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- 148 Posts. Joined 8/2012
- Location: Anderson South Carolina
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Hello PappaPig. Here is my opinion for what it is worth. Others may have different advice. This is only my opinion. To make the calculation you need to know the amperage at 220/208 volts.
Hello. I had a a blonde moment there. And I ain't blonde. If you change the voltage the amperage will change when using the same resistance. If I remember my electronics classes correctly. I really don't see a calculation to answer your question. As a VERY RUFF GUESS I'd say using the same resistance and halving the voltage I'd assume around half the wattage. BUT! half the voltage with the same resistance MAY pull more amps. I'm afraid this one is an install it and see if it gets hot enough. This is only my opinion and I have been wrong before. Others may know more and have different advice. Sorry I couldn't help further. Good luck.
Hello. Yeah, I would not dispute that. I went to electronics school back in the '80's ( long time ago ). Without all the calculation B.S. I would agree about 50% output more or less. I think that wattage may be a LITTLE low but I won't argue with calculations. You pay for kilowatts of electricity so you should be ok there. Good luck.
First a question. Is the 208v 5,300 watt element a single phase element or a 3 phase element? Makes a difference.
For anyone interested, here is the science behind using a 220v element in a 110v circuit. You would at first think that since you have cut the voltage in 1/2 the wattage is cut in 1/2 also. Nope, don't work that way. For the purpose of this illustration, let's assume the 5,300 watt is a single phase element (although 3 phase is more common in 208v circuits - the math would be the same but the wiring is totally different).
The load is a resistive load and as such it is governed by Ohm's law. Current is directly proportional to the voltage and inversely proportional to the resistance. As the voltage goes down so does the current. Current formula is I=P/E or 5,300 watt heater at 208 volts = 5,300/208 = 25.48 amps. Once we know that, we can calculate the Resistance by using R = E/I = 208/25.48 = 8.16 ohms of resistance in the heater. Now we have all the data we need to take the 5,300 watt heater element and calculate the wattage if it is powered on a 110v circuit. Current (amps) can also be expressed as I = E/R, Amps = Volts / Resistance. Using our 110v circuit and the 8.16 ohms resistant heating element you get 110/8.16 = 13.48 amps at 110v. So now we can answer the question. Yes a 220 volt heater will run on a 110 volt circuit but at a reduced wattage. Using the formula for wattage (power), P = E x I = 110v x 13.48amps = 1,482.84 watts or 28% of the original 5,300 watt rating (all calculations assume 100% efficiency which never occurs so the real world number would be less than that).
Here is the Ohm's law wheel for reference. Amps (current) = I, Volts (electromotive force) = E, Power (watts) = P and Resistance = R.
I was afraid of that. 3 phase is not my specialty, but it is also not plug and play with a single phase system either. If it were me, I would just get a new single phase element (either 110 or 220v depending on what wattage you need).
Here is why....
See the difference in the wiring? On the left is the typical smoker & PID setup (the SSR from the PID would be the "Thermostat SPST" in that drawing). If you look at the diagram on the right, and assuming you have 110v with one "hot" and one "neutral" with 110v potential between them, in essence you end up only using 1/3 of the heating element. This skews the formula I posted earlier even further, so in essence your 5,300 watt element will likely perform more like a 494 watt element (1/3 of the overall element is being used on top of the wattage reduction from dropping the voltage from 208 to 110). There are many other variations of 3 phase wiring, but this the simple version. It just get's more complex from here. Like I said, ditch the 3 phase element and go with a single phase one.
Hello. Good job on the help dward51. Been 30+ years since I had to use those calculations. I kept thinking way into the nite there was a way but for the life of me I just couldn't remember. AND I should have caught the 208 volt- 3 phase issue. Wife says she can't wait till I get completely senile. Says she is gonna convince me I really don't like John Wayne movies, I like musicals. Thanks for the post. Saved for future reference.
If each element is fed from a separate leg of the 3 phase and back to the neutral, then it is not running the element as a 3 phase element and may not be 208v either. Given I'm totally unfamiliar with the type of heater you are describing, you may want to check that particular unit with someone who works on that type of circuit regularly. I don't think it would work though.
I do know you there is a way to connect single phase loads to a 3 phase circuit. You may in fact have multiple single phase elements wired into a 3 phase "device". Double check the specs on that. If the element has only two wire connection points, I'm pretty sure it's a single phase element. Can you post the make and model number of the elements?
Ah ha...... They are indeed single phase elements.
Below is cut and pasted from the heater element page you posted. They are 208v, 60Hz, 1 phase elements. So the math is correct. On 110v you will get just shy of 1,500 watts of heat. If you already have one of these it should work on 110v power. Just connect to the hot and neutral wires of a standard 110v circuit using high temp wire and crimp female spade connectors and you are good to go from what I see. It does not matter which wire goes to which terminal connection on that heater.
It would be the rough equivalent of the 1,500 Brinkman element that is often used in this type of smoker cabinet for both hot and cold smoking. The common wiring of PID and SSR units found in many builds and threads on this site should work just fine with that element.
If you are going to use two of them, you need to go with a 220v feeder circuit. But instead of running them on 220v, split the L1 and L2 each to the neutral return for two 110v lines.
That may sound confusing, but here is the reason. You are talking at around 3,000watts for two of them. That's around 27amps on 110v which is more than 99% of the 110v wiring and outlets available. It would take a 30amp RV type 110v outlet (with proper gauge wiring) to support a 27 amp draw. Simplest thing to do is run a 220v line on a 220v breaker to the smoker. Then split out the L1 and L2 which normally have a 220v potential between them into two circuits of L1 to neutral, and L2 to neutral (110v each) with a single heating element on each circuit. You would need two SSR's but they can be triggered from a single PID and you can install a switch in the SSR triggering line and only run one heater when you don't want to hot smoke. In essence this is what the oven the elements came from is doing, except it's splitting the phases instead of the L1/L2.
Does this make sense?