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Blackburn 5

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tulum

Industrial
Jan 13, 2004
335
This post has to do with blackburns protective relaying text - 2nd edition...I know you guys have it!

On page 37, 2.4 Blackburn says that:

For either the wye or the delta connection the following basic equations apply;

S3p = sqrt(3)*VLL*IL (VA)... 2.3

VLL = sqrt(3)*VLN<+30... 2.4

IL = S3p/(sqrt(3)*VLL)...2.5

Doesn't 2.4 only apply to wye?



 
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It applies to both, although you might have a hard time measuring the line-neutral voltage on a delta system. Every system has a neutral point. If you look at the phasor diagram of a delta system, you can draw a set of three Vl-n phasors inside the triangle.
 
dpc,

I have a hard time with the subject of delta systems. I have not done much work with transformer connections so my question might seem basic.... however...

I thought in a delta VLL = VLN, not VLL = sqrt(3)*VLN as in a wye system?

Some of the texts I have, clearly state this?

Please respond, this concept has confused me for awhile?
 
I think if Vll=Vln, the system has a problem, such as a line-ground fault.

For any delta set of phasors, there is a corresponding set of wye phasors that has a neutral point.

If you consider a three-phase, three-wire load as a black box, it is not possible to determine whether the load is delta or wye connected by external measurements. You can model it either way, with appropriate impedances and get identical results.

For a delta system, the neutral is a bit of a fiction, but think of it as the resulting neutral point if the three Vll phasors are converted to equivalent Vln phasors.
 
tulum:

All 3 phase system analysis is done based on per phase values. As far as the analysis goes you can think of a 3 phase system coming out of a black box.

The formula for 3 phase system is equally applicable for both wye and delta system. In otherwords, power per phase remains the same whether its a wye [(V/sqrt3)*I)] or delta [V*(I/sqrt3)]system. Taking all Line values. Total power is 3 times the per phase values.

Now the current measured in wires coming out of the 'black box' is always the line current and voltage measure between lines is always L-L (reagrdless of N is there or not).

For the purpose of system analysis it is 'conventional' to assume a wye connection at the soruce and use per phase voltage value and leaving the current as is as it will be same as the phase currents. I guess you can anlayze the system by divding the line current by sqrt 3 and leaving voltage unchanged.

Notice the difference of angle +30, in equation 2.4. If you resolve the 3 phase delta phasors in equivalent wye-V-N phasors, you will see that there is a 30 degree phase shift. This is also true for a wye system, represented as delta, the sign of angle will depend on which one is taken as reference.

I am not sure if this helps or confuses you more or I am singing to a quire.
 

In fault and loadflow calculations, it takes some acquired intuition to know when to apply [&radic;]3—but most often applies to balanced systems. “Crash courses” seem to sometimes assume that this is magically known by students, or so instructors may presume—possibly for convenience and cramped time frames. For delta situations, when to know neutral shift applies and when not can be a big leap.

I guess one comparison is three identical 120V lamps applied to a 208V 4-wire wye source with a physical neutral connection versus a 208V 3-wire delta system where neutral is artificial—the electrical center of the delta triangle. If lamp impedances are not equal with balanced voltage or have unequal impedances with a balanced voltage, neutral shift occurs. ‘Neutral-to-ground’ potential/current in those cases is other than zero.
 
Comment on tulum (Industrial) Mar 11, 2004 marked ///\\This post has to do with blackburns protective relaying text - 2nd edition...I know you guys have it!

On page 37, 2.4 Blackburn says that:
///Before continuing the discussion of the per unit method, a review of some general relations between circuit quantities applicable to all three-phase power systems is in order. This will focus on the two basic types of connections, wye and delta, as shown in Fig. 2.1.\\\ For either the wye or the delta connection the following basic equations apply;

S3p = sqrt(3)*VLL*IL (VA)... 2.3

VLL = sqrt(3)*VLN<+30... 2.4

IL = S3p/(sqrt(3)*VLL)...2.5

Doesn't 2.4 only apply to wye?
///Yes, indeed, it does apply to wye only, since the text above, which I added, refers to Fig. 2.1 in the book. This figure delineates applicability of the above equations. Equation 2.4 is clearly applicable to wye in the figure. Although, the wye connection presented in the figure might theoretically be moved inside the delta connection, the figure does not show it there for clarity and proper physical interpretation. Physically, e.g. for the transformer, each connection, wye and delta, represents a different winding on the transformer core.\\
 
Comment on tulum (Industrial) Mar 11, 2004 marked ///\\
I thought in a delta VLL = VLN, not VLL = sqrt(3)*VLN as in a wye system?
///Normally, VLL=VLN is never valid in the three phase system. If the wye is inserted inside delta, then VLL=sqrt(3)*VLN will apply, not VLL=VLN. This is clear geometrically.\\Some of the texts I have, clearly state this?
///Which one and where?\\Please respond, this concept has confused me for awhile?
///Please notice the Blackburn book reviews the posted relations; it does not explain them as appropriate textbooks do. See suitable textbooks for the better understanding of delta vs wye relationships, e.g.
William D. Stevenson "Elements of Power System Analysis," Third Edition, McGraw-Hill Book Company, 1975,
page 28 Figure 2.16 "Voltages in a balanced three-phase circuit: (a) voltages to neutral; (b) relation between a line voltage and voltage to neutral",
and statement:
"The fact that the magnitude of line-to-line voltages of a balanced three-phase circuit is always equal to sqrt3 times the magnitude of the line-to-neutral voltages is very important.\\
 
I think this boils down to definition of the term "neutral". In the case of a delta system there is no physical neutral. In that sense VLN is not defined for a delta system. If you define neutral as the geometric center of the delta triangle, then I suppose you can argue that 2.4 is true for a delta. This would not be true in the case of LG phasers unless neutral is at ground potential.
 
Thanks everyone for responding. I am definitly learning from this forum and appreciate the input.

Jbartos:

I have looked up the text books that I thought said VLL = VLN, and think I may have misunderstood from day 1... They state VLL=Vp (where p is phase)in a delta. Which from the discussion above, and the phasor representation of the delta is not the same thing (i.e. Vp does not = VLN in a delta)..

To recap the Delta:

VLL = Vp;
ILL = sqrt(3)*Ip< +/- 30deg;
S3p = sqrt(3)*VLL*ILL;

And theoretically, we can say that VLL = sqrt(3)*VLN, which is geometrically located at the center of the delta.

Where does ILN fit into all of this?


Furthermore, if I center tap a winding in the delta, this does not change the theoretical value of VLN... it only changes the value of Vphase....i.e. theoretically VLN can still be used for calculation purpose, like finding phase shift between primary and secondary windings of a transformer etc. All the center tap does is make available new voltage levels.

Am I on the right track?










 
To take another step on my last post...

Lets say we have a solidly grounded Y feeding a splitter, and the CU bus in the splitter was spaced with Line to Line distances between phases, and Line to Neutral distances between ground.

Now if I hook up a delta ungrounded source to this splitter are the distances still OK?

As of now my last post is more important for me to understand (please respond to it first), this post is just for curiosity sake...

 
ILN=Ip as you've defined it.

Center tapping and grounding a delta connected winding. can confuse the issue. The IEEE definition of neutral puts it at the center of a three phase system so your statement above is correct. Lots of electricians will call the center tap the neutral. Better to call it the grounded conductor.

A problem (or benefit) with ungrounded systems is that the first ground fault will not cause overcurrent tripping. The system will and should continue to operate in this corner grounded condition until repair is made. All insulation and clearances should then be based on phase to phase values. Lots of discussion over these systems here, try searching.
 
tulum:

Your first post is answered by stevenal. Your last post, if I understand correctly, you are talking about spacing of the buses from insulating point of view.

The answer is yes, the Line -Line voltage applied to your bus system still has to be same (rated system voltage)regardless of it comes from a delta source or wye-source and grounded or ungrounded. Bus spacing are related to the voltage between them, not the configuration of their source's windings. Think of that "Black box" I mentioned before.

Except that if you have any L-N loads then they will not work very well on a delta source:).

 
Comment on tulum (Industrial) Mar 12, 2004 marked ///\\Jbartos:
I have looked up the text books that I thought said VLL = VLN, and think I may have misunderstood from day 1... They state VLL=Vp (where p is phase)in a delta.
///Vp is sometimes used in a generic term for phase voltage as opposed to neutral voltage Vn. However, Vp alone is an ambiguous since there are Vpp (phase to phase) and Vpn (phase to neutral).\\ Which from the discussion above, and the phasor representation of the delta is not the same thing (i.e. Vp does not = VLN in a delta)..
///If Vp is not meant Vpn, then it is not equal to VLN)
To recap the Delta:

VLL = Vp;
///It is clearer to write VLL=Vpp\\ILL = sqrt(3)*Ip< +/- 30deg;
///ILL is rarely used. IL is proper and sufficient designation. Ip is also ambiguous. It is not a good idea to mix subscripts L with p. Either, use consistently p or L but not both in the same place.\\S3p = sqrt(3)*VLL*ILL;
///It is better to use IL instead of ILL since the ILL is very confusing with the internal delta current flowing from Line to Line (or phase to phase).\\And theoretically, we can say that VLL = sqrt(3)*VLN,
///This is a regular designation.\\which is geometrically located at the center of the delta.
///The center of delta is often called a neutral. Or, actually, the neutral is located in the middle of the delta\\\

Where does ILN fit into all of this?
///Normally, ILN is used for wye where it is the same as IL, i.e. IL=ILN.
When it comes to delta and some imaginary wye inserted inside the delta, then ILN will flow from a delta corner, or phase, to neutral N.\\
Furthermore, if I center tap
///The center tap is often used for the single phase transformer center on its secondary side. When it comes to three phase transformers, it is preferable to use neutral N rather than the centertap.\\ a winding in the delta, this does not change the theoretical value of VLN... it only changes the value of Vphase....
///It changes value of Vpp=VLL to Vpn=VLN.\\i.e. theoretically VLN can still be used for calculation purpose, like finding phase shift between primary and secondary windings of a transformer etc.
///Yes, correct.\\ All the center tap does is make available new voltage levels.
///Center tap, preferably named a neutral, makes available VLN=Vpn and established very frequent low voltage system, e.g. 208V/120V 3phase, 4-wire, where VLL=208V, VLN=120V.\\Am I on the right track?
///A good textbook would really help you a lot, since it tends to explain these details.\\
 

Per jbartos' suggestion, IEEE Std 141-1993 ...Electric Power Distribution for Industrial Plants {Red Book} may be a useful reference.
 
Comment: IEEE Standards are good if one has a basic background either from a suitable course work or self-educated from textbooks (or from the Web? nowadays).
 
I think that the equation (2.4 Blackburn) applies to both delta and wye systems. The situation with some delta systems is that you cannot get the neutral point physically. Of course, all those expressions apply to balanced steady state conditions.
 
Comment on the previous posting: The Blackburn limited the equation 2.4. applicability to the Fig. 2.1, he presented. However, I agree with you that the equation is also applicable to delta for various purposes.
 
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