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Room A
2.4
26
2.8
Room B
2.8
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
2.8
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
2.8
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: RVMH100GBM0
Indoor unit: AVMKH026EA0 x 2,AVMKH035EA0 x 2,AVMCH052EA0 x 1,AVMCH070EA0 x 1
2) Indoor unit combination total capacity index
26x2 + 35x2 + 52x1 + 70x1 = 244, (244/280) x 100% = 87%
3) Result: Because it is within 50~100%, it is 'Right' selection.
(4) Real function data with indoor unit combination
1) For the 87% combination, calculate the cooling capacity of outdoor unit (RVMH100GBM0).
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 (RVMH100GBM0) 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).
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 E-D/B(chapter4-1)-E<03759 3/21/02 7:39 PM Page 3