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Table of Contents
ELEN90074 - Introduction to Power Engineering
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Introduction
Power systems need to generate electrical energy economically and with minimal ecological disturbance. The aim it to transfer this energy with maximum efficiency and reliability and deliver electrical energy to consumers at virtually fixed voltage and frequency.
Frequency
50 and 60 Hz are used in power transmission. Lower frequencies can not be used due to flicker in filament lamps. Iron losses increase in proportion to frequency, as do leakage reactances. Capacitive reactance between lines reduce with frequency. Interference with telephone lines increase with frequency. Higher frequencies enable smaller motors, generators and transformers. The current frequencies were decided a while ago and are kept for compatibility.
Market power systems
Traditionally “set and forget” systems, bulk energy made and distributed to consumers using predictive modelling. Limited reactive ability. Nowadays there is generators, transmitters, distributers and suppliers.
As green technologies become more available, this model is being shook up as consumers generate their own energy. This is transitioning the power system to a more distributed one, where there is not just a single directional flow as in the traditional model. This requires much more information flow, being dubbed a “smart grid”.
Phasors
Notation
$x(t)$ in lowercase is instanteous values. $X_{max}$ in uppercase and with a subscript represents the maximum value of $x(t)$. $|X|$ in uppercase and in vertical bats is the magnitude of $x(t)$ and is the root mean square (RMS) values and is equal to $X_{max}/\sqrt{2}$. $X$ in uppercase and alone is the phasor representation of $x(t)$.
Introduction
In AC circuits, voltages and currents are ideally sinusoidal (frequency assumed constant). These signals are characterised by two parameters: a maximum angle and a phase angle. The goal of phasor analysis is the simplify the analysis of constant frequency AC systems. Assuming voltages and currents in sinusoidal state are constant, we can use a phasor, taken from Euler's identity. $$e^{j\theta}=\cos\theta+j\sin\theta$$ $$v(t)=V_{max}\cos(\omega t+\theta_v)=\sqrt{2}|V|\cos(\omega t+\theta_v)$$ $$V=|V|e^{j\theta_v}=|V|\angle\theta=|V|\cos\theta_v+j\sin\theta_v$$
Impedance can also have a phasor representation, as shown in the AC variant of Ohm's law. $$V=ZI$$ $$|V|\angle\theta_v=|Z|\angle\theta_z\cdot |I|\angle\theta_i$$ With purely resistive loads, the current is in phase with the voltage. A capacitor has impedance with phase $+\frac{\pi}{2}$ (current leads voltage) and an inductor has impedance with phase $-\frac{\pi}{2}$ (current lags voltage). When combining impedances, we need to treat the real and complex parts separately.
In AC circuits, power consumption is $p(t)=v(t)i(t)$, the product of two sinusoidal functions. The component associated with the resistive load is the active or real power, P. The component associated with the reactive (capacitive or inductive) load is the reactive power, Q. The active power can be found with only the resistive load, $P=|V||I|=I|Z|\cos\theta_z=I^2R$. The apparent power is defined as the product of the RMS voltage and current, $|S|=|V||I|$. The complex power can be quantified as $S=P+jQ=V*I$. The power factor is the ratio of active to apparent power, $pf=\frac{P}{|S|}$. For impedance elements, it is simply the cosine of the impedance angle, $pf=\cos\theta_z$. The pf is lagging if $\theta_z>0$ (I lags V, inductive) and leading if $\theta_z<0$ (I leads V, capacitive). Introducing inductors and capacitors can help correct the power factor, and reduce the overall current, reducing strain on the system.
Three Phase AC Circuits
A balanced 3 phase system has three voltage sources of equal magnitude but shifted $120^\circ$ apart. There is an equal load exerted on each phase and equal impedance on the lines connecting the generators to the loads. Bulk power systems are almost exclusively three phase. Single phase is used primarily in low voltage, low power settings such as residential and some commercial.
$3\phi$ systems have some advantages:
- Can transmit more power for the same amount of wire
- More constant torque on the machines
- Less material used for same power rating
- Machines start more easily
If we have three phases connected to identical loads, we can connect their return wires and find that there is no current in the return. This is a star or Wye (Y) configuration and there is a shared neutral at the meeting points of the sources and loads. A three phase system has three 'active' conductors and a fourth neutral carrying a nominal zero volts. The active conductors ideally have the same voltage and current magnitudes but maintain a $120^\circ$ phase separation. For a positive phase sequence (a-b-c), $V_{an}$ leads $V_{bn}$ leads $V_{cn}$. A negative phase sequence is a-c-b. A balanced system can be specified by a single phasor voltage and current: $$V_{an}=|V_{an}|\angle\theta_{an}$$ $$V_{bn}=|V_{an}|\angle\theta_{an}-120^\circ=a^2V_{an}$$ $$V_{an}=|V_{an}|\angle\theta_{an}+120^\circ=aV_{an}$$ $$V_{an}+V_{bn}+V_{cn}=(1+a^2+a)V_{an}=0$$ When the voltages are balanced, it is much easier to refer to line-to-line voltages. $$V_{ab}=V_{an}-V_{bn}=|V_{an}|(1\angle\theta_{van}-1\angle(\theta_{van}-120^\circ))=\sqrt{3}|V_{an}|\angle(\theta_{van}+30^\circ)=\sqrt{3}V_{an}*1\angle 30^\circ$$ All the line to line voltages are $120^\circ$ out of phase with each other. For the currents, it is the line-neutral voltages divided by the line impedance. The sum of the line currents are 0, and they are $120^\circ$ out of phase. In calculations involving a wye circuit, line-neutral voltages are used.
Delta ($\Delta$) connected loads have each load feeding into each other in a triangle, with the three phases connecting to the corners of the triangle. As a result, the line to line current is what runs through each load. $I_a=I_{ab}-I_{ca}=\sqrt{3}I_{ab}*1\angle 30^\circ$ A Delta connected load draws a higher current for the same voltage and loads, so is used in higher power situations. In calculations involving a delta circuit, line-to-line voltages are used.
The fundamental relationship defining phasors is: $$|V_L|=\sqrt{3}|V_{LN}|;\theta_{V_{L}}=\theta_{V_{LN}}+30^\circ$$ $$|I_\Delta|=|I_{LN}|/\sqrt{3};\theta_{I_\Delta}=\theta_{I_{LN}}+30^\circ$$ In a wye connection, the line current (line-to-neutral) is what is going through the impedance. In a delta connection, the line current (line-to-line) is what is going through the impedance. Phase voltage and current refers to the load, line voltage and current refers to the lines.
Electromagnetism
Electrical circuit theory is a set of abstractions useful for engineering circuits (KCL, KVL, Tellegen). Electromagnetic theory is useful for making physical predictions about electromagnetic interactions (Maxwell's equations). Trying to model from individual charges makes the analysis of networks difficult, although they are linked.
Vector calculus reminder
Electromagnetic fields are studied in vector systems, and utilise vector calculus and coordinate systems (rectangular, cylindrical, spherical). The gradient of a scalar ($\nabla f$) is the direction of maximum increase of the scalar field. The divergence of a vector field ($\nabla F$) represents source or sink of vector field flux, as it is non-zero at a source or sink. The curl of a vector fields ($\nabla\times F$) is non-zero only if the vector field circulates at the point and its direction maximises the elemental path integral. If $\vec{F}=\nabla f$, then $\oint_{path}\vec{F}\cdot d\vec{l}=0$, so the field is conservative and the path integral through the field is equal to the value of the scalar potential at the end, less the scalar potential at the start. Gauss' theorem states that: $$ \iint_{surf}\vec{F}\cdot dS\vec{a_n}=\iiint_{vol}\nabla\cdot\vec{F}\cdot d\text{vol}$$ Stokes' theorem: $$\oint_{path}\vec{F}\cdot d\vec{l}=\iint_{surf}(\nabla\times\vec{F})\cdot dS\vec{a_n}$$
Electrostatic forces
We can express the force on stationary charges excepted on each other: $$\vec{F}_{Qq}=\frac{1}{4\pi\epsilon_0}\frac{qQ}{R^2}\vec{a}_r=-\vec{F}_qQ$$ The force on one charge is equal and opposite to the force experienced from the other, and the direction of the force is in the opposite direction to the charge. $\epsilon_0$ is a constant of proportionality, the permittivity of free space. The interaction can be modelled by the Coloumb electric field: $$\vec{E}_Q=\frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}\vec{a}_r$$ The force experienced by a charge in the field is equal to $\vec{F}_{Qq}=q\vec{E}_Q$. The Coloumb electric field is the gradient of scalar potential, and thus is conservative. The principle of superposition holds when a charge is in the presence of multiple others.
Charges in motion
When charges are in motion, such as due to a current, magnetic effects begin to manifest. The Lorentz force is the relativistic transformation of a Coloumb force. To a stationary observer, the force exerted on a charge $Q$ at the origin with velocity $\vec{u}$, from a charge $q$ moving at velocity $q$ at radius $R$: $$\vec{F}_{Qq}=q(\vec{E}_Q+\vec{v}\times\vec{B}_Q)$$ $$\vec{E}_Q=\frac{1}{4\pi\epsilon_0}\frac{Q\vec{a}_n}{R^2\sqrt{1-u^2/c^2}}$$ $$\vec{B}_Q=\frac{\mu_0}{4\pi}\frac{Q(\vec{u}\times\vec{a}_\rho)}{R^2\sqrt{1-u^2/c^2}}$$ Here it is important to note that $\vec{F}_{Qq}\neq-\vec{F}_{qQ}$. The $B$ term represents the magnetic force, and is typically much smaller than the electric force, although it scales up due to the number of moving conductors.
The magnetic field from a current in a wire, at distance $d$ from the wire, is: $$\vec{B}=\frac{\mu_0I}{2\pi d}\mathbf{a}_z$$ The force on a charge travelling at a distance of $R$, travelling at $\vec{v}_q$ parallel to a straight fixed conductor with current $I=-\lambda_nv_n$ experiences force: $$\vec{F}=qv_q\frac{\mu_0I}{2\pi R}\vec{a}_\rho=q(v_q\vec{a}_z)\times\frac{\mu_0I}{2\pi R}(-\vec{a}_\phi)=q(\vec{v}_q\times\vec{B})$$ The magnetic field circulates around the current without an electric field as the conductor is electrically neutral.
The Biot-Savart Law relates the magnetic field and current: $$d\vec{B}=\frac{\mu_0}{4\pi}\frac{(Id\vec{l})\times\vec{a}_r}{r^2}=\frac{\mu_0}{4\pi}\frac{I(\cos\alpha) d\alpha}{R}(-\vec{a}_\phi)$$ Current is the bulk flow of electrons, moving at drift velocity: $$\vec{v}_n=\frac{Id\vec{l}}{|d\vec{l}|ANe}$$ We can then write Ampere's law as (A is the conductor cross sectional area, N is conduction electron density, e is electronic charge): $$d\vec{F}=Id\vec{l}\times\vec{B}=(|d\vec{l}|ANe)\vec{v}_n\times\vec{B}$$ Giving the total force as: $$\vec{F}=I\oint_{path}d\vec{l}\times\vec{B}$$ If the magnetic field is uniform, there is no net force but there could be a torque.
Induction
Faraday's law of induction relates the a moving conductor in a constant magnetic field and the induced current. $$\oint_{path}(\vec{v}\times\vec{B})\cdot d\vec{l}=-\frac{d\phi}{dt}=-\iint_{surf}\frac{\partial\vec{B}}{\partial t}\cdot dS\vec{a}_n$$ This induced electric field varies with time, and as such is different to a Coulomb field.
Maxwell's equations
Maxwell's equations describe electromagnetic phenomena, with electric fields having sources and magnetic fields having none. They cover the electric dipole moment density, arising from the polarisation of atoms with non-congruent positioning of consistent charges which are charge balance neutral. They also cover the magnetic dipole moment density, arising form the circulating charges in the atoms of the material being charge balance neutral. The source fields are specified across all space as functions of position and time; they are zero in free/empty space. $$\nabla\times\vec{E}=-\frac{\partial\vec{B}}{dt}$$ $$\nabla\cdot\vec{D}=\rho$$ $$\vec{D}=\epsilon_0\vec{E}+\vec{P}$$ $$\nabla\times\vec{H}=\vec{J}+\frac{\partial\vec{D}}{\partial t}$$ $$\nabla\cdot\vec{B}=0$$ $$\vec{B}=\mu_0(\vec{H}+\vec{M})$$ Appropriate boundary conditions are needed for these coupled differential equations, in addition to charge continuity. $$\nabla\cdot\vec{J}=\frac{d\rho}{dt}$$ Inside homogeneous isotropic linear stationary media: $$\vec{P}=\chi\epsilon_0\vec{E}\implies\vec{D}=\epsilon_0(1+\chi)\vec{E}=\epsilon_0\epsilon_r\vec{E}$$ $$\vec{M}=\chi_m\vec{H}\implies\vec{B}=\mu_0(1+\chi_m)\vec{H}=\mu_0\mu_r\vec{H}$$ This gives us the equations in integral form for stationary media: $$\oint_{path}\vec{E}\cdot d\vec{l}=\iint_{surf}(\nabla\times\vec{E})\cdot dS\vec{a}_n=-\frac{d}{dt}\iint_{surf}\vec{B}\cdot dS\vec{a}_n$$ $$\iint_{surf}\vec{D}\cdot dS\vec{a}_n=\iiint_{vol}\rho dvol$$ $$\oint_{path}\vec{H}\cdot d\vec{l}=\iint_{surf}(\nabla\times\vec{H})\cdot dS\vec{a}_n=\iint_{surf}(\vec{J}+\frac{\partial\vec{D}}{\partial t})\cdot dS\vec{a}_n$$ $$\iint_{surf}\vec{B}\cdot dS\vec{a}_n=0$$
Electric fields and particle collisions
Conductors are materials that allow the flow of current due to their unfilled outer shells, allowing for the sharing and movement of electrons. Insulators (or dielectrics) are materials with tightly held electrons and filled outer shells with a large band gap to the next shell, stopping the movement of electrons. Semi-conductors have filled outer shells but have a small band gap, permitting electron movement in some conditions. An electrical current is the net flow of conductors through a surface per unit time. The current density at a point is the net transport of free charge at that point and magnitude: $$|\vec{J}|=\lim_{\Delta S\to 0}\frac{\Delta I}{\Delta S}$$ The current density is such that it can have non-zero components due to negative and positive charges, but is an average of the speed of the charges. There is a conservation of charge, such that the net charge must remain constant.
Drift velocity
Ohm's law relates the electric field and the current density with the conductivity of the material; $$\vec{J}=\sigma\vec{E}$$ Ohm's law represents the friction-like forces acting on moving charges as they collide with their carrier material. Thus an electric field is required to maintain unaccelerated motion and thus a steady current density. The average velocity increment due to the electric field is: $$\Delta\vec{v}=e\vec{E}\Delta t/m_e$$ This results in an aggregate drift velocity $\Delta\vec{v}$. Collisions result in the transfer of the velocity increment energy to the conductor, causing it to heat up (Joule loss).
Energy and charge density
The average energy per unit time per unit volume (power) is: $$P=\mathcal{N}\frac{\frac{1}{2}m_e|\Delta\vec{v}|^2}{\Delta t}=\frac{\mathcal{N}e^2|\vec{E}|^2\Delta t}{2m_e}=\sigma*\vec{E}\cdot\vec{E}=\vec{J}\cdot\vec{E}=\frac{|\vec{J}|^2}{\sigma}$$ In a material, overall electrically neutral, but with charge separation and an internal electric field, we can characterise the charge density at point p as: $$\rho(t;p)=\nabla\cdot\vec{D}=\epsilon\nabla\cdot\vec{E}=\frac{\epsilon}{\sigma}\nabla\cdot\vec{J}=-\frac{\epsilon}{\sigma}\frac{d}{dt}\rho(t;p)$$ At a fixed point in space, the solution is: $$\rho(t;p)=e^{-t/\tau}\rho(t_0;p)$$ This decays to zero as time increases. This means that more than a Coulomb electric field to sustain a current, as the charges will eventually redistribute to produce 0 charge density.
EMF
A sustained current can only exist on a closed path (circuit) with a non-coulomb electric field. The work done (EMF) by an electric field on a unit positive charge as it traverses the path is: $$\mathcal{V}=\oint_{path}\vec{E}\cdot d\vec{l}$$ The Coulomb field cannot contribute to EMF as it is conservative, so the path integral is zero. Magnetic fields do contribute as they are circulating and non-conservative: $$\mathcal{V}=-\frac{d}{dt}\iint_{surf}\vec{B}\cdot d\vec{S}$$ In a battery, the chemical reaction sustains the electric field to provide the EMF. The total electric field will consist of the sum of conservative and non-conservative components. In a conductor carrying no current, the net electric field must be zero; the Coulomb electric field balances the conservative parts. In a circuit, the potential difference between two points is the path integral of the Coulomb field by definition: $$\varphi(B)-\varphi{A}=V_{BA}=-\oint_{path}\vec{E}_C\cdot d\vec{l}$$ This is consistent with Kirchhoff's voltage laws. For a closed loop with a battery providing the EMF: $$V_{+-}=IR+L\frac{dI}{dt}$$ Where $RI$ accounts for the resistance in the circuit, and $L\frac{dI}{dt}$ accounts for the rate of change of flux cutting the circuit (inductance).
Magnetic flux and inductance
If we were to consider two loops, where the time-varying magnetic flux in a stationary loop flows into the other loop, we produce a current in the second loop. The flux in loop a and linking loop b is: $$\Phi_{ab}=\oint\vec{A}_a\cdot d\vec{l}_b=\oint_b\left(\frac{\mu_0I_a}{4\pi}\oint_a\frac{d\vec{l}_a}{r}\right)\cdot d\vec{l}_b=M_{ab}I_a$$ Where $M_{ab}=\frac{\mu_0}{4\pi}\oint_a\oint_b\frac{d\vec{l}_a\cdot d\vec{l}_b}{r}$ and $r$ is the distance between $d\vec{l}_a$ and $d\vec{l}_b$. It follows that the EMF induced in circuit b by a change in $I_a$ is: $$\mathcal{V}_{ab}=\oint_b\vec{E}_{I_a}\cdot d\vec{l}_b=-\frac{d\Phi_{ab}}{dt}=-M_ab\frac{dI_a}{dt}-I_a\frac{dM_{ab}}{dt}=-M_{ab}\frac{dI_{ab}}{dt}$$ A device comprising two circuits designed to have mutual inductance is a transformer.
Where a current in a wire links with its own flux, we call this inductance. $$\Phi_{aa}=\frac{\mu_0}{4\pi}\oint_{a1}\oint_{a2}\frac{d\vec{l}_{a1}\cdot d\vec{l}_{a2}}{r}$$ $$L_a=\frac{d\Phi_{aa}}{dI_a}=\Phi_{aa}/I_a$$ Where there is a nearby loop which intersects a fraction of the flux: $$\Phi_{ab}=k_{ab}\Phi_{aa}$$ $$M_{ab}=\Phi_{ab}/I_a=k_{ab}\Phi_{ab}/I_a=k_{ab}L_a=M_{ba}=k_{ba}L_b=M$$ $$M^2=k_{ab}k_{ba}L_aL_b$$ Thus the mutual inductance is $M=k\sqrt{L_aL_b}$, there the coupling coefficient has magnitude $|k|=|k_{ab}k_{ba}|^\frac{1}{2}$. Where there is a coil of wire instead of a single loop, the length of the wire can considered to be N times the length of a single loop.
The EMF is $\mathcal{V}=-\frac{d\Lambda}{dt}$, and the flux linkage is $\Lambda=N\iint_{\times-sec}\vec{B}\cdot dS\vec{a}_n=N\oint\vec{A}\cdot d\vec{l}$. The self-inductance of a wound loop is: $$L_a=\frac{\Lambda}{I_a}=\frac{N_a\oint\vec{A}\cdot d\vec{l}}{I_a}=\frac{N_a}{I_a}\oint\oint\frac{\mu_0N_aId\vec{l}_{a1}}{4\pi r}\cdot d\vec{l}_{a2}=N_a^2\frac{\Phi_aa}{I_a}$$ The mutual inductance is given by $M_{ab}=\Lambda_{ab}/I_a=k_{ab}\Lambda_{ab}/I_a$, where the coefficient $k_{ab}$ can now be bigger than one, however $|k_{ab}k_{ba}|\leq 1$. In a solenoid of n turns per meter and axis along $\vec{a}_z$, if the solenoid bears current $I$, then the flux density inside is $\vec{B}=\mu_0nI\vec{a}_z$, with quick decay before it ends. Where one solenoid is inside another, the flux of the inner solenoid linked to the outer is $\Phi_{io}=\pi b^2\mu_0\frac{N_i}{l_i}I_i$ and thus: $$M_{io}=\frac{\Lambda_{io}}{I_i}=\frac{N_i\Phi_{io}}{I_i}=\pi b^2\mu_0\frac{N_iN_o}{l_i}$$ This is the same as the mutual inductance from the outer to the inner.
For a single loop connected to a battery, we can express the potential difference and power dissipated as; $$V_{+-}=IR+L\frac{dI}{dt}=IR+\frac{d\Lambda}{dt}$$ $$P=IV_{+-}=I^2R+I\frac{d\Lambda}{dt}=I^2R+IL\frac{dI}{dt}=I^2R+\frac{d}{dt}\left(\frac{1}{2}LI^2\right)$$ The $I^2R$ term is being dissipated as heat and $\frac{d}{dt}\left(\frac{1}{2}LI^2\right)$ is the storage of magnetic energy $\mathcal{W}_m=\frac{1}{2}LI^2=\frac{1}{2}I\Lambda$. If there are two coupled circuits, the energy stored is: $$\frac{1}{2}(I_a\Lambda_a+I_b\Lambda_b)=\frac{1}{2}L_aI_a^2+\frac{1}{2}L_bI_b^2+MI_aI_b$$ We can further express the stored magnetic energy as: $$\mathcal{W}_m=\frac{1}{2}=\frac{1}{2}\oint_{\ell}|\vec{H}||d\vec{l}|\iint_{\mathcal{A}}|\vec{B}|dS=\frac{1}{2}\iiint_{\text{all space}}\vec{H}\cdot\vec{B}d\text{vol}$$ This gives the magnetic energy density: $w_m=\frac{1}{2}\vec{H}\cdot\vec{B}$.
Magnetic fields and flux density
All atoms containing orbiting and spinning electrons can give rise to magnetic dipoles as a quantum-mechanical effect. In magnetic materials, the individual atomic magnetic dipoles can be oriented to manifest as a net macroscopic magnetic moment, said to be magnetised. This is weak except for in ferromagnetic materials, where the manifestation exhibits non-linear characteristics. In ferromagnetic materials, several spontaneous aligned domains form and can be aligned by an external magnetic field. For the material with per unit volume magnetic dipole density $\vec{M}$, in the presence of a magnetic flux density $\vec{B}$, the magnetic field strength within the material is: $$\vec{H}=\frac{1}{\mu_0}\vec{B}-\vec{M}$$ For homogeneous isotropic linear materials (non-ferromagnetic): $$\vec{M}=\chi_m\vec{H}\implies\vec{B}=\mu_0(1+\chi_m)\vec{H}=\mu_0\mu_r\vec{H}$$
In ferromagnetic materials, the fields $\vec{H}$ and $\vec{B}$ are not always aligned and can differ depending on the value of $\vec{H}$ and the past of the material. Magnetisation curves for $\vec{B}$ and $\vec{H}$ typically exhibit a non-linear S shape, with saturation of $\vec{M}$ at high field strengths and hysteresis. Soft magnetic materials have high max permeability, being the slope of the B-H curve, and narrow hysteresis. When displacement currents are negligible, $\nabla\times\vec{H}=\vec{J}$, where $\vec{J}$ is the free-charge current density.
By Ampere's circuital law $\oint_{path}\vec{H}\cdot d\vec{l}=\iint_{surf}\vec{J}\cdot dS_{\vec{a}_n}$, which is the free-charge current linking the path. If we consider a coil of wire with N turns: $$NI=\oint\vec{H}\cdot d\vec{l}=\int_0^{2\pi}\vec{H}\cdot(rd\phi\vec{a}_n)$$ Application of ACL along a path of radius r within the core yields field strength $\vec{H}=\frac{NI}{2\pi r}\vec{A}_\phi$ and thus flux density $\vec{B}=\mu_r\mu_0\frac{NI}{2\pi r}\vec{a}_\phi$. The presence of the ferromagnetic core scales the flux density by $\mu_r$, and can be as large as $10^6$.
Magnetic circuits
Magnetic materials can be used to direct and contain magnetic fields, although there is some leakage flux perpendicular to the surface of the material. This leakage can be worse than the gain from using a magnetic material of the core isn't designed properly. If we were to consider a ring of radius b with N turns of wire around it, we can find the core flux as: $$\Phi=\frac{\mu NI}{2\pi r_{av}\pi b^2}=\frac{NI}{2\pi r_{av}/(\mu\pi b^2)}$$
We can define the magnetomotance (MMF) as $\mathcal{F}=NI$, so the core flux is $\Phi=\mathcal{F}/\mathcal{R}$, where $\mathcal{R}=\frac{2\pi r_{av}}{\mu\pi b^2}$ is the reluctance of the magnetic circuit ($\mathcal{R}=\ell/\mu\mathcal{A}$, Ohm's law). The quantity $\mathcal{P}=1.\mathcal{R}$ is the permeance of the circuit. Assuming no flux leakage, fringing air gap and uniformity is the flux density and depending of permeability on the field strength, a linear magnetic circuit theory cna be established. Considering a magnetic circuit with a block with an air gap and a coil of wire; $$\mathcal{F}=NI=\oint\vec{H}\cdot d\vec{l}=H_c\ell_c+H_g\ell_g=\Phi\ell_c/\mu_c\mathcal{A}+\Phi\ell_g/\mu_g\mathcal{A}=\Phi(\mathcal{R}_c+\mathcal{R}_g)$$ When the core permeability is very high, the circuit reluctance is dominated by the air gap reluctance $\mathcal{R}_g$. A conservation of flux density property holds at branch point in a magnetic circuit, which is a Kirchhoff flow law to compliment the previous Kirchhoff drop law. We can characterise the work done by a magnetic field over a period of time T as: $$W=\int_0^TIVdt=\int_0^T\frac{H\ell}{N}Vdt=\ell\mathcal{A}\int_0^TH\frac{dB}{dt}dt=\ell\mathcal{A}\int_{B(0)}^{B(T)}HdB$$ This work is losses to the hysteresis effects in the material and moving on different sides of the S curve.
Ferromagnetic core materials also tend to be good conductors, so a current is induced and cause currents and conduction loss. These currents oppose the change in the external magnetic fields, reducing the overall magnetic field. Use of thin laminated cores yields long narrow eddy current paths with net reduction in the adverse effects. If we were to consider an iron sheet of thickness $h$ (greater than the skin depth) and conductivity $\sigma$ that is immersed in an oscillating flux $\vec{B}=\vec{B}_z\exp(j\omega t)\vec{a}_z$ parallel to the sheet's surface. Faraday's law of induction $\nabla\times\vec{E}=-j\omega\vec{B}_z$, where the steady state electric field is $\vec{E}=\vec{E}_{xyz}\exp(j\omega t)$. Taking x's origin to be the centre of the sheet, we can find $E_{y}=-j\omega B_zx$ and $J_y=\sigma\vec{E}_y=-j\omega B_z\sigma x$. The local instantaneous power dissipated is: $$P(x)=|\vec{J}|^2/\sigma=\omega^2|B_z|^2\sigma x^2$$ And the average power dissipated across the sheet is: $$P_{av}=\frac{1}{h}\int_{-\frac{h}{2}}^{\frac{h}{2}}P(x)dx=\frac{1}{12}\omega^2|B_z|^2h^2\sigma$$ The power dissipated increases the further from the centre you are, and is proportional to the thickness of the sheet. The linkage is $\Lambda=N\Phi=N\mathcal{A}\vec{B}\exp(j\omega t)$, and the potential difference across the coil terminals is $V=\vec{V}\exp(j\omega t)$, where $\vec{V}=j\omega N\mathcal{A}\vec{B}$. Assuming the total volume of the core is $\ell\mathcal{A}$, the total power dissipated is $P_{total}=\frac{\ell\mathcal{A}}{12}\omega^2|\vec{B}|^2h^2\sigma=\frac{|\vec{V}|^2}{R_{core}}$, where $R_{core}=\frac{12\mathcal{A}}{\ell}\frac{N^2}{\sigma h^2}$ is equivalent to the core resistance.
Transmission line and transformer modelling
Transmission lines
A transmission line is a pair of parallel wires with four terminals, forming two ports. We can get voltage and current characteristics by applying a voltage or current source and shorting or opening opposing terminals.
Resistance
The series resistance and inductance is found by connecting a current source across one port and shorting the other. The potential difference across the terminals and the current in the wire is related by: $$V_1=IR+L\frac{dI}{dt}\sim \vec{V}_1\exp(j\omega t)=(R+j\omega L)\vec{I}\exp(j\omega t)$$ This assumes the variation is sufficiently slow that that the speed of light doesn't affect transmission speed and that the current is the same throughout the circuit. Assuming uniform current density (no skin effect), the resistance of the two wire transmission line is: $$R=\frac{2l}{\sigma\pi a^2}$$ Where $a$ is the radius of the line, $\sigma$ is the line conductivity and $l$ is the line length.
Inductance
The inductance is more tricky, requiring a number of assumptions. We assume the wire is long enough that we can approximate it as infinitely long, but that the propagation along the line is fast. We can find the magnetic fields of the two lines, being the same but offset by the distance between them, $d$. We use that to find the flux linkage per unit length. $$\varphi_b=\lim_{X\to\infty}\frac{\mu_0}{2\pi}\left(\left(\frac{1}{4}+(\ln(X-d)-\ln(a))\right)I_b+(\ln(X)-\ln(d))I_a\right)I_t$$ Similarly the flux per unit linkage of current $I_t$ is: $$\varphi_t=\lim_{X\to\infty}\frac{\mu_0}{2\pi}\left(\left(\frac{1}{4}+(\ln(X)-\ln(a))\right)I_t+(\ln(X-d)-\ln(d))I_a\right)I_b$$ The energy stored per unit length is: $$\frac{1]{2}(\varphi_bI_b+\varphi_tI_I)=\frac{\mu_0}{2\pi}\left(\frac{1}{4}+\ln\left(\frac{d}{a}\right)\right)I^2$$ For a line of length $l$, the inductance is: $$L=l\frac{\mu_0}{\pi}\left(\frac{1}{4}+\ln\left(\frac{d}{a}\right)\right)$$
Capacitance
The shunt capacitance is found by connecting a voltage source across one port and leaving the other port unconnected. The separation of charges from the lines results in a Coulomb electric field, with per unit length charge $q$ and $-q$. The electric field between the lines is: $$\vec{E}_c(x)=\left(\frac{ql}{\epsilon_0}\frac{1}{w\pi xl}+\frac{ql}{\epsilon_0}\frac{1}{2\pi(d-x)l}\right)\vec{a}_x=\frac{q}{2\pi\epsilon_0}\left(\frac{1}{x}+\frac{1}{d-x}\right)\vec{a}_x$$ The potential difference between the lines is: $$V_1=-\int_{a}{d-a}\vec{E}_c\cdot(-ds\vec{a}_x)=\frac{q}{2\pi\epsilon_0}2\ln\left(\frac{d-a}{a}\right)=\frac{1}{\kappa}q=V_2$$ If $d>>a$, $\kappaa\approx\pi\epsilon_0/\ln(d/a)$. We can get the following relationship; $$\frac{dV_1}{dt}=\frac{dV_2}{dt}=\frac{1]{\kappa}\frac{dq}{dt}=\frac{1}{\kappa l}I_1=\frac{1}{C}I_1$$ $$C=\l\kappa=l\frac{\pi\epsilon_0}{\ln(d/a)}$$ As a result, for short lines at low voltages and steady state analysis at 50 Hz, the line capacitance is usually negligible. For medium length lines and high-voltage lines and steady state analysis at 50 Hz, the lumped capacitance should be included. For long lines or high-frequency analysis, a distributed parameter model is needed.
Three phase
In three-phase systems, we can use similar analysis so find the per unit length flux linkage: $$\varphi_1=\frac{\mu_0}{2\pi}\left(\ln\left(\frac{1}{0.78a}\right)I_1+\ln\left(\frac{1}{d_{12}}\right)I_2+\ln\left(\frac{1}{d_{13}}\right)I_3\right)$$ The lines are sometimes transposed to yield the same average characteristics per phase in balanced conditions: $$\varphi_{kav}=\lambda_sI_k+\lambda_mI_{(k+1)\mod 3}+\lambda_mI_{(k+2)\mod 3}$$ $$\lambda_s=\frac{\mu_0}{2\pi}\ln\left(\frac{1}{0.78a}\right), \lambda_m=\frac{\mu_0}{2\pi}\ln\left(\frac{1}{(d_{12}d_{23}d_{13})^\frac{1}{3}}\right)$$ We can get the per phase characteristics as: $$R_s=\frac{l}{\sigma\pi a^2},L_s=l\frac{\mu_0}{2\pi}\ln\left(\frac{1}{0.78a}\right),L_m=l\frac{\mu_0}{2\pi}\ln\left(\frac{1}{(d_{12}d_{23}d_{13})^\frac{1}{3}\right)$$
Transformers
A transformer has two ports and consists of two coils magnetically coupled, usually from being wound on a common magnetic core. One port is often called the primary and the other the secondary. Time-varying port currents give rise to time-varying flux linkages, induced electric fields according to Faraday's law and corresponding Coulomb electric fields consistent with the port coils and Ohm's law. We will assume the coils have infinite conductance, infinite permeability and no leakage.
Ideal model
The role of the core is to contain the flux, with its permeability relating its ability to do so. Since the winding conductance is assumed infinite, and by Ohm's law the Coulomb field in each could must balance the induced electric field due to time-variation of the flux linkage: $$V_p=-\int_{p-}^{p+}\vec{E}_{C_p}\cdot d\vec{\lambda}_p\approx N_p\frac{d\Phi}{dt}$$ $$V_s=-\int_{s-}^{s+}\vec{E}_{C_s}\cdot d\vec{\lambda}_p\approx N_s\frac{d\Phi}{dt}$$ The ideal port relationship is thus: $$\frac{V_s}{V_p}=\frac{N_s}{N_p}$$
Since the permeability is assumed infinite, the magnetic field strength in the core must be zero, otherwise flux density would be infinite, so by Ampere's circuit law: $$0\approx\vec{H}\ell=N_pI_p+N_sI_s$$ From this we get the following ideal port current relationship: $$\frac{I_s}{I_p}=-\frac{N_p}{N_s}$$ If the secondary could were wound in the opposite direction (ACW), the voltage polarity of the secondary port would be reversed and the current relationship sign would be flipped. The winding is represented with dots on the model, with the dot denoting the higher potential terminal when the voltage is positive. The positive secondary current flows out of the positive terminal, and the positive primary current flows into the positive terminal.
When a load is connected to the secondary: $$V_p=\frac{N_p}{N_s}V_s=\frac{N_p^2}{N_s^2}ZI_p=\hat{Z}I_p$$ This can increase or decrease the apparent impedance seen by the primary. Auto-transformers are when the two windings are electrically connected. $$V_p=\left(1+\frac{N_p}{N_s}\right)V_s$$ $$I_s=\left(1+\frac{N_p}{N_s}\right)I)p$$ $$S_{auto}/S=\frac{(V_p+V_s)I_p}{V_pI_p}=1+\frac{N_s}{N_p}$$ The last equation represents the power rating advantage.
Real model
A more practical model accounts for winding conductance, finite permeability, leakage of flux and core losses from hysteresis and eddy currents. For finite winding conductance, $\vec{E}_{C_p}=\frac{\vec{J}_p}{\sigma}-\vec{E}_{I_p}$, where $|\vec{J}_p=\frac{I_p}{\mathcal{A}_w}$, and $\mathcal{A}_w$ is the winding conductor area. $$V_{p}=\int_{p-}^{p+}\left(\frac{\vec{J}_p}{\sigma}-\vec{E}_{I_p}\right)\cdot d\vec{l}_p\approx R_pI_p+N_p\frac{d}{dt}(\Phi_{core}+\Phi_{p,leak})$$ Where $R_p=\frac{l_p}{\sigma\mathcal{A}_w}$ is winding resistance and $\Phi_{p,leak}$ is the flux that does not link with the primary coil. This is the same for the secondary, with the sign of $-R_sI_s$ instead of $R_pI_p$. This gives the electric field as: $$E_p=V_p-R_pI_p-N_p\frac{d\Phi_{p,leak}}{dt}=V_p-R_pI_p-L_{p,leak}\frac{dI_p}{dt}=N_p\frac{d\Phi_{core}}{dt}$$ $$E_s=V_s+R_sI_s+N_s\frac{d\Phi_{s,leak}}{dt}=V_s+R_sI_s+L_{s,leak}\frac{dI_s}{dt}=N_s\frac{d\Phi_{core}}{dt}$$ It follows that $E_p/E_s=N_p/N_s$, which suggests that a two port circuit model with a resistor and inductor in series with both ends of the transformer account for winding resistance and flux linkage.
When the core has non-zero reluctance, hysteresis and finite conductance, the ideal current relationship no longer holds. If the secondary port is left open and the primary is driven by a sinusoidal source, the movement around the hysteresis curve will alter the signal and introduce harmonics. Ignoring the harmonics in the non-zero, no-load primary current, we can find a linear relationship between the magnetising current and primary electric field: $$R_p(t)=N_p\frac{d}{dt}(\Phi_{core}\exp(j\omega t))=j\omega N_p\frac{N_pI_m}{\mathcal{R}_c}\exp(j\omega t)=\omega\frac{N_p^2}{\ell_c/\mu_0\mu_r\mathcal{A}}I_m\exp(j(\omega t+\pi/2))$$ This adds a resistor and inductor in series with the ideal transformer, in addition to the resistor and inductor in series from before. In practice, we tend to experimentally determine the values of these resistors and inductors.
Electromechanical energy conversion
Electrical power
Instantaneous power flow into a port is $V(t)I(t)$. When the power is positive, the potential energy dropped across the electrical field is absorbed, it can be stored in electric and magnetic fields, dissipated as heat or transferred. When the power is negative, the port exports (supplies) energy. We can characterise the instantaneous power flow for LTI systems as: $$V(t)=|V|\cos(\omega t+\angle V)$$ $$I(t)=|I|\cos(\omega t+\angle I)$$ $$V(t)I(t)=|\mathbf{V}||\mathbf{I}|(\cos^2(\omega t)\cos(-\angle \mathbf{I})+\frac{1}{2}\sin(2\omega t)\sin(\angle\mathbf{I}))$$ With $\phi=\angle\frac{V}{I}=-\angle I$ and $T=2\pi/\omega$, the average power flow is: $$P=\frac{1}{T}\int_{-T/2}^{T/2}|V||I|(\cos^2\omega t)\cos(\phi)+\frac{1}{2}\sin(2\omega t)\sin(\phi))dt=\frac{1}{2}|V|I|\cos(\phi)$$ This results in real power. We can also see that there is a reactive power component: $$jQ=j\frac{1}{2}|V||I|\sin{\phi}$$ From this we can observe that: $$\frac{1}{2}\mathbf{VI}=P+jQ=\mathbf{S}$$ The reactive power corresponds to energy being stored and released in electric and magnetic fields, not doing any real work. As the potential across a port corresponds to the rate of change of flux linkage $V=\frac{d\Lambda}{dt}$, we can characterise instantaneous power as: $$P=I(t)\frac{d\Lambda}{dt}$$
Kinetic power
Moving masses can store energy as kinetic energy: $$e_m=\frac{1}{2}mv^2$$ When a force acts on the mass, there is a change in energy: $$e_m(t_1)-e_m(t_0)=\int_{t_0}^{t_1}\frac{de_m}{dt}(t)dt=\int_{t_0}^{t_1}f(t)v(t)dt$$ The corresponding instantaneous power flow is: $$f(t)v(t)=f(t)\frac{dx}{dt}(t)$$ A body with moment of inertia $J_m$ and speed of rotation $\Omega=\frac{d\Theta}{dt}$ stores kinetic energy: $$e_m=\frac{1}{2}J_m\Omega^2$$ Application of a torque leads to a change in kinetic energy $T=J_m\frac{d\Omega}{dt}$: $$e_m(t_1)-e_m(t_0)=\int_{t_0}^{t_1}T(t)\Omega(t)dt$$ The instantaneous power flow is $T(t)\Omega(t)=T(t)\frac{d\Theta}{dt}(t)$.
Elemental lengths ($|d\vec{l}|$) of a current carrying conductor in a magnetic field experience a force according to Ampere's law: $$d\vec{F}_B=Id\vec{l}\times\vec{B}$$ Integrating around the conductor yields total force. The torque about a point is; $$d\vec{T}_B=\vec{r}\times Id\vec{l}\times\vec{B}$$ Integrating around the loop gives total torque. The action of the torque on the body to which the loop is attached, about the axis of rotation, constitutes instantaneous power flow $T_B\frac{d\Theta}{dt}$ out of the corresponding mechanical port.
Transducers
Transducers convert energy from one form to another. The energy domains may be different or the same. An example is a transformer for electrical to electrical, or a motor for electrical to mechanical. We tend to model with an ideal system, and augment the ports to account for non-ideal characteristics such as above with the transformer.
If we consider an electromechanical transducer which:
- Converts related to magnetic forces
- Has negligible losses in energy dissipation (can be accounted for later)
- The electrical port is related to the change in flux linkage (no conduction losses)
The power in is $P_e=V(t)I(t)-I(t)\frac{d\Lambda}{dt}(t)$ and the power out is $P_m=T(t)\frac{d\Theta}{dt}(t)$, the lossless magnetic storage is $\frac{dW}{dt}=P_e-P_m$. The energy density of the magnetic field is $\frac{1}{2}\vec{H}\cdot\vec{B}$ and depends on geometry (possibly port position $\Theta$). The electric port current $\vec{I}$ contributes to the magnetic field $\vec{H}$, which can be expressed as a function of $\Lambda$ and $\Theta$. The stored magnetic energy (from integrating over all space) is a function of state, $(\Lambda,\Theta)$, evolving over time. By assumption the energy isn't dissipated, so the energy in a state is path independent. $$W_s=W(\Lambda(s),\Theta(s))=\int_0^sI\frac{d\Lambda}{dt}-T\frac{d\Theta}{dt}dt=W_0+\int_{\Lambda(0)}^{\Lambda(s)}I(\lambda,\Theta(0))d\lambda-\int_{\Theta(0)}^{\Theta(s)}T(\Lambda(s),\theta)d\theta$$ It follows that $\frac{\partial W}{\partial\Lambda}=I$ and $\frac{\partial W}{\partial\Theta}=-T$.
An energy-based approach to modelling involves:
- Identifying mechanisms for energy dissipation and associating these with the ports of a lossless magnetic energy storage system
- Determining the energy storage dependence on state
- Using $T=-\frac{dW}{d\Omega}$ to determine torque
The electrical loss comes from wire resistances and can be modelled as a lumped system connected to the port.
It can be convenient to work with the alternative state $(I,\Theta)$. This is possible when there is a linear relationship between flux linkage and current; $$\Lambda=L(\Theta)I+M(\Theta)$$ The co-energy is defined as: $\hat{W}(I,\Theta)=\Lambda I-W(\Lambda,\Theta)$, whereby: $$\frac{d\hat{W}}{dt}=\Lambda\frac{dI}{dt}+T\frac{d\Theta}{dt}$$ $$\frac{d\hat{W}}{d\I}=\Lambda$$ $$\frac{d\hat{W}}{d\Theta}=-\frac{dW}{d\Theta}=T$$
Three phase rotating machines
Consists of a rotor, turning within a stator housing 3 windings with 3 corresponding electrical ports. Synchronous machines have a fourth electrical port for DC excitation. The winding can be configured to have any even number of magnetic poles. Asynchronous/induction motors have shorted rotor winding.
Basic principles
The stator flux lines of a phase produce a square wave of flux with respect to angle from the wire. Additional windings turn this into a fundamental standing wave. Adding in time and space separated windings $\frac{2\pi}{3}$ radian apart produces a rotating wave of magnetic flux density within the air gap. The wave rotates at the same frequency as the stator currents. The air gap field strength at position $\theta$ is: $$KI_p\left(\cos(\omega t)\cos(\theta)+\cos\left(\omega t-\frac{2\pi}{3}\right)\cos\left(\theta-\frac{2\pi}{3}\right)+\cos\left(\omega t+\frac{2\pi}{3}\right)\cos\left(\theta+\frac{2\pi}{3}\right)\right)=\frac{3K}{2}I_p\cos(\omega t-\theta)$$ $$KI=\frac{4}{\pi}\frac{N}{2g}I$$ g is the gap length and K is a constant related to the magnetic field strength and current. Interaction with the field due to rotor windings gives rise to torque. In a two pole synchronous machine, the angle is $\theta(t)=\omega t+\delta$, and in an asynchronous/induction machine the angle is: $\theta(t)=(1-s)\omega t+\delta$. The current in each stator and rotor coil is linked by magnetic flux due to its own current and those in other coils. For a uniform air gap two pole rotor the flux linkage $\Lambda_k$ of current $I_k$ is: $$\Lambda_a=LI_a+MI_b+MI_c+M_f\cos(\theta)I_f$$ $$\Lambda_f=M_f\cos(\theta)I_a+M_f\cos\left(\theta-\frac{2\pi}{3}\right)I_b+M_f\cos\left(\theta+\frac{2\pi}{3}\right)I_c+L_fI_f$$ a, b, and c represent the stator windings and are $\frac{2\pi}{3}$ radian apart, f represents the rotor windings. For salient pole rotors, the air gap is non-uniform and thus the self/mutual inductance of the stator is also dependent on angle.
Synchronous machines
If a synchronous machine operates in steady state so that the stator currents are balanced, the rotor current is constant and the angle is $\theta(t)=\omega t+\delta$, then the torque is $T=-\frac{3}{2}M_fI_fI_p\sin(\delta)$. The torque is constant resulting in constant mechanical power: $P_m=T\frac{d\theta}{dt}=-\frac{3}{2}\omega M_fI_fI_p\sin(\delta)$. The maximum torque is at $\delta=\pm\frac{\pi}{2}$, above this we lose synchronicity. When $\frac{\pi}{2}<\delta\leq0$ the system is motoring, and when $0<\delta<\frac{\pi}{2}$ the system is generating. Delta is the spacial offset of the rotor to the stator. At a synchronous rotor speed $\omega$ and fixed torque angle $\delta$, with constant $I_f$, the internal voltage induced is: $$E_{kf}=\frac{d}{dt}(L_{kf}I_f)=-\omega M_fI_f\sin(\omega t+\delta)$$ Where k is one of the phases. For a balanced phasor current: $$\Lambda_k=(L-M)I_k+M_f\cos(\omega t+\delta_k)I_f$$ Making the corresponding port stator voltage: $$V_k=\frac{d\Lambda}{dt}=(L-M)\frac{dI_k}{dt}-\omega M_fI_f\sin(\omega t+\delta_k)=(L-M)\frac{dI_k}{dt}-E_{kf}$$ $$\mathbf{V}_k=j\omega(L-M)\mathbf{I}_k+\mathbf{E}_{kf}=jX_s\mathbf{I}_k+\mathbf{E}_{kf}$$ By augmenting the port with an inductor and resistor, we can model leakage resistance with an inductor and armature resistance with a resistor. The electrical power a synchronous generator can supply scales linearly with the field current/internal voltage. For a fixed field current, the power for a port is $P_k=\frac{1}{2}\mathcal{R}(\mathbf{V}_k\mathbf{I}_k)$, where $\mathbf{I}_k=\frac{1}{jX+R}(\mathbf{V}_k-\mathbf{E}_{kf})$, so when the resistance is negligible the power is: $$P_k=\frac{1}{2}\frac{|V_k||E_{kf}|}{X}\sin(\angle V_k-\angle E_{kf})$$ The mechanical power is equal to the sum of the real power at three stator ports.
Connected machines
Connecting two machines together results in the real power flow from 2 to 1 being: $$-\frac{1}{2}\frac{|E_2||E_1|}{X_1+X_2}\sin(\angle E_1-\angle E_2)$$ Increasing the power angle results in less voltage at the connecting port.
Salient poles
Salient poles are often used in low speed machines of wide diameter. The stator self and mutual inductances are dependent on position. Park's (d-q-0) transformation is a rotor dependent transformation that can be used to get constant flux linkage and current relationship.
Induction machines
In an induction machine, the rotor windings are shorted (wound rotor or squirrel cage). Rotor currents are induced by motion of rotor conductors relative to stator sourced air gap magnetic flux wave. When the rotor speed equals the stator speed, there is no magnetic field with respect to the rotor and no current/torque is produced. Slip is the speed difference as a fraction of synchronous speed: $$s=(\frac{d\theta_r}{dt}-\omega)/\omega\implies\frac{d\theta_r}{dt}=(1-s)\omega$$ Frequency of induced rotor currents is $\omega_r=s\omega$. Superimposed on the rotor motion is an air gap flux rotating at $\omega$.
Interaction of the rotor and stator flux wave gives torque $T=-K\sin\delta_r$, where $\delta_r$ is the spacial phase difference. Torque is produced at zero speed, so sometimes synchronous machines use squirrel cages for starting. The rotor induces a counter emf on the stator phases.
At the slip frequency, the rotor impedance is: $$\mathbf{E}_{kf}/\mathbf{I}_k=R_r+j\omega_rL_r=R_r+j\omega sL_r$$ The power transfer from the stator over the air gap is $P<sub>g</sub>=$3I_r^2\frac{R_r}{s}$. The real power loss in the rotor is $P_r=3I_r^2R_r$. The mechanical power is $P_m=P_g-P_r=(1-s)P_g$. The torque is $T=\frac{3}{\omega}I_r^2R_r/s$.