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Electrical Installation 2 1
Overcurrent Protection(Note: All the mentioned tables in this course refer to, unless otherwise specified, Low
Voltage Electrical Installation Handbook, by Johnny C.F. Wong, Edition 2004)
Chapter 6
Electrical Installation 2 2
General
Purpose– Safety of Personnel (Shock) and Property (Fire Hazards)
– Maintain reliable life of equipment and systems Overcurrent
– a current exceeding the rated value of a circuit or the current-carrying capacity of a conductor
– Overload
– Fault• Short-circuit fault
• Earth fault
– This part, we are concerned with the short-circuit fault only.
Electrical Installation 2 3
Devices for Overcurrent Protection
Examples are:
– Fuses (HBC/HRC)
– Miniature circuit breakers (MCBs)
– Combined MCB and RCD (RCBOs)
– Moulded case circuit breakers (MCCBs)
– Air circuit breaker + IDMTL relay
Electrical Installation 2 4
Devices for Overcurrent Protection
Protection for the NEUTRAL conductor is NOT required for TT and TN systems– 100% Neutral should be used
– Protection already provided by the live conductor protective device
– Neutral link (not protective device)
– If the neutral breaks, the live supply must break too
– LOSS OF NEUTRAL must be avoided to eliminate the risk of raising the potential of the load star point to dangerous level
Electrical Installation 2 5
Protection against Overload
Main purpose is to avoid sustained temperature that causes deterioration of insulation
e.g. only a short duration of overload current is allowed to flow in a motor circuit - the starting duration should be short. Otherwise larger cables shall be installed
Electrical Installation 2 6
Selection of Overload Protective Device
design current Ib nominal current or rated current In lowest CCC, Iz
Electrical Installation 2 7
Position of Overload Protective Device
At the point where there is a reduction of Iz (CCC) such as– CSA of conductor is reduced
– Worsening of environmental condition
– Change of cable type or installation method Overload protective device and fault current protective
device may be the same device and may be 2 different devices
Electrical Installation 2 8
Overload Protection of Conductors in Parallel
The Iz in this case is the sum of Iz of the individual cables provided they are in accordance with the conditions for parallel running cables.
Standard ring final circuits are not in this context.
Electrical Installation 2 9
Omission of Overload Protective Device
Overload current is unlikely to flow Refer to Fig. 6.5 for illustration
Electrical Installation 2 10
Omission of Overload Protective Device
Unexpected loss of supply is more dangerous than overloading of circuit
Refer to Fig. 6.6 for illustration
Electrical Installation 2 11
Omission of Overload Protective Device
CT secondary circuit should not be broken. If this is the case, dangerous high voltage will appear at the CT secondary side
Refer to Fig. 6.7 for illustration
Electrical Installation 2 12
Omission of Overload Protective Device
Protection is afforded by electricity supplier’s protective device (not normally accepted by power companies in Hong Kong)
Refer to Fig. 6.8 for illustration
Electrical Installation 2 13
Protection against Fault Current
Cause - Insulation failure, faulted switching operation and invariably associated with arcs
Effect - Thermal and mechanical stress produced in conductors, associated support and plant components
Fault current protection is to prevent this
Electrical Installation 2 14
Protection for Maximum prospective fault current, Isc
Maximum prospective fault current, Isc
– 3-phase : calculation based on symmetrical fault impedance,
Isc = Up / Z
where Up = phase voltage Z = phase conductor impedance at supply source– 1-phase : calculation based on line-neutral impedance at 20oC,
Isc = Up / (Z + Zn)
where Zn = neutral conductor impedance at supply source– The above should base on fault appeared just after the protective device
– Breaking capacity of fault current protective devices should exceed the max. prospective fault current, Isc
Electrical Installation 2 15
Minimum Prospective Fault Current, I
Minimum prospective fault current, I
– Calculation bases on total phase-neutral impedance values, up to the remote end
I = Up / (Z + Zn+ Z1 + Z2)
where Z1 = phase conductor impedance at consumer side
Z2 = neutral conductor impedance at consumer side
– Significant in determining fault disconnection time, t
Electrical Installation 2 16
Protection for Minimum Prospective Short Circuit, I
Basic equation to satisfy– k2S2 > I2t
Where– k - a constant associated with the type of conductor + in
sulation– S - Cross-sectional Area (CSA) of conductor– I - minimum prospective fault current (fault occur at re
mote end)– t - disconnection time– I2t - let-through energy
Electrical Installation 2 17
Guidelines in fault current protection
Max. prospective 3-ph symmetrical short-circuit at the l.v. source of supply provided by the supply company is 40kA.– All fuses and MCCBs at source of energy must have breakin
g capacity > 40kA
– Fault current protective devices with smaller breaking capacities are generally acceptable if they are backed up by fuses to BS88-2.1 or BS88-6 (Backup protection will be discussed later in Chapter 10)
The further away from the source of supply, the smaller the prospective short circuit current.
Electrical Installation 2 18
Fault Current Protection in General
Example: The following single phase circuit is protected by 63A BS88 fuse, the prospective short circuit current at the fuse is known to be 3 kA. A connected load, with circuit distance 87m from the fuse, is to be supplied by using 16mm2 1/C PVC copper cable. Please check whether the fuse can provide short circuit protection for the cable.
ZSource voltage Up
Zn
Z1
Z2
Load
63A fuse
Source Installation side
1.68 Ω / km
Electrical Installation 2 19
Fault Current Protection in General
At fuse position, it is given that the 1-Ф prospective short circuit current is 3 kA,
i.e. Isc = Up / (Z + Zn) Z + Zn = 3000 / 220 = 0.073 Ω
The total impedance from the fuse to the remote load end, Z1 + Z2 = 2 x 87m x 1.68 Ω/km = 0.292 Ω
So, the minimum short circuit current at the load end, I = Up / (Z + Zn+ Z1 + Z2) = 220 / (0.073 + 0.292) = 603 A
Electrical Installation 2 20
Fault Current Protection in General
Whether k2S2 > I2t ??
From I-t characteristic of BS88 fuse, t = 0.18 s when I = 603 A
PVC copper cable is used k = 115
S = 16 mm2
k2S2 = 1152 x 162 = 3,385,600 A2S
I2t = 6032 x 0.18 = 65,450 A2S
k2S2 > I2t O.K
Electrical Installation 2 21
Fault Current Protected by Overload Protective Device
The protective device is assumed to be adequate if it– satisfies conditions for overload protective device. That is, w
e sizes cable and protective device by using the principle
Ib ≤ In ≤ Iz ; and
– Breaking capacity of protective device ≥ Maximum prospective fault current, Isc
This is the most common way to protect a circuit, since only ONE protective device is needed.
Electrical Installation 2 22
Position of fault current protective device
Normally placed at or before the point where a reduction in the conductor’s current-carrying capacity (Iz) occurs. Such change may be due to a change in:– cross-sectional area, method or installation, type of cable or
conductor, or in environmental conditions
Electrical Installation 2 23
Fault current protection of conductors in parallel
A single device may provide protection against fault current for conductors in parallel provided the parallel conductors are in accordance with Section 5.8
Electrical Installation 2 24
Omission of short-circuit protective devices
Conductor between a transformer and its control panel Refer to Fig. 6.18 for detailed illustration