How to reduce mains voltage from 245 to 220V or less?

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I'm puzzled. Having laid out some facts that make farting about with mains voltages quite un-necessary, those facts are ignored to continue to further the needless discussion? I suspect I'm being a killjoy, just getting in the way of the need to fiddle with high power mains devices (… in the interests of safety? That's ironic isn't it).
 
peebee said:
I'm puzzled. Having laid out some facts that make farting about with mains voltages quite un-necessary, those facts are ignored to continue to further the needless discussion? I suspect I'm being a killjoy, just getting in the way of the need to fiddle with high power mains devices (… in the interests of safety? That's ironic isn't it).
Click to expand...
You’re absolutely right.

Resistors and dummy loads to drop voltage is ridiculous.
 
SteveH said:
I think the only safe way to do that would be to have it locally fused e.g something similar to https://www.toolstation.com/hook-up-unit/p92936

Personally I'd prefer to have that on a dedicated (16a or 20a fused) radial circuit from the consumer unit to avoid potentially tripping the existing ring, but really this is something best advised by an electrician with up-to-date knowledge of the regs.
Click to expand...
Yes a consumer unit spurred off a ring would work, although not entirely ideal as per regs.
 
No nothing is going to seriously drop your current draw and still give you the required heat.Its all about I2R.I used to use at work a 30A variac it was the size of a car wheel,Of course the company paid for it.

The way forward is to have a "cooker" supply installed that meets the IEEE regs.The downside is the new 30amp supply will be rated for 100% duty cycle, Since liquids are involved also an earth leakage trip should be installed.Therefore the job might be a bit pricey.

I think it important when posting on forums to ONLY give safe advice.
I would hate to give anybody shortcut ideas that could lead to loss of property or worse life.

BTW. Electrical engineering is my lifelong buisiness,wine making is only my hobby.
 
kelper said:
I could use a 30W 2 Ohm resistor in series.
Click to expand...
I'd suggest you don't do any of the things you've suggested @kelper; and I don't mean to patronise you in any way...

It's very simple. With both elements on, your power is a constant at 3.3kW, if you increase your voltage, you decrease your current - likewise: decrease your voltage, you'll increase your current.

This is why long line transmission is such a high voltage. It decreases the current to allow for lesser CSA conductors.
 
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Ghillie said:
How can you possibly put a 30W resistor (this is the maximum amount of power it can tolerate) in series with 3.3kW of inductive load?
Click to expand...
But it's not inductive! The 2R resistor will dissipate 338W 9corrected thanks to SteveH)while the heating elements will now dissipate 3kW and the current drawn will be less than 13A

Ghillie said:
It's very simple. With both elements on, your power is a constant at 3.3kW, if you increase your voltage, you decrease your current - likewise: decrease your voltage, you'll increase your current.
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This is incorrect for a resistive load.
 
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peebee said:
I'm puzzled. Having laid out some facts that make farting about with mains voltages quite un-necessary, those facts are ignored to continue to further the needless discussion? I suspect I'm being a killjoy, just getting in the way of the need to fiddle with high power mains devices (… in the interests of safety? That's ironic isn't it).
Click to expand...
I have explained that, with my mains voltage being 245V, the fuse is overheating if I run both elements. It's my thread so please ignore it. acheers.
 
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kelper said:
The 2R resistor will dissipate 26W
Click to expand...

P = I^2 x R

So a 2 ohm resistor will dissipate 338w at 13A

A series resistor may be a reasonable approach in low voltage and/or low current applications but IMHO it's a very bad idea and a potential fire risk in a high power application like this.
 
kelper said:
But it's not inductive! The 2R resistor will dissipate 26W while the heating elements will now dissipate 3kW and the current drawn will be less than 13A


This is incorrect for a resistive load.
Click to expand...
It's been a long time since I was at college so may well be talking rubbish.

Your elements are an inductive load. Your power resistor will be, obviously, a resistive load. If you intend to put it in series with the inductive elements then you will need to accurately calculate circuit impedance for your results to mean anything; and for your resistor to be the correct size.

But in all honesty, just run the right cable and have adequate and correct circuit protection in place. If you want to fit power resistors to drop voltages and make things more difficult then that's your choice;)

If you fit a resistor, as a current limiter, you're basically doing the same as trying to power something with a very long extension lead. Ultimately (without getting caught up in unnecessary technicalities) your wort will take forever to heat up.

If in doubt... Stay safe.
 
A heating element is resistive. It may have an inductive component but this will be small. The inductance or capacitance in a domestic setting is irrelevant since the PF will always be close to unity. All the resistor 'sees' is the current. This thread has helped me get my thoughts straight so you have all helped, even those who think I'm nuts.

I will install a 300W 2R resistor in series. (corrected wattage, thanks SteveH).
 
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kelper said:
A heating element is resistive. It may have an inductive component but this will be small. The inductance or capacitance in a domestic setting is irrelevant since the PF will always be close to unity. All the resistor 'sees' is the current. This thread has helped me get my thoughts straight so you have all helped, even those who think I'm nuts.

I will install a 100W 2R resistor in series. I've picked 100W as it can easily dissipate 30W without a heat sink. All wires will be properly insulated. If successful I will build an enclosure (ventillated) with a 13A socket and a cable with 13A plug. This will save me a lot of money.
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Good luck, let us know the outcome!
 
SteveH said:
P = I^2 x R

So a 2 ohm resistor will dissipate 338w at 13A

A series resistor may be a reasonable approach in low voltage and/or low current applications but IMHO it's a very bad idea and a potential fire risk in a high power application like this.
Click to expand...
Thanks. I had a brain fart.
 
Johncrobinson has given the best advice and with the required electrical knowledge IMO but if you do go your own way please please be careful as electrics are not like water you don't just mop it up it kills. Good luck my friend and I hope it all ends well and safe whatever you do
 
Fit a triac-based phase-angle controller just before the kettle elements and set the "level" to wherever you're happy. Rule of thumb losses in the triac are about 1W per amp, reduced in proportion to the amount of "on-time" spent by the triac per cycle.

Here's mine: https://www.thehomebrewforum.co.uk/...a-cheap-digital-boiler-power-regulator.74112/

Power needed for decent 23 litre boil should be around 1600W
 
kelper said:
I have explained that, with my mains voltage being 245V, the fuse is overheating if I run both elements. It's my thread so please ignore it. acheers.
Click to expand...
Okay, I'll ignore the thread, but only if you stop challenging me and "I have explained that …" is a challenge!:p

… I know. But 245V is within limits (from Wikipedia: "In the UK and Australia the nominal supply voltage is 230 V +10%/−6%"). So what you are trying to do most likely isn't to correct failings in the brewing system you have (unless it operates dangerously out of spec), it is probably to correct failings in the overheating circuit breakers you are using. That could be a sobering thought? These are not uncommon occurrences as we are awash with cheap Chinese components these days (if it wasn't the Chinese, we'd find someone else to manufacture the stuff for peanuts).
 
peebee said:
Okay, I'll ignore the thread, but only if you stop challenging me and "I have explained that …" is a challenge!:p

… I know. But 245V is within limits (from Wikipedia: "In the UK and Australia the nominal supply voltage is 230 V +10%/−6%"). So what you are trying to do most likely isn't to correct failings in the brewing system you have (unless it operates dangerously out of spec), it is probably to correct failings in the overheating circuit breakers you are using. That could be a sobering thought? These are not uncommon occurrences as we are awash with cheap Chinese components these days (if it wasn't the Chinese, we'd find someone else to manufacture the stuff for peanuts).
Click to expand...
I have no wish to "challenge" you. But please stop lecturing me on UK voltages. Mine is 245V. The equipment is rated at 3.3kW at 240V and fitted with a Shuko plug which I'm sure you know, is 16A. I don't worry about the wiring or MCB in my house, since it's on a 32A ring. But I do take it seriously that this CE-marked unit draws 15A and that this causes my 13A plug to get very, very hot.
 

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