MWBC= more heat or Less heat?

[480sparky]

In diagram "B" - Shouldn't the arrows next to the 60 and 20 be reversed???.........

Don't ask me.... it's not my drawing.

But my guess is no, they are the right direction. 40 amps going through the A leg, and 20 through the neutral, which equals the 60 amps going back out the B leg.



BTW, there are no asteroids in the rings of Saturn.

 

[LarryFine]

I guess what I don't see is why with the MWBC there isn't 30 amps coming back on the neutral with 2 hot conductors of different phases when the load is 15 amps on each.

Okay, read this (which I coincidentally wrote for another Dennis): http://forums.mikeholt.com/showpost.php?p=708650&postcount=4

 

[LarryFine]

In diagram "B" - Shouldn't the arrows next to the 60 and 20 be reversed???

`No, but all three will be reversed during the next half-cycle.

 

[PetrosA]

Okay, read this (which I coincidentally wrote for another Dennis): http://forums.mikeholt.com/showpost.php?p=708650&postcount=4

That was a great post! I've had difficulty imagining the neutral point no matter how hard I tried, but that did it for me. Thanks!


Dennis,

Phased is probably the wrong word, but what I mean is this - if you put equal loads on three phases (ex. three 100W bulbs in pigtails off a three-pole breaker) you can create (or recreate) the neutral point by wirenutting the three white leads on the pigtails together and the bulbs will all light up with the same and correct intensity, without being connected to the neutral bar. What's happening when the three 100W bulbs are tied together is that the current can flow back and forth evenly between the three phases, meeting up at the so-called neutral point. If you replace one of the 100W bulbs with a 60W bulb you would lose your neutral point due to the imbalance of the load and (I think...) the voltage through the 60W bulb would increase. If someone can explain what exactly would happen please chime in, I'm curious.

 

[hurk27]

That was a great post! I've had difficulty imagining the neutral point no matter how hard I tried, but that did it for me. Thanks!


Dennis,

Phased is probably the wrong word, but what I mean is this - if you put equal loads on three phases (ex. three 100W bulbs in pigtails off a three-pole breaker) you can create (or recreate) the neutral point by wirenutting the three white leads on the pigtails together and the bulbs will all light up with the same and correct intensity, without being connected to the neutral bar. What's happening when the three 100W bulbs are tied together is that the current can flow back and forth evenly between the three phases, meeting up at the so-called neutral point. If you replace one of the 100W bulbs with a 60W bulb you would lose your neutral point due to the imbalance of the load and (I think...) the voltage through the 60W bulb would increase. If someone can explain what exactly would happen please chime in, I'm curious.

In a single phase service this would be true, but I don't think this would be the case in a 3 phase Y service, each bulb would see 104 volts, we tend to forget that a 4 wire Y service the neutral is always current carrying, the two 100 watt bulbs have already balanced across the two 208 volt phases, and the 60 watt will just increase this voltage ever so slightly at the common point, the fun happens if you change the 60 watt bulb to exceed the two 100 watt bulbs like a 400 watt bulb, then the 100 watt bulbs will increase in voltage at a more proportional rate.
I wish I knew the math better but maybe someone will chime in on this part.

 

[hurk27]

In diagram "B" - Shouldn't the arrows next to the 60 and 20 be reversed??? I always thought that electrical waves reacted similar to waves behind your boat when you are turning tight circles.

Sometimes they double their size, and other times they cancel out (pulling youngsters on a tube is one of my favorite hobbies)_ I've always wanted to get three boat drivers going in a circle (like a three phase motor ) to see what would happen??? Then run the tuberider/victim through the apex of waves. My unofficial record in getting a fatboy-little brother- over 12ft high

back to the subject --- in diagram B -- at the neutral point after load "A" and load "B" where there would be a wire-nut, 20 amps would 'magically disappear' or cancel out???? Similar to water waves canceling out at the point of impact???

You must remember the circuit in Ed's diagram is fed by a single phase transformer with a center tap, the current flow does not cancel out as you would think like in a case of being out of phase 180 degrees, but rather in-phase from one end of the transformer winding to the other, you have two series circuits, the supply is two 120 volt windings in series connected together with a tap (neutral), and two loads in series connected together with a tap (neutral), so some of the current in the 60 amp load passes through the 40 amp load (40 amps) and the extra passes through the neutral (20 amps), thus why the arrows are facing that way

 

[LarryFine]

If you replace one of the 100W bulbs with a 60W bulb you would lose your neutral point due to the imbalance of the load and (I think...) the voltage through the 60W bulb would increase. If someone can explain what exactly would happen please chime in, I'm curious.

the fun happens if you change the 60 watt bulb to exceed the two 100 watt bulbs like a 400 watt bulb, then the 100 watt bulbs will increase in voltage at a more proportional rate.
I wish I knew the math better but maybe someone will chime in on this part.

Ding-dong! Chime-time.

Okay, with a 3ph wye, three bulbs connected in a why will (in theory) read 0v between the supply common and the load common. There will be 120v across each bulb.

Change one or two loads, and the load neutral point will leave 0v, with the greatest voltage now measured across the smallest load, which is also the highest resistance.

If you connect the load neutral to the supply neutral, the voltages will be 120v again, with the neutral current being whatever is required to keep the load neutral at 0v.

 

[LarryFine]

You must remember the circuit in Ed's diagram is fed by a single phase transformer with a center tap, the current flow does not cancel out as you would think like in a case of being out of phase 180 degrees, but rather in-phase from one end of the transformer winding to the other, you have two series circuits, the supply is two 120 volt windings in series connected together with a tap (neutral), and two loads in series connected together with a tap (neutral), so some of the current in the 60 amp load passes through the 40 amp load (40 amps) and the extra passes through the neutral (20 amps), thus why the arrows are facing that way

Inhale, man! :grin:

 

[hurk27]

Ding-dong! Chime-time.

Okay, with a 3ph wye, three bulbs connected in a why will (in theory) read 0v between the supply common and the load common. There will be 120v across each bulb.

Change one or two loads, and the load neutral point will leave 0v, with the greatest voltage now measured across the smallest load, which is also the highest resistance.

If you connect the load neutral to the supply neutral, the voltages will be 120v again, with the neutral current being whatever is required to keep the load neutral at 0v.

Ok lets do this one step at a time, lets say we connect two 100 watt bulbs in series from A leg to B leg, the voltage across this circuit is 208, which will put 104 volts across each bulb to the common (neutral) point.
now lets add a 60 watt bulb from C leg to this common point, this will rise the voltage toward the 120 volts balance of the circuit in whole that would be achieved if all the bulbs were 100 watt, putting in a less wattage bulb say a 20 watt, would move the voltage toward the original voltage of 104 volts, removing the bulb would put it back at 104 volts. now exceed the 100 watts of the two other bulbs and there voltage will rise as they can no longer keep balance.
The above will always be true as long as two of the loads are the same and the third is less than one of the two in balance

Yep I know almose 1:30 and my mind is out for the count lol

 

[76nemo]

Wah,...hehehehe

 

[LarryFine]

Ok lets do this one step at a time, lets say we connect two 100 watt bulbs in series from A leg to B leg, the voltage across this circuit is 208, which will put 104 volts across each bulb to the common (neutral) point.

Okay so far.

now lets add a 60 watt bulb from C leg to this common point, this will rise the voltage toward the 120 volts balance of the circuit in whole that would be achieved if all the bulbs were 100 watt, putting in a less wattage bulb say a 20 watt, would move the voltage toward the original voltage of 104 volts, removing the bulb would put it back at 104 volts.

Agreed. The 60 and 20 watt bulbs have a higher resiance than the 100's, and would see greater than 120v.

now exceed the 100 watts of the two other bulbs and there voltage will rise as they can no longer keep balance.

I still agree, and this agrees with my post. The greater-than-100-watt bulb will have a lower resistance than the 100's, will have less than 120v across it, and the 100's will see greater than 120v

The above will always be true as long as two of the loads are the same and the third is less than one of the two in balance

As I stated, the load(s) with the higher resistance will see greater voltage(s).

 

[infinity]

Ok lets do this one step at a time, lets say we connect two 100 watt bulbs in series from A leg to B leg, the voltage across this circuit is 208, which will put 104 volts across each bulb to the common (neutral) point.

You're saying that they're not connected to the neutral correct? Otherwise each lamp will have 120 volts across it.

 

[Smart $]

Inhale, man! :grin:

Man, I hope you're going to let him exhale sometime soon... he's probably past dark blue and into shades of purple by now :grin:

 

[hardworkingstiff]

Probably what I need to do is understand the reasoning WHY there is no current on the neutral and that's what I can't see. Formulas work and I accept it but don't get it...

The way I get it for single phase circuits is as follows:

A-leg and B-leg share a neutral.

Current flows from - to +. If we were to look at the current flow in the conductors at a point in time when we had a peak sine wave then one of the legs would be - in relation to neutral and the other would be + in relation to neutral. Current would flow from the - leg through the load towards neutral. At the same time, current would be flowing from the neutral to the + leg. Since the load neutrals are connected together to form the MWBC, the current from the - leg that's flowing towards the neutral would be used by the leg that is positive (flowing from the neutral to the + leg.

If the loads are equal, then all of the current flowing from the - leg will be used by the + leg and no current needs to flow back through the MWBC neutral. If the currents are 10 and 8, then the neutral will have to carry the unbalanced load of 2-amps.

When people mistakenly put both hots on the same phase, then both hots are + to neutral at the same time causing the neutral to carry the current from both legs.

I hope this makes as much sense to y'all as it does to me, lol. :smile:

 

[Dennis Alwon]

I hope this makes as much sense to y'all as it does to me, lol. :smile:

Actually Lou that helps alot in understanding WHY the neutral load is the difference between A and B. To bad that sine wave is another voodoo thing I have to accept. :grin:

Thanks all. Larry your explanation was great but not quite what I was looking for but I am not sure I can explain myself any better. Are you sure you didn't write that for me? I certainly remembered it and actually had it saved in a word doc on my puter.

Thanks again for helping me

Your Honorable Denseness--- Dennis

 

[hardworkingstiff]

To bad that sine wave is another voodoo thing I have to accept. :grin:

Lol, that's funny Dennis! I know what you mean.

 

[ActionDave]

A 240V circut- power comes in on A goes makes water hot and returns on B.
Got it.
A 120V circut- power comes in on A lights a 100W bulb and returns on neutral/grounded/center tap, or comes in on B lights a 100W bulb and returns the same way.
Got it.
MWBC- power comes in on A and B, each light their own 100W bulb, bypass the neutral and return on each other?
I don't got it.

 

[Dennis Alwon]

MWBC- power comes in on A and B, each light their own 100W bulb, bypass the neutral and return on each other?
I don't got it.

On a MWBC it does not bypass the neutral. It returns on the neutral for both circuits but it returns at 180 degrees ( another voodoo concept) from each other which makes the resulting load on the neutral zero. If there were 10 amps on phase "A" and 5 amps on phase "B" then the neutral would have a load of the difference of A and B or 5 amps.

 

[iwire]

On a MWBC it does not bypass the neutral. It returns on the neutral for both circuits but it returns at 180 degrees ( another voodoo concept) from each other which makes the resulting load on the neutral zero.

I have to disagree, with equal loading on both sides of the neutral it does in fact 'bypass' the neutral. The neutral could be removed with no change at all in the circuit operation.

 

[Dennis Alwon]

I have to disagree, with equal loading on both sides of the neutral it does in fact 'bypass' the neutral. The neutral could be removed with no change at all in the circuit operation.

So explain to me how the circuit would work with a MWBC feeding 2 100 watt bulbs with no neutral connected. What am I missing. No neutral makes it 220 volts and the neutral has to be connected to the socket to work. :-?

Now I am confused.