Afriso EURO-INDEX KAV Mode D'emploi page 39

Maintenance
14
Example:
The following values have been determined for a system:

Maximum level h

Current level in the tank h

Adjusted value KAV = 2.5 m
p = 0.084 bar/m x 2.5 m
g,max
p = 0.21 bar
g,max
p = 0.084 bar/m x 1.5 m
3
p = 0.13 bar
3
The test with the tester yields a pressure of -0.12 bar.
Criterion:

-0.12 bar + 0.21 bar - 0.13 bar = -0.04 bar

-0.04 bar < 0; criterion met, i.e. anti-siphon protection available.
If the result is > 0, the criterion is not met; this could be due to the fol-
lowing reasons:
1. Accumulation of air in the oil-carrying pipes.

De-aerate the lines.
2. Adjusted value KAV less than h

Correct the adjusted value at KAV.
3. Anti-siphon valve defective.

Replace the anti-siphon valve.
= 2.5 m
max
= 1.5 m
3
.
max
KAV