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CIM ApparentPower vs QUDT ComplexPower #43

Closed VladimirAlexiev closed 2 months ago

VladimirAlexiev commented 2 months ago
cim:ApparentPower  a qudt:QuantityKind ;
  qudt:applicableUnit cim:UnitSymbol.VA;
  skos:broader quantitykind:ComplexPower. # ApparentPower is a sub-concept of ComplexPower
cim:UnitSymbol.VA skos:exactMatch unit:V-A

But is it true?

Questions:

griddigit-ci commented 2 months ago

Looks good, but we need to dive in definitions. The complex power can be expressed as P+jQ where P is the active and Q reactive power. the S - apparent power we exchange is the module of this

the complex power can be expressed as sqrt(P^2+Q^2)<ctan(Q/P), where the < is the angle part. I think we do not want the angle but just the module

We need to see an example of ComplexPower in QUDT maybe that will get clearer

VladimirAlexiev commented 2 months ago

QUDT refers to https://www.electropedia.org/iev/iev.nsf/display?openform&ievref=131-11-39, and has these details (in addition to those above):

  qudt:latexDefinition "$\\underline{S} = \\underline{U}\\underline{I^*}$, where $\\underline{U}$ is voltage phasor and $\\underline{I^*}$ is the complex conjugate of the current phasor."^^qudt:LatexString ;
  qudt:latexSymbol "$\\underline{S}$"^^qudt:LatexString ;

So it also calls it S. Even though it talks of complex numbers, the unit V-A applies to magnitudes, not to vectors. So it has to be the same thing.

VladimirAlexiev commented 2 months ago

Answered in https://github.com/qudt/qudt-public-repo/issues/970: QUDT has ApparentPower (I overlooked it). Fixed that section.

steveraysteveray commented 2 months ago

By the way, @VladimirAlexiev, this topic happens to be what I used in explaining how we compute qudt:applicableUnit triples. See https://github.com/qudt/qudt-public-repo/wiki/Advanced-User-Guide#4-computing-applicable-units-for-a-quantitykind