Required capacity of receiver tank
Selection of compressors for additional installations
Amount of drained water

Required capacity of receiver tank


This is a calculation example for determining rough estimate of required receiver tank capacity.

t1+t2=

Load and unload period (loading time + unloading time)

Example: In case of SGIV, t1+t2 ≧ 0.5 min ≧(30 seconds or more)

Qs=

Discharge air flow of compressor

Pa=

Suction absolute pressure (atmospheric pressure = 0.101 Mpa)

D=

Difference between upper pressure limit (PH) and lower pressure limit (PL)

x=

Load ratio (air consumption rate against discharge air flow of compressor on above Qs)

* The tank capacity becomes maximum when the load ratio is 50%.

Vr=

Capacity of receiver tank

Disclaimer: Calculation results are provided for purpose of reference only, and we will not bear any responsibility for any outcome arising from using such calculation.

Selection of compressors for additional installations


This is a calculation example for determining a rough estimate of how much air flow rate must be increased by installing additional compressors, to maintain higher pressure when a large amount of air consumption is reducing pressure.

V2 ≧ V1*(

P2

P1

-1)

P1

Current line pressure

P2

Desired sustained level of pressure

V1

Total air flow rate of existing compressors

V2=

Required air flow rate for additionally installed compressors

Disclaimer: Calculation results are provided for purpose of reference only, and we will not bear any responsibility for any outcome arising from using such calculation.

Amount of drained water


The maximum amount of water vapor in the air (amount of saturated water vapor) increases as temperature elevates and decreases as pressure elevates. Compressing air, therefore, causes draining of water.

Condition: 100% road operation

a

Discharge air flow of compressor

b

Compressor operating time

1) Moisture content in the air suctioned by compressor

Suctioned air temperature

c

Amount of saturated water vapor

d

Relative humidity

e=c*d

Moisture content in the suctioned air (per m3/min)

f=a*b*e*60/1000

Moisture content in the air suctioned by compressor

2) Moisture content in the output of dryer

Atmospheric dew point (refer to below chart for conversion from pressure dew point)

g

Amount of saturated water vapor

h=a*b*g*60/1000

Total moisture content in the output of dryer

Example: The pressure dew point for 0.39 Mpa is about 6°C if the atmospheric pressure dew point is-15°C.

3) Amount of drained water at dryer output

f-h

Disclaimer: Calculation results are provided for purpose of reference only, and we will not bear any responsibility for any outcome arising from using such calculation.