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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
80 ft
2
× ceiling height 8.8 ft. = room volume 704 ft.
3
), the
graph at right shows that the minimum room volume
should be 497 ft.
3
(floor area 56 ft.
2
) for refrigerant of 147
oz. Accordingly, openings such as louvers are required for
this room.
<Determination by calculation>
Overall refrigerant charge amount for the air conditioner: oz
(Minimum room volume for indoor unit: ft.
3
)
=
147 (oz) + 123 (oz)
= 0.38 (oz/ft.
3
) > 0.3 (oz/ft.
3
)
704 (ft.
3
)
Therefore, openings such as louvers are required for this
room.
● Obtain charge amount for each tubing size
Note that the charge amounts per 3.3 ft. are different for each liquid tubing size.
ø3/8" (ø9.52) → LA + LB + LC + LD : 212 ft. × 0.602 oz/ft. = 127 oz
ø1/4" (ø6.35) → 1 + 2 + 3 + 4 + 5 : 75 ft. × 0.279 oz/ft. = 20 oz
Total 147 oz
Additional refrigerant charge amount is 147 oz.
Be sure to check the limit
density for the room in which
the indoor unit is installed.
CAUTION
● Example of each tubing length
Main tubing Distribution joint tubing
LA = 131 ft. Indoor side
LB = 16 ft. 1 = 16 ft. 4 = 20 ft.
LC = 16 ft. 2 = 16 ft. 5 = 16 ft.
LD = 49 ft. 3 = 7 ft.
LN
L1
L2
LCLB
LA
Main tube of unit
1st branch
model 0752 model 0952 model 1252 model 1852
model 1852
Unit distribution tube
1
2 3
n–1
n
Example:
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)
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07-056 Mini_ECOi_II_NA 3/19/07 2:40 PM Page 16