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Posted: 2020-07-03 08:50 AM . Last Modified: 2024-04-08 03:44 AM
With the recent upgrade to v8 what is the best way to model a rack that has a device connected directly to it's own breaker and not via a rack mounted PDU (rPDUs)?
For example I have a rack with servers and a blade chassis in. The servers are connected to the two rPDUs (A and B) whilst the chassis is connected directly to a breaker. If I put another set of rPDUs in I would get double the potential capacity on the rack overlay which would distort planning.
(CID:110007526)
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Posted: 2020-07-03 08:50 AM . Last Modified: 2024-04-08 03:44 AM
Hi James,
Unfortunately currently it is not possible/supported (in DCO) to model rack mounted devices that can be power-cabled from another source than racks own pdus. Hopefully it's okay with you if I go ahead and capture this as a feature request, eg. feature for connecting a rack-mounted device(s) directly to a breaker out side the rack.
Kind regards
(CID:110007905)
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Posted: 2020-07-03 08:51 AM . Last Modified: 2024-04-08 03:44 AM
Jef Faridi PLease add this as a feature request. Is there any advice from Schnieder on how best to model this situation? I can't really see another way other than adding a new set of PDUs to the rack.
(CID:110007921)
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Posted: 2020-07-03 08:51 AM . Last Modified: 2024-04-08 03:44 AM
If you use the RackPDU workaround, then you can achieve correct remaining capacity of the rack by setting a design limit on the additional RackPDU(s) matching the estimated load of the connected server, the only down side is that the total power limit on the rack is too high.
Regards
Gert
PS: you find the design limit setting In the power property page of RackPDU itself (i.e. not the rack property page).
(CID:110008088)
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Posted: 2020-07-03 08:51 AM . Last Modified: 2024-04-08 03:44 AM
Gert Lehmann Thanks, I'll give it a go.
(CID:110008505)
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Posted: 2020-07-03 08:51 AM . Last Modified: 2023-10-20 05:03 AM
This question is closed for comments. You're welcome to start a new topic if you have further comments on this issue.
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