# Cable Voltage Drop and IEEE 141

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One of the frequent electrical issues we have to deal with from time-to-time is the effect cable sizing has on loads, or in reverse on source voltages.

IEEE 141 is quite a useful document worth buying as it discusses many issues including this issue as shown in Figure 3-11

One of the other problems we face in the industry is sometimes being a little "free" with our terminology and hence leading to some confusion when others refer to what ever we are trying to describe e.g. my other posts regarding to the fact that there is no such thing as "negative Power Factor"!! What we probably mean to say is it is a Lagging Power Factor  ... but some would argue that negative Power Factor means reverse power flow which it mathematically it doesn't.!  The fact that those two interpretations of negative meaning leading/lagging versus power being forward or reverse are 90 degrees different is a simple confirmation that negative Power Factor is a term we should not use because it does not exist!  But that is another page for you to seek out

But getting back to voltage drop ...

As you can see in IEEE 141 Fig 3-11, as the load voltage is at angle "zero" and the cable vectors are added to it,

IEEE 141 have approached the issue of voltage drop due to cable impedance from the point of knowing the voltage and current at the load and needing to determine what the voltage is at the source end of the cable. Having presented the graphical vector analysis in that sense, they are really showing the voltage RISE along the cable towards the source! But we can ignore the semantics of that as we are essentially interested in the difference in voltage end to end! :)

However lets consider this from the source end where we can measure the supply point values (the grid) with the intent of sizing the cable to keep the load voltage within limits.

In this scenario we want to know the affect of line impedance on reducing the voltage seen at the load end of a cable.

This analysis starts at the source where we measure Voltage and Current as follows:

ItemMagnitude component

Angle Component
(degrees)

Resistive ComponentReactive Component

Voltage:

|Vs|

<Vs

1000.00

0.00

Current:

20.172

-30.00

We can therefore deduce the source is supplying

VA

20172

PFsource

0.86603

PF directionInductive / Lagging

The total impedance seen at the source is dictated by Ohm's Law: Z = V/I
When we divide vectors we divide the magnitude and subtract the angle
We can therefore deduce that the total impedance seen at source is

Impedance seen at source:

|Ztotal|

<Ztotal

49.57

30.00

From that we can calculate to total resistive and inductive impedances connected to the source

Rtotal

Xtotal

42.93

24.79

We know the line impedance from the cable data:

Zcable

Rcable

Xcable

0.20

0.50

Which we can convert to a total value of impedance

Zcable

|Zcable|

<Zcable

0.54

68.20

Ohms Law demands that the voltage drop across an impedance = Vd = I x Z

When we multiply vectors we multiply the magnitude and add the angle.

Hence we know the voltage drop from end-to-end (i.e. as would be measured by putting a voltmeter connected across the two ends of the cable) is

Vdrop down the cable end-to-end:
(this is the value of IZ on the IEEE 141 diagram)

|Vcable|

<Vcable

10.86

38.20

Now going back to the impedances, since we know the value of the total resistance and reactance seen by the load and we know the resistance and recatance of the line we can calculate the resistance and reactance of the load

42.73

24.29

Which means that the total load impedance is

49.15

29.61

Ohms Law demands that the voltage drop across the load impedance = Vload = Iline x Zload

When we multiply vectors we multiply the magnitude and add the angle

Hence we know what the voltage across the load is at its terminals

991.49

-0.39

Here we have an interesting conundrum when we compare the magnitude of the phase-to-neutral voltage at the source with the magnitude of the phase-to-neutral voltage at the load

Recalling that at source:

Vsource

|Vs|

<Vs

1000.00

0.00

So if we used a voltmeter at each end to measure the phase-neutral voltage at each end, we would see there is a voltage difference equal to |Vs| - |Vload|

Difference in measured voltages:

|Vdiff|

8.51

This is very different to the magnitude of the voltage drop end-to-end of the cable as calculated above!!

|Vd|

10.86

If we now refer to IEEE 141, the approximation formula associated with Fig 3-11 is presented as

V = IRCos(ϕ) + IXCos(ϕ)

ϕ is the phase angle between the voltage and current as seen at the load

Annoyingly this is not the angle seen at the source!! And since we are measuring everything at the source we are seemingly a bit "stuck"

So we now have to calculate that as the difference between <Vload and <Iload

29.61

and just to verify that by calculating from the derived values above of Rload and Xload

29.61

Now we can calculate the approximation of the voltage drop as power IEEE 141

V = IRCos(ϕ) + IXCos(ϕ)

Vest

8.49

So we have now have three different views of this system

|Vest|

8.49

IEEE value

|Vdiff|8.51Measuring the phase-to-neutral voltages at each end
|Vcable|10.86Measuring the voltage end-to-end of the cable

As mentioned above the IEEE formula relies on us knowing the phase angle ϕ of the load

We have deduced this by just using the Volatge and Current measurements at the source and knowing the impedance of the cable

How else can we know ϕ of the load ?

If the load is a single piece of equipment, e.g. a single motor, we have a chance as that is usually rated in terms of Watts, VA and Power Factor in some combination.

In this case we can determine the Watts and vars as follows and hence the value of ϕ

20000.00

9882.42

Hence the value of ϕ is as used above

29.61

But if you have multiple pieces of plant and machinery as the load, you have no chance of knowing the value of load ϕ in order to use the IEEE 141 formula without full complex impedance analysis of all the loads in parallel !

So in the spirit of the IEEE diagram, what is the full formula to know the exact voltage at the source for a given voltage at the load?

{Vs – Vload} = ( I x Zcable )

Zcable = ( Rcable + j Xcable )

Since Iload and Zcable are complex quantities we need to expand that as

{Vs – Vload} = { [ (Iload cosϕ) + j (Iload x sinϕ) ] x [ Rcable + j Xcable ] }

={ [ (Iload cosϕ) x Rcable ]    +    [ (Iload cosϕ) x j Xcable ]    +    [ j (Iload x sinϕ) ] x Rcable ]    +    [ j (Iload x sinϕ)] x j Xcable ] }

={ [ Iload x cosϕ x Rcable ] – [ Iload x sinϕ x Xcable ] }    +    j { [ Iload x cosϕ x Xcable ]+[ Iload x sinϕ x Rcable ] }

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