Question About Unbalanced Load Amperage On A Delta System

Thread Starter

PGB1

Joined Jan 15, 2013
110
Good Morning Everyone!

The question below is something that I certainly used to know the answer to, but the answer has apparently fallen out of my brain.
I think it belongs in this section because it is education related. (I wasn't sure if it belonged here or in Physics)

It relates to the amperage load when connecting two heating elements on a 3-Phase delta supply.
The only way I can describe my question is by the examples below. I'll start with a Wye system to hopefully make my question clear. (Yeah, sure....)

For simplicity, the amp draw on the sample loads will be 20 each, even though the voltage is different on each example. There are no real parts involved, just hypothetical.

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WYE SYSTEM WE KNOW THE ANSWER-
Picturing 2 purely resistance loads (for example heating elements) of 20 amps per item on a 120/208 volt Wye system as:

Element One is wired across L-1 & L-2

Element Two is wired across L-2 & L-3

Reading the amperage between L-1 & the element, we will have 20 amps.

Reading the amperage between L-3 & the element, we will have 20 amps.

Reading the amperage between L-2 & where both elements are connected, we will have 40 amps.

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BUT ON A DELTA SYSTEM = ?
Picturing 2 purely resistance loads (for example heating elements) of 20 amps per item on a 125/240 volt Delta system, where

L-1 & L-3 to ground are each 125 volts
L-2 to ground is 208 volts

L-1 to L-2 reads 240 volts
L-1 to L-3 reads 240 volts
L-2 to L-3 reads 240 volts

Element One is across L-1 and L-2
Element Two is across L-3 and L-2

Reading the amperage between L-1 and the element, we will read 20 amps.

Reading the amperage between L-3 and the element, we will read 20 amps.


THE QUESTION-
But, reading the amperage between L-2 and where both elements are tied together, will we have 20 amps, 40 amps or
20 amps times the square root of 3?

Thank You All for putting up with this "I should remember this" question. I appreciate your help!
Paul
 

Thread Starter

PGB1

Joined Jan 15, 2013
110
Thank You WBahn for taking time to reply.


I apologize, but today is one of the all to frequent cloudy brain days. (Age is sometimes quite inconvenient.) I also apologize for using L-1, L-2, L-3. I should have named Phase A, B, C to be more conventional.

QUOTE FROM Y FIRST POST-
Reading the amperage between L-2 & where both elements are connected, we will have 40 amps.
Why do you believe that this is the case?


I intended to type that I believe that we would have 20 amps at L-2 where both elements (loads) tie together. At least in the WYE system...
------------------------------------------------
I probably went into confusion mode because if this were a single phase supply and I had 20 amp load between L1 & Neutral and 20 amp load between L2 and Neutral, the neutral current would be 20.
And if it were a 3-Phase system with 20 amp loads from each of the three phases to neutral, the neutral current would read zero. (If I remember properly the math...)
However, this isn't load-across-neutral. Both loads connect to phase L2.

I (hope I still) remember the formula for unbalanced loads in three phase: (I being amperage L being the phase)
IL = Square Root Of 1Asq + IBsq + ICsq - (IA x IB) - (IB x IC) - (IC x IA)

But, since we are not using a neutral in my example, what would the math be for the L-2 amperage?
Would it be the same for Delta and Wye, since Delta has L-2 at higher voltage to ground?
Would we look at the loads as two single phase loads since there are only 2 loads and find L-2 current is equal to either one of the (balanced) loads?

Thanks Again for refreshing my brain. Please pardon if these are ignorant questions. Perhaps tomorrow, or next week, the answer to my question will be crystal clear as if I never forgot how to reason them out in my mind.

I appreciate your help & the education!
Paul
 
Last edited:

WBahn

Joined Mar 31, 2012
24,974
Thank You WBahn for taking time to reply.


I apologize, but today is one of the all to frequent cloudy brain days. (Age is sometimes quite inconvenient.) I also apologize for using L-1, L-2, L-3. I should have named Phase A, B, C to be more conventional.

QUOTE FROM Y FIRST POST-
Reading the amperage between L-2 & where both elements are connected, we will have 40 amps.
Why do you believe that this is the case?


I intended to type that I believe that we would have 20 amps at L-2 where both elements (loads) tie together. At least in the WYE system...
------------------------------------------------
I probably went into confusion mode because if this were a single phase supply and I had 20 amp load between L1 & Neutral and 20 amp load between L2 and Neutral, the neutral current would be 20.
Why do you thing that this would be the case?

Take a single phase supply and put two equal loads in series between L1 and L2 such that they draw 20 A. How much current is flowing in the neutral? None! It's not even connected. What is the voltage at the junction of the two loads? Zero volts, by symmetry. So now connect the neutral to that junction. How much current will flow in it? Zero, because there's no voltage there.
 

Thread Starter

PGB1

Joined Jan 15, 2013
110
Thank You Again for helping me out. After reading your above reply, it all is coming back to me (slowly, but surely).

To take this further, if I connect two 20 amp loads to a 3 phase delta supply (rough sketch attached) as-
Load One is across Phases A & B, Load Two is across Phases B & C (Each is line-To-Line. None go to neutral.)

I realize that an ammeter connected between Load One and the transformer Phase A tap will read 20 amps and a meter connected between Load Two and transformer Phase C tap will read 20 amps. But, since it is an unbalanced load on the transformer, will a meter connected between both loads and transformer Phase B tap read 20 or 40? From the math I remember, I believe it will read 20 amps.

Did I get this correct or goof it up again?
 

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WBahn

Joined Mar 31, 2012
24,974
Thank You Again for helping me out. After reading your above reply, it all is coming back to me (slowly, but surely).

To take this further, if I connect two 20 amp loads to a 3 phase delta supply (rough sketch attached) as-
Load One is across Phases A & B, Load Two is across Phases B & C (Each is line-To-Line. None go to neutral.)

I realize that an ammeter connected between Load One and the transformer Phase A tap will read 20 amps and a meter connected between Load Two and transformer Phase C tap will read 20 amps. But, since it is an unbalanced load on the transformer, will a meter connected between both loads and transformer Phase B tap read 20 or 40? From the math I remember, I believe it will read 20 amps.

Did I get this correct or goof it up again?
The two 20 A readings have a phase associated with them. Your meter is only showing the amplitude (probably scaled as RMS) with no information about the phase.

When two such signals are connected and add together, the result can be anywhere from A-B to A+B, depending on the phase relationship.

You need to do the math -- taking the phase into account.
 

Thread Starter

PGB1

Joined Jan 15, 2013
110
Thanks Once Again!
I found & studied my formula books, did the math for both Delta and Wye with the unbalanced loads. The results were quite interesting.

By coincidence, I was working in a Delta system and had the opportunity to do a quick test of an un-balanced load on a low voltage (240 ph-ph) 3-phase branch circuit. (Balanced loads are easy for me because that is most of my experience. Unbalanced was what caused my original post.)
The loads were pure resistance loads, so I assumed my Phase Voltage would be just about the same as Line Voltage.

After doing the math, I tested.
My amperages at Phase A & Phase C were just about as-expected. Surprisingly, my amperage reading at Phase B was also just about as expected. (A & C were each 36.67 amps with 8800 watt loads at 240 vac- 6.55 Ohm). B was 63.94 amps. (Amps Line-Line times 1.73) I account for the small difference between math & as-tested to be voltage drop with the loads running, meter variances, etc. It was less than 2 amps on each.

It would be fun to calculate an unbalanced induction load, such as a motor or lighting transformer, and test how my math worked out.

Thank You Again for helping. You are a great teacher!
Paul
 
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