What is a branch of a circuit

Methods for calculating the voltage drop with detailed examples

Voltage drop considerations

The first consideration for voltage drop is that under steady state conditions of normal load, the voltage across the utility equipment must be adequate.

Methods for calculating the voltage drop with detailed examples

Fine-print notes in the NEC recommend sizing leads and branching circuits so that the maximum voltage drop in both is no more than 3% with the total voltage drop for feeders and Branch circuits does not exceed 5% in order to ensure the efficiency of the operation .

In addition to the steady-state conditions, a voltage drop under transient conditions with sudden high-current short-term loads must be taken into account.

The most common loads of this type are motor inrush currents during starting. These loads cause a voltage drop in the system due to the voltage drop in conductors, transformers and generators under the high current. This voltage dip can have numerous adverse effects on equipment in the system, and equipment and conductors must be designed and dimensioned to minimize these problems.

In many cases it is starting engines with reduced tension required to reduce the inrush current.

  • Voltage drop formulas
    • Approximate method
    • Exact method # 1
    • Exact method # 2
  • Voltage drop tables
  • Calculations

Voltage drop formulas

Let's see two common methods of calculating voltage drop - approximate and accurate methods:

1. Approximate method

Voltage drop E VD = IR cos? + IX sin θ, with abbreviations being the same as "Exact Method" below.

2. Exact method # 1

When the transmission end voltage and load PF are known.

where from:

  • E. VD - Voltage drop, line to neutral, volts
  • E. s - Source voltage, line to neutral, volts
  • I. - Line (load) current, amps
  • R. - Circuit (branch, feeder) resistance, ohm
  • X - Circuit (branch, feeder) reactance, ohm
  • cosθ - Power factor of the load, decimal
  • sinθ - Reactive factor of the load, decimal

When the receiving end voltage, load current and power factor (PF) are known.

E. R. is the receiving end voltage.

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2. Exact method # 2

When receiving or transmitting, mVA and its power factor are known with a known transmit or receive voltage.

or

where from:

  • E. R. - Receipt of the mains voltage in kV
  • E. S. - Sending the line voltage in kV
  • MVA R. - Reception of three-phase MVA
  • MVA S. - Sending three-phase mVA
  • Z - Impedance between and receiving ends
  • γ - The angle of the impedance Z
  • R. - Receiving PF
  • S. - Send end PF, positive in case of follow-up

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Voltage drop tables

Tables for calculating voltage drop for copper and aluminum conductors, in either magnetic (steel) or non-magnetic (aluminum or non-metallic) conduction, are shown below. These tables give a voltage drop per ampere per 100 feet (30 m) of circuit length.

The circuit length from the starting point to the end point of the circuit is independent of the number of conductors.

The tables are based on the following conditions:

Condition # 1

Three or four individual conductors in a line, randomly lying. For three-core cables, the actual voltage drop is roughly the same for small wire sizes and high power factors. The actual voltage drop is with larger conductor sizes and lower power factors of 10 to 15% less .

Condition # 2

Voltage drops are phase-to-phase, for three-phase, three-wire or three-phase, four-wire 60 Hz circuits. For other circuits, multiply the voltage drop specified in the tables by the following correction factors:

Correction factors table:

Three-phase, four-wire, phase-to-neutral× 0, 577
Single-phase, two-wire× 1.155
Single-phase, three-wire, phase-to-phase× 1.155
Single-phase, three-wire, phase-to-neutral× 0, 577

Condition # 3

Voltage drops are for a Conductor temperature of 75 ° C. They can be used for conductor temperatures between 60 ° C and 90 ° C with reasonable accuracy (within ± 5%). However, correction factors in Table 1 can be applied if desired. The values ​​in the table are in Percent of the total voltage drop .

  • At a Conductor temperature from 60 ° C - SUBTRACT the percentage from Table 1.
  • If the conductor temperature is 90 ° C - add the percentage from table 1.

Table 1 - Temperature correction factors for voltage drop

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Calculations

To calculate the voltage drop:

  1. Multiply the current in amps by the length of the circuit in feet to get amp-feet. The circuit length is the distance from the point of origin to the load end of the circuit.
  2. Divide by 100.
  3. Multiply by the correct voltage drop value in tables. The result is a voltage drop.

example 1

A 460V, 100HP motor running at 80% PF draws 124A full load current. It is fed by three 2/0 copper conductors in steel pipes. The feeder length is 150 feet (46 m) .

What is the Voltage drop in the feeder? How tall is he percentage voltage drop?

  • 124 A × 150 feet (46 m ) = 18,600 A-feet
  • Divided by 100 = 186
  • Table: 2/0 copper, magnetic line,
    80% PF = 0, 0187
    186 x 0.0187 = 3, 48v drop
    3, 48 / 460 × 100 = 0.76% drop

Conclusion: 0.76% voltage drop is very acceptable. (See NEC Article 215 which indicates that a voltage drop of 3% or less on an infeed is acceptable.)

To choose the minimum conductor size:

  1. Determine the maximum desired voltage drop in volts.
  2. Divide the voltage drop by 2 (amps x switch feet).
  3. Multiply by 100.
  4. The next lower voltage drop value can be found in the tables in the correct column for line type, line and power factor. Read the ladder size for this value.
  5. If this results in an oversized cable, check the lug sizes for molded case circuit breakers and fuse links. If the allowable size of the tab is exceeded, go to the next higher rating.

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Example # 2

A three-phase four-wire lighting feeder on a 208V circuit is 250 feet (76.2 m) long . The load is 175 A at 90% PF . It is desirable to use aluminum 7-wire to be used in an aluminum pipe.

Which leader is required to limit the voltage drop to 2% phase-to-phase?

  • VD = 2/100 × 208 = 4, 16V
  • 4, 16 / (175 × 250) = 0, 0000951
  • 0, 0000951 × 100 = 0, 00951
  • In the table, under aluminum conductors, non-magnetic conductors, 90% PF, the next lower value is 0.0091. Conductor required is 12 500 kcmil.
    (Size 4/0 THW would have a sufficient current carrying capacity, but the voltage drop would be too high. )

Table 2 - voltage drop volts per ampere per 100 feet (30 m); Three-phase, phase-to-phase

Table 2 - Voltage Drop - Volts per ampere per 100 feet (30 m); Three-phase, phase-to-phase

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Reference // Power distribution systems from EATON

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