V-phase

[Jabbajaws]

Hi Everyone,

I have recently read an article in Professional Electrician magazine, regarding a fixed unit that can be installed, called V-phase. From what l can see this unit is placed between the electric meter and the Consumer Unit.

It restricts the incoming supply voltage to 220V and is supposed to be more efficient, saving the owner of the installation up to £100 per year on electrical energy.

Has anyone come across this unit or installed one anywhere?

Is it worth it in the long run? And does it actually do what it states? or is it merely a gimmick? The average supply voltage in my area for domestic use tends to be around 240-245V.

Could lowering the supply voltage to 220V have any problematic effects?

 

[GrannySmiff]

I can't understand why it would make your electricity bills cheaper, if the voltage goes down, the current will go up.

so by watts law if you are using 500 W at 230 V the current will be 2.17 A, and if you go to 220 V the the current will rise to 2.27.

As you can see the current has increased, but its the wattage used is how the bill is calculated right?

Woould really like to understand what differnce this would make, not because I am intrested in buying one but out of uinderstanding electricity a little bit better.

Plus how would they make this work would it be some sort of capacitor/inductor can't work out which one would could limit the supply voltage?.

Hmmmm.

 

[Jabbajaws]

Im not exactly sure to be honest how it works. Ive seen it advertised in a my monthy edition of Professional Electrician and was curious.

And it seems to make sense, what you say with regards to watts.

Im sure that the unit advertised will itself have some form of advanced electronics, to be able to do what it says, in my opinion.

I'll do some research and report back when ive learned more...

 

[Graeme Harrold]

This the one then........... V-Phase

Looks like it works like a restrictor keeping supply voltage to an otimum level for connected equipment. Bit like a car running at 56mph where optimum fuel consumption is

 

[Jabbajaws]

Thank You Graeme,

that is indeed the exact same V-Phase l am talking about.

Have you any comment on what GrannySmiff pointed out? Regarding a higher current value being used, in relation to the watts value. To be quite honest l have wondered about this before and would like to find out the answer. Im sure that theres a mathematical explaination that im missing in all this.

Are you or anyone else able to explain this to me as im dumbfounded at the mo...

 

[Sintra]

What I know on the subject is that appliances are designed to run on an optimum voltage ie 220V.
Running them on a higher voltage reduces efficiency and life span.
Appliances running at optimum voltage can be 13 - 17 % more efficient.
The car at 56 mph is an excellent comparison.

 

[Phil]

I had a chat to the guys from v phase at the elex show last month. nice looking unit the same size as a consumer unit. easy to wire up too. i think i may be investing some time soon

 

[GrannySmiff]

I was thinking about it myself and I presume it must be some sort of step down transformer, but as voltage fluctuates keeping it a steady 220 would require some form of alternating transformer if one exists???.

Also on safety point on an old install you will be upping the current and could cause nuisance tripping or worse maybe?.

 

[saveloy]

I was quite interested in this unit. However one thing that puts me off a bit is the amount of circuits you CAN'T connect to it

Circuits that must NOT be connected to the VX1:
Electric shower circuit.
Immersion heater.
Dedicated electric cooker circuit.
Circuits feeding high power tools.
Circuits where individual loads exceeding
13A rating will be connected.

A lot of high current circuits where the most savings could be made. Still think its a good concept though.

 

[Doug]

I can't understand why it would make your electricity bills cheaper, if the voltage goes down, the current will go up.

so by watts law if you are using 500 W at 230 V the current will be 2.17 A, and if you go to 220 V the the current will rise to 2.27.

As you can see the current has increased, but its the wattage used is how the bill is calculated right?

Would really like to understand what difference this would make, not because I am interested in buying one but out of understanding electricity a little bit better.

Plus how would they make this work would it be some sort of capacitor/inductor can't work out which one would could limit the supply voltage?.

Hmmmm.

This is not the correct way of looking at it. It is not the power consumption (or wattage) that is fixed, but rather it is the resistance that is fixed.The power rating is merely the power that will be drawn at a particular voltage. If you reduce the voltage, then you also reduce the current drawn and hence reduce the power consumed.

The relevant equation is Ohm's Law, V = I x R . If R is fixed, then reducing V will also reduce I. You can then use P = I x V to calculate power consumption.

This is why electric showers often have two power ratings listed. They can have a lower power rating calculated at the correct nominal voltage of 230V and also a more impressive power consumption using 240V.

Think what would happen if you connected a shower to a 1.5 V battery. Using my theory of a fixed resistance, the battery would supply such a low current that nothing would happen. With GrannySmiff's theory of a fixed power consumption, an enormous current would flow, frying the wires.

Hope this makes sense.

Is it worth it in the long run? And does it actually do what it states? or is it merely a gimmick?

I can see that having a steady voltage could be a good thing. I'm sceptical though about the voltage reduction. If things are designed for 230 or 240 V, then why use 220V? Do people really want the lights a bit dimmer?

The regs teach that voltage drop is a bad thing. How can you sign off on an EIC that you have checked for voltage drop if you have built in a big drop at the origin?

What happens when the unit packs up and the house is in darkness for a few days until a service engineer can visit?

 

[AEC]

Sanity at last.

 

[GrannySmiff]

I wish it did make sense Doug, and its not that I am going against what you are saying I am just new to this.

SO what I don't understand is

The relevant equation is Ohm's Law, V = I x R . If R is fixed, then reducing V will also reduce I. You can then use P = I x V to calculate power consumption.

When you say the resistance is fixed what do you mean, that if we say my kettle uses 2200W which it says it does at the nominal voltage of 230 the current will be 9.56, and with 220V I will be 10A.

Now what part is the fixed resistance, can we say that the load is the fixed resistance? if so which measurement would we use for R?

Thanks if you can help me understand this as I like to have a fundamental understanding of why.

 

[Doug]

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. The rating of 2.2kW only applies at 230V. At 220V, the kettle is rated at 2.01kW.

The calculations could be simplified by combining P=IxV and V=IR to give R=V(squared)/P

At 230V, R = 230x230/2200 = 24.05 Ohms

At 220V, P = 220x220/24.05 = 2012 Watts

At 200V, P = 200x200/24.05 = 1663 Watts

 

[Jabbajaws]

Cheerz Doug,

Thats the missing link im after...

 

[albi]

v-phase are holding a seminar and installers corse in durhan (castle edan) on the 5 & 6 of may so i will tell you all about the ins and outs of it later its always worth a look