This vlog is based on a previous blog we posted called “Temperature versus Heat Loss in a Heater”. In this video, Thermal Corporation Engineer, Kyle Otte, explains how to calculate the exact wattage needed for any type of industrial heater for your heating process.
How Wattage Is Calculated
The Rate of Energy Consumption equation above is used to calculate the wattage, where
E is the energy required for the process (or the total watts needed)
This case study was the first of many to come, recorded on May 24, 1996, by Thermal Corporation engineer, Jim Dixon. Over the last fifty years, the great team of Thermal Corporation employees has helped many countless customers with heating applications. One of the byproducts of being an established company is having an extensive library of past solutions to lean on.
When we entered the quote into our tech sheet program, the program was unable to choose an internal resistance wire for these specific cartridge heater values. This caused us to ask three questions:
Why is this a problem?
If the program could pick a wire, why would this probably not be a good heater?
What would we recommend?
Think About the Formula for Wattage
To begin, think about the formula for wattage.
Wattage = Current x Voltage
Heaters that have low wattage and relatively high voltage are typically a problem. If wattage is low and voltage is high, then what is the current? The current would be very low. How do you make a current very low?
I (Current) = E (Voltage) / R (Resistance) OR R = E / I
If the current is low and the voltage is high, what does R need to be? R needs to be very high. How do you make the resistance high? Think of electricity as water flowing through a heater. What diameter water hose has high resistance to the flow of water? A very small diameter hose will resist the flow of water. Thus, we have to use a very small wire and a LOT of it. In this case, we did not have any wire on hand that was small enough to make this particular case work.
Why Would This Not Be a Good Heater?
This leads into our second question: if we did have a small enough wire, why would this not be a good heater?
Remember that as heaters age, an oxide coating builds up on the surface. Temperature cycling accelerates this. Under a certain set of circumstances, the nickel-oxide will build up a constant rate. At some point, the oxide coating cuts off the path for electricity to flow- just like cholesterol clogging up an artery.
The oxide coating on each wire is the same thickness. So, which size do you think would fail first? If you thought the small wire, then you are correct.
What Would We Recommend?
Now, to address the question of what should be recommended. First of all, to cause the wire size to increase, what needs to be changed and how? The answer is the resistance must decrease.
W = E2 / R
If W does not change, and R goes down, what happens to E? E must also go down, as well. This means the customer must go to a lower voltage.
208 Volts, 3 phase power, is common in small industrial plants.
If 208V is a voltage from one hot-line to another hot-line, what is the voltage from a hot-line to the ground?
Line to Ground Voltage = Line to Line Voltage / √3 = 120V
Therefore, wherever there is 208V there is probably 120V, also. If we recommend 120 volts, how much does it help the resistance?
Initially: R = E2 / W 2082 / 75 = 577Ω
Now: R = E2 / W 1202 / 75 = 192Ω
Thus, resistance is decreased by a factor of 3. This is what we recommended to our customer.
Have a question about a heating application?
Contact the Thermal Corporation engineers! We can help you solve any issue or any question you may have regarding your heating application. Get in touch with us by email at engineering@thermalcorp.com, by phone at (800) 633-2962 x152, or chat with us using the instant chat feature located at the bottom of the page.
Written by Jim Dixon Edited by Shelby Reece and Kyle Otte Date Published: 08.23.2019 Last Updated: 08.23.2019
A customer phoned in one day about an issue with two heaters. The two heaters were wired in series on a barrel. One of the heaters, however, was getting red hot and the other was not. He did not know why. Our first thoughts were that maybe the heaters were accidentally wired in parallel rather than in series. After checking with the customer, we found that the heaters were wired correctly. But, why was one of them getting so hot?
One heater was 610W at 230V. The other was 1260W at 230V. The customer applied 460V across these two heaters when they were wired in series.
Hint: to successfully wire two heaters in series, they must be equal wattage AND equal voltage, i.e. they must have equal resistance.
Now to see what happens when we calculate the resistance of each heater:
Now when we apply 460V across these two heaters that are wired in series the circuit looks like this:
Written by Jim Dixon and Shelby Reece Edited by Kyle Otte Date Published: 07.02.2019 Last Updated: 09.03.2019
A customer contacted our engineers a while back about an issue they were having with an industrial heater that was getting poor life.
The Case:
The customer said he was using 480V 3-phase power and was bringing out two legs. He had been using a heater rated for 600W at 240V. The power that was actually applied to the heater was 277V, but he was getting poor heater life.What was going on?
The photo above is a diagram of the two lines he was bringing out.
The customer was applying 277V to a 600W 240V heater.
What was the actual wattage generated?
The formula below described to our engineers what was happening.
Conclusion
So, in this case, he was actually generating 800W, which is 1/3 more wattage than the original designed wattage of 600W.
The watt density was increased by 33% also, hence the shorter life.
Written by Jim Dixon and Shelby Reece
Edited by Kyle Otte
Date Published: 06.19.2019
Last Updated: 11.21.2019
We often get questions regarding how much power (or wattage) is needed to maintain a process at a certain temperature. That is a complex question. In this article, we will discuss the general concepts that are used to determine wattage needs.
Energy Balance
First, we start with a simple energy balance: Ein = Eout
Ein is the rate of energy coming into the system. This is the wattage that is generated by the heater when an electrical current is passed through a resistance element inside the heater. Every heater is stamped with a wattage and a voltage. So, energy coming into the system is a variable that can be controlled. Eout is the rate at which energy leaves the system. Eout is often what we want to determine because we can control Ein (the rate of energy coming in) and set it equal to this value. This consists of energy that is input into the process and leaves the system, e.g., plastic is melted, formed into a part, and then ejected from a machine. The heat that is absorbed by that plastic leaves the system. Eout also consists of power that is lost to the surroundings and is not used by the process. We call this waste heat. So, we need to determine both of those components. Power lost to the atmosphere can be approximated using heat loss graphs that are readily available online. In most cases, a wide variety of surface conditions, temperatures, and insulation thicknesses are plotted on these graphs. The end result of this calculation should yield a rate of energy loss (typically in watts per square inch) that is lost to the surroundings.
Rate of Energy Consumption
Next, we’ll look at the rate of energy consumption of the process. This is often much more complicated to determine.
For Example…
Let’s look at a process where we heat a chunk of polypropylene plastic from a temperature of 100°C to melting temperature (175°C). We are going to process one of these blocks of plastic every 15 minutes. The correct equation is:
where, E = power required in watts m = mass(kg) (25 in our example) cp = specific heat (Joules/kg°C) (1,920 in our example) = change in temperature (°C) (75 in our example) t = time in seconds (900 in our example)
Using the numbers above, we calculate that E = 4,000 which means we need 4,000 watts for this process. We can now add this 4,000W to the heat lost to the surroundings that we found on the approximate heat loss curve. That will give us a total steady-state wattage that we need to keep our process functioning.
Some Important Notes
Start up time is often an important consideration for selecting heaters.
Using the simple example above, if we wanted to heat the same mass of polypropylene from room temperature to melting temperature in the same amount of time, we would need more wattage.
No matter what the calculations say, it is a good idea to add 20% to this power requirement as a “safety factor” when sizing heaters. Why?
It is always good to allow for some error in the calculations.
Heater lifetime is improved if the heater does not run 100% of the time. The ideal run time, using a short cycle time, is 75% – 80%.
Written by Kyle Otte Edited by Shelby Reece Date Published: 04.17.2019 Last Updated: 01.29.2020
We received a question from a customer the other day regarding how to check whether or not a heater is operating correctly on the line. This is a great question! But, how do you actually know if a heater is operating correctly or not? Our answer is this: The best and safest way to check a single phase heater that is still on a machine is to…
Turn off power to the heater and disconnect the lead wires.
Find the number engraved on the heater specifying the watts and volts. (See photo)
Measure the resistance between the two lead wires or two post terminals, depending on the heater.
The resistance should be close to the voltage squared divided by the wattage (for you mathematicians out there, here is a formula: R = V^2 / W)
For example: You have a 500 W – 240 V heater.
R = 240^2 / 500
R= 115.2 omhs
Standard tolerances allow for resistance to measure between -10% and +5% of this total when the heater is at room temperature. If the heater is still at operating temperature expect resistances in the range of -5% to +10% of the calculated total.
Listed below are some tips for extending the life of cartridge heaters.
#1 Avoiding Contamination
When using a release agent, such as Thermal Corporation’s Ease-Aid, to ease heater removal from its hold, be sure to wait until agent is bone dry before inserting the heater into the hole. If the heater is inserted before the agent is dry, some of the liquid will usually be pushed towards the lead end and then will soak into the heater through the lead insulation or through the lava or ceramic plug at that end. When that happens, the heater can be expected to fail as soon as the power is applied. (Choosing the lead option of Teflon wire insulation and a Teflon plug can reduce the likelihood of water entering the heater, but this does not provide a fully hermetic seal, and the lead end of the heater must be restricted to 400°F).
Moisture, oil, and other liquids on the lead wires can be wicked into the heater and cause early failure.
Oil or other organic material on the lead end cap of the heater will carbonize at elevated temperature, causing a short from the leads to the sheath.
#2 Avoiding Over-Temperature
A loose fit of the cartridge heater in its hole will reduce the heater’s life time because the heat generated is not transferred efficiently to the object or material being heated, causing the heater to run at a higher temperature to transfer its energy. The higher the operating temperature, the shorter the lifetime. A rule-of-thumb for the fit is to make the hole diameter no more than .005 inches greater than the diameter of the heater.
Choose the lowest wattage heater that will maintain the desired operating temperature of the part being heated and still provide a short enough start-up time. Choosing a heater with a higher wattage than required will result in the controller turning the heater on and off to maintain the desired temperature and a higher operating temperature during on-time. These conditions will shorten the heater life.
If used as an immersion heater, the type of fluid and its velocity passing over the heater are important factors. Ask for assistance from a Thermal Applications Engineer.
#3 Excessive Cycling
The way that Thermal Corporation evaluates the lifetime of their own cartridge heaters and those from competitors is to cycle the units from 150°F to 1,400°F and count the cycles to failure. Cycling reduces lifetime because the surface of the element wire oxidizes rapidly at higher temperatures; if the higher temperature is maintained, the oxide coating actually protects the wire from further oxidation, but if the wire temperature is reduced substantially, the oxide coating breaks off due to contraction and exposes fresh metal to more oxidation. With continuous cycling, the wire diameter is eventually reduced, and the resistance of the element is increased to the point that it becomes too hot. At that point, the element wire either melts and breaks open the circuit, or it causes the insulation over the wire to break down, causing a short to the sheath.
Edited by Shelby Reece Date Published: 07.03.2018 Last Updated: 09.06.2019
4.5″ Diameter x 1.5″ Wide, 480V/3 Phase, 700 Watts
Introduction
Virtually all industrial power is generated as 3 Phase power. At the generating plant, it is generated as three phase power, because it is MUCH more economical to generate power as three phase. The power is then transmitted as three phase power. Just look at power transmission lines. You always see the lines in sets of three. Sometimes the heaters use all 3 phase voltages and sometimes they use one phase of the power. A single phase heater can run off of one of the phases of the 3 Phase power, connected either between two of the phases or between one of the phases and a neutral return path (0 Volts).
Why would you want to build the band heater as a 3 Phase heater?
You want to build a band heater as a 3 Phase heater to reduce the current in the wires going to the heater. When you carry the power through three wires rather than the same power through two wires, the current is less. Why do you want the current to be less? Higher current requires larger wire which costs more. Also, failures frequently occur where connections are made and the higher the current in a connection, the greater the likelihood of a failure. Some of these connections are inside the heater.
Generally, you can start considering making the band heater a three phase heater when the total wattage gets in the range of 2000 watts. You probably should make the band heater a three phase heater when the wattage reaches 5000 watts and almost definitely be a three phase when the heater reaches 10,000 watts.
So, what happened?
What happens sometimes is that someone is told that the heater will be running off of three phase power. That does not necessarily mean the heater has to be a three phase heater. All single phase power comes from three phase power. The question is how much is the current or wattage? In the heater we are talking about, the current is 700 W/480 v = 0.69 amps. Thus, even connecting the heater as a single phase heater, the current is 0.69 Amperes. This is just a little more than a 60 watt light bulb. Making this band heater a three phase heater would make the current even less. A three phase heater is more complex to build than a single phase heater. Thus, there are more things that can go wrong and it cost more. Why choose a heater that is more expensive and less reliable?
In Conclusion…
All power starts out as three phase power. Just because the power source is three phase does not mean the heater has to be a three phase heater. To answer the question, does the heater need to be three phase, look at the current or wattage. Don’t order a heater as a three phase unless it really needs to be one.
Written by Jim Dixon Edited by Shelby Reece Date Published: 10.30.2017 Last Updated: 09.06.2019
Can you connect two heaters in series if they are rated for the same voltage?
Sometimes. Basically, a heater stamped with Z watts and Y volts is just a resistor. The resistance is constant. If you apply Y volt to the heater, it produces X watts of heat. If you apply a different voltage, the heater produces a different wattage. The resistance never changes.
If you have two heaters that are rated for the same voltage, they will have the same resistance ONLY if they are ALSO the same wattage. That is, if the voltage and wattage are the same, then the resistance will be the same. Putting two equal resistances in series will cause them to divide the voltage equally. If you put two 1000 Watt/ 240 volt heaters if series across 480 volts each will get half of the 480 volts or 240 volts.
What about two heaters with the same voltage but different wattages?
Now, if we have two heaters that have the same voltage but different wattages, they will have different resistances. If we put two resistances that are different in series, they do not divide the voltage in equally. The larger resistance will get a larger portion of the voltage. Thus, consider if we have two heaters with the same voltage(240V), one has 4000 watts and the other has 2000 watts. The lower wattage will have a resistance of: R = 240V2 / 2000W = 28.8 ohms. Thus, it will get more than 50% of the voltage. The lower resistance heater will get more voltage and will produce more wattage. The higher wattage heater will have a resistance of: R = 240V2 / 4000W = 14.4 ohms.
Connecting these two heaters in series across 480V would give the following results: the current through each heater would be I = 480V / 43.2 ohms = 11.1 amps. The amperage will be equal through each heater since they are connected in series, and the resistance does not change. So, the higher wattage heater will actually produce W = 14.4 ohms x 11.1A2 = 1774W (versus a rating of 4,000W). While the lower wattage heater will actually produce W = 28.8 ohms x 11.1A2 = 3548W (versus a rating of 2000W). The lower resistance heater will get more voltage and will product more wattage. The higher wattage heater will have the smaller resistance and will product a smaller wattage. Frequently, the heater with the lower wattage will produce so much heat that it may burn up.
Example
The classic example is where we have a two piece heater. Each half of the heater has half the wattage and the same voltage. Thus, these can be connected in series and they will divide the voltage equally.
Conclusion
Thus, you CAN connect two heaters in series ONLY if they have the same voltage AND the same wattage.
The resistance = ((voltage) x (voltage))/ wattage
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Written by Jim Dixon and Kyle Otte Edited by Shelby Reece Date Published: 04.28.2016 Last Updated: 09.06.2019
is the change in temperature (°C)
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