Page 71 - Gast Regenair Regenerative Blower

Reference

REGENAIR Regenerative Blowers
®

Air Density Varies In Inverse Proportion To Absolute Temperature
For example, a blower is to handle 150 °F air at 40 inches of water pressure. What pressure (standard air) blower is required?
Let:
P – pressure hot air (40 inches of water) Temperature in Degrees % Of Change In Air
1 of Fahrenheit Density Compared to 70°
P – pressure standard air
2 100 –5
AT – absolute temperature hot air (150+460=610°) 90 –4
1
AT – absolute temperature standard air (68+460=528°) 70 0
2 60 +2
P = P X (AT /AT ) 50 +4
2 1 1 2 40 +6
30 +8
P = 40 X (610/528) = 40 X 1.15 = 46 inches of water 20 +10
2 10 +13
If a blower is capable of delivering 30 inches of water pressure with standard air, what pressure will it develop handling 150 °F inlet
air?
P = P X (AT /AT ) P = 30 X (528/610) = 30 X.866 = 26 inches of
1
1
1
2
2
water pressure
Relation Of Density To Inlet Volume
At high altitudes it is frequently specified that a specific blower must be capable of handling a given volume of “standard air”. For
example, a blower is to operate at 5000 feet and is to handle 500 CFM of standard air. To determine the equivalent volume of air
the blower must handle at the higher altitude:
Let:
V – volume of standard air (500 CFM) V – volume of thinner air
1 2
Hg – barometric pressure sea level (29.92) Hg – barometric pressure at altitude (24.89 for 5000 feet)
1 2
V =V X (Hg1/Hg2) V = 500 X (29.92/24.89) = 601 CFM of air at 5000 feet altitude
2 1 2
Pressure Varies In Direct Proportion To Density
For example, a blower operating at 80 inches of water with standard air is to be used to handle air having a specific gravity of 0.8.
What pressure does the blower create when handling the air?
Let:
Pa – air pressure Pg – gas pressure
SG – specific gravity of gas

Pg = Pa X SG Pg = 80 X 0.8 = 64 inches of water

Horsepower Changes As The Cube Of The Speed Ratio
For example, a blower is operating at a speed of 3500 RPM and requiring 5 horsepower. If the speed is reduced to 3000 RPM,
what is the new required horsepower?
Let:
HP – original horsepower
1
HP – new horsepower
2
RPM – original speed
1
RPM – new speed
2
HP = HP X (RPM /RPM ) HP = 5 X (3000/3500) = 5 X .630 = 3.15 horsepower
3
3
2 1 2 1 2

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