Page Comparison

Voltage:

|Vs|

<Vs

1000.00

0.00

Current:

|Iload|

<Iload

20.172

-30.00

We can therefore deduce the source is supplying

VA

20172

PFsource

0.86603

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

 Zcable+load

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

Rload = Rtotal - Rcable

Xload = Xtotal - Xcable

Load Impedance: 

Rload

Xload

42.73

24.29

Which means that the total load impedance is

Load Impedance:

|Zload|

<Zload

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

 Vload

|Vload|

<Vload

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

 Angle of the load:

29.61

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

 Angle of the load:

29.61

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

Vest

8.49

So we have now have three different views of this system

  |Vest|

8.49

IEEE value

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?

Zcable =  (Rcable + j Xcable)

Iload = (Iload cosϕ) + j (Iload x sinϕ)

Vs = Vload + (Iload x Zcable)

{Vs – Vload} = (I x Zcable)

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]}

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 

 VAload

20000.00

var_load

Hence the value of ϕ is as used above

<load (ϕ)

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 + ( Iload x Zcable )

{Vs – Vload} = ( I x Zcable )

Zcable = ( Rcable + j Xcable )

Iload = ( Iload cosϕ ) + j ( Iload x sinϕ )

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 ] }