liebert 225kva PDU
- Thread starter jcapp
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- Lockport, IL
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- Semi-Retired Electrical Engineer
Re: liebert 225kva PDU
In most instances, when you say ?per,? someone is going to multiply something to get a total. If you have four boxes, and if there are five books per box, then you have 4 times 5, or 20 books. But this math does not work for current. If you have three legs, and if you have 625 amps per leg, then the total current is 625 amps. The total current is not 3 times 625 amps. So long as you keep this in mind, I will try not to cringe at the phrase ?per leg.?
Steve has the right answer, but you asked the wrong question. I cringe when I see the phrase ?per leg? used in connection with current. It?s ok for conversational use, but please be careful to make clear your intended meaning.
In most instances, when you say ?per,? someone is going to multiply something to get a total. If you have four boxes, and if there are five books per box, then you have 4 times 5, or 20 books. But this math does not work for current. If you have three legs, and if you have 625 amps per leg, then the total current is 625 amps. The total current is not 3 times 625 amps. So long as you keep this in mind, I will try not to cringe at the phrase ?per leg.?
Re: liebert 225kva PDU
For your example question: the answer is yes if all the loads are 1 phase 120V (line to neutral loads).
For the UPS:
On each leg (or phase), you can place 625 amps of line to neutral loads (120V loads). In this case, you would have 625A * 120V * 3 legs = 225KVA.
OR, if your loads are line - line, you will have two loads connected to each line (One from A-B and one from A-C. Same with phases B and C.) In this case, EACH line-line load should draw a maximum of 360 Amps (625/ sqrt(3)). This is because the addition of the current drawn from two loads will be sqrt(3) times the current of each load. The total power drawn could be calculated by: 208V * 625A * sqrt(3) = 225 KVA.
An easier way to understand the total power in this case would be: 208V (across each load) * 360A (the current through each load) * 3 (total number of loads). It is still 225 KVA.
Hope that helps.
Steve
For your example question: the answer is yes if all the loads are 1 phase 120V (line to neutral loads).
For the UPS:
On each leg (or phase), you can place 625 amps of line to neutral loads (120V loads). In this case, you would have 625A * 120V * 3 legs = 225KVA.
OR, if your loads are line - line, you will have two loads connected to each line (One from A-B and one from A-C. Same with phases B and C.) In this case, EACH line-line load should draw a maximum of 360 Amps (625/ sqrt(3)). This is because the addition of the current drawn from two loads will be sqrt(3) times the current of each load. The total power drawn could be calculated by: 208V * 625A * sqrt(3) = 225 KVA.
An easier way to understand the total power in this case would be: 208V (across each load) * 360A (the current through each load) * 3 (total number of loads). It is still 225 KVA.
Hope that helps.
Steve
- Location
- Lockport, IL
- Occupation
- Semi-Retired Electrical Engineer
Re: liebert 225kva PDU
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So if you want to use a 100 amp breaker, the current in each phase can be as high as 100 amps. The power you will be delivering is found by 100 amps times 208 volts times the square root of three, or 36 KVA. This will work, but you will only be using 36 of your available 225 KVA.
No, load per leg would be 100 amps, and the total is 100 amps (not 300). You are trying to add an ?amp? to an ?amp,? and are looking for an answer in ?amps.? When you deal with three phase, the math does not work that way. Let me try a simple mechanical analogy:
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- <font size="2" face="Verdana, Helvetica, sans-serif">If you have three people each holding a garden hose over a tank, and if each hose is sending 100 gallons per minute into the tank, then you have ?100 gpm per hose, 300 gpm total.? The math works this way because every gallon leaving one hose goes into the tank, and none of the water leaving one hose goes backwards into one of the other two hoses.</font>
So if you want to use a 100 amp breaker, the current in each phase can be as high as 100 amps. The power you will be delivering is found by 100 amps times 208 volts times the square root of three, or 36 KVA. This will work, but you will only be using 36 of your available 225 KVA.
- Location
- Lockport, IL
- Occupation
- Semi-Retired Electrical Engineer
Re: liebert 225kva PDU
Looks like Dereck and I had the same thing to say, and he beat me to the draw.
Looks like Dereck and I had the same thing to say, and he beat me to the draw.
Re: liebert 225kva PDU
my question was unclear. I understand that I have only 225kva of power available. I was looking in
terms of overcurrent protection. When I supply subs, I can load each leg to the value of the protection. But cannot exceed the total value of the source. Correct?
my question was unclear. I understand that I have only 225kva of power available. I was looking in
terms of overcurrent protection. When I supply subs, I can load each leg to the value of the protection. But cannot exceed the total value of the source. Correct?
- Location
- Lockport, IL
- Occupation
- Semi-Retired Electrical Engineer
Re: liebert 225kva PDU
True. But a better way to say it is the other way around. The purpose of the protection is to protect the cable. You have to have a protection setpoint that is no higher than the capabilities of the conductor.
Correct.
jafro
Member
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- Fort Saskatchewan, Alberta, Canada
Re: liebert 225kva PDU
I don't mean to be rude or anything but your profile says you are an inspector. I am also an inspector and this should be well within your technical expertise. Please explain
Jafro
I don't mean to be rude or anything but your profile says you are an inspector. I am also an inspector and this should be well within your technical expertise. Please explain
Jafro
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