If you haven’t read many of my blogs then this may be a surprise. I like to use videos to embellish the typed word. I find this is an effective way and often gives better understanding when available. Today’s discussion is nothing short of benefiting from a video.
We’ve shared before that there are three types of heat transfer, more if you go into sub-categories of each. These types are Convection, Conduction, and Radiation. If you want a better understanding of those, feel free to check out Russ Bowman’s blog here. Thanks to the US Navy’s nuclear power school, he is definitely one of the heat transfer experts at EXAIR. If you are a visual learner like myself, check out the video below.
The Application Engineering team at EXAIR handles any call where customers may not understand what EXAIR product is best suited for their application. A good number of these applications revolve around cooling down a part, area, electrical cabinet, or preventing heat from entering those areas. Understanding what type of heat transfer we are going to be combating is often helpful for us to best select an engineered solution for your needs.
Other variables that are helpful to know are:
Part / cabinet dimensions
Material of construction
External ambient temperature
If a cabinet, the internal air temperature
Maximum ambient temperature
Amount of time available
Area to work with / installation area
Understanding several of these variables will often help us determine if we need to look more towards a spot cooler that is based on the vortex tube or if we can use the entrained ambient air to help mitigate the heat transfer you are seeing.
If you would like to discuss cooling your part, electrical cabinet, or processes, EXAIR is available. Or if you want help trying to determine the best product for your process contact us.
When you have two objects and they are of different temperatures, we know from experience that the hotter object will warm up the cooler one, or conversely, the colder object will cool down the hotter one. We see this everyday, such as ice cooling a drink, or a fan cooling a person on a hot day.
The Second Law of Thermodynamics says that heat (energy) transfers from an object of a higher temperature to an object of a lower temperature. The higher temperature object has atoms with higher energy levels and they will move toward the lower energy atoms in order to establish an equilibrium. This movement of heat and energy is called heat transfer. There are three common types of heat transfer.
Heat Transfer by Conduction
When two materials are in direct contact, heat transfers by means of conduction. The atoms of higher energy vibrate against the adjacent atoms of lower energy, which transfers energy to the lower energy atoms, cooling the hotter object and warming the cooler object. Fluids and gases are less heat conductive than solids (metals are the best heat conductors) because there are larger distances between atoms. Solids have atoms that are closer together.
Heat Transfer by Convection
Convection describes heat transfer between a surface and a liquid or gas in motion. The faster the fluid or gas travels, the more convective heat transfer that occurs. There are two types of convection: natural convection and forced convection. In natural convection, the motion of the fluid results from the hot atoms in the fluid moving upwards and the cooler atoms in the air flowing down to replace it, with the fluid moving under the influence of gravity. Example, a radiator puts out warm air from the top, drawing in cool air through the bottom. In forced convection, the fluid, air or a liquid, is forced to travel over the surface by a fan or pump or some other external source. Larger amounts of heat transfer are possible utilizing forced convection.
Heat Transfer by Radiation
Radiation refers to the transfer of heat through empty space. This form of heat transfer does not require a material or even air to be between the two objects; radiation heat transfer works inside of and through a vacuum, such as space. Example, the radiation energy from the sun travels through the great distance through the vacuum of space until the transfer of heat warms the Earth.
EXAIR‘s engineered compressed air products are used every day to force air over hot surfaces to cool, as well as dry and/or blow off hot materials. Let us help you to understand and solve your heat transfer situations.
“Nothing happens until something moves.” -Albert Einstein
These five words are the foundation on which the science of physics is built upon. This statement not only applies to the things we can see, but to those we can’t…like heat transfer.
OK; technically, we CAN visually observe the EFFECTS of heat transfer…that’s called “reading a thermometer.” But the actual mechanism of heat transfer takes place at a molecular level, and concerns the rate of motion of those molecules: the higher the rate of molecular motion, the higher the heat of the material. Hence, the higher the rate of CHANGE of that molecular motion, the higher the heat transfer rate is.
All you need for heat transfer to occur is a difference in temperature between two materials. Contact, or even proximity, helps…but not always. More on that in a minute. And keeping at least one of the materials in motion can help maintain the temperature differential. We’ll unpack that a little more too.
Let’s start with the three ways that heat is transferred…what they are, and how they work:
What it is: The transfer of heat between materials that are in physical contact with each other.
How it works: If you’ve ever touched a hot burner on a stove, you’ve successfully participated in the process of conduction heat transfer.
What it is: The transfer of heat through a fluid medium, enhanced by the motion of the fluid.
How it works: If you’ve ever boiled water in a pan on a hot stove burner, you’ve successfully participated, again, in the process of conduction heat transfer (as the burner heats the pan) AND convection (as the heated water in the bottom of the pan both transfers heat through its volume, and moves to the surface.)
What it is: Remember what I said earlier about how you don’t always need contact or proximity for heat transfer? Well, this is it…the transfer of heat through empty space, via electromagnetic waves.
How it works: If you didn’t actually TOUCH the hot stove burner, but felt your hand getting hot as it hovered, then you’ve successfully participated in the process of radiation heat transfer. OK; it’s a little convection too, since the air between the burner and your hand was also transferring some of that heat. The best example of STRICTLY radiation heat transfer I can think of is the sun’s rays…they literally pass through 93 million miles of empty space, and make it quite warm on a nice sunny day here on Earth.
Regardless of how material, or an object, or a system receives heat, engineered compressed air products can be used to efficiently and effectively remove that heat. For the record, they employ the principles of both conduction and convection. If you’d like to discuss a heat transfer application, and the way(s) that an EXAIR product can work in it, give me a call.
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.
q = m * Cp * (T2 – T1)
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.
h = 1.08Qs(T2 – T1) OR h = 0.33Qs(T2 – T1)
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:
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
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.