Rotating MMF Wave from Three Phase Winding Stator MMF of Three Phases (1) θb θa θc 2π θ be = θ ae − 3 2π θ ce = θ ae + 3 Stator MMF of Three Phases (2) Fa = Fb = Fc = P θa ) 2 ∑ Fah Fah = Fh ia cos(hθ ae ) = Fh ia cos(h ∑ Fbh P 2π 2π Fbh = Fh ib cos hθ ae − F i h θ cos = − h b 2 a 3 3 ∑ Fch P 2π 2π Fch = Fh ic cos hθ ae + = Fh ic cos h θ a + 3 2 3 h =1, 3, 5... h =1, 3, 5... h =1, 3, 5... Total MMF from stator winding: Fstator = Fa + Fb + Fc = P 2hπ F i i i h ( ) cos cos( θa ) + + + ∑ h a b c 3 2 h =1, 3, 5... (ib − ic )sin 2hπ sin(h P θ a ) 3 2 Stator MMF of Three Phases (3) For h = 1, 5, 7, 11 … cos 2 hπ 1 =− 3 2 2hπ 3 = 3 2 For h = 1, 7, 13 … sin For h = 5, 11, 17 … 2 hπ 3 sin =− 3 2 For h = 3, 9, … cos 2 hπ =1 3 sin 2 hπ =0 3 For Y connected windings without neutral return: ia + ib + ic = 0 Stator MMF of Three Phases (4) Fstator P P 3 3 (ib − ic )sin(h θ a ) + = ∑ Fh ia cos(h θ a ) + 2 2 2 2 h =1, 7 ,13... P P 3 3 ( ) F i h i i h cos( ) sin( ) θ θ − − ∑ h a a b c a 2 2 2 2 h = 5 ,11,17... Note that the third harmonic components have been eliminated if the machine is Y connected without neutral return. Consider balanced three phase sinusoidal currents: = ia i pk cos(ωet + θi ) 2π ) = ib i pk cos(ωet + θi − 3 2π ) = ic i pk cos(ωet + θi + 3 ⇒ 3 3 = i pk sin(ωet + θi ) (ib − ic ) 2 2 Note: ipk may change with time. It’s a constant Ipk at steady state. Stator MMF of Three Phases (5) Fstator h P 3 − ω θ a + θi ) + F i t h cos( ∑ h pk e 2 2 = 1,7,13... h wave rotating in forward direction (counterclockwise) P 3 + ω θ a + θi ) F i t h cos( ∑ h pk e 2 5,11,17... 2 wave rotating in backward direction (clockwise) From: P θa θa = ωet h θ a ω= t ω t e e ω ω hP 2 2 / e h 2ω ⇒ ωh =e hP angular velocity of the hth harmonic Physical Understanding of Wave (1) z f (t − ) vp z f (− ) vp z f (t − ) vp = f (− z − v pt vp ) 0 z z1 = v p t z f (t − ) : stands for a wave propagating in + z direction with velocity v p . vp “forward wave” Likewise: z g (t + ) : stands for a wave propagating in − z direction with velocity v p . vp “backward wave” Physical Understanding of Wave (2) Plot u + ( z , t ) = A cos(ωt − kz ) A 0 -A z Double Layer Lap Winding Scheme Double Layer Lap Winding Scheme (1) Step 1. Find the nominal coil span in slots S S= −1 If S/P is an integer c P S If S/P is not an integer Sc = max fix ,1 P Step 2. Calculate the relative electric angle of In slot of all coils The relative electrical angle (expressed in the range of -180o to 180o ) of the kth slot is: θ slot = (k ) mod (k − 1)γ + 180 ,360 − 180 Of course, the relative electrical angle of In slot of the kth coil is also: θ coil = (k ) mod (k − 1)γ + 180 ,360 − 180 Step 3. Readjust In slot angles if their magnitude are greater than 90o For coil angles having a magnitude greater than 90o , the coil direction is reversed; thereby changing the angle by 180o. Double Layer Lap Winding Scheme (2) Step 4. Picking up (S/m) coils for Phase A For the angles calculated in last step, pick up those (S/m) close to 0o as phase A coils. Step 5. Calculate the slot offset to wind other phases From 360 mod Soff γ ,360 = m ⇒ Soff P 360 γ= 2 S S = 2 (1 + mn ) = 2q(1 + mn ) mP where n is a integer value that makes Soff also an integer. Phase B will start from slot (mod(Soff ,S)+1). Step 6. Check to find out whether the winding is valid. The winding is valid if all slots contain two coil sides each. If it’s not valid, go back to Step 4 and pick up coils for Phase A. Example 1: 3 Phase, 4 Pole, 15 Slot Machine (1) Step 1. Find the nominal coil span in slots 15 = Sc max = fix 4 ,1 3 = q 15 = 1.25 3× 4 Step 2. Calculate the relative electric angle of In slot of all coils Step 3. Readjust In slot angles if their magnitude are greater than 90o Example 1: 3 Phase, 4 Pole, 15 Slot Machine (2) Step 4. Picking up (S/m) coils for Phase A Step 5. Calculate the slot offset and wind other phases Soff = 2q(1 + mn ) = 2 × 1.25 × (1 + 3 × 1) = 10 Phase B will start from slot 11 Phase C will start from slot mod(11+10, 15) = 6. Example 1: 3 Phase, 4 Pole, 15 Slot Machine (3) Step 6. Check and find out the winding is valid. Example 2: 3 Phase, 10 Pole, 12 Slot Machine (1) Step 1. Find the nominal coil span in slots 12 = Sc max = fix 10 ,1 1 = q 12 = 0.4 3 × 10 Step 2. Calculate the relative electric angle of In slot of all coils Step 3. Readjust In slot angles if their magnitude are greater than 90o Example 2: 3 Phase, 10 Pole, 12 Slot Machine (2) Step 4. Picking up (S/m) coils for Phase A Step 5. Calculate the slot offset and wind other phases Soff = 2q(1 + mn ) = 2 × 0.4 × (1 + 3 × 3) = 8 Phase B will start from slot 9 Phase C will start from slot mod(9+8, 12) = 5 Example 2: 3 Phase, 10 Pole, 12 Slot Machine (3) Phase C – Group 1 5 6 4Phase B – Group 2 3 5 4 6 3 7 7 Phase A – Group 2 2 2 8 Phase A – Group 1 1 8 9 1 10 9 Phase B – Group 1 10 12 11 11 12 Phase C – Group 2 Example 2: 3 Phase, 10 Pole, 12 Slot Machine (4) Slot Number Phase A 1 In 2 Out & Out 3 In Phase B Phase C Out Out 4 In & In 5 Out 6 In Out & Out 7 Out 8 In & In 9 Out In In 10 Out & Out 11 In 12 Step 6. Check and find out the winding is valid. Out In & In Slot Pole Combination for Three Phase Machine S/P Ratios for Three Phase Machine Winding Diagram Example 1 3 Phase, 24 Slot, 2 Pole, 5/6 Pitch Three Phase Windings 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 B2 A1 C2 B1 A2 C1 Phase A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Phase B 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Phase C 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Example 2 3 Phase, 36 Slot, 4 Pole, 8/9 Pitch Three Phase Windings 1 2 3 4 B2 5 6 A1 7 8 9 C2 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 B1 C1 A2 Phase A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 Phase B 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 Phase C 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36