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AC Voltage Drop Calculation in Cable
Engineering and workplace issues. topic
Posted by TPS on 12 May, 2013 - 12:36 am
Dear Sir,
I need help for calculating the voltage drop in following cases:

Brief:
We plan to draw 110VAC/50Hz @ 150A UPS single phase power from Point A to Point B, which is approx 600 meters away.

2 Core cable(s) is/are proposed to be laid over the ground and will be drawn through cable trays.

One Core of cable will be connected to Phase and 2nd Core will be connected to Neutral.

The peak summer temperature range in our area is between 35-45 Deg C.
The minimum temp in winters can be between 10-15 Deg C.

We have two cable options: Copper cables 2c 300 Sq mm OR 2c 400 Sq. mm with following specs:

2c 300Sq. mm:
DC Resistance in ohms/km @ 20 Deg C: 0.0601
AC Resistance in ohms/km @ 70 deg C: 0.073
Reactance @ 50Hz in ohms/km : 0.086
Max Current in Air: 465A

2c 400Sq. mm:
DC Resistance in ohms/km @ 20 Deg C: 0.047
AC Resistance in ohms/km @ 70 deg C: 0.059
Reactance @ 50Hz in ohms/km : 0.086
Max Current in Air: 530A

Request help on follow Questions:
1) What will be the voltage drop if 2c 300 Sq. mm cable is used?

2) What will be the voltage drop in 2c 400 Sq. mm cable is used?

3) Will laying parallel multiple cables reduce the voltage drop?

4) Does Sq. mm size imply size of each core or it indicates the total cross section area comprising of both cores?

Thanks
TPS


Posted by connolly on 12 May, 2013 - 12:35 pm
From: http://en.wikipedia.org/wiki/Voltage_drop

Voltage drop = 2 * length in meters * resistance per km * current in amps / 1000 meters per km

A: VD = 2 * 600 * .073 * 150 / 1000 = 13.1 volts

B: VD = 2 * 600 * .059 * 150 / 1000 = 10.6 Volts

The above numbers are approximate, not taking into account that resistance varies with temperature; but 150 amps is an approximation as well.

C: Yes, a second set of conductors cut the resistance in half, so according to the voltage drop formula, they would also cut the voltage drop in half.

D: The size should be the size of each conductor, not the total size of conductors in the cable.


Posted by Dibin P kuriakose on 24 October, 2013 - 7:28 am
In this formula,
Voltage drop = 2 * length in meters *
resistance per km * current in amps /
1000 meters per km

Is this 2 represents the core of cable?
if it is 3core cable does the formula change to

Voltage drop = 3 * length in meters *
resistance per km * current in amps /
1000 meters per km

Thanks & Regards,
Dibin P.Kuriakose
dibinpk@yahoo.co.in


1 out of 1 members thought this post was helpful...
Posted by Phil Corso on 27 October, 2013 - 12:53 am
Dibin ...

can you explain why you believe the circuit-length multiplier '3' should be used instead of '2'?

Regards,
Phil Corso


Posted by Curt Wuollet on 12 May, 2013 - 3:13 pm
Assumptions: 150A AC, 50 Hz:
The two core thing in the UK generally means two wires insulated from each other, the area would then be for each.
Other places might be different.
We'll use 70 C as you shouldn't count on ambient and it would then be worst case.

.073 X .6 X 150 = 6.57 per wire so 13.14 total
.059 X .6 X 150 = 5.31 " 10.62 total

the .6 is 600m in km .

The first is greater than 10%, the second a little less.

I would use bigger wire or multiples if allowed.
Multiples are often not allowed and not considered best practice because
a slight difference in connections can unbalance the currents widely.

That's the basic idea, correct for any wrong assumptions.

Regards
cww


Posted by TPS on 14 May, 2013 - 1:59 pm
Thank you so much for the replies. Just wanted to know if there is any specific reason for ignoring the reactance of cable.

Regards
TPS


Posted by connolly on 14 May, 2013 - 7:43 pm
I assumed the "R" was resistance. Turns out it's a resistance factor, per the NEC code. Reading further into the wikipedia article, total impedance should be taken into account.

When accounting for reactance, I calculate the impedance of the first cable to be: 0.677 ohms, and the second to be 0.0626 ohms.

((0.073 ohms/km*0.6km)^2 + (0.086 ohms/km*0.6km)^2)^0.5 = .677

((0.059 ohms/km*0.6km)^2 + (0.086 ohms/km*0.6km)^2)^0.5 = .626

Those are for the length from the source to the load, so the total impedance is twice that, and the voltage drop formula would be

VD = 2*I*Z = 2*150 * 0.677 = 20.3 V
VD = 2*I*Z = 2*150 * 0.626 = 18.8 V

Honestly, I don't think it matters what the voltage drop is; 600 meters is a very long distance to be trying to transmit 150 amps at 110 Volts. I would look a second UPS at the load, or using transformers to transmit the power at a higher voltage.


Posted by Gerald Beaudoin on 15 May, 2013 - 12:59 pm
I agree entirely with the suggestion to use transformers at both ends of the run and use a higher transmission voltage. certainly common practice around here. 240vac to 600vac at the source, then back down from 600 to whatever.


Posted by connolly on 14 May, 2013 - 8:42 pm
To answer your question a little better -

If you ignore reactance and do the calculation with only resistance, that will give you the actual voltage drop.

The issue with reactance is that the voltage will lag the current.
I think it works out that when the current is at the peak of its sine wave, the voltage would still be climbing, and the effective voltage would be the second calculation I did.

You could negate the inductive reactance of the cable by adding an equal amount of capacitive reactance.


Posted by gajendra ds on 17 September, 2013 - 11:18 pm
Dear All,

I kindly seek a quick reply for this query.

I have a 4 core cable. one of the leads is connected to the phase other 3 left open.

But if you measure the voltage at all of the 4 leads with respect to neutral of connected supply its showing voltage. its 220V for bundled full length but its 30, 38 and some for half meter cable.

So whether the problem is with cable or its a normal electrical phenomena. I am just confused. Please reply fast

Thanks in advance

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