Technibel CAFM1V Series Technical Data Manual page 27

2. System Design
Example:
Main tube of unit
1st branch
Unit distribution tube
2
G Example of each tubing length
Main tubing Distribution joint tubing
LA = 40 m Indoor side
LB = 5 m
LC = 5 m
LD = 15 m
G
Obtain charge amount for each tubing size
Note that the charge amounts per 1 meter are different for each liquid tubing size.
ø9.52 → LA + LB + LC + LD
ø6.35 →
+
1 2
Additional refrigerant charge amount is 4.238 kg.
CAUTION
Checking of limit density
Density limit is determined on the basis of the size of a
room using an indoor unit of minimum capacity. For
instance, when an indoor unit is used in a room (floor
× ceiling height 2.7 m = room volume
area 7.43 m 2
20.06 m ), the graph at right shows that the minimum
3
room volume should be 14.1 m
refrigerant of 4.238 kg. Accordingly, openings such as
louvers are required for this room.
<Determination by calculation>
Overall refrigerant charge amount for the air conditioner: kg
(Minimum room volume for indoor unit: m
4.238 (kg) + 3.5 (kg)
=
3
20.06 (m )
Therefore, openings such as louvers are required for
this room.
LA
LB LC
2
1
model 7 model 9
= 5 m = 6 m
1 4
= 5 m = 5 m
2 5
= 2 m
3
: 65 m × 0.056 kg/m = 3.64 kg
: 23 m × 0.026 kg/m = 0.598 kg
+ + +
3 4 5
Be sure to check the limit densi-
ty for the room in which the
indoor unit is installed.
3 (floor area 5.2 m 2 ) for
3
)
) ≥ 0.3 (kg/m
= 0.39 (kg/m 3 3 )
Design of Mini Multiset
L1
L2
LN
n
model 18
3
n–1
model 12 model 16
Total 4.238 kg
3
m
40
108.0
Range below the
35
94.5
density limit
2
m
of 0.3 kg/m
30
81.0
(countermeasures
not needed)
25
67.5
54.0 20
15
40.5
27.0 10
13.5 5
0 0
2 - 20
3
Range above
the density limit
3
of 0.3 kg/m
(countermeasures
needed)
10 20 30
kg
Total amount of refrigerant