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Example of Installation
- Total power cable length L = 328 ft(100 m), Running current of each units 1[A]
- Total 10 indoor units were installed
ELCB
or MCCB+
ELB
0 ft (0 m)
• Apply following equation.
n
Coef×35.6×L
Σ
(
1000×A
k=1
❈ Calculation
• Installing with 1 sort wire.
0.0039 inch
(2.5 mm
2
-2.2[V]
220[V]
• Installing with 2 different sort wire.
0.0062 inch
(4.0 mm
2
-1.4[V]
220[V]
AM0**JNZDCH/AA IM ENG_V1.5_07082015
9[A]
10[A]
Indoor unit1
32.8 ft (10 m)
×i
k
k
10 % of input
) <
voltage[V]
k
)
0.0039 inch
(2.5 mm
)
2
2
2
-2.0[V]
-(2.2+2.0+1.8+1.5+1.3+1.1+0.9+0.7+0.4+0.2)=-11.2[V]
)
0.0062 inch
(4.0 mm
)
2
2
2
-1.2[V]
-(1.4+1.2+1.8+1.5+1.3+1.1+0.9+0.7+0.4+0.2)=-10.5[V]
Indoor unit2
65.6 ft (20 m)
············ 0.0039 inch
(2.5 mm
) ············
2
2
208.8[V](Within 198 V~242 V)
············ 0.0039 inch2 (2.5 mm2) ············
209.5[V](Within 198 V~242 V)
24
1[A]
Indoor unit10
328 ft (100 m)
it's okay
it's okay
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