Closed VladimirAlexiev closed 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
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.
Answered in https://github.com/qudt/qudt-public-repo/issues/970: QUDT has ApparentPower
(I overlooked it). Fixed that section.
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
But is it true?
unit:V-A
applies toquantitykind:ComplexPower
ComplexPower
: units apply to magnitudes, not to vectorsComplexPower
as "under sinusoidal conditions, is the product of the phasor U representing the voltage between the terminals of a linear two-terminal element or two-terminal circuit and the complex conjugate of the phasor I representing the electric current in the element or circuit"ApparentPower
as "Product of the RMS value of the voltage and the RMS value of the current."Questions:
skos:broader
or maybe evenskos:exactMatch
ApparentPower
as a new QuantityKind to QUDT?