Convection or Infrared?
Like many other products that must be heated, an electric motor stator has areas of high mass and materials of construction that are sensitive to overheating. What is the best way to heat such a part without creating defects?
Follow one method of selecting the heating method for a process or product. Like many other products that must be heated, an electric motor stator has areas of high mass and materials of construction that are sensitive to overheating. What is the best way to heat such a part without creating defects?
At the beginning of every heating process, the question that should be asked is, “What is the best heating method for this part or process?” Many heating options are available, and this article will compare two: gas convection and quartz infrared heating. Both heating methods are effective, but of course, neither is effective for every type of part. This article also will address some of the questions that arise at the beginning of a project and offer insight into how to answer these questions. The same review process can be used for most other heating applications.
To allow some comparisons, an AC electric motor stator will be used as a sample part throughout the article. For this example, the reference part will weigh approximately 30 pounds and is produced at a rate of 30 parts per hour. It was selected as a representative part because motor stators typically are coated with resin or epoxies. The parts are preheated prior to coating and then go through a final post-heat/cure stage.
So, what is the “right” heat processing method for this type of product? A number of factors need to be evaluated to answer that question. They include:
How much product will be processed?
How will the part be conveyed?
What are the main differences between convection and infrared heating for heating this part?
How much time will it take to heat this part effectively?
Production Rate and Heating System Size.To start the analysis, the weight of each part and total weight to be processed per hour is used to determine the heat load. Calculated based on an hourly rate and described in BTUs per hour, the heat load equation is:
BTUs = Weight/Hour x Specific Heat x Change in Temperature
The results of this calculation are used to help size the heating system.
Method of Part Conveying.How the parts will be held, loaded and conveyed into and past the heat source must be determined next. Having good access to the part for airflow or good line-of-sight to the part for the infrared energy will optimize part heating.
In the case of the AC motor stator, holding and rotating the part in front of the heat source is the key to having both heating methods work. By rotating the stator, the heat is exposed to the part at an average exposure rate rather than heating only one area of the part. If the part is stationary, then control of the infrared heaters or convection-heated air is critical to avoid overheating the part in any particular location.
Differences Between Convection and Infrared Heating of the Part.The characteristics of convection (hot air) heating and infrared (radiant) heating influence how the part is heated. To understand the difference, consider this simple analogy: Supposed you see a black, painted metal chair on a warm summer day. If the metal is in the shade, the warm breeze may heat it to the ambient temperature. This is like convection. Then, if you move the chair to a sunny location, you will soon notice that the seat and other parts in direct sun have gotten much hotter. The radiant sunlight is like infrared heating. It heats what it sees — and areas in the shade are not heated.
Returning to the AC motor stator example, for convection heating, the part shape and orientation are less important (than with infrared heating) because the heat is distributed by the moving air and can circulate around the part from different angles. With convection heating, the oven is set at a constant temperature, and high velocity air distributes the heat to and past the parts. The more frequently the air is turned over in the heat zone, the more uniform the temperature will be throughout the heat zone. Likewise, the more frequently the air is turned over in the heat zone, the faster the part heats up to the desired temperature.
With convection, heating of the part takes place by the air moving past the stator and the energy being absorbed by the part. Overheating is not normally an issue as long as there is good airflow because the airflow distributes the heat and reduces the likelihood of hot spots. Figure 1 shows how heating via convection can create a hot side during heating.
It is frequently argued that one downside of convection heating is that the heating rate is mass dependent and often slower in comparison to other heating methods. With gas convection heating, it is critical that a portion of the air be exhausted due to the products of combustion (gas) burning so as to not create an unsafe atmosphere inside the heating chamber.
When infrared heat is chosen as the heat source, the stator will require a direct line-of-sight from the emitter to the part for optimal heating. The emitter power rating and the distance from the emitter to the part will determine the rate at which the stator will be heated.
A unique factor of infrared heating is that the color of the part will affect its heating rate. This is due to the part’s emissivity, which is a measure of an object’s ability to emit or absorb infrared energy as compared to a black body. The motor stator will heat at different rates due to the steel lamination core and the copper winding’s varying emissivity. The difference in the heatup rate between the two materials can be seen in figure 2.
Heating with infrared energy can be the most effective, efficient source of heating, but it also can cause problems when the surface temperatures become too hot. Infrared relies on conduction heating to transfer heat from the hotter outer surface into the interior of the part. The only way to increase the heating rate once an emitter style is chosen is to have a higher outer surface temperature that will conduct more energy into the protected areas of the part. If not controlled, these temperatures can rise quite high and are representative of the black thermocouple shown in figure 2. Control of the temperature normally takes place by monitoring the part temperature itself and not the air temperature. This is because, theoretically, the air does not get heated (note the red thermocouple in figure 2).
If adequate time is available for heating the stator, both heating methods will work as long as there is control of the heat being applied. When it is desired that the time be minimized and the heatup rate maximized, the differences will emerge.
Time Required to Heat the Product.If the answer to the question, “How much time will it take to heat the product?” is not known, it can be determined by conducting a heat profile of the part. This is done through the use of a datalogger, thermocouples and the part. Part temperatures are taken by mounting thermocouples onto the part in strategic positions and recording the data while the part is being heated.
Two different heat profiles were made using a reference stator. As previously noted, figure 1 shows a convection heating process and figure 2 shows an infrared heating process. The graph of the heating curves uses six thermocouples that read the temperature of each of six locations:
The stator steel surface.
Within the stator steel body (buried in the steel).
Three separate locations of the copper winding.
The ambient air temperature.
In both figures, the heating profile shows three periods:
A preheat zone.
A simulated coating time in ambient air.
A post-heat zone.
The times and temperatures vary primarily during the curing, where the infrared heating shows a much higher surface temperature during the entire heating cycle.
It also is important to note that the maximum temperature of the convection heating curve never exceeds the maximum oven setting. In fact, for the time shown, it stays below that setting.
By comparison, the infrared heating profile sees a significantly higher temperature on the surface and in the buried thermocouples of the stator steel. Perhaps surprisingly, the copper winding, which has less mass than the stator steel body, does not heat as high as one would expect. This is due to the emissivity difference between copper and steel.
The resulting difference in the heatup rates can be either positive or negative depending upon the desired outcome.
The convection-heated part may take longer to reach the desired temperature based on part mass, but it never exceeds that temperature. The entire part is thoroughly heated using the convection heating for the time allotted.
The infrared heat cure reaches much higher temperatures in the post-heat zone, which could meet some processes’ goals. However, in the case of the motor stator, the excessive heat creates a nonuniformly heated part with temperatures that can damage some of the nonmetallic materials used in the motor assembly.
As noted, the desired outcome of the heating zone determines the best heating method for a given product. In the case of curing the coatings on the motor stator, both uneven heating and overheating can result in defective parts. If convection heating is selected, care must be taken to avoid hot spots and ensure the part is heated adequately, which may require a longer heating time. If infrared heating is selected, care must be taken to avoid hot spots and overheating.
In conclusion, you may be wondering, “What heating method is the most efficient?” From an energy-usage standpoint, using electric infrared heat rather than a gas-fired convection oven reduces some of the exhaust requirements because the products of combustion from the gas burning do not need to be exhausted from the chamber.
To compare costs, an average rate of power usage for the infrared heating vs. the cost of the gas consumed in the oven would need to be made based on the process’s specific operating temperatures, operating time, product size, quantity and weight. The local gas (therm) rate and power (KWhr) costs are used to compare the BTUs consumed during heating and to determine which method is most efficient for the parts being used. As it happens, AC electric motor stators are cured using both methods today.