25
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 15 m
3
×
ceiling height 2.7 m = room volume 40.5
m
3
), the graph at right shows that the minimum room
volume should be 70.3 m
3
(floor area 26 m
2
) for refriger-
ant of 21.093 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
3
)
=
21.093 (kg)
= 0.52 (kg/m
3
)
≥
0.3 (kg/m
3
)
40.5 (m
3
)
Therefore, openings such as louvers are required for
this room.
40.5
54.0
20
30
40
50
60
70
80 kg
67.5
81.0
94.5
108.0
121.5
135.0
148.5
162.0
175.5
189.0
202.5
216.0
229.5
243.0
256.5
270.0
m
2
m
3
Total amount of refrigerant
Min. indoor floor area
(when the ceiling is 2.7 m high)
Min. indoor volume
Range below the
density limit of
0.3 kg/m
3
(Countermeasures
not needed)
Range above the
density limit of
0.3 kg/m
3
(Countermeasures
needed)
15
20
25
30
35
40
45
50
55
60
65
70
75
80
85
90
95
100
●
Obtain narrow tubing size from Tables 1-11a, 11b, 12, 13 and 16.
Main tubing
LO =
φ
15.88 m (Total capacity of outdoor unit is 50.4 kW)
LA =
φ
19.05 m (Total capacity of indoor unit is 88.4 kW)
The longest tubing length in this example
LB =
φ
19.05 m (Total capacity of indoor unit is 77.2 kW)
(LM = 40 + 5 = 45 m)
LC =
φ
15.88 m (Total capacity of indoor unit is 66.0 kW)
LD =
φ
15.88 m (Total capacity of indoor unit is 52.0 kW)
LE =
φ
12.7 m (Total capacity of indoor unit is 38.0 kW)
LF =
φ
9.52 m (Total capacity of indoor unit is 22.0 kW)
Distribution joint tubing
Outdoor side
A:
φ
9.52
B:
φ
9.52
C:
φ
9.52 (from outdoor unit connection tubing)
Indoor side
1:
φ
9.52
2:
φ
9.52
3:
φ
9.52
4:
φ
9.52
5:
φ
9.52
6:
φ
9.52
7:
φ
9.52 (from indoor unit connection tubing)
●
Obtain charge amount for each tubing size
Note that the charge amounts per 1 meter are different for each narrow tubing size.
φ
19.05
→
LA + LB
: 45m
×
0.259 kg/m = 11.655
φ
15.88
→
LO + LC + LD
: 22 m
×
0.185 kg/m = 4.07
φ
12.7
→
LE
: 10 m
×
0.128 kg/m = 1.28
φ
9.52
→
LF +
A – C +
1 – 7 : 73 m
×
0.056 kg/m = 4.088
Total 21.093 kg
Additional refrigerant charge amount is 21.093 kg.
Be sure to check the limit densi-
ty for the room in which the
indoor unit is installed.
CAUTION
04-396 Inverter-W_a-p50 ARGO 12/21/04 9:38 AM Page 25
