How will a 2000 watt element act using a 15 amp GFI outlet? The controller is PID controlled, 2000 watt element, controller wired for 120v 20 amp. Would like to know if it is worth it to plug this into 15 amp outlet and if there are any problems other than time to get to temperature.
You’ll trip the breaker at max load, and it also is a fire hazard.
I just looked at the load it would be. 15 amps wont work for this, guess I need to get a 1650 watt element for this to work with 15 amp outlet.
Thank you for the info.
I use 1650W on a 15A circuit for my RIMS. Works great. …but then again, a RIMS only fires momentarily to maintain a SV, and rarely at 100%.
An electrical element is a resistive load. If one knows a element’s rated power, one can determine it’s resistance. An element’s resistance does not change.
element_resistance = voltage * voltage / wattage
A 2000W, 120V element has a resistance in the range of:
element_resistance = 120 * 120 / 2000 = 7.2 ohms
element_wattage = 120 * 120 / 7.2 = 2000W
While wattage is amperage multiplied by voltage, the equations above take advantage of Ohms law, which states that E (voltage) = I (current, a.k.a amps) * R (resistance). If we want to determine the I (amps) for a given voltage and resistance we divide voltage by resistance. The part of the equation equation above that divides 120 by 7.2 gives us the current draw for this element.
How much current (amps) does the element above draw?
I (amps) = E (voltage) / R (resistance)
I = 120 / 7.2 = 16.67 amps, which means that one cannot plug this element into a 15A circuit because it will blow the breaker.
The National Electrical Code (NEC) states that no more than 80% of a circuit’s rated current carrying capacity can be drawn continuously. Eighty percent of 15A is 12A. A 1650W element draws 13.75A (1650 / 120 = 13.75). However, that part of the code is for continuous current draws lasting more than 3 hours. As long as the current draw is not for more than 3 hours continuous operation, an element of up to 1800W (120 * 15 = 1800) can be operated on a 15A circuit. A directly immersed 1800W element is capable of bringing 6 to 7 gallons of wort to a boil; however, it will not be fast.
One last thing, the reason why using a 240V-rated element on a 120V-rated circuit requires one to divide the rated is wattage by 4 can be seen if we use the equation above. A voltage or 240 is 2 * 120; thus, the square of 240 is 4 times the square of 120.
element_resistance = (2 * 120) * (2 *120) / 5500
We can factor out the 2s, resulting in an equation that demonstrates this physical property:
element_resistance = 2 * 2 * 120 * 120 / 5500 = 4 * 120 * 120 / 5500 = ~10.47 Ohms
The element’s resistance does not change when operated on a 120V circuit.
element_wattage = 120 * 120 / 10.47 = 1375W
We can verify this calculation by dividing 5500 by 4, which equals 1375
15 amp AC circuit at 110 V will not provide 1650 W because the voltage of AC is not at 110 all the time. The actual max power is ~ 1155 W.
Huh? No. 110, which in rarely is 110, it’s 120Vac, is at 120Vac +/- 3-5% at all times. Maximum power from a 120Vac/15 amp circuit is 1800 watts, but you should stay at 80% with a constant load. That would be 1450 watts.
American household voltage has not been 110V since the 30s when it was changed to 115VAC. The standard household voltage was changed from 115 to 120VAC in the sixties. The wall voltage in my marital home ran between 125 and 127 VAC, but we were close to the step down transformer. Most communities are fed 7,200VAC, which is stepped down by ground or pole-mounted transformers to center-tapped 240VAC to each home. The voltage potential between the outside legs is 240VAC. The voltage potential between the center tap (a.k.a. white neutral wire) and either outside leg is 120VAC. If you can find a two 120VAC circuits on two different legs (the breakers for the two circuits are on different sides of the load center, a.k.a. breaker box), you can use the hot (black) wires from the two circuits to form a 240VAC circuit. You treat one black hot wire as the traditional red wire in a 240VAC circuit. You do not need the neutral white wire. If they are 15A circuits, you will have a 240VAC 15A circuit. If they are both 20A circuits, you will have a 240VAC 20A circuit. If one 120VAC circuit is 15A and the other 120VAC circuit is 20A, you will have 240VAC 15A circuit.
Wow, you went a little deep on this one, but dead on. Must be an engineer, my wife tells me to shut up when I go too deep.
I am half electrical engineer and half computer scientist. Electrical engineers think I am a computer scientist and computer scientists think I am an electrical engineer. 
Yeah, you will trip a 15amp breaker. DO NOT replace a 15 amp breaker with a 20amp breaker, ever. The wire used in a 15 amp circuit (usually 14 gauge copper) cannot carry a 20amp load. To do so will create a fire hazard.
That is not correct. Almost all conductors have a resistance that increases with increasing temperature. The current will decrease as the element gets hot. If you have a material where the resistance decreases with temperature you can get a runaway condition with the current increasing until the element burns out.
The rest of the Mark’s analysis is OK.
AC is alternating current which means it is mostly not at max voltage. Max power for from a 120 VAC 15 amp circuit is 12015sqrt (2)/2 = 1272 W.
I don’t know how you came up with those numbers, but 120Vac is a RMS value, in other words it’s a dc equivalent. Very basic electrical theory. 120Vac RMS x 1.414 (square root of 2) = @170 volts. That’s peak voltage. Peak to peak is double that or 340 volts. All those numbers are usually not referenced, just the RMS value.
AC values, unless otherwise stated are RMS values. That includes dielectric ratings. In a resistive circuit, RMS values can be exchanged with DC values. Once inductive and capacitive components are added, the calculations change.
An easier way to calculate the max amperage draw from a 2000 watt element is to use the P=IE fomula (Electrical Power (P) = Electrical current (I) x Electrical voltage (E)). To refactor the equation to find current use P/E = I
So for a 2000 watt element at a voltage of 120 volts the solution would be 2000/120 = 16.7 amps which exceeds the rating of the 15 amp GFCI breaker and the current limitations of the house wiring.
If you measure electrical voltage with a voltmeter, and see 120 volts, you are measuring the RMS voltage (Root Mean Square voltage) because the voltmeter does not have the balistics to measure the voltage at the peak of the AC sine wave (you need an oscilloscope to do that). So using 120 volts is accurate to determing if you will pop the breaker with a 2000 watt element (which you will).
As was mentioned by MnWayne, DO NOT put a bigger breaker in place of the 15 amp GFCI breaker since the breaker is sized for the house wiring. 14 gauge wire is only rated for 15 amps. If you draw more than 15 amps through the wiring you will cause it to heat up which could cause an electrical fire in your brewery and house. Use a 1650 watt element and you will only draw 13.75 anps which is well within the capacity of the GFCI breaker. I do that here for my RIMS system and it works fine.
Also remember that if the heating element draws a calculated of say 5 amps at 240 volts, you will draw twice the current at 120 volts (10 amps) to maintain the same power usage. That is why you always calculate the current draw at 120 volts before using a 240 volt electrical heating element in a 120 volts environment. These elements are used in electric water heaters and have the power usage expressed at 240 volts which the water heater runs to minimize current draw when heating the water.
RMS voltage does not apply in this rating.
What I meant by that statement is that the resistance of an element is constant with respect to voltage and current for wattage calculations in properly designed circuits (i.e., the resistance of the element does not automatically scale down when using an element designed to operate on a 20A circuit on a 15A circuit). The resistance of the element is only going to drop so much in a properly designed circuit.
Yes, resistance changes with respect to temperature. Thermal runaway is why one cannot operate most electric elements dry. The phase change of water from liquid to vapor (latent heat of vaporization) dissipates heat, keeping the liquid cool enough that the element does not reach runaway temperature. If the liquid in which the element is submersed does not make a phase change from liquid to vapor, then the temperature of the liquid would continue to rise until the element reaches the temperature where thermal runaway occurs.
Finally, yes, electronic components have de-rating temperatures; therefore, operating temperature has to be factored into component selection and system tolerances.
I did not know that!
Another advantage for using a higher voltage and lower current to produce the same amount of power is that there is less voltage sag. The voltage drop across a conductor is equal to the current drawn multiplied by the resistance of the conductor. If we double the voltage, we halve the current draw. Larger diameter conductors made from the same material have less resistance.
Yes, and that is why long-distance transmission lines operate at very high voltages. The loss in transformers to step down the voltage to something usable is less than the loss in the long lines. If and when we ever get superconducting transmission lines that will change but that is a long time in the future.