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119152-02 PDF预览

119152-02

更新时间: 2024-02-12 16:30:14
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其他 - ETC 风机
页数 文件大小 规格书
57页 2219K
描述
145MM240V 1200 BLDC BLOWER

119152-02 数据手册

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Physical Laws for Blower Applications  
In the following formulae these symbols are used:  
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.  
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.)  
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)  
SG-Specific Gravity (ratio of density of gas to the density  
of air)  
b-Altitude (location of blower)  
c-Temperature of inlet air  
Basic Fan Laws Chart  
VARIABLE  
VOLUME  
PRESSURE  
HORSEPOWER  
WHEN SPEED CHANGES Varies DIRECT with Speed Ratio Varies with SQUARE of Speed Ratio Varies with CUBE of Speed Ratio  
2
3
RPM2  
( )  
RPM1  
RPM2  
RPM2  
CFM2 = CFM1  
P2 = P1  
HP2 = HP1  
( )  
( )  
RPM1  
RPM1  
WHEN DENSITY CHANGES  
Does Not Change  
Varies DIRECT with Density Ratio  
Varies DIRECT with Density Ratio  
D2  
D2  
P2 = P1  
HP2 = HP  
( )  
1 ( )  
D1  
D1  
Volume  
The Volume changes in direct ratio to the speed.  
V = Original Volume (1000 CFM)  
1
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?  
V = New Volume  
2
RPM1 = Original Speed (3500 RPM)  
RPM2 = New Speed (3000 RPM)  
1
1
RPM2  
3000  
V = V  
= 1000 x  
= 1000 x .857 = 857 CFM  
( )  
(3500 )  
2
1
RPM1  
Pressure  
Pressure (barometric) varies in direct proportion  
to altitude.  
Let: V = Volume of standard air (1000 CFM)  
1
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?  
V = Volume of thinner air  
2
Hg1 = Barometric pressure sea level (29.92)  
Hg2 = Barometric pressure 6000’ (23.98)  
29.92  
Pressure = 3 x  
= 3.75 or 3 3/4 PSI  
Hg1  
Hg2  
29.92  
23.98  
23.98  
V = V1 x  
= 1000 x  
= 1248 CFM  
2
If it is desired to determine what pressure a 3 PSI (standard  
air) blower will deliver at 6000 feet –  
The pressure changes as the square of the  
speed ratio.  
23.98  
Pressure = 3 x  
= 2.4 or about 2 1/2 PSI  
29.92  
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?  
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.  
P1 = Original Pressure (5 PSI)  
P2 = New Pressure  
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?  
RPM1 = Original Speed (3500 RPM)  
RPM2 = New Speed (3000 RPM)  
RPM2  
2
2 = 5 x .735 = 3.68 PSI  
3000  
P2 = P1  
= 5 x  
( ) (3500 )  
RPM1  
5 0  
AE99  

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