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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 161 ft.
2
× ceiling
height 8.8 ft. = room volume 1417 ft.
3
), the graph at right
shows that the minimum room volume should be 2455 ft.
3
(floor area 279 ft.
2
) for refrigerant of 418 oz.
<Determination by calculation>
Overall refrigerant charge amount for the air conditioner: oz
(Minimum room volume for indoor unit: ft.
3
)
=
418 (oz)
= 0.29 (oz/ft.
3
) < 0.3 (oz/ft.
3
)
1417 (ft.
3
)
Therefore, openings such as louvers are not required for this
room.
00
57
114
170
227
284
341
398
454
0
500
1000
1500
2000
2500
3000
3500
4000
400200 600 800 1000 1200
Total amount of refrigerant
Min. indoor volume
Min. indoor floor area
(when the ceiling is 8.8 ft. high)
ft.
3
ft.
2
oz
Range above
the density limit of
0.3 oz/ft.
3
(countermeasures
needed)
Range below
the density limit of
0.3 oz/ft.
3
(countermeasures
not needed)
● Obtain liquid tubing size from Tables 1-9, 10, 11, 12 and 15.
Main tubing
LA = ø5/8 (ft.) (Total capacity of indoor unit is 179,400 BTU/h) The longest tubing length in this example
LB = ø1/2 (ft.) (Total capacity of indoor unit is 131,600 BTU/h) (LA = 131 ft.)
LC = ø3/8 (ft.) (Total capacity of indoor unit is 83,800 BTU/h)
Distribution joint tubing
Outdoor side A: ø3/8 (ft.) B: ø3/8 (ft.) (from outdoor unit connection tubing)
Indoor side 1: ø3/8 (ft.) 2: ø3/8 (ft.) 3: ø3/8 (ft.) 4: ø3/8 (ft.) (from indoor unit connection tubing)
● Obtain charge amount for each tubing size
Note that the charge amounts per 1 meter are different for each liquid tubing size.
ø5/8 (ft.) → LA : 131 ft. × 1.99 oz/ft. = 261 oz
ø1/2 (ft.) → LB : 16 ft. × 1.38 oz/ft. = 22 oz
ø3/8 (ft.) → LC + A – B + 1 – 4 : 225 ft. × 0.602 oz/ft. = 135 oz
Total 418 oz
Additional refrigerant charge amount is 418 oz.
Be sure to check the limit density for
the room in which the indoor unit is
installed.
CAUTION
07-007 W-2WAY_II_NA 1/23/07 6:09 PM Page 18