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Rabu, 11 April 2012

Voltage Drop calculation


Generators

Generators supplying non-linear loads must be derated due to the additional losses caused by harmonic currents.
The level of derating is approximately 10% for a generator where the overall load is made up of 30% of non-linear loads. It is therefore necessary to oversize the generator.

Uninterruptible power systems (UPS)

The current drawn by computer systems has a very high crest factor. A UPS sized taking into account exclusively the RMS current may not be capable of supplying the necessary peak current and may be overloaded.

Transformers

  • The curve presented below (see Fig. M9) shows the typical derating required for a transformer supplying electronic loads


FigM09.jpg














Fig. M9: Derating required for a transformer supplying electronic loads

Example
  • If the transformer supplies an overall load comprising 40% of electronic loads, it must be derated by 40%.
  • Standard UTE C15-112 provides a derating factor for transformers as a function of the harmonic currents.



Typical values:
  • Current with a rectangular waveform (1/h spectrum (1)): k = 0.86
  • Frequency-converter current (THD ≈ 50%): k = 0.80
(1) In fact, the current waveform is similar to a rectangular waveform. This is the case for all current rectifiers (three-phase rectifiers, induction furnaces).

Asynchronous machines

Standard IEC 60892 defines a weighted harmonic factor (Harmonic voltage factor) for which the equation and maximum value are provided below.
Example
A supply voltage has a fundamental voltage U1 and harmonic voltages u3 = 2% of U1, u5 = 3%, u7 = 1%. The THDu is 3.7% and the MVF is 0.018. The MVF value is very close to the maximum value above which the machine must be derated. Practically speaking, for supply to the machine, a THDu of 10% must not be exceeded.

Capacitors

According to IEC 60831-1 standard, the rms current flowing in the capacitors must not exceed 1.3 times the rated current.
Using the example mentioned above, the fundamental voltage U1, harmonic voltages u5 = 8% (of U1), u7 = 5%, u11 = 3%, u13 = 1%, i.e. total harmonic
distortion THDu equal to 10%, the result is  , at the rated voltage. For a voltage equal to 1.1 times the rated voltage, the current limit  is reached and it is necessary to resize the capacitors.

Neutral conductors

Consider a system made up of a balanced three-phase source and three identical single-phase loads connected between the phases and the neutral (see Fig. M10).


FigM10.jpg









Fig. M10: Flow of currents in the various conductors connected to a three-phase source


Figure M11 shows an example of the currents flowing in the three phases and the resulting current in the neutral conductor.
In this example, the current in the neutral conductor has an rms value that is higher than the rms value of the current in a phase by a factor equal to the square root of 3. The neutral conductor must therefore be sized accordingly.



FigM11.jpg




























Fig. M11: Example of the currents flowing in the various conductors connected to a three-phase load (In = Ir + Is + It)

Figure G27 below gives formulae commonly used to calculate voltage drop in a given circuit per kilometre of length.
If:
  • IB: The full load current in amps
  • L: Length of the cable in kilometres
  • R: Resistance of the cable conductor in Ω/km
 for copper

 for aluminium
Note: R is negligible above a c.s.a. of 500 mm2
  • X: inductive reactance of a conductor in Ω/km
Note: X is negligible for conductors of c.s.a. less than 50 mm2. 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
- Motor power:
   - At start-up: cosφ = 0.35
   - In normal service: cosφ = 0.8
  • Un: phase-to-phase voltage
  • Vn: phase-to-neutral voltage
For prefabricated pre-wired ducts and bustrunking, resistance and inductive reactance values are given by the manufacturer.


CircuitVoltage drop (ΔU)
in voltsin %
Single phase: phase/phase
Single phase: phase/neutral
Balanced 3-phase: 3 phases (with or without neutral)











Fig. G27: Voltage-drop formulae

Simplified table

Calculations may be avoided by using Figure G28, 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 cosφclose to 0.8, or lighting with a cosφclose to 1.
  • Type of cable; single-phase or 3-phase
Voltage drop in a cable is then given by:
K x IB x L
K is given by the table,
IB is the full-load current in amps,
L is the length of cable in km.
The column motor power “cosφ = 0.35” of Figure G28 may be used to compute the voltage drop occurring during the start-up period of a motor (see example no. 1 after the Figure G28).


c.s.a. in mm2Single-phase circuitBalanced three-phase circuit
Motor powerLightingMotor powerLighting
Normal serviceStart-upNormal serviceStart-up
CuAIcos φ = 0.8cos φ = 0.35cos φ = 1cos φ = 0.8cos φ = 0.35cos φ = 1
1.52410.630209.425
2.514.46.418125.715
49.14.111.283.69.5
6106.12.97.55.32.56.2
10163.71.74.53.21.53.6
16252.361.152.82.0512.4
25351.50.751.81.30.651.5
35501.150.61.2910.521.1
50700.860.470.950.750.410.77
701200.640.370.640.560.320.55
951500.480.300.470.420.260.4
1201850.390.260.370.340.230.31
1502400.330.240.300.290.210.27
1853000.290.220.240.250.190.2
2404000.240.20.190.210.170.16
3005000.210.190.150.180.160.13
Fig. G28: Phase-to-phase voltage drop ΔU for a circuit, in volts per ampere per km

Examples

Example 1 (see Fig. G29)
A three-phase 35 mm2 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: 
Table G28 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. 
This value is less than that authorized (8%) and is satisfactory.
  • Voltage drop during motor start-up:
Δ Ucable = 0.52 x 500 x 0.05 = 13 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. 
ΔU distribution board = 14 V
ΔU for the motor cable = 13 V
ΔU total = 13 + 14 = 27 V i.e. 

a value which is satisfactory during motor starting.



FigG29.jpg



























Fig. G29: Example 1

Example 2 (see Fig. G30)
A 3-phase 4-wire copper line of 70 mm2 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 mm2 c.s.a. copper 20 m long, and each passing 20 A.
It is assumed that the currents in the 70 mm2 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:  
Figure G28 shows 0.55 V/A/km
ΔU line = 0.55 x 150 x 0.05 = 4.125 V phase-to-phase
which gives: phase to neutral.
  • Voltage drop in any one of the lighting single-phase circuits:
ΔU for a single-phase circuit = 18 x 20 x 0.02 = 7.2 V
The total voltage drop is therefore
7.2 + 2.38 = 9.6 V

This value is satisfactory, being less than the maximum permitted voltage drop of 6%.


FigG30.jpg

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