
Qs – Flow (SCFM)                  OR                   Qs – Flow (NM^3/hr)
ΔT – Temperature (oF)                                  ΔT – Temperature (oC)

 

As an example, we have an aluminum part that came out of a baking oven at 400 oF (204 oC), and we want to cool the part down to 100 oF (38 oC) for handling.  If we give the part a mass of 20 lbs. (9.1 Kg), we can determine how long we would need to cool the part. The specific heat, Cp, of aluminum is 0.22 BTU/Lbm oF or Kcal/Kg oC. Applying this to Equation 1, we get the following:

q = 20 lbs * (0.22 BTU/lb/oF) *(400 oF – 100 oF)

q = 1,320 BTU

Or

q = 9.1 Kg * (0.22 Kcal/Kg/ oC) * (204 oC – 38 oC)

q = 332 Kcal

This tells us how much heat we would need to remove in order to handle. To keep going along with this example, we will use the 120021 Super Air Amplifier. With the large amplification level, it has a flow of 436 SCFM (740 NM^3/hr) at 6” away. To produce that volume, it only uses 8.1 scfm of compressed air at 80 psig. There are a couple of things that we should consider with our estimation. Ideally, we will want to be at a distance where the velocity will be between 1,700 to 2,500 fpm (8.6 to 12.7 mps). This gives us the best velocity for the maximum cooling rate. Depending on certain situations, you may have to add a little more time for cooling if the velocity is too high or too low. With the 120021 Super Air Amplifier, a distance of 18” from your target will give you a velocity close to 1,850 fpm (9.4 mps). The other consideration is the rate of heat loss. The bigger the temperature difference, the faster it will cool. As you get near the target temperature, the rate change becomes smaller. With that, I usually take the average temperature as an estimate. Using our example above, it would be (400 + 100)/2 = 250 oF, or (204 + 38)/2 = 121 oC. The other estimation will be the temperature of the ambient air. It will be cooler as it first hits the target and then heat up. If we add roughly 7 oF (4 oC) to the ambient air temperature because of the velocity, then the ambient temperature would become 68 + 7 = 75 oF (20 + 4 = 24 oC). With Equation 2, we will get the following:

h = 1.08 * 436 SCFM *(250 oF – 75 oF)

h = 80,404 BTU/hr

Or

h = 0.33 * 740 NM^3/hr * (121 oC – 24 oC)

h = 23,687 Watts or 20,372 Kcal/hr

With Equation 1 and Equation 2, we can estimate the amount of time needed to cool the part. In having to remove 1,320 BTU at a rate of 80,404 BTU/hr, the equation will give us 1,320 BTU/ (80,404 BTU/hr) = 0.016 hr or 1 minute.  In metric units, 332 Kcal/(20,372 Kcal/hr) = 0.016 hr or 1 minute. This means that we will have to keep the target in the air stream for about 1 minute. Depending on the geometry of the part, the angle of the amplifier, and the speed of the conveyor, we may need multiple Super Air Amplifiers. As I mentioned before, these are estimations, but it will help in getting an idea for your project. You can always contact the Application Engineers at EXAIR for any help.

Cooling with Air Amplifiers

Cooling with Air Amplifiers. The Super Air Amplifiers are mounted across the top of this connecting rod/piston assembly. 

John Ball
Application Engineer
Email: johnball@exair.com
Twitter: @EXAIR_jb

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