Cable Voltage Drop and IEEE 141
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One of the frequent electrical issues we have to deal with from timetotime 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 311
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 311, 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:
Item  Magnitude component  Angle Component  Resistive Component  Reactive Component 

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  
PF direction  Inductive / Lagging  
The total impedance seen at the source is dictated by Ohm's Law: Z = V/I  
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 endtoend (i.e. as would be measured by putting a voltmeter connected across the two ends of the cable) is  
Vdrop down the cable endtoend:  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 phasetoneutral voltage at the source with the magnitude of the phasetoneutral 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 phaseneutral 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 endtoend of the cable as calculated above!!  
 Vd  
 10.86  
If we now refer to IEEE 141, the approximation formula associated with Fig 311 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 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  8.51  Measuring the phasetoneutral voltages at each end  
Vcable  10.86  Measuring the voltage endtoend 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  
VAload  20000.00  
PFload  0.86939  
Watt_load  17387.86  
var_load  9882.42  
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 ] } 
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