Guest Posted September 22, 2005 Report Share Posted September 22, 2005 Hi, do you know how to calculate the derating of a capcitor bank? If a 440V 100kVar capacitor bank installed in a 380V electrical system ? How much of the capacitor bank will be derated? Is it right if I calculate like that 100k x 380 / 440 ? =86.363kVar ?? Thanks in advance Stephen Link to comment Share on other sites More sharing options...
marke Posted September 22, 2005 Report Share Posted September 22, 2005 Hello Stephen The capacitor provides an impedance that is frequency dependant. If you change the frequency, then the impedance changes and this will affect the current. If the frequency does not change, then the impedance is constant and the current will reduce in proportion to the voltage reduction.As KVAR is a product of both voltage and current, the KVAR wil reduce by the voltage reduction squared. Best regards, Mark Empson | administratorSkype Contact = markempson | phone +64 274 363 067LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | AuCom | Soft Starters Link to comment Share on other sites More sharing options...
Guest Posted September 23, 2005 Report Share Posted September 23, 2005 Dear Marke, Thanks. For the above example, = 100kVar - (440-380) (440-380))= 100kVar - 3.6kVar= 96.4kVar Right? Stephen Link to comment Share on other sites More sharing options...
marke Posted September 23, 2005 Report Share Posted September 23, 2005 Hello Steven Assuming both ratings are at 50Hz, KVAR = 100 x (380/440) x (380/440) Best regards, Mark Empson | administratorSkype Contact = markempson | phone +64 274 363 067LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | AuCom | Soft Starters Link to comment Share on other sites More sharing options...
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