參數(shù)資料
型號(hào)： 150072-02
英文描述： 79MM MOTOR
中文描述： 79毫米電機(jī)
文件頁(yè)數(shù)： 48/57頁(yè)
文件大?。?/td> 2228K
代理商： 150072-02
50
AE99
Physical Laws for Blower Applications
In the following formulae these symbols are used:
P-Pressure in pounds per square inch (PSI) or inches of
mercury column (inches Hg)
CFM-Volume in cubic feet per minute
RPM-Speed in revolutions per minute
D-Density in pounds per cubic foot (lbs./cu. ft.)
H-Height of air or gas column (ft.)
SG-Specific Gravity (ratio of density of gas to the density
of air)
“Standard Air”-Air at 68O F (absolute temperature 528O) and
29.92” Hg. (barometric pressure at sea level). The density of
such air is 0.075 lbs./cu. ft. and the specific volume is 13.29
cu. ft./lb. The specific gravity is 1.0.
The outlet pressure of a blower depends on the condition of
the air or gas at the inlet. The inlet condition is influenced by:
a-Specific gravity (the ratio of density of the gas to density
of standard air)
b-Altitude (location of blower)
c-Temperature of inlet air
Basic Fan Laws Chart
Volume
The Volume changes in direct ratio to the speed.
Example – A blower is operating at 3500 RPM and delivering
1000 CFM. If the speed is reduced to 3000 RPM, what is the
new volume?
V1 = Original Volume (1000 CFM)
V
2 = New Volume
RPM1 = Original Speed (3500 RPM)
RPM2 = New Speed (3000 RPM)
V
2
= V
1 (
RPM
2
)1 = 1000 x (
3000 )1 = 1000 x .857 = 857 CFM
RPM
1
3500
Pressure
VARIABLE
VOLUME
PRESSURE
HORSEPOWER
WHEN SPEED CHANGES Varies DIRECT with Speed Ratio
Varies with SQUARE of Speed Ratio
Varies with CUBE of Speed Ratio
CFM
2
= CFM
1
( RPM2 )
P
2
= P
1
( RPM2 )2
HP
2
= HP
1
( RPM2 )3
RPM1
WHEN DENSITY CHANGES
Does Not Change
Varies DIRECT with Density Ratio
P
2
= P
1
(D2 )
HP
2
= HP
1
(D2 )
D1
Pressure (barometric) varies in direct proportion
to altitude.
Example – A blower is to operate at an elevation of 6000 feet
and is to deliver 3 PSI pressure. What pressure (standard air)
blower is required?
Pressure = 3 x
29.92
= 3.75 or 3 3/4 PSI
23.98
If it is desired to determine what pressure a 3 PSI (standard
air) blower will deliver at 6000 feet –
Pressure = 3 x
23.98
= 2.4 or about 2 1/2 PSI
29.92
When a blower is to operate at a high altitude it is frequently
specified that the blower be capable of handling a given
volume of “standard air”. It is then necessary to determine
the equivalent volume of air at the higher altitude.
Example – A blower is to operate 6000 feet altitude and is to
handle 1000 CFM of standard air. What is the CFM of air the
blower must handle at 6000 feet altitude?
Let: V1 = Volume of standard air (1000 CFM)
V2 = Volume of thinner air
Hg
1 = Barometric pressure sea level (29.92)
Hg2 = Barometric pressure 6000’ (23.98)
V2 = V1 x
Hg1 = 1000 x 29.92 = 1248 CFM
Hg
2
23.98
The pressure changes as the square of the
speed ratio.
Example – A blower is operating at a speed of 3500 RPM and
delivering air at 5.0 PSI pressure. If the speed is reduced to
3000 RPM, what is the new pressure?
P1 = Original Pressure (5 PSI)
P
2 = New Pressure
RPM1 = Original Speed (3500 RPM)
RPM2 = New Speed (3000 RPM)
P
2
= P
1
(
RPM
2
)
2
= 5 x
(
3000 )2 = 5 x .735 = 3.68 PSI
RPM
1
3500
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