Overcurrent Protection

Download Report

Transcript Overcurrent Protection

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
1
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 currentcarrying 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
2
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
3
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
4
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
5
Selection of Overload Protective
Device

design current Ib  nominal current or rated current In 
lowest CCC, Iz
Electrical Installation 2
6
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
7
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
8
Omission of Overload Protective
Device


Overload current is unlikely to flow
Refer to Fig. 6.5 for illustration
Electrical Installation 2
9
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
10
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
11
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
12
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
13
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
14
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
15
Protection for Minimum
Prospective Short Circuit, I


Basic equation to satisfy
– k2S2 > I2t
Where
– k - a constant associated with the type of conductor +
insulation
– S - Cross-sectional Area (CSA) of conductor
– I - minimum prospective fault current (fault occur at
remote end)
– t - disconnection time
– I2t - let-through energy
Electrical Installation 2
16
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
breaking 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
17
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.
Source
Installation side
Source voltage Up
Z
Z1
63A fuse

1.68 Ω / km
Zn
Load
Z2
Electrical Installation 2
18
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
19
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
20
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,
we 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
21
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
22
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
23
Omission of short-circuit
protective devices


Conductor between a transformer and its control panel
Refer to Fig. 6.18 for detailed illustration
Electrical Installation 2
24