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Discharge Valve Springs —
When 5H compressors
are used for booster applications where discharge pressure is
below 10 psig, the standard discharge valve springs furnished
with the machine should be replaced with an equal number of
lighter weight springs, Part Number 5H41-1801.
No change in discharge valve springs is recommended for
5F compressors.
Water-Cooled Heads —
Standard 5F,H compressors
are not equipped with water-cooled heads but they are avail-
able on special order. Water cooling of heads is generally not
necessary in R-12 or R-502 booster applications. For applica-
tions with R-22 involving high compression ratios, 5 or above,
5F,H booster compressors should be equipped with water-
cooled heads.
Motor Selection Data —
In staged refrigeration sys-
tems, the high stage compressor starts first and runs until low
stage pressure has been reduced to a predetermined level
before the low stage machine starts. With direct staged arrange-
ments, the high stage machine draws gas from the evaporator
through low stage machine bypass during this initial period.
Size of the selected motor must be related to the maximum
condition at which booster compressor can operate.
Compressor may run under heavy loads during periods of
high suction pressure, especially on starting when system is
warm. To handle these situations the motor must be sized larger
than the actual balanced operation brake horsepower indicates,
or special attention must be paid to operation of the system
when starting initially. Tables 25-27 give balanced brake horse-
power values at 1750 rpm.
If the system is to operate only at a fixed low temperature, it
is possible to avoid oversizing of motors providing careful op-
eration is followed when the system is first put in operation.
On applications requiring reduction from ambient condi-
tions to some extremely low temperature, the compression
system will be operated at high suction pressures for consider-
able periods of time. General practice is to drive the high stage
compressor with a motor that will operate compressor at the
highest expected evaporator temperature. This is generally the
“air conditioning” rating of unit. For intermediate or low stage
compressors, it is generally sufficient to size motor to take care
of double the balance load indicated horsepower plus friction
horsepower.
Also consider compressor starting torque requirements
when selecting motor for a booster compressor. Starting torque
of a motor only large enough to provide required normal
operating bhp for booster applications may not be large enough
to start the compressor. Recommended minimum motor sizes
shown in Table 28 have been selected to assure adequate
starting torque. Actual motor size selected is usually larger,
depending on the maximum bhp conditions under which the
compressor will run during pulldown or other abnormal operat-
ing periods.
It is good practice to select motors with allowance for 10%
voltage reduction unless there is a certainty that this cannot
occur.
Compressor Starting Torque —
Required compres-
sor starting torque is dependent on the discharge pressure as
well as the pressure differential occuring during start-up.
Maximum expected torque required during the starting period
for 5F,H compressors, used as boosters, is shown in Table 28 at
2 saturated discharge temperatures.
Selection Procedure —
Selection of a 5F,H booster
compressor requires that the load, saturated suction tempera-
ture, saturated discharge temperature, type of system and
refrigerant are known.
After the saturated intermediate temperature is determined
from Fig. 21, the booster rating (Tables 25-27) can be entered
and the compressor selected. Low stage load is then multiplied
by the “R” factor from Table 23 to obtain high stage compres-
sor load. With this information, the Compressor Ratings tables
on pages 7-15, and page 17 can be entered and the high-stage
compressor selected.
SELECTED EXAMPLE:
Given:
Refrigeration Load. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5.7 tons
Saturated Suction Temperature. . . . . . . . . . . . . . . . . . . . . –60 F
Saturated Condensing Temperature . . . . . . . . . . . . . . . . . . .80 F
Open-Type Intercooler
Refrigerant 22
Find:
Compressor size and motor size.
Solution:
1. Figure 21 indicates an optimum saturated intermediate
temperature of –2 F. Allow a 1 degree or 2 degree drop
from the booster compressor to intercooler and from the
intercooler to the high stage compressor.
Booster Saturated Suction Temperature = –60 F
Booster Saturated Discharge Temperature = 0° F
2. At –60 F suction and 0° F discharge, the 5H60 booster
compressor has a capacity of 6.8 tons with 12.1 bhp input
at 1750 rpm.
The safety factor at 1750 rpm:
This is satisfactory from Fig. 19 and a 5H60 compressor
is selected.
3. Indicated hp (ihp) = bhp – Friction hp (fhp)
Where bhp is given in Table 26 and fhp is given in
Table 28.
Indicated hp (ihp) = 12.1 – 3.07 = 9.03
Recommended minimum hp
= (2 x ihp) + fhp
= (2 x 9.03) + 3.07 = 21.13
Tentatively select a 25-hp motor. Assume that low stage
will never start against a saturated discharge higher than
30 F. At 30 F discharge, Table 28 indicates a starting
torque of 54 lb-ft. Therefore, a normal starting torque
25-hp motor is selected.
4. With –60 F suction and 0° F discharge, Table 23 indicates
an “R” value of 1.303. Therefore, the high stage load is:
1.303 x 6.8 = 8.86 tons (actual load)
5. Allowing a 1 degree drop from the intercooler, the high
stage saturation suction temperature is –3 F.
Allowing a 2 degree drop between the compressor and
condenser, the high stage saturated discharge temperature
= 80 + 2 = 82 F.
6. Referring to the 5F,H Compressor Ratings table, 5F60 at
1450 rpm (using multiplier in compressor capacity notes)
has a capacity of 9.21 tons at –3 F suction and 82 F dis-
charge (through interpolation). The 5F60 is selected and
requires 13.0 bhp at 1450 rpm.
7. Assume that maximum load during pulldown occurs at
50 F suction and 90 F discharge. For this condition, the
rating tables (using the multiplier in Step 6) indicate
15.8 bhp, thus a 20-hp motor is selected.
(
6.8
)
x 100 – 100 = 19.3 or 20%
5.7