48 PROBLEM SOLUTIONS: Chapter 4 Problem 4.1 ω m = 100 π/30 = 40π rad/sec part (b): 60 Hz; 10π rad/sec part (c): 100 5/6 = 1000 r/min Problem 4. The voltages in the remaining two phases can be expressed as V 0 cos (ωt π/3) and V 0 cos (ωt +π/3). Problem 4.3 It is an induction motor. parts (b) and (c): It sounds like an 8-pole motor supplied by 60 Hz. Problem 4.4 part (b):
49 part (c): part (d): Problem 4.5 Under this condition, the mmf wave is equivalent to that of a single-phase motor and hence the positive- and negative-traveling mmf waves will be of equal magnitude. Problem 4.6 The mmf and flux waves will reverse direction. Reversing two phases is the procedure for reversing the direction of a three-phase induction motor. Problem 4.7 F 1 = F max cos θ ae cos ω e t = F max (cos (θ ae ω t ) + cos (θ ae + ω t )) F = F max sin θ ae sin ω e t = F max and thus (cos (θ ae ω t ) cos (θ ae + ω t )) F total = F 1 + F = F max cos (θ ae ω t )
50 Problem 4.8 For n odd β/ β/ π/ π/ cos (nθ)dθ cos (nθ)dθ = sin (nθ ) For β =5π/6, sin ( nθ )= 0.97 n = 1 0 n = 3 0.6 n = 5 Problem 4.9 Rated speed = 100 r/min part (b): I r = πgb ag1,peak(poles) 4µ 0 k r N r = 113 A part (c): Φ P = ( ) lrb ag1,peak =0.937 3 Wb Problem 4.10 From the solution to Problem 4.9, Φ P =0.937 Wb. V rms = ωnφ =8.4 kv Problem 4.11 From the solution to Problem 4.9, Φ P =0.937 Wb. V rms = ωk wn a Φ =10.4 kv Problem 4.1 The required rms line-to-line voltage is V rms =13.0/ 3=7.51 kv. Thus Vrms N a = ωk w Φ Problem 4.13 The flux per pole is = 39 turns Φ=lRB ag1,peak =0.0159 Wb The electrical frequency of the generated voltage will be 50 Hz. The peak voltage will be
51 V peak = ωnφ = 388 V Because the space-fundamental winding flux linkage is at is peak at time t = 0 and because the voltage is equal to the time derivative of the flux linkage, we can write v(t) =±V peak sin ωt where the sign of the voltage depends upon the polarities defined for the flux and the stator coil and ω = 10π rad/sec. part (b): In this case, Φ will be of the form Φ(t) =Φ 0 cos ωt where Φ 0 =0.0159 Wb as found in part (a). The stator coil flux linkages will thus be λ(t) =±NΦ(t) =NΦ 0 cos ωt = ± 1 NΦ 0(1 + cos ωt) and the generated voltage will be v(t) = ωφ 0 sin ωt This scheme will not work since the dc-component of the coil flux will produce no voltage. Problem 4.14 Similarly, we can write F a = i a [A 1 cos θ a + A 3 cos 3θ a + A 5 cos 5θ a ] = I a cos ωt[a 1 cos θ a + A 3 cos 3θ a + A 5 cos 5θ a ] F b = i b [A 1 cos (θ a 10 )+A 3 cos 3(θ a 10 )+A 5 cos 5(θ a 10 )] = I a cos (ωt 10 )[A 1 cos (θ a 10 )+A 3 cos 3θ a + A 5 cos (5θ a + 10 )] and F c = i c [A 1 cos (θ a + 10 )+A 3 cos 3(θ a + 10 )+A 5 cos 5(θ a + 10 )] = I a cos (ωt + 10 )[A 1 cos (θ a + 10 )+A 3 cos 3θ a + A 5 cos (5θ a 10 )] The total mmf will be
5 F tot = F a + F b + F c = 3 I a[a 1 cos (θ a ωt)a 5 cos (5θ a + ωt)] = 3 ( I a[a 1 cos (θ a ωt)a 5 cos 5 θ a +( ωt ) 5 ) ] We see that the combined mmf contains only a fundamental space-harmonic component that rotates in the forward direction at angular velocity ω and a 5 th space-harmonic that rotates in the negative direction at angular velocity ω/5. Problem 4.15 The turns must be modified by a factor of ( )( ) 18 100 = 9 4 1400 14 =0.64 Problem 4.16 Φ p = 30E a N(poles)n =6.5 mwb Problem 4.17 ( Φ p = poles ) B peak lr = ( ) 1.5 0.1 (.095/) = 1.5 mwb 4 N ph = V rms poles (30/ 3) 4 = = 43 turns πfme k w Φ p π 60 0.95 0.015 part (b): From Eq. B.7 Problem 4.18 L = 16µ 0lr πg ( ) kw N ph =1. mh poles Φ p = V rms πnph =10.8 mwb B peak = Φ p lr =0.53 T
53 part (b): I f = πb peakg µ 0 k r N r =0.65 A part (c): L af = λ a,peak Vrms /ω = =0.69 H I f I f Problem 4.19 No numerical solution required. Problem 4.0 Φ peak = ( ) Dl B peak poles F r,peak = 4k rn r I r,max π poles ( ) poles Φ peakf r,peak =4.39 10 6 N m T peak = π P peak = T peak ω m = 88 MW Problem 4.1 Φ peak = ( ) Dl B peak poles F r,peak = 4k rn r I r,max π poles ( ) poles Φ peakf r,peak =16.1 N m T peak = π P peak = T peak ω m =6.06 kw Problem 4. dm af dm bf T = i a i f + i b i f = Mi f (i b cos θ 0 i a sin θ 0 )
54 This expression applies under all operating conditions. part (b): T =MI 0 (cos θ 0 sin θ 0 )= MI 0 sin (θ 0 π/4) Provided there are any losses at all, the rotor will come to rest at θ 0 = π/4 for which T = 0 and dt/ < 0. part (c): part (d): T = MI a I f (sin ωtcos θ 0 cos ωt sin θ 0 ) = MI a I f sin (ωt θ 0 )= MI a I f sin δ v a = R a i a + d dt (L aai a + M af i f ) = I a (R a cos ωt ωl aa sin ωt) ωmi f sin (ωt δ) Problem 4.3 v b = R a i b + d dt (L aai b + M bf i f ) = I a (R a sin ωt + ωl aa cos ωt)+ωmi f cos (ωt δ) T = MI f (i b cos θ 0 i a sin θ 0 ) = MI f [(I a + I /) sin δ +(I /) sin (ωt + δ)] The time-averaged torque is thus Problem 4.4 <T >= MI f (I a + I /) sin δ T = i a dl aa + i b dl bb + i a i b dl ab = I a I f M sin δ +I al sin δ + i a i f dm af + i b i f dm bf part (b): Motor if T>0, δ>0. Generator if T<0, δ<0. part (c): For I f = 0, there will still be a reluctance torque T =I a L sin δ and the machine can still operate.
55 Problem 4.5 v = f λ =5 m/sec part (b): The synchronous rotor velocity is 5 m/sec. part (c): For a slip of 0.045, the rotor velocity will be (1 0.045) 5 = 3.9 m/sec. Problem 4.6 ) (π ) ( ) p 4 k w N ph = 1.45 ( )( ) 9.3 10 3 (π 3 4 I rms = B peak ( g µ 0 )( 3 µ 0 Problem 4.7 Defining β = π/wavelength ) ( 7 0.91 80 ) = 18 A Φ p = w π/β 0 B peak cos βxdx = wb peak β = 1.48 mwb part (b): Since the rotor is 5 wavelengths long, the armature winding will link 10 poles of flux with 10 turns per pole. Thus, λ peak = 100Φ p =0.148 Wb. part (c): ω = βv and thus V rms = ωλ peak =34.6 V, rms