Getting 240 volts from 208 volts

wrote:


--------- the two (exact) integers 2 and 2 are identical but this doesn't mean that the calculations involved to reach them are the same. The above argument doesn't apply. If you want to eliminate round off errors in calculations then store the numbers between calculations -surely you are not entering numbers with so many digits- My calculator can display the number of sig figs that I want but the arithmetic done is to whatever floating point precision that is involve. That doesn't mean that the results are more accurate. ---------

--------- Your confidence is misplaced in that it has no relationship to reality. when you say 120V, you are saying that it is a voltage between 119.5V and 120.5 V That is 3 significant figures. It is not the same as 120.0000 V. Similarly, even though your calculator multiplies this by root(3) and gives 207.8460969...., the result is accurate only to 3 significant figures or 208V +/_ 0.5V. The rest is a tribute only to the capability of your calculator. This is something that engineering schools try to drum into their students. Solutions to problems that show answers like 207.8460 .... lose marks if the original data doesn't justify this accuracy. In addition, how often does a nominal 120V actually appear at the outlet? You are working with nominal values In rare cases, such as ill conditioned matrices, there is a need for double precision arithmetic. ------

---------- So what is the confidence in the 114?

------- No. Suppose that you have your Zig-zag secondary with a resistive load so that both voltage and current are shifted 30 degrees (in the same direction) from the reference (say phase a primary -consider a star primary for simplicity). Now, if the secondary current is reflected into the primary as you indicate then there would be a problem - you have secondary windings from each pair of phases tied in series. That correctly implies a common current in these two windings. However there is another phase a winding in series with a c phase winding and the current in this is actually lagging by 30 degrees (accounting for winding polarity) Now, on the basis of secondary currents reflecting back to the primary, this implies that the primary current of phase a is both leading and lagging. It also implies that all 3 phase currents are in phase - double contradiction. An easier argument is that the transformer doesn't produce or absorb (ideally) any power or reactive (conservation of energy)so that for a given secondary P and Q out, there must be the same P and Q input. If the loads are balanced then that would also mean that the pf of the input is the same as the output. The answer to this quandary is that one must look at ampere turns, not amperes.
Check it out in the simpler case of a single phase transformer 1200/240 V with a center tapped secondary. If you have a load of 5A, unity pf at 120V on one side and a load of 10A at 0.8 pf lag on the other leg. what is the primary current? It cannot be 0.5A ,1.0 pf and 1.0A at 0.8 pf. Considering each secondary separately you would have a=Np/Ns  Np*Ip =Ns* (Is1 +Is2)= Ns* (14.3 @ -24.8 degrees) or 1.43 A at 0.91 pf lag Check from power Ps1 0*5 `0 watts Ps2 0*8 0 watts Qs1 =0 Qs20*6 r0 Vars total 1560 watts, 720VAR or 1720VA at 0.91 pf lag and a current of 1.43A
For the 3 phase scheme this is also true. The primary amp turns must balance the amp-turns of the two secondary windings in each phase. The currents in these windings are 60 degrees apart and the number of turns is the same in each. Then considering that one of the phase a secondaries has a current leading by 30 degrees, the other phase a secondary has a current effectively lagging by 30 degrees so the sum of the ampere turns seen by the primary is NI1 +NI2 =N(root(3)I at angle 0 so for a balanced secondary load at unity pf, the primary current which must balance the total secondary ampere turns- not the amperes in the secondary windings) is at unity pf.
--

Don Kelly snipped-for-privacy@shawcross.ca
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| wrote: |> |> | Get rid of all the extra decimal places that your calculator is giving |> you- |> | they are actually meaningless as 120V (implied +/- 0.5V is not the same |> as |> | 120.0000000V (implied +/-0.0000005 V) |> |> I use the extra digits for a purpose. First, it ensures that accumulated |> errors do not happen. Secondly, it helps determine if numbers are |> "matches" |> that suggest simplified formulas. I've seen cases where NOT having them |> can lead to suggested matches that really aren't just because certain |> paths |> of arithmetic come close but don't hit the mark. For example I ran into |> these numbers: |> 138.390751136 |> 138.564064606 |> They look close. But they have different meaning. They just happen to be |> close. But there is no mathematical equivalency to how they were |> calculated. |> But if some calculation had come up with: |> 138.564064605 |> then the chance that it would be equivalent to the 2nd of the first 2 is |> very high. The use of irrational numbers and a variety of different |> formulas |> can result in "number crossings" that may or may not have significance. | --------- | the two (exact) integers 2 and 2 are identical but this doesn't mean that | the calculations involved to reach them are the same. The above argument | doesn't apply. If you want to eliminate round off errors in calculations | then store the numbers between calculations -surely you are not entering | numbers with so many digits- My calculator can display the number of sig | figs that I want but the arithmetic done is to whatever floating point | precision that is involve. That doesn't mean that the results are more | accurate.
The calculations _may_ be equivalent. Or they may not be. But I know from experience that when irrational numbers are involved (like most square roots) the chance of a false positive prefect match is very low.
|> When I see a reference to "208v" I figure it is probably the voltage |> involved |> line to line when the line to neutral voltage is "120v". The confidence |> is |> high, but it is not a certainty. But if I see "207.846v" then the |> confidence |> is way up. Even more so with "207.84609690826527522329356v". | --------- | Your confidence is misplaced in that it has no relationship to reality. | when you say 120V, you are saying that it is a voltage between 119.5V and | 120.5 V That is 3 significant figures. It is not the same as 120.0000 V. | Similarly, even though your calculator multiplies this by root(3) and gives | 207.8460969...., the result is accurate only to 3 significant figures or | 208V +/_ 0.5V. The rest is a tribute only to the capability of your | calculator. This is something that engineering schools try to drum into | their students. Solutions to problems that show answers like 207.8460 .... | lose marks if the original data doesn't justify this accuracy. | In addition, how often does a nominal 120V actually appear at the outlet? | You are working with nominal values | In rare cases, such as ill conditioned matrices, there is a need for double | precision arithmetic.
When you say that "120V" means a voltage between "119.5V" and "120.5V" (which I don't necessarily agree with), what are you saying when you say "119.5V"? Are you saying "a voltage between 119.45V and 119.55V" ??
Range has to be defined. It is so often _defaulted_ to be defined as have a precision step above or below. But that is not always reality. A range might be 3/4 of a precision step above or below.
"120V" might be a range from "114V and 126V".
| ------ |> |> Yes I understand that voltage can vary. And I understand my arithmetic is |> more accurate than the voltage. |> |> |> | Noting that your 120V is likely to be in the range 114-126V normally- |> then |> | what is the point? |> |> The point is correct numeric identification. If the line to neutral |> voltage |> can vary from 114 volts to 126 volts, then the line to line voltage can |> vary |> from 197.45379206285201146 volts to 218.23840175367853898 volts :-) |> |> If I see "197v" somewhere, I don't know if it comes from just dropping |> "200v" |> down by 3 volts or what. But if I see "197.45379v" then I have a very |> high |> confidence that it was derived from 114 times ths square root of three. | ---------- | So what is the confidence in the 114?
That depends on what I'm relating the confidence to. There would be a high level of confidence if I am relating 114 to 197.45379 divided by the square root of three. But it certainly is possible for the 114 to come about for other reasons.
| No. Suppose that you have your Zig-zag secondary with a resistive load so | that both voltage and current are shifted 30 degrees (in the same direction) | from the reference (say phase a primary -consider a star primary for | simplicity). Now, if the secondary current is reflected into the primary as | you indicate then there would be a problem - you have secondary windings | from each pair of phases tied in series. That correctly implies a common | current in these two windings. However there is another phase a winding in | series with a c phase winding and the current in this is actually lagging | by 30 degrees (accounting for winding polarity) Now, on the basis of | secondary currents reflecting back to the primary, this implies that the | primary current of phase a is both leading and lagging. It also implies | that all 3 phase currents are in phase - double contradiction. | An easier argument is that the transformer doesn't produce or absorb | (ideally) any power or reactive (conservation of energy)so that for a given | secondary P and Q out, there must be the same P and Q input. If the loads | are balanced then that would also mean that the pf of the input is the same | as the output. | The answer to this quandary is that one must look at ampere turns, not | amperes. | | Check it out in the simpler case of a single phase transformer 1200/240 V | with a center tapped secondary. If you have a load of 5A, unity pf at 120V | on one side and a load of 10A at 0.8 pf lag on the other leg. what is the | primary current? It cannot be 0.5A ,1.0 pf and 1.0A at 0.8 pf. | Considering each secondary separately you would have a=Np/Ns  | Np*Ip =Ns* (Is1 +Is2)= Ns* (14.3 @ -24.8 degrees) or 1.43 A at 0.91 pf lag | Check from power | Ps1 0*5 `0 watts Ps2 0*8 0 watts Qs1 =0 Qs20*6 r0 Vars | total 1560 watts, 720VAR or 1720VA at 0.91 pf lag and a current of 1.43A | | For the 3 phase scheme this is also true. The primary amp turns must | balance the amp-turns of the two secondary windings in each phase. The | currents in these windings are 60 degrees apart and the number of turns is | the same in each. | Then considering that one of the phase a secondaries has a current leading | by 30 degrees, the other phase a secondary has a current effectively lagging | by 30 degrees so the sum of the ampere turns seen by the primary is NI1 | +NI2 =N(root(3)I at angle 0 so for a balanced secondary load at unity pf, | the primary current which must balance the total secondary ampere turns- not | the amperes in the secondary windings) is at unity pf.
I see now we weren't taking about the same thing. I was only looking at each single phase load. You're showing the current lead/lag cancel out when the same thing is down in a balanced three phase system.
Consider the typical "2 legs of 208Y/120" service where an autotransformer derives the missing leg:
A Y A Y \ -> \ \ / \ N Z Y ==> N Z / ------> / / B N B
This is certainly not very well balanced.
Now you get a voltage between X and Z at a phase angle of the phase that is missing. Put a load between X and Z. What phase angle do you get on the incoming A-N and B-N? What pf? 0.5? Leading or lagging?
I could also do:
A Y A Y \ -> \ / \ / \ N Z Y A ==> W N Z / ------> / --> / B N W
Hey, now I have the 240 volts I wanted, and with a neutral right in between. Of course I could have gotten it from either 120 volt leg with a 1:2 step up. But at least I am putting half the current on A and half on B. It just has a lousy power factor.
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|---------------------------------------/----------------------------------|
| Phil Howard KA9WGN (ka9wgn.ham.org) / Do not send to the address below |
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