Service manual

K DCI
PXD DCI
LS DCI
PERFORMANCE DATA
5.14 Model Correction Factors (F
M
)
Model
Capacity Power input
Cooling Heating Cooling Heating
WNG DCI
1.00 1.00 1.00 1.00
ECF DCI
1.03 1.07 1.01 1.10
PXD DCI
TBD TBD TBD TBD
LS DCI
TBD TBD TBD TBD
15.15 Calculation Example
Outdoor Unit
Quattro DCI
Indoor Combination
WNG9+WNG12+ECF12+WNG18
Operation Mode
Cooling Mode
Conditions Indoor
22°CDB/15°WB
Conditions Oudoor
30°CDB
Tubing length
20m+10m+5m+25m
Cooling Capacity calculation:
C
A-D
[KW] = Nominal x F
M
x F x F
TC
Total System Capacity [KW] (TC) = C
A
+ C + C
C
+ C
B D
Indoor Unit
Nom’ Cooling
Capacity
[KW]
Model
Factor
(F
M
)
Condition
Factor
(F
C
)
Tubing(L)
Factor
(F
T
)
Corrected Capacity
[KW], (C
A-D
)
Room A – WNG9 1.43 1.00 0.92 0.95 C
A
= 1.43x1.00x0.92x0.95=1.25
Room B – WNG12 1.91 1.00 0.92 0.985 C
B
= 1.91x1.00x0.92x0.985=1.73
Room C – ECF12 1.91 1.03 0.92 1.00 C
C
= 1.91x1.03x0.92x1.00=1.81
Room D – WNG18 2.87 1.00 0.93 0.93 C
D
= 2.87x1.00x0.93x0.93=2.48
Total TC =1.25+1.73+1.81+2.48=7.27
Cooling Power Input calculation:
P
A-D
[KW] = Nominal x F
M
x F
C
x F
T
Total System Power Input [W] (TP) = P
A
+ P
B
+ P
C
+ P
D
Indoor Unit
Nom’ Cooling
Power Input
[W]
Model
Factor
(F
M
)
Condition
Factor
(F
C
)
Corrected Power Input [W]
(P
A-D
)
Room A – WNG9 1.00 0.88 P
A
= 602.5 x 1.00 x 0.88 = 530
Room B – WNG12 1.00 0.88 P
B
= 602.5 x 1.00 x 0.88 = 530
Room C – ECF12
2,410 / 4 = 602.5
1.01 0.88 P
C
= 602.5 x 1.01 x 0.88 = 535
Room D – WNG18 1.00 0.86 P
D
= 602.5 x 1.00 x 0.86 = 518
Total TP = 530 + 530 + 535 + 518 =
2,113
5-26
Revision 0