Generation, transmission and distribution, as well as power supplied to industrial and commercial customers uses a 3 phase system.

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1 Three-phase Circuits Generation, transmission and distribution, as well as power supplied to industrial and commercial customers uses a 3 phase system. Where 3 voltages are supplied of equal magnitude, but 120 out of phase. an 0 an bn cn cn I c ab cn bn bn 120 I b θ I a an bn Why? fewer conductors, less loss a) Economic advantages (at higher power, lower $ / kw than single-phase) b) Operational advantages (rotating machinery; constant instantaneous power, constant torque developed at shaft and less vibration) 77

2 (For s.s. analysis) We will mostly be dealing with balanced 3-phase systems (quantities of equal magnitude but displaced by 120 with balanced loads). Loads can be naturally balanced, like 3 phase machinery / loads or designed to be balanced (all large loads 3 balanced, 1 are averaged to balance out). Distribution to an office building is 3-phase, while the individual loads (outlets, lights, PC s etc) are 1-phase. In this case, during the design phase, care is taken to balance the load (e.g. one phase per floor/ branch circuit). 78

3 Generation of 3-phase voltages (Only scratch the surface, more details in ECE 331, 432, 433 ). 3 synchronous generators are commonly used for power generation (see next page). The rotor with dc current excitation is moved by the prime mover (steam or hydro turbine etc). The dc current creates a magnetic field, which is now rotating by the force of the prime mover, thus establishing a time-varying magnetic field. This magnetic field flux (magnetic field lines) links the stationary stator windings, and a voltage is induced according to Faraday s law: e dλ dt N d dt λ flux linkage associated with a winding N number of turns total winding flux 79

4 The generator is designed to ensure sinusoidal induced voltages. Since the three stator winding axes are 120 displaced ( in space) the resulting three voltage waveforms are also displaced by 120 (in time). The 3 windings can be connected in either wye ( Υ ) or delta ( Δ ). Of course the 3 voltages are of the same frequencies, We represent them as phasors where the reference of time zero is chosen for convenience. 80

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9 an, bn, cn Have equal magnitudes Displaced by 120deg Source Imp. Z ga Z gb Z gc an cn n Z ga bn I a I b Z gb Z la a line impedance b Z lb B Z B Z In a balanced system I a + I b + I c 0 I n n and N are at the N same potential Z la Z lb Z lc Z Z B Z C Z gc I n Z C I c c Z lc C 85

10 Typical Building Power Distribution Scheme Delta/Wye Distribution Transformer For Top Floor 3 Distribution Panel Load N G N G Power Feeder Branch Circuit Three Phase Bus Delta/Wye Distribution Transformer For First Floor 3 Distribution Panel Load N N G G Power Feeder Branch Circuit Three Phase Supply From Utility Mains Three Phase Stepdown Transformer In Building Basement B C 86

11 Note that the phase sequence a c, b, is also possible (negative sequence). lso note that: an + bn + cn 0, which also implies for the instantaneous voltages: van ( t) + vbn ( t) + vcn ( t) 0. (i.e. 3 sinusoids of the same magnitude and frequency, but displaced 120, add up to zero.) This also means that a balanced 3-phase set can be described by 2 quantities since c a b oltage sources and loads can be connected in wye or delta, i.e. we have the following combinations: Source Load Line 87

12 Let s consider a Υ / Υ system in more detail a Z la I a Z ga Z I b an bn cn n Z gb b Z lb B Z B N Z gc I n Z C I c c Z lc C 88

13 For a balanced 3 circuit: 1. an by, bn, cn are a set of balanced 120 ). 2. Z ga Z gb Z gc (Source impedance is same) 3 voltages (equal magnitude displaced 3. Z la Zlb Zlc (Line impedance is same) 4. Z Z B ZC (Load impedance is same) 5. I a, I b, Ic are a set of balanced 3 currents. I a + Ib + Ic 0 In neutral current 0 and the neutral can be omitted. Balanced 3-phase circuits are more easily analyzed by considering only one phase. ( if Δ load converted to a Υ ) to get neutral point N In analyzing 3 circuits, first construct 1 equivalent, converting all Δ ' s to Υ ' s ( Δ Z ). Υ Z 3 89

14 Per phase equiv. a' Z ga a Zla Δ Z Z Υ 3 cn I c a'n I a Z I b θ I a an n N bn Now it is easier to see that I a ' a n I Z + + a ga Z la Z Z p θ, and I Z most likely lagging Other two currents are ' I θ 120 ± 120 displaced ' I θ b n c n I b, I c Z Z p Z p Z 90

15 With this Υ circuit, we have worked with phase voltages (referred to the neutral) and line currents (equal to the phase currents in a wye configration) However, oftentimes the wye neutral point is not accessible. We can t use the phase voltages. This of course is true for Δ circuits as well. So we need to define quantities, which are accessible from the terminals. The line (or line-to-line) voltage,, ab bc ca cn ab Where ab an bn L 30 bc L ( ) L ca L ( ) L Summarize line vs. phase. phase notation or p line voltage, or line-to-line voltage bn an B N bn 91

16 Wye: line Delta line I C B C I I I B B N phase phase N C I I N B I B B phase voltage θ line voltage 3 ( ) L (0 line current phase current I I θ I L θ θ + 30 phase voltage line voltage Δ 30 L 30 phase current I I Δ θ Δ B θ Δ θ Δ line current Δ Δ Δ 3I ( θ 30 ) I ( θ 30 ) I Note Z y Z Δ /3 L Z Δ ) 30 I B I C I I B + I C I BC I 92

17 Example: The magnitude of the phase voltage of an ideal balanced threephase Y-connected source is The source is connected to a balanced Y-connected load by a distribution line that has an impedance of 1+j8Ω/. The load impedance is 119+j27Ω/. The phase sequence of the source is abc. Use the a-phase voltage of the source as the reference. Specify the magnitude and phase angle of the following quantities: (a) the three line currents, (b) the three line voltages at the source, (c) the three phase voltages at the load, and (d) the three line voltages at the load ref. Υ a n I a 1Ω j8ω all per phase N Υ 119Ω j27ω 93

18 Polar form of a) The 3 line currents, Υ connected, line current phase current I I I a b c j35 (sum of Z s) 32 ( ) ( ) b) the 3 line voltages at the source L ab bc ca 3 ( (4000) ) ( ) ( ) c) the 3 phase voltages at the load I ( j27) (or 4000 (1 + j8) N BN CN a d) the 3 line voltages at the load B BC C ( ) a I ) 94

19 Power in Balanced 3 Circuits (So far we have discussed only voltage and current in 3 systems.) Start with Balanced Υ load - will be using all rms quantities. I a I bb I cc B + C Z B BN Z Z C + N + N CN We can calculate the average power associated with each phase by using the same techniques as used for 1. I B N a line to line, or just line phase, or line to neutral voltage phase and line current 95

20 The average power associated with phase- is: P P P B C (if not rms, have 2 1 here) ( θ Z L phase angle of the phase quantities, angle of the load impedance per phase) I cos( θ θ ) (see power summary) N, rms a, rms v i (only magnitude, since we are only after a real number) θ Z L BN I bb cos( θ vb θ ib ), θ Z L I cos( θ θ ) CN cc vc ic Where N BN CN, voltage magnitudes same ( I I I I a bb cc current magnitudes same ( 120 apart) 120 apart) θ θ Z L θ v θ i θ vb θ ib θ vc θ ic 96

21 Thus power delivered to each phase of the load is the same. P P P P I cosθ θ θ ) B C ( v i The total average power delivered to the balanced Υ connected load P T (not a function of time) P 3 I cosθ constant 3 ve power per phase Important in motors where the shaft power is a constant and there are no torque pulsations. It is very useful to express in terms of rms line quantities (accessible!) 97

22 Where L rms line voltage( B in diagram) I rms linecurrent phasecurrent L P T 3 L 3 I L cosθ 3 LI L cosθ phase angle between the phase voltage and current NOT between L and I L The same holds true for Reactive Power Q. The reactive power Q associated with each phase is: Q I sinθ Q T 3Q 3 L I L sineθ The complex power S associated with each phase is: S I * P + jq ST 3 S 3L I L magnitude θ 98

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