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Ah yes I see now Doug thanks for explaining, should have got that by myself as I know its in there somewhere.
 
As far as I know from reading the info on the site there is a bypass, this will activate and the voltage will return to normal levels in certain circumstances, but if the whole unit malfunctions? Guess if your that interested to know ask the manufacturer.
 
It seems to me by looking at the site the kettle would take longer to reach boiling point thus using the same amount of power in the long run, same with central heating. I think there would be no saving with this sort of appliance.
 
I'm not convinced on this sort of thing.
Different types of appliances will react differently to the reduction in voltage. But simply:

Resistive loads (eg convector heaters)
Their output will be reduced. The only time this will be of benefit is if the heater was oversize for the room anyway.
If thermostatically controlled, the exact same amount of power will be used, as the heater will need to be on longer (Duty cycle) to achieve the set temperature.

Lighting loads (filament and magnetic ballast discharge)
Light output will be reduced, colour temperature will shift.
Lamp life will increase in most cases.

Motor loads (Fridge, Freezer)
A small reduction in available torque...but the appliance should have been designed to work correctly within the whole voltage range... so a small efficiency gain is possible

Equipment with linear power supply (Hi-Fi Amplifier, portable audio, alarm panel, etc)
Small efficiency gains are possible with items containing linear regulators.
But the maximum output of the amplifier will be reduced as the power supply for that stage generally does not have any regulation.

Equipment with switching power supplies (PC, TV, Freeview/Sky, advanced microwaves, HF Lighting ballasts,...)
THIS IS THE KILLER.
Most switching power supply designs are wide-range (90-264V), to use on any electrical system in the world.
They do exhibit the constant power characteristic.
So, as you reduce the input voltage, you increase the input current.
In this case, there can be no benefit from reducing the voltage. The higher current will cause more voltage-drop losses.

My next door neighbours just moved to a new house and wanted a few little jobs doing. They asked me if something was wrong with the electrics at the new place as the kettle was taking longer to boil and the shower was pathetic (funnily enough exact same model and rating)
Old house: 243V. New house 226V.

Does anyone remember those magnets that you could stick onto your car's fuel line to increase mpg lol?
 
Actually just seen this for the First time today on TV and it sounds similar to what some of the American Electrical firms are doing by Installing a Capacitor next to the meter to reduce the Electricity bill (I'll look for the Youtube vid of it)
 
Just had some experience of these.

Basically, it will only work on current up to 8A, and anything over it will switch into bypass mode, thus through putting standard incoming voltage.

Once the load is reduced, it will switch back in.

So, in other words, it will just about do the lighting.

Also, if the unit fails, it will also go into bypass, but if you install the unit remotely, then you wont know until you check the front indicator is lit.
 
lol so your basically paying around 300 quid just to save on Lighting energy when everyone is more or less using Energy saving Light bulbs anyway.

This wasn't the Exact vid I was looking for but it's similar[video=youtube;_h30X6EEsn0]http://www.youtube.com/watch?v=_h30X6EEsn0&feature=related[/video]
 
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I believe it might save you money on the low consumption devices that don't automatically compensate for a voltage reduction.(like computers do). Low power heating will just take longer. They claim bulbs will last longer which may well be true and this will save some money, but only if people don't put higher power ones to get the light back. Small motors will be more efficient as the extra volts over 220 just results in resistive heating. It cuts out when demand goes up although they are talking about making a more powerful one.

I doubt that I could sell one confidently to a good customer on the money saving argument though.........Needs to go with a CU change but will double the cost......
 
Strange but true, these units do work and as previousley stated provides optimome voltage , with regards to voltage drop calcs ( unless im the only one ) should be at 230/ 400v ( DNO state 230v + 10% - 6% )
There are a number of manufacturers supplying units for commercial use ( Claude Lyons ) for 1.
 
I've done a bit of research into v-phase on another thread with the same arguments. I'm still not convinced on the estimated money savings.

All appliances made from 1990 with a CE mark are designed to operate on 220v so that's where this comes from. But people will just increase the watts of their light bulbs (luminaries), going from let's say 243v 40w down to 220v, you would see a difference. I didn't know that they don't work above 8A? So It's all about lights, radials with <8A and I don't think this will be worth the cost. May be worth while on new & rewire properties to design each circuit to use less current on each but we will end up with 15 way c/u's. If these units were free and we can only charge for installation it would still cost ÂŁ120-ÂŁ180 and would people go for it in large numbers to save money? The jury is out on this one for a while.
 
So basically BS then. Any type of resistive heating will take longer at a reduced voltage, and due to heat loss will end up using more energy. The majority of electronic items around the home use SMPSU's to comply with EU efficiency regulations, therefore will draw the same power at 220/230/240V, and this is likely to be one of the main areas where most people would expect/want to save money.

Incandescent lamps have a very non-linear voltage to brightness relationship, so a reduced input voltage has a significant effect on the brightness - perhaps enough for people to turn on another light. The colour rendering and temperature will also be affected negatively.
 
GrannySmiff, using your example a kettle is rated at 2200 W at 230 V
rearranging P=IxV gives I=P/V
I = 2200/230 = 9.57 Amps
so, at 230V, the kettle draws 9.57A

To find the resistance of the kettle, use R= V/I = 230/9.57 = 24.03 Ohms
If the voltage is reduced to 220V, I = V/R = 220/24.03 = 9.15 Amps
the power consumption at 220V is P = IxV = 9.15x220 = 2013 Watts

So, if the voltage is reduced then current is reduced and less power is consumed.

But the kettle will take longer to boil. ;)
 

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