On a single phase 120/240v service what is the correct way to figure your amperage? For example :With a Amp meter I have 60 amps on one leg and 40 amps on the other leg.What would be my total amp load on that service? I have been told you never add amps so what is the correct way to figure it ? Thanks !
Figuring Single Phase Amps
- Thread starter dsmith411
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shamsdebout
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Any measurement on the neutral? What is the service size? If I had to guess what system was drawing I would take the average of current on each leg (50A), rational: in a balanced system each leg should have the same current, so you can't add them up to get total current draw. With the info you presented that is all I can come up with. I think it is a good question I ran into this once and I was surprised at the various answers, my boss who is a PE told me to add them up, my co-worker who is an EE, and EC, took the opportunity to show how wrong my boss was.
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Welcome to the Forum.
I'm uncertain what you are trying to ascertain.
What you have is your "amps" per phase. This is important as your main breaker will be looking at each phase and if either exceeds it's limit it should open.
In terms of "load", the figure is most often expressed in watts or kilowatts in order to size transformers, etc.
You actually have to have a handle on both numbers.
I'm uncertain what you are trying to ascertain.
What you have is your "amps" per phase. This is important as your main breaker will be looking at each phase and if either exceeds it's limit it should open.
In terms of "load", the figure is most often expressed in watts or kilowatts in order to size transformers, etc.
You actually have to have a handle on both numbers.
shamsdebout
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I should have waited on you to respond.
gar
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dsmith411:
From a perspective of the loading on the transformer and thus temperature rise of the transformer consider the following:
1. Power dissipation in the transformer determines its temperature rise. More important is the absolute internal maximum hot spot temperature of the transformer. This is the sum of the ambient temperature and the temperature rise. A transformer will usually be defined as the temperature rise and some assumed maximum ambient temperature.
2. The power dissipation in each half of the secondary is the current in that half squared times the resistance of that half.
3. The power dissipated in the primary is the primary current squared times the primary resistance. Note: the primary current could be less than the reflected equivalent of the sum of the secondary currents depending upon the phase relationship of the secondary currents. Also there is some primary excitation current for the transformer core.
4. There are core losses in the transformer that add to the power dissipation, but these are somewhat independent of load current.
What does all this mean. Suppose the transformer has two identical secondary coils, and is rated for 10 KVA. Then you should not load either secondary with more than 5 KVA.
Suppose you put a 10 KVA load on one secondary and 0 KVA on the other side. This doubles the current in that one side and therefore raises the power dissipation by a factor of 4. That secondary will all most certainly burn out. The primary won't care because it is still within its specification.
What do I mean by reflected current? A 1 to 1 transformer with one secondary will have a primary current that approximately equals the secondary current. Put 10 times the turns on the primary as compared to the secondary, then the primary current is approximately 1/10 the secondary current. Reflected current to the primary is the current seen from the primary side resulting from current in the secondary.
Edit: As an approximation the current at the primary is the sum of the secondary currents times the turns ratio or divided depending upon which way you define turns ratio. You simply use logic to determine whether to divide or multiply, or you make a rigorous definition of how you define turns ratio.
.
dsmith411:
From a perspective of the loading on the transformer and thus temperature rise of the transformer consider the following:
1. Power dissipation in the transformer determines its temperature rise. More important is the absolute internal maximum hot spot temperature of the transformer. This is the sum of the ambient temperature and the temperature rise. A transformer will usually be defined as the temperature rise and some assumed maximum ambient temperature.
2. The power dissipation in each half of the secondary is the current in that half squared times the resistance of that half.
3. The power dissipated in the primary is the primary current squared times the primary resistance. Note: the primary current could be less than the reflected equivalent of the sum of the secondary currents depending upon the phase relationship of the secondary currents. Also there is some primary excitation current for the transformer core.
4. There are core losses in the transformer that add to the power dissipation, but these are somewhat independent of load current.
What does all this mean. Suppose the transformer has two identical secondary coils, and is rated for 10 KVA. Then you should not load either secondary with more than 5 KVA.
Suppose you put a 10 KVA load on one secondary and 0 KVA on the other side. This doubles the current in that one side and therefore raises the power dissipation by a factor of 4. That secondary will all most certainly burn out. The primary won't care because it is still within its specification.
What do I mean by reflected current? A 1 to 1 transformer with one secondary will have a primary current that approximately equals the secondary current. Put 10 times the turns on the primary as compared to the secondary, then the primary current is approximately 1/10 the secondary current. Reflected current to the primary is the current seen from the primary side resulting from current in the secondary.
Edit: As an approximation the current at the primary is the sum of the secondary currents times the turns ratio or divided depending upon which way you define turns ratio. You simply use logic to determine whether to divide or multiply, or you make a rigorous definition of how you define turns ratio.
.
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skeshesh
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- Los Angeles, Ca
It depends on what you're trying to accomplish. From your question I'm guessing you're asking if you were filling out a regular panel schedule and were trying to indicated "total current" at the bottom what would you write. I'd have to agree that I've often seen the average value written in that spot. Similar to how you would add all the load from three phases A,B & C in a 120/208, and divide by 208*1.732 (or 120*3), you would add the two phases in a 240V panel and divide by 240(or 120*2). Be careful to understand that what you're getting is average current draw per phase. As mentioned by Gar for purpose of power system analysis and load summaries you would need to look at that panel, how much its total load is in kVA and many other factors such as where it's fed from, supplying Xformer, distance from supply, other loads fed from same supply, etc.
K8MHZ
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- Electrician
You have 100 amps at 120 volts.
Since services are rated at 240 volts you might like to think like you now have 50 amps at 240 volts.
However.....
The OCPDs are in each 120 volt leg, so it is possible that transposing like I did above can get you in trouble.
Your measurements would max out a 60 amp 240 volt OCPD even though it seems you only have 50 amps at 240 volts. What matters is the max going through either leg, which in your case is 60 amps.
Theoretically, you can max out a 100 amp 240 volt OCPD with 100 amps at 120 volts, so long as the entire load is on one leg.
This little mathematical nuance is why it is important to balance your loads as much as possible.
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