Quincy Rotary Screw Sales Manual
Useful Formulas
1) Estimating horsepower
requirements at pressures other than rated pressure.
In most rotary
screw compressors, increasing or decreasing the pressure settings will have
similar effects. The rule of thumb
is:
For every I
PSIG change from rated pressure, the brake horsepower required will change 0.5%
from the rated BHP. Increase the pressure by 10 PSIG and the BHP will go up 5%.
Decrease the pressure by 20 PSIG and the BHP will go down 10%.
2) Estimating volume flow
rates at pressures other than rated pressure.
Changes in discharge pressure from the rated
pressure of the compressor will result in
changes in the overall compression ratio. These
compression ratio changes will cause
changes in the volumetric efficiency of the
compressor that will result in changes in capacity. The rule of thumb is:
For every 10
PSIG change from rated discharge pressure, the CFM capacity of the compressor
will change 0.4% from the rated capacity. Reducing pressure from 110 PSIG to
100 PSIG will result in a capacity increase offour tenths of one percent.
Increasing the pressure by 10 PSIQ will cause a reduction in capacity of about
four tenths of one percent.
3) Estimating power costs.
To estimate power costs, you will need to know the
following:
a) What is the cost per KWH?
b) How many hours per year does the compressor run?
c) At what capacity will the compressor run or how
many hours will the compressor run at
various
load levels?
d) What are the brake horsepower requirements of the
compressor at the required load
levels?
e) What is the motor efficiency?
It is important to use actual CFM requirements to
figure the load level of the compressor. Do not base power cost calculations on
comments like, “About half the time we run at full load and about half the time
we run at 70% of full load.” Full load for one machine may not be the same as
full load for another machine. Always determine the exact air requirement in
order to provide the customer with a power cost calculation that approximates
his situation.
Motor efficiencies vary from horsepower to
horsepower and from manufacturer to manufacturer within horsepower ranges. The
only way to accurately figure power costs will be to use the motor efficiency
number on the nameplate of the actual motor being used.
With
the above information in hand annual power costs can be estimated by using the
following formulas:
1) kW = BHP x .746 / Motor Efficiency
EXAMPLE – Find the kW of a 100 HP, nominal
Efficiency motor running at a 95 HP load.
KW = 95 x .746 / .93 = 76.2
2) Cost per Hour = KWH x Power Cost in $
EXAMPLE – Find the cost per hour to operate the
compressor in the above example assuming a cost of 7 cents. Cost Per Hour = 76.2 x 0.07 = $5.334
To
find the annual power costs, calculate the cost per hour of operating at the
various anticipated load levels and multiply by the anticipated number of hours
that the machine will operate at those load levels.
4) Estimating additional capacity required to raise system from one
pressure to a higher pressure.
To calculate the additional capacity required you need to know:
1) Current CFM capacity (total) of all compressors feeding the
system.
2) Current System Press (PSIG)
3) Desired System Press (PSIG)
4) Ambient absolute press (PSIA)
The formula for this calculation is:
(Desired
absolute system pressure divided by Current Absolute system Press) times
Current CFM Capacity equals Actual Capacity required to achieve desired system
pressure.
EXAMPLE:
What is the additional
capacity required to maintain a 100 PSIG system pressure at sea level in a
system that now operates at 91 PSIG using 500 cfm?
(Desired Pressure / Current Pressure) x
Current Capacity
(114.7 / 105.7) x 500 = Actual
Capacity
Required
1.085 x 500 = 542.5 cfm
5) Estimating BTU heat
rejection of air-cooled rotary screw air compressors.
Heat
transfer in rotary screw compressors is dependent on a number of factors
concerning efficiencies of lubricant coolers and aftercoolers and the rate of
radiant cooling. The only way to arrive
at exact heat rejection rates is to actually test the compressor under
anticipated operating conditions.
Approximate heat rejection rates of rotary screw compressors in standard
plant operating conditions for use in designing heat recovery packages or use
in sizing additional plant air conditioning requirements can easily be
calculated. The rule of thumb is:
The total
BTU’s per minute of heat rejected by a rotary screw air compressor is equal to
the brake horsepower being used times 42.41.
A 100 BHP compressor would have a total heat load of 4,241
BTU/minute. Of this total about 8% is
rejected as radiant heat. Of the
remaining 92% about 85% (78.2% of total) is rejected through the lubricant
cooler and about 15% (13.8% of total) is rejected through the aftercooler.
Calculating BHP Requirements For Less Than Full Load Operation
Modulating single (All and two-stage (Ingerso1l-Rand) rotary
screw compressors:
To
calculate the horsepower required to produce an amount of compressed air that
is less than the full load capacity of a modulating rotary screw compressor,
convert the desired CFM demand level
into a percentage of the full load capacity of the compressor. If the demand on
a 750 CFM compressor was 600 CFM, the
percentage would be 80%. Find the Percent of Capacity in the chart below and
note the multiplier next to it. Multiply the drive motor BHP by the number
corresponding to the demand percentage. This will give you an estimate of the
brake horsepower required to meet the specified air demand.
MODULATING
|
Percent of |
Full Load |
Percent of |
Full Load |
Percent of |
Full Load |
|
|
100% |
1 |
80% |
0.9577 |
60% |
0.9033 |
|
|
99% |
0.9981 |
79% |
0.9554 |
59% |
0.9001 |
|
|
98% |
0.9962 |
78% |
0.9529 |
58% |
0.8968 |
|
|
97% |
0.9942 |
77% |
0.9505 |
57% |
0.8935 |
|
|
96% |
0.9923 |
76% |
0.948 |
56% |
0.8902 |
|
|
95% |
0.9903 |
75% |
0.9455 |
55% |
0.8868 |
|
|
94% |
0.9883 |
74% |
0.943 |
54% |
0.8833 |
|
|
93% |
0.9863 |
73% |
0.9404 |
53% |
0.8798 |
|
|
92% |
0.9842 |
72% |
0.9378 |
52% |
0.8762 |
|
|
91% |
0.9821 |
71% |
0.9351 |
51% |
0.8725 |
|
|
90% |
0.98 |
70% |
0.9324 |
50% |
0.8687 |
|
|
89% |
0.9779 |
69% |
0.9297 |
|
|
|
|
88% |
0.9758 |
68% |
0.927 |
|
||
|
87% |
0.9736 |
67% |
0.9242 |
|
||
|
86% |
0.9714 |
66% |
0.9213 |
|
||
|
85% |
0.9692 |
65% |
0.91 84 |
|
||
|
84% |
0.967 |
64% |
0.9155 |
|
||
|
83% |
0.9647 |
63% |
0.9125 |
|
||
|
82% |
0.9624 |
62% |
0.9095 |
|
||
|
81% |
0.9601 |
61% |
0.9064 |
|
||
If,
in the above example, the BHP listed for the 750 CFM compressor was 163, then
the BHP required at the 600 CFM level (80% of full load) would be 163 x .9577
or 156.1 BHP.
Variable displacement rotary screw compressors with built-in
clearance volume (Turn Valve and Spiral Valve:
To
calculate the horsepower required to produce an amount of air that is less than
the full capacity of a variable displacement rotary screw compressor with
built-in clearance volumes, follow the preceding example to determine the
percentage of full load capacity. Then use the following table to determine the
BHP consumed at the desired load level.
|
Percent of |
Full Load |
Percent of |
Full Load |
Percent of |
Full Load |
|
100% |
1 |
80% |
0.8629 |
60% |
0.7448 |
|
98% |
0.9853 |
78% |
0.8503 |
|
|
|
97% |
0.9781 |
77% |
0.844 |
|
|
|
96% |
0.9709 |
76% |
0.83 79 |
|
|
|
95% |
0.9638 |
75% |
0.83 17 |
|
|
|
94% |
0.9567 |
74% |
0.8256 |
|
|
|
93% |
0.9497 |
73% |
0.8 195 |
|
|
|
92% |
0.9427 |
72% |
0.8135 |
|
|
|
91% |
0.9358 |
71% |
0.8076 |
|
|
|
90% |
0.9289 |
70% |
0.8016 |
|
|
|
89% |
0.922 1 |
69% |
0.7958 |
|
|
|
88% |
0.9153 |
68% |
0.7899 |
|
|
|
87% |
0.9086 |
67% |
0.7841 |
|
|
|
86% |
0.9019 |
66% |
0.7784 |
|
|
|
85% |
0.8953 |
65% |
0.7727 |
|
|
|
84% |
0.8887 |
64% |
0.767 |
|
|
|
83% |
0.8822 |
63% |
0.7614 |
|
|
|
82% |
0.8757 |
62% |
0.7558 |
|
|
|
81% |
0.8693 |
61% |
0.7503 |
|
|
Variable displacement rotary screw compressors without built-in clearance
volume
(PowerSync®):
To
calculate the horsepower required to produce an amount of air that is less than
the full capacity of a variable displacement rotary screw compressor without built-in
clearance volumes, follow the preceding example to determine the percentage of
full load capacity. Then use the following table to determine the BHP consumed
at the desired load level.
POWER
$YNC
|
Percent of |
Full Load |
Percent of |
Full Load |
Percent of |
Full Load |
||||||||
|
100% |
1 |
80% |
0.8266 |
60% |
0.683 |
||||||||
|
99% |
0.9905 |
79% |
0.8187 |
59% |
0.6766 |
||||||||
|
98% |
0.9811 |
78% |
0.811 |
58% |
0.6701 |
||||||||
|
97% |
0.9718 |
77% |
0.8033 |
57% |
0.6638 |
||||||||
|
96% |
0.9626 |
76% |
0.7956 |
56% |
0.6575 |
||||||||
|
95% |
0.9535 |
75% |
0.7881 |
55% |
0.6512 |
||||||||
|
94% |
0.9445 |
74% |
0.7806 |
54% |
0.645 |
||||||||
|
93% |
0.9355 |
73% |
0.7732 |
53% |
0.6389 |
||||||||
|
92% |
0.9267 |
72% |
0.7659 |
52% |
0.6328 |
||||||||
|
91% |
0.9179 |
71% |
0.7586 |
51% |
0.6268 |
||||||||
|
90% |
0.9092 |
70% |
0.75 14 |
50% |
0.6209 |
||||||||
|
89% |
0.9006 |
69% |
0.7443 |
|
|||||||||
|
88% |
0.892 |
68% |
0.7372 |
|
|||||||||
|
87% |
0.8836 |
67% |
0.7302 |
|
|||||||||
|
86% |
0.8752 |
66% |
0.7233 |
|
|||||||||
|
85% |
0.8669 |
65% |
0.7164 |
|
|||||||||
|
84% |
0.8587 |
64% |
0.7096 |
|
|||||||||
|
83% |
0.8505 |
63% |
0.7029 |
|
|||||||||
|
82% |
0.8425 |
62% |
0.6962 |
|
|||||||||
|
81% |
0.8345 |
61% |
0.6896 |
|
|||||||||
Two-stage rotary screw compressor with variable displacement first
stage (Suilair):
To calculate the horsepower required to produce an amount of air that is less than the full
capacity of a two-stage rotary screw compressor with a variable displacement
first stage, follow the preceding
example to determine the percentage of full load capacity. Then use the
following table to determine the BHP consumed at the desired load level.
|
Percent of |
Full Load |
Percent of |
Full Load |
Percent of |
Full Load |
|
100% |
1 |
80% |
0.9154 |
60% |
0.8167 |
|
99% |
0.996 |
79% |
0.9108 |
59% |
0.8113 |
|
98% |
0.992 |
78% |
0.9062 |
58% |
0.8058 |
|
97% |
0.988 |
77% |
0.9016 |
57% |
0.8003 |
|
96% |
0.984 |
76% |
0.897 |
56% |
0.7947 |
|
95% |
0.9799 |
75% |
0.8923 |
55% |
0.789 |
|
94% |
0.9758 |
74% |
0.8875 |
54% |
0.7833 |
|
93% |
0.9717 |
73% |
0.8828 |
53% |
0.7775 |
|
91% |
0.9633 |
71% |
0.8731 |
|
90% |
0.9591 |
70% |
0.8682 |
|
89% |
0.9549 |
69% |
0.8633 |
|
88% |
0.9506 |
68% |
0.8583 |
|
87% |
0.9463 |
67% |
0.8532 |
|
86% |
0.942 |
66% |
0.8482 |
|
85% |
0.9376 |
65% |
0.8431 |
|
84% |
0.9333 |
64% |
0.8379 |
|
83% |
0.9288 |
63% |
0.8327 |
|
82% |
0.9244 |
62% |
0.8274 |
|
81% |
0.9199 |
61% |
0.8221 |
1. COMP RPM= motor
pulley dia x motor rpm
comp. pulley dia.
2. MOTOR PULLEY p. d.= comp pulley dia x comp
rpm
motor r p m
3. COMP PULLEY p. d.= motor pulley dia x motor
rpm
comp rpm
4. MOTOR RPM
= Comp pulley dia x comp rpm
motor
pulley p. d.
5. FREE AIR = Piston Displacement x volumetric efficiency
6. REQUIRED PISTON DISPLACEMENT= free air
vol. eff.
7. PISTON DISP. IN CU. FT. MIN. = Cyl. Bore in IN. x Cyl bore x stroke in IN. x rpm
2200
8. CU FT COMPRESSED AIR= Cu. Ft. free Air x 14.7
2200
9. CU. FT. Free AIR = cu. Ft. free air x (psig + 14.7)
14. 7
10. CU FT. Free Air Required to Raise Rec. from
0 Gauge to final Pressure=
vol.
of Rec. in cu. Ft. x psig
(atmospheric
press) p.s.i.a
11. CU. FT of Free Air Req’d to raise Rec from
some press. Greater than 0 to a final press.
Vol. Of rec in cu. Ft. x (final psig –
initial psig)
(atmospheric press) p.s.i.a.
12. Piston Speed in Ft. per Min.= 2 x Stroke (in IN.) x rpm
12
13. GALLONS =
CU. FT.
.134
14. CU. FT. = gallons
x .134
15. Total Force in Lbs. of Air Cylinder = Area of the Cylinder Dia. X PS I G of
in sq inches air
press used
16.
CFM of Free Air req’d to operate=
Vol of Cyl X
Cycles (Gage Press + 14.7
Cylinder (Single
Acting in cu ft Per Min X (14.7)
17. PUMP UPTIME (MIN) = V (tank size in gal )
x (final tank press — initial tank press
7.48 x atmos. Press. (psia ) x pump delivery (cfm)
Piston displacement for multi-stage compressors — only
the low pressure cylinders are considered
—