240v debate....

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mivey

Senior Member
Interstate 95 has two lanes running from North to South and two lanes from South to North...
Two cars are headed north on I-95 towards Fayetteville. Both could be headed home or both headed away from home or one toward home and one away from home, depending on your definition of home.

They are both still headed North but "away from" or "toward" home is a relative term. That is simply the way voltages are defined because they are relative to a reference and that reference does not have to be a specific one, even if that one might make an analysis easier in some cases.
 

mivey

Senior Member
It is possible but much less likely, to me that tells me in most cases 24 volts is safer than 120, which is safer than 240.

'Safer' is a relative term, not a black and white one. :)
I agree.
Bob, I call the two statements nonsense, in the sense that energized electrical equipment is not safe, and voltage levels do not change that statement.
I agree with that also but, strictly speaking, it is relative like Iwire said. He is not saying 120 volts is "safe".
It is the same sense in which I would declare that a pistol is never unloaded
But would you rather get shot with a 22 cal or 50 cal pistol?
that a stop sign is not capable of causing a driver to stop his car
Would you rather get hit by a peddle car or a Mack truck?
and that it is unwise to stand under a load being lifted by a crane.
Would you rather stand under a bag of inflated balloons being lifted by a crane or a loaded shipping container?

We are surrounded by dangerous circumstances. We improve our chances of coming home at the end of the day by recognizing the dangers, and by taking specific precautions to mitigate the dangers. You are far more aware of that than I could ever be, since your chosen profession puts you closer to such dangers than mine does. But the basic notion of ?this voltage level is safer than that voltage level? simply runs contrary to my notion of the meaning of safety.
Would you rather be accidently exposed to a live 120 volt circuit or a live 500 kV circuit?
 

mivey

Senior Member
There is a difference between having a chosen representation for a system, and defining the system. All too often, people forget that it is a chosen representation, and they try to define the system based on the representation. This is where you run into problems.
The problem is when you look at how a voltage was derived and let that define everything after that. There is more than one way to create usable voltage sources.

Saying that the two waveforms are the equivalent of being 180? out of phase is not the same as saying the two wave forms are 180? out of phase.
Potentials 101.

This chosen representation works only because we have a symmetrical waveform. The symmetrical waveform allows you to represent one of them as though it had an 8 ms time delay with a magnitude reversal. However, if the waveform was not symmetrical, you cannot represent them with this time delay. There is no time delay. It is just a chosen representation.

Let's say your mechanical generator has some sort of flaw that introduces an artifact into the waveform every 360 degrees. This artifact is going to appear in both of your output waveforms at the same instant in time, yet your chosen representation says the two waveforms are time shifted by 180?. This is a situation where the chosen representation fails.
A mechanical bump on a normal 3-phase generator will produce an "artifact" on all three phases at the same time.
There is nothing wrong with using whatever chosen representation someone wants. It's their choice. The problem arises when someone forgets that it is simply a chosen representation, and tries to define the system as though the representation was real.
The choice of voltage reference is a choice, not some given just because of the way the voltages were created.

To me, it seems blindingly obvious...Saddens me.
I agree. We learned this in basic physics and again in Electric 101.

A 180? phase difference is an 8ms time delay at 60Hz (or 10ms @50 Hz). If this assertion that a 180? phase shift truly existed...transformed spike would allegedly occur in the secondary A-phase at To, but in the B-phase at To + 8ms...and it cannot be defined with this artificial time delay.
Who says there must be a time delay? Time delaying one signal through two stages is one way to create a 3-phase source (I have a circuit that does exactly that for lab experiments). The other way is to physically displace the windings on a common/linked shaft (where all waveforms have the same beginning time).

The reason why you are seeing this as one 170 peak + and 170 peak ? is due to the manner in which you are connecting the two channels of your meter to the system not because there are two phases that are 180 degrees out of each other.

Just what type of phase system is it you are working with? Is it a two phase system or a single phase system? If you say a single phase then there is only one phase and nothing that is 180 degrees out of anything.
Polyphase system definitions are another topic. Here we are discussing the definition of a voltage. It is a fundamental understanding of voltage potentials.
 

Smart $

Esteemed Member
Location
Ohio
:eek:hmy:... you guys really must like to argue.

It is what it is, and does what it does... no matter how you describe it. The true question is whether you understand it.


Given a black box with three terminals marked:

----------------- L1 ---------
120VAC---------------------|
----- ------------ N-------240VAC
120VAC--------------------_|
----------------- L2 ---------

What camp of thought is correct? Those who understand it will say both.
 
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Rick Christopherson

Senior Member
Who says there must be a time delay?
Who? Apparently you are. Didn't you learn in your "potentials 101" class that a phase shift is a time delay? :dunce:

If you know how to make a transformer to give you a 180? phase shift time delay, I would like to buy one. If 180? and 8ms is possible, then 120? and 5ms should be a piece of cake. We could put all of those pesky phase converter manufacturers out of business by selling these magical transformers of yours. We'll be rich! :thumbsup:
 

mivey

Senior Member
Who? Apparently you are. Didn't you learn in your "potentials 101" class that a phase shift is a time delay? :dunce: If you know how to make a transformer to give you a 180? phase shift time delay, I would like to buy one. If 180? and 8ms is possible, then 120? and 5ms should be a piece of cake. We could put all of those pesky phase converter manufacturers out of business by selling these magical transformers of yours. We'll be rich! :thumbsup:
Cute. I guess. Go back and review Signals & Systems. A phase difference does not necessarily mean the same thing as a time shift. A bit off topic anyway since we look at steady state anyway and assume, for the most part, a balanced system with symmetrical waveforms.
 

Rick Christopherson

Senior Member
Cute. I guess. Go back and review Signals & Systems. A phase difference does not necessarily mean the same thing as a time shift. A bit off topic anyway since we look at steady state anyway and assume, for the most part, a balanced system with symmetrical waveforms.
How you "look at it" is entirely your choice with no right or wrong method. However, this discussion was about how some people have mistakenly transposed how they look at it into how they define it.

You can look at it as though it's a phase shift, because that works 95% of the time. However, when you mistakenly define it as a phase shift, then you do 2 things: (1) you force everyone to look at it as a phase shift, and (2) the analysis fails when it's not symmetrical over time.

The whole point of the discussion is education. People have become so accustomed to looking at it as a phase shift, that they have lost sight that this is only a tool for viewing it. They have taken this to mean that it is phase shifted, not simply being viewed as phase shifted.
 
I wanted to run a pump at 240v vs 120v, for obvious reasons, voltage drop, wire size, etc. But... I get this a lot from people, 240v is more dangerous. So I'm looking for a better way to explain to people that its just as safe as a 120v circuit but with better benefits.

What debate?

The higher the voltage is the more severe the effect of the electrocution on the body will be.

More convenient and economical for the distribution? Yes.

Neither is debateable.....:p
 

Besoeker

Senior Member
Location
UK
The reason why you are seeing this as one 170 peak + and 170 peak ? is due to the manner in which you are connecting the two channels of your meter to the system not because there are two phases that are 180 degrees out of each other.
It's what you would see and do see on an oscilloscope using the the neutral as the common point. Each half 0f the 120V-0-120V HAS to be displaced180deg from the other to get 240V in total. Another point you might like to ponder. Suppose you load each 120V half with say, a 10 ohm resistive load. Each will have a current of 12A, and, being a resistive load, that current would be in phase with the voltage driving it. What about the current in the neutral? It's zero in such circumstances. If the currents were not 180deg apart and of equal magnitude they would not cancel. Make the currents resistive but unequal, say one is 12A and the other 11A, and the neutral carries the difference, one amp in this case. That can't happen unless you have a 180deg phase difference.

Just what type of phase system is it you are working with? Is it a two phase system or a single phase system? If you say a single phase then there is only one phase and nothing that is 180 degrees out of anything.
That's come up here before and also elicited many responses. If you were using just the 240V with nothing connected to the centre tap, then it would be a single phase load.
Connect the neutral and have different loads on each half and you end up with two phases.

I mentioned earlier in the thread that we make hexaphase rectifiers particularly high current rectifiers. This arrangements takes the secondary windings of a three-phase primary transformer and each winding has a centre tap. The center taps of the windings are connected together. The other ends of the windings produce six voltages equally displaced at 60 deg intervals over a full cycle. Six phases. Hence hexaphase.

By the way I think I got it down pretty good.
I hope what I have posted has been helpful.
But I'm tempted to quote Jeremiah 5:21 or Matthew 7:6.
;)
 

jim dungar

Moderator
Staff member
Location
Wisconsin
Occupation
PE (Retired) - Power Systems
If you'd like, I'll try to find a bit of time to capture real-time waveforms on a storage 'scope. Would that convince you?
Given the terminal designations of L1-N-L2, what will you waveforms look like if you use L1 as your common reference point and then plot the L1-N voltage and the L1-L2 voltage? Are the voltages in-phase or out of phase?
What is the phasing of the waveforms if you did not use a common reference but instead measured L1-N and then keeping both probes in the same relationship moved both of them to measure N-L2?

When you measure L1-N and L2-N voltages as 120V, as you have been doing, what is the mathematical formula you use for 'summing' them to get L1-L2 = 240V,
V1n+V2n, V1n-V2n, V1n+Vn2, V1n-Vn2, Vn1+Vn2, Vn1-Vn2, Vn1+Vn2, Vn1-Vn2 (have I forgotten any possibilities)?
Do you use the same formula to determine L2-N if you measured L1-N and L1-L2?
 
Given the terminal designations of L1-N-L2, what will you waveforms look like if you use L1 as your common reference point and then plot the L1-N voltage and the L1-L2 voltage? Are the voltages in-phase or out of phase?
What is the phasing of the waveforms if you did not use a common reference but instead measured L1-N and then keeping both probes in the same relationship moved both of them to measure N-L2?

When you measure L1-N and L2-N voltages as 120V, as you have been doing, what is the mathematical formula you use for 'summing' them to get L1-L2 = 240V,
V1n+V2n, V1n-V2n, V1n+Vn2, V1n-Vn2, Vn1+Vn2, Vn1-Vn2, Vn1+Vn2, Vn1-Vn2 (have I forgotten any possibilities)?
Do you use the same formula to determine L2-N if you measured L1-N and L1-L2?
L1-N - 120V
L1-L2 - 240V
In phase since they are using the same reference point.

N- L1 - 120V
N - L2 -120V
180* out of phase with each other when referenced to N

The easiest way for me to think of these is vectorial, rotating imagery, which actually mimics the physical occurance how the magnetic flux rotates and induces current that drives voltage.
 

Besoeker

Senior Member
Location
UK
Hmmmm so contrary to popular myth, they still read and perhaps even know Scripture on the other side of the Big Pond :eek:hmy::jawdrop:
Don't know about the myth.
My experience is that is more commonly done so on the west side of the pond.
 

Besoeker

Senior Member
Location
UK
Given the terminal designations of L1-N-L2, what will you waveforms look like if you use L1 as your common reference point and then plot the L1-N voltage and the L1-L2 voltage? Are the voltages in-phase or out of phase?.
In phase but that is not comparing the two 120V halves with respect to each other.

What is the phasing of the waveforms if you did not use a common reference but instead measured L1-N and then keeping both probes in the same relationship moved both of them to measure N-L2?
With measuring both at the same time you have no way of determining phase relationship.

When you measure L1-N and L2-N voltages as 120V, as you have been doing, what is the mathematical formula you use for 'summing' them to get L1-L2 = 240V,
V1n+V2n, V1n-V2n, V1n+Vn2, V1n-Vn2, Vn1+Vn2, Vn1-Vn2, Vn1+Vn2, Vn1-Vn2 (have I forgotten any possibilities)?
Do you use the same formula to determine L2-N if you measured L1-N and L1-L2?
Potential difference. That's all.
 

david luchini

Moderator
Staff member
Location
Connecticut
Occupation
Engineer
When you measure L1-N and L2-N voltages as 120V, as you have been doing, what is the mathematical formula you use for 'summing' them to get L1-L2 = 240V,
V1n+V2n, V1n-V2n, V1n+Vn2, V1n-Vn2, Vn1+Vn2, Vn1-Vn2, Vn1+Vn2, Vn1-Vn2 (have I forgotten any possibilities)?
Do you use the same formula to determine L2-N if you measured L1-N and L1-L2?

The formula for V1-2 given measured voltages of V1n and V2n is KVL.

V1-2 = V1n-V2n

Of course KVL could also be correctly written as:

V1-2 = V1n+Vn2, or
........= -Vn1-V2n, or
........= -Vn1+Vn2.

But whichever way it is written, it still follows KVL.
 

jim dungar

Moderator
Staff member
Location
Wisconsin
Occupation
PE (Retired) - Power Systems
In phase since they are using the same reference point.

180* out of phase with each other when referenced to N

So, the reference point is absolutely critical when describing the relationship between the two waveforms.
 

steve66

Senior Member
Location
Illinois
Occupation
Engineer
I wanted to run a pump at 240v vs 120v, for obvious reasons, voltage drop, wire size, etc. But... I get this a lot from people, 240v is more dangerous. So I'm looking for a better way to explain to people that its just as safe as a 120v circuit but with better benefits.

Of course 120V is a little safer than 240V. But with that same logic, these are even safer:
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jim dungar

Moderator
Staff member
Location
Wisconsin
Occupation
PE (Retired) - Power Systems
But whichever way it is written, it still follows KVL.

So, the relationship of the measurements is critical when describing the system.

So saying, two waveforms in phase V1n and Vn2 which add together to V12, is a perfectly valid statement.
 

jim dungar

Moderator
Staff member
Location
Wisconsin
Occupation
PE (Retired) - Power Systems
Given the terminal designations of L1-N-L2, what will you waveforms look like if you use L1 as your common reference point and then plot the L1-N voltage and the L1-L2 voltage? Are the voltages in-phase or out of phase?
In phase but that is not comparing the two 120V halves with respect to each other.

That is the whole point. To say they are out phase without including the reference point is technically mis-leading.


With measuring both at the same time you have no way of determining phase relationship.
So the phase relationship of the waveforms is dependent on the orientation of the probes.
 
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