Voltage Drop Calc, Different Conductors In The Same Circuit

[zimmerDN]

How do you proceed with the voltage drop calculation of a circuit that has various conductor sizes, material, lengths and 1 vs 2 conductors per phase?

Do you start with the known supplied voltage and do the calculation for each section separately, use the calculated voltage result from the previous calc for next section?

Here is a scenario; These numbers are made up, but is the method correct?

600V Supply, 50A Load
100' of 3C #4 AWG Aluminum, 600V * 0.5% = 597V
250' of 3x1C #6 AWG Copper, 597V * 0.7% = 592V
400' of 2x3C #2 AWG Aluminum 592V * 0.2% = 590V

Therefore the final voltage at the load is 590V

 

[waross]

The simple solution with a known current is to calculate the ZR voltage drop of each section and then add the various voltage drops together.
If the current drops when the voltage drops the solution is a little more difficult.
The seemingly simple problem:
100 Volts, 10 Amps, Load Impedance 5 Ohms may be estimated by considering the circuit as a number of series impedances and calculating the voltage drop across each circuit impedance.
BUT This assumes that the X:R ratio of each length of conductor an the X:R ratio of the load are all equal.
A rigorous solution requires the resolution of each length of the conductor and of the load into their respective resistance and reactive components.
But in the real world, for compliance with the voltage drop requirements of the code in North America consider the following method.
1> Determine the impedance of the conductor sections from the voltage drop tables in the code book. The tables use an assumed X:R ratio for the cable to take into account the inductive reactance of the cables.
2> Use the load current calculated with the full applied voltage (600V).
3> Calculate the voltage drop of each section of conductors.
4> Add the voltage drops of the individual sections for the total voltage drop.
5> Use the total voltage drop and the full applied voltage (600V) to calculate the PU voltage drop.
This method, while not exact is conservative in the majority of instances and as such will satisfy the majority of AHJs.
Sources of error:
1> Most loads are voltage dependent and the current changes as the applied voltage changes.
For loads where the current drops with a drop in applied voltage the actual voltage drop will be less than the calculated voltage drop.
Motors are an exception but with conductors sized for the code maximum 3% drop, the change will be in the order of 3% of 3%.
2> The voltage drop across an impedance is a directed quantity. In the instance that the phase angles of each impedance (cable sections and the load) are the same the total voltage drop will be as calculated.
If the phase angles are not the same, then the total voltage drop will be less than the simple arithmetic sum of the individual voltage drops.
3> The voltage drops given in the code tables are representative and are more accurate than simple resistance values but the inductive reactance of a cable depends on the arrangement and the spacing of the cables and may not be as stated in the tables.
4> The voltage drops given in the tables are accurate for only one load power factor and a load power factor other than the assumption on which the tables are based will cause errors.
BUT Most of the errors result from differing phase angles so that the end result will be conservative in almost all instances. As the code states the maximum voltage drop allowable, an actual voltage drop that is slightly less than the code maximum is allowable.
Ps/ Your original post does not make it clear if the PU voltage drops are based on the full voltage or on the voltage at the start of each section of cable. For this reason you may wish to consider calculating the individual drops as voltage and then expressing the total drop as a PU of the full voltage (600V).

Bill
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