GMV DC Inverter VRF
38
Diameter of liquid
pipe
(mm)
Φ
28.6
Φ
25.4
Φ
22.2
Φ
19.05
Φ
15.9
Φ
12.7
Φ
9.52
Φ
6.35
kg/m
0.680
0.520
0.350
0.250
0.170
0.110
0.054
0.022
(2)
Σ Refrigerant charging amount B of every module
Refrigerant charging amount B of every
module (kg)
②
Module capacity(kW)
IDU/ODU rated capacity
collocation ratio C
①
Quantity of
indoor unit
22.4
28
33.5
40
45
50.4
56
61.5
50%
≤C≤70%
＜
4
0
0
0
0
0
0
0
0
≥
4
0.5
1
1
1
1
0.5
1
1.5
70%
＜
C
≤90%
＜
4
0.5
1
1
2
2
1.5
2
2
≥
4
1
1
1
2
2
2.5
3
3.5
90%
＜
C
≤
105%
＜
4
1
1
1
2
2
2.5
3
3.5
≥
4
2
2
2
4
4
4
5
5
105%
＜
C
≤
135%
＜
4
2
2
2
3
3
3.5
4
4
≥
4
3.5
4
4
5
5
5.5
6
6
NOTES!
①
Rated capacity configuration rate of indoor unit and outdoor unit C = sum of indoor unit rated
cooling capacity / sum of outdoor unit rated cooling capacity.
②
If all indoor units are all fresh air indoor units, the added refrigerant amount for each module B
is 0kg.
③
If all fresh air indoor units are mixed with the general VRF indoor units, charge the refrigerant
according to the refrigerant-charging method of the general indoor unit.
For example1:
Outdoor unit consists of one 28kW module and one 45kW module. Five 14kW duct type units are
used as indoor units.
IDU/ODU rated capacity collocation ratio C= 14.0×5/ (28.0+45.0) =96%. The quantity of included
IDUs is more than 4 sets. Please refer to the above table.
Additional refrigerant quantity B for 28kW module is 2.0 kg.
Additional refrigerant quantity B for 45kw module is 4 kg.
So, ∑Refrigerant charging amount B of every module=2.0+4=6 kg.
Suppose the Pipeline charging amount A=∑Liquid pipe length × refrigerant charging amount of
every 1m liquid pipe=20kg.
Total refrigerant charging amount R=20+6=26kg.
For example 2:
Outdoor unit is a 45kW module and the indoor unit is a 45kW fresh air unit. The quantity (B) of
refrigerant added to this module is 0kg.
So,
∑Refrigerant charging amount B of every module= 0kg.
Suppose the Pipeline charging amount A=∑Liquid pipe length × refrigerant charging amount of
every 1m liquid pipe = 5kg.
