Calculating expected current in a 3-phase delta circuit 3

[PaulKraemer]

Hi,

I have three electric heaters with resistance 82.3 ohms each, wired in a delta configuration as shown on the left below. As I see it, each 480 VAC line-to-line voltage is applied across a parallel circuit as shown on the right below.

I calculate the total resistance of the circuit on the right as 54.87 ohms. The formula I = E/R = 480/54.87 = 8.75 amps leads me to believe that my expected current draw on each line would be 8.75 amps, but I am not entirely confident about this.

I am not sure if the delta-configuration introduces any complexity that would make my simple calculation invalid.

If anyone can tell me if my calculation is correction and if not, how/where I went wrong, I would greatly appreciate it.

Thanks in advance,
Paul

 

[sushilksk]

After rearranging your diagram:

Now, you may re-calculate the currents in each resistor, and total phase current.

 

[PaulKraemer]

Hi sushilksk,

Thank you for your response. Your rearranged drawing makes me believe that with line-to-line voltage of 480 VAC, the current through each resistor (indicated by I1, I2, and I3 below) would be 5.83 amps.

What I am still not clear on is how to determine what the currents will be at I[sub]A[/sub], I[sub]B[/sub], and I[sub]C[/sub]. The illustration makes it look like I[sub]B[/sub] is feeding both I1 and I2, but being this is alternating current and not always flowing in the same direction, I do not think that I[sub]B[/sub] = I1 + I2. Am I right?

I know in 3-phase circuits, a factor of 1.73 sometimes comes into play. Would I[sub]B[/sub] = (I1 + I2) / 1.73 ?

Thanks again,
Paul

 

[IRstuff]

see:

https://www.watlow.com/resources-and-support/engineering-tools/3phase-delta-wye-calculator

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

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[mparenteau]

Kirchhoff's current law?

Mike

 

[waross]

For a balanced load the line current is 1.73 times the phase current.
For unbalanced loads you must take the directed sums of the phase currents.

I understand the thinking behind your question.
The effect that you suspect applies not to the load current but to the current through the transformer windings.
When a single phase load is applied to a delta transformer bank, all three phases contribute to the current.
The in phase winding contributes 50% of the current and the out of phase transformers contribute 50% of the current.
At first glance this may look like over unity, but you must consider the phase angles.
The current through the out of phase windings is at a power factor of 50% so one winding supplies 50% of the kW and the other two windings together supply the other 50% of the kw.
Now if we add another equal single phase load on another phase, the same division of current occurs.
On the common phase, the interaction of the various phase angles of the currents resolves into a current of 1.73 times the individual currents.
The relationship between the line currents and the transformer winding currents are not straightforward when unbalanced loads are applied to a delta transformer bank.
For instance the load limits of a single phase load on a delta transformer bank are somewhat counter-intuitive.

For example consider a 150 KVA, delta transformer.
The rated balanced loading at three phases is 50 KVA per phase.
That will result in rated current through each transformer winding.
The maximum single phase load that may be applied to the same transformer bank is 100 KVA.
That will result in rated current through each transformer winding.
An interesting side note: the losses of a delta transformer supplying a 100 KVA single phase load will be the same as the losses of the same transformer supplying a 150 KVA three phase load.
Note: In this discussion the loads are assumed to be at unity power factor.




Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[waross]

Kirchhoff's current law?

Always valid.
Different phase angles may complicate the calculation but the instantaneous currents always follow Kirchhoff's current law.
With different phase angles, the directed sums of the currents follow Kirchhoff's current law.
However, when the currents are at different phase angles, they cannot be considered arithmetically.

As an example:
In a simple three wire single phase circuit I once measured 11 Amps, 9 Amps and 7 Amps.
By a simple arithmetic application of Kirchhoff's current law I should have seen 11 Amps, 9 Amps and 2 Amps.
On investigation I discovered that one of the circuits was running at a very low power factor.
When the resulting phase angle was considered, Kirchhoff's current law was completely valid.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[PaulKraemer]

Hi Bill, IRStuff, mparenteau, and sushilksk,

In response to Bill, my 3-phase delta circuit is actually not a transformer. It is just three heaters of the same resistance. I agree that the diagram that sushilksk sent (below) accurately represents my connections:

With this being the case, can I assume that this is a balanced load, and therefore my line current I[sub]A[/sub] = I[sub]B[/sub] = I[sub]C[/sub] = 5.83 * 1.73 = 10.1 amps? This is what I come up with using the Watlow calculator referenced by IRStuff as shown below:

What confuses me about this Watlow calculator is that it only shows the line-to-line voltage applied across the points I have labeled "A" and "B" on the delta, whereas in reality, I have line-to-line voltage applied across A-B, A-C, and B-C.

Am I using this calculator correctly?

Thanks again - I really appreciate your help.

Best regards,
Paul

 

[waross]

Actually we had a delta confusion a couple of days ago in another thread.
Delta implies a delta connected supply.
For loads, consider it a line to line connected load.

When two directed vectors with a unit value of one and at a displacement of 120 degrees are added the arithmetic sum is two units,
but the directed sum is root three units.
Just use 1.73 times the load current for line to line loads on different phases.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[itsmoked]

Paul; The Watlow dorks left out one of the voltages in their diagram. The LINE currents should, indeed, be 10.1A

Keith Cress
kcress -

http://www.flaminsystems.com

 

[PaulKraemer]

Thanks Bill,

Just to make sure I am clear, when you say "just use 1.73 times the load current for line to line loads on different phases", am I correct to assume that in the diagram below, what I have calculated as 5.83 amps (480 VAC / 82.3 ohms) is the correct value to use for "load current". If this is correct, I believe this would indicate my expected current on each line (indicated by I[sub]A[/sub] , I[sub]B[/sub], and I[sub]C[/sub] would be 5.83 amps * 1.73 = 10.1 amps. This is the same value I came out with using the Watlow tool, so I am fairly confident that I am on the right track. I just want to make sure. I really appreciate your help. Thanks and best regards, Paul

 

[waross]

Hi Paul:
To avoid minimize rounding errors lets run 480 Volts / 82.3 Ohms x root three on the calculator.
= 10.1018759129169‬ Amps. 10.1 is close enough for me.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[waross]

Hi Keith.
I tried the Watlow calculator for open delta. You had to know the answer before hand to understand what question was being answered.
I'm going to lose that link.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[waross]

Looks as if you have added (or subtracted) a little rounding error.
Rounding errors may change the number but they won't change the actual current.
When looking for the best accuracy I try to use the full range of the calculator rather than rounded or truncated values.
For a quick answer I will use 1.73.
When I want to avoid cumulative rounding errors I will use Root 3 as calculated by the calculator; (1.732050807568877‬‬‬‬)


Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[sushilksk]

We can slove like this:

Now vector calculation I1 = IR1 - IR3 = 5.83 - 5.83 (120˚) = 10.09 A

 

[IRstuff]

Bill, that looks more like pi than the square root of 3 ;-)

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[IRstuff]

The phasor calculation has to have phase offsets in both IR1 and IR2 and sign convention has to be rigorous.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

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[che12345]

Dear Mr. PaulKraemer
I hope the attachment would offer some help to your problem.

Che Kuan Yau (Singapore)

 

[waross]

IR, ouch. corrected.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[sushilksk]

It should be i[sub]12[/sub] - i[sub]31[/sub]