Voltage Drop Calculations in 400 V systems

T

tahir ayub

I would be grateful if someone could assist me in finding the method of calculating voltage drop at the point of utilisation from the source in a 400 V 3 phase distribution system. The load is connected accross two phases. The neutral is not utilised in the distribution.

PhilCorso

Can you provide some informaton regarding the "load(s)?"

Regards,
Phil Corso, PE
(Boca Raton, FL)

J

Johan Bengtsson

Voltage drop is quite easy to calculate by ohms law:

Voltage drop for a wire:
U=R*I

In your case you should first find out the current you have in your wire (ie the size of the load)

Next part is to get the wire resistance to do this you need:
- Wire lenght (l)
- Wire area (A)
- Material and the resistivity for that material (ro)

Resistance is then
R = ro * l / A

if you have l in meters and A in square millimeters then ro for copper is 0.0172

This will of course give you the voltage drop for each wire, and since the current need to go thru two wires you have to double this to get the total voltage drop.

/Johan Bengtsson

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Box 252, S-281 23 H{ssleholm SWEDEN
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PhilCorso

In response to Johan's Jun 8, 11:03 am reply on the subject:

What you propose is a misleading oversimplification. The two loads are connected to a three-phase, 400 V, source. Hence, each of the three phase conductors will carry a different current. I'm sure you are not recommending that the two "phase' loads be treated as individual 400 V single-phase loads.

Thus, the final solution will involve calculations for an unbalanced
three-phase system. Of course, this presumes that Mr. Ayub's original question was related to voltage drop in the three supply conductors.

Regards,
Phil Corso, PE
(Boca Raton, FL)

J

Johan Bengtsson

First of all: The original post states that "The load is connected across two phases" This makes the third phase unused completely and the current will go thru one phase, thru the load, and back in the other used phase. The third phase will of course not have any voltage drop at all since there is no current.

Second: It does actually not really matter much how the load is balanced between the phases, this way of calculating actually works anyway if you apply it individualy on each phase. What you need it the current thru each phase if you have a balanced three phase load this will be equal and the voltage drop will be equal too. If you (as in this case) have a load connected to two of the phases those two will have equal current (and the third no current). If the load is unbalanced (in some other way) you will have different currents and different voltage drops in all three phases, but the formula still aplies to each phase.

This introduces a small approximation of course. Since the current probably is calculated assuming no voltage drop it might not be as expected. This would normally not be a problem since you would not like a voltage drop big enough
to seriously affect the final voltage (and thereby the final current). If it actually is a concern the easiest way is to do a second calculation based on the calculated voltage drop from the first and this will in most cases be good enough (or you have much too thin wires!)

Besides, depending on what the load is you might not know what the current is when the voltage don't match the specifications. If it is a purely resistive load where the resistance doesn't change when the voltage is changed it is easy to calculate, but if it isn't then you probably either need more information about the load or simply enough connect everything and measure the voltage drop or current.

Once the current thru each wire (phase or neutral) is calculated or you have the current from some other source (measured, or whatever) the formula applies to the voltage drop for that wire.

Is there something in this you don't agree to?

/Johan Bengtsson

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P&L, Innovation in training
Box 252, S-281 23 H{ssleholm SWEDEN
Tel: +46 451 49 460, Fax: +46 451 89 833
E-mail: [email protected]
Internet: http://www.pol.se/
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PhilCorso

Johan:

This time I agree... to disagree!

The last time we got into a agree/disagree contest was over the definition of the word "almost!" I would prefer to debate you about a more substantive topic. However, for the time being I wish to disagree over the definition, nay, interpretation of the originator's words "two phases" (Note plural).

I interpreted them to mean, for example, one load connected between 'R' and 'S' , the other between 'S' and 'T' or ''R' and 'T'. And, if I've interpreted you correctly, your interpretation is that both loads are between just one phase (note singular), for example 'R' and 'S'.

Rather than continue 'ad infinitum' with the disagreement, why don't we wait for Mr. Ayub's reply to my initial question regarding specification of the load(s)!

And may the best interpreter win!

Regards,
Phil Corso, PE
(Boca Raton, FL)

J

Johan Bengtsson

Ok, we'll wait.

My interpretation was that there is one load, connected between for example 'R' and 'S', that effectively making it a one phase 400V load, but using the three phase connector since it needs the higher voltage availiable between phases rather than between phase/netutral.

Besides that I still say that if you have any chance of calculating / measuring the current thru each phase the load could be as balanced / unbalanced between phases as you want, you can still caclulate the voltage drop for each phase by using the current and the wire length, area and material.

/Johan Bengtsson

----------------------------------------
P&L, Innovation in training
Box 252, S-281 23 H{ssleholm SWEDEN
Tel: +46 451 49 460, Fax: +46 451 89 833
E-mail: [email protected]
Internet: http://www.pol.se/
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