Gree GMV-615WM/B-X

D.C. Inverter Multi VRF Modular
30
How much additional refrigerant should be charged
Total refrigerant charging amount R= Pipeline charging amount A + ∑charging amount B of
every module
(1) Pipeline charging amount
Pipeline charging amount A= ∑Liquid pipe length×refrigerant charging amount of every 1m
liquid pipe
Diameter of liquid
pipe (mm)
Φ28.6
Φ25.4
Φ22.2
Φ19.05
Φ15.9
Φ12.7
Φ9.52
Φ6.35
kg/m
0.680
0.520
0.350
0.250
0.170
0.110
0.054
0.022
(2) ∑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
included IDUs
22.4
28.0
33.5
40.0
45.0
50.4
56.0
61.5
50%≤C≤70%
＜4
0
0
0
0
0
0
0
0
≥4
0.5
0.5
0.5
0.5
0.5
0.5
1.0
1.5
70%＜C≤90%
＜4
0.5
0.5
1
1.5
1.5
1.5
2.0
2
≥4
1
1
1.5
2
2
2.5
3.0
3.5
90%＜C≤105%
＜4
1
1
1.5
2
2
2.5
3.0
3.5
≥4
2
2
3
3.5
3.5
4.0
4.5
5
105%＜C≤135%
＜4
2
2
2.5
3
3
3.5
4
4
≥4
3.5
3.5
4
5
5
5.5
6
6
For example1:
The OUD is composed of 3 modules: GMV-280WM/B-X, GMV-400WM/B-X, and
GMV-450WM/B-X. The IDUs are made up of 8 sets of GMV-ND140PLS/A-T.
IDU/ODU rated capacity collocation ratio C= 140×8/(280+400+450)=108%. The quantity of
included IDUs is more than 4 sets. Please refer to the above table.
Refrigerant charging amount B for GMV-280WM/B-X module is 3.5kg
Refrigerant charging amount B for GMV-400WM/B-X module is 5.0kg
Refrigerant charging amount B for GMV-450WM/B-X module is 5.0kg
So, ∑Refrigerant charging amount B of every module=3.5+5.0+5.0=13.5kg
Suppose the Pipeline charging amount A=∑Liquid pipe length×refrigerant charging amount of
every 1m liquid pipe=25kg
Total refrigerant charging amount R=25+13.5=38.5kg
For example 2:
The outdoor unit is made up of 1 module: GMV-450WM/B-X. The indoor unit is 1 set of fresh
air indoor unit: GMV-NX450P/A(X4.0)-M. In this case, the quantity (B) of refrigerant added to this
module is 0kg.
So,
∑
B (Quantity of refrigerant added to each module) = 0kg