Sulzer ABS PC 441 - How To Calculate Overflows By Using Constants And Exponents; Pump Alternation

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3.3.1 Howtocalculateoverowsbyusingconstantsandexponents
• In Settings / Pump pit / Calc. overow/youcantypeintheconstantandexponentsmanually.
Therearetwodi󰀨erentexponentsandtwoconstantswhichcanbesetinPC441,dependingonthemanufac-
turer and nature of the weir.
Thoseconstantsshallnormallybeprovidedbythemanufactures.Ifyoudon’thavethee2andc2values,you
cansete2andc2to0(zero),onlyusetheleftsideoftheequation.Forthebasicweirtypesthec2constant
canbesetto0(zero).
Overow==h
e1
c1 + h
e2
c2 [m
3
/s]
Type of weir Exp Constant
Thompson 30°
2.5 0.373
Thompson 45°
2.5 0.569
Thompson 60°
2.5 0.789
Thompson 90°
2.5 1.368
Straight weir 1 m 1.5 1.76
Forstraightweirswithawidthotherthan1m,multiplytheconstantwiththewidthinmeters.Ex.c=b*1.76(b
in meters)
NOTE!  If”Locked on inow”ischosen,thePC441taketheoverowtobethelastcalculationofinowinthepitminus
thecapacityofthepumpswhicharerunning.
3.4 Pump alternation
PC441hasseveraldi󰀨erentmethodsinordertoalternatepumps.
1. Normal alternation
Pumpsarestartedalternatelyaccordingtoarotatingschedule.Thepumpthatstartedrstinthepump
cycle,willstartlastonthecycle.Inthiswaytherunningtimeisdividedequallybetweenalternatingpumps.
Pumps that are not activated for alternation start and stop on their own start and stop levels.
One can choose between alternation at each pump stop or when all pumps are stopped.
Alternateateachpumpstopmethodispreferredifthenormalinowtothepitissohighthatthepumps
don’thavethecapacitytoemptyit.Ifalternatewhenallpumpsstopmethodisselectedinthissituation,the
issuecouldarisethatonepumpisalwaysrunning,hencenoalternationwilltakeplace.
Alternatewhenallpumpsstopmethodispreferredifthepumpshavethecapacitytoemptythepitatnormal
inow.Thenallpumpsstopandthestart/stoplevelsalternate.
Example 1 Continuous high inflow. A single pump can’t empty the pit.
Startlevelpump1=2.0m
Startlevelpump2=3.0m
Stoplevelpump1=1.0m
Stoplevelpump2=1.5m
Method used: Alt. each pump stops Alt. when all pumps stop
Pit level increase
At level 2.0 m Pump 1 start Pump 1 start
At level 3.0 m Pump 2 start Pump 2 start
Pit level decrease
At level 1.5 m
Pump 2 stop
Pump 2 stop
Pit level increase
At level 3.0 m Pump 2 start Pump 2 start
Pit level decrease
At level 1.5 m
Pump 1 stop
Pump 2 stop
Pit level increase
At level 3.0 m Pump 1 start Pump 2 start
Ifalternatewhenallpumpsstopmethodisused,pump1willneverstop