Specifications

24
1-16. Example of Tubing Size Selection and Refrigerant Charge Amount
Additional refrigerant charging
Based on the values in Tables 1-11a, 11b, 12, 13 and 16, use the narrow tubing size and length, and calculate the
amount of additional refrigerant charge using the formula below.
= [366
× (a) + 259 × (b) + 185 × (c) + 128 × (d) + 56 × (e) + 26 × (f)] × 10
–3
(a) : Narrow tubing Total length of φ22.22 (m) (d) : Narrow tubing Total length of φ12.7 (m)
(b) : Narrow tubing Total length of
φ19.05 (m) (e) : Narrow tubing Total length of φ9.52 (m)
(c) : Narrow tubing Total length of
φ15.88 (m) (f) : Narrow tubing Total length of φ6.35 (m)
● Charging procedure
Be sure to charge with R410A refrigerant
in liquid form.
1. After performing a vacuum, charge with refrigerant from the narrow tubing side. At this time, all valves must
be in the “fully closed” position.
2. If it was not possible to charge the designated amount, operate the system in Cooling mode while charging
with refrigerant from the wide tubing side. (This is performed at the time of the test run. For this, all valves
must be in the “fully open” position.)
Charge with R410A refrigerant in liquid form.
With R410A refrigerant, charge while adjusting the amount being fed a little at a time in order to prevent
liquid refrigerant from backing up.
● After charging is completed, turn all valves to the “fully open” position.
● Replace the tubing covers as they were before.
Required additional
refrigerant charge (kg)
CAUTION
1. R410A additional charging absolutely must be done
through liquid charging.
2. The R410A refrigerant cylinder has a gray base
color, and the top part is pink.
3. The R410A refrigerant cylinder includes a siphon
tube. Check that the siphon tube is present. (This is
indicated on the label at the top of the cylinder.)
4. Due to differences in the refrigerant, pressure, and
refrigerant oil involved in installation, it is not possi-
ble in some cases to use the same tools for R22 and
for R410A.
1 2 3 4 5 6 7
LO
LA
LB
LC
LD
LE
LF
A
B
C
Outdoor unit
08
model
10
model
10
model
106 model 106 model 140 model 140 model
140 model
140 model 73 model
Main unit
Example:
● Example of each tubing length
Main tubing Distribution joint tubing
LO = 2 m LD = 15 m Outdoor side Indoor side
LA = 40 m LE = 10 m A = 1.5 m 1 = 30 m 5 = 2 m
LB = 5 m LF = 10 m B = 1.5 m 2 = 5 m 6 = 6 m
LC = 5 m C = 2 m 3 = 5 m 7 = 5 m
4 = 5 m
04-396 Inverter-W_a-p50 ARGO 12/21/04 9:38 AM Page 24