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Example:
8 HP
LA
LC
LB
1
2
type
type
22
● Example of each tubing length
Main tubing
Distribution joint tubing
LA = 40 m
Indoor side
LB = 5 m
1 = 5 m
LC = 5 m
2 = 5 m
LD = 15 m
3 = 2 m
● 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 : 65 m × 0.056 kg/m = 3.64 kg
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
Be sure to check the limit density for the room in which the
indoor unit is installed.
LD
5
type
3
4
56
type
type
28
36
45
4 = 6 m
5 = 5 m
Total 4.238 kg
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 area 8.00 m
ceiling height 2.7 m = room volume 21.6 m
room volume should be 23.3 m
for refrigerant of 10.538 kg (6.3 kg + 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) + 6.3 (kg)
=
21.6 (m
)
3
Therefore, openings such as louvers are required for this room.
2
3
m
m
85
229.5
80
216.0
75
202.5
70
189.0
65
175.5
60
162.0
55
148.5
50
135.0
45
121.5
40
108.0
35
94.5
30
81.0
25
67.5
20
54.0
15
40.5
10
27.0
5
13.5
0
0.0
0
10
17
3
), the minimum
3
(10.238 kg ÷ 0.44 kg/m
= 0.49 (kg/m
)
0.44 (kg/m
3
3
Range below the
density limit of
0.44 kg/m³
(Countermeasures
not needed)
Range above the
density limit of
0.44 kg/m³
(Countermeasures
needed)
20
30
40
50
60
70
Total amount of refrigerant
2
×
3
)
3
)
)
80
90 100
kg
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