Cable rating calculation21
The Regulations indicate the following symbols
for use when selecting cables:
Iz
|
is the current carrying capacity of the cable in
the |
It
|
is the tabulated current for a single circuit at an ambient temperature of 30°C |
Ib
|
is the design current, the actual current to be
carried by the cable |
In
|
is the rating of the protecting fuse or circuit
breaker |
I2
|
is the operating current for the fuse or circuit breaker (the current at which the fuse blows or the circuit breaker opens) |
Ca
|
is the correction factor for ambient
temperature |
Cg
|
is the
correction factor for grouping |
Ci
|
is the correction factor for thermal insulation. |
The correction factor for protection by a semi-enclosed
(rewirable) fuse is not given a symbol but has a fixed value of
0.725.
Under all circumstances, the cable current carrying capacity
must be equal to or greater than the circuit design current and the rating of
the fuse or circuit breaker must be at least as big as the circuit design
current. These requirements are common sense, because otherwise the cable would
be overloaded or the fuse would blow when the load is switched
an.
To ensure correct
protection from overload, it is important that the protective device operating
current (I2) is not bigger than 1.45 times the current carrying capacity of the
cable (Iz). Additionally, the rating of the fuse or circuit breaker (In) must
not be greater than the the cable current carrying capacity (Iz) It is important
to appreciate that the operating current of a protective device is always larger
than its rated value. In the case of a back-up fuse, which is not intended to
provide overload protection, neither of these requirements
applies.
To select a cable for a particular application, take the
following steps: (note that to save time it may be better first to ensure that
the expected cable for the required length of circuit will] not result in the
maximum permitted volt drop being exceeded {4.3.11}).
1. - Calculate
the expected (design) current in the circuit (Ib)
2. - Choose the
type and rating of protective device (fuse or circuit breaker) to be used
(In)
3. - Divide the
protective device rated current by the ambient temperature
----- correction factor (Ca) if ambient temperature differs from 30°C
----- correction factor (Ca) if ambient temperature differs from 30°C
4. - Further
divide by the grouping correction factor (Cg)
5. - Divide
again by the thermal insulation correction factor (CI)
6. - Divide by
the semi-enclosed fuse factor of 0.725 where applicable
7. - The result
is the rated current of the cable required, which must be chosen
----- from the appropriate tables {4.6 to 4.9}.
----- from the appropriate tables {4.6 to 4.9}.
Observe that one should divide by the correction factors,
whilst in the previous subsection we were multiplying them. The difference is
that here we start with the design current of the circuit and adjust it to take
account of factors which will derate the cable. Thus, the current carrying
capacity of the cable will be equal to or greater than the design current. In {4.3.7} we
were calculating by how much the current carrying capacity was reduced due to
application of correction factors.
{Tables 4.6 to 4.9}
give current ratings and volt drops for some of the more commonly used cables
and sizes. The Tables assume that the conductors and the insulation are
operating at their maximum rated temperatures. They are extracted from the
Regulations Tables shown in square brackets e.g. [4D1A]
The examples below will illustrate the calculations, but do
not take account of volt drop requirements (see
{4.3.11}).
An immersion heater rated at 240 V, 3 kW is to be installed
using twin with protective conductor p.v.c. insulated and sheathed cable. The
circuit will be fed from a 15 A miniature circuit breaker type 2, and will be
run for much of its 14 m length in a roof space which is thermally insulated
with glass fibre. The roof space temperature is expected to rise to 50°C in
summer, and where it leaves the consumer unit and passes through a 50 mm
insulation-filled cavity, the cable will be bunched with seven others. Calculate
the cross-sectional area of the required cable.
First calculate the design current Ib
| Ib= |
P
|
= 3000A =
|
12.5A |
-U
|
240
|
The ambient temperature correction factor is found from {Table 4.3} to be 0.71. The group
correction factor is found from {Table 4.4} as 0.52. (The
circuit in question is bunched with seven others, making eight in
all).
The thermal insulation correction factor is already taken
into account in the current rating table (4D2A ref. method 4] and need not be
further considered. This is because we can assume that the cable in the roof
space is in contact with the glass fibre but not enclosed by it. What we must
consider is the point where the bunched cables pass through the insulated
cavity. From {Table 4.5} we
have a factor of 0.89.
The correction factors must now be considered to see if more
than one of them applies to the same part of the cable. The only place where
this happens is in the insulated cavity behind the consumer unit. Factors of
0.52 (Cg) and 0.89 (CI) apply. The combined value of these (0.463), which is
lower than the ambient temperature correction factor of 0.71, and will thus be
the figure to be applied. Hence the required current rating is
calculated:-
| Table 4.6 - Current ratings and volt drops for unsheathed single core p.v.c. insulated cables | ||||||||
Cross sectional area
|
In conduit in thermal insulation
|
In conduit in thermal insulation
|
In conduit on wall
|
In conduit on wall
|
Clipped direct
|
Clipped direct
|
Volt drop
|
Volt drop
|
(mm²)
|
(A)
|
(A)
|
(A)
|
(A)
|
(A)
|
(A)
|
(mV/A/m)
|
(mV/A/m)
|
2 cables
|
2 cables
|
2 cables
|
2 cables
|
3 or 4 cables
| ||||
1.0
|
11.0
|
10.5
|
13.5
|
12.0
|
15.5
|
14.0
|
44.0
|
38.0
|
1.5
|
14.5
|
13.5
|
17.5
|
15.5
|
20.0
|
18.0
|
29.0
|
25.0
|
2.5
|
19.5
|
18.0
|
24.0
|
21.0
|
27.0
|
25.0
|
18..0
|
15.0
|
4.0
|
26.0
|
24.0
|
32.0
|
28.0
|
37.0
|
33.0
|
11.0
|
9.5
|
6.0
|
34.0
|
31.0
|
41.0
|
36.0
|
47.0
|
43.0
|
7.3
|
6.4
|
10.0
|
46.0
|
42.0
|
57.0
|
50.0
|
65.0
|
59.0
|
4.4
|
3.8
|
16.0
|
61.0
|
56.0
|
76.0
|
68.0
|
87.0
|
79.0
|
2.8
|
2.4
|
| Table 4.7 - Current ratings and volt drops for sheathed multi-core p.v.c.-insulated cables | ||||||||
Cross sectional area
|
In conduit in thermal insulation
|
In conduit in thermal insulation
|
In conduit on wall
|
In conduit on wall
|
Clipped direct
|
Clipped direct
|
Volt drop
|
Volt drop
|
(mm²)
|
(A)
|
(A)
|
(A)
|
(A)
|
(A)
|
(A)
|
(mV/A/m)
|
(mV/A/m)
|
-
|
2 core
|
3 or 4 core
|
2 core
|
3 or 4 core
|
2 core
|
3 or 4 core
|
2 core
|
3 or 4 core
|
1.0
|
11.0
|
10.0
|
13.0
|
11.5
|
15.0
|
13.5
|
44.0
|
38.0
|
1.5
|
14.0
|
13.0
|
16.5
|
15.0
|
19.5
|
17.5
|
29.0
|
25.0
|
2.5
|
18.5
|
17.5
|
23.0
|
20.0
|
27.0
|
24.0
|
18.0
|
15.0
|
4.0
|
25.0
|
23.0
|
30.0
|
27.0
|
36.0
|
32.0
|
11.0
|
9.5
|
6.0
|
32.0
|
29.0
|
38.0
|
34.0
|
46.0
|
41.0
|
7.3
|
6.4
|
10.0
|
43.0
|
39.0
|
52.0
|
46.0
|
63.0
|
57.0
|
4.4
|
3.8
|
16.0
|
57.0
|
52.0
|
69.0
|
62.0
|
85.0
|
76.0
|
2.8
|
2.4
|
Iz =
in =
15 A = 32.4 A
From {Table 4.7}, 6 mm² p.v.c. twin with protective conductor
has a current rating of 32 A. This is not quite large enough, so 10 mm²with a
current rating of 43 A is indicated. Not only would this add considerably to the
costs, but would also result in difficulties due to terminating such a large
cable in the accessories.
A more sensible option would be to look for a method of
reducing the required cable size. For example, if the eight cables left the
consumer unit in two bunches of four, this would result in a grouping factor of
0.65 (from {Table 4.4}). Before applying this, we must
check that the combined grouping and thermal insulation factors (0.65 x 0.89 =
.0.58) are still less than the ambient temperature factor of 0.71, which is the
case.
| Table 4.8 - Current ratings of mineral insulated
cables clipped direct | ||||||
Cross-sectional area
|
Volt
|
p.v.c. sheath
2 x single or twin |
p.v.c. Sheath 3 core
|
p.v.c. Sheath
|
Bare sheath
|
Bare sheath
|
(mm²)
|
(A)
|
(A)
|
(A)
|
(A)
|
(A)
| |
1.0
|
500v
|
18.5
|
16.5
|
16.5
|
22.0
|
21.0
|
1.5
|
500v
|
24.0
|
21.0
|
21.0
|
28.0
|
27.0
|
2.5
|
500v
|
31.0
|
28.0
|
28.0
|
38.0
|
36.0
|
4.0
|
500v
|
42.0
|
37.0
|
37.0
|
51.0
|
47.0
|
1.0
|
750v
|
20.0
|
17.5
|
17.5
|
24.0
|
24.0
|
1.5
|
750v
|
25.0
|
22.0
|
22.0
|
31.0
|
30.0
|
2.5
|
750v
|
34.0
|
30.0
|
30.0
|
42.0
|
41.0
|
4.0
|
750v
|
45.0
|
40.0
|
40.0
|
55.0
|
53.0
|
6.0
|
750v
|
57.0
|
51.0
|
51.0
|
70.0
|
67.0
|
10.0
|
750v
|
78.0
|
69.0
|
69.0
|
96.0
|
91.0
|
16.0
|
750v
|
104.0
|
92.0
|
92.0
|
127.0
|
119.0
|
| Table 4.9 - Volt drops for mineral insulated cables | ||||
Cross-sectional area
|
Single-phase p.v.c. Sheath
|
Single-phase bare
|
Three-phase p.v.c. Sheath
|
Three-phase bare
|
(mm²)
|
(mV/A/m)
|
(mV/A/m)
|
(mV/A/m)
|
(mV/A/m)
|
1.0
|
42.0
|
47.0
|
36.0
|
40.0
|
1.5
|
28.0
|
31.0
|
24.0
|
27.0
|
2.5
|
17.0
|
19.0
|
14.0
|
16.0
|
4.0
|
10.0
|
12.0
|
9.1
|
10.0
|
6.0
|
7.0
|
7.8
|
6.0
|
6.8
|
10.0
|
4.2
|
4.7
|
3.6
|
4.1
|
16.0
|
2.6
|
3.0
|
2.3
|
2.6
|
This leads to a cable current rating of 15
A = 25.9 A
This is well below the rating for 6 mm² of 32 A, so a
cable of this size could be selected.
Example
4.2
There is no correction factor for the presence of the glass
fibre, so the calculation of Iz will be exactly the same as Example 4.1 at 32.4
A.
This time reference method I (clipped direct) will apply to
the current rating {Table 4.7}. For a two core cable, 4.0 mm², two core has a
rating of 36 A, so this Will be the selected size.
It is of interest to notice how quite a minor change in the
method of installation, in this case clipping the cable to the joists or battens
clear of the glass fibre, has reduced the acceptable cable size.
Example
4.3
This time the value of the acceptable current carrying
capacity Iz will be different because of the need to include a factor for the
rewirable fuse as well as the new ambient temperature and grouping factors for
the rewirable fuse from {Tables 4.3 and 4.4}.
Iz = in
= 15 A = 45.7 A
In this case, the cable is in contact with the glass fibre,
so the first column of {Table 4.7} of current ratings will apply. The acceptable
cable size is 16 mm² which has a current rating of 57 A.
This cable size is not acceptable on the grounds of high cost
and because the conductors are likely to be too large to enter the connection
tunnels of the immersion heater and its associated switch. If the cables leaving
the consumer unit are re-arranged in two groups of four, this will reduce the
grouping factor to 0.65, so that the newly calculated value of Iz is 36.6 A.
This means using 10 mm² cable with a current rating of 43A (from {Table
4.7}), since 6 mm² cable is shown to have a current rating in these
circumstances of only 32 A. By further rearranging the cables leaving the
consumer unit to be part of a group of only two, Cg is increased to 0.8, which
reduces Iz to 29.7 A which enables selection of a 6 mm² cable.
Should it be possible to bring the immersion heater cable out
of the consumer unit on its own, no grouping factor would apply and Iz would
fall to 23.8 A, allowing a 4 mm² cable to be selected.
Where the cable is not in contact with glass fibre there will
be no need to repeat the calculation of Iz, which still has a value of 29.7 A
provided that it is possible to group the immersion heater cable with only one
other where it leaves the consumer unit. This time we use the 'Reference Method
1 (clipped direct)' column of the current rating {Table 4.7}, which shows that 4
mm² cable with a current rating of 36 A will be satisfactory.
Examples 4.1, 4.2 and
4.3 show clearly how forward planning will enable a more economical and
practical cable size to be used than would appear necessary at first. It is, of
course, important that the design calculations are recorded and retained in the
installation manual.
Example
4.4
The first step is to calculate the line current of the
motor.
Input =
output = 7.5 x 100 kW = 8.82
kW
Line current Ib = P
= 8.82 x 103 A =
15.3 A
We must now select a suitable fuse. {Fig 3.15} for BS 88 fuses shows
the 16 A size to be the most suitable. Part of the run is subject to an ambient
temperature of 50°C, where the cable is also part of a group of three, so the
appropriate correction factors must be applied from {Tables
4.3 and 4.4}.
Iz =
In =---------- 16A
= 34.2A
Note that the grouping factor of 0.70 has been selected
because where the cable is grouped it is clipped to a metallic cable tray, and
not to a non-metallic surface. Next the cable must be chosen from {Table 4.8}.
Whilst the current rating would be 15.3 A if all of the cable run were clipped
to the wall, part of the run is subject to the two correction factors, so a
rating of 34.2 A must be used. For the clipped section of the cable (15.3 A),
reference method I could be used which gives a size of 1.0 mm² (current rating
16.5 A). However, since part of the cable is on the tray (method 3) the correct
size for 34.2 A will be 4.0 mm², with a rating of 37 A.
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