Equivalent AC power of existing DC power amp readings?

Im dealing with a TELECT DUAL FEED 6/6 CIRCUIT BREAKER PANEL 23" PDU [https://telexpresslive.com/009-6212-2100-200a-dual-feed-6-6-circuit-breaker-panel-23-white/] ​which runs via 2AWG power cables to a large inhouse Valere Rectifier (Flatpacks). And then 10 AWG cables run power to the equipment in the cabinet. (routers) The Telect PDU is loaded with four 60 amp breakers PER SIDE. I can see the A and B feeds of this rectifier are consuming the following DC amps, which I determined by using a clip on multimeter with 400amp option on the 2AWG power cables that run from the Valere to the Telect. (4 cables total)

I checked this against 2 cabinets, which was necessary to make sure the DC amp metrics are consistent:

Cabinet 1 - (4 cables)
17.2 on black left, -17.9 on red left. -11.3 on red right, 15.0 on black right.

Cabinet 2- (4 cables)
17.0 black left, -16.5 on red left, -12.3 on red right, and 14.8 on black right.

We buy a lot of TDK LAMBDA HFE250048 power modules from Digikey, which fit into a 1U rectifier. We load these up in customer cabinets who don’t have a huge inhouse Valere rectifier like we do. The power modules by TDK Lambda hfe250048 are going into a 1U Rectifier Shelf. 1U is just to tell you that it only takes up one unit of rackmount realestate. The part number of this shelf im calling a 1U rectifier, though unimportant, is hfe2500s1u.

My question is… I gave metrics of DC AMPS above per CABINET. And I need the equivalent AC power required for the 1U rectifier we will be using for that cabinet to produce the DC feeds. Can someone help me with this translation?? I’m not going to have a Valere to work with eventually, and I need to hook these cabinets up to TDK Lambda rectifiers, not the Valere.

My reason for making this thread is 1) How many HFE250048 power modules [https://www.digikey.com/product-detail/en/tdk-lambda-americas-inc/HFE250048/285-2166-ND/2810343] do I need to meet the needs of my DC amp metrics above? 2) Whats the proper way to communicate total DC Amps of a cabinet? I don’t think left red left black right red right black is very practical…

Finally, if helpful, I will share a verbatim request given to me, which is the reason I started this thread.

Could you provide the power consumption of (1) cabinet of routers in the lab (Both the A and B feeds). There should be an ampmeter to measure the DC amps. Could you provide that information, as well as the equivalent AC power required for the 1U rectifier we will be using for that cabinet to produce the DC feeds?

Hello @anonymous

Please let me know if I am not answering this correctly. When looking at the HFE2500-48 power supply the input can range from 85-265VAC and has a Max current of 15A with 100VAC and 12A with 230VAC.

Max Current out will depend on the Vin. This is given on page 2 of the datasheet.

I also might be missing something in translation here. If you have a specific DC voltage and Amperage you need I can speak with my TDK product manager and make sure we have the correct unit for you. The only thing I am seeing is Amp rating and that leaves many options depending on voltage.
-Robert

Hi,
Thanks for quickly replying.

I already have a few dozen of these HFE250048 power modules stocked. They’re the ones we plan on using and have used on past builds. I specified the DC amps our pdu is producing from our in house rectifier. Again, I don’t want to use the in house rectifier. I’m using the lambda portable rectifier. Does this help you recommend the quantity of power modules I should ship to fulfill power requirements? My problem is, I don’t know how to communicate the amps I read off the meter. I gave you red left and right / black left and right. How else can I communicate the reading?

A single HFE250048 has an output capacity of up to 52A, which exceeds the reported sum of currents on both sides of the dual-feed breaker panel by a substantial margin; one should be able to supply the total load at the time it was measured. Note that if the two sides of the panel are not interconnected, the current flow should be same in both red and black leads; that this does not appear to be the case is most likely an indication of load variation in between the moments when the measurements were taken.

The worst-case load scenario should inform your decisions about AC-DC converter capacity. If you don’t know what your worst-case draw is in actual practice based on the specifications of the powered equipment, the sum of the installed circuit breaker capacity would be a safe guide, though likely overkill.

Matters of redundancy are also a consideration: these supplies are hot-swappable and designed to be run in parallel arrays, such that the installation of one more module than required to supply the load allows a single failed module to be replaced without interruption of service; this is known as N+1 redundancy.

The precise architecture of the system in question isn’t altogether clear, but in a scenario where one is using separate AC sources from different utility suppliers, separate rectification equipment would of course be needed for each.

Estimation of AC input power requirements is based on the conservation of energy principle and estimated efficiency of the converter: There’s roughly (18 + 15)A * 48V = 1.6kW going out and the converters are about 90% efficient, indicating that approximately 1.6kW/0.9 = 1.8kW total was being drawn from the AC input sources used as the measurements were being made. One would of course want to base specifications for AC service feeds based on the worst-case scenario, rather than conditions measured at some random point in time.

Note that a person can come to very different conclusions depending on how the problem is stated and what one chooses to use as worst-case values. If the reported values represent worst-case load conditions and no redundancy is needed, one module may be sufficient. If one wants to be able to run (8) 60A breakers to capacity simultaneously with N+1 redundancy on the other hand, that would require 12 modules…

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