Voltage Drop Calculation

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Voltage drop

This article is the first of a series in the selection of electrical cables.

Cable selection is an important part of electrical installation design. Selecting the correct size of cable could lower initial cost, lower operating cost, better voltage regulation notwithstanding the safety factor.

Cable parameters are provided in values per unit length, usually ohms per kilometer(Ω/km) for IEC or miles(Ω/mi) for North American Standards.

The following table illustrate the typical parameters for industrial cables



Table 1. 600/1000 V. Cu/EPR/CSP/GSWB run on open trays or enclosed in air. 1 to 6 x single, 3 or 4 cores
Nominal
conductor
area (mm2)		 Single cores
in trefoil		 3 cores Approximate
armouring
resistance
Resistance
at 90oC
(ohm/km)		 Reactance
at 50Hz		 Resistance
at 90oC
(ohm/km)		 Reactance
at 50Hz		 GSWB at 60oC
3–4 cores
(ohm/km)
1.5 15.6 0.185 15.6 0.118 46.2
2.5 9.64 0.173 9.64 0.111 51.3
4 5.99 0.163 5.99 0.108 60.3
6 3.97 0.153 3.97 0.105 30.5
10 2.35 0.148 2.35 0.0983 36.7
16 1.48 0.134 1.48 0.0933 23.1
25 0.936 0.125 0.936 0.0892 28.1
35 0.674 0.121 0.674 0.0867 10.43
50 0.499 0.118 0.499 0.0858 11.81
70 0.344 0.112 0.344 0.0850 13.61
95 0.271 0.108 0.271 0.0825 10.87
120 0.214 0.106 0.214 0.0808 11.92
150 0.175 0.105 0.175 0.0808 7.38
185 0.140 0.105 0.140 0.0808 8.15
240 0.108 0.103 0.108 0.0800 8.94
300 0.087 0.101 0.087 0.0800 10.10
400 0.069 0.0992 0.069 0.0795 10.00

Derivation of Voltage Drop Formula

A short cable voltage drop in AC systems is provided by the formula

Vd = IR cosØ + IX sinØ

where:
Vd = Voltage drop per phase, volts
Ø = Load power factor angle
IR cosØ = Voltage drop component on the cable resistance
IX sinØ = Voltage drop component on the cable reactance

You might be wondering where this formula came from. Actually, this formula is just an approximation. We shall going through the process of deriving this formula to better understand it.

Considering the following vector diagram:

Voltage - Current Phasor Diagram
Figure 2 Voltage - Current Phasor Diagram

where:
Vs = Sending end voltage per phase
Vr = Receiving end voltage per phase
I = Load current
R = Cable resistance
X = Cable reactance
Ø = Load power factor angle
AB = IR cosØ
BE = CF = IR cosØ
BC = EF = IX sinØ
DF = IX cosØ

AC = AB + BC = AB + EF
AC = IR cosØ + IX sinØ

DC = DF - CF = DF - BE
DC = IX cosØ - IR sinØ

Vs = OD = √(OA + AB + BC)2 + (DF - BE)2

Unless the cable is very long, the imaginary axis component of the voltage is very small compared to the real axis component.

(OA + AB + BC) >> (DF - BE)

thus the sending end voltage will be

Vs = (OA + AB + BC)

Vs = Vr + IR cosØ + IX sinØ volts/phase

From the above formula, the voltage drop on a per phase basis will be

Vd = Vs - Vr

Vd = IR cosØ + IX sinØ volts/phase per unit length

That is how the voltage drop formula was derived.

Normally, voltage drop is expressed as a percentage of the sending end line-to-line voltage. The formula will be

voltage drop

where:
V = sending end line-to-line voltage

R = rl, where r is the unit resistance in ohms/km
X = xl, where x is the unit reactance in ohms/km
as published in cable data publications.

For Part 2, we shall be providing real world examples on voltage drop calculations.

References:

  • Handbook of Electrical Engineering For Practitioners in the Oil, Gas and Petrochemical Industry
    Alan L. Sheldrake
  • Electrical Engineer's Reference Book Sixteenth edition
    M. A. Laughton