Cram for Exam Vol 2 Question 50 - 3-ph capacitor bank

[akyip]

Hi guys,

I'm a little confused at one particular question (question 50) from Cram for PE Power Exam volume 2, about 3-phase capacitor bank. The question is:

"A 3-phase wye-connected capacitor bank is installed at a 69-KV substation. The capacitor bank consists of 3 single-phase capacitor banks. Each bank consists of 4 parallel and 7 series capacitor units. Each capacitor unit is 5 KV and 100 KVAR. What is the nominal power rating in KVAR for the 3-phase capacitor bank?"

My understanding is that the KVARs of each capacitor unit simply get added up, regardless of whether the capacitors are in series or in parallel. So my initial take was:

Q 1-ph = (4 + 7) x 100 KVAR = 1100 KVAR

Q 3-ph = 3 x Q 1-ph = 3 x 1100 KVAR = 3300 KVAR

However, the solution states:

Q = 3 x 4 x 7 x 100 KVAR = 8400 KVAR

Why multiply the 4 and 7, if the problem states that each bank has 4 parallel and 7 series capacitor units? Am I missing something or not understanding something correctly?

 

[Orchid PE]

Why multiply the 4 and 7, if the problem states that each bank has 4 parallel and 7 series capacitor units? Am I missing something or not understanding something correctly?

I think it's because each phase has 4 capacitors in parallel, and then 7 of those parallel units in series. Imagine 7 rows and 4 columns of capacitors per phase.

You are correct the kvar just adds together regardless of connection.

 

[Cram For The PE]

Hi guys,

I'm a little confused at one particular question (question 50) from Cram for PE Power Exam volume 2, about 3-phase capacitor bank. The question is:

"A 3-phase wye-connected capacitor bank is installed at a 69-KV substation. The capacitor bank consists of 3 single-phase capacitor banks. Each bank consists of 4 parallel and 7 series capacitor units. Each capacitor unit is 5 KV and 100 KVAR. What is the nominal power rating in KVAR for the 3-phase capacitor bank?"

My understanding is that the KVARs of each capacitor unit simply get added up, regardless of whether the capacitors are in series or in parallel. So my initial take was:

Q 1-ph = (4 + 7) x 100 KVAR = 1100 KVAR

Q 3-ph = 3 x Q 1-ph = 3 x 1100 KVAR = 3300 KVAR

However, the solution states:

Q = 3 x 4 x 7 x 100 KVAR = 8400 KVAR

Why multiply the 4 and 7, if the problem states that each bank has 4 parallel and 7 series capacitor units? Am I missing something or not understanding something correctly?

It is as Chattaneer PE has stated. I can see why you did what you did though. Perhaps there is a better wording for this. Any suggestions?

 

[akyip]

It is as Chattaneer PE has stated. I can see why you did what you did though. Perhaps there is a better wording for this. Any suggestions?

I see. It is indeed the wording that got to me. Based on what the solution actually says, I myself would revise the question to say "Each bank consists of 4 parallel sets of 7 capacitor units." That, to me, is one way to say 4 x 7 capacitor units (for each bank).