three-phase power transmission, circuits, etc.

Thank you, MrAl.

MrAl said:

Most of the delta's i have seen only have an earth ground, which means it does not connect to any winding. It looks like it is totally separate from the windings.
The delta can be loaded line to line or line to 'fake' neutral. The fake neutral is just the point where the three loads connect together like those three capacitor looking things in your drawing. The voltage then is obviously lower because it would be line to 'neutral' rather than line to line.

Click to expand...

Please have a look here. The voltage across terminals "A" and "B" is same as that between "A" and 'fake neutral' (red dot) or as that between "B" and 'fake neutral'. In my humble opinion, I don't think there would be any difference between line-to-line or line-to-'fake neutral' voltage. Do I have this correct? Please let me know. Thanks.

Q:
Suppose we have a 240V single phase generator. We will assume that the generator is not grounded; there are only live and neutral wires. (Technically, I believe that both wires should be considered live). Anyway, suppose that someone accidentally touches live wire, will he get electric shock?

If you have time, please also help me with Q2 and Q3 from my previous post. Thank you.

Regards
PG

Hi,

Actually if the loads are all the same, then the line to fake neutral is the line to line voltage divided by the square root of 3. So if each line to line voltage was 208v then each line to neutral voltage would be close to 120v, just like any three phase system.

I am not sure what would give you the idea that the line to neutral voltage would be the same as the line to line voltage. Perhaps you can explain why you thought that.

Thanks, MrAl.

Please have a look here. Don't you think that there is a problem? For example, when terminals "a" and "c" are connected to the central fake neutral point, then the voltage source Vca will get shorted out? Thank you.

PG1995 said:

Thanks, MrAl.

Please have a look here. Don't you think that there is a problem? For example, when terminals "a" and "c" are connected to the central fake neutral point, then the voltage source Vca will get shorted out? Thank you.

Click to expand...

Hi,

SO DONT DO DAT

Seriously, there is no reason to do that. If you connect all the phases together they all short out, leading to no voltage and blown fuses.

If the delta is loaded with a wye, the load creates the fake neutral. The fake neutral is not created first, then the load applied, but first the load is applied and the fake neutral is a byproduct of that connection.
For example, replace all the wires (graphical lines) you have drawn from each phase to fake neutral with resistors of say 100 ohms each (all equal). Then do the analysis over again. The resistors would go in place of those things that look like capacitors in your other drawing. You can use equal capacitors too, but resistors will be simpler for any analysis. So you will end up having one resistor on each phase a,b,c going to the fake neutral, and you can then measure line to 'neutral' voltages from each phase to the fake neutral.
Try it, it's interesting to think about

BTW as well as the 1/sqrt(3) relationship between line to line and line to neutral voltages there is also a 30 degree phase difference. What i like to do is ground one phase (like B) and then solve for the fake neutral voltage and phase shift referenced from phase voltage VAB with all load resistors equal. If all load resistors are not equal then we have an unbalanced load condition which will lead to a more complex solution.

Read this: https://en.wikipedia.org/wiki/High-leg_delta

Thank you, MrAl, KISS.

MrAl, I think that I need to step back to clarify few things.

Q1:
Please have a look here and kindly help me.

The text says that when a neutral wire is present ln=-(la+lb+lc). It means that the neutral wire carries the current from all lines. When the neutral is absent then la+lb+lc=0.

I was curious to see that how the phase currents, la, lb, and Ic, are affected when neutral is present in the example on left. I tried to do some calculations and not sure if the result is correct. My work is shown below and I'm very doubtful about it. Do you think that it's correct?

Q2:
Please have a look here. Think of it as secondary of a three phase transformer. What difference would it make if the system is floating and not grounded at all?

Q3:
Please have a look here. Now I'm somewhat sure that those line segments in green represent capacitors. But now I'm confused about the role of those capacitors. It's alternating current which can pass thru a capacitor therefore those the sources can get shorted out thru those capacitors unless the reactance of capacitors is quite large.

Thank you for the help.

Regards
PG

Hello again,

Lets start with question Q1.

If your answer is 53.57 amps at 60 degrees then i dont think that is correct.
I suppose you should also state if that is an RMS or Peak value current too.

The simplest method i can think of is to transform the sources into their complex form, and then divide by each impedance to get the current, then sum. So taking each source to be a complex number VA=a1+b1*j, VB=a2+b2*j, VC=a3+b3*j we can write:
In=-(VA/ZA+VB/ZB+VC/ZC)

so you see this is a little simpler both intuitively and mathematically.

Try that or else just try your own method over again and see what you can come up with. I believe the correct answer is:
44 Amps RMS at 68 degrees

but these are rounded, so see if you can come up with the exact result before rounding and we can compare.
Multiply 44 times sqrt(2) to get the peak current value.

BTW the sources transform from V to -a+b*j using these equations:
|V|=sqrt(a^2+b^2)
angle(V)=atan(b/-a)

and when you have the right amplitude and angle the last two equations are satisfied. 'a' is made negative because that is the real part and it is always negative for either 120 degrees or -120 degrees.

Thank you, MrAl.

Please have a look here. I went thru it again and corrected some important mistakes but unfortunately I believe that the end result is still wrong. You can check the solved example from the book and how they verify the answer. I might be applying the method wrongly.

By the way, in this example my aim is not exactly to solve this example. I just wanted to know how the line currents are affected when a neutral is present. Thanks.

Regards
PG

Hi PG,
been out of radio range for a while but I see you are still looking at power transmission.
The use of star or delta depends on where you are in the network.
generally, the use of star is confined to residential or end user applications. Similarly, delta is confined almost universally to high voltage distribution.
part of the design criteria is based on fault current limiting. In a big system, the amount of energy flow can be enormous and the use of delta assists in the peak energy flowing during the fault. With the subscribers distribution system, the use of multi phase supply for a domestic residence is not required from the point of view of the energy demand. However, with the domestic users, the number of various loads gives rise to the 3 phase system becoming unbalanced. The use of the neutral wire allows the phase voltages to remain balanced. The introduction of an earth link from neutral to ground is a safety measure. The earth link is made at each distribution transformer and as well at each subscribers switchboard. If the phase wire is earthed by some fault or other, then the customer supply fuse will blow. The neutral wire carries no current when the system is balanced but will carry the phase current when only one of the phases is carrying current. So the neutral wire must be of the same gauge as each phase conductor.
You did raise queries about single phase systems with grounded wires etc, but the reality is that for efficient generation, the system is 3 phase and single phase supply will be a part of a bigger 3 phase system, so there will be a neutral wire as well as the phase wire and the star point will be earthed and each subscriber's switchboard will have an earth link connected to the neutral wire. The earth wire will carry no current under normal operating conditions.
There is also a difference in the harmonic frequencies transmitted through a supply transformer. I think the delta star configuration has a better attenuation of harmonics compared to delta/delta. I note you are working through the maths on 3 phase systems, and this is interesting academically. Just be careful when you read books on the subject that you dont get sidetracked by the US system. The IEC is pretty much the rule setter for the Multiple Earthed Neutral system. US practice is to keep going with what they have which is not a MEN system.
My post 2 (?) was a fairly complete explanation but maybe there was some confusion in what I wrote.

PG1995 said:

Thank you, MrAl.

Please have a look here. I went thru it again and corrected some important mistakes but unfortunately I believe that the end result is still wrong. You can check the solved example from the book and how they verify the answer. I might be applying the method wrongly.

By the way, in this example my aim is not exactly to solve this example. I just wanted to know how the line currents are affected when a neutral is present. Thanks.

Regards
PG

Click to expand...

Hi again,

I am consistently getting the same results, but they differ from yours, so let me ask you a few questions.

First, you write:
120+j*5*I1=0

but you still get:
I1 = 24 at angle -90 degrees

If you solve that first equation you really get +90 degrees.

So question 1:
Why do you write the equation like that rather than like this:
120-j*5*I1=0

or like this:
120=j*5*I1
?

Question 2:
How did you get an angle of -90 degrees from your equation:
120+j*5*I1=0
?

To see how the currents change you have to be able to analyze both the circuit without the neutral wire and also the circuit with the neutral wire.
The neutral wire diverts some of the current away or toward the center node.

Here is a handy formula for calculating the neutral current in the direction you have shown:
In=
-Ampl*(2*(j*zbb+zba)*(j*zcb+zca)-sqrt(3)*j*(j*zab+zaa)
*(j*zcb+zca)-(j*zab+zaa)*(j*zcb+zca)+sqrt(3)*j*(j*zab+zaa)*
(j*zbb+zba)-(j*zab+zaa)*(j*zbb+zba))/(2*(j*zab+zaa)*
(j*zbb+zba)*(j*zcb+zca))

This formula is for a balanced source and unbalanced load. The impedance loads on the three phases are as follows:
ZA=zaa+zab*j
ZB=zba+zbb*j
ZC=zca+zcb*j
Ampl=voltage of any phase

For your example we have:
zaa=0, zab=5 (inductor)
zba=10, zbb=0 (resistor)
zca=0, zcb=-10 (capacitor)
Ampl=120 (all three source voltage amplitudes)

We would insert these seven values wherever they appear in the formula, then calculate the real part and imaginary part and that would give us a single complex number:
In=a+b*j
and from that we can easily get the amplitude and phase angle.

Because three of the seven variables are all zero, we can simplify for this one example and get:
Real=(Ampl*(zcb-sqrt(3)*zba))/(2*zba*zcb)
Imag=(Ampl*(2*zba*zcb+sqrt(3)*zab*zcb-zab*zba))/(2*zab*zba*zcb)

and this would give us a quicker calculation, but only for this example with the inductor, capacitor, and resistor loads on those same phases, although we could still change the values of the three components to generate other similar examples.

Hi,

Here is another simplification.

If we let w=1, then we have:
L=ZL/j=5
C=-j/ZC=1/10
R=R=10
Ampl=Ampl=120

where you can see L retained the inductor impedance without the 'j', and C is the reciprocal of the cap impedance without the "-j".

Then we can calculate the amplitude from the above values of Ampl, R, L, and C from:
In=Ampl*sqrt((C^2*L^2+C*L+1)*R^2+(sqrt(3)*C*L^2+sqrt(3)*L)*R+L^2)/(L*R)

and phase shift from:
Ph=atan((C*L*R+2*R+sqrt(3)*L)/(sqrt(3)*C*L*R+L))

This is probably a pretty quick way to check answers.
Again this only holds for one inductor on phase A, one resistor on phase B, and one capacitor on phase C, and all phase voltages are equal.

Thank you, rumpfy, MrAl.

rumpfy: I wouldn't make any query about your post because at the moment I'm more interested in knowing where I'm going wrong with my calculation. Your post was a good one as was your last one, i.e. post #4.

MrAl: I have been thru this again. Everything looks quite okay now. I have corrected the mistake your indirectly pointed out in addition to correction of some other mistakes related to plus/minus signs but surprisingly I'm still getting the wrong answer. As you can see that I have used mesh analysis and I believe that at some point I'm applying the method wrongly. It would be nice if you could help me to trace out the issue. I see that you have come up with a simpler method to solve the problem but I believe that I should first find out the mistake(s) in my own procedure. Thank you for the help.

Best regards
PG

PG1995 said:

Thank you, rumpfy, MrAl.

rumpfy: I wouldn't make any query about your post because at the moment I'm more interested in knowing where I'm going wrong with my calculation. Your post was a good one as was your last one, i.e. post #4.

MrAl: I have been thru this again. Everything looks quite okay now. I have corrected the mistake your indirectly pointed out in addition to correction of some other mistakes related to plus/minus signs but surprisingly I'm still getting the wrong answer. As you can see that I have used mesh analysis and I believe that at some point I'm applying the method wrongly. It would be nice if you could help me to trace out the issue. I see that you have come up with a simpler method to solve the problem but I believe that I should first find out the mistake(s) in my own procedure. Thank you for the help.

Best regards
PG

Click to expand...

Hello again,

Yes I understand completely why you want to stick with the original method as that is the thing you are trying to understand. By all means, stick with that before moving on.

Your original set of equations in this new attachment look 100 percent correct. The set i used are as follows:

eq1: Va-j*5*I1=0
eq2: -Vb-10*(I2-I3)=0
eq3: -Vc+Vb-10*(I3-I2)+j*10*I3=0

and these look exactly like your set. And then we have:

ia: i1
ib: i3-i2
ic: -i3

and to find the final solution:
In=-ia-ib-ic

These equations give me the same results i have been getting all along using another method, so it looks like your only error is in the algebra in the steps that come after the first set of three equations. In particular, there may be a problem with one of the signs in equation 3, but i'll leave this for you to find. If something still doesnt work out i'll take another look at your work after those three equations ok?

I am confident you will find the error, but at least you know how to write the initial equations and that is a very good sign

MrAl said:

but at least you know how to write the initial equations and that is a very good sign

Click to expand...

And it shows that yours and Steve's (and many others') efforts didn't go fruitless altogether!

Hi again,

He he, that is good to know for me and most likely for Steve too, thanks very much.

I just took another quick look and it appears that your calculations for equation 3 later on look ok, it was hard to tell because sometimes you write your plus signs "+" with a very weak vertical line. Maybe you can stay more consistent with your font weight when writing and that will make it easier for someone to check over.

But i do see maybe an error with the final calculation for I3, which looks like the imaginary and real parts somehow got swapped? You can check that, but one thing that stands out more strikingly is that ib is not equal to -I2. That's because there are two currents that combine to produce ib.

See if you can find the error with the calculation if I3, and the calculation of ib. I am sure you are very close to getting this now.

Hi MrAl

I have been thru it again and the error had occurred where I wrote Ib=-I2 when it should have been Ib=I3-I2. Thank you for helping me. Now we can talk about other things.

Now you can see here that how the line currents are affected by presence of a neutral. The presence of a neutral makes line currents comparatively quite small.

Regards
PG

Hi again,

Well that looks much much better, just one more little thing...
The amplitude looks correct now, but i see the phase shift is still not correct yet.
Recall i mentioned something about the real and imaginary parts swap in one calculation. Maybe check that out and see if you can correct it. The true phase shift should come out somewhat close to 70 degrees, plus or minus say 3 degrees, just so you have something to compare your calculations with.

I did not calculate the circuit with no neutral wire yet. I'll get to that next and we can compare notes on that too

MrAl said:

I did not calculate the circuit with no neutral wire yet. I'll get to that next and we can compare notes on that too

Click to expand...

Thanks. But the calculation in this solved example here was already done with no neutral.

Regards
PG

Hello again,

Ok yes, and i repeated the calculations just to compare anyway.

Also, another interesting method is to use superposition which gives us:
Vn=(Za*Zb*VC)/(Zb*Zc+Za*Zc+Za*Zb)+(Za*Zc*VB)/(Zb*Zc+Za*Zc+Za*Zb)+
(Zb*Zc*VA)/(Zb*Zc+Za*Zc+Za*Zb)

and this is the voltage at the common of all three loads which we called the "fake neutral" earlier.

Once we have that Vn voltage, then the currents in each phase are:
Ia=(Va-Vn)/Za
Ib=(Vb-Vn)/Zb
Ic=(Vc-Vn)/Zc

That's just another way of doing it so we have a way to check our results.

Thanks, MrAl.

But right now we are trying to understand something else or perhaps I'm missing something. You can see here that line currents are really reduced when a neutral is present for the circuit we were discussing. From this, personally, I can conclude only one thing that a neutral is a must (at least for a star system) if you don't want to 'waste' too much current! Do I have it correct?

Moreover, we were discussing fake neutral from delta system's point of view because start system has already got a 'real' neutral. Thank you.

Best regards
PG