5-28
PERFORMANCE DATA
SM TQDCI 1-A.1 GB
5.14
Model Correction Factors (F
M
)
Model
Capacity
Power input
Cooling
Heating
Cooling
Heating
FLO DCI
1.00
1.00
1.00
1.00
K DCI
1.03
1.07
1.01
1.10
SX DCI
TBD
TBD
TBD
TBD
BS DCI
TBD
TBD
TBD
TBD
15.15 Calculation
Example
Outdoor Unit
Quattro-80 DCI
Indoor Combination
FLO 9+FLO 12+K 12+FLO 18
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
C
x F
T
Total System Capacity [KW] (TC) = C
A
+ C
B
+ C
C
+ C
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 – FLO 9
1.43
1.00
0.92
0.95
C
A
= 1.43x1.00x0.92x0.95=
1.25
Room B – FLO 12
1.91
1.00
0.92
0.985
C
B
= 1.91x1.00x0.92x0.985=
1.73
Room C – K 12
1.91
1.03
0.92
1.00
C
C
= 1.91x1.03x0.92x1.00=
1.81
Room D – FLO 18
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 – FLO 9
2,410 / 4 = 602.5
1.00
0.88
P
A
= 602.5 x 1.00 x 0.88 = 530
Room B – FLO 12
1.00
0.88
P
B
= 602.5 x 1.00 x 0.88 = 530
Room C – K 12
1.01
0.88
P
C
= 602.5 x 1.01 x 0.88 = 535
Room D – FLO 18
1.00
0.86
P
D
= 602.5 x 1.00 x 0.86 = 518
Total
TP
= 530 + 530 + 535 + 518 =
2,113
CONTENTS
