Mitsubishi Electric PUHY-RP-Y(S)JM-B Data Book page 84

7. UNIT SELECTION
(2) To connect indoor units using non-standard pipes
Design conditions
<Cooling>
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
<Heating> Indoor design dry-bulb temperature: 21ºC
Outdoor design wet-bulb temperature: 5ºC
Heating load: 14.5 kW for each of the two rooms
<Miscellaneous information>
Main piping: ø25.4 x 45 m, Branch piping: ø15.88 x 5 m (Equivalent indoor and outdoor piping length: 50 m)
A. Calculating the cooling load
a) The thermal load of 13 kW per room and the indoor unit return air temperature 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 26 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:
Outside air temperature 36ºCDB
Piping length: 50 m
Main piping: ø25.4 x 45 m
The standard capacity Qs is corrected for indoor/outside air temperatures, piping length, and pipe diameter to obtain the
maximum outdoor unit capacity Qm as follows: Qm = 28 x 0.99 x 0.97 x 0.98 = 26.3 kW. Because this value does not
meet the thermal load Qi (27 kW), a larger 300 model outdoor unit needs to be selected.
d) For the 300 model, the outside air temperature correction factor is 0.99, the pipe diameter correction factor for a 45 m
pipe with a diameter of ø25.4 is 0.98, and the piping length correction factor is 0.96 (read from the diagram for the unit
whose total indoor unit capacity N is 250).
Where the standard outdoor unit capacity Qs is 33.5 kW, the maximum outdoor unit capacity Qm can be obtained as
follows: Qm = 33.5 x 0.99 x 0.98 x 0.96= 31.2 kW. This value is greater than the thermal load Qi (27 kW), and the
maximum capacity Qm satisfies the capacity requirements.
e) Compare the thermal loads on the indoor unit side, using the maximum outdoor unit capacity apportioned to each
indoor unit and taking the indoor unit return air condition correction factors into consideration. The correction factor for
the return air temperature at 18.5ºC is 0.99 (at the standard outside dry-bulb temperature of 35ºC), and the pipe
diameter correction factor for a 5 m pipe with a diameter of ø15.88 connected to a 125-model unit is 0.99. These
values can be plugged into the following formula to obtain the capacity.
31.8 kW x 125/250 x 0.99 x 0.99=15.5 kW.
The result shows that the capacity (15.5 kW) is greater than the thermal load of 13.5 kW, and based on this result, two
125 model indoor units and one 300 model outdoor unit can be selected.
B. Calculating the heating load
Next, we will calculate the heating load, using the models that are selected based on the cooling load calculation.
a) The standard capacity Qs of the tentatively selected (during cooling load calculation) 300 model outdoor unit is 37.5
kW.
b) The correction values obtained from the air temperature correction factor graph and the piping length correction factor
graph for the 300 model outdoor units are as follows:
Outside air temperature 5ºCWB
Piping length: 50 m
Main piping: ø25.4 x 45 m
The standard capacity Qs is corrected for indoor/outside air temperatures and piping length to obtain the maximum
outdoor unit capacity Qm as follows: Qm = 37.5 x 1.00 x 0.975 x 0.975 x 0.98 = 34.9 kW.
The result shows that the Maximum outdoor unit capacity Qm (34.9 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: 34.9 x 125/250 x 0.96 x 0.99 = 16.5 kW (where the correction factor for the indoor unit return air temperature
of 21ºCDB is 0.96 (standard outside temperature at 6ºC), and the pipe diameter correction factor for a 5 m branch pipe
with a diameter of ø15.88 connected to a 125 models is 0.99. The result shows that the unit capacity 16.5 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
Capacity correction factor 0.99 (at the standard indoor wet-bulb temperature of 19ºC)
Capacity correction factor: 0.97
Pipe diameter correction factor: 0.98
Capacity correction factor 1.00 (at the standard indoor dry-bulb temperature of 20ºC)
Defrost correction factor is 0.975
Capacity correction factor: 0.975
Pipe diameter correction factor: 0.98
OUTDOOR UNITS 81