Voltage Drops; ETAP vs formula, please help?

[vagojr]

Your inquiry is not clear. Need SLD. And again, etap uses load flow algorithm, which is an iterative process. Even if you use "exact" formula, it will never come close. Load flow is a lot more "exact" than the "exact" formula. I have been saying this and you are still insisting to prove it otherwise. Never compare it in the first place.


Sent from Mars

Thank you rian0201 for your interest.
Voltage drops behave different with constant power loads than with constant impedance loads. Such that the known formula seems to be more accurate to constant power loads, and surprissingly different with constant impendance loads which current increases when the voltage increases... it would not be only a ETAP matter.
Regards and thank you all for your kind and very useful attention in this forum.

 

[steve66]

Thank you rian0201 for your interest.
Voltage drops behave different with constant power loads than with constant impedance loads. Such that the known formula seems to be more accurate to constant power loads, and surprissingly different with constant impendance loads which current increases when the voltage increases... it would not be only a ETAP matter.
Regards and thank you all for your kind and very useful attention in this forum.

The voltage drop formula's should work in either case. With a constant power load, the impedance changes as necessary to keep the power constant. If you account for that (use the worst case), your voltage drop should be worst case.

Its one thing if you are trying to understand the equations and figure out why there are fractions of a percent differences. It's another thing if you are just trying to make sure the voltage drop won't be a problem. If that's the case, do you really need to worry about a fraction of a percent difference?

 

[Smart $]

...

• VD at MCC by Formula: 1.44%
• VD at MCC by ETAP (V=100% at switchgear) = 0,55%
• VD at MCC by ETAP (V=90% at switchgear) = 0,88%

...
Its one thing if you are trying to understand the equations and figure out why there are fractions of a percent differences. It's another thing if you are just trying to make sure the voltage drop won't be a problem. If that's the case, do you really need to worry about a fraction of a percent difference?

I agree... but the OP indicates a difference of about 0.9%, which is noteworthy if you are only permitted 5% overall.

 

[Phil Corso]

The following Table illustrates V-Drop using %-Voltage Rgulation Method:
E-702, XFMR, %-VR	Prj 1668	MHF, T-M	Date/Time: 16.11.26
kVA	Prim'y, kV	Second'y, V	%Zt	X/R	LF = P-unit Xfmr Cap'y
300	6.60	433	4.75	3.43
Loads	kVA	Xfmr LF	PF	Cosq	Sinq	%V-Drop	V-Drop
Running	150	0.50	0.85	0.85	0.53	1.79	7.73
Starting	900	3.00	0.30	0.20	0.95	13.9	60.0
Equivalent	1,020	3.40	0.39	0.39	0.92	15.5	67.2
Regards, Phil Corso

 

[rian0201]

Thank you rian0201 for your interest.
Voltage drops behave different with constant power loads than with constant impedance loads. Such that the known formula seems to be more accurate to constant power loads, and surprissingly different with constant impendance loads which current increases when the voltage increases... it would not be only a ETAP matter.
Regards and thank you all for your kind and very useful attention in this forum.

I agree, thats why in ETAP there is a option to input on what type of loads you have.

You can even specify percentage of types of loads. (50% power 50% z for example).

You can even set it to balanced or unbalanced load. Or even exponential loads.

The end point is that voltage drop is voltage drop.


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