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
Derating Calculation of Capacitor Bank
Started by
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, Sep 22 2005 03:29 PM
3 replies to this topic
#2
marke
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Posted 22 September 2005 - 06:43 PM
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,
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,
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#3
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Posted 23 September 2005 - 04:07 PM
Dear Marke,
Thanks. For the above example,
= 100kVar - (440-380) (440-380))
= 100kVar - 3.6kVar
= 96.4kVar
Right?
Stephen
Thanks. For the above example,
= 100kVar - (440-380) (440-380))
= 100kVar - 3.6kVar
= 96.4kVar
Right?
Stephen
#4
marke
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Posted 23 September 2005 - 08:46 PM
Hello Steven
Assuming both ratings are at 50Hz,
KVAR = 100 x (380/440) x (380/440)
Best regards,
Assuming both ratings are at 50Hz,
KVAR = 100 x (380/440) x (380/440)
Best regards,
Mark Empson | administrator
Skype Contact = markempson | phone +64 274 363 067
LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | AuCom | Soft Starters
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