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Megabreak® - Guida di manutenzione
Guida di manut
enzione
7
Controllare la condizione generale dell’interruttore, se ci sono parti rotte, usurate o danneggiate.
- Coperchio frontale e protezione IP
- Etichette di identificazione
- Display e mostrina dell’unità di protezione
- Connessioni
- Cella estraibile
- Struttura dell’interruttore
Controllare la compatibilità tra gli accessori elettrici e la tensione utilizzata.
Disconnettere l’interruttore dall’alimentazione, sia di potenza che dei circuiti ausiliari.
Eseguire un ciclo di apertura/chiusura.
Aprire l’interruttore e scaricare le molle (OFF-DISCHARGED).
In caso di interruttore estraibile rimuovere l’interruttore dalla cella
Rimuovere il coperchio frontale
3. Procedura
preliminare
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
versione fissa
versione
estraibile
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
Se non diversamente spe-
cificato nelle procedure di
manutenzione, tutte le ope-
razioni devono essere ese-
guite senza alimentazione
(potenza e circuiti ausiliari)
e con l’interruttore in posi-
zione di OFF-DISCHARGED.
Se nella procedura ci sono
diverse indicazioni, alla fine
della procedura l’interrut-
tore deve comunque essere
riportato in condizioni di
sicurezza: nessuna alimen-
tazione in potenza e nei
circuiti ausiliari e posizione
OFF-DISCHARGED.
Содержание Megabreak T803LE1250
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