LG Multi F A2UW14GFA0 Engineering Product Data Book page 285

3. Distributor type MULTI F DX (3 phase)
3) Calculation of actual system capacity
① Outdoor unit rated capacity
Q [from specification table]
odu(rated)
② Outdoor unit capacity at Ti, To temperature.
Q [from capacity table]
odu(Ti, To)
③ Outdoor unit capacity coefficient factor
F = Q
(Ti, To) odu(Ti, To)
④ Piping correction factor [from capacity voefficient factor table]
F for main piping length or elevation
main (length, elevation)
F
branch (length, elevation)
⑤ Individual indoor unit combinational capacity
Q = Q
idu (combi) odu(rated)
⑥ Individual indoor unit actual capacity
Q = Q
idu (actual) odu(combi)
Example)
• Outdoor unit model : A7UW42LFA0 [FM41AH U32]
• Total indoor units combination : 7k(A room) + .... + 9k = 46k
• Outdoor temperature : 39°CDB
• Indoor temperature : 19.5°CWB (A room)
• Main piping length (O/D to BD) : 40m
• Branch piping length (BD to I/D) : 10m (A room)
• Indoor unit actual cooling capacity in A room?
- Q : 10.9 kW
odu (rated)
- Q :
odu (Ti, To)
1) Combination ratio is calculated from capacity index in Btu/h
Total indoor capacity index / Outdoor unit cooling capacity
= 46 / 42 = 110%
2) From capacity table at T
∴ Q
odu (Ti, To)
- F
(Ti, To)
- F
main (length, elevation)
- F
branch (length, elevation)
- Q
idu(combi)
∴Q
idu(actual)
/ Q
odu(rated)
for branch piping length or elevation
x Q / Q
idu(rated) idu(rated-total)
x F x F x F
main (length, elevation)
(Ti, To)
, T
i o
= 10.64 kW
: 10.64 kW / 10.9 kW = 97.6%
= 89.7% (40m)
= 98.0% (7k, 10m)
7k
[7k] = 10.9kW x ––––– = 1.66 kW
46k
[7k] = 1.66 x 0.976 x 0.897 x 0.98
= 1.42 kW
branch (length, elevation)
Heat pump _ 34
50Hz/R410A