capacitor bank for 15 K.W load

helloo friends..i want to design a capacitor bank with 19000 uf/400 V capacitor ..please sugget me what precaution should i take while designing..like no of capacitors should i use..

What is your actual DC bus voltage? At a minimum at 400 V @ 15KW, the current will be 38A, so the ripple current could be 3 times this, say 100A. So the secret is to keep the capacitors cool and to share the ripple current equally among them. So as a ball park figure, using 200 MFD caps , you need 19 X 5 = 195 caps ( I would use 200). Another point to bear in mind is that if a capacitor goes short circuit, it might explode or if it does not you will have to unsolder them one by one until you find it. So I would wire , say 5 in a star configuration (equal length leads), with a red 4mm socket and a black 4mm socket as the centre or star points. Using thick flex with 4mm red and black plugs, connect from the star points to the bus. All these leads should be the same length. A decent 4mm plug and socket should handle 10A! Remember cooling!! !!
Frank

chuckey said:

What is your actual DC bus voltage? At a minimum at 400 V @ 15KW, the current will be 38A, so the ripple current could be 3 times this, say 100A. So the secret is to keep the capacitors cool and to share the ripple current equally among them. So as a ball park figure, using 200 MFD caps , you need 19 X 5 = 195 caps ( I would use 200). Another point to bear in mind is that if a capacitor goes short circuit, it might explode or if it does not you will have to unsolder them one by one until you find it. So I would wire , say 5 in a star configuration (equal length leads), with a red 4mm socket and a black 4mm socket as the centre or star points. Using thick flex with 4mm red and black plugs, connect from the star points to the bus. All these leads should be the same length. A decent 4mm plug and socket should handle 10A! Remember cooling!! !!
Frank

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thanks frank ...i have DC voltage of 320 volt..can u tell me shud i put a resistance across each capacitor to discharge them when they are not in use..so that leakage current shud not flow through dielectric and i can increase the life of capacitor

First with a bus voltage of 320 V @ 15 KW, I = 15000/32 = 500 A!! the purpose of the resistor across the capacitor (bleed resistor) is to discharge the capacitor, so that when the unit is switched off (for working on), there are no dangerous voltages stored on the capacitors. The lower the value of resistor, the quicker the capacitor discharges, but the more power wasted (and heat generated). to get from the 320 to 50V in time T, this is when C X R = .5T (about), so if you think 10 Seconds is a reasonable time, then C X R = 5 Seconds so R = 5/200 M ohms or 25 k, the power dissipated in each resistor will be 320 X 320/25K = 4 W, there are 195 of them so that is 800 W ( fan required).
Frank

Whoops!, Its 50A not 500. I was thinking about the heat and reckon that it is better to use a few contact cooled resistors mounted on the same heat sink as the diodes. also the time constant thing, I was out on this too, I think you need about half the resistance value, which only makes the heat given off double, its now 1600 W. As the whole equipment is of the lethal kind, It should all be behind locked covers, so you could arrange that if any cover is removed a micro switch is operated which switches the discharge resistor(s) in which need to be lower in value but can be a lot smaller as the 1600 W is only dissipated during the capacitor discharge.
I think you will need a soft start else the charging currents for this mega capacitor will blow your mains fuses.
High power equipment normally has a tailor made switching arrangement interlocked so when you need to work on the inside of it, the system , makes you switch it off, isolate the mains, earth down dangerous voltages then finally allows you to gain access to the keys to open the cabinets. I think you need this.
Frank

The lower the value of resistor, the quicker the capacitor discharges, but the more power wasted (and heat generated). to get from the 320 to 50V in time T, this is when C X R = .5T (about), so if you think 10 Seconds is a reasonable time, then C X R = 5 Seconds so R = 5/200 M ohms or 25 k, the power dissipated in each resistor will be 320 X 320/25K = 4 W, there are 195 of them so that is 800 W ( fan required).

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Did you ever see an inverter or similar device with a similar passive discharge circuit involving a ridiculous 5% efficiency loss?

If a fast discharge is required for some reasons (2 min discharge time to a save voltage level would be accepted as industry standard), you'll place a normally closed contactor and preferably a fail-safe PTC power resistor.

For the other points involved with the post, I would prefer a more complete specification (inverter type, switching frequency, expected capacitor form factor). Generally there's a choice between a few large dedicated power electronics or a bank of smaller (e.g. 470 uF) capacitors.

" Did you ever see an inverter or similar device with a similar passive discharge circuit involving a ridiculous 5% efficiency loss? " Have you ever seen 19,000MF with 320V on them?
The way this chap is asking basic questions, leads me to believe that he has not worked on high voltage/high power equipment before.? It looks like he is using a half wave rectifier from the mains to need this sort of capacitance ( I wonder what the conduction angle and peak currents are ?). What he should be doing is to go down the three phase star and delta secondaries, which give you 4% ripple without any capacitors.
Frank

chuckey said:

First with a bus voltage of 320 V @ 15 KW, I = 15000/32 = 500 A!! the purpose of the resistor across the capacitor (bleed resistor) is to discharge the capacitor, so that when the unit is switched off (for working on), there are no dangerous voltages stored on the capacitors. The lower the value of resistor, the quicker the capacitor discharges, but the more power wasted (and heat generated). to get from the 320 to 50V in time T, this is when C X R = .5T (about), so if you think 10 Seconds is a reasonable time, then C X R = 5 Seconds so R = 5/200 M ohms or 25 k, the power dissipated in each resistor will be 320 X 320/25K = 4 W, there are 195 of them so that is 800 W ( fan required).
Frank

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Frank..suppose i have 10 capacitors of value 19000uF/400V connetcted in parallel..they are charged with a 228 V AC volt that approx 320 volt pk to pk DC voltage...when capacitors are fully charged upro 320 volt ..then for how much time i could draw the load of 5 K.W..waiting ur reply frank...

load = 5KW, =V X V/R , R = 320 X 320/5000 = 100,000/5000 = 20 ohms. So putting your 20 ohm load onto your .19F capacitor will cause it to discharge, so after 3.8 Secs the voltage will have fallen to .37 X 320 V ~ 100V.
Frank

chuckey said:

load = 5KW, =V X V/R , R = 320 X 320/5000 = 100,000/5000 = 20 ohms. So putting your 20 ohm load onto your .19F capacitor will cause it to discharge, so after 3.8 Secs the voltage will have fallen to .37 X 320 V ~ 100V.
Frank

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how the ..number 0.37 come..???

The total stored energy is 0.5 CV², less than 10 kWs.

The .37 come from the exponential decay of the voltage on a capacitor, Search WWW.Wikipedia.org for a detailed explanation.
Frank

Hii...Frank how r u..,,,when capacitors are fully charged should we disconnect the charging voltage to the capacitors or not...if yes then why ..??

You have not said what you are going to do with all these charged up capacitors. If you do not disconnect the capacitor charging circuit then you have to make sure that it can handle the added current of the load but it might simplify the discharge switching.
Frank