Bticino Megabreak T803LE1250, Руководство по обслуживанию

Страница: 43 / 76

Megabreak® - Maintenance guide
M
ain
tenanc
e guide
43
Check the general condition of breaker, if there are broken parts, ageing or damaged parts
- front covers and IP protection
- identification labels
- Protection Unit display and enclosure
- Connections
- Chassis
- Structure of breaker
Check the compatibility between electrical accessories and the voltage used
Disconnect the breaker from the supply, both the power and auxiliary circuit
Perform a close/open cycle
Open the breaker and discharge the springs (OFF-DISCHARGED)
In case of draw-out breaker, remove the breaker from chassis
Remove the front cover
3. Preparatory
work
dischar
ged
10
5
2
0.7
0.4
5
0.6
0.5
Ir
2
0.9
8
0.8
10
xIr
3 Im
6 4 5
s
xIn
0.1
0.0
0.0
1
20
5 10 @6
Ir
30
tr
30 20
0.3
tm
0.2
0.3
0.2 0.1
I>1.0
5 I>.90
xIr
ON
N
0.0 0
0.02
0.0
4
0.0 6
0.0 8 0.1
I 2 t= k@
12I r
MEM
=O
FF
s
1 .
. 5
Ii
Icw
4
3
2
8
6
10
12
15
xIn
discharged
discha
rged
10
5
2
0.7
0.4
5
0.6
0.5
Ir
2
0.9
8
0.8
10
xIr
3 Im
6 4 5
s
xIn
0.1
0.0
0.0
1
20
5 10 @6
Ir
30
tr 30
20
0.3
tm
0.2
0.3
0.2 0.1
I>1
.05 I>.9
0
xIr
ON
N
0.0 0
0.0 2
0.04
0.0 6
0.0
8 0.1
I 2 t=k @1
2Ir
ME
M= OFF
s
1 .
. 5
Ii
Icw
4
3
2
8
6
10
12
15
xIn
discharged
discharged
10
5
2
0.7
0.4
5
0.6
0.5
Ir
2
0.9
8
0.8
10 xIr
3 Im 6
4 5 s
xIn
0.1 0.0
0.01
20
5
10 @6I r
30 tr
30 20
0.3
tm
0.2
0.3
0.2
0.1
I>1.05
I>.90
xIr
ON
N
0.0 0 0.0
2 0.0
4
0.06
0.0 8 0.1
I
2
t=k
@1 2Ir
ME M=O
FF
s
1.
.5
Ii
Icw
4
3
2
8
6
10 12
15 xIn
10
5
2
0.7
0.4
5
0.6
0.5
Ir
2
0.9
8
0.8
10
xIr
3
I
xIn
0.
20
5
10 @6
Ir
30
tr
30
20
I>1.05
I>.9
0
xIr
ON
N
0.00
0.02
0.04
0.06
0.08
0.1
ME
M=
OFF
s
1.
.5
Ii
Icw
4
3
2
8
6
10
12
15
xIn
10
5
2
0.7
0.4
5
0.6
0.5
Ir
2
0.9
8
0.8
10 xIr
3
Im 6
4 5
s
xIn
0.1
0.0
0.01
20
5
10 @6I r
30
tr 30
20
0.3 tm
0.2
0.3
0.2
0.1
I>1.05
I>.9
0
xIr
ON
N
0.0 0
0.02
0.0
4
0.0 6
0.08
0.1
I
2
t=k@
12Ir
MEM=OF
F
s
1.
.5
Ii
Icw
4
3
2
8
6
10 12
15
xIn
Fixed version
Draw-out
version
discha
rged
10
5
2
0.7
0.4
5
0.6
0.5
Ir
2
0.9
8
0.8
10
xIr
3
Im 6
4 5
s
xIn
0.1
0.0
0.01
20
5
10 @6 Ir
30 tr
30 20
0.3 tm
0.2
0.3
0.2
0.1
I>1.0 5 I>.90
xIr
ON
N
0.00
0.0
2
0.04
0.0
6
0.0 8 0.1
I
2
t=k@
12I
r
MEM =OF
F
s
1
.
. 5
Ii
Icw
4
3
2
8
6
10 12
15
xIn
discharged
discharg
ed
10
5
2
0.7
0.4
5
0.6
0.5
Ir
2
0.9
8
0.8
10
xIr
3
Im 6
4 5 s
xIn
0.1
0.0
0.0
1
20
5 10 @6
Ir
30 tr
30 20
0.3 tm
0.2
0.3
0.2
0.1
I>1 .05 I>.9
0
xIr
ON
N
0.0
0 0.0
2
0.04
0.0
6
0.0 8 0.1
I 2 t=k
@1 2Ir
ME M=
OF F
s
1
.
. 5
Ii
Icw
4
3
2
8
6
10
12
15
xIn
discharged
discharged
10
5
2
0.7
0.4
5
0.6
0.5
Ir
2
0.9
8
0.8
10
xIr
3 Im
6 4 5
s
xIn
0.1
0.0
0.0 1
20
5 10 @6 Ir
30
tr 30
20
0.3 tm
0.2
0.3
0.2 0.1
I>1.0 5 I>.90
xIr
ON
N
0.0 0
0.0 2
0.0 4
0.06
0.0
8 0.1
I 2 t=k @1
2Ir
MEM
=O FF
s
1.
.5
Ii
Icw
4
3 2
8
6
10
12
15
xIn
10
5
2
0.7
0.4
5
0.6
0.5
Ir
2
0.9
8
0.8
10
xIr
3 I
xIn
0.
20
5
10 @6Ir
30 tr
30
20
I>1.0
5
I>.90
xIr
ON
N
0.00
0.02
0.04
0.06
0.08 0.1
MEM=OFF
s
1.
.5
Ii
Icw
4
3
2
8
6
10
12
15
xIn
10
5
2
0.7
0.4
5
0.6
0.5
Ir
2
0.9
8
0.8
10
xIr
3 Im
6 4 5
s
xIn
0.1
0.0
0.01
20
5 10 @6I
r
30
tr
30 20
0.3
tm
0.2
0.3
0.2
0.1
I>1.05
I>.9 0
xIr
ON
N
0.0
0 0.02
0.0 4
0.0
6
0.08 0.1
I
2 t=k @12
Ir
MEM=OFF
s
1.
.5
Ii
Icw
4
3 2
8
6
10
12
15
xIn
disc
harge
d
10
5
2
0.7
0.4
5
0.6
0.5
Ir
2
0.9
8
0.8
10
xIr
3 Im
6 4 5
s
xIn
0.1
0.0
0.01
20
5
10 @6 Ir
30 tr
30 20
0.3
tm
0.2
0.3
0.2
0.1
I>1.0
5 I>.9
0
xIr
ON
N
0.0
0 0.02
0.0
4
0.0 6
0.0
8 0.1
I
2
t=k@
12Ir
MEM=
OFF
s
1 .
.5
Ii
Icw
4
3
2
8
6
10 12
15
xIn
discharged
discharge
d
10
5
2
0.7
0.4
5
0.6
0.5
Ir
2
0.9
8
0.8
10
xIr
3 Im
6 4 5
s
xIn
0.1
0.0
0.01
20
5
10 @6 Ir
30 tr
30 20
0.3 tm
0.2
0.3
0.2
0.1
I>1
.05 I>.9
0
xIr
ON
N
0.0
0 0.02
0.0
4
0.0 6
0.0 8 0.1
I 2 t=k
@1 2Ir
ME M= OFF
s
1.
.5
Ii
Icw
4
3
2
8
6
10
12
15
xIn
discharged
discharged
10
5
2
0.7
0.4
5
0.6
0.5
Ir
2
0.9
8
0.8
10 xIr
3 Im 6
4 5 s
xIn
0.1 0.0
0.01
20
5 10 @6I r
30
tr 30 20
0.3 tm
0.2
0.3
0.2 0.1
I>1 .05 I>.90
xIr ON
N
0.00
0.0 2
0.0 4
0.0 6
0.0
8 0.1
I 2 t=k @1 2Ir
ME M=O FF
s
1.
.5
Ii
Icw 4
3 2
8 6 10 12
15
xIn
10
5
2
0.7
0.4
5
0.6
0.5
Ir
2
0.9
8
0.8
10
xIr
3
I
xIn
0.
20
5
10 @6 Ir
30 tr
30
20
I>1.05
I>.90
xIr
ON
N
0.00
0.02
0.04
0.06
0.08
0.1
ME
M=OFF
s
1
.
. 5
Ii
Icw
4
3
2
8
6
10
12
15
xIn
10
5
2
0.7
0.4
5
0.6
0.5
Ir
2
0.9
8
0.8
10 xIr
3
Im 6 4 5
s
xIn
0.1
0.0
0.01
20
5 10 @6I
r
30
tr 30
20
0.3
tm
0.2
0.3
0.2
0.1
I>1.05
I>.9
0
xIr
ON
N
0.0 0
0.02
0.0 4
0.0
6
0.08 0.1
I
2 t=k@ 12Ir
MEM=OFF
s
1.
.5
Ii
Icw
4
3
2
8
6
10
12
15 xIn
Unless otherwise specified in
the maintenance procedures,
all operations must be
carried out without supply
(power and aux circuit) and
with breaker in position OFF-
DISCHARGED.
If in procedure there are
different indications, at
the end of procedure the
breaker must anyway be in
the safe condition: no supply
in power and aux circuit and
OFF-DISCHARGED