Current Calc: 2 Phase 3 Wire

[mivey]

Never mind. I see you were talking about the real component.

 

[mivey]

Never mind. I see you were talking about the real component.

BTW, I was responding to jghrist. And I follow what you were saying about the sign.

...At this point I wish I hadn't said anything.

Never hurts to be correct the error in case a newbie reads it later. Clarity is a good thing. It also keeps us from being sloppy (like my first post in the thread).

 

[jghrist]

BTW, I was responding to jghrist. And I follow what you were saying about the sign.Never hurts to be correct the error in case a newbie reads it later. Clarity is a good thing. It also keeps us from being sloppy (like my first post in the thread).

Except that Post #8 really was wrong. I'm not sure why the angles get screwed up in the math of dividing P by V vectorially. It might require getting into complex conjugates. I don't want to go there.

 

[bob]

Originally Posted by dkarst
Sorry but the value you arrived at above for the combined leg is incorrect, you cannot multiply the phase current by 1.73 to get line current in this unbalanced delta case. The correct value is 2.72 A

I would really appreciate if you could elaborate on this.

Original post

Consider a purely resistive heater circuit; 1 - 600W between phases 1 & 2 and 1 ? 900W between phases 2 & 3; for a total of 1500W.

You only have load between phase 1-2 and phase 2-3. there is no load between 1-3. Since you do not have a 3 phase load the 1.73 factor does not come into play. This is an unbalanced single phase load.

 

[mivey]

Except that Post #8 really was wrong. I'm not sure why the angles get screwed up in the math of dividing P by V vectorially. It might require getting into complex conjugates. I don't want to go there.

Since we are just having fun at this point:

I12* = 600/480@0? = 1.25@0? = 1.25 + j0 => I12 = 1.25 - j0

I23* = 900/480@-120? = 1.875@120? = -0.9375 + j1.6238 => I23 = -0.9375 - j1.6238

I2 = I23 - I12 = (-0.9375 - j1.6238) - (1.25 - j0) = (-0.9375 - 1.25) + j(-1.6238 + 0) = -2.1875 - j1.6238 = 2.7243@-143.41?

For those playing along:

Instantaneous power, P(t), is always v(t) * i(t) but of little interest most of the time. The average power, Pave, is Vrms * Irms * cos(?) or 0.5 * Vmax * Imax * cos(?) where cos(?) gives the power factor and ? is the angle difference between the voltage and current waveform at any instant in time.

The complex conjugate is used when we use phasor notation for voltages and currents and is due to our convention of angle directions where we say that a lagging current has a negative phase angle when the voltage has a zero phase angle.

Using the complex conjugate then gives us a positive angle for the power:
Irms@-A?* x Vrms@0? = Irms@A? x Vrms@0? = IrmsVrms@A?

 

[74one]

... Since you do not have a 3 phase load the 1.73 factor does not come into play. This is an unbalanced single phase load.[/QUOTE]

I would agree if addition to the missing load between 1-3, leg #2 was disconnected from the power source but this is not the case.

 

[mivey]

...Since you do not have a 3 phase load the 1.73 factor does not come into play. This is an unbalanced single phase load.

I would agree if addition to the missing load between 1-3, leg #2 was disconnected from the power source but this is not the case.

It is actually two unbalanced single-phase loads (or were you disputing the 1.73 part?)

 

[74one]

It is actually two unbalanced single-phase loads (or were you disputing the 1.73 part?)

I would say this is unbalanced 3phase circuit, and if we are not truing figure out current thru leg #2 only at the given moment, then I believe this is correct formula for leg #2: Leg#2=1500w/480v*1.73=1.8A

 

[mivey]

I would say this is unbalanced 3phase circuit

There is a difference when discussing the source and the load. A source can play multiple roles, a defined load does not.

...and if we are not truing figure out current thru leg #2 only at the given moment, then I believe this is correct formula for leg #2: Leg#2=1500w/480v*1.73=1.8A

Of course, but watch the grouping/calc order: 1500w/480v/1.73=1.8A
or 1500w/(480v*1.73)=1.8A

 

[Marc L]

Do parden if this is a stupid question because I've been out of school for 25 years and never worked as an engineer:

Shouldn't the 2 vectors be added together instead?

Original quote by mivey:

I2 = I23 - I12

The magnitude of the current in the common phase wire has got to match the magnitude of the vector sum of the currents in the first two legs.

Sherlock

 

[dkarst]

Of course, but watch the grouping/calc order: 1500w/480v/1.73=1.8A
or 1500w/(480v*1.73)=1.8A

Now I'm confused, I agree with your post a few back (and mine ~15 back) that the current in common (#2) leg is 2.72A, not 1.8 A ? You're not saying it is 1.8A, correct?

 

[mivey]

Do parden if this is a stupid question because I've been out of school for 25 years and never worked as an engineer:

Shouldn't the 2 vectors be added together instead?



The magnitude of the current in the common phase wire has got to match the magnitude of the vector sum of the currents in the first two legs.

Sherlock

It is just a sign thing. One current was headed towards the node and one was headed away from the node so we subtracted one to get the sum.

 

[mivey]

Different scenario

Different scenario

Now I'm confused, I agree with your post a few back (and mine ~15 back) that the current in common (#2) leg is 2.72A, not 1.8 A ? You're not saying it is 1.8A, correct?

The 1.8 was for a "what if was not an open delta but a closed delta" scenario. Not the same as what the OP asked.

 

[dkarst]

The 1.8 was for a "what if was not an open delta but a closed delta" scenario. Not the same as what the OP asked.

Thank you, I missed the "what if" and thought we had a new answer to the OP question...

 

[mivey]

Thank you, I missed the "what if" and thought we had a new answer to the OP question...

I hope not. I can barely see my screen for the White-Out.

 

[74one]

What would be your calculations for the current in the leg #2 if we assume: V12=480V @ 90?, V23=480V @ -30?, V31=480V @ 210??

 

[dtonjes]

Come on guy!! Simple two single phase loads. 3.125 amps!!! Y-delta makes no difference, SINGLE PHASE LOADS. Ah, but what would the VOLTAGE be?

 

[mivey]

What would be your calculations for the current in the leg #2 if we assume: V12=480V @ 90?, V23=480V @ -30?, V31=480V @ 210??

Using the direction conventions given in #18 (important to keep the signs consistent), and the method from #25:
I12* = 600/480@90? = 1.25@-90? = 0 - j1.25 => I12 = 0 + j1.25

I23* = 900/480@-30? = 1.875@30? = 1.6238 + j0.9375 => I23 = 1.6238 - j0.9375

I2 = I23 - I12 = (1.6238 - j0.9375) - (0 + j1.25) = (1.6238 - 0) + j(-0.9375 - 1.25) = 1.6238 - j2.1875 = 2.7243@-53.4132?

You can also solve for the impedances for another method:
R12 = V12^2 / P12 = 480^2 / 600 = 384 ohms
R23 = V23^2 / P23 = 480^2 / 900 = 256 ohms
|I12| = sqrt(P12 / R12) = 1.25 => I12 = 1.25@90?
|I23| = sqrt(P23 / R23) = 1.875 => I23 = 1.875@-30?
I2 = I23 - I12 = 2.7243@-53.4132?

 

[mivey]

74One, I looked back at some of your prior posts and I may have misunderstood you. Please clarify your position.

Are you saying we should use the sqrt(3) if we weren't working with an unbalanced load?

The current formula for the balanced 3-phase case is:
I = VA/sqrt(3)/V_LL
or for a unity pf:
I = W/sqrt(3)/V_LL

If you are using this formula like in your #15 for the OP case, you are wrong.

 

[Smart $]

Using the direction conventions ... (important to keep the signs consistent) ...

Your sign conventions suck...JMHSO.

Please spare me any explanatory rhetoric you may consider in response. I'm well aware of how changing value signs amidst calculations can achieve the desired result... but that does not mean your application of sign changing is acceptable in all math circles.