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13. Installation
▶ Decide the power cable specification and maximum length within 10 % power drop among indoor units.
• coef : 1.55
Lk : Distance among each indoor unit [m (inch)], Ak: Power cable specification[mm
ik : Running current of each unit [A]
Example of Installation
▶ Total power cable length L = 100 m (328.1 ft), Running current of each units 1[A]
▶ Total 10 indoor units were installed
MCCB+
ELCB
ELB
Or
0 m (0 ft)
f Apply following equation
n
∑(
k=1
❋ Calculation
• Installing 1 type of wire .
2 .5 mm
220 [V]
• Installing with 2 different sort wire .
4 .0 mm
220 [V]
n
Coef×35.6×Lk×ik
∑(
1000×Ak
k=1
10[A]
Indoor unit
1
10 m (32 .8 ft)
Coef×35 .6×Lk×ik
1000×Ak
(0 .0039 inch
2
) 2 .5 mm
2
(0 .0039 inch
2
-2 .2 [V]
-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]
2
(0 .0062 inch
2
) 4 .0 mm
2
(0 .0062 inch
-1 .4 [V]
-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]
) < 10 % of input voltage[V]
9[A]
Indoor unit
2
20 m (65 .6 ft)
) < 10 % of input voltage[V]
2
)
· ··········· 2 .5 mm
2
(0 .0039 inch
· ··········· 2 .5 mm
2
2
)
2
(inch
2
)]
1[A]
Indoor unit
10
100 m (328 .1 ft)
2
) ············
208 .8 [V] (Within 187 V~253 V)
it's okay
(0 .0039 inch
2
) ············
209 .5 [V] (Within 187 V~253 V)
it's okay
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