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06-062 Mini ECO-i II for Technibel
Example:
Main tube of unit
1st branch
Unit distribution tube
G
Example of each tubing length
Main tubing
Distribution joint tubing
LA = 40 m
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 + 3 + 4 + 5 : 23 m × 0.026 kg/m = 0.598 kg
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
3
), the graph at right shows that the minimum
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)
=
20.06 (m
3
)
Therefore, openings such as louvers are required for
this room.
2/8/06
4:32 PM
LA
LB
LC
1
2
model 74R model 94R model 124R model 184R
Indoor side
1 = 5 m
4 = 6 m
2 = 5 m
5 = 5 m
3 = 2 m
: 65 m × 0.056 kg/m = 3.64 kg
Be sure to check the limit density
for the room in which the indoor
unit is installed.
3
(floor area 5.2 m
3
) ≥ 0.3 (kg/m
= 0.39 (kg/m
3
3
)
Page 20
L1
L2
LD
5
model 254R
3
4
Total
4.238 kg
3
m
40
108.0
35
94.5
2
m
30
81.0
67.5
25
2
) for
20
54.0
15
40.5
10
27.0
13.5
5
0
0
)
20
Range below the
density limit
3
of 0.3 kg/m
(countermeasures
not needed)
Range above
the density limit
3
of 0.3 kg/m
(countermeasures
needed)
10
20
30
Total amount of refrigerant
kg
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