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2
JOHNSON CONTROLS
5276178-JIM-C-0118
TABLE 12 - REFRIGERANT FACTORY CHARGE R-410A
UNIT (TONS)
MODEL
CHARGE
SYSTEM #1
SYSTEM #2
SYSTEM#3
SYSTEM #4
25
WO/HGBP
13LB 8OZ
12LB 8OZ
12LB 8OZ
12LB 8OZ
25
W/HGBP
14LB
12LB 8OZ
12LB 8OZ
12LB 8OZ
30
WO/HGBP
22LB 15OZ
20LB 7.5OZ
22LB 4OZ
20LB 6OZ
30
W/HGBP
23LB 7OZ
20LB 7.5OZ
22LB 4OZ
20LB 6OZ
40
WO/HGBP
20LB 8OZ
18LB 8OZ
18LB 8OZ
20LB
40
W/HGBP
21LB
18LB 8OZ
18LB 8OZ
20LB
CFM, STATIC PRESSURE, AND POWER –
ALTITUDE AND TEMPERATURE CORRECTIONS
The examples below will assist in determining the
airflow performance of the product at altitude.
Example 1: What are the corrected CFM, static
pressure, and BHP at an elevation of 5,000 ft. if
the blower performance data is 6,000 CFM, 1.5
IWC and 4.0 BHP?
Solution: At an elevation of 5,000 ft the indoor
blower will still deliver 6,000 CFM if the rpm
is unchanged. However, Table 34 must be used
to determine the static pressure and BHP. Since
no temperature data is given, we will assume an
air temperature of 70°F. Table 13 shows the cor-
rection factor to be 0.832.
Corrected static pressure = 1.5 x 0.832 = 1.248 IWC
Corrected BHP = 4.0 x 0.832 = 3.328
Example 2: A system, located at 5,000 feet of eleva-
tion, is to deliver 6,000 CFM at a static pressure
of 1.5". Use the unit blower tables to select the
blower speed and the BHP requirement.
Solution: As in the example above, no temperature
information is given so 70°F is assumed.
The 1.5" static pressure given is at an elevation of
5,000 ft. The first step is to convert this static
pressure to equivalent sea level conditions.
Sea level static pressure = 1.5 / .832 = 1.80"
Enter the blower table at 6000 SCFM and static
pressure of 1.8". The rpm listed will be the same
rpm needed at 5,000 ft.
Suppose that the corresponding BHP listed in the
table is 3.2. This value must be corrected for
elevation.
BHP at 5,000 ft = 3.2 x .832 = 2.66
Example 3: Plotting fan performance using the charts
provided on pages 42, 45 & 48.
Refer to figure 14, Fan Performance 40 Ton Example
determine internal unit static loss. In this case the
loss is approximately 1.6" IWC.
Plot the fan performance at cooling sea level 0' el-
evation. Design conditions: 40 ton unit producing
14,000 CFM @ 1.5 ESP with additional static
losses for an economizer and 6", 65% air filters.
Economizer additional static loss =.28"
IWC Filter additional static loss 6", 65%=
= .46" IWC
Add the external static pressure and any additional
losses to the internal static loss:
External static loss =1.5"
IWC Economizer additional static loss
=.28" IWC Filter additional static loss 6",
65% = .46" IWC Internal unit static
loss = 1.6" IWC Total
system static =
3.84 IWC
TABLE 13 - ALTITUDE CORRECTION FACTORS
AIR
TEMP
ALTITUDE (FEET)
0
1000
2000
3000
4000
5000
6000
7000
8000
9000
10000
40
1.060
1.022
0.986
0.950
0.916
0.882
0.849
0.818
0.788
0.758
0.729
50
1.039
1.002
0.966
0.931
0.898
0.864
0.832
0.802
0.772
0.743
0.715
60
1.019
0.982
0.948
0.913
0.880
0.848
0.816
0.787
0.757
0.729
0.701
70
1.000
0.964
0.930
0.896
0.864
0.832
0.801
0.772
0.743
0.715
0.688
80
0.982
0.947
0.913
0.880
0.848
0.817
0.787
0.758
0.730
0.702
0.676
90
0.964
0.929
0.897
0.864
0.833
0.802
0.772
0.744
0.716
0.689
0.663
100
0.946
0.912
0.880
0.848
0.817
0.787
0.758
0.730
0.703
0.676
0.651
Содержание 40 25 TON
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