●Obtain liquid tubing size from Tables
Main tubing
LA = ø5/8 (ft.) (Total capacity of indoor unit is 179,400 BTU/h) LB = ø1/2 (ft.) (Total capacity of indoor unit is 131,600 BTU/h) LC = ø3/8 (ft.) (Total capacity of indoor unit is 83,800 BTU/h)
The longest tubing length in this example (LA = 131 ft.)
Distribution joint tubing |
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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.
CAUTION
Be sure to check the limit density for
the room in which the indoor unit is
installed.
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.
Min. indoor volume
ft.3
4000 |
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3500 |
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3000 |
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2500 |
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2000 |
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1500 |
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1000 |
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500 |
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0 |
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0 | 200 | 400 | 600 | 800 | 1000 | ||||||
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ft.2
454
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341 | area | ||
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284 | floor | ||
227 | |||
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| 114 | Min. | |
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1200 |
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(when the ceiling is 8.8 ft. high)
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