dc power plant load calculation
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MarineTech
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- Camarillo, CA
There are several ways to calculate the conversion loss. However, to me NEC in its application, is Power Factor centric as related to apparent power.
So do you know the pf of the conversion equipment?
pf is a simple way to model steady state AC power.
So do you know the pf of the conversion equipment?
pf is a simple way to model steady state AC power.
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Useful as it is for load calculations, for some purposes such as sizing of neutral, etc, the harmonic or distortion power factor is very important when looking at potentially non-linear loads such as rectifiers. The PF of the conversion equipment should give you some information in this area, but it would be best to know separately any phase versus distortion PF.
Also, totally separate from the PF, the conversion equipment will have an efficiency factor which relates DC output power to the real AC input power.
Depends a bit on rectifier configuration* and whether it is an uncontrolled rectifier (diodes) or a phase controlled rectifier (SCRs).
*Six pulse, 12-pulse, hexaphase? Does it have an IPT? Air natural, forced air, water cooled?
The most recent we did, earlier this year, was a 7kA hexaphase job. The fourth in a row for the same customer as it happens. Water cooled.
I have just quoted for a 2kA job for that same customer. The more economic arrangement was a conventional six pulse configuration. The semiconductor cost goes up. The transformer cost goes down.
petersonra
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- Northern illinois
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- engineer
about the only thing that can be said for sure without a lot more information is that it will require a service of at least 48V*3000A or 144 kVA. That works out to 173A.
For guessing purposes, I would guess at a 400A service.
For guessing purposes, I would guess at a 400A service.
about the only thing that can be said for sure without a lot more information is that it will require a service of at least 48V*3000A or 144 kVA. That works out to 173A.
For guessing purposes, I would guess at a 400A service.
If it's a plain six pulse rectifier (B6U) then the ac supply current would be about 2450A.
To get an unregulated 48Vdc you would need about 38Vac. So a bit over 160kVA.
At 480V,close to 200A.
I don't know what other rules you have to factor in.
And I don't know if it is a B6U arrangement.
JoeStillman
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- West Chester, PA
You really need more info. There will probably be more than 3000A of rectifier load because the plants are typically designed to be able to recharge the batteries while supporting the 3000A load.
You need to know how many rectifiers there are and what the AC input amps are for each. For estimating purposes, I would allow at least another 10% for battery charging. No matter what the connected DC load is, you will have to support all rectifiers at full output at some point.
You need to know how many rectifiers there are and what the AC input amps are for each. For estimating purposes, I would allow at least another 10% for battery charging. No matter what the connected DC load is, you will have to support all rectifiers at full output at some point.
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Thanks all for responses, found the equipment cut-sheet, total of six rectifiers, 30A input each/480V.
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