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U Department of Electrical I Engineering ECE529: Session 44 An Overview of HVDC Transmission Systems ECE 529 Spring 2009 HVDC Transmission Systems 1/24 U Department of Electrical I Engineering ECE529: Session 44 Steady-State HVDC Converter Representation • Steady state equivalent circuit I dc R dc Inverter V V dcr dci Rectifier f (α ,Idc , |V |) ac • Have fast, direct control over α (ﬁring delay angle) • Vdc = Vdo ∗ cosα (ﬁring delay angle) where Vdo = const ∗ |VLL| • Some control of |Vac| with tap changing transformer • DC current indirectly controlled by changing α HVDC Transmission Systems 2/24 U Department of Electrical I Engineering ECE529: Session 44 Basic Six-Pulse Converter • Based on line commutated, current source converter • Thyristors used as devices • Converter with stiff current source on dc side • Stiff voltage source on ac side (turns off thyristors) • Basic 6-pulse bridge: Smoothing Reactor Ls I A dc + 1 3 5 ean (t) Xc - - + - ebn (t) V - - dc ecn (t) + + B 4 6 2 C Transformer Inductance HVDC Transmission Systems 3/24 U Department of Electrical I Engineering ECE529: Session 44 Basic Six-Pulse Converter • Initially assume: 1) Ideal ac sources, 2) ideal switches, 3) Xc = 0, and 4) Ls → ∞ source) • AC side of converter has an ideal voltage source, dc side of converter has an ideal current source • Apply Kirchhoff’s Current Law: i1 + i3 + i5 = Idc (one switch always closed) i2 + i4 + i6 = Idc • Apply Kirchhoff’s Voltage Law: ean + ebn + ecn = 0 (balanced 3 phase set) • Since Xc = 0, only one switch in (1,3,5) can be closed with a switch in (2,4,6) HVDC Transmission Systems 4/24 U Department of Electrical I Engineering ECE529: Session 44 Basic Six-Pulse Converter (cont.) • Allowable combinations: 1 with (2 or 6) (4 shorts dc bus) 3 with (2 or 4) 5 with (4 or 6) 2 with (1 or 5) 4 with (1 or 3) 6 with (3 or 5) • Need to determine a switching sequence • Start from assumption of positive phase sequence • Typical current waveforms: ia | | ib | | ic | | HVDC Transmission Systems 5/24 U Department of Electrical I Engineering ECE529: Session 44 Basic Six-Pulse Converter (cont.) • Possible sequences: Top three switches: 1-3-5-1 or 1-5-3-1 Bottom three switches: 4-6-2-4 or 4-2-6-4 + − • Assume: Vdc = Vdc - Vdc + − Switch # Vdc Switch # Vdc 1 ean(t) 4 ean(t) 3 ebn(t) 6 ebn(t) 5 ecn(t) 2 ecn(t) HVDC Transmission Systems 6/24 U Department of Electrical I Engineering ECE529: Session 44 Basic Six-Pulse Converter (cont.) • Positive sequence (α = 0, 1-3-5-1 and 4-6-2-4) 10.00 6.00 2.00 Voltage (V) -2.00 -6.00 -10.00 0.0 6.67 13.33 20.00 26.66 33.33 Time (mS) • Negative sequence (α = 0, 1-5-3-1 and 4-2-6-4) 10.00 6.00 2.00 Voltage (V) -2.00 -6.00 -10.00 0.0 6.67 13.33 20.00 26.66 33.33 Time (mS) HVDC Transmission Systems 7/24 U Department of Electrical I Engineering ECE529: Session 44 Basic Six-Pulse Converter (cont.) I dc 5 1 3 5 I dc 6 2 4 • Phase currents: + − Switch Combination Vdc=Vdc - Vdc 1-6 eab= ean - ebn = Vdc 1-2 eac= ean - ecn • Look at the line voltages: 3-2 ebc= ebn - ecn 3-4 eba = ebn - ean 5-4 eca= ecn - ean 5-6 ecb= ecn - ebn √ • If α = 0, then Vdc = 3 2 π |VLL | = 1.35|VLL| We deﬁne this as Vdo HVDC Transmission Systems 8/24 U Department of Electrical I Engineering ECE529: Session 44 Controlled Firing of Thyristors • Now add a ﬁring delay (α) for the thyristors. Same delay for all 6 switches 10.00 6.00 2.00 Voltage (V) -2.00 -6.00 -10.00 0.0 6.67 13.33 20.00 26.66 33.33 Time (mS) HVDC Transmission Systems 9/24 U Department of Electrical I Engineering ECE529: Session 44 Controlled Firing of Thyristors Ê π +α √ √ π +α • Vdc = 3 − π +α π 6 2|VLL |cos(θ)dθ = 3 2 π |VLL |sin(θ) |− π +α 6 6 6 √ • Then Vdc = π |VLL |cosα 3 2 √ • Deﬁne Vdo = π |VLL| 3 2 • Therefore Vdc = Vdo cosα • α = 0 → diode bridge Vdc = Vdo 0 ≤ α < 90 → rectiﬁer Vdc > 0 α = 90 → P =0 Vdc = 0 90 < α ≤ 180 → inverter Vdc < 0 • Current does not reverse HVDC Transmission Systems 10/24 U Department of Electrical I Engineering ECE529: Session 44 Commutation Overlap • Now add source inductance (Lc = 0) Ls I dc 1 3 5 + Lc + V V - dc r dc i 4 6 2 - HVDC Transmission Systems 11/24 U Department of Electrical I Engineering ECE529: Session 44 Current Transfer Between Switches • Current does not fall to zero immediately in ac side inductance • Temporarily create line to line short Ls 1 3 Xc Xc ean (t) ebn (t) ecn (t) α µ Xc I dc 1 3 2 Ls 0 HVDC Transmission Systems 12/24 U Department of Electrical I Engineering ECE529: Session 44 Current Transfer Between Switches (cont.) • What happens if α gets to big (i.e. α ⇒ 180◦)? I dc 1 1 0 This is called a commutation failure • Thyristor 3 fails to turn on and thyristor 1 fails to turn off • This is more common if Lc is large, which is the case looking into a “weaker” ac system • Normally corrects during next interval, although often have a second failure when thyristor 5 turns on, “double commutation failure” HVDC Transmission Systems 13/24 U Department of Electrical I Engineering ECE529: Session 44 Output Voltage During Commutation + • Switch 1 contribution: Vdc1 = ean - Lc di1 dt + • Switch 3 contribution: Vdc3 = ebn - Lc di3 dt + + • During overlap we see the average between Vdc1 &Vdc3 + + Vdc1 +Vdc3 + So Vdc = 2 = ean+ebn - 2 Lc 2 di1 dt + di3 dt di1 +di3 • i1 + i3 = Idc, so dt =0 di1 +di3 But since its a linear network: dt = di1 dt + di3 = 0 dt eac+ebc • So: Vdc = Vdc − ecn = ean+ebn − ecn = + 2 2 HVDC Transmission Systems 14/24 U Department of Electrical I Engineering ECE529: Session 44 Average DC Voltage with Overlap √ √ √ • Recall: Vdo = = π |VLL | 3 2 = π |Vφ | 3 6 π |Em | 3 3 where Em is peak line to neutral voltage • Then we ﬁnd: 3 α+µ 3 α+ π √ 3 π Vdc = Emcosθdθ + 3|Em|cos(θ − )dθ π α 2 α+µ 6 √ • Leading to: Vdc = 3 3 E [cosα + cos(α + µ)] 2π m • Or Vdc = V2 [cosα + cos(α + µ)] do HVDC Transmission Systems 15/24 U Department of Electrical I Engineering ECE529: Session 44 Average DC Current • Start out with Lc = 0 and α = 0 for now Fundamental Current Component 120 210 330 360 0 30 150 180 • Firing delay simply adds a phase shift to the current (always lagging), and cosα = cosφ E E an an 30 90 I a Ia • Fundamental Component π π √ 2 2 2 3 2 3 i1pk = iacos(θ)dθ = Idc cos(θ)dθ = Idc π −π 2 π −π 3 π √ • |I1RMS| = 6 π Idc HVDC Transmission Systems 16/24 U Department of Electrical I Engineering ECE529: Session 44 Average DC Current √ • Then i1(t) = 2 π3 Idccos(ωt − α) • Also: P = 3I1RMSVPcosφ = VdcIdc √ So: 3I1RMSVPcosφ = π VPcosαIdc 3 6 √ • So: |Ia1RMS| = 6 I π dc as expected √ • During overlap: Idc = Ic = 3Em 2ωLc = 2Xc eLL • i3(t) = Ic(cosα − cosωt) with α ≤ ωt ≤ α + µ where ωt = α + µ at the end of the commutation interval • So average current is: Idc = Ic(cosα − cos(α + µ)) √ 3 |Vp | • Also: Ic = 3Em 2ωLc = 2 Xc |VLL = √2X| c HVDC Transmission Systems 17/24 U Department of Electrical I Engineering ECE529: Session 44 Average DC Circuit Equations • We have the following equations: Vdo Vdc = [cosα + cos(α + µ)] 2 Idc = Ic(cosα − cos(α + µ)) √ 3 2 Vdo = |VLL| π |VLL| πVdo Ic = √ = 2Xc 6Xc • Substitute for the cos(α + µ) in the Vdc equation • Then Vdc = Vdo cosα − Vdo Idc 2Ic • Where Vdo = 2Ic Vdo πVdo = π Xc = Rc (called the commutating “resistance”) 3 2 6X c • So Vdc = Vdo cosα − IdcRc HVDC Transmission Systems 18/24 U Department of Electrical I Engineering ECE529: Session 44 Average DC circuit • Rc represents a current dependent voltage drop due to overlap • Rc does not represent any energy dissipation! • So using Vdc = Vdo cosα − IdcRc we get: R line Rc Rc V cos α V cos α do do RECTIFIER INVERTER HVDC Transmission Systems 19/24 U Department of Electrical I Engineering ECE529: Session 44 Inverter Operation • α + µ + γ = π Covers positive half cycle of voltage • γ is deﬁned as the extinction angle • γo is minimum extinction angle for proper turn-off Typical values: 15◦ → 20◦ • So α + µ ≤ 180◦ − γo gives limits for control settings • Replace α with 180◦ − µ − γ in averaged equations * note: cos(180◦ − θ) = −cos(θ) HVDC Transmission Systems 20/24 U Department of Electrical I Engineering ECE529: Session 44 Inverter Operation (cont.) • Generate equations in terms on γ instead of α Vdo Vdc = [cosα + cos(α + µ)] 2 Vdo Vdc = [cos(180◦ − µ − γ) + cos(180◦ − γ)] 2 Vdo Vdc = − [cos(γ + µ) + cos(γ)] 2 Idc = Ic(cosα − cos(α + µ)) Idc = Ic(cosγ − cos(γ + µ)) • Sign reversal in voltage equation expected for inverter HVDC Transmission Systems 21/24 U Department of Electrical I Engineering ECE529: Session 44 Effect of Overlap on Power Transfer • Pac = 3I1RMSVp cosφ √ 3 6Vp • Pdc = Idc ∗Vdo cosα+cos(α+µ) 2 = Idc π cosα+cos(α+µ) 2 √ • I1RMS = Idc π6 • Then cosφ = cosα+cos(α+µ) = Vdo 2 Vdc • Note: overlap equations change if µ > 60◦, covered in Kimbark, Direct Current Transmission: Volume I. Wiley, 1971. HVDC Transmission Systems 22/24 U Department of Electrical I Engineering ECE529: Session 44 Transformer Loading DC System AC system x (p.u.) • Xc = ωLc where X → 12 − 20% • ZB = VB I B 2 3VBφ 2 VLL VBφ • Xc = ZBX = IB X = 3VBφ IB X , this is B MVA3φB 2 VBL • Xc = MVAB3φ X √ • IB = I1RMS = 6 I π dcB √ • So MVAB3φ = 3VBφIB = 3VBφ 6 π IdcB = Vdo IdcB R HVDC Transmission Systems 23/24 U Department of Electrical I Engineering ECE529: Session 44 Transformer Loading (cont.) • Need to use true RMS (transformers sees all harmonic components) Ê 2π • IRMS = 1 3 2 I dθ π o dc = Idc 2 3 • So MVAB3φ = 3IRMSVφ = πVdo IdcB 3 R 2 √ VBL • Then Xc = MVAB3φ X with Vdo = 3 2 π VBL πVdo • Then ZB = 6IdcB • Then from Vdc = Vdo cosα − Idc π Xc we get 3 Vdc = Vdo cosα − Idc π (ZBX ) 3 Vdc = Vdo cosα −Vdo X IIdcB 2 dc • Leading to a “per unit” expression: Vdc Vdo = cosα − X Idc 2 HVDC Transmission Systems 24/24