Power per phase - 2 pole circuit, why half?

[g1ant]

Hi,

With reference to this thread - http://forums.mikeholt.com/showthread.php?t=79044&page=2

Part of the answer states:
(6) For a single phase, 2-pole, 208V load, assign half that kva to each of the two phases to which the load is connected.

Could anyone explain the theory/maths behind this statement please.

I am currently working on a 2-pole trace heating system and am trying to work out this exact question.

Thanks in advance,
Matt

 

[GoldDigger]

Ultimately, you will get to the point of converting the power drawn to current in the phase wires.
If you add up 100% of the power divided by the phase to neutral voltage for each load on both of the phase wires and then divide by the phase to neutral voltage you will get twice the actual current that will be measured on the phase lines.
If you split the wattage evenly between the two phases present in the connection and then apply the three phase power formula you will get the correct line currents.
Another way of looking at it is that you do it because it works and gives the right answer.

 

[g1ant]

Hi Golddigger,

Thanks alot for the response.

I guess my next question therefore is:

Why, in a two pole system, do you get (when measured) half the current in each phase conductor to the single resistive load? Or in reference to your answer - Why split the wattage evenly? What is the theory behind this?

If this was supplied with a single pole circuit (V line to neutral) then you would see 100% of the current in the single pole as it feeds the load (and the electricity flows from live to neutral in the circuit). Which way then does the current flow in a two pole circuit? Is it constantly changing direction?

Thanks again,
Matt

 

[GoldDigger]

Hi Golddigger,

Thanks alot for the response.

I guess my next question therefore is:

Why, in a two pole system, do you get (when measured) half the current in each phase conductor to the single resistive load? Or in reference to your answer - Why split the wattage evenly? What is the theory behind this?

If this was supplied with a single pole circuit (V line to neutral) then you would see 100% of the current in the single pole as it feeds the load (and the electricity flows from live to neutral in the circuit). Which way then does the current flow in a two pole circuit? Is it constantly changing direction?

Thanks again,
Matt

Well, in a two pole load in a single phase 120/240 system you do NOT get half the current in each phase conductor. Instead you get half the *voltage* between each phase conductor and ground.
That means that if you divide the full kVA by 240 you get the correct currentvand if you divide half the kVA by 120 you get the same value.
It really depends on what you are going to do with the total kVA number you add up for each phase.

 

[kwired]

Hi Golddigger,

Thanks alot for the response.

I guess my next question therefore is:

Why, in a two pole system, do you get (when measured) half the current in each phase conductor to the single resistive load? Or in reference to your answer - Why split the wattage evenly? What is the theory behind this?

If this was supplied with a single pole circuit (V line to neutral) then you would see 100% of the current in the single pole as it feeds the load (and the electricity flows from live to neutral in the circuit). Which way then does the current flow in a two pole circuit? Is it constantly changing direction?

Thanks again,
Matt

GD mostly answered your question, I will just reaffirm it and maybe in slightly different manner.

With two pole load you still have all of current flowing through the load, it is power that you are only considering half per "phase conductor" of the supply.

Say you had 30 KVA balanced across all three phases. Each phase conductor is seeing 10 KVA.

Now add another 10 KVA between A and B phase.

If you were to add 10 to the total panel schedule for both A and B you would have a total of 50 for the panel - but you only actually have 40 total KVA

Starting to get clearer?

 

[Smart $]

Hi,

With reference to this thread - http://forums.mikeholt.com/showthread.php?t=79044&page=2

Part of the answer states:
(6) For a single phase, 2-pole, 208V load, assign half that kva to each of the two phases to which the load is connected.

Could anyone explain the theory/maths behind this statement please.

I am currently working on a 2-pole trace heating system and am trying to work out this exact question...

...

Why, in a two pole system, do you get (when measured) half the current in each phase conductor to the single resistive load? Or in reference to your answer - Why split the wattage evenly? What is the theory behind this?

...

Welcome to the forum, Matt. :thumbsup:

FWIW, your link went to the wrong thread (for me). Here's a link to the post you quoted...
http://forums.mikeholt.com/showthread.php?t=54480&p=395522#post395522

The statement quoted is in reference to line current calculations on a 3? system, where the individual load currents are combined. Let's use 10A 1? loads as an example. The current through the two wires going to an individual load is 10A.

Now let's say you have 3 connected AB, BC, and CA. There will be two connected to leg A (AB and CA). Because the currents are out of phase, their currents combined is 17.32A. Assuming a 208/120V system...

10A ? 208V ? 3 = 6240VA
6240VA ? (208 ? 1.732) = 17.32A per leg

Where the system is 208/120, you multiply 17.32A times 120V and you get 2078.4VA. Each load by itself is 10A times 208V for 2080VA. The difference is rounding error. But you can see the VA on one leg is the same as one load... or half of each of the two connected loads (1/2 times 2 is 1). The reason the VA is halved is because not all load calculations have exactly balanced loads. What if you had three loads at 10, 15, and 20A. While summing half the VA of each connected load per leg is not completely accurate, it is a conventionally-accepted method.

 

[charlie b]

FWIW, your link went to the wrong thread (for me). Here's a link to the post you quoted...
http://forums.mikeholt.com/showthread.php?t=54480&p=395522#post395522

I thought the text quoted at the begining of this thread looked familiar. It was my comment from over 12 years ago! Be careful what you post on the Internet my friends. It sticks around forever. :lol: