IV. Three-Phase Induction Machines. Induction Machines

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1 IV. Three-Phase Induction Machines Induction Machines 1

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14 Example 1: A 480V, 60 Hz, 6-pole, three-phase, delta-connected induction motor has the following parameters: R 1 =0.461 Ω, R 2 =0.258 Ω, X 1 =0.507 Ω, X 2 =0.309 Ω, X m =30.74 Ω Rotational losses are 2450W. The motor drives a mechanical load at a speed of 1170 rpm. Calculate the following information: i. Synchronous speed in rpm ii. slip iii. Line Current iv. Input Power v. Airgap Power vi. Torque Developed vii. Output Power in Hp viii. Efficiency This machine has no iron loss resistance, so the equivalent circuit is as follows: i. Synchronous speed is given by: ii. Slip is given by Using the rpm equation, iii. Now, phase current is given by s = ( )/1200 =

15 where phase impedance is given by Using the above equation, Z in = j3.84 Ω And noting that the machine is delta connected, V 1 = V LL = 480V iv. Input power is given by: I 1 = j17.4 A. I 1 =46.6 A, θ = Therefore I L = = 80.6 A P in = 62.2 kw v. Airgap power is the input power minus stator losses. In this case the core losses are grouped with rotational loss. Therefore P gap = 62.2 kw = 59.2 kw P gap vi. Torque developed can be found from where synchronous speed in radians per second is given by giving τ = 471 Nm Vii.Output power in horsepower is the output power in Watts divided by 746. (there are 746 W in one Hp). and 15

16 Therefore output power in Watts is: P out = 55.3kW Viii. Efficiency is given by η = 55.3/62.2 = 88.9% 16

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21 Using the induction machine power and torque equations it is possible to produce the torque speed curve shown below. Operating Regions The torque-speed curve brakes down into three operating regions: 1. Braking, n m < 0, s > 1 Torque is positive whilst speed is negative. Considering the power conversion equation it can be seen that if the power converted is negative (from P = τ ω) then the airgap power is positive. i.e. the power is flowing from the stator to the rotor and also into the rotor from the mechanical system. This operation is also called plugging. This mode of operation can be used to quickly stop a machine. If a motor is travelling forwards it can be stopped by interchanging the connections to two of the three phases. Switching two phases has the result of changing the direction of motion of the stator magnetic field, effectively putting the machine into braking mode in the opposite direction. 2. Motoring, 0 < n m < n s, 1 > s > 0 Torque and motion are in the same direction. This is the most common mode of operation. 21

22 3. Generating, n m > n s, s < 0 In this mode, again torque is positive whilst speed is negative. However, unlike plugging, indicates that if the power converted is negative, so is the air gap power. In this case, power flows from the mechanical system, to the rotor circuit, then across the air gap to the stator circuit and external electrical system. Motoring Torque Characteristic The motoring region of the induction machine torque-speed curve is the region of greatest interest. Starting torque Torque at zero speed Typically 1.5 times the full-load torque Pull-up torque The minimum torque developed by the motor while accelerating from zero speed Greater than full-load torque; less than starting torque Breakdown torque The maximum torque that the motor can develop Typically 2.5 times the full-load torque Normal operation At full-load, the motor runs at n rpm Rotor speed decreases slightly from synchronous speed with increasing load torque Motor will stall when the load torque exceeds the breakdown torque 22

23 The torque equation Using the equation Multiple solutions of the above equation for torque at different slips can be made simpler by simplifying the equivalent circuit model. Consider the diagram below: The stator part of the equivalent circuit (together with the magnetising branch) can be replaced by a Thevenin equivalent circuit. In the Thevenin circuit, the stator phase voltage has been replaced by its Thevenin equivalent, 23

24 and the impedances have been replaced by Thevenin equivalent impedances. Incorporating the Thevenin model into the circuit model results in the Thevenin equivalent circuit model of an induction machine. In the above circuit, the calculation of rotor current is greatly simplified The above expression for rotor current can be squared and substituted into the torque equation Using the above equation, the variation of torque with slip can be plotted directly. Note that if power or efficiency calculations are needed, the full equivalent circuit model should be used (not the Thevenin version). since synchonous speed is constant, maximum torque occurs at the same slip as maximum airgap power. Considering the Thevenin circuit, and applying maximum power transfer theory, maximum airgap power and maximum torque will occur when RR 2 ss = RR TTTT 2 + (XX TTTT + XX 2 ) 2 24

25 Re-arranging it is possible to obtain the slip for maxiumum torque, or pullout torque. SS mmmmmm = RR 2 2 RR TTTT + (XX TTTT + XX 2 ) 2 Substituting the pullout slip into the Thevenin torque equation: TT mmmmmm = 2 3VV TTTT 2 2ωω ssssssss RR TTTT + RR TTTT + (XX TTTT + XX 2 ) 2 The Starting torque 2 The starting torque is proportional to VV TTTT TT ssssssssss = 3VV 2 TTTT RR 2 ωω ssssssss [(RR TTTT +RR 2 ) 2 +(XX TTTT +XX 2 ) 2 ] The starting torque will be reduced if the stator and rotor leakage inductances are increased The starting torque will be reduced if the stator frequency is increased When the rotor resistance is increased, the starting torque will first increase and then decrease. TT ssssssssss mmmmmm = TT mmmmmm when RR 2= XX 1 + XX 2 25

26 Equivalent Circuit Model Analysis Example A 480V, 60 Hz, 6-pole, three-phase, delta-connected induction motor has the following parameters: R 1 =0.461 Ω, R 2 =0.258 Ω, X 1 =0.507 Ω, X 2 =0.309 Ω, X m =30.74 Ω Rotational losses are 2450W. The motor drives a mechanical load at a speed of 1170 rpm. Find: i. Thevenin circuit parameters and Thevenin voltage ii. Pullout slip iii. Pullout Torque iv. Start Torque Solution: Using Matlab or Excel (or another computer program) plot the torque speed curve for slip in the range 0 to 1 i. Thevenin circuit parameters and Thevenin voltage: The Thevenin voltage is the voltage applied to the rotor assuming that the rotor current is zero. Thevenin impedance is the impedance of the stator part of the circuit, seen from the rotor, assuming that the stator supply is short circuited. i. Substituting the equivalent circuit parameters in to the above equations gives: V TH = V, R TH = 0.452Ω, X TH = 0.313Ω i. Pullout slip The slip at which maximum torque occurs can be found from maximum power transfer theory. Maximum torque and maximum airgap power occur at the same slip, therefore maximum torque occurs when 26

27 ii. Pullout Torque Pullout torque can be found by substituting the above pullout slip into the Thevenin torque equation or from the maximum torque equation directly Substituting into the above equation: i. Start Torque Start torque can be found by setting s=1 in the above equation for torque. 27

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29 Investigating the torque-speed curve it is apparent that the rotor circuit resistance has significant impact on speed at which maximum torque occurs. The plots below illustrate two cases, with low rotor resistance on the left and high rotor resistance on the right. Example 1: A 460-V, 25-hp, 60-Hz, four-pole, Y-connected wound-rotor induction motor has the following impedances in ohms per phase referred to the stator circuit: RR 11 = ΩΩ RR 22 = ΩΩ XX 11 = ΩΩ XX 22 = ΩΩ XX MM = ΩΩ a) What is the maximum torque of this motor? At what speed and slip does it occur? b) What is the starting torque of this motor? c) When the rotor resistance is doubled, what is the speed at which the maximum torque now occurs? What is the new starting torque of the motor? Solution: The Thevenin voltage of this motor is VV TTTT = VV φφ XX MM = RR 1 2 +(XX 1 +XX MM ) 2 (266 VV)(26.3 ΩΩ ) ( ΩΩ ) 2 +( ΩΩ ΩΩ ) 2 = 255.2VV The Thevenin resistance is RR TTTT RR 1 XX MM XX 1 +XX MM ΩΩ ( ΩΩ) ΩΩ ΩΩ 2 = 0.590ΩΩ The Thevenin reactance is XX TTh XX ΩΩ (a) The slip at which maximum torque occurs is given by Equation (2): RR SS mmmmmm = ΩΩ = (0.590ΩΩ ) 2 +(1.106ΩΩ ΩΩ ) 2 = RR 2 TTTT +(XX TTTT +XX 2 ) 2 This corresponds to a mechanical speed of nn mm = (1 ss)nn ssssssss h = ( )(1800 rr/mmmmmm) = 1444 rr mmmmmm 29

30 The torque at this speed is TT mmmmmm = 2 3VV TTTT 2ωω ssssssss RR TTTT + RR 2 TTTT +(XX TTTT +XX 2 ) 2 = 3(255.2 VV) 2 2(188.5 rrrrrr /ss)[(0.590ωω) 2 +(1.106ΩΩ ) 2 ] = 229 NN. mm (b) The starting torque of this motor is found by setting s=1 in Equation (1) : TT ssssssssss = 2 3VV TTTT RR 2 ωω ssssssss [(RR TTHH + RR 2 ) 2 + (XX TTTT + XX 2 ) 2 ] 3(255.2 VV) 2 ( ΩΩ) = (188.5 rrrrrr ss)[(0.590ωω ΩΩ) 2 + (1.106ΩΩ ) 2 = 104 NN. mm ] (c ) If the rotor resistance is doubled, then the slip at maximum torque doubles, too. Therefore, ss mmmmmm = and the speed at maximum torque is The maximum torque is still The starting torque is now nn mm = (1 ss)nn ssssssss = ( )(1800 rr mmmmmm) = 1087 rr mmmmmm TT mmmmmm = 229 NN. mm 3(255.2VV) 2 (0.664ΩΩ) TT ssssssssss = (188.5 rrrrdd ss)[(0.590ωω ) 2 + (1.106ΩΩ ) 2 = 170NN. mm ] 30

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33 When the switch S is in the START position, the stator windings are connected in STAR When the motor picks up speed, say 80% of its rated value, the switch S is thrown quickly to the RUN position which connects the stator windings in DELTA. In induction motors that are designed to operate with delta stator connection it is possible, during starting, to reduce the phase voltage by switching to Y- connection During Y- connection, the phase voltage V s becomes so the phase current, for same slip, I sy, is reduced 3 times 33

34 Now the line current in Δ connection I lδ is So the line current is three times smaller for Y- connection. The torque is proportional to phase voltage squared Therefore, the Y-delta starting is equivalent to in torque. a reduction of phase voltage and a 3 to 1 reduction 34

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45 Fig.1 Slip ring IM Fig.2 Torque-slip curves 45

46 Fig.3 Torque-speed curves 46