Connecting a neutral to a 3-Phase Wye Load

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mull982

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What happens if you connect a nuetral to the center point of of a three phase wye connected load? Will you create three individual single phase loads? For instance if you connect a nuetral to a wye connected motor?

I there are three heater coils rated for 120V each physically located in parallel with one end of each coil connected to a nuetral what happens if you remove this nuetral? Does the load then just act as a three phase load. I'm assuming that the loads arent 100% balanced. The voltage supply in this case is a 120/208 source.
 
mull982 said:
What happens if you connect a nuetral to the center point of of a three phase wye connected load? Will you create three individual single phase loads? For instance if you connect a nuetral to a wye connected motor?
Click to expand...

In the case of a 3-phase motor, no current would flow on the neutral because the line to line loads are balanced. So it would be a useless conductor that does nothing.
 
mull982 said:
What happens if you connect a nuetral to the center point of of a three phase wye connected load? Will you create three individual single phase loads? For instance if you connect a nuetral to a wye connected motor?

I there are three heater coils rated for 120V each physically located in parallel with one end of each coil connected to a nuetral what happens if you remove this nuetral? Does the load then just act as a three phase load. I'm assuming that the loads arent 100% balanced. The voltage supply in this case is a 120/208 source.
Click to expand...

In the heater coil example if you were to remove the nuetral connection you would infact put 208V across each heater coil causing some damage.
 
mull982 said:
What happens if you connect a nuetral to the center point of of a three phase wye connected load? Will you create three individual single phase loads? For instance if you connect a nuetral to a wye connected motor?

I there are three heater coils rated for 120V each physically located in parallel with one end of each coil connected to a nuetral what happens if you remove this nuetral? Does the load then just act as a three phase load. I'm assuming that the loads arent 100% balanced. The voltage supply in this case is a 120/208 source.
Click to expand...
If the loads were equal nothing would change. If they were not equal the voltages would float dependant on the load.
 
Strahan said:
In the heater coil example if you were to remove the nuetral connection you would infact put 208V across each heater coil causing some damage.
Click to expand...

Even if you disconnect the nuetral the threee coils will still be tied together on one end of each coil forming a wye connection just without the nuetral wire. I'm not sure I see why this would cause 208V across each coil.

quogueelectric said:
If the loads were equal nothing would change. If they were not equal the voltages would float dependant on the load.
Click to expand...

I understand that if the loads were equal nothing would change due to the three currents canceling out in the nuetral. If the loads were not equal could you explain the floating voltages that you mentioned? What would happen with the currents?
 
First off, the loads are NOT in parallel as the OP says. They are connected in a wye, right?

To calculate the voltage on the floating neutral, we employ Kirchoff's current equation:

(Va - Vn)/Za + (Vb -Vn)/Zb + (Vc - Vn)/Zc = 0

Now we solve for Vn, but this gets a little messy, so I will leave that exercise up to the student. Phasors you know!
 
Strahan said:
In the heater coil example if you were to remove the nuetral connection you would infact put 208V across each heater coil causing some damage.
Click to expand...

I think that with 3 coils connected in a wye and 208 volts line to line, each coil would see 104 volts. The OP is asking two different questions. First for the motor connected in wye and then a neutral added, see Peters' response. Second, if 3 heater coils are connected in parallel from line to neutral in a 120/208 system and then the neutral is lost to the panel but is still connected to each coil then i believe each coil would see 104 volts, but I think the 3 phase factor would kick in for total power used.
 
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rattus said:
First off, the loads are NOT in parallel as the OP says. They are connected in a wye, right?

To calculate the voltage on the floating neutral, we employ Kirchoff's current equation:

(Va - Vn)/Za + (Vb -Vn)/Zb + (Vc - Vn)/Zc = 0

Now we solve for Vn, but this gets a little messy, so I will leave that exercise up to the student. Phasors you know!
Click to expand...

Yes the loads are connected in wye

I'm going to try to work my way through this equation. In the mean time just so I verbally understand what we are saying is that the current will flow into each leg and then out on the other two legs and not through the nuetral because it doesn't exist. Because the current balance wont be equal on all of the legs the nuetral point will float to some voltage.

To calculate the total power on the three phase load would se just add the power or watts across each of the three unbalanced heating elements?
 
mull982 said:
Even if you disconnect the nuetral the threee coils will still be tied together on one end of each coil forming a wye connection just without the nuetral wire. I'm not sure I see why this would cause 208V across each coil.

Yes I agree in my earlier post I stated I typed before my brain was in gear I was thinking of something totally different.:wink:

I understand that if the loads were equal nothing would change due to the three currents canceling out in the nuetral. If the loads were not equal could you explain the floating voltages that you mentioned? What would happen with the currents?
Click to expand...

I believe if the loads were not equal which in my opinion is always the case maybe not enough to hurt anything but you would see different voltage dropped across each load according to the resistance (impedance) of that load.
 
mull982 said:
Yes the loads are connected in wye

I'm going to try to work my way through this equation. In the mean time just so I verbally understand what we are saying is that the current will flow into each leg and then out on the other two legs and not through the nuetral because it doesn't exist. Because the current balance wont be equal on all of the legs the nuetral point will float to some voltage.

To calculate the total power on the three phase load would se just add the power or watts across each of the three unbalanced heating elements?
Click to expand...

Mull, look at this way: At any instant, the sum of the currents entering the floating neutral node equals the sum of the currents leaving that node.

Yes, the total average power is simply the summation of the averages powers in the individual loads.
 
rattus said:
Mull, look at this way: At any instant, the sum of the currents entering the floating neutral node equals the sum of the currents leaving that node.
Click to expand...

Ok yeah that makes sense. So even if the phases are unbalanced and there is no neutral the current split betwwen the other two phases depending on their impedance. The unbalanced circuit will work however there will be a voltage on the nuetral. This is the point I am trying to figure out.

rattus said:
To calculate the voltage on the floating neutral, we employ Kirchoff's current equation:

(Va - Vn)/Za + (Vb -Vn)/Zb + (Vc - Vn)/Zc = 0

Now we solve for Vn, but this gets a little messy, so I will leave that exercise up to the student. Phasors you know!
Click to expand...

I'm having trouble trying to sove for the equation you posted. Using an example of a 480/277 system lets say I have three phase load connected in wye with unbalanced loads and have the following:

Va = 277@0deg
Vb = 277@-120deg
Vc = 277@-240deg
Za = 100ohms
Zb = 10ohms
Zc = 1ohms

From this I would come up with the following currents:

Ia = 2.77@0deg A
Ib = 27.7@-120degA
Ic = 277@-240degA

So I'm guessing that the 2.77A which is flowing into A will be split among the B anc C brances after the node, and so forth for the other phases. I'm sure I'm missing something to do with the phasors here, but this is where I'm getting stuck.


As another examle lets say I have (3) identical heaters rated for 277V. Lets say they these heaters were designed to be individual single phase heaters and each have one hot leg and a neutral connected to them. Lets say we then want to make these three individual (1) three phase heater and connect them all together in wye. Theoretically if they are all the same heater they will be balanced and we will not need a neutral for any unbalanced current. However if there is a slight unbalance in any of the heaters, then we would be in the situation above and would either need a neutral.

The reason this topic came up for me is that we have a similar wye connected heater arrangement in the field which is calling for a nuetral and one is not avaliable. So I am therefore faced with getting a transformer to derive this neutral or hooking it up without a neutral. I'm wanting to learn what happens in the latter case witout this neutral.
 
mull982 said:
Ok yeah that makes sense. So even if the phases are unbalanced and there is no neutral the current split betwwen the other two phases depending on their impedance. The unbalanced circuit will work however there will be a voltage on the nuetral. This is the point I am trying to figure out.



I'm having trouble trying to sove for the equation you posted. Using an example of a 480/277 system lets say I have three phase load connected in wye with unbalanced loads and have the following:

Va = 277@0deg
Vb = 277@-120deg
Vc = 277@-240deg
Za = 100ohms
Zb = 10ohms
Zc = 1ohms

From this I would come up with the following currents:

Ia = 2.77@0deg A
Ib = 27.7@-120degA
Ic = 277@-240degA

So I'm guessing that the 2.77A which is flowing into A will be split among the B anc C brances after the node, and so forth for the other phases. I'm sure I'm missing something to do with the phasors here, but this is where I'm getting stuck.


As another examle lets say I have (3) identical heaters rated for 277V. Lets say they these heaters were designed to be individual single phase heaters and each have one hot leg and a neutral connected to them. Lets say we then want to make these three individual (1) three phase heater and connect them all together in wye. Theoretically if they are all the same heater they will be balanced and we will not need a neutral for any unbalanced current. However if there is a slight unbalance in any of the heaters, then we would be in the situation above and would either need a neutral.

The reason this topic came up for me is that we have a similar wye connected heater arrangement in the field which is calling for a nuetral and one is not avaliable. So I am therefore faced with getting a transformer to derive this neutral or hooking it up without a neutral. I'm wanting to learn what happens in the latter case witout this neutral.
Click to expand...

Mull, since Vn is no longer zero in your example, your currents are wrong. Likewise the phase angle of Vn is not zero, therefore your angles are wrong as well. Messy problem best solved with a computer program.

But look at it this way:

Start with a balanced, continuous load.

Now open Za. The remaining loads are in series across 480V. 240V across each load. No big deal.

Now short Za. Each load now sees 480V. Not good.

Could it be that the neutral is required for some control circuitry?
 
rattus said:
Mull, since Vn is no longer zero in your example, your currents are wrong. Likewise the phase angle of Vn is not zero, therefore your angles are wrong as well. Messy problem best solved with a computer program.
Click to expand...

When working with phasors I usually draw then in AutoCad to compute vaules dealing with phasors. Is there a way I can draw these phasors to see the resultant Vn or is this strictly a hand calculation?


rattus said:
Could it be that the neutral is required for some control circuitry?
Click to expand...
[/QUOTE]

No there is no control circuit that would use this neutral. Therefore I am being asked if we need to use this neutral. My response is, if the loads are balanced then there will be no neutral current anyway. If the loads happen to be unbalanced then then current will still flow through heaters and they will supply same kW rating except we will now have a floating voltage on the neutral. Either way, it will still work. Is this a correct explanation?
 
mull982 said:
When working with phasors I usually draw then in AutoCad to compute vaules dealing with phasors. Is there a way I can draw these phasors to see the resultant Vn or is this strictly a hand calculation?





No there is no control circuit that would use this neutral. Therefore I am being asked if we need to use this neutral. My response is, if the loads are balanced then there will be no neutral current anyway. If the loads happen to be unbalanced then then current will still flow through heaters and they will supply same kW rating except we will now have a floating voltage on the neutral. Either way, it will still work. Is this a correct explanation?
Click to expand...

Mull, problem is that you can't draw a phasor until you know its value. The only things we know are Va, Vb, and Vc relative to the system neutral.

We must solve for the voltage at the floating node, and that takes some trig and complex numbers. Not really hard, but tedious. If I were doing it, I would use a spread sheet.

If the loads are unbalanced, some will see undervoltage, some will see overvoltage, therefore, the power dissipation of each will change.

For a slight imbalance, I see no problem. I presume there will be an EGC though?
 
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Check me:

Check me:

OK, I did it. I computed the voltage of the floating neutral node. I get

Vn' = 232Vrms @ 125 degrees.

Someone check me please.
 
rattus said:
OK, I did it. I computed the voltage of the floating neutral node. I get

Vn' = 232Vrms @ 125 degrees.

Someone check me please.
Click to expand...
I got 237V, >125. But I didn't spend much time at it, didn't check my numbers. Your's could well be right.

Here is the rest of my numbers:
Ia = 5A >-25
Ib = 43A >-90
Ic = 46A >95

Vza = 455V
Vzb = 434V
Vzc = 46V

The interesting part is Za and Zb have over 400V on them and Zc has 46V.

Mull -
The model is the interesting part. If I get a chance tomorrow, I'll make a sketch.

cf
 
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Cold Fusion said:
I got 237V, >125. But I didn't spend much time at it, didn't check my numbers. Your's could well be right.

Here is the rest of my numbers:
Ia = 5A >-25
Ib = 43A >-90
Ic = 46A >95

Vza = 455V
Vzb = 434V
Vzc = 46V

The interesting part is Za and Zb have over 400V on them and Zc has 46V.

cf
Click to expand...

Close enough for rough work.

Yes indeed, some elements would see an overvoltage, and those elements would soon fail leaving Zc open.
 
rattus said:
OK, I did it. I computed the voltage of the floating neutral node. I get

Vn' = 232Vrms @ 125 degrees.

Someone check me please.
Click to expand...

Did you use the example voltages and impedances that I posted to arrive at this answer?

Can you plaese give me the first step in arriving at this solution. I'm stuck and hopefully all I need is a hint to head in the right direction.
 
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