1
3
. Installation
▶ Decide the power cable specification and maximum length within 10 % power drop among indoor units.
n
∑(
k=1
) < 10 % of input voltage[V]
Coef×35.6×Lk×ik
1000×Ak
• coef : 1.55
Lk : Distance among each indoor unit [m (inch)], Ak: Power cable specification[mm
2
(inch
2
)]
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
10[A]
Indoor unit
10
20 m (65 .6 ft)
10 m (32 .8 ft)
0 m (0 ft)
9[A]
1[A]
100 m (328 .1 ft)
Indoor unit
2
Indoor unit
1
ELCB
MCCB+
ELB
Or
f
Apply following equation
n
∑(
k=1
Coef×35 .6×Lk×ik
) < 10 % of input voltage[V]
1000×Ak
❋
Calculation
•
Installing 1 type of wire .
220 [V]
-2 .2 [V]
·
··········· 2 .5 mm
2
(0 .0039 inch
2
) ············
-(2 .2+2 .0+1 .8+1 .5+1 .3+1 .1+0 .9+0 .7+0 .4+0 .2)=-11 .2 [V]
208 .8 [V] (Within 187 V~253 V)
it's okay
2 .5 mm
2
(0 .0039 inch
2
) 2 .5 mm
2
(0 .0039 inch
2
)
•
Installing with 2 different sort wire .
220 [V]
-1 .4 [V]
·
··········· 2 .5 mm
2
(0 .0039 inch
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]
209 .5 [V] (Within 187 V~253 V)
it's okay
4 .0 mm
2
(0 .0062 inch
2
) 4 .0 mm
2
(0 .0062 inch
2
)
-2 .0 [V]
