208v watt calculations

Status
Not open for further replies.
B

bf111d

Member
I have a 2 pole breaker serving a server cabinet with 208v single phase servers. We are measuring amps on each pole of the breaker. The server cabinet watt load calculation i assume is the I(avg between poles)*208=W.

On another cabinet there is a 3pole 208v breaker serving a power strip that breaks down through local breakers 3 individual 208v single phase circuits. I assume 1 phase AB, 2 phase BC, 3 phase AC. Again measuring amps at each main breaker pole what is the calculation for total watts? Is the first part of my post off also and each pole calculation being I*120 and then adding the 2 poles for total watts? Little confused.
 
BF, If your single phase loads are 2700 watts @ 208v then its 2700w divided by 208v. If you connecting them as you said in the 3 phase power supply , they are connected in a delta config.Then its 2700w divided by 208 x 1.732.The phase current will be 22amps @ 3 phase and 12.9 amps single phase.
Rick
 
bf111d said:
I have a 2 pole breaker serving a server cabinet with 208v single phase servers. We are measuring amps on each pole of the breaker. The server cabinet watt load calculation i assume is the I(avg between poles)*208=W.
Click to expand...

If the load is the same for each phase then you are correct. You could use
120 volts and then add the two together.

bf111d said:
On another cabinet there is a 3pole 208v breaker serving a power strip that breaks down through local breakers 3 individual 208v single phase circuits. I assume 1 phase AB, 2 phase BC, 3 phase AC. Again measuring amps at each main breaker pole what is the calculation for total watts? Is the first part of my post off also and each pole calculation being I*120 and then adding the 2 poles for total watts? Little confused.
Click to expand...
If the loads are equal (or close) the calculation is 208 x amps x 1.73 = watts.
If the loads are not balanced then the calculation is 120 x amps = watts
and add the three together.
 
Yes the loads are truly single phase and not balanced. They come from a common 3 pole breaker at the PDU source but then are distributed to 3 individual breakers at the server cabinet to feed 3 separate loads. It sounds like the rule of amps*120v=watts then adding the 3 for the total KW load in the cabinet applies. Please let me know if this is it. Thanks for everyone's help.

Brad
 
We have had threads on this exact topic going to many, many posts. See, for example:
http://www.mikeholt.com/code_forum/showthread.php?t=81200
http://www.mikeholt.com/code_forum/showthread.php?t=79827

The short answer: if, on a feeder supplying many loads, you measure the amps on each leg, multiply by the line to neutral voltage of that leg (120V nominal in your case) and add up all three values, you will get an approximate load that is good enough for most purposes. If you do this with individual 208V loads and then add everything up, your results will be about 15% high. Neither of these answers will be 'exact', and the above threads wound their way into the math pretty far.

-Jon
 
bob said:
If the loads are not balanced then the calculation is 120 x amps = watts and add the three together.
Click to expand...
That is not strictly true. But it does give you the best available approximation that you can get, without going deeply into mathematics.

I would like to caution you against making any habit of ?adding the three together,? when you are dealing with three phase systems. You can?t add three phase currents as though they were the same type of thing. They are not. They are out of phase from each other, such that the current is at its positive peak in one phase, while at the same moment it is below zero in another phase. In a balanced system, the three phase currents are 120 degrees apart. In an unbalanced system, the phase angles can be anything, and the math required to accurately add them up is very complicated. The math is called, ?Symmetrical Components.? That particular technique is usually reserved for graduate-level college classes.

Here is a simple proof that your suggested method is a reasonable approximation:

You suggested that if the phase currents are not balanced, then you get watts from,
? Watts equals (120 times Ia) plus (120 times Ib) plus (120 times Ic).

One of those fancy ?properties? from early algebra tells us that this is the same as saying,
? Watts equals 120 times (Ia plus Ib plus Ic).

In reality, the best way to get ?watts? in an unbalanced system is to guess. You are going to guess wrong, of course, but you can get close. As your guess, take an average of the three phase currents. Then multiply that value by 208 volts, and again by the 1.732 factor. Thus,
? Watts equals 208 times one third of (Ia plus Ib plus Ic), and then times 1.732.

If you combine the three numerical terms, 208, and 1/3, and 1.732, you are left with 120. That brings us back to,
? Watts equals 120 times (Ia plus Ib plus Ic).
 
Kevin,

The data center client wants to monitor the server cabinet watt load against the cabinet watt rating to assure that the available cooling equipment around the cabinet can handle the load under all circumstances. The circuit amp monitoring also provides assurance that there is 2N redundancy with the "sister circuit".

I wish I knew about this forum years ago. Great stuff.
 
charlie b said:
That is not strictly true. But it does give you the best available approximation that you can get, without going deeply into mathematics.
I would like to caution you against making any habit of ?adding the three together,? when you are dealing with three phase systems. You can?t add three phase currents as though they were the same type of thing.
Click to expand...
The OP wanted to know how to caculate the watts, not the amps.
 
bob said:
The OP wanted to know how to caculate the watts, not the amps.
Click to expand...
I know. It was a general comment about adding things in three phase systems. You will note that each of the bulleted items in my "proof" begins with "Watts equals."
 
Hey Brad,

Where I work we have been fighting the same battle. We use the HP 10000 series racks 42U. We could put anywhere between 10 to 13 HP DL380 servers in these racks. The configuration will depend on what the project is, we may have a some HP 580's and network switches in there somewhere. We started with the G2 family which has redundant 400W power supplies so we started with two 30 amp 208V single phase circuits which worked out great at about 2 amps per server. I like to try and do my math at 80% but my boss doesn't always like it that way. My 30 amp circuit at 24 amps worked out to be 12 servers, and life was good. Now we are at G4 and G5 class DL380's and they have up to 575W redundant power supplies, that is 575W / 208V= 2.76Amps per server now, and that is if we just install the 380's. A G4 or G5 DL580, DL585 will have 800W power supplies (3.8 Amps). We were having trouble keeping up with the change of technology. We went from a rack rating of 5-6KW up to 10KW. We now run two 30Amp 208V 3-Phase circuits to each rack, inside the rack we have two of what HP calls PDUs. Look at HP's web site Part# 370960-D71 This allows us to monitor the load on each line of the circuit with a digital read out.. Works great. I am sure we will be changing again...
 
Status
Not open for further replies.
Top