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Example of Installation
-
Total power cable length L = 100(m), Running current of each units 1[A]
-
Total 10 indoor units were installed
10[A]
ELCB
or MCCB+
ELB
0[m]
• Apply following equation.
n
Coef×35.6×L
Σ
(
1000×Ak
k=1
❈ Calculation
• Installing with 1 sort wire.
2.5[mm
2
]
-2.2[V]
220[V]
-(2.2+2.0+1.8+1.5+1.3+1.1+0.9+0.7+0.4+0.2)=-11.2[V]
• Installing with 2 different sort wire.
4.0[mm
]
2
-1.4[V]
220[V]
-(1.4+1.2+1.8+1.5+1.3+1.1+0.9+0.7+0.4+0.2)=-10.5[V]
9[A]
Indoor unit1
10[m]
×i
k
k
10% of input
) <
voltage[V]
2.5[mm
2
]
-2.0[V]
4.0[mm
]
2
-1.2[V]
Indoor unit2
20[m]
············ 2.5[mm
2
] ············
208.8[V](Within 198V~242V)
it's okay
············ 2.5[mm
] ············
2
209.5[V](Within 198V~242V)
1[A]
Indoor unit10
100[m]
it's okay
21
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