3.Protective Devices, Cable Selection, Electrical Supply, and Electrical Generation.pptx
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Unit 203 Electrical Installations Technology– Session
Exam date: 28/04/2025
Protective Devices, Cable Selection, Electrical Supply
Electricity Supply
What does it comprise of?
The electrical supply industry comprises of
Generation
Transmission
Distribution
Voltage between Transmission and Distribution
Transmission
Distribution
Electrical Supply – Components
Component parts of the electrical distribution network include:
Sub-Stations
Pylons
Power Stations
Cables
Insulators
Transformers
Protective Device
Protective devices are equipment applied to electric power systems to detect abnormal and intolerable conditions and to initiate appropriate corrective actions. These devices include lightning arresters, surge protectors, fuses, and relays with associated circuit breakers.
Protective Devices – Faults
When a fault occurs, it will cause either an overcurrent or an earth leakage.
An overcurrent is: A current exceeding the rated value. For conductors, the rated value is the current‑carrying capacity.
This can be either an overload current or a fault current.
Earth leakage is: Electric current appearing in a protective conductor, such as leakage current or electric current resulting from an insulation fault.
Please answer the following questions:
What is the job of an
MCB
RCB
SPD
RCBO
MCCB
Isolating Switch
Cartridge Fuse
Semi-Enclosed Fuse
When / where would we use one of these?
What are the advantages and disadvantages of these?
Challenge: What is the British standard for these items?
BS 3036 Semi-Enclosed Fuse
Available sizes are:
5A (white)
15A (blue)
20A (yellow)
30A (red)
45A (green)
BS 3036 Semi-Enclosed Fuse
Advantages
Simple to check if blown
Low cost to replace fuse element
No moving parts.
Disadvantages
Danger of being repaired with wrong size wire
Deteriorate with age
Circuit cannot be quickly restored
Cannot break large fault currents
Danger if replaced on faulty circuit (melting wire)
Fusing factor of around 1.8–2.0 means that they cannot be guaranteed to operate up to twice the rated current that is flowing. As a result, cables protected by them must have a larger current‑carrying capacity.
BS 88-3:2010 Cartridge Fuse
BS 88-3:2010 Cartridge Fuse
Advantages
Disadvantages
BS 88‑2:2010 fuses
(replacing BS 88‑2 and BS 88‑2.1)
BS 88‑2:2010 fuses
(replacing BS 88‑2 and BS 88‑2.1)
Advantages
No mechanical moving parts
Declared rating is very accurate
Operation is very quick
With gM fuses you can distinguish between a persistent fault and a transient fault such as the large starting current taken by motors
Reliable – it can break large current safely
Fusing factor 1.25 – 1.70.
Disadvantages
Expensive
BS 1362 cartridge fuses
Circuit breaker BS EN 60898
Circuit breakers’ two means of tripping:
Thermal trip that operates relatively slowly and is ideal for detecting overload currents
Magnetic tripping device that operates very quickly and is ideal for detecting fault currents.
Circuit breaker to BS EN 60898
Advantages
Tripping characteristics, and therefore circuit protection, are set by the installer
Circuit protection difficult to interfere with
The circuit provides discrimination
A faulty circuit may be easily and quickly restored by an unskilled operator
Disadvantages
They contain mechanical moving parts
Circuit breaker to BS EN 60898
Circuit breaker to BS EN 60898 – On-Site Guide
Residual current device (RCD) BS EN 61008
In order to detect much smaller leakage currents that could still be lethal to life, an RCD must be used.
An RCD compares the current flowing out through the line conductor with the current returning through the neutral; if the current exceeds a predetermined value, the device will trip and disconnect the circuit.
The rated value is referred to as the IΔn and is usually rated in mA.
Residual current device (RCD) BS EN 61008
Residual current operated circuit breaker with integral overcurrent protection (RCBO) BS EN 61009
An RCBO is a combination of a thermal‑magnetic circuit breaker and an RCD.
It allows earth fault protection to be restricted to a single circuit.
With most devices, two additional wires must be connected in order for this device to function.
RCBOs are available in types B and C but not in type D.
Question Time – You have 5 mins to work this out on your own.
A 6.5kW, 230V shower unit is to be wired in a domestic bathroom some 18m away from the mains consumer unit. A general-purpose thermoplastic PVC insulated and sheathed flat twin with cpc cable will be clipped to the side of the 100mm ceiling joists over much of its length with one other similar cable in a roof space which, it is anticipated, will reach 35 °C in the summer and where thermal insulation is installed up to the top of the joists. Assuming a TN‑S supply, calculate the minimum cable size to carry the current if the circuit is to be protected by:
A semi-enclosed fuse to BS 3036
a Type B CB to BS EN 60898.
How do we know what size cable to use? There's so many options!
The size of a cable to be used for an installation depends upon:
the current rating of a cable under defined installation conditions
the maximum permitted drop in voltage, as defined by BS 7671:2018 Section 525 with values in Appendix 4, section 6.4
satisfying earth fault loop impedance requirements specified in BS 7671:2018 Regulation Tables 41.2, 41.3 and 41.4.
Factors that we need to consider?
The factors which influence the current rating are the:
design current, Ib – the cable must carry the full load current
type of cable – PVC, MIMS, copper conductors or aluminium conductors
installed conditions – eg clipped to the surface or installed with other cables in trunking
surrounding temperature – cable resistance increases as temperature increases and insulation may melt if the temperature is too high
type and size of protection – for how long will the cable have to carry fault current?
Current Carrying Capacity
In order to comply with BS 7671:2018, the following relationship must be complied with:
Ib ≤ In ≤ Iz
where: Ib = design current of circuit
In = rated current or current setting of protective
device
Iz = current carrying capacity of a cable for
continuous service under the particular installation
conditions concerned
It ≥ Iz
where: It = tabulated current carrying capacity of a cable.
Steps to determine the size of cable
Correction Factors
Let's revisit this!
A 6.5kW, 230V shower unit is to be wired in a domestic bathroom some 18m away from the mains consumer unit. A general-purpose thermoplastic PVC insulated and sheathed flat twin with cpc cable will be clipped to the side of the 100mm ceiling joists over much of its length with one other similar cable in a roof space which, it is anticipated, will reach 35 °C in the summer and where thermal insulation is installed up to the top of the joists. Assuming a TN‑S supply, calculate the minimum cable size to carry the current if the circuit is to be protected by:
A semi-enclosed fuse to BS 3036
a Type B CB to BS EN 60898.
Design Current (A)
Correction Factors (A)
Correction Factors formula (A)
The installation method is from Table 4A2 Reference Method 100.
From Column 2 of Table 4D5 a 16mm2 cable, having a rating (It) of 57 amperes, is required to carry this current.
Volt Drop (A)
Now test for volt drop. The maximum permissible volt drop is 5% of the nominal supply voltage:
From Table 4D5 Column 8 the volt drop per ampere metre for a 16mm2 cable is 2.8mV. Therefore, the volt drop for this cable length and load is equal to:
Since this is less than the maximum permissible value of 11.5 volts, a 16mm2 cable satisfies the current carrying capacity and voltage drop requirements, and is therefore the chosen cable when semi-enclosed fuse protection is used.
Correction Factors (B)
Ca ambient temperature; the correction factor for 35C is 0.94 from Table 4B1 of Appendix 4.
Cc cable is installed above ground so Cc = 1.
Cd as the cable is installed above ground, this factor does not apply.
Cf since protection is by a CB, the factor is 1
Cg the cable is grouped with one similar cable so we have a factor of 0.8 from Table 4C1 of Appendix 4.
Ci thermal insulation is in contact with one side of the cable and therefore current carrying capacities are tabulated in Appendix 4.
Cs as the cable is installed above ground, this factor does not apply.
Correction Factor Formula (B)
The installation method is from Table 4A2 Reference Method 100.
From Column 2 of Table 4D5 a 10mm2 cable, having a rating (It) of 45 amperes, is required to carry this current.
Volt Drop (B)
From Table 4D5 Column 8 the volt drop per ampere metre for a 10mm2 cable is 4.4mV. Therefore, the volt drop for this cable length and load is equal to:
Now test for volt drop. The maximum permissible volt drop is 5% of the nominal supply voltage.
Since this is less than the maximum permissible value of 11.5 volts, a 10mm2 cable satisfies the current carrying capacity and voltage drop requirements, and is therefore the chosen cable when circuit breaker type B protection is used.
