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• Decide the power cable specication and maximum length within 10% power drop among indoor units.
n
Σ
k=1
(
Coef×35.6×L
k×ik
) < 10% of input voltage[V]
1000×Ak
Coef: 1.55
✴ L
k: Distance among each indoor unit[m]
A
k: Power cable specication[mm
2
]
ik: Running current of each unit[A]
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]
Indoor unit10
20[m]10[m]0[m]
9[A] 1[A]
100[m]
Indoor unit2Indoor unit1
ELCB
or
MCCB+
ELB
• Apply following equation.
n
Σ
k=1
(
Coef×35.6×L
k×ik
) <
10% of input
voltage[V]
1000×A
k
❈ Calculation
• Installing with 1 sort wire.
220[V]
-2.2[V] -2.0[V]
············ 2.5[mm
2
] ············
208.8[V](Within 198V~242V)
-(2.2+2.0+1.8+1.5+1.3+1.1+0.9+0.7+0.4+0.2)=-11.2[V]
it's okay
2.5[mm
2
] 2.5[mm
2
]
• Installing with 2 dierent sort wire.
220[V]
-1.4[V] -1.2[V]
············ 2.5[mm
2
] ············
209.5[V](Within 198V~242V)
-(1.4+1.2+1.8+1.5+1.3+1.1+0.9+0.7+0.4+0.2)=-10.5[V]
it's okay
4.0[mm
2
] 4.0[mm
2
]
NASA_Big Duct_AM@@@FNHDEH@_IM_03864A-05_EN.indd 23 2016-04-01 오후 5:29:44