Specifications

• 22 • Application Guide • BALTIC - 0108
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owneeded,percentageoffreshairandexternalstaticpressure(toovercomesystemlosses,egduc
-
twork,diffusers.)
D. Accessories needed
Step 2 : Cooling Capacity
A.Preselecttheequipmentusing‘generaldata’intables3.1-3.2tondunitsclosetotherequiredcapacity.
B. Size the equipment using the ‘cooling performance’ in tables 4.1-4.40 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5.1-5.20withtherequiredairowandstaticpressure.(Donot
forgettoaddthepressuredropforaccessoriesintable5.24)
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4.2-4.40.
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 4.41-4.43, electric heater in table 4.44, and gas burner in table 4.45.
(*) : 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 table 6.1
A. Heat pump unit or humidity control pack.
Pa=P(Unit+Deltakitindooroptional+Extractionfan+Electricheater+gas+DeltaPLn)
la=la(Unit+Deltakitindooroptional+Extractionfan+Electricheater+gas+DeltaILn)
ld/la(base)=Table6.1
Id=la(base)xId/la(base)+la(Deltakitindooroptional+Extractionfan+Electricheater+gas)
B. Cooling unit
P1,la1(summeroperation)=P,la(Unit+Deltakitindooroptional+Extractionfan)
P2(winteroperation)=P(0,2+kitindoorstd+Deltakitindooroptional+Extractionfan+Electricheater)
la2(winteroperation)=la(0,5+kitindoorstd+Deltakitindooroptional+Extractionfan+Electricheater)
Pa=max(P1;P2)
Ia=max(Ia1;Ia2)
ld/la(base)=Table6.1
ld=la(base)xId/la(base)+la(Deltakitindooroptional+Extractionfan+Electricheater+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 BAC 035
will give 35,2 kW gross at nominal
operating conditions.
B. Table 4.12 shows that a
BAC 035 has a gross cooling
capacity of 34,8 kW.
C. Table 5.19 shows that econo-
miser and 36 kW electric heater
will add 36 + 88 Pa to the external
static specied, giving a total of
324 Pa.
The table 5.6 shows that fan drive
kit ‘k8’ (2,2 kW) is required for a
BAC 035 providing 6 300 m3/h at
300 Pa.
The net capacity is therefore
34,8 kW - 2,2 kW = 32.6 kW
The table 5.4 shows that fan drive
kit ‘k8’ (2,2 kW) is required for a
BAC 035 providing 6 300 m
3
/h
at 300 Pa.
The net capacity is therefore
35,1 kW - 2,2 kW = 32,9 kW
Step 4
A. Table 6.1 shows that an BAC
035 (cooling unit) With 36
kW Electric hea-
ter + KIT '8'
Ia1=28,9+1,4=30,3 A
P1=16,6+0,8=17,4 kW
Ia2=0,5+3,4+1,4+50=55,3 A
P2=0,2+1,9+0,8+36=38,9kW
P2>P1 so P=P2=38,9 kW
Ia2>Ia1 so Ia=Ia2=55,3 A
Id/Ia=3,1
Id=28,9x3,1+1,4+50=141 A
PERFORMANCES SELECTION PROCEDURE