Heat pump
50Hz/R410A _ 34
3. Distributor type MULTI F DX (3 phase)
Part 2 Product data_Outdoor units
3) Calculation of actual system capacity
①
Outdoor unit rated capacity
Q
odu(rated)
[from specification table]
②
Outdoor unit capacity at Ti, To temperature.
Q
odu(Ti, To)
[from capacity table]
③
Outdoor unit capacity coefficient factor
F
(T
i,
To)
= Q
odu(Ti, To)
/ Q
odu(rated)
④
Piping correction factor [from capacity voefficient factor table]
F
main (length, elevation)
for main piping length or elevation
F
branch (length, elevation)
for branch piping length or elevation
⑤
Individual indoor unit combinational capacity
Q
idu (combi)
= Q
odu(rated)
x Q
idu(rated)
/ Q
idu(rated-total)
⑥
Individual indoor unit actual capacity
Q
idu (actual)
= Q
odu(combi)
x F
(Ti, To)
x F
main (length, elevation)
x F
branch (length, elevation)
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
odu (rated)
: 10.9 kW
- 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
i
, T
o
∴ Q
odu (Ti, To)
= 10.64 kW
- F
(Ti, To)
: 10.64 kW / 10.9 kW = 97.6%
- F
main (length, elevation)
= 89.7% (40m)
- F
branch (length, elevation)
= 98.0% (7k, 10m)
7k
- Q
idu(combi)
[7k] = 10.9kW x ––––– = 1.66 kW
46k
∴Q
idu(actual)
[7k] = 1.66 x 0.976 x 0.897 x 0.98
= 1.42 kW
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