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Calculating loads on 120/208V system
|
|
| MammalianFish@gmail.com 2007-09-21, 3:25 am |
| I'm a system engineer responsible for specifying power quantities for
my systems. There are a lot of ways that are accepted in my industry
to do this that I know to be incorrect, so I'd like to hear from
someone with more specific expertise.
The systems are comprised of a combination of 120V resistive loads,
and 208V resistive and inductive loads, although the devices that have
inductive loads actually operate at various voltages, their power
supplies accept 208VAC. Most of the power available is 208/120VAC
60hz, usually in 400A increments.
What a lot of people do is add up the load P, divide by operating
voltage and divide by 3, so if we had 10 500w loads at 120V and 5 at
208V, it would look something like
((5000/110)+(2500/203))/3 = 20A needed.
I can't imagine that's correct. What I have done based on advice and
reading is divide the load P by 1.73, a relevant power factor and the
supply voltage, so this:
(5000+2500)/(1.73*0.9*203) = 24A needed
But I can imagine that being wrong, also. What is really the way to do
this?
Thanks in advance.
| |
| Gerald Newton 2007-09-21, 3:25 am |
| On Sep 20, 8:01 pm, MammalianF...@gmail.com wrote:
> I'm a system engineer responsible for specifying power quantities for
> my systems. There are a lot of ways that are accepted in my industry
> to do this that I know to be incorrect, so I'd like to hear from
> someone with more specific expertise.
>
> The systems are comprised of a combination of 120V resistive loads,
> and 208V resistive and inductive loads, although the devices that have
> inductive loads actually operate at various voltages, their power
> supplies accept 208VAC. Most of the power available is 208/120VAC
> 60hz, usually in 400A increments.
>
> What a lot of people do is add up the load P, divide by operating
> voltage and divide by 3, so if we had 10 500w loads at 120V and 5 at
> 208V, it would look something like
>
> ((5000/110)+(2500/203))/3 = 20A needed.
>
> I can't imagine that's correct. What I have done based on advice and
> reading is divide the load P by 1.73, a relevant power factor and the
> supply voltage, so this:
>
> (5000+2500)/(1.73*0.9*203) = 24A needed
>
> But I can imagine that being wrong, also. What is really the way to do
> this?
>
> Thanks in advance.
As an electrician, I thought we always added all the loads in watts to
find the total watts.
Then we used the P=pf*1.73*I*E for 3-phase. However, for single phase
the 1.73 is removed. But for 208/120 volts this sounds like a three
phase system.
I=P/(pf*1.73*E) where I is the curent in each phase.
Which looks like what you did.
I have written some JavaScript programs on this at:
http://www.electrician2.com/calculators/theory.html
and at
http://www.electriciancalculators.com/
| |
| Michael Moroney 2007-09-21, 1:25 pm |
| MammalianFish@gmail.com writes:
>The systems are comprised of a combination of 120V resistive loads,
>and 208V resistive and inductive loads, although the devices that have
>inductive loads actually operate at various voltages, their power
>supplies accept 208VAC. Most of the power available is 208/120VAC
>60hz, usually in 400A increments.
>What a lot of people do is add up the load P, divide by operating
>voltage and divide by 3, so if we had 10 500w loads at 120V and 5 at
>208V, it would look something like
>((5000/110)+(2500/203))/3 = 20A needed.
>I can't imagine that's correct. What I have done based on advice and
>reading is divide the load P by 1.73, a relevant power factor and the
>supply voltage, so this:
>(5000+2500)/(1.73*0.9*203) = 24A needed
>But I can imagine that being wrong, also. What is really the way to do
>this?
1) Add the power: 5000+2500 = 7500 watts.
2) Divide by 3 (legs): 7500/3=2500
3) Divide by phase-neutral voltage: 2500/120 = 20.8A
4) Divide by power factor, unless power in 1) is VA not watts.
Round up according to NEC/local rules and future expansion.
This assumes the loads are balanced, but in your case they are not, since
neither 10 nor 5 are evenly divisible by 3. Of the 10 500 watt 120V
loads, you'll have to load the phases 4-3-3 for example.
| |
| MammalianFish@gmail.com 2007-09-21, 8:25 pm |
| >
> 1) Add the power: 5000+2500 = 7500 watts.
> 2) Divide by 3 (legs): 7500/3=2500
> 3) Divide by phase-neutral voltage: 2500/120 = 20.8A
> 4) Divide by power factor, unless power in 1) is VA not watts.
>
> Round up according to NEC/local rules and future expansion.
>
> This assumes the loads are balanced, but in your case they are not, since
> neither 10 nor 5 are evenly divisible by 3. Of the 10 500 watt 120V
> loads, you'll have to load the phases 4-3-3 for example.
Just out of curiosity, why doesn't the operating voltage matter? It
would seem that 208V loads would ultimately draw less amperage than
120V loads. No?
Thanks,
Moo
| |
|
| In article <1190417101.918448.158150@g4g2000hsf.googlegroups.com>,
MammalianFish@gmail.com says...
>
> Just out of curiosity, why doesn't the operating voltage matter? It
> would seem that 208V loads would ultimately draw less amperage than
> 120V loads. No?
120*sqrt(3)=208
208 three-phase is three 120V legs. He's accounted for that
in 2) and 3).
--
Keith
| |
| Gerald Newton 2007-09-21, 8:25 pm |
| On Sep 21, 7:19 am, moro...@world.std.spaamtrap.com (Michael Moroney)
wrote:
> MammalianF...@gmail.com writes:
>
> 1) Add the power: 5000+2500 = 7500 watts.
> 2) Divide by 3 (legs): 7500/3=2500
> 3) Divide by phase-neutral voltage: 2500/120 = 20.8A
> 4) Divide by power factor, unless power in 1) is VA not watts.
>
> Round up according to NEC/local rules and future expansion.
>
> This assumes the loads are balanced, but in your case they are not, since
> neither 10 nor 5 are evenly divisible by 3. Of the 10 500 watt 120V
> loads, you'll have to load the phases 4-3-3 for example.- Hide quoted text -
>
> - Show quoted text -
208 divided by 1.732 = 120 volts
So I= 2500/120 equals I =2500/(208/1.732)
or I=2500*1.732/208
I = 20.8 amperes
or I = 7500/(208*1.732)
I = 20.8 amperes
Then 7500/(208*1.732) is identical to 2500*1.732/208
Why?
Multiply both sides by 1.732 and you get 7500/208 ==7500/208 since
1.732*1.732 = 3.
They are identities.
So why all the multiplying by three, just use the standard three phase
equation.
Of course this assumes a balanced load, for unbalanced loads the
neutral will carry the unbalance.
| |
| MammalianFish@gmail.com 2007-09-22, 3:25 am |
|
>
> Then 7500/(208*1.732) is identical to 2500*1.732/208
> Why?
> Multiply both sides by 1.732 and you get 7500/208 ==7500/208 since
> 1.732*1.732 = 3.
> They are identities.
> So why all the multiplying by three, just use the standard three phase
> equation.
> Of course this assumes a balanced load, for unbalanced loads the
> neutral will carry the unbalance.
That makes a lot of sense, thank you.
So when the system is out of phase and there is voltage in the
neutral, is the copper in the neutral system electrified in the same
way as the phases, or is it different somehow? Meaning, would I
measure 120V from the Neutral to Ground?
| |
| Gerald Newton 2007-09-22, 3:25 am |
| On Sep 21, 6:59 pm, MammalianF...@gmail.com wrote:
>
> That makes a lot of sense, thank you.
>
> So when the system is out of phase and there is voltage in the
> neutral, is the copper in the neutral system electrified in the same
> way as the phases, or is it different somehow? Meaning, would I
> measure 120V from the Neutral to Ground?
"electrified" where did you come up with this?
Somehow, I think you must be pulling a leg, so to speak.
Of course you measure 120 volts to the neutral. 208 volts from phase
to phase and 120 volts from each phase to the neutral.
The current changes with the load, not the voltage. This assumes what
we electricians work with - steady state conditions and not an
engineering analysis that would require Thevenin's Theorem. It also
does not include voltage drop.
| |
| Stuart 2007-09-22, 1:25 pm |
| In article <1190429996.499464.272340@g4g2000hsf.googlegroups.com>,
<MammalianFish@gmail.com> wrote:
> So when the system is out of phase and there is voltage in the
> neutral, is the copper in the neutral system electrified in the same
> way as the phases, or is it different somehow? Meaning, would I
> measure 120V from the Neutral to Ground?
The neutral is normally connected to ground at some point, usually your
local sub-station. You should NOT be able to measure 120V between neutral
and ground unless you have a dangerous fault.
Depending on the resistance of the neutral cable between your measurement
point and the connection to ground, and the current flowing through it, it
will develop a potential to earth but this should not normally be more
than a few hundreds of millivolts.
--
Stuart Winsor
From is valid but subject to change without notice if it gets spammed.
For Barn dances and folk evenings in the Coventry and Warwickshire area
See: http://www.barndance.org.uk
| |
| Bill Shymanski 2007-09-22, 1:25 pm |
|
<MammalianFish@gmail.com> wrote in message
news:1190347269.147828.208400@k79g2000hse.googlegroups.com...
> I'm a system engineer responsible for specifying power quantities for
> my systems. There are a lot of ways that are accepted in my industry
> to do this that I know to be incorrect, so I'd like to hear from
> someone with more specific expertise.
>
Really? What state/province did you get your licence in? An odd sort
of question to ask on a news group.
Bill
| |
| Gerald Newton 2007-09-22, 1:25 pm |
| On Sep 22, 8:11 am, Stuart <SW_NOS...@dsl.pipex.com> wrote:
> In article <1190429996.499464.272...@g4g2000hsf.googlegroups.com>,
> <MammalianF...@gmail.com> wrote:
>
>
> The neutral is normally connected to ground at some point, usually your
> local sub-station. You should NOT be able to measure 120V between neutral
> and ground unless you have a dangerous fault.
>
> Depending on the resistance of the neutral cable between your measurement
> point and the connection to ground, and the current flowing through it, it
> will develop a potential to earth but this should not normally be more
> than a few hundreds of millivolts.
>
> --
> Stuart Winsor
>
> From is valid but subject to change without notice if it gets spammed.
>
> For Barn dances and folk evenings in the Coventry and Warwickshire area
> See:http://www.barndance.org.uk
The supply was not specified, but for 208 volts and 120 volts on the
same system we encounter this in the field about 99 percent of the
time on a three phase 4-wire 208/120 volt solidly grounded wye system
where the neutral is connected at XO at the transformer and is also
connected through a grounding electrode conductor to earth. This is a
very common system that any licensed journeyman finds familar. We
then measure 120 volts from each of the three phases to the neutral
and 208 volts from phase to phase. This is a very common
configuration and I cannot undrstand why there is any confusion with
this.
| |
|
| In article <1190429996.499464.272340@g4g2000hsf.googlegroups.com>,
MammalianFish@gmail.com says...
>
>
> That makes a lot of sense, thank you.
>
> So when the system is out of phase and there is voltage in the
> neutral, is the copper in the neutral system electrified in the same
> way as the phases, or is it different somehow? Meaning, would I
> measure 120V from the Neutral to Ground?
>
No, neutral is the "grounded conductor". It will remain close (IR
drop, where I is the imbalance in the legs) to ground. The neutral
current is the vector sum of the current in the three phases. In a
balanced system it is zero so no current flows in the neutral, thus
zero voltage across the neutral conductor.
--
Keith
| |
| Stuart 2007-09-22, 8:25 pm |
| In article <1190479174.802236.276190@19g2000hsx.googlegroups.com>,
Gerald Newton <electrician@electrician2.com> wrote:
> This is a
> very common system that any licensed journeyman finds familar. We
> then measure 120 volts from each of the three phases to the neutral
> and 208 volts from phase to phase. This is a very common
> configuration and I cannot undrstand why there is any confusion with
> this.
Indeed
--
Stuart Winsor
From is valid but subject to change without notice if it gets spammed.
For Barn dances and folk evenings in the Coventry and Warwickshire area
See: http://www.barndance.org.uk
| |
| Gerald Newton 2007-09-22, 8:25 pm |
| On Sep 22, 1:37 pm, krw <k...@att.bizzzz> wrote:
> In article <1190429996.499464.272...@g4g2000hsf.googlegroups.com>,
> MammalianF...@gmail.com says...
>
>
>
>
>
>
>
>
> No, neutral is the "grounded conductor". It will remain close (IR
> drop, where I is the imbalance in the legs) to ground. The neutral
> current is the vector sum of the current in the three phases. In a
> balanced system it is zero so no current flows in the neutral, thus
> zero voltage across the neutral conductor.
>
> --
> Keith- Hide quoted text -
>
> - Show quoted text -
Sorry to rebut your claim, but I just returned from the IAEI NW
Section meeting where the New Definition of neutral in the 2008 NEC
was discussed. The NEC correlating committee assigned a specific task
group to examine the necessity of such a definition in the Code. A
definition of a neutral conductor was developed as a result of work by
the NEC TCC. The new definition given in Article 100 is: "Neutral
conductor is defined as the conductor connected to the neutral point
of a system that is intended to carry current under normal
conditions." The next point made on page 32 of the Analysis of Change
Book states, "Grounded conductors are not always neutral conductors."
The illustration shows a grounding electrode conductor as an example.
The definition of Neutral Point has been revised and is now defined as
the common point on a wye-connection in a polyphase system, or
midpoint on a single-phase, 3-wire system, or mid-point of a single-
phase portion of a 3-phase delta system, or a midpoint of a 3-wire
direct current system.
A grounded conductor is still defined as a system or circuit conductor
that is intentionally grounded.
So a neutral may be a grounded conductor but a grounded conductor is
not always a neutral but is always grounded. Nowhere in the
definition of a neutral conductor or neutral point does it say the
neutral or neutral point must be grounded.
Now this certainly means more confusion in the field. But it looks as
if a ungrounded system can have both a neutral point and a neutral
conductor that are not grounded, and a separate grounded conductor
that is not a neutral.
| |
|
| In article <1190506132.897285.159280@y42g2000hsy.googlegroups.com>,
electrician@electrician2.com says...
> On Sep 22, 1:37 pm, krw <k...@att.bizzzz> wrote:
>
> Sorry to rebut your claim, but I just returned from the IAEI NW
> Section meeting where the New Definition of neutral in the 2008 NEC
> was discussed. The NEC correlating committee assigned a specific task
> group to examine the necessity of such a definition in the Code. A
> definition of a neutral conductor was developed as a result of work by
> the NEC TCC. The new definition given in Article 100 is: "Neutral
> conductor is defined as the conductor connected to the neutral point
> of a system that is intended to carry current under normal
> conditions." The next point made on page 32 of the Analysis of Change
> Book states, "Grounded conductors are not always neutral conductors."
> The illustration shows a grounding electrode conductor as an example.
It's not a claim. You, and the NEC, are welcome to redefine terms
[*], then show how brilliant you are. The electrons don't know any
differently.
> The definition of Neutral Point has been revised and is now defined as
> the common point on a wye-connection in a polyphase system, or
> midpoint on a single-phase, 3-wire system, or mid-point of a single-
> phase portion of a 3-phase delta system, or a midpoint of a 3-wire
> direct current system.
>
> A grounded conductor is still defined as a system or circuit conductor
> that is intentionally grounded.
Ok, so why would you not ground the neutral point in a Wye?
....sounds dangerous.
> So a neutral may be a grounded conductor but a grounded conductor is
> not always a neutral but is always grounded. Nowhere in the
> definition of a neutral conductor or neutral point does it say the
> neutral or neutral point must be grounded.
Again...
> Now this certainly means more confusion in the field.
Isn't that their intention? ;-)/2
> But it looks as
> if a ungrounded system can have both a neutral point and a neutral
> conductor that are not grounded, and a separate grounded conductor
> that is not a neutral.
Again, YOY would you float the neutral? Sounds dangerous.
--
Keith
| |
| Gerald Newton 2007-09-22, 8:25 pm |
| On Sep 22, 4:36 pm, krw <k...@att.bizzzz> wrote:
> In article <1190506132.897285.159...@y42g2000hsy.googlegroups.com>,
> electric...@electrician2.com says...
>
>
>
>
>
>
>
>
>
>
>
>
>
> It's not a claim. You, and the NEC, are welcome to redefine terms
> [*], then show how brilliant you are. The electrons don't know any
> differently.
>
>
>
> Ok, so why would you not ground the neutral point in a Wye?
> ...sounds dangerous.
>
>
> Again...
>
>
> Isn't that their intention? ;-)/2
>
>
> Again, YOY would you float the neutral? Sounds dangerous.
>
> --
> Keith- Hide quoted text -
>
> - Show quoted text -
How about the high impedance grounded system. We have a neutral that
is not grounded in a system that is commonly used and considered safe.
| |
|
| In article <1190508807.763571.171580@d55g2000hsg.googlegroups.com>,
electrician@electrician2.com says...
> On Sep 22, 4:36 pm, krw <k...@att.bizzzz> wrote:
>
> How about the high impedance grounded system. We have a neutral that
> is not grounded in a system that is commonly used and considered safe.
>
How about it? How does it work? Why don't you let us in on the
theory behind it, making it safe? Why? What keeps neutral at a
reasonable voltage, say, if one of the legs drops? Call me
skeptical.
--
Keith
| |
| Gerald Newton 2007-09-23, 1:25 pm |
| On Sep 23, 7:30 am, krw <k...@att.bizzzz> wrote:
> In article <1190508807.763571.171...@d55g2000hsg.googlegroups.com>,
> electric...@electrician2.com says...
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
> How about it? How does it work? Why don't you let us in on the
> theory behind it, making it safe? Why? What keeps neutral at a
> reasonable voltage, say, if one of the legs drops? Call me
> skeptical.
>
> --
> Keith- Hide quoted text -
>
> - Show quoted text -
The IEEE Green book has ample diagrams of high impedance grounding in
the last part of the book.
The theory is simple; we install impedance in the ground return path
between XO and ground. For 3-phase 480/277 volt wye systems this is
usually a 55 ohm resistor but can be an inductor.
The resistor allows charging current to return, but limits the ground
fault current to about 5 amperes (277/55). A ct is place on the
ungrounded neutral that runs from XO to one end of the resistor. The
ct sends a signal to a relay set to energize at 3 to 5 amperes or at
some value just above the charging current which is roughly 1 ampere
for each system 1000 KVA . The relay sends a signal to the control
room.
The first ground fault does not take the system down but allows
maintenance persons time to troubleshoot and repair. However, I have
seen such ground faults allow to exist for months at the Valdez Marine
Terminal on the Trans Alaska pipeline. While the ground fault existed
the other phase conductors measure 480 volts to ground instead of 277
volts. In Valdez we experienced a second ground fault in this system
on a separate phase which then gave us a phase to phase fault. The
400 ampere breaker did not trip because the second ground fault was
some distance from the first ground fault and this terminal has one of
the greatest engineering blunders of the century. Fluor engineers
permitted almost all 480 volt, 5 kv, and 13.8 kv cables to be
installed in the 14 miles of cable tray in 1975 when the cables DID
NOT HAVE EQUIPMENT GROUNDING CONDUCTORS INSTALLED IN THEM. Equipment
grounding is done through an elaborate system of metal bonding and
grounding electrode bonding. Believe me this is true. Almost 20
percent of the nation's oil has been going through this terminal for
about 28 years and the electrical system has a major grounding
anomaly. That second ground fault I am referring to took place near a
crude tank in a Class 1, Division 1 hazardous location! It really
shook some people up to see that phase to phase arc and not see a
circuit breaker trip.
The reasons the cables do not have equipment grounding conductors
vary, but one is after the error was discovered and the cables were on
site, if the error had been corrected Arco would have had to file for
bankruptcy. The oil had to flowing in the pipeline by a certain date
and that was more critical than reordering the correct cables.
Please excuse me for being off topic here but then these threads go
all over the place. The bottom line is there are systems where the
neutral is not grounded and where there is a grounded conductor that
is supposed to be safe. These are primarily found in industrial
locations where a ground fault detection system has been installed.
| |
|
| From: electrician@electrician2.com (Gerald=A0Newton)
On Sep 22, 1:37 pm, krw <k...@att.bizzzz> wrote:
In article <1190429996.499464.272...@g4g2000hsf.googlegroups.com>,
MammalianF...@gmail.com says...
Then 7500/(208*1.732) is identical to 2500*1.732/208 Why?
Multiply both sides by 1.732 and you get 7500/208 =3D=3D7500/208 since
1.732*1.732 =3D 3.
They are identities.
So why all the multiplying by three, just use the standard three phase
equation.
Of course this assumes a balanced load, for unbalanced loads the neutral
will carry the unbalance.
That makes a lot of sense, thank you.
So when the system is out of phase and there is voltage in the neutral,
is the copper in the neutral system electrified in the same way as the
phases, or is it different somehow? Meaning, would I measure 120V from
the Neutral to Ground?
No, neutral is the "grounded conductor". It will remain close (IR drop,
where I is the imbalance in the legs) to ground. The neutral current is
the vector sum of the current in the three phases. In a balanced system
it is zero so no current flows in the neutral, thus zero voltage across
the neutral conductor.
--
=A0=A0=A0=A0=A0=A0Keith- Hide quoted text -
- Show quoted text -
Sorry to rebut your claim, but I just returned from the IAEI NW Section
meeting where the New Definition of neutral in the 2008 NEC was
discussed. The NEC correlating committee assigned a specific task group
to examine the necessity of such a definition in the Code. A definition
of a neutral conductor was developed as a result of work by the NEC TCC.
The new definition given in Article 100 is: "Neutral conductor is
defined as the conductor connected to the neutral point of a system that
is intended to carry current under normal conditions." The next point
made on page 32 of the Analysis of Change Book states, "Grounded
conductors are not always neutral conductors." The illustration shows a
grounding electrode conductor as an example.
The definition of Neutral Point has been revised and is now defined as
the common point on a wye-connection in a polyphase system, or midpoint
on a single-phase, 3-wire system, or mid-point of a single- phase
portion of a 3-phase delta system, or a midpoint of a 3-wire direct
current system.
A grounded conductor is still defined as a system or circuit conductor
that is intentionally grounded.
So a neutral may be a grounded conductor but a grounded conductor is not
always a neutral but is always grounded. Nowhere in the definition of a
neutral conductor or neutral point does it say the neutral or neutral
point must be grounded.
Now this certainly means more confusion in the field. But it looks as if
a ungrounded system can have both a neutral point and a neutral
conductor that are not grounded, and a separate grounded conductor that
is not a neutral.
--------
As per your NEC explanation: I agree It is a mistake to call the
neutral, ground., hence the person that taught me wiring referred to the
Neutral as Common or Common Ground, so as to regard It as the conductor
common on the line circuit.In other words the conductor in Common with
the Hot conductor from the source.....
Just like in your revised NEC explanation., I guess it's a old timers
thing to be so specific, but it helps to curtail making wiring mistakes
- Earth Ground isn't Common & is Never Neutral., but something like you
mention; Neutral Ground is always Common but Never Earth.I suspect the
term "Common Ground" came from there to differentiate it from Earth
Ground.
Needless to say it's all good to Test your branch circuits, we'll still
obtain +/-120 to ground or neutral., so it's easy for untrained handlers
to mistake the terms for there use., as long as they remember that
Neutral is from the Source & not the Earth.
Roy Q.T.
Urban Technician
[I don't make em, I just fix em]
| |
| HapticZ 2007-09-25, 9:25 am |
| floating neutrals?
perhaps xformer isolated and within a separated enclosure to prevent faulted
grounding? this seems inherently dangerous a definition.
ground is generally a physical hard connection somewhere to "earth/terra
firma" the great return loopback to the nuclear plant.
neutral is, i thought it was anyway, a localised pathway among the power
circuits at the devices/installation themselves, rather than a "faux ground"
where standards diverge we open many problematic situations for dangerous
installs and site specific alterations. this requires again specialized
maintenance and education of technicians and subsequent add-ons later.
basic standard and un-ambiguous design eliminates potentially dangerous
confusion and death.
"Gerald Newton" <electrician@electrician2.com> wrote in message
news:1190506132.897285.159280@y42g2000hsy.googlegroups.com...
> On Sep 22, 1:37 pm, krw <k...@att.bizzzz> wrote:
phase[color=darkred]
>
> Sorry to rebut your claim, but I just returned from the IAEI NW
> Section meeting where the New Definition of neutral in the 2008 NEC
> was discussed. The NEC correlating committee assigned a specific task
> group to examine the necessity of such a definition in the Code. A
> definition of a neutral conductor was developed as a result of work by
> the NEC TCC. The new definition given in Article 100 is: "Neutral
> conductor is defined as the conductor connected to the neutral point
> of a system that is intended to carry current under normal
> conditions." The next point made on page 32 of the Analysis of Change
> Book states, "Grounded conductors are not always neutral conductors."
> The illustration shows a grounding electrode conductor as an example.
>
> The definition of Neutral Point has been revised and is now defined as
> the common point on a wye-connection in a polyphase system, or
> midpoint on a single-phase, 3-wire system, or mid-point of a single-
> phase portion of a 3-phase delta system, or a midpoint of a 3-wire
> direct current system.
>
> A grounded conductor is still defined as a system or circuit conductor
> that is intentionally grounded.
>
> So a neutral may be a grounded conductor but a grounded conductor is
> not always a neutral but is always grounded. Nowhere in the
> definition of a neutral conductor or neutral point does it say the
> neutral or neutral point must be grounded.
>
> Now this certainly means more confusion in the field. But it looks as
> if a ungrounded system can have both a neutral point and a neutral
> conductor that are not grounded, and a separate grounded conductor
> that is not a neutral.
>
>
| |
| Gerald Newton 2007-09-25, 5:25 pm |
| On Sep 25, 2:04 am, "HapticZ" <hapt...@sbcglobal.net> wrote:
> floating neutrals?
>
> perhaps xformer isolated and within a separated enclosure to prevent faulted
> grounding? this seems inherently dangerous a definition.
>
> ground is generally a physical hard connection somewhere to "earth/terra
> firma" the great return loopback to the nuclear plant.
>
> neutral is, i thought it was anyway, a localised pathway among the power
> circuits at the devices/installation themselves, rather than a "faux ground"
>
> where standards diverge we open many problematic situations for dangerous
> installs and site specific alterations. this requires again specialized
> maintenance and education of technicians and subsequent add-ons later.
>
> basic standard and un-ambiguous design eliminates potentially dangerous
> confusion and death.
>
> "Gerald Newton" <electric...@electrician2.com> wrote in message
>
> news:1190506132.897285.159280@y42g2000hsy.googlegroups.com...
>
>
>
>
> phase
>
>
>
>
>
>
>
>
>
>
>
> - Show quoted text -
A high impedance grounded neutral is not an ungrounded or "floating"
neutral. It is a neutral purposely grounded through a resistor or
inductor. This way the charging current (or leakage current) can
return, but ground fault current is limited to several amperes. This
is primarily used for a wye wound secondary. For a 480/277 volt 3-
phase system this does not permit the 277 volts to be used for power
or lighting. If 277 volts is required for power a separate isolation
480/480/277 volt delta/wye transformer is used with a solidly grounded
neutral. For ungrounded delta systems a zig zag transformer is used
to allow charging current to return and to prevent high voltages on
the system in case of an arcing ground fault in the 480 volt system.
These are common configurations used in the oil industry and mine
mills in Alaska.
Traditionally, I used to think of the neutral as the grounded
conductor that carried the unbalanced current in a 3-phase 4-wire
system. However, the new definition broadens the definition to where
we now can call the grounded white conductor in a 240/120 volt single
phase residential wiring system the neutral conductor.
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