volt drop calcs redundant?

[johnboy6083]

Good morning ladies and gents. I had a thought last night. It hurt.
Anyway I took some painkillers, and carried on thinking.

These v phase voltage optimisation units reduce your voltage to around 220v. Does this mean that in houses without them, then calculating Vd is no longer a concern. On long runs, the cable going up a size because of the Vd calc, can cost a fortune, but I'd people are fitting these units at the origin of the installation to lower the voltage anyway, then?
Also does it say in the manufacturers info whether you have to carry out an assessment of the installation before? If you have large loads, and the cable calc was tight, then lowering the vOltage will increase the current drawn. Not by much, but surely the calcs are there for a reason. Also what about Zs. Lowering the voltage decreases the max Zs. Im aware that the manufacturers have a safety margin, but I don't like the idea of eating into that.

Just some thoughts.

John

 

[JUD]

Good morning ladies and gents. I had a thought last night. It hurt.
Anyway I took some painkillers, and carried on thinking.

These v phase voltage optimisation units reduce your voltage to around 220v. Does this mean that in houses without them, then calculating Vd is no longer a concern. On long runs, the cable going up a size because of the Vd calc, can cost a fortune, but I'd people are fitting these units at the origin of the installation to lower the voltage anyway, then?
Also does it say in the manufacturers info whether you have to carry out an assessment of the installation before? If you have large loads, and the cable calc was tight, then lowering the vOltage will increase the current drawn. Not by much, but surely the calcs are there for a reason. Also what about Zs. Lowering the voltage decreases the max Zs. Im aware that the manufacturers have a safety margin, but I don't like the idea of eating into that.

Just some thoughts.

John

I used to think this but it's wrong. The only constant in the circuit is the resistance. As current = voltage ÷ resistance the current will actually decrease with the voltage. Say you have 10Ω at 230V you current will be 23A (230 ÷ 10 = 23). Reduce the voltage to 220V and you current becomes 22A (220 ÷ 10 = 22).

You might be getting mixed up with transformers (less voltage = more current, more voltage = less current etc.)

 

[malcolmsanford]

Current also is w/v so if you had a 3kw kettle at 230v then 3000/230 = 13.04 amps like wise 3000/220 = 13.63 amps

 

[JUD]

Current also is w/v so if you had a 3kw kettle at 230v then 3000/230 = 13.04 amps like wise 3000/220 = 13.63 amps

But the wattage is not a constant. It is the product of the voltage x the current. P = V² ÷ R

230² ÷ 10 = 5290W (5290 ÷ 230 = 23)
220² ÷ 10 = 4840W (4840 ÷ 220 = 22)

 

[malcolmsanford]

Ahhhhhhhhhhhh but Jud that is in DC .............................:wink_smile:

And also as the element in the kettle gets hotter so the resistance will alter ...and so on and so ................


I was being devil advocate really, but bottom line IMO they still don't work

 

[telectrix]

exactly, so the kettle will take longer @220v. it still takes the same amount of energy to heat a specific volume of water by a specific rise in temperature. i forget whose law it was, something to do with calories.

 

[JUD]

exactly, so the kettle will take longer @220v. it still takes the same amount of energy to heat a specific volume of water by a specific rise in temperature. i forget whose law it was, something to do with calories.

Ohms law and Joules law I think

 

[JUD]

Ahhhhhhhhhhhh but Jud that is in DC .............................:wink_smile:

??????????

Are you saying ohms/joules law doesn't apply to AC

 

[malcolmsanford]

??????????

Are you saying ohms/joules law doesn't apply to AC

Jud I didn't say that and I was having a little joke with you mate hence the wink, but you know, that the ohms law formula and our eternal triangle is for DC systems, and will only apply to AC systems if the load is purely resistive, which is very unlikely due to capacitance, induction etc.

Yes there are formulas, which you quoted , that are used in AC systems, the ones you used are when the voltage and current are aligned in phase.

As I said mate I was trying to have a joke with you about, obviouslyfailing miserably

 

[JUD]

:lol: No worries

 

[JUD]

malcolm, in the cases where you had inductance/capacitance would you not just replace R (resistance) with Z (impedance) in the formula.

Z = √(R² + X²)
P = V² ÷ Z

 

[Guest123]

malcolm, in the cases where you had inductance/capacitance would you not just replace R (resistance) with Z (impedance) in the formula.

Z = √(R² + X²)
P = V² ÷ Z

Absolutely, in AC circuits we can simply substitute (R) for (Z) where the (Z) value is known or can be calculated. Stands to give a more accurate result IMO.

 

[Rockingit]

Just please, dear God, don't let me hear the words 'vector' and 'phasor' here in the pub. Sin is a different matter, that's what pubs are for.

 

[Guest123]

'vector' and 'phasor'

Aren't they from N-Dubz???

:50:

 

[GLENNSPARK]

exactly, so the kettle will take longer @220v. it still takes the same amount of energy to heat a specific volume of water by a specific rise in temperature. i forget whose law it was, something to do with calories.

might have been rayner`s law.....or was it waller`s?...hmm...

 

[Knobhead]

You’ve got to go back to Newton, but it’s been tweaked a fair bit since.