Page 17 - Application Guide 04-2003 / Rooftop BALTIC Series - Page 17
Step 1 : Input
Calculate the total and sensible loads of the area to be conditioned at design conditions.
A. Total cooling load in kW
B. Summer design condition
C. Air flow needed, percentage of fresh air and external static pressure (to overcome system losses, eg
ductwork, diffusers.)
D. Accessories needed
Step 2 : Cooling Capacity
A. Preselect the equipment using ‘general data’ in tables 3.1-3.2 to find units close to the required capacity.
B. Size the equipment using the ‘cooling performance’ in tables 4.1-4.21 to match the cooling loads at design
conditions.
C. To establish the net capacity, the supply fan motor power should be subtracted.
Review the indoor fan performance in tables 5.1-5.14 with the required air flow and static pressure. (Do not
forget to add the pressure drop for accessories in table 5.19)
Step 3 : Heating Capacity
A. Heat pump (
*
)
The selection procedure is the same as that undertaken for cooling.
Preselect equipment in “General data” in tables 3.1-3.2.
Obtain the gross heating capacity at design condition (winter conditions) from tables 4.2-4.20.
Obtain the net capacity by adding the supply fan power (selected above) to the gross capacity.
B. Other Heating
Select hot water coil in tables 4.22-4.24, electric heater in table 4.25, and gas burner in table 4.26.
(*) : This procedure doesn't take into account the impact of defrost in the heating performance. Depending on
the outdoor moisture and temperature condition, the defrost operation might reduce the heat pump capacity.
Step 4 : Electrical data
Data from tables 6.1-6.2
A. Heat pump unit or humidity control pack.
Pa = P(Unit+Delta kit indoor oExtraction fan+Electric gas)
la = la(Unit+Delta kit indoor oExtraction fan+Electric gas)
ld/la(base) = Tables 6.1-6.2
Id=la(base) x Id/la(base)+la(Delta kit indoor oExtraction fan+Electric gas)
B. Cooling unit
P1, la1 (summer operation) = P, la(Unit+Delta kit indoor oExtraction fan)
P2 (winter operation) = P(0,3+kit indoor std+Delta kit indoor oExtraction fan+Electric heater)
la2 (winter operation) = la(0,75+kit indoor std+Delta kit indoor oExtraction fan+Electric heater)
Pa = max (P1; P2)
Ia = max (Ia1;Ia2)
ld/la(base)=Tables 6.1-6.2
ld=la(base) x Id/la(base)+la(Delta kit indoor oExtraction fan+Electric gas)
EXAMPLE
Step 1
A. 32kW
B. 35°C outdoor temperature,
24°C DB, 19°C WB entering air
condition (room return air)
C. 6 300 m
3
/h at 200Pa
D. Economiser and 36 kW electric
heater.
Step 2
A. Table 3.1 shows that BCK 035
will give 34,6 kW gross at nomi-
nal operating conditions.
B. Table 4.12 shows that a
BCK 035 has a gross cooling
capacity of 34,0 kW.
C. Table 5.19 shows that
economiser and 36 kW electric
heater will add 23 + 88 Pa to the
external static specified, giving a
total of 311 Pa.
The table 5.6 shows that fan drive
kit ‘k8’ (2,2 kW) is required for a
BCK 035 providing 6 300 m
3
/h at
300 Pa.
The net capacity is therefore
34,0 kW - 2,2 kW = 31,8 kW
Step 4
A. Table 6.1 shows that an BCK 035
(cooling unit) With 36 kW
Electric KIT '8'
Ia1=37,3+2,3=39,6 A
P1=18,9+1,3=20,2 kW
Ia2=0,75+2,5+2,3+50=55,55 A
P2=0,3+1,4+1,3+36=39,0 kW
P2>P1 so P=P2=39,0 kW
Ia2>Ia1 so Ia=Ia2=55,5 A
Id/Ia=3,3
Id=37,3x3,3+2,3+50=175,4 A
PERFORMANCES
selection procedure
