Single and three phase current capacity and voltage drop differences.

[bbudding]

Hi All,

From what I understand, working out the design current - Ib for single phase is quite easy (max appliance power/voltage).

For three phase, do I take the max power, divide by 3 then divide by 230 to get the Ib?

(We definitely use 230V for everything right?)

Now for voltage drop, We use the same Ib for both single and three phase voltage drop? But when we find the voltage which it drops, with single phase we divide by 230 to find the percentage dropped, and for three phase we divide by 400V to find the percentage dropped?

Thanks for your help!

Ben

 

[Lucien Nunes]

If you like, you can visualise a balanced 3-phase load as three independent star-connected loads each of one third the power, operating from 230V as you mention, to discover the line current. However the convention is to divide by sqrt(3) and use 400V which will give the same numerical result. The reason is that many applications such as motors do not have or need a neutral connection so there will not be 230V present anywhere. You didn't mention load power factor, so it's worth recalling that the figure for power required for calculating the line current is the apparent power in kVA rather than the real power in kW.

Moving on to the voltage drop, note that for fully balanced conditions there is no drop in the neutral because there is no neutral current, but for one phase loaded to the same current there is an equal drop in the neutral as in the line. Therefore one needs to know the worst case unbalance to know the worst case drop. A 3-phase DB that serves a few large single-phase loads might be at its worst when only one load is operational and the other two phases are very light, whereas one feeding balanced loads only or a large diverse array of single-phase ones might be at its worst with everything up full.

 

[pc1966]

Be aware that the 3-phase voltage drop calculation uses the current per phase as computed above, and the mV/A/m value tabulated for the cable, but it is for the nominal 400V system. See also this thread:

Understanding 3 phase volt drop.......

I am trying to understand 3 phase volt drop. I understand single phase VD. Lets say we have a 70 meter run of 35mm, 5 core cable from the supply to the 3 phase consumer unit. From the 3 phase consumer unit we have 1 x 16A triple pole MCB (so that's 16A per phase). which has a run of 12metres...

www.electriciansforums.net

 

[bbudding]

If you like, you can visualise a balanced 3-phase load as three independent star-connected loads each of one third the power, operating from 230V as you mention, to discover the line current. However the convention is to divide by sqrt(3) and use 400V which will give the same numerical result. The reason is that many applications such as motors do not have or need a neutral connection so there will not be 230V present anywhere. You didn't mention load power factor, so it's worth recalling that the figure for power required for calculating the line current is the apparent power in kVA rather than the real power in kW.

Moving on to the voltage drop, note that for fully balanced conditions there is no drop in the neutral because there is no neutral current, but for one phase loaded to the same current there is an equal drop in the neutral as in the line. Therefore one needs to know the worst case unbalance to know the worst case drop. A 3-phase DB that serves a few large single-phase loads might be at its worst when only one load is operational and the other two phases are very light, whereas one feeding balanced loads only or a large diverse array of single-phase ones might be at its worst with everything up full.

Hi!
That's very helpful. We're trying to work out basic voltage drop to solar inverters. If we take for example a 10kW 3 phase inverter. The Ib would be 10,000/root3/400 = 14.49A, but if I do it my way its 10,000/3/230 it gives 14.49A, which is odd. It's almost the same. But not quite. Shall I just stick to the root3 method to be sure. Yes - we're not considering power factor here.
Many thanks,
Ben

 

[pc1966]

The Ib would be 10,000/root3/400 = 14.49A, but if I do it my way its 10,000/3/230 it gives 14.49A, which is odd. It's almost the same. But not quite.

You also must remember the 400V and 230V and not quite srqt(3) in ratio, they are closes "human" rounded choices!

 

[bbudding]

Hi!
That's very helpful. We're trying to work out basic voltage drop to solar inverters. If we take for example a 10kW 3 phase inverter. The Ib would be 10,000/root3/400 = 14.49A, but if I do it my way its 10,000/3/230 it gives 14.49A, which is odd. It's almost the same. But not quite. Shall I just stick to the root3 method to be sure. Yes - we're not considering power factor here.
Many thanks,
Ben

Correction I mean 14.43A using the root method

 

[bbudding]

You also must remember the 400V and 230V and not quite srqt(3) in ratio, they are closes "human" rounded choices!

ah ok - 230 is rounded. That's great I think I got it now. Many thanks

 

[pc1966]

ah ok - 230 is rounded. That's great I think I got it now. Many thanks

I think technically it is the 400V that is "rounded" as the ESQCR actually gives the nominal value UK LV supply as 230V L-E here:

The Electricity Safety, Quality and Continuity Regulations 2002

These Regulations replace the Electricity Supply Regulations 1988 (S.I. 1988/1057) and all subsequent amendments (S.I. 1990/390, S.I. 1992/2961, S.I. 1994/533, S.I. 1994/3021 and S.I. 1998/2971).

www.legislation.gov.uk