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

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.

--------- 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 1.43AFor 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.

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Don Kelly snipped-for-privacy@shawcross.ca

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Don Kelly snipped-for-privacy@shawcross.ca

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