2. Distributor type MULTI F DX (1 phase)
36
_ Heat pump
50Hz/R410A
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
(Ti, To)
=
Q
odu(Ti, To)
/
Q
odu(rated)
④
Piping correction factor [from capacity coefficient 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 : 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)
:
11.2 kW
- 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
i
, T
o
∴
Q
odu (Ti, To)
= 10.71 kW
-
F
(Ti, To)
: 10.71 kW / 11.2 kW =
95.6%
-
F
main (length, elevation)
=
89.8%
(40m)
-
F
branch (length, elevation)
=
98.0%
(7k, 10m)
7k
-
Q
idu(combi)
[7k] = 11.2kW x ––––– = 1.63 kW
48k
∴
Q
idu(actual)
[7k] = 1.63 x 0.956 x 0.898 x 0.98
= 1.37 kW
