Synchronous machines Synchronous generator (alternator): transforms mechanical energy into electric energy; designed to generate sinusoidal voltages and currents; used in most power plants, for car alternators, etc. Synchronous motor: transforms electric energy into mechanical energy; used for highpower applications (ships, original TGV…) Saliant poles Turbo-alternator Rotor (inductor) : 2p poles with excitation windings carrying DC current; nonlaminated magnetic material θ! = ω / p Stator: polyphase (e.g. 3-phase) winding in slots; laminated magnetic material ... 01/03/2018 Synchronous machines 1 No-load characteristic Evolution of the voltage Ev in a stator phase vs. intensity of the excitation current Ie, for a given rotation speed and with no generated stator current • The rotor winding, carrying the DC current Ie and rotating at speed ω/p, produces in the airgap a sliding m.m.f. Fe (as seen from the stator). • Fe generates a magnetic flux density Br (with the same phase) in the airgap, which induces sinusoidal e.m.f.s Ev in the stator windings, with a phase lag of π/2. speed θ! = constante constant ⎧vitesse E v = f (Ie ) with avec ⎨ ⎩I = 0 average characteristic average characteristic E v = k E θ! Φ v (Ie ) Magnetic flux produced by the inductor and seen by the stator winding due due remanent magnetization Non linearity with hysteresis Synchronous machines 2 Vector diagram with load Diagram of magnetomotive forces and magnetic flux densities Br (4a) resulting m.m.f. (3) = (1)+(2) m.m.f. from I (2) (4b) m.m.f. from Ie (1) Fe = γ Ie and FI = δ I (1) The rotor winding, carrying the DC current Ie and rotating at speed ω/p, produces in the airgap a sliding m.m.f. F e (as seen from the stator).. (2) The polyphase current I in the stator winding produces a sliding m.m.mf. F I (in phase with I). (3) The resulting m.m.f. is F r = F e+F I. (4) F r generates a magnetic flux density Br (with the same phase) in the airgap,, which induces sinusoidal e.m.f.s in the stator windings, with a phase lag of π/2. Synchronous machines 3 Vector diagram with load Stator leakage flux (seen by the stator but not coming from the rotor) e.m.f. induced by leakage flux eλ ( t ) = −λ ∂ t i1 − λ m ∂ t i 2 − λ m ∂ t i3 Slot leakage flux = −(λ − λ m ) ∂ t i1 because i1 + i 2 + i3 = 0 Eλ = − j X f I with X f = ω (λ − λ m ) Stator leakage reactance End-winding leakage flux E t = Er − j Xf I Diagrams Fe = γ Ie U = Er − j Xf I − R I total e.m.f. and FI = δ I Fr = γ Ie + δ I Resistance of stator winding def Ir = Fr δ = Ie + I γ γ Equivalent excitation current Synchronous machines E r ≡ no-load e.m.f. produced by Ir 4 Potier diagram ‘Which excitation current Ie should one impose in the synchronous machine to reach the functioning point corresponding to a given voltage U and current I in the stator, with a phase shift of ϕ between U and I?’ E r ≡ no-load e.m.f produced by the equivalent current Ir (2) (4) Er = U + R I + j Xf I (5) Ie = I r − (2) (3) δ I γ (3) (0) (1) Synchronous machines 5 Reaction Demagnetizing reaction Magnetizing reaction The m.m.f. is smaller than the no-load m.m.f. (Ir < Ie) The m.m.f. is larger than the no-load m.m.f. (Ir > Ie) Capacitive behaviour of the load (I in front of U) Inductive behaviour of the load (I lagging behind U) Synchronous machines 6 Zero power factor characteristic Evolution of the stator voltage U as a function of the excitation current Ie, for a given rotation speed and stator current, with a zero power factor ⎧vitesse speed θ! = constante constant ⎪ constant U = f (Ie ) with avec ⎨I = constante ⎪cos ϕ = 0 (ϕ = ± π / 2) ⎩ Example: ϕ = π/2 δ Ie ≈ I r + I γ U ≈ E r − Xf I Synchronous machines 7 Short-circuit characteristic Evolution of the stator current as a function of the excitation current Ie, for a given rotation speed and with the stator windings in short-circuit Er = U + R I + j Xf I with ⎧vitesse speed θ! = constante constant I = f (Ie ) with avec ⎨ ⎩U = 0 U=0 Er = R I + j Xf I E r ≈ Xf I I≈ β δ Xf + β γ Ie δ Ie ≈ I r + I γ Non saturated machine E r = β Ir Synchronous machines 8 Simplified vector diagram Behn-Eschenburg’s method – Synchronous reactance Xs When the magnetic materials are not saturated, the combined effect of the reaction and of stator leakage fluxes can be taken into account thanks to a single parameter: the synchronous reactance E v Er constant = = constante Ie I r Equal angles OAB and A’OB’ Similar triangles OAB and OA’B’ A, B and C colinear E v = U + R I + j Xs I Synchronous reactance Synchronous machines 9 Experimental determination of Xs Behn-Eschenburg’s method – Synchronous reactance Xs U=0 ⇒ ⇒ E v = (R + j Xs ) Icc R + j Xs = Ev Icc with Xs ≈ R << Xs E v ( Ie ) Icc (Ie ) Approximation when magnetic materials are saturated! Synchronous machines 10 Exterior characteristic Alternator exterior characteristic Evolution of the voltage U on a given stator phase as a function of the current Iin this phase, when the alternator drives a load characterized by a constant power factor, at constant speed and excitation ⎧vitesse speed θ! = constante constant ⎪ constant U = f (I) with avec ⎨Ie = constante ⎪cos ϕ = constante constant ⎩ Synchronous machines 11 Network connection Need for interconnection of electric power plants Daily peak Economical organization of power production + Stability of the network despite local defects Daily load Synchronization of an alternaltor on an ideal (infinitely powerful) AC network Large number of production units in parallel ⇒ constant voltage and frequency The current should be zero when the connection is made → 4 conditions I= Ev − U R + j Xs 1. 2. 3. 4. same pulsation ω (correct rotation speed) same amplitudes for E v and U (adjusting Ie) no phase shift between E v and U identical phase ordering (in a 3-phase system) Synchronous machines 12 Behaviour with load Active electric power Static stability Static instability Electromagnetic power Internal angle Torque Pelm ≈ P = 3 U I cos ϕ C= P 3p = U I cos ϕ ω/ p ω C= 3p U E v sin δint ω Xs Xs I cos ϕ = E v sin δint Internal angle After network synchronization, there is no exchanged current. Then : • If mechanical power is provided to the alternator, E v gets ahead of U ⇒ δint increases (until the equilibrium of the electromagnetic and mechanical torques) • If a braking torque is applied to the alternator, E v gets behind U ⇒ δint decreases (negative torque) Synchronous machines The variations of the rotor mechanical angle Δδmec are proportional to the variations of the internal (electric) angle Δδint Δδmec = Δδint p 13 Behaviour with load Static stability Internal angle δint < 0 δint > 0 Unstable Increasing the mechanical (breaking) torque leads to an increase of the absolute value of δint , and a decrease in the absolute value of Celm ⇒ unstable alternator Unstable synchronous motor Increasing the mechanical (breaking) torque leads to an increase of δint , and a decrease of Celm ⇒ unstable static stability Increasing the mechanical (breaking) torque leads to an increase of the absolute value of δint et of Celm ⇒ stable. The equilibrium is reached when the two torques are equal. Increasing the mechanical (breaking) torque leads to an increase of δint and Celm ⇒ stable. The equilibrium is reached when the two torques are equal. Synchronous machines 14 Behaviour with load Power diagram MB = Xs I cos ϕ = Xs X X 3 U I cos ϕ = s P = s Pelm 3U 3U 3U Active power P Alternator O' B = Xs I sin ϕ = Xs X 3 U I sin ϕ = s Q 3U 3U Motor Reactive power Q Synchronous machines 15 V-curves (Mordey curves) under-excited machine Evolution of the stator current I as a function of the excitation current Ie of a synchronous machine connected to an ideal network, at constant active power Over-excited machine "V" Constant active power ϕ > 0 ... ϕ>0 Network= inductive load Machine = capacitif system static stability limit Synchronous machines 16
