Calculation of voltage drop in steady load conditions
Use of formulae
Figure G29 below gives formulae commonly used to calculate voltage drop in a given circuit per kilometre of length (copper cable with XLPE insulation).
where:
I_{B} = The full load current in amps
L = Length of the cable in kilometres
R = Resistance of the cable conductor in Ω/km
- [math]\displaystyle{ R=\frac{23.7\ \Omega\ mm^2/km}{S\left ( c.s.a\ in\ mm^2 \right )} }[/math] for copper^{[1]}
- [math]\displaystyle{ R=\frac{37.6\ \Omega\ mm^2/km}{S\left ( c.s.a\ in\ mm^2 \right )} }[/math] for aluminium^{[1]}
- Note: R is negligible above a c.s.a. of 500 mm^{2}
X = inductive reactance of a conductor in Ω/km
- Note: X is negligible for conductors of c.s.a. less than 50 mm^{2}. In the absence of any other information, take X as being equal to 0.08 Ω/km.
φ = phase angle between voltage and current in the circuit considered, generally:
- Incandescent lighting: cosφ = 1
- Led lighting: cosφ > 0.9
- Fluorescent with electronic ballast: cosφ > 0.9
- Motor power:
- At start-up: cosφ = 0.35
- In normal service: cosφ = 0.8
U_{n} = phase-to-phase voltage
V_{n} = phase-to-neutral voltage
For prefabricated pre-wired ducts and busways (busbar trunking systems), resistance and inductive reactance values are given by the manufacturer.
Circuit | Voltage drop (ΔU) | |
---|---|---|
in volts | in % | |
Phase/phase | [math]\displaystyle{ \Delta U=2I_B\left ( R \cos\phi + X \sin\phi \right )L }[/math] | [math]\displaystyle{ \frac {100 \Delta U}{Un} }[/math] |
Phase/neutral | [math]\displaystyle{ \Delta U=2I_B\left ( R \cos\phi + X \sin\phi \right )L }[/math] | [math]\displaystyle{ \frac {100 \Delta U}{Vn} }[/math] |
Balanced 3-phase: 3 phases (with or without neutral) | [math]\displaystyle{ \Delta U=\sqrt 3I_B\left ( R \cos\phi + X \sin\phi \right )L }[/math] | [math]\displaystyle{ \frac {100 \Delta U}{Un} }[/math] |
Simplified table
Calculations may be avoided by using Figure G30, which gives, with an adequate approximation, the phase-to-phase voltage drop per km of cable per ampere, in terms of:
- Kinds of circuit use: motor circuits with cosφ close to 0.8, or lighting with a cosφ close to 1.
- Type of circuit; single-phase or 3-phase
Voltage drop in a cable is then given by: K x IB x L
K = given by the table,
IB = the full-load current in amps,
L = the length of cable in km.
The column motor power “cosφ = 0.35” of Figure G30 may be used to compute the voltage drop occurring during the start-up period of a motor (see example no. 1 after the Figure G30).
Copper cables | Aluminium cables | ||||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|---|
c.s.a. in mm^{2} |
Single-phase circuit | Balanced three-phase circuit | c.s.a. in mm^{2} |
Single-phase circuit | Balanced three-phase circuit | ||||||||
Motor power | Lighting | Motor power | Lighting | Motor power | Lighting | Motor power | Lighting | ||||||
Normal service |
Start-
up |
Normal service |
Start-up | Normal service |
Start-up | Normal service |
Start-up | ||||||
cos ϕ
= 0.8 |
cos ϕ
= 0.35 |
cos ϕ
= 1 |
cos ϕ
= 0.8 |
cos ϕ
= 0.35 |
cos ϕ
= 1 |
cos ϕ
= 0.8 |
cos ϕ
= 0.35 |
cos ϕ
= 1 |
cos ϕ
= 0.8 |
cos ϕ
= 0.35 |
cos ϕ
= 1 | ||
1.5 | 25.4 | 11.2 | 32 | 22 | 9.7 | 27 | |||||||
2.5 | 15.3 | 6.8 | 19 | 13.2 | 5.9 | 16 | |||||||
4 | 9.6 | 4.3 | 11.9 | 8.3 | 3.7 | 10.3 | 6 | 10.1 | 4.5 | 12.5 | 8.8 | 3.9 | 10.9 |
6 | 6.4 | 2.9 | 7.9 | 5.6 | 2.5 | 6.8 | 10 | 6.1 | 2.8 | 7.5 | 5.3 | 2.4 | 6.5 |
10 | 3.9 | 1.8 | 4.7 | 3.4 | 1.6 | 4.1 | 16 | 3.9 | 1.8 | 4.7 | 3.3 | 1.6 | 4.1 |
16 | 2.5 | 1.2 | 3 | 2.1 | 1 | 2.6 | 25 | 2.50 | 1.2 | 3 | 2.2 | 1 | 2.6 |
25 | 1.6 | 0.81 | 1.9 | 1.4 | 0.70 | 1.6 | 35 | 1.8 | 0.90 | 2.1 | 1.6 | 0.78 | 1.9 |
35 | 1.18 | 0.62 | 1.35 | 1 | 0.54 | 1.2 | 50 | 1.4 | 0.70 | 1.6 | 1.18 | 0.61 | 1.37 |
50 | 0.89 | 0.50 | 1.00 | 0.77 | 0.43 | 0.86 | 70 | 0.96 | 0.53 | 1.07 | 0.83 | 0.46 | 0.93 |
70 | 0.64 | 0.39 | 0.68 | 0.55 | 0.34 | 0.59 | 120 | 0.60 | 0.37 | 0.63 | 0.52 | 0.32 | 0.54 |
95 | 0.50 | 0.32 | 0.50 | 0.43 | 0.28 | 0.43 | 150 | 0.50 | 0.33 | 0.50 | 0.43 | 0.28 | 0.43 |
120 | 0.41 | 0.29 | 0.40 | 0.36 | 0.25 | 0.34 | 185 | 0.42 | 0.29 | 0.41 | 0.36 | 0.25 | 0.35 |
150 | 0.35 | 0.26 | 0.32 | 0.30 | 0.23 | 0.27 | 240 | 0.35 | 0.26 | 0.31 | 0.30 | 0.22 | 0.27 |
185 | 0.30 | 0.24 | 0.26 | 0.26 | 0.21 | 0.22 | 300 | 0.30 | 0.24 | 0.25 | 0.26 | 0.21 | 0.22 |
240 | 0.25 | 0.22 | 0.20 | 0.22 | 0.19 | 0.17 | 400 | 0.25 | 0.22 | 0.19 | 0.21 | 0.19 | 0.16 |
300 | 0.22 | 0.21 | 0.16 | 0.19 | 0.18 | 0.14 | 500 | 0.22 | 0.20 | 0.15 | 0.19 | 0.18 | 0.13 |
Examples
Example 1
(see Fig. G31)
A three-phase 35 mm^{2} copper cable 50 metres long supplies a 400 V motor taking:
- 100 A at a cos φ = 0.8 on normal permanent load
- 500 A (5 In) at a cos φ = 0.35 during start-up
The voltage drop at the origin of the motor cable in normal circumstances (i.e. with the distribution board of Figure G29 distributing a total of 1,000 A) is 10 V phase-to-phase.
What is the voltage drop at the motor terminals:
- In normal service?
- During start-up?
Solution:
- Voltage drop in normal service conditions:
[math]\displaystyle{ \Delta U\%=100\frac {\Delta U}{Un} }[/math]
Table Figure G30 shows 1 V/A/km so that:
ΔU for the cable = 1 x 100 x 0.05 = 5 V
ΔU total = 10 + 5 = 15 V = i.e. [math]\displaystyle{ \frac {15}{400} \times 100 = 3.75\% }[/math]
This value is less than that authorized (8%) and is satisfactory.
- Voltage drop during motor start-up:
ΔUcable = 0.54 x 500 x 0.05 = 13.5 V
Owing to the additional current taken by the motor when starting, the voltage drop at the distribution board will exceed 10 Volts.
Supposing that the infeed to the distribution board during motor starting is 900 + 500 = 1,400 A then the voltage drop at the distribution board will increase approximately pro rata, i.e.
[math]\displaystyle{ \frac{ 10 \times 1,400}{1,000} = 14V }[/math]
ΔU distribution board = 14 V
ΔU for the motor cable = 13 V
ΔU total = 13.5 + 14 = 27.5 V i.e.
[math]\displaystyle{ \frac{27.5}{400} \times 100 = 6.9\% }[/math]
a value which is satisfactory during motor starting.
Example 2
(see Fig. G32)
A 3-phase 4-wire copper line of 70 mm^{2} c.s.a. and a length of 50 m passes a current of 150 A. The line supplies, among other loads, 3 single-phase lighting circuits, each of 2.5 mm^{2} c.s.a. copper 20 m long, and each passing 20 A.
It is assumed that the currents in the 70 mm^{2} line are balanced and that the three lighting circuits are all connected to it at the same point.
What is the voltage drop at the end of the lighting circuits?
Solution:
- Voltage drop in the 4-wire line:
[math]\displaystyle{ \Delta U \%=100\frac{\Delta U}{Un} }[/math]
Figure G30 shows 0.59 V/A/km
ΔU line = 0.59 x 150 x 0.05 = 4.4 V phase-to-phase
which gives:
[math]\displaystyle{ \frac{4.4}{\sqrt 3}= 2.54 V }[/math] phase to neutral.
- Voltage drop in any one of the lighting single-phase circuits:
ΔU for a single-phase circuit = 19 x 20 x 0.02 = 7.6 V
The total voltage drop is therefore
7.6 + 2.54 = 10.1 V
[math]\displaystyle{ \frac{10.1V}{230V} \times 100 =4.4\% }[/math]
This value is satisfactory, being less than the maximum permitted voltage drop of 6%.