Ee 212 Ac Induction Motors

E E 212 AC MOTORS INDUCTION MOTORS Like D. C. Motors, A. C motors also comprise of stator and rotor both of which carry windings. The rotor core is laminated and its conductors often consist of un-insulated copper/aluminum bars in semi-enclosed slots. In some induction motors, these conductors are short-circuited at the ends by rings or places. Such a rotor is also known as Cage or short-circuited rotor. On the other hand, a wound rotor machine employs slip-rings and normally has equal number of rotor and stator poles. Principle of Action: The air gap between the stator and rotor is uniform and made as small as possible. When current flows in the stator, the stator currents produce flux. This flux easily crosses the narrow air gap in induce/generate on e.m.f. in the rotor conductors. The e.m.f. is maximum in regions under the poles (maximum flux density). Assuming an anticlockwise rotation of the flux as shown below:

N Rotor Х Force on conductor

Stator p

Force on conductor

Rotation of the flux

S

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The directions of the generated emf, in the rotor conductors is determined by the right hand rule. These e.m.f., circulate a current which strengthens the flux density on one side of the rotor conductor. This consequently exerts a force on the rotor tending to rotate it in the direction of the rotating flux. This summaries one great advantage three-phase a.c. motors have over single-phase a.c. motors; they are self-starting and this is due to the fact that 3 phase flux is rotating (also has a constant value of 1.5 times the maximum flux per phase). The higher the speed of the rotating field relative to the rotor winding, the higher the e.m.f. induced in the latter and vice-versa. Thus, as the rotor speed increases, the speed of the rotating flux relative to the rotor conductors decreases and hence the induced e.m.f. decreases. The speed at which the flux/field rotates is called the Synchronous speed. As the speed of the rotor increases to its synchronous value, the rotor conductors would appear to be stationery relative to the rotating flux Under this situation, no e.m.f/current will be in the rotor conductors and consequently no torque develops on the rotor. Hence the rotor does not continue to rotate at synchronous speed. As the rotor speed falls below the synchronous speed, the current/emf and hence torque increases until the torque is enough to rotate any connected load and also overcoming the rotor losses. Definition: The speed of the rotor relative to that of the rotating flux (synchronous speed) is termed as “Slip”. Rotor speed/ Slip B Rotor speed Slip

O Thus for torque OA;

Where AD = AB-AC, i.e. BC = AD

C D

A A

Torque NM

AB = synchronous speed AC = Rotor speed AD = Slip

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Slip is expressed as a fraction or as a percentage of the synchronous speed. Therefore, p.u slip = slip (rpm)/sunchronous speed (rpm) = AD/AB. = (Synchronous speed – Rotor speed)/Synchronous speed. Let s = p.u slip, N = synchronous speed and Nr = rotor speed; By definition, s = (N-Nr)/N or (N-Nr)/N x 100% At full-load, s is very small and thus induction motors are regarded practically as a constant speed machine and varying its speed is economically expensive. This is a disadvantage. 3-Phase Rotating Flux Consider a 3-phase arrangement of which each phase is assumed to produce a sinusoidal space distribution of flux with maximum value Фmax If the phases are A, B and C, then the respective fluxes are  A,  B, and  C respectively. These three phases are 1200 (electrical degrees) apart; hence if ωs is the synchronous speed, then,  A = Фmax sin ωst ,  B, = Фmax sin(ωst - 1200) , and  C = Фmax sin(ωst - 2400) If the angle θ around the air-gap is taken with its origin on the axis of one phase say phase A, then for any point at angle θ from the origin, the fluxes are given by  Asin θ,  Bsin(θ - 1200) , and  Csin(θ - 2400). The total flux is given by ФTotal =  Asin θ +  Bsin(θ - 1200) +  Csin(θ - 2400). = Фmax sin ωst sin θ + Фmax sin(ωst - 1200) sin(θ - 1200) + Фmax sin(ωst 2400)sin(θ - 2400) = Фmax [sin ωst sin θ + sin(ωst - 1200) sin(θ - 1200) + sin(ωst - 2400)sin(θ 2400)] Using the identity 2sinAsinB = cos(A-B) – cos(A+B), where A = f(θ) and B = f(ωs) ФTotal =½ Фmax [cos(θ - ωst) – cos(θ + ωst) + cos(θ - ωst) – cos(θ + ωst - 2400) + cos(θ - ωst) – cos(θ + ωst - 4800)]

Generally, cosα + cos(α - 2400) + cos(α - 4800) = 0 Thus, ФTotal =3 x ½ Фmax cos(θ - ωst) or 1.5 Фmax cos(θ - ωst)

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The above equation shows that the total flux has constant amplitude of 1.5 Фmax, is a sinusoidal function of the phase angle θ and rotates around the air-gap in synchronism with the supply frequency at a speed ωs. For a general case of a p-pole pair machine supplied at f Hz, the speed Ns known as the synchronous speed will be given by Ns = 60f/p revolutions per minute. Rotor Constants a)

Rotor emf:

When an induction motor is stationery, the stator and rotor windings form an equivalent transformer. If Es = stator e.m.f. and ER = the rotor e.m.f., then ER = (Nr/ Ns)x Es = nEs where n = equivalent rotor-stator turns ratio. When motor runs, the induced e.m.f in the rotor becomes less, since the relative movement between the rotor conductors and the rotating field is low. The induced e.m.f is proportional to this movement and hence proportional to slip, s. When motor is rotating: Rotor emf per phase = Er = sER = nsEs. b)

Rotor frequency The rotor emf is induced by an alternating flux and the rate at which this flux cuts (passes) the conductors is the slip speed, (N s – N). Thus the frequency of the rotor emf is: fr = (Ns – N) x P/60 = [(Ns – N)/Ns] x (Nsp)/60 But (Ns – N)/Ns = s and (Nsp)/60 = f Hence fr = s.f

c)

Rotor Resistance

The resistance of the rotor winding is independent of the frequency and the slip. Thus ignoring the temperature effect, the rotor resistance per phase remains unchanged. d)

Rotor reactance

Rotor reactance depends on the frequency of the rotor current. At standstill, the reactance per phase X = 2π fL When running, reactance per phase Xr = 2π fr L

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But fr = s.f Hence Xr = 2π x sf L = s x 2π f L; or Xr = s.X e) Rotor Impedance Rotor impedance per phase = Zr = [R2 + (sX)2]½ Since at stand still, s = 1, then the respective impedance (Standstill impedance) is Z = [R2 + X2]½ f) Rotor Current The rotor current per phase at stand still (I) and when running at slip(s) (I r) are given by: I = ER/Z = n.Es/[R2 + X2]½ and Ir = Er/Zr = n.sEs/[R2 + (sX)2]½ Relationship between the Rotor I2R Loss and the Rotor Slip The following figure shows the flow of power supplied to the motor stator to the rotor shaft Input power to stator winding

Power transferred to the Rotor via the magnetic field of air gap

Total mechanical power developed by the Rotor

Useful mechanical power obtained from the Rotor shaft

Stator core loss

Core loss in the rotor core.

Windage losses

I2R loss in the stator winding

I2R loss in the Rotor windings

Frictional losses

Let T = torque in NM on the rotor by rotating flux; n = synchronous speed in rps The power transferred from the stator to rotor = 2πnT Watts. If nr = rotor speed in rps, the total mechanical power developed by the rotor is 2πnrT Watts.

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Assuming the rotor core losses are negligible and 0, 2 Then the total I R loss in the rotor ≈ (Power transferred from stator to rotor) – (total mechanical power developed by the rotor). ∴ I2R loss = 2πT (n-nr) Watts ∴

Total rotor I2R loss = 2πT (n-nr) = n-nr = slip s. Input power to rotor 2πnT n

Or Rotor I2R loss = s x Input power to rotor Examples Example 1 A 3-ph 50 Hz induction motor has 4 poles and runs at a speed of 1440 rpm when the torque developed by the rotor is 70NM. Calculate a) the total input to the rotor in kW; b) the rotor I2R loss in watts. Solution: f = 50 Hz, p = 2 (4 poles), N = 1440 rpm and T = 70NM By definition, Ns =60f/p = (60 x 50)/2 = 1500 rpm Therefore, slip s = (Ns – N)/Ns = (1500 – 1440) ÷ 1500 = 0.04 pu Mechanical power at the rotor = 2πNrT/60 = (2π x 1440 x 70) ÷ 60 = 10560 Watts Let Pi = input power to the rotor, Pi - Pmechanical = I2R = s x Pi Hence, Pi = Pm /(1-s) = 10560 ÷ (1- 0.04) = 11kW Rotor copper loss = I2R = s x Pi = 0.04 x 11000 = 440 Watts. Example 2 Determine the efficiency and output kW of a 3-ph, 400V induction motor running light on load with a slip of 0.04 and taking a current of 50A at a p.f of 0.86. When running light at 400V, the motor has an input current of 15A and power taken is 2000W of which 650W represents the frictional, windage and rotor core losses. The resistance per phase of the stator windings (delta connected) is 0.5Ω. Solution At no load: Running light: Losses:

s = 0.04, V = 400V and Ir = 50A at p.f of 0.86 Iin = 15A, Pin = 2000W, V = 400V 650W, R = 0.5Ω/ph

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Stator input power = √3 VL IL cosφ = √3 x 400 x 50 x 0.86 = 29791 (≈ 30000)Watts. Stator losses = I2R = 502 x 0.5 = 1250 Watts/ph Frictional and other losses = 650Watts Power input to the rotor = 2000 – 650 = 1350 W ∴ Rotor I2R loss = s x Pinput to rotor = 0.04 x 1350 = 54W Total loss = 54 + 650 + (1250 x 3) = 4454W Efficiency, η, = Pout/Pin = 1 – (Total losses/Pin) = 1- (4454/30000) pu or 85.2%

=

0.852

Power output = Power input – Total losses = 29800 – 4454 = 25.34 kW Example 3 A 3-ph 50 Hz, 6-pole induction motor has a slip of 0.04pu when the output is 20kW. The frictional loss is 250W. Calculate: a) the rotor speed; b) the rotor I2R loss Solution: f = 50Hz, Pout = 20000W, frictional loss = 250W, p = 3 (6 poles) and s = 0.04 Let N = synchronous speed and Nr = rotor speed Thus, N = 60f/p = 60x50/3 = 1000 rpm. From (N – Nr)/N = slip s; then, Rotor speed Nr = N(1-s) = 1000(1-0.04) = 960 rpm. Let P = power input into the rotor Pm = mechanical power output and frictional loss, Then, Rotor copper loss = I2R = P - Pm But also, I2R = sP Thus P = Pm /(1-s) = (20000 + 250) ÷ (1 – 0.04) = 21093.75 W Therefore, the rotor copper loss = s x P = 0.04 x 21093.75 = 844W. Torque of Induction Motors Let k = number of rotor phases; Er = rotor e.m.f; Ir = rotor current; and Фr = phase difference between Er and Ir.

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The power generated in the rotor = k x Er x Ir cosФr Watts. But cosФr = R/[R2 + (sXo)2]½, Er = sEo and Ir =sEo/[R2 + (sXo)2]½ Thus the power generated in the rotor = (k x s2 Eo2x R)/[R2 + (sXo)2] All this power is dissipated as I2R loss in the rotor windings. Since power input to the rotor = 2πnT Watts,then, I2R loss = s x 2πnT Thus s x 2πnT = (k x s2 Eo2x R)/[R2 + (sXo)2] Or T = (k/2πn) x (sEo2x R)/[R2 + (sXo)2] = C. (sEo2x R)/[R2 + (sXo)2] since n and k are constants for any supply. Also, Eo ∝ Ф, Thus T ∝ (sEo2x R)/[R2 + (sXo)2] ∝ (s Ф 2x R)/[R2 + (sXo)2] Variation of Torque with Slip For a given supply voltage, Ф and Eo remain constant. So, Torque ∝ (sR)/[R2 + (sXo)2] where Xo is always > R of rotor windings. For illustration, take values of R = 1Ω and Xo = 8Ω and calculate the value of (sR)/[R2 + (sXo)2] for various values of slip ,s, between 1 (standstill) and 0 (synchronous speed). Afterwards take other set of values of R = 2, 3, 4, ……., 8Ω and Xo = 8Ω for various values of s. Analysis of Torque ∝ (sR)/[R2 + (sXo)2] For small values of s, R2 >>> (sXo)2 Thus, Torque ∝ (sR)/[R2] ∝ s/R or T ∝ s, since R is constant. This is a linear relationship and approximates to a straight line. When s is high i.e. approaching 1pu, (sXo)2 >>> R2 Thus, Torque ∝ (sR)/[(sXo)2] ∝ R/sXo or T ∝ 1/s since R and Xo are all constants. This gives an inverse relationship. Torque-Slip curves for an Induction Motor Torque Tmax

R = 8; Xo = 8

R = 4; Xo = 8 R = 2; Xo = 8

8

O

R = 8; Xo = 8

s=1

slip

Example 4 In a certain 8pole, 50Hz induction machine, the rotor resistance per phase is 0.04Ω and the maximum torque occurs at a speed of 645rpm. Assuming the flux is constant at all loads, determine the percentage of the maximum torque (i) at starting; (ii) when the slip is 3%. Solution Poles = 8; and so p = 4, 645rpm

f = 50Hz

R = 0.4/Ωph Tmax is when speed =

Ns = (50 x 60)/4 = 750rpm Hence when running at 645rpm, slip ,s, = (750 – 645)/750 = 0.14 Using T = (k.s.Eo2 R) ÷ [R2 + (sX)2], Tmax = (k1 x 0.14 x 0.04) ÷ (2 x 0.042) where k1 = k. Eo2, and R = (sX) = 1.75k1 NM Using R = sXo, then Xo = R/s = 0.4/014 = 0.286Ω (i) At starting, s = 1 and so Tstarting = (k1 x 0.04 x 1) ÷ [0.042 + 0.2862] = 0.48k1 NM Percentage of maximum torque at starting = Tstarting/Tmax = 0.48k1/1.75k1 = 0.274 or 27.4% (ii) When slip s = 3% = 0.03pu, Then T = (k1 x 0.03 x0.04) ÷ [0.042 + (0.03 x 0.286)2] = 0.717k1 NM Percentage of maximum torque at s = 3% = (0.717k1/1.75k1) = 40.97%

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