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Earth fault loop impedance
Electrical Engineering (EE 123)
Malayan Colleges Mindanao
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Earth fault loop impedance
When an earth fault occurs at the far end of a cable it is possible that the armouring, cable gland and the frame of the consumer equipment can be raised to a dangerous potential with respect to electric shock exposure to human operators. This subject has been given considerable attention over the last 20 years, and is well documented in for example IEC60364. The international documentation concentrates on low voltage fixed and portable equipment protected by fuses and miniature circuit breakers. See also Chapter 13. BS7430 (1998), sub-section 3, defines the earth fault loop impedance Zloop in relation to the various types of earthing systems, as follows. For TN systems: Zloop = Znez + Zsec + Zc + Za + Zbond + Zmr (9) Table 9. Cable and fuse currents and time data
Cable current (kA) Time (seconds) 400 A fuse operating current (kA) 429 0 11. 191 0 9. 135 0 7. 60 0 5. 42 1 4. 19 5 3.
Figure 9 Earth loop impedance diagram.
For TT and IT systems:
Zloop = Znez + Zsec + Zc + Za + Zbond + Zer (9)
These components are shown in Figure 9.
Where Znez is the impedance of the neutral earthing device at the source winding. Zsec is the positive sequence impedance of the source. For a transformer this includes both windings and the upstream impedance. For a generator this will be the sub-transient impedance. Zc is the impedance of one phase conductor of the particular cable. Za is the impedance of its armouring of the particular cable. This impedance can be taken as purely resistance (Ra). Zbond is the impedance of the earthing terminals and bonding conductors at the sending end of the cable. Zmr is the metallic return path impedance of a TN system. This impedance can be taken as purely resistance, but will usually be low enough to ignore. For offshore installations the multiple series and parallel branches of steel work in three dimensions will render such an impedance as almost zero. Zer is the impedance of the earth return path of the ground for a TT or IT system. It will be approximately the total impedance of the ground rod or grid at the source and the ground rod or grid at the consumer. The impedance will be almost purely resistance, being typically a fraction of an ohm for damp soil conditions to several ohms for dry sandy and rocky soils. See also the international standard IEEE 80, 1986, section 12.
Vph is the nominal phase-to-neutral voltage of the source. If is the fault current at the far end of the feeder cable where the point of fault occurs.
equipment is not considered in the following discussions and calculations. See also BS7430. c) Portable equipment: Portable equipment is not necessarily hand-held equipment, it may be too heavy to carry or lift by one person. d) Hand-held equipment: Hand-held equipments are usually light-weight tools such as drilling machines, sanding machines etc., that are held in one or both hands.
e) Disconnection time: The standard recommends two nominal disconnection times 0 and 5 seconds. The time of 0 seconds is based on a nominal phase-to-neutral voltage of approximately 240 Vac, where as the time of 5 seconds is invariant of voltage.
Where the distribution circuit feeds a stationary item of equipment, not socket outlets and not portable equipment, the disconnection time may be taken as 5 seconds. This applies to motors. The nominal time of 0 seconds is intended for circuits supplying socket outlets, regularly moved portable equipment and Class 1 hand-held equipment. For voltages (Vph) different from 240 Vac, the disconnection time (tdis) of 0 seconds becomes approximately related as,
with a lower limit of 0 second.
tdis
1 log
600 seconds Vph
The maximum value of Zloop can be determined from the following information,
- The network phase-to-neutral voltage Vph. The operating current that causes the supply protective device to disconnect the consumer in the specified time tdis. This can be found from the manufacturer’s data.
9.4.3.2 Worked example A 37 kW 415 V induction motor is protected by fuses at the motor control centre. The route length of the motor feeder cable is 200 metres. The supply frequency is 50 Hz. The MCC is fed by one 250 kVA, 4% impedance, transformer. Assume an X/R ratio of the transformer of 10. The motor running efficiency at full-load is 92% and its power factor is 0. The starting to running current ratio is 7, and the starting power factor is 0. The cable is routed in air that has an ambient temperature of 40◦C. The conductor maximum temperature is 90◦C. The insulation material is EPR and the armouring is galvanised steel wire braid. Assume the cable data in Tables 9 and 9. for 3-core cables is applicable. The motor fuse data are shown in Figure 8 for 100 A, 125 A and 160 A fuses. The permissible volt-drops in the cable for running and starting are 3% and 15% respectively. Find the most appropriate cable and fuses for the motor. Determine whether or not an earth leakage current relay should be used at the motor control centre. Assume a TN earthing arrangement. Replace the steel wire braid armour with round steel wires (GSWA) and reduce the metallic return path impedance Zmr to 0 ohm, and compare the effect on the hazardous shock voltage.
= √3 × 347 = 0 ohms/phase
Therefore the 4% leakage impedance
Zsec = |Rsec + j Xsec|= 0 × 0 = 0 ohms/phase or 0 + j0. ohms/phase b) Find the motor impedances for the starting and running conditions: Full-load current Iflmr of the motor Sm = √3 × V Efficiency × Power factor 37000 = √3 × 415 × 0 × 0 = 65 amps The 100% impedance Zmrn of the motor
V
= √3 I
flmr
415 = √3 × 65 = 3 ohms/phase at a power factor of 0.
The starting impedance Zms of the motor Zmm
= Starting to running current ratio 3. = 7 = 0 ohms/phase at a power factor of 0. c) Find the conductor and armouring impedances for various cables that may be suitable. From Tables 9 and 9 the impedance data for the circuit temperature and power frequency conditions are shown below. d) Since the motor is fixed equipment of the Class 1 type, the disconnection time tdis is 5 seconds. The fuse operating currents at 5 seconds are shown in Table 9. e) Find the I-squared-t parameters for the cable conductors with respect to the fuse operating current (Is) at 5 seconds. From (9)
Amin =
Is√
K
= Is × 0.
Table 9. Cable data for the worked example Nominal Current 3-core conductor GSWB armouring
conductor area (mm2)
rating at 40◦C (amps)
Table 9. Cable data for the worked example
Fuse rating Operating current Minimum conductor Nearest practical (amps) at 5 sec (amps) CSA Amin (mm2) CSA above Amin (mm2) 100 500 7 10 125 650 10 16 160 850 13 16 200 1100 17 25
Where K = 143 for EPR insulation and Amin is the smallest conductor cross-sectional area. The resulting Amin for the four sizes of fuses are given in Table 9: Hence all the cables 16 mm2 to 70 mm2 in c) that suit the motor full-load current will be adequately protected by the fuses in the range given. f) Calculate the volt-drops for running and starting the motor. From (9) for ‘running’ conditions, cos Ø = 0 and sin Ø = 0 Assume the cable conductor area is 16 mm2, Therefore,
R = 0 ohms and X = 0 ohms
∆Vrun
√365(0 × 0 + 0 × 0)
415 = 7%.
From (9) for ‘starting’ conditions,
cos Ø = 0 and sin Ø = 0.
∆Vstart
√3460(0 × 0 + 0 × 0)
415 = 30%.
Table 9 shows the volt-drop results for all the available cables. g) Check if the earth loop impedance of the motor circuit is greater than that allowed by the fuses. The motor circuit is shown in Figure 9. Where 415 Vph = Zloopc Zsec = 0 + j 0 ohms/phase Table 9. Volt-drop in the motor feeder cable for the worked example
Nominal conductor Running volt-drop Starting volt-drop Accept or reject area (mm2) (%) (%) 16 7 30 Reject 25 4 20 Reject 35 3 15 Reject 50 2 12 Accept
Fuse rating should be below the rating of the cable since its primary purpose is to protect the cable. Hence the largest fuse should be 125 A for a 50 mm2 cable. (If a larger fuse is needed the cable size would need to be increased.) h) Calculate the electric shock voltage From Figure 9 the shock voltage Vshock is,
(Za + Zmr)Vph shock Zsec + Z + Z + Zmr
For the 16 mm2 cable
V
= 4 + 1 415 = 227 volts
i) Replace the braided armour with round steel wires. Assume the resistances of the armour wires to be 0, 0, 0, 0 and 0 ohms for the 200 m route length. Repeat the calculations of g).
For a 16 mm2 cable,
Za + Zc = 0 + j 0 + 0 = 1 + j 0. Zloopc = Za + Zc + Zsec + 0.
= 1 + j 0 + 0 + j 0 + 0. = 1 + j 0.
Magnitude of Zloopc = 1 ohms The resulting Zloopc calculated from the circuit for each cable is given below, For the 16 mm2 the shock voltage is, V = 0 + 0 415 = 175 volts
If a 50 mm2 cable and a fuse rating of 125 amps are chosen as recommended in g) then the circuit earth loop impedance is still too high by a ratio of about 1:1. Hence an earth leakage circuit breaker should still be used for this motor circuit. The hazardous shock voltage is still too high. j) Now replace the fuses with moulded case circuit breakers and show whether or not the situation is improved. From Figure 7 it can be seen that a typical MCCB for motor application operates in its inverse region for times equal to 5 seconds. Repeating d) but for MCCBs gives the following limits for Zloopf,
Table 9. Earth loop impedance results for the worked example with steel wire armouring
Nominal conductor area (mm2)
Za Zc Zloopc magnitude (ohms)
Vshock
16 1 + j 0 1 175. 25 0 + j 0 0 181. 35 0 + j 0 0 191.
Earth fault loop impedance
Course: Electrical Engineering (EE 123)
University: Malayan Colleges Mindanao
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