switching between 2 sockets on a timer

[califauna]

gas would be ideal but the outlay to buy all new appliances would be too much. Idea for the future b ut not for now.

Cant seem to find and elements for this tea urn, let alone lower power ones.

Anyone know of a way to safely reduce the power consumption of an electectrical item? Increase the electrical resistance of the circuit path of the appliance somehow? Its not digital so i think it would run on reduced voltage.

Would a non digital 240 v kettle run on a 12v power supply? The power consumption of the kettle would be reduced by a factor of 20 (240 / 12 = 20), but would it damage the kettle?

 

[big-all]

reduce the volts by 20 multiply the amps by 20
10 amps at 240v is 100 at 12v 12watts at 12v=1 amp
240watts at 240v=1 amp

 

[Mursal]

So you have a 2800W (240v) kettle, and you want to drive it at 12volts. Well I get 7watts from the element, so it will take a while to boil the water.

2800W/240 ≈ 11.7A
240v/11.7A ≈ 20 Ohms
=> (12v x 12v) / 20 Ohns ≈ 7Watts

(7W/2800W) x 100 ≈ 0.25% of the original output

 

[califauna]

reduce the volts by 20 multiply the amps by 20
10 amps at 240v is 100 at 12v 12watts at 12v=1 amp
240watts at 240v=1 amp

Im confused. Im quite sure these appliances dont have switching circuits inside them to adjust according to a changing voltage, so I dont see how the current can go up if the voltage goes down.

At 240v the curretn draw is 11 amps.

i=v/r

r=v/i

r=240/11 = 20 ohms

so if the voltage is changed to 12 then it would be:

(i=v/r)

i=12/20 so i= 0.6 Amps

So I calculate that the amps at 12v for the same appliance would be 0.6.

This would mean the appliance is drawing just a fraction of what it should, so it proably wont work, but Im guessing it wouldnt damage it.

 

[bernardgreen]

You can reduce the power consumption of a resistive heater working on AC by putting a capacitor in series with it.

A rough calculation suggests that 100 u F ( microFarad ) capacitor would bring the current down to about 7 amps , about 1.6 Kw

A suitable capacitor of that size is not cheap, around £30 pounds.

If the capacitor fails open circuit the heater gets no power. if it fails short circuit the heater gets full power which is safe.

 

[califauna]

You can reduce the power consumption of a resistive heater working on AC by putting a capacitor in series with it.

A rough calculation suggests that 100 u F ( microFarad ) capacitor would bring the current down to about 7 amps , about 1.6 Kw

A suitable capacitor of that size is not cheap, around £30 pounds.

If the capacitor fails open circuit the heater gets no power. if it fails short circuit the heater gets full power which is safe.

I just looked online and 100 micro farad capaciors seem to cost about 3 pounds for a pack of about 5. THeyre not 30 pounds each. Am I looking at the wrong type?

Not sure Ill go this route, but out of curiosity, should it go on the positive side, or nuetral?

Would some kind of resistor also work?

 

[Mursal]

A resistor wired in series will work, but the 240volts will be shared across the 2 loads (element (50v) and resistor (190V)) so you get power loss (wastage) across the resistor.The voltage levels across each one will depend on the their size (in ohms). The result will be a resistor that heats up and wastes precious power.

But if you put two 240volt heating elements in series, the current will be halved and the overall heating effect (wattage)will be halved. So that might be an option? Just take twice as long to heat the water?

240v/41 Ohms = 6 Amps

 

[califauna]

A resistor wired in series will work, but the 240volts will be shared across the 2 loads (element (50v) and resistor (190V)) so you get power loss (wastage) across the resistor.The voltage levels across each one will depend on the their size (in ohms). The result will be a resistor that heats up and wastes precious power.

OK got it. thanks for the explanation.

 

[Mursal]

I had a wee think about it ......

See Edit 1 above?
Provided you have room for another or different (higher resistance) element

 

[califauna]

I had a wee think about it ......

See Edit 1 above?
Provided you have room for another or different (higher resistance) element

Great idea. DOnt think I could get another element in there though. I cant seem to find any for that model, and I dont know how I would connect it and fit it in there.

I have another smaller water urn (1500 watts), again just an element and mechanical thermostat (not electronic). Could I just connect the whole appliance in series before the water urn

 

[Mursal]

Yes

If you could use the small one on its own you would have the full 1500watts (6A) output?
Both in series around 990watts (4A) in total
Large one (2800W) giving out 337W and the small one 653W

So that leaves you with lots of options, and no wasted power

All high voltage so be extra careful as the neutral (usually 0V) from the first element will be the new live to the second element. Not an issue now, but could be if someone else had to go faultfinding.

 

[bernardgreen]

I just looked online and 100 micro farad capaciors seem to cost about 3 pounds for a pack of about 5. THeyre not 30 pounds each. Am I looking at the wrong type?

Yes

Almost certainly you are looking at electrolyic capacitors which are not suitable for the task. For the power reduction you need a different type.

http://uk.rs-online.com/web/p/polypropylene-film-capacitors/7441726/ being the first one I found.

 

[big-all]

just a though

if this is a commercial operation subject to regulation it might be worth checking whats allowed as they will be non standard ??

 

[ban-all-sheds]

If the capacitor fails open circuit the heater gets no power. if it fails short circuit the heater gets full power which is safe.

There is also the safety aspect to consider of how spectacularly the capacitor fails....

 

[ban-all-sheds]

So you have a 2800W (240v) kettle, and you want to drive it at 12volts. Well I get 7watts from the element, so it will take a while to boil the water.

The specific heat of water is 4.19J/gK.

1W = 1 joule/sec, so a 2800W kettle delivers 2800 per second, and a 7W one delivers 7 per second.

If we assume 1l of water to be raised by 90°C we need to put in 377,100 joules (assuming no heat losses during the process).

So with that assumption (which will be false in practice), a 2800W kettle will do it in 135 seconds. In practice a bit longer, but not much, as heat will be going in much faster than it is being lost,

A 7W one will take 15 hours. In practice I'd wager that "never" would be what happens.