Design
D - 3
1. System Configuration
(2) Calculation of actual performance
● Indoor units up to a maximum capacity of 200% (24 units) of the outdoor unit capacity can be connected.
● The performance of a multi air conditioner system varies with temperature conditions, the piping length, the difference
between highest and lowest points, etc. The model type should be selected having taken each correction factor into account.
1 Actual performance of a multi-type air conditioning system depending on installation conditions
1) Indoor unit cooling capability=( k
1
Outdoor unit rated cooling capacity)×(k
3
Indoor unit rated cooling capacity)
÷(❋
5
Total rated cooling capacity of the indoor units)
×(❋
7
Correction factor for temperature and connected capacity, from the performance characteristics)
×(❋
8
Correction factor for piping length)
2) Indoor unit heating capability=(k
2
Outdoor unit rated heating capacity)×(k
4
Indoor unit rated heating capacity)
÷(❋
6
Total rated heating capacity of the indoor units)
×(❋
7
Correction factor for temperature and connected capacity, from the performance characteristics)
×(❋
8
Correction factor for piping length)
❋
1
Outdoor unit rated total cooling capacity (see the outdoor unit specification table) is the cooling capacity under JIS
conditions (indoor side: 27
o
CDB, 19
o
CWB, outdoor side: 35
o
CDB)
❋
2
Outdoor unit rated total heating capacity (see the outdoor unit specification table) is the heating capacity under JIS
conditions (indoor side: 20
o
CDB, outdoor side: 7
o
CDB , 6
o
CWB)
❋
3
Read the rated cooling capacity of the applicable indoor unit from the indoor unit specification table.
❋
4
Read the rated heating capacity of the applicable indoor unit from the indoor unit specification table.
❋
5
Read the rated cooling capacity of the applicable indoor unit from the indoor unit specification table, and obtain the total
for all units.
❋
6
Read the rated heating capacity of the applicable indoor unit from the indoor unit specification table, and obtain the total
for all units.
❋
7
Read the percentage data at the required temperature from the relevant capacity table in “Model basic data table” for the
outdoor unit, and divide it by 100.
❋
8
Correction factor for piping length
Design the effective refrigerant piping and the difference in height between the outdoor unit and indoor units (with the indoor
unit as standard, plus when the outdoor unit is higher, minus when the indoor unit is higher). Read the correction factor as a
percentage from the “Performance change rate due to refrigerant piping length” for the outdoor unit, and divide by 100.
2 Example of calculation of actual performance
<Example calculation conditions>
• Indoor units: one of each type 28, 36, 45, 71, 112,
• Outdoor unit: 1 no. type 224 (standard specification)
• Indoor/outdoor temperatures: cooling (indoor wet bulb temperature 22
o
CWB, outdoor air temperature 33
o
CDB)
Heating (indoor dry bulb temperature 22
o
CWB, outdoor air temperature 3
o
CDB)
• Difference in height between indoor and outdoor units: Outdoor unit is higher, indoor unit is lower, maximum difference in
height is 50m
• Refrigerant effective piping length: 121m
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