Circutor computer SMART III-Fast Series Manuale Utente Scaricare il pdf (Pagina 83)

CT Ratio

(Ip / Is)

Power of the smallest stage at 440 V (in kvar)

2�5 5�0 7�5 10�0 12�5 15�0 20�0 25�0 30�0 40�0 50�0 60�0 75�0 80�0

4000/5 0�04 0�05 0�07 0�08 0�10 0�11

Forothervoltagesorconditionsnotincludedinthetable,thevalueofC/Kcanbeobtainedby

meansofasimplecalculation�

Calculating the C/K Factor

TheequationforcalculatingtheC/Kfactoris:

⋅⋅

where Ic:isthesmallestcapacitorcurrent�

K:thecurrenttransformertransformationratio�

TocalculateIcitisnecessarytoknowthereactivepowerofthesmallestcapacitorQandthe

networkvoltageV�

ThetransformationratioKiscalculatedas:

sec

prim

IK /=

whereIprim :isthenominalcurrentofthetransformerprimary�

Isec:isthecurrentofthetransformersecondary�

Example: In a 400 V unit the smallest capacitor is of 60 kvar with a current transformer having

a ratio of 500/5, and the calculation would be made as follows:

Current of the smallest capacitor Ic:

6,86

4003

60000

⋅

K Factor

1005/500 ==K

The C/K value is: 0.866.

If the power of 60 kvar is referenced at 440 V, it should be multiplied by Vgrid

/440

in which case the C/K value of the previous example would be 0.72.

IftheC/Kisconguredlowerthantheactualvalue,connectionsanddisconnec-

tions would occur continuously with few load variations (the system performs

moreoperationsthannecessary)�

IftheC/Kisconguredhigher,theregulatorrequiresahigherdemandforreac-

tivepowerinordertoswitchandperformfeweroperations�

Instruction Manual

Computer SMART III FAST

1 2 ... 78 79 80 81 82 83 84 85 86 87 88 ... 107 108

Nessun commento

Circutor computer SMART III-Fast Series Manuale Utente Pagina 83