When selecting blowers, there are several fundamental laws that guide the process. Sometimes several laws apply to an application, so it may be necessary to combine equations to determine the best blower for the job. Also, keep in mind that the outlet pressure of blowers depends on the inlet condition of the air or gas. Altitude, inlet temperature, and specific gravity (the ratio of density of the gas to density of standard air) all influence inlet conditions.

In the following cases, "standard air" is treated as air at 68°F or an absolute temperature of 528°R. Its density and barometric pressure at sea level are 0.075 lb/ft ^{3 }and 29.92 in.-Hg, respectively. Specific volume of standard air is 13.29 ft ^{3 }/lb and specific gravity is 1.0.

**LAW #1 **

*The volume that a blower delivers varies directly with speed. *For example, consider a blower operating at speed *S *= 3,000 rpm and delivering 1,500 cfm. If the speed slows to 2,500 rpm, the new volume becomes:

*V*2 = *V*1 (*S*2 /*S*1)

= 1,500(2,500/3,000)

= 1,250 cfm

**LAW #2 **

*Barometric pressure varies directly with altitude. *For instance, a blower operates at an elevation of 6,000 ft (barometric pressure = 23.98 in.-Hg) and must deliver 3 psi. The required pressure (standard air) blower is:

*Pb *= *Pout*(P1/P2)

= 3(29.92/23.98)

= 3.75 psi

Designers commonly specify that blowers operating at high altitudes must handle a given volume of "standard air" making it necessary to determine the equivalent volume of air at the higher altitude. For example, if a blower operates at 6,000 ft above sea level and handles 1,000 cfm of standard air, the volume of air the blower must handle at this altitude is:

*V*2 = *V*1 (*P*1/P2)

= 1,000(29.92/23.98)

= 1,248 cfm

**LAW #3 **

*Pressure changes with the square of speed ratio. *For instance, consider a blower operating at 3,000 rpm and delivering air at 4.0 psi. If the speed is reduced to 2,500 rpm, the new pressure is:

*P*2 = *P*1(S2/S1) ^{2}= 4(2,500/3,000) ^{2 }

= 2.78 psi

**LAW #4 **

*Air density varies inversely with absolute temperature, *where 0°F = 460°R. So, if a blower must handle 300°F air at 4 psi, the required blower pressure (standard air) is: *P2 = P1 (T1/T2) *

= 4(760/528)

= 5.75 psi

If, on the other hand, a blower delivers 4 psi with standard air, the pressure it develops when handling 300°F inlet air is: *P*1 = *P*2(T2/T1)

= 4(528/760)

= 2.78 psi

**LAW #5 **

*Pressure varies directly with density. *For example, if a 4-psi (standard air) blower handles gas with a specific gravity *SG *of 0.5. The blower creates the following pressure when handling the gas:

*Pg *= *Pa*(SG)

= 4(0.5)

= 2 psi

**LAW #6 **

*Horsepower changes with the cube of speed ratio. *For instance, a blower operates at 3,000 rpm and requires 40 hp. Reducing speed to 2,500 rpm means the new required horsepower is:

*HP*2 = *HP*1 (*S*2 /*S*1) ^{3}= 40(2,500/3,000) ^{3}= 23.1 hp

**LAW #7 **

*Horsepower varies directly with specific gravity *(ratio of gas density to air density). For example, if a standard-air blower requires a 20-hp motor, the horsepower required for the blower to handle a gas with a specific gravity of 0.5 is:

*HP*2 = *HP*1(SG)

= 20(0.5) = 10 hp

This information supplied by Ametek Inc., Kent, Ohio*. *