Modifying 110v device to run on 240v

Forgive my simplistic approach.

By inserting the diode, no voltage will be over the element for half the cycle. This will allow the element to dissipate the heat into the water, before the current flows again. So I would guess that even if you had too high a voltage(within reason - not 240V) the element should survive as it is not just the rated wattage but also the rate of dissipation of the heat.
Ensure a diode that can handle the current.

 

Samon, you've already worked out that P = V^2 / R. Assuming R is roughly constant, P is proportional to V squared.

That means that doubling the voltage will quadruple the power. This happens because doubling the voltage (for a fixed resistance) also doubles the current, and power is voltage multiplied by current.

So without the diode, using a 120V-rated element at 230V will double the current (which is probably a problem in itself) and quadruple the power. If you add the diode, the mean power will be twice what it should be.

 

As a matter of interest, a friend of mine South Africa(230V) put a big capacitor over the supply to his US TV etc and it worked for him. So maybe that is another solution.

 

You would need a very large and expensive capacitor for this power. Use a transformer, also large, expensive and heavy.

You could use a series resistance but this will dissipate the same power as the device unless the diode bodge is also added.

 

The trouble is that we don't know the efficiency of heat dissipation of the element to the water at the different water temps, so calculation is not possible. I also noted that the calculation of power changes with different wave forms/phases, so maybe the calculations are not correct.
For the better electronics guys here it is a bit like calculating heatsinks - unfortunately I am no good at this and do it by trial and error.

The mop is useless unless it works in your area. The diode idea is fairly cheap and, unless it is a very expensive mop, worth trying. There won;t be a market for it where you are.

As a final note: In remote controls the IR transmitter LEDs are driven with up to 250ma (rated at 45ma) for short bursts to get the required range and left to cool between bursts. They are typically off for 25ms between bursts.

Good luck and let us know what happens. This is a fairly common problem I have seen asked a few times before, but never answered or reported back.

 

I agree with Steve. You would be dissipating twice as much power, which is a VERY significant factor, and your peak current would be twice as high, which would stress everything in the system.

Remember that the 110VAC system already operates at twice the current of a 230VAC system (for a given amount of power), which is why fuses and wiring in 110VAC systems have higher current ratings than the 230VAC system in your house.

The instantaneous peak power that element will draw (at the voltage peaks) at 230VAC is FOUR times what the element, and other parts of the unit, are designed to handle.

So I'm afraid you've lost this battle. But you've learned a few things!

If you decide to continue, do plenty of testing outdoors, using an industrial RCD/ELCB/GFI (whatever you call it) and keep it away from anything flammable.

Edit: Also, read your fire insurance documentation. It will definitely not cover you if you use the diode idea; it may specifically mention running equipment designed for a different mains voltage as an exclusion to the policy, since it is a known cause of fires (and at low power levels as well)!

 

@Steve @Kris

Thanks for the explanations, I again learned a lot and appreciate it.

The only bit I am still not clear on is the heat dissipation. How do we know that the thermal efficiency of the transfer between the element and the water is at its limit. If it is not, it might be able to dissipate more heat than it does at 110V. I would guess that, as it is a home appliance, there is a significant safety factor. What is true regardless, is that the unit will, if the element does not blow, produce steam at a much larger volume than designed and could become a 'steam bomb'. Hopefully the thermal fuses will protect against this or some safety valves.

From all the discussion, it seems the best solution would be to replace the mop with a local model. If the same model is sold locally, you might be able to get a properly rated element and replace it, but this will most likely cost as much as replacing the mop and other components might still fail.

 

I don't think that makes sense. Given that nearly all the power drawn by the element is converted into heat energy, and that most of that energy can't go anywhere except into the water being heated, I don't see what "thermal efficiency" even means in this context.

I tried to explain why, but I accidentally closed the tab and lost the explanation. Probably just as well, because I don't know much about the physics of heat. My barely informed opinion was that doubling the power dissipation in the heating wire (or whatever it was) would increase the temperature difference across the insulating material within the element tube and possibly damage it, and that the tube temperature would also increase due to the thermal resistance between the tube and the water due to the bubbles of steam.

No doubt there are folks here that really understand this stuff and can give an accurate explanation.

 

Thanks for the explanations.

 

One way of reducing the power of the element would be to use phase angle control, like a light dimmer but MUCH fatter.

The peak current will be very high, the radio interference could be severe and fuses/trips may object.

 

Is it possible to get a local electrical shop to wind a new element on the original former?