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School of Electrical Systems Engineering ABD RAHIM 2008.

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Presentation on theme: "School of Electrical Systems Engineering ABD RAHIM 2008."— Presentation transcript:

1 School of Electrical Systems Engineering ABD RAHIM 2008

2 simply means of an electro-mechanical switch which is opened and closed to stop and start the motor School of Electrical Systems Engineering ABD RAHIM 2008

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4 School of Electrical Systems Engineering ABD RAHIM 2008 - Low cost and simple -Torque too high – causes snatch - Torque too low – motor stalls - Motor can stall in transition

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10 School of Electrical Systems Engineering ABD RAHIM 2008 Breakdown torque peak is shifted to zero speed by increasing rotor resistance. Wound rotor induction motor.

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13 School of Electrical Systems Engineering ABD RAHIM 2008 Advantages : Reduced starting current Reduced starting torque Less mechanical stress Improved control of acceleration and deceleration (to apply an adjustable voltage to the motor and increase this voltage gradually over a user-selectable acceleration period)

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15 School of Electrical Systems Engineering ABD RAHIM 2008 Equivalent circuits … INDUCTION MOTORS -S-Similar circuit with Transformer’s -T-There are no internal voltage source E A because it doesn’t have any independent field circuit. -T-The only different is that the secondary (rotor part) is moveable thus it will affect the E R, and impedance R R and jX R rotorstator

16 School of Electrical Systems Engineering ABD RAHIM 2008 Rotor circuit model… INDUCTION MOTORS -So does the frequency of the induced voltage : Rotor reactance : But, statorrotor -T-The voltage magnitude and frequency of induced voltage on the rotor E R directly proportional to the slip of the motor. -E-E LR - voltage at Locked Rotor condition

17 School of Electrical Systems Engineering ABD RAHIM 2008 Rotor circuit model.. INDUCTION MOTORS -With above circuit, rotor current IR is: Z R, eq

18 School of Electrical Systems Engineering ABD RAHIM 2008 Final Rotor equivalent circuit …. INDUCTION MOTORS

19 School of Electrical Systems Engineering ABD RAHIM 2008 Per-phase equivalent circuit …. INDUCTION MOTORS (Chapman ) (Rashid) or

20 School of Electrical Systems Engineering ABD RAHIM 2008 Torque & Power-flow diagram …. INDUCTION MOTORS or P SCL = 3Is 2 Rs 3 phase Source – Y connected P core =3V m 2 /R m ≈3V s 2 /R m P RCL =3I r 2 R r P AG or P g =3I r 2 R r /s P conv or P dev =P AG -P RCL =(1-s)P AG τ ind or τ dev =P conv /ω m P out =P conv -P F&W -P misc

21 School of Electrical Systems Engineering ABD RAHIM 2008 Power Loss AreaEfficiency Improvement 1. Fixed loss (iron) Use of thinner gauge, lower loss core steel reduces eddy current losses. Longer core adds more steel to the design, which reduces losses due to lower operating flux densities. 2. Stator I2R Use of more copper & larger conductors increases cross sectional area of stator windings. This lower resistance (R) of the windings & reduces losses due to current flow (I) 3 Rotor I2R Use of larger rotor conductor bars increases size of cross section, lowering conductor resistance (R) & losses due to current flow (I) 4 Friction & Winding Use of low loss fan design reduces losses due to air movement 5. Stray Load Loss Use of optimized design & strict quality control procedures minimizes stray load losses

22 School of Electrical Systems Engineering ABD RAHIM 2008 Torque & Power-flow diagram …. INDUCTION MOTORS Stated that, τ ind =P conv /ω m =(1 - s)P AG / (1 - s)ω sync τ ind =P AG / ω sync Substitution … Thus, τ load =P out / ω m While,

23 School of Electrical Systems Engineering ABD RAHIM 2008 Example 7.2: INDUCTION MOTORS Ans: a)38.6kW b)37.9kW c)37.3kW d)88%

24 School of Electrical Systems Engineering ABD RAHIM 2008 Example 7.3: INDUCTION MOTORS a) ns:1800rpm,ws:188.5rad/s,nm:1760rpm, wm:184.4rpm b) Is:18.88L-33.6A c)0.833lagging d) Pconv:11585W,Pout:10.485W e)Tind:62.8Nm, Tload:56.9Nm f) n:83.7%

25 School of Electrical Systems Engineering ABD RAHIM 2008 Per-phase equivalent circuit …. INDUCTION MOTORS (Rashid) In real world: Value of X m is very large (R m can be omitted) X m 2 >> (R s 2 + X s 2 )  Vs ≈Vm Thus, for circuit simplification, the magnetizing reactance X m can be moved-out to the stator winding.

26 School of Electrical Systems Engineering ABD RAHIM 2008 INDUCTION MOTORS The input impedance of the motor becomes: Power factor angle:

27 School of Electrical Systems Engineering ABD RAHIM 2008 INDUCTION MOTORS Rotor current (rms): P AG or P g =3I r 2 R r /s Previously noted, And developed torque τ dev or, τ ind =P dev / ω m = P AG / ω sync Thus, --(15-18)

28 School of Electrical Systems Engineering ABD RAHIM 2008 INDUCTION MOTORS

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30 School of Electrical Systems Engineering ABD RAHIM 2008 Extended Torque-speed characteristic.. The curve is nearly linear up to full load (Rr >> Xr, thus Ir,Br and Tind rise ∞ increasing slip). Pullout torque or breakdown torque = max possible handling torque. If speed driven faster than ns the Tind reverses and motor becomes generator. Plugging = reversing of magnetic field rotation by switching any two stator phases. The reversed Tind will stop the motor rapidly and rotate it to reverse direction. s

31 At start: high current and low “pull-up” torque School of Electrical Systems Engineering ABD RAHIM 2008 INDUCTION MOTORS At start: high current and low “pull-up” torque At 80% of full speed: highest “pull-out” torque and current drops At full or synchronous speed: torque and stator current are zero Torque-speed characteristic..

32 School of Electrical Systems Engineering ABD RAHIM 2008 LRT is higher than 100% of the FLT Starting current at LRC) may reach 1000% of FLC Once rotor starts to rotate the torque may decrease a bit for certain classes of motors to a value known as the pull up torque. breakdown torque is due to the larger than normal 20% slip. Slip will be only a few percent during normal operation. Any motor torque load above the breakdown torque will stall the motor. The torque, slip, and current will approach zero for a “no mechanical torque” load condition. This condition is analogous to an open secondary transformer.

33 School of Electrical Systems Engineering ABD RAHIM 2008 INDUCTION MOTORS Characteristics for NEMA designs…. All motors, except class D, operate at %5 slip or less at full load. Class B (IEC Class N) motors are the default motor to use in most applications. With a starting torque of LRT = 150% to 170% of FLT, it can start most loads, without excessive starting current (LRT). Efficiency and power factor are high. It typically drives pumps, fans, and machine tools. Class A starting torque is the same as class B. Drop out torque and starting current (LRT)are higher. This motor handles transient overloads as encountered in injection molding machines. as encountered in injection molding machines.

34 School of Electrical Systems Engineering ABD RAHIM 2008 INDUCTION MOTORS Characteristics for NEMA designs…. Class C (IEC Class H) has higher starting torque than class A and B at LRT = 200% of FLT. This motor is applied to hard-starting loads which need to be driven at constant speed like conveyors, crushers, and reciprocating pumps and compressors. Class D motors have the highest starting torque (LRT) coupled with low starting current due to high slip ( 5% to 13% at FLT). The high slip results in lower speed. Speed regulation is poor. However, the motor excels at driving highly variable speed loads like those requiring an energy storage flywheel. Applications include punch presses, shears, and elevators. Class E motors are a higher efficiency version of class B. Class F motors have much lower LRC, LRT, and break down torque than class B. They drive constant easily started loads.

35 School of Electrical Systems Engineering ABD RAHIM 2008 INDUCTION MOTORS 1)1.67% 2)48.6Nm 3)2900rpm 4)29.5kW


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