3
ΙΛ
Room A
2.4
26
2.8
Room B
2.6
26
2.8
Room C
3.2
35
3.77
Room D
3.4
35
3.77
Room E
4.8
52
5.6
Room F
7.2
70
7.54
Location
Load (kW)
Unit size
Capacity(kW)
Room A
2.4
26
2.66
Room B
2.6
26
2.66
Room C
3.2
35
3.57
Room D
3.4
35
3.57
Room E
4.8
52
5.31
Room F
7.1
70
7.15
Location
Load (kW)
Unit size
Capacity(kW)
Location
Load (kW)
Room A
2.4
Room B
2.6
Room C
3.2
Room D
3.4
Room E
4.8
Room F
7.2
1-4. Example for unit selection with cooling load
(1) Given condition
1) Design condition <Cooling: Indoor 20°C(WB), Outdoor 33°C(DB)>
2) Cooling load
3) Power supply unit: 3 Phase 380V 50Hz
4) Pipe length: 30m
(2) Indoor unit selection
Select the suitable capacity for condition of 'Indoor 20°C(WB), Outdoor 33°C(DB)' using
indoor unit capacity table. The selected result is as follow.
(3) Outdoor unit selection
1) Assume the indoor unit and outdoor unit combination as follow.
Outdoor unit: RVMH100GAM0
Indoor unit: AVMKH026EA0 x 2,AVMKH035EA0 x 2,AVMCH052EA0 x 1,AVMCH070EA0 x 1
2) Indoor unit combination total capacity index
2.6x2 + 3.5x2 + 5.2x1 + 7.0x1 = 24.4, (24.4/28.0) x 100% = 87%
3) Result: Because it is within 50~130%, it is 'Right' selection (Within 100% for the Middle East).
(4) Real function data with indoor unit combination
1) For the 87% combination, calculate the cooling capacity of outdoor unit (RVMH100GAM0).
27.14kW ← 90% (Indoor temperature: WB 20°C, Outdoor temperature: DB 33°C)
24.12kW ← 80% (Indoor temperature: WB 20°C, Outdoor temperature: DB 33°C)
Therefore, 26.23 = 24.12 + {(27.14-24.12)/10}x7: Calculated in 87%
2) Outdoor unit (RVMH100GAM0) cooling capacity: 26.23kW ← 87%
(Indoor temperature: WB 20°C, Outdoor temperature: DB 33°C)
3) Capacity change factor with pipe length (30m): 0.95(30m) (Refer to page IV-45.)
4) Each cooling capacity
AVMKH026EA0: 26.23 x 26/244 x 0.95 = 2.66(kW)
AVMKH035EA0: 26.23 x 35/244 x 0.95 = 3.57(kW)
AVMCH052EA0: 26.23 x 52/244 x 0.95 = 5.31(kW)
AVMCH070EA0: 26.23 x 70/244 x 0.95 = 7.15(kW)
DVM TM(chapter4-1)_E<03759 6/14/02 2:52 PM Page 3