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51
AE99
The Air Density varies in inverse proportion to the
absolute temperature.
Example – A blower is to handle 200O F air at 3 PSI pressure.
What pressure (standard air) blower is required?
Let: P
1 = Pressure hot air (3 PSI)
P2 = Pressure standard air
AT1 = Absolute temperature hot air (200+460=660
O F)
AT
2 = Absolute temperature standard air (68+460=528
O F)
P2 = P1 x
AT1
= 3 x
660
= 3.75 or 3 3/4 PSI
AT2
528
A blower is capable of delivering 3 PSI pressure with
standard air. What pressure will it develop handling 200O F
inlet air?
P
1 = P2 x
AT
2
= 3 x
528
= 2.4 or about 2 1/2 PSI
AT
1
660
Pressure varies in direct proportion to the density.
Example – A 3 PSI (standard air) blower is to be used to
handle gas having a specific gravity of 0.5. What pressure
does the blower create when handling the gas?
Let: Pa
= Air pressure (3 PSI)
Pg
= Gas pressure
SG
= Specific gravity of gas (0.5)
Pg = Pa x SG = 3 x .5 = 1.5 PSI
If we are required to handle a gas having a specific gravity of
0.5 at 1.5 PSI pressure, we can determine the standard air
pressure blower as follows:
Let: Pa =
Pg
=
1.5
= 3 PSI
SG
.5
Horsepower
The horsepower changes as the cube of the speed
ratio.
Example – A blower is operating at a speed of 3500 RPM and
requiring 50 horsepower. If the speed is reduced to 3000
RPM, what is the new required horsepower?
HP
1 = Original Horsepower (50)
HP2 = New Horsepower
RPM
1 = Original Speed (3500 RPM)
RPM2 = New Speed (3000 RPM)
HP
2
= HP
1
x
(
RPM
2
)
3
= 50 x
(
3000 )3=50x.630=31.5horsepower
RPM
1
3500
Horsepower vs. Specific Gravity & Ratio of Density.
The horsepower varies in direct proportion to the specific
gravity (ratio of density of gas to density of air).
Example – A standard air blower requires a 10 HP motor.
What horsepower is required when this blower is to handle a
gas whose specific gravity is 0.5?
HP = 10 x 0.5 = 5 horsepower
It is possible that several of the above modifications may be
required on one installation. Therefore, it may be necessary
to use various combinations of these formulae.