
#1
Jan3113, 03:37 AM

P: 38

Good morning,
in circuit theory I know that reacting power arise from phasors and represents a power which can't be used, because not delivered to any load, but continuously flows back and forth between the load and the generator with a zero mean during one period. I can't understand very well, anyway, the meaning of this power in a field context, with electromagnetic sources in free space. Let's consider a hertzian dipole, which has several fields component with several dipendence from the distance [itex]r[/itex] (in spherical coordinates). [itex]H_{\varphi} = \displaystyle \frac{I_0 h}{4 \pi} e^{jkr} \left( \displaystyle \frac{jk}{r} + \frac{1}{r^2} \right) \sin \theta \\ E_r = \displaystyle \frac{I_0 h}{4 \pi} e^{jkr} \left( \displaystyle \frac{2 \eta}{r^2} + \frac{2}{j \omega \epsilon r^3} \right) \cos \theta \\ E_{\theta} = \displaystyle \frac{I_0 h}{4 \pi} e^{jkr} \left( \displaystyle \frac{j \omega \mu}{r} + \frac{\eta}{r^2} + \frac{1}{j \omega \epsilon r^3} \right) \sin \theta[/itex] Components proportional to [itex]1/r^2[/itex] and [itex]1/r^3[/itex] are called near field components; components proportional to 1/r are called farfield components. [itex]I_0[/itex] is the phasor of the current in the dipole and [itex]h[/itex] its length. I calculate the power as flux of the Poynting vector through a surface [itex]S[/itex]: [itex] P = \displaystyle \oint_S \mathbf{E} \times \mathbf{H}^* \cdot d\mathbf{S} [/itex] where the dipole is in the origin and [itex]S[/itex] is a sphere with radius [itex]r[/itex] centered in the origin too. I find for the power [itex]P[/itex] a real part, which is the power that goes away from the dipole, and an imaginary part, which is the reactive power and is only stored near the dipole. In the waveguide theory, reactive power is carried by modes that cannot propagate, that is, modes that have a purely imaginary propagation constant: they attenuate exponentially along the waveguide. But now the reactive power is carried by a field that propagate, because it has the [itex]e^{jkr}[/itex] term like the "far field" components: how can it is possible? Moreover, what about the meaning of this reactive power? Books say that it is due to the [itex]1/r^2[/itex] and [itex]1/r^3[/itex] components, which are related to *static* fields. So, should I state that reactive power is that carried from static fields? A static field stores an energy and I can *use* this energy if I place a charge in the field, because the field will move the charge, making a work: but this is a sort of active power, a suitable power, isn't it? Thank you for having read and sorry for my confusion. Bye! Emily 



#2
Jan3113, 11:29 AM

Sci Advisor
PF Gold
P: 11,353

I feel that the term 'reactive power' is not strictly appropriate because no energy is being transferred. I think that the term 'reactive energy' is a better one because it is an energy build up between the inductive and capacitive elements in the structure. It takes a finite time, after switchon for the waves / oscillating fields to build up  this can be either an antenna or a transmission line feeding into a load, where there are mismatches along the way. When the dipole (or any other antenna structure) is at resonance, the reactive energy is maximum and the power source 'sees' just a resistance (the radiation resistance).




#3
Feb113, 04:29 AM

P: 38

Emily 



#4
Feb113, 05:42 AM

Sci Advisor
PF Gold
P: 11,353

Reactive power with electromagnetic sources in free spaceFor a line, Z_{0} =√(L/C) The time I was referring to is the time for the waves to reach a steady state of energy flow in a transmission line  taking into account any reflections there might be at interfaces, due to mismatches. The 'standing waves' need to establish themselves and this will take a number of journeys along the section of line. The 'charge / discharge' times of reactive elements is included in the transmission equations, I think. I don't know how relevant this is, actually, to your post, now I think about it. 



#5
Feb113, 05:59 AM

P: 38

Now I would like to know more about the question: reactive power is that carried from static fields? I observed that a static field stores an energy and I can *use* this energy if I place a charge in the field, because the field will move the charge, making a work. Which is the difference between the energy provided by a static field and the energy provided by an electromagnetic field in dynamic conditions? Emily 



#6
Feb113, 07:00 AM

Sci Advisor
PF Gold
P: 11,353

I don't think the term "reactive" applies to static fields, does it? For a static situation, you will either have a reactance of zero or infinity. It's more 'potential energy' that is involved in that case, I think. Also, does a 'static field' actually carry energy (implies from place to place?)
But in your OP, you are referring to changing fields  EM waves associated with a Hertzian dipole. I think my post about transmission lines is, in fact, relevant in as far as the quadrature E and H fields in the near field region must take time to build up to the steady state. These fields store Energy and not Power (you don't store power, by definition because power is rate of energy transfer). Because they exist at a distance from the dipole, they take time to establish themselves. The inphase components of the field are the ones involved with radiating power. (I don't think this is being too pedantic, is it?) 



#7
Feb213, 10:59 AM

P: 38




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