Another long one from me ... I hope that it covers all the bases and is a more thorough approach to the subject. It is my first design work so constructive criticism welcome. I want to build on a solid foundation. This one is for a friend.
TN-C-S Supply
Install power and lighting to 2, 3.6 m wide x 2.4 m 'deep' wooden sheds set 1 m apart under a common roof, and an IP66 rated socket outdoors. The requirement is a double socket and a 5 ' T5 fluorescent batten in each shed. The sockets are to be used to run power tools and a ~ 1200 W lawn mower, and the lighting is primarily to be able to find things during the long winter evenings in the N of Scotland.
The sheds are behind the house and the opposite side of a path alongside the house ~ 5 m diagonally from the door of a wooden 'lean-to', used as a garden shed, built on the back of an annex to the main porch. These additions to the side of the house are 2.1 m deep. The House CU is located upstairs in the middle of the property, and the nearest easily accessible point of the electrical installation to the sheds is the last socket on a Schneider B16 RCBO protected 2.5 mm2 radial circuit. The socket is flush mounted in the wood paneling of the back of the porch annex that covers the original ~ 2 ' thick ashlar masonry of the Victorian stone built house and is situated at a height of about 0.5 m.
It was inconvenient to the customer, when I took a look at their requirements, to switch off the power and disconnect the earthing conductor in order to measure Ze so I took a Zs measurement at the end of the radial of 0.47 Ohms with an indicated supply voltage of 238 V. Investigation, though not conclusive, suggests that this circuit feeds only 2 sockets, one by the front door and the last socket in the circuit, already identified.
Unfortunately, the loft has boarding over the insulation and I was unable to inspect the cable installation method. If the standards of the time of building were followed, and some of the structure was visible, it will have a minimum of ~ 6 " x 2 1/2 " joists and the cable will have been clipped to the joists or laid on top of a lath and plaster ceiling or laid over the insulation. I suspect that the first or last option is most likely. Furthermore, BS 7671, APPENDIX 4, TABLE 4D5 records a CCC of 17 A for 2.5 mm2 using method 101#. This suggests to me that the installation is designed and installed IAW current regulations.
The existing circuit, on the basis of the customer's measurements and 'informed guesswork' as to its route, for design purposes is assumed to comprise:
- 6.0 m running from the CU up into the loft immediately above and in, on or below insulation through the loft space to the wall of the property.
- 5.6 m drop from the loft to socket level, assumed to be 0.5 m from ground level, running in the void between the internal rubble stone masonry and lath and plaster room linings.
- 0.6 m through the wall thickness.
- 2.1 m running horizontally from the socket in the porch to the socket in the porch annex between the face of the original house and the wooden panelling of the porch annex.
Total approximate run from CU to end of radial circuit:
6.0 + 5.6 + 0.6 + 2.1 = 14.3 m
To validate assumptions, the resistance of 2.5 mm2 Twin + CPC cable for the Line-CPC circuit, R1 + R2, is 19.51 mOhms / m and that for the Line-Neutral circuit, R1 + Rn, is 14.82 mOhms / m; Table I1 in Appendix I, On-Site Guide.
- R1 + R2 = 14.3 x 19.51 / 1000 = 0.279 Ohms
- R1 + Rn = 14.3 x 14.82 / 1000 = 0.212 Ohms
Check against measured value for Zs:
Zs = Ze + (R1 + R2)
Ze ~ 0.47 - 0.28 = 0.19 Good value for TN-C-S.
According to table 4D5, voltage drop for 2.5 mm2 Twin + CPC cable is 18 mV / A / m. Circuit voltage drop at OCPD nominal current of 16 A is therefore:
14.3 x 16 x 18 / 1000 = 4.1 V
If the additional circuit voltage drop becomes a problem, replacing the existing cable with 4.0 mm2 Twin + CPC cable reduces the voltage drop to 11 mV / A / m according to table 4D5. Therefore the voltage drop for this section of the new circuit becomes:
14.3 x 16 x 11 / 1000 = 2.5 V
Proposed circuit extension comprises:
- 0.5 m of 4.0 mm Twin + CPC cable from socket at the end of the existing radial circuit to a Metal Clad 13 A 1-gang Double Pole (DP) switched Fused Connection Unit (FCU) with neon, fused @ 13 A; table 4D5 applies.
- 0.3 m of 4.0 mm 2-Core or 3-Core Steel Wire Armoured (SWA) cable through SWA gland from FCU to horizontal run; BS 7671, APPENDIX 4, TABLE 4D4A and TABLE 4D4B apply.
- 2.1 m of 4.0 mm 2-Core or 3-Core SWA cable in horizontal run, clipped direct, along back wall of porch annex in lean-to shed.
- 0.2 m of 4.0 mm 2-Core or 3-Core SWA cable from horizontal run to ground level.
- 0.6 m of 4.0 mm 2-Core or 3-Core SWA cable to trench depth.
- 4.7 m of 4.0 mm 2-Core or 3-Core SWA cable at trench depth from corner of porch annex to nearest corner of right hand shed.
- 0.6 m of 4.0 mm 2-Core or 3-Core SWA cable from trench depth to ground level.
- 1.3 m of 4.0 mm 2-Core or 3-Core SWA cable from ground level through floor into shed and through SWA cable gland into Metal Clad 13 A 1-gang Double Pole (DP) switch.
According to table 4D5, voltage drop for 4.0 mm2 Twin + CPC cable is 11 mV / A / m. First section, 0.5 m of new circuit voltage drop at OCPD nominal current of 16 A is therefore:
0.5 x 16 x 11 / 1000 = 0.1 V
According to table I1, the resistance of 4.0 mm2 Twin + CPC cable for the Line-CPC circuit, R1 + R2, is 12.02 mOhms / m. Additional R1 + R2:
0.5 x 12.02 / 1000 = 0.006 Ohms
Total run of SWA cable in new circuit, measurements taken by self, is approximately:
0.3 + 2.1 + 0.2 + 0.6 + 4.7 + 0.6 + 1.3 = 9.8 m.
According to table 4D4B, voltage drop for 4.0 mm2 SWA cable is 11 mV / A / m. Circuit voltage drop at nominal current of 13 A fuse in FCU is therefore:
9.8 x 13 x 11 / 1000 = 1.4 V
According to table I1, the resistance of 4.0 mm2 SWA cable for the Line-CPC circuit, R1 + R2, is 9.22 mOhms / m. Additional R1 + R2:
9.8 x 9.22 / 1000 = 0.090 Ohms
This assumes 4.0 mm2 Line and CPC. CPC will be provided by the cable armour giving a considerably lower actual value of R1 + R2.
Total circuit voltage drop from installation origin to proposed supply point in shed using existing wiring is therefore:
4.1 + 0.1 + 1.4 = 5.6 V
Assuming that a 5' T5, high efficiency, fluorescent tube lamp units is used for lighting in each shed, both fed from a single Metal Clad 13 A 1-gang Double Pole (DP) switched Fused Connection Unit (FCU) with neon, fused @ 3 A. Their power consumption will be 38 W each and they will therefore draw a current of (2 x 38) / (230 x 0.95) = 0.35 A. The maximum cable run allowed for the lighting circuit will depend on the voltage drop at the supply point in the shed of 5.6 V and the choice of lighting circuit cable. According to table 4D5 the voltage drop for standard lighting circuit cable, 1.5 mm2 Twin + CPC, is 29 mV / A / m. The maximum cable run available within the limits of table 4Ab is:
(6.555 - 5.6) x 1000 / (29 x 0.35) = 94 m
This is clearly adequate for the wiring of 2 individually switched lights, one in each 3.6 m x 2.4 m shed spaced 1m apart under a common roof.
If the nominal current of the 3 A fuse in the FCU is used to calculate the maximum cable run then it becomes (6.555 - 5.6) x 1000 / (29 x 3) = 11 m. This distance makes installation very 'tight' but it is highly unlikely that current consumption will reach this level even if a 500 W halogen spotlight were fitted to provide garden 'lighting'.
According to table I1, the resistance of 1.5 mm2 SWA cable for the Line-CPC circuit, R1 + R2, is 30.20 mOhms / m. Additional R1 + R2 based on likely cable routing:
11.4 x 30.20 / 1000 = 0.344 Ohms
Ze for lighting circuit verification:
0.47 + 0.006 + 0.090 + 0.344 = 0.91 Ohms
Assuming that 2.5 mm2 Twin + CPC cable is used for the power cabling in the shed and the nominal current of the 13 A fuse in the FCU at the house end of the SWA cable is the limiting current for the calculation. According to table 4D5, voltage drop for 2.5 mm2 Twin + CPC cable is 18 mV / A / m. Therefore the maximum cable run available within the limits of table 4Ab is:
(10.925 - 5.6) x 1000 / (18 x 13) ~ 23 m
This is adequate for the wiring of a 2-gang 13 A socket in each shed and an IP66 rated outside socket in any location on the outside of either shed.
According to table I1, the resistance of 1.5 mm2 SWA cable for the Line-CPC circuit, R1 + R2, is 19.51 mOhms / m. Additional R1 + R2 based on likely cable routing:
10.2 x 19.51 / 1000 = 0.199 Ohms
Ze for power circuit verification:
0.47 + 0.006 + 0.090 + 0.199 = 0.77 Ohms
TN-C-S Supply
Install power and lighting to 2, 3.6 m wide x 2.4 m 'deep' wooden sheds set 1 m apart under a common roof, and an IP66 rated socket outdoors. The requirement is a double socket and a 5 ' T5 fluorescent batten in each shed. The sockets are to be used to run power tools and a ~ 1200 W lawn mower, and the lighting is primarily to be able to find things during the long winter evenings in the N of Scotland.
The sheds are behind the house and the opposite side of a path alongside the house ~ 5 m diagonally from the door of a wooden 'lean-to', used as a garden shed, built on the back of an annex to the main porch. These additions to the side of the house are 2.1 m deep. The House CU is located upstairs in the middle of the property, and the nearest easily accessible point of the electrical installation to the sheds is the last socket on a Schneider B16 RCBO protected 2.5 mm2 radial circuit. The socket is flush mounted in the wood paneling of the back of the porch annex that covers the original ~ 2 ' thick ashlar masonry of the Victorian stone built house and is situated at a height of about 0.5 m.
It was inconvenient to the customer, when I took a look at their requirements, to switch off the power and disconnect the earthing conductor in order to measure Ze so I took a Zs measurement at the end of the radial of 0.47 Ohms with an indicated supply voltage of 238 V. Investigation, though not conclusive, suggests that this circuit feeds only 2 sockets, one by the front door and the last socket in the circuit, already identified.
Unfortunately, the loft has boarding over the insulation and I was unable to inspect the cable installation method. If the standards of the time of building were followed, and some of the structure was visible, it will have a minimum of ~ 6 " x 2 1/2 " joists and the cable will have been clipped to the joists or laid on top of a lath and plaster ceiling or laid over the insulation. I suspect that the first or last option is most likely. Furthermore, BS 7671, APPENDIX 4, TABLE 4D5 records a CCC of 17 A for 2.5 mm2 using method 101#. This suggests to me that the installation is designed and installed IAW current regulations.
The existing circuit, on the basis of the customer's measurements and 'informed guesswork' as to its route, for design purposes is assumed to comprise:
- 6.0 m running from the CU up into the loft immediately above and in, on or below insulation through the loft space to the wall of the property.
- 5.6 m drop from the loft to socket level, assumed to be 0.5 m from ground level, running in the void between the internal rubble stone masonry and lath and plaster room linings.
- 0.6 m through the wall thickness.
- 2.1 m running horizontally from the socket in the porch to the socket in the porch annex between the face of the original house and the wooden panelling of the porch annex.
Total approximate run from CU to end of radial circuit:
6.0 + 5.6 + 0.6 + 2.1 = 14.3 m
To validate assumptions, the resistance of 2.5 mm2 Twin + CPC cable for the Line-CPC circuit, R1 + R2, is 19.51 mOhms / m and that for the Line-Neutral circuit, R1 + Rn, is 14.82 mOhms / m; Table I1 in Appendix I, On-Site Guide.
- R1 + R2 = 14.3 x 19.51 / 1000 = 0.279 Ohms
- R1 + Rn = 14.3 x 14.82 / 1000 = 0.212 Ohms
Check against measured value for Zs:
Zs = Ze + (R1 + R2)
Ze ~ 0.47 - 0.28 = 0.19 Good value for TN-C-S.
According to table 4D5, voltage drop for 2.5 mm2 Twin + CPC cable is 18 mV / A / m. Circuit voltage drop at OCPD nominal current of 16 A is therefore:
14.3 x 16 x 18 / 1000 = 4.1 V
If the additional circuit voltage drop becomes a problem, replacing the existing cable with 4.0 mm2 Twin + CPC cable reduces the voltage drop to 11 mV / A / m according to table 4D5. Therefore the voltage drop for this section of the new circuit becomes:
14.3 x 16 x 11 / 1000 = 2.5 V
Proposed circuit extension comprises:
- 0.5 m of 4.0 mm Twin + CPC cable from socket at the end of the existing radial circuit to a Metal Clad 13 A 1-gang Double Pole (DP) switched Fused Connection Unit (FCU) with neon, fused @ 13 A; table 4D5 applies.
- 0.3 m of 4.0 mm 2-Core or 3-Core Steel Wire Armoured (SWA) cable through SWA gland from FCU to horizontal run; BS 7671, APPENDIX 4, TABLE 4D4A and TABLE 4D4B apply.
- 2.1 m of 4.0 mm 2-Core or 3-Core SWA cable in horizontal run, clipped direct, along back wall of porch annex in lean-to shed.
- 0.2 m of 4.0 mm 2-Core or 3-Core SWA cable from horizontal run to ground level.
- 0.6 m of 4.0 mm 2-Core or 3-Core SWA cable to trench depth.
- 4.7 m of 4.0 mm 2-Core or 3-Core SWA cable at trench depth from corner of porch annex to nearest corner of right hand shed.
- 0.6 m of 4.0 mm 2-Core or 3-Core SWA cable from trench depth to ground level.
- 1.3 m of 4.0 mm 2-Core or 3-Core SWA cable from ground level through floor into shed and through SWA cable gland into Metal Clad 13 A 1-gang Double Pole (DP) switch.
According to table 4D5, voltage drop for 4.0 mm2 Twin + CPC cable is 11 mV / A / m. First section, 0.5 m of new circuit voltage drop at OCPD nominal current of 16 A is therefore:
0.5 x 16 x 11 / 1000 = 0.1 V
According to table I1, the resistance of 4.0 mm2 Twin + CPC cable for the Line-CPC circuit, R1 + R2, is 12.02 mOhms / m. Additional R1 + R2:
0.5 x 12.02 / 1000 = 0.006 Ohms
Total run of SWA cable in new circuit, measurements taken by self, is approximately:
0.3 + 2.1 + 0.2 + 0.6 + 4.7 + 0.6 + 1.3 = 9.8 m.
According to table 4D4B, voltage drop for 4.0 mm2 SWA cable is 11 mV / A / m. Circuit voltage drop at nominal current of 13 A fuse in FCU is therefore:
9.8 x 13 x 11 / 1000 = 1.4 V
According to table I1, the resistance of 4.0 mm2 SWA cable for the Line-CPC circuit, R1 + R2, is 9.22 mOhms / m. Additional R1 + R2:
9.8 x 9.22 / 1000 = 0.090 Ohms
This assumes 4.0 mm2 Line and CPC. CPC will be provided by the cable armour giving a considerably lower actual value of R1 + R2.
Total circuit voltage drop from installation origin to proposed supply point in shed using existing wiring is therefore:
4.1 + 0.1 + 1.4 = 5.6 V
Assuming that a 5' T5, high efficiency, fluorescent tube lamp units is used for lighting in each shed, both fed from a single Metal Clad 13 A 1-gang Double Pole (DP) switched Fused Connection Unit (FCU) with neon, fused @ 3 A. Their power consumption will be 38 W each and they will therefore draw a current of (2 x 38) / (230 x 0.95) = 0.35 A. The maximum cable run allowed for the lighting circuit will depend on the voltage drop at the supply point in the shed of 5.6 V and the choice of lighting circuit cable. According to table 4D5 the voltage drop for standard lighting circuit cable, 1.5 mm2 Twin + CPC, is 29 mV / A / m. The maximum cable run available within the limits of table 4Ab is:
(6.555 - 5.6) x 1000 / (29 x 0.35) = 94 m
This is clearly adequate for the wiring of 2 individually switched lights, one in each 3.6 m x 2.4 m shed spaced 1m apart under a common roof.
If the nominal current of the 3 A fuse in the FCU is used to calculate the maximum cable run then it becomes (6.555 - 5.6) x 1000 / (29 x 3) = 11 m. This distance makes installation very 'tight' but it is highly unlikely that current consumption will reach this level even if a 500 W halogen spotlight were fitted to provide garden 'lighting'.
According to table I1, the resistance of 1.5 mm2 SWA cable for the Line-CPC circuit, R1 + R2, is 30.20 mOhms / m. Additional R1 + R2 based on likely cable routing:
11.4 x 30.20 / 1000 = 0.344 Ohms
Ze for lighting circuit verification:
0.47 + 0.006 + 0.090 + 0.344 = 0.91 Ohms
Assuming that 2.5 mm2 Twin + CPC cable is used for the power cabling in the shed and the nominal current of the 13 A fuse in the FCU at the house end of the SWA cable is the limiting current for the calculation. According to table 4D5, voltage drop for 2.5 mm2 Twin + CPC cable is 18 mV / A / m. Therefore the maximum cable run available within the limits of table 4Ab is:
(10.925 - 5.6) x 1000 / (18 x 13) ~ 23 m
This is adequate for the wiring of a 2-gang 13 A socket in each shed and an IP66 rated outside socket in any location on the outside of either shed.
According to table I1, the resistance of 1.5 mm2 SWA cable for the Line-CPC circuit, R1 + R2, is 19.51 mOhms / m. Additional R1 + R2 based on likely cable routing:
10.2 x 19.51 / 1000 = 0.199 Ohms
Ze for power circuit verification:
0.47 + 0.006 + 0.090 + 0.199 = 0.77 Ohms
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