Limiting Amp Draw to 1500w Heating Element

Hey Guys

I have a 1500w heating element that is rated for 120V and I would like to know if it is possible to reduce the current to it. Right now it is 1500w/120V = 12.5amps, but I would prefer it to be below 10amps. This heater operates under PID control with SSR to control temperature of water tank. An image of the heater is attached below.

The reason I'd like to reduce the power/current of the element is so I can make room to power other items on this same circuit without tripping the breaker. I tried purchasing a simple voltage regulator that would reduce the voltage to the heater and this achieved my goal. But the switching power supply caused so much electrical noise that my temperature sensor in the tank could not read anymore.

Does anyone have any suggestions?

heating element.jpg

heating element.jpg

What you are showing should work, I do not believe your breaker will see the current peaks. You could also use an autotransformer, this would be the better solution.

You could look into a Variac, that will lower the AC voltage.
I found at a thrift shop that I use on occasion. I mostly got it to reform capacitors for old tube audio amps.
Example

You just reduced the average power with phase angle control which fooled the breaker. Sounds like the thermal side of the breaker is higher trip point than the magnetic trip rating. This is common on residential breakers but where you are in the world will have some effect on the trip curves, making it hard to predict where it will actually trip.

The current is only reduced if the firing angle is less than 90 degrees. Otherwise, if the firing angle is at least 90 degrees, the full peaks of voltage at the tops and bottom of the sine wave appear and the current is at maximum where I=E/R.

If you want a variable voltage, reduce the voltage with a variac (ac variable transformer). A step-down transformer can provide a fixed voltage reduction or you must lower the wattage of the heater.

There’s no magic.

At the risk of going off on a tangent, do you know what percentage of time your heater is on to maintain the highest possible setpoint at the lowest possible ambient temperature? Yes, that may seem like an impossible question but it’s not.

Perhaps an easier question would be the obvious one: How did you come to choose a 1500W heater?

Would connecting only 2 of the 3 elements do the trick? (1000W, 8.33A) ... omit the middle element.

Or connect one element and then two elements in series.

or use one of these if you can't decide ...

Thanks for the replies!

gilshultz:
What you are showing should work, I do not believe your breaker will see the current peaks. You could also use an autotransformer, this would be the better solution.

What I was using did reduce the power of the heating element like I wanted but the switching of the voltage regulator caused so much electrical noise that I couldn't get temperature readings in my tank. But I am thinking the transformer will not have this problem.

CrossRoads:
You could look into a Variac, that will lower the AC voltage.
I found at a thrift shop that I use on occasion. I mostly got it to reform capacitors for old tube audio amps.
Example
20amp 110v Variac Transformer Variable AC Power Regulator 0-130v 20a Metered US for sale online | eBay

I think this will do the trick that is giving lower AC voltage without the electrical noise. But it is pretty bulky. I can go for this if there is nothing smaller, but this will probably not fit in the set-up I have right now.

WattsThat:
You just reduced the average power with phase angle control which fooled the breaker. Sounds like the thermal side of the breaker is higher trip point than the magnetic trip rating. This is common on residential breakers but where you are in the world will have some effect on the trip curves, making it hard to predict where it will actually trip.

The current is only reduced if the firing angle is less than 90 degrees. Otherwise, if the firing angle is at least 90 degrees, the full peaks of voltage at the tops and bottom of the sine wave appear and the current is at maximum where I=E/R.

If you want a variable voltage, reduce the voltage with a variac (ac variable transformer). A step-down transformer can provide a fixed voltage reduction or you must lower the wattage of the heater.

There’s no magic.

At the risk of going off on a tangent, do you know what percentage of time your heater is on to maintain the highest possible setpoint at the lowest possible ambient temperature? Yes, that may seem like an impossible question but it’s not.

Perhaps an easier question would be the obvious one: How did you come to choose a 1500W heater?

The voltage could be fixed, could I make a smaller step down transformer myself? The heater stays on for 1hr straight max. Usually it is switching between on and off but really most of the day it is off. I have the 1500W heater because it is the smallest that I could find for a DIY espresso machine I built controlled by an arduino. The heater is used in the boiler, picture attached below, and came all the way from Europe to US.

I can't just use any heater that will fit in the hole either. It has to seal well with a teflon gasket and withstand about 3barg pressure.

dlloyd:
Would connecting only 2 of the 3 elements do the trick? (1000W, 8.33A) ... omit the middle element.

Wish I could do something like this but only have one port for heating element in the copper boiler.

I meant this (1000W, 8.33A) ...

dlloyd:
I meant this (1000W, 8.33A) ...

I see, my bad. This would solve my problem but the heater picture I took online and wasn't my actual heater. The heater in the picture with your annotations has six poles but mine only has 4. Actual set-up shown below. Is there any way I could modify it?

If you put them in series you get 750 watts, 6.25 Amps.

EDITED: If you remove just 1 jumper, you'll have 750W 6.25A. Will 1/2 power be enough to bring up the temp to your setpoint? If it does, but you need faster heat-up, then you might need a switch or relay to enable second element and disconnect any additional load.

Maybe your supplier has a 1000W element for this?

Sure hope there is a pressure relief valve... and run dry prevention.

How big is that tank? Liters, gallons, whatever. Seems odd that it takes an hour to heat water to about 100C with 1500 watts.

Winding step down transformers isn’t going to be cost effective or safe.

WattsThat:
How big is that tank? Liters, gallons, whatever. Seems odd that it takes an hour to heat water to about 100C with 1500 watts.

An hour is too long or too short? 1.5 kW = 5.4 MJ per hour. That is about 16 litres of water from 20 to 100 °C. That could be enough if the coffee is strong.

dlloyd:
EDITED: If you remove just 1 jumper, you'll have 750W 6.25A. Will 1/2 power be enough to bring up the temp to your setpoint? If it does, but you need faster heat-up, then you might need a switch or relay to enable second element and disconnect any additional load.

Maybe your supplier has a 1000W element for this?

Thanks for these suggestions. This sounds like a pretty good solution! I believe I will give the process the full 1500W when heating up from cold temperatures and then when it gets close to normal operating temperatures, switch to 1/2 power and run the PID control. I am not sure which is better though running in series or just removing a jumper.
The tank size is about 9L, and right now there isn't insulation around the boiler. So I should probably add this and it will cut down heating time. I bet it is losing quite a bit of heat in its current state. I think heating up in under an hour is not a bad goal because I just have the machine on a program that heats up before I need it.
Also to WattsThat's question there is a pressure relief valve set to 2.5 barg.

Running in series will reduce the current even more: to 1/4 of the original power. Is this what you want/need? Do you know what power is needed to keep the water hot?

Each element is rated at 750W, 120V, 6.25A, 19.2Ω

Series: 375W (187.5W + 187.5W), (60V + 60V), 3.125A, (19.2Ω + 19.2Ω)
Single: 750W
Parallel: 1500W (750W + 750W), 120V, (6.25A + 6.25A), (19.2Ω / 2)

Yes, adding insulation is a good idea ... will make a huge improvement.

Smajdalf:
Running in series will reduce the current even more: to 1/4 of the original power. Is this what you want/need? Do you know what power is needed to keep the water hot?

I think 750W would be better than the 375W but I may do some experimenting with this. The reason is it will be more complicated to actively switch from running the elements in parallel to series versus keeping them in parallel and just cutting one element out of operation with a relay. I think that both 375W and 750W operation will be able to keep the boiler at temperature though. If I am losing 375W due to natural convection then I am in trouble haha.

If you're using the Arduino PID library but would like an AutoTune function, you might want to give QuickPID a try. I've only tested the AutoTune function with an RC filter having 0.1 to 20 sec time constant, but sometime in the future I intend to test this on a unique re-flow oven (just an idea at this point).