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8. UNIT SELECTION
OUTDOOR UNITS
8-2. Calculation samples
(1) To connect indoor units using the standard pipes
A. Calculating the cooling load
B. Calculating the heating load
C. Calculating the capacity requirement
Design conditions
<Cooling>
<Heating>
<Miscellaneous information>
a) The thermal load (13.5 kW/room), indoor unit return air temperature correction factor, and piping length correction
factor are used to calculate the required indoor unit capacity, based on which a 125 model of indoor unit is tentatively
selected. (Because the total thermal load is 27 kW, the air temperature correction diagram for the 250 model outdoor
unit will be used.)
b) Because the total capacity of indoor units (N) is 250, the 250 model outdoor unit is tentatively selected.
Based on the above, the standard capacity Qs will be 28 kW.
c) The correction values obtained from the air temperature correction factor graph and the piping length correction factor
graph for the 250 model outdoor units are as follows:
d) For the 300 model units, the outside air temperature correction factor is 0.99. The piping length correction factor that
can be obtained from the diagram for the system whose total indoor unit capacity is 250 would be 0.96.
Because the standard outdoor unit capacity Qs is 33.5 kW, the maximum outdoor unit capacity Qm can be obtained
using the following formula: Qm = 33.5 x 0.99 x 0.96 = 31.8 kW. The result 31.8 kW is greater than the Qi (27 kW), so
the maximum capacity Qm meets the capacity requirement.
e) Compare the thermal loads on the indoor units, using the maximum outdoor unit capacity apportioned to each indoor
unit and taking the indoor unit return air temperature correction factors into consideration. The correction factor for the
return air temperature at 18.5ºCWB is 0.99 (at the standard outside dry-bulb temperature of 35ºC), and this value can
be plugged into the following formula to obtain the capacity.
32.1 kW x 125/250 x 0.99 = 15.8 kW
The result shows that the capacity (15.8 kW) is greater than the thermal load (13.5 kW), and based on this result, two
125 model indoor units and one 300 model outdoor unit can be selected.
a) The standard capacity (Qs) of the tentatively selected 300 model outdoor unit is 37.5 kW.
b) Use this value in the following formula to obtain the maximum outdoor unit capacity Qm as shown below:
Qm = 37.5 x 1.00 x 0.965 x 0.975 = 35.2 kW, (where the outside air temperature is 5ºCWB, capacity correction factor
is 1.00 (at the standard indoor dry-bulb temperature of 20ºC), defrost correction factor is 0.965, piping length is 50 m,
and capacity correction factor is 0.975).
The result shows that the Maximum outdoor unit capacity Qm (35.2 kW) exceeds the heating load Qi (29 kW).
c) You can now check to see if this value will meet the capacity requirement for each indoor unit by using the following
formula: 35.2 x 125/250 x 0.96 = 16.8 kW (where the indoor unit return air temperature correction factor at 21ºCDB is
0.96) (standard outside temperature at 6ºC). The result shows that the unit capacity (16.8 kW) exceeds the thermal
load for each room (14.5 kW).
Based on the above calculation, the following indoor and outdoor units can be selected.
Indoor units: 125 model x 2 units
Outdoor unit: RP300 model
a) The cooling capacity of the tentatively selected outdoor unit apportioned to each indoor unit is 15.8 kW, which exceeds
the rated cooling capacity of the 125 model unit (14.0 kW). The actual cooling capacity under the specified conditions is
calculated as follows: 14.0 x 0.99 x 0.97 = 13.4 kW (where the return air temperature correction factor at the standard
outside temperature of 35ºCDB is 0.99 and the piping length correction factor is 0.97).
b) The heating capacity of the tentatively selected outdoor unit apportioned to each indoor unit is 16.8 kW, which exceeds
the rated cooling capacity of the 125 model unit (16.0 kW). The actual heating capacity under the specified conditions is
calculated as follows: 16.0 x 0.95 x 0.975 = 14.8 kW (where the return air temperature correction factor at the standard
outside temperature of 6ºCWB is 0.95 and the piping length correction factor is 0.975).
If the standard capacity Qs is corrected for indoor/outside air temperatures and piping length, the maximum outdoor unit
capacity Qm can be obtained as follows: Qm = 28 x 0.99 x 0.97 = 26.8 kW.
The result shows that the thermal load Qi (27 kW) exceeds the unit capacity Qm, so a larger 300 model outdoor unit
needs to be selected.
Main piping: ø28.58 x 45 m, Branch piping: ø19.05 x 5 m (Equivalent indoor and outdoor piping length: 50 m)
Indoor design dry-bulb temperature: 26ºC/Indoor design wet-bulb temperature: 18.5ºC
Outdoor design dry-bulb temperature: 36ºC
Cooling load: 13.5 kW for each of the two rooms
Indoor design dry-bulb temperature: 21ºC
Outdoor design wet-bulb temperature: 5ºC
Heating load: 14.5 kW for each of the two rooms
Outside air temperature 36ºCDB
Piping length: 50 m Capacity correction factor 0.99 (at the standard indoor wet-bulb temperature of 19ºC)
Capacity correction factor: 0.97
Next, you will calculate the heating load and unit capacity requirements, using the models that were selected in the
previous section.