LECTURE 5 PER-PHASE CIRCUITS AND MAGNETICS (1)

Transcription

1 ECE 330 POWER CIRCUITS AND ELECTROMECHANICS LECTURE 5 PER-PHASE CIRCUITS AND MAGNETICS (1) Aknowledgment-These handouts and leture notes given in lass are based on material from Prof. Peter Sauer s ECE 330 leture notes. Some slides are taken from Ali Bazi s presentations Dislaimer- These handouts only provide highlights and should not be used to replae the ourse textbook. 1/29/2018

2 PER-PHASE EQUIVALENTS Soure: blogspot.om 1/29/2018 2

3 PER-PHASE EQUIVALENTS In balaned three phase iruits, it is preferable to work with per-phase equivalents and then onvert the variables to three-phase quantities. I a Z Line V an Z Load 1/29/2018 3

4 Δ-Y CONVERSION PER-PHASE EQUIVALENTS Per-Phase equivalent iruits are very onvenient for analyzing three-phase iruits. For a Y-soure the load ould be either Y or Δ. The load seen between two phases, e.g., a and b, an be expressed as: Z / /2Z 2Z Z 3Z Y Y a Z Δ b Z Δ Z Δ a Z Y n Z Y Z Y b 1/29/2018 4

5 PER-PHASE EQUIVALENTS The per-phase iruits an then be shown as follows: S a + I 1 V ab - b,1 S V I ab * a S V Z Y,1,1 2 ab Z Δ S a + I a V an - n * Y,1 VanIa V Z 2 an Y Z Y 1/29/2018 5

6 PER-PHASE EQUIVALENTS For either load, the voltage and load an be transformed to a Y-per-phase iruit, or Δ-per-phase iruit, and the S should be the same. Z 10 j5 Vab 208 Example:, V S Y,1 V ab 2 2 an 3 Vab 2 V Z Z Y Z 3 o kva 1/29/2018 6

7 EXAMPLE 2.17 The following two three-phase loads are onneted in parallel aross a three-phase 480 V wye-onneted supply. 1/29/2018 7

8 EXAMPLE 2.17 Load 1: 24 kw at 0.8 PF lag (wye-onneted) Load 2: 30 kva at 0.8 PF lead (delta-onneted) I 2 Find the line urrents L1 and I L for eah of the two loads, total omplex power ST and total line urrent. Take V an as referene. Triangle method: 1/29/2018 8

9 EXAMPLE S T 1 = (0.8 j 0.6) VA = 24,000 j18,000 VA 3 S T 2 = 3010 (0.8 j 0.6) = 24,000 j18,000 VA S T = S T 1S T 2 48,000 W 1/29/2018 9

10 EXAMPLE 2.17 Line urrent: ( 3) V I os = 24,000 L L1 I L1 24, 000 = (480)( 3)(0.8) = A I L1 = sine the urrent is lagging 3) V I = 30,000 I = 30,000 / ( 3) A L L 2 L 2 1/29/

11 EXAMPLE 2.17 I L 2 = sine the urrent is leading. QT = QT 1QT 2 = = 0 48, 000 ( 3) V L I L = 48,000, I L = 57.7 A (480)( 3) Per-phase equivalent method 1/29/

12 EXAMPLE 2.17 Phase-to-neutral voltage is 480 / 3 = V = (277.13) L1(0.8) L1= I I A I L1 = sine the PF is lag 3 (1010 )(0.8) = (277.13)( L2)(0.8) L2 = I I A I L2 = A sine the PF is leading I L = I L1 I L 2 = = A 1/29/

13 MAGNETIC CIRCUITS Maxwell s Equations: Ampere s Law: The magneti field in any losed iruit is proportional to the eletri urrent flowing through the loop. Faraday s Law: C C H. dl J.nda N I The line integral of the eletri field around a losed loop is S db E. dl. n da dt equal to the negative of the rate of hange of the magneti flux S through the area enlosed by the loop.. 1/29/

14 Conservation of Charge: Gauss s Law: MAGNETIC CIRCUITS S J. n da 0 The net magneti flux out of any losed surfae is zero. (for a magneti dipole,in any losed surfae the magneti flux inward toward the south pole will equal the flux outward from the north pole). S B.nda 0 1/29/2018 Soure:study.om 14

15 MAGNETIC CIRCUITS What do these symbols mean? 1/29/2018 Integral over a losed ontour C: Surfae S define by C: S Integral over a losed surfae S: Length of the ontour C: H is the magneti field intensity (A.turns/m) B is the magneti flux density (Tesla or Wb/m 2 ) E is the eletri field (V/m) J is the urrent density (A/m 2 ) n is the normal vetor to S. dl C C S 15

16 STATIC MAGNETIC CIRCUIT In stati magneti iruits, there are no moving members. The most important devie in this ategory is the transformer. We use Ampere s urrent law (ACL) and Gauss s law (GL) for magneti fields to derive useful flux-urrent terminal relations. The analysis is helpful in the design of indutors, and study of transformers. 1/29/

17 MAGNETIC CIRCUIT - TOROID Current flowing in a ondutor produes a magneti field. Voltage produes an eletri field. Common example: Soure: eletronishub.org i(t) v(t) 1/29/ A r o r N 1 H + v(t) - H, B i(t) i(t) in i(t) out 7

18 MAGNETIC CIRCUIT - TOROID Applying Ampere s law where is the mean length of the ore. H. d Ni we get H Ni an be approximated as 2π(r o +r 1 )/2. Assuming a linear relationship between B and H where B H and μ is the permeability (H/m). μ = μ r μ o where μ r is the relative permeability and μ o is the permeability of free spae. μ o = 4π x10-7 H/m. 1/29/

19 MAGNETIC CIRCUIT - TOROID The flux density in the ore is Sine B is the flux density (Wb/m 2 ), then the flux is A A B Define the magnetomotive fore (mmf) as Define the relutane to be B 1/29/ N i. mmf Ni ampere turns / weber A H Ni. N i 9

20 Magneti Ciruit Equivalent Then, The permeane is Ni mmf 1/ Wb / ampere turn It has similarity to a resistive iruit with the following equivalenes Eletri Ciruit Magneti Ciruit Ni Voltage mmf Current Flux Resistane Relutane Condutane Permeane 1/29/

21 MAGNETIC VS. ELECTRIC CIRCUITS The following analogies hold: Differenes: Leakage. μ is not perfetly onstant. Saturation KVL and KCL analogous to total MMFs aross a loop add up to zero, and total flux entering or leaving a node is zero. 1/29/

22 INDUCTANCE Faraday s law an be written as: C d E. d B. n da. dt S E is (V/m) and is (m) => left side is voltage. Define the flux linkage as N B A. Then, d d NBA d 2 A di () t v ( t ) N N. dt dt dt dt Define the indutane 2 A L N l, then di() t v( t) L. dt 1/29/

23 MAGNETIC CIRCUIT TOROID WITH AIR GAP Some magnetis have air gaps that store energy. Bak to the Toroid example but with air gap. The gap length is g and its permeability is 7 4x10 H / m A H. d Ni H H Ni g g i(t) v(t) N r o r 1 H Soure: softsolder.om 1/29/

24 MAGNETIC CIRCUIT TOROID WITH AIR GAP Therefore, B H Ni and B H g o g o Ni g. Assuming all flux passes through the air gap, A B A B. g g g Then, Ni g ( g ) A A g Ni g 1/29/

25 FRINGING Fringing ours when the flux jumps around the air gap to join the other side of the ore. b a Fringing an be simply modeled by having A g >A Two methods to aount for this: - Empirial approximation - A is given as perentage times g A ab Ag ( a g )( b g ) A, 1 g ka k 1/29/ A

26 EXAMPLE Ni 400 At, 6 m, 0.1 m, A 1 m g 2 A A 4 g 1.1, r 10 H/ m, Find the flux. H H Ni g g i(t) v(t) N r o A 1 A x 10 Wb o g A Ni r 1 H 1/29/