Samsung AM Series - Air Filter Cleaning

14
Example of Installation
- Total power cable length L = 100(m), Running current of each units 1[A]
- Total 10 indoor units were installed
 Apply following equation.
∑ (
Coef×35.6×Lk×ik
) <
10% of input
voltage[V]
1000×Ak
n
k=1
h 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 different 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
]
Power supply MCCB ELB or ELCB Power cable Earth cable Communication cable
Max : 242V
Min : 198V
X A
X A, 30mmA
0.1 s
2.5mm
2
2.5mm
2
0.75~1.5mm
2
 Decide the capacity of ELCB(or MCCB+ELB) by below formula.
 Power supply cords of parts of appliances for outdoor use shall not be lighter than polychloro-
prene sheathed exible cord.
(Code designation IEC:60245 IEC 57 / CENELEC: H05RN-F or IEC:60245 IEC 66 / CENELEC: H07RN-F )
The capacity of ELCB(or MCCB+ELB) X [A] = 1.25 X 1.1 X ∑Ai
T X : The capacity of ELCB(or MCCB+ELB).
T
∑
Ai : Sum of Rating currents of each indoor unit.
T Refer to each installation manual about the rating current of indoor unit.
 Decide the power cable specification and maximum length within 10% power drop among
indoor units.
T coef: 1.55
T Lk: Distance among each indoor unit[m], Ak: Power cable specification[mm
2
]
ik: Running current of each unit[A]
∑ (
Coef×35.6×Lk×ik
) <
10% of input voltage[V]
1000×Ak
n
k=1
10[A]
Indoor
unit10
20[m]10[m]0[m]
9[A] 1[A]
100[m]
Indoor unit2Indoor unit1
ELCB
Or
MCCB+
ELB
Specication of electronic wire
Wiring work