LG Multi F A2UW14GFA0 Engineering Product Data Book page 246

Heat pump 50hz/r410a
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2. Distributor type MULTI F DX (1 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)
④ Piping correction factor [from capacity coefficient factor table]
F
main (length, elevation)
F
branch (length, elevation)
⑤ Individual indoor unit combinational capacity
Q
= Q
idu (combi)
⑥ Individual indoor unit actual capacity
Q
= Q
idu (actual)
Example)
• Outdoor unit model : A7UW40GFA0 [FM40AH UO2]
• Total indoor units combination : 7k(A room) + .... + 9k = 48k
• 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
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
= 48 / 40 = 120%
2) From capacity table at T
∴ Q
- F
(Ti, To)
- F
main (length, elevation)
- F
branch (length, elevation)
- Q
idu(combi)
∴Q
36 _ Heat pump
50Hz/R410A
/ Q
odu(Ti, To)
odu(rated)
for main piping length or elevation
for branch piping length or elevation
x Q
/ Q
odu(rated)
idu(rated)
idu(rated-total)
x F
x F
main (length, elevation)
odu(combi)
(Ti, To)
: 11.2 kW
:
, T
i
o
= 10.71 kW
odu (Ti, To)
: 10.71 kW / 11.2 kW = 95.6%
= 89.8% (40m)
= 98.0% (7k, 10m)
7k
[7k] = 11.2kW x ––––– = 1.63 kW
48k
[7k] = 1.63 x 0.956 x 0.898 x 0.98
idu(actual)
= 1.37 kW
x F
branch (length, elevation)

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