60 p. 2. A 200hp 600V, 60 Hz 3-phase induction motor has start code F. What line current should be expected at starting? 4 marks.

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1 EE 004 Final Solution : Thi wa a hr exam. A 60 Hz 4 pole -phae induction motor rotate at 740rpm. a) What i the lip? mark b) What i the peed o rotation o the rotor magnetic ield (in rpm)? mark The motor ha 4 pole, 60 Hz: n 0 n nm 60 = = 800rpm = = p n 800 = 0.0 The rotor magnetic ield rotate at ynchronou peed: 880 rpm. A 00hp 600V, 60 Hz -phae induction motor ha tart code F. What line current hould be expected at tarting? 4 mark Code F ha max kva =5.6 time rated hp. S max = = 0 kva 0 0 S = VLLIL IL = = 078A max 600. A pole ynchronou generator produce a voltage at 50Hz. What i the rotor peed? mark For a ynchronou machine, rotor peed = ynchronou peed. 0 nm = n = = 000r pm p 4. Lit 4 type o ingle phae induction motor in order o cot, with mot expenive irt. 4 mark Any 4 o: i) Capacitor Start Capacitor un ii) Capacitor Start iii) Permanent Split Capacitor (PSC) iv) Split Phae v) Shaded Pole

2 EE 004 Final Solution : Thi wa a hr exam 5. Two induction motor have identical deign except or rotor bar conductivity. Motor one ha a higher conductivity than motor two. Decribe the relative dierence between maximum torque and the lip at which maximum torque occur. 4 mark Conductivity i the reciprocal o reitivity. Thereore motor ha a lower rotor reitance. otor reitance ha no impact on maximum torque, o both motor have the ame maximum torque. Maximum torque occur when = TH + j( XTH + X ). Thereore τmax i proportional to τ max rotor reitance. Motor ha a lower lip or maximum torque than motor. 6. Name one method o limiting the tart current o a cage rotor induction machine. Decribe briely how it reduce the tart current. 4 mark Take your pick rom: Y tart: reduce motor phae voltage Added line reitance or reactance: increae total impedance o the ytem, reducing current Autotranormer tart: reduce motor upply voltage Power Electronic (ot tart or variable peed drive) reduce motor phae voltage 7. A tator o a imple ac machine ha phae, 4 pole and lot. Each lot contain a coil with 50 turn and the coil or each phae are connected in erie. The phae are Y connected. A rotating magnetic ield produce lux o 0.070Wb per pole and rotate at 800 rpm. Calculate the induced line-line terminal voltage. 8 mark phae, 4 pole, lot mean lot per pole per phae. There are coil per phae, in erie. np For each coil E = π Ncφ e and at 800 rpm e = = 60Hz 0 E coil = 9V With erie connection o coil, E phae = E coil = 866V Finally, the phae are Y connected. ELL = Ephae = V

3 EE 004 Final Solution : Thi wa a hr exam 8. A 0MVA,.kV, 60Hz ynchronou generator (Y connected) ha a rated power actor o 0.8 lagging and ynchronou reactance, X S, o = 0.9 Ω/phae. a i negligible. a) Sketch the phaor diagram or rated operation 4 mark b) Calculate the magnitude o the induced voltage, E and load angle, δ 4 mark c) Calculate the load angle, power actor and armature current i the excitation (E) i reduced to 70% o the rated value while the output power i maintained 7 mark Thi i a generator, o E lead V. The power actor i lagging, o I lag V and i no I a a phaor ince a i negligible: Ecoδ > V. There 6 S 0 0 The machine i Y connected: I L =I A: I A = = = 549A V LL. 0 I = I coθ jinθ = j0.6 Since the power actor i lagging: ( ) ( ) Now E = V + jix. 0 E = + j549( 0.8 j0.6) 0. 9 E = j779v E = E =.kv δ = E = 9.9 VE I the power i maintained whilt E i reduced, then rom P = inδ, E inδ = contant X To ind the load angle: E inδ = inδ = in9.9 inδ = E 0.7 δ = 9. Now we can ue E to ind I : E = E coδ + jinδ = V + jix ( ) ( co j in 9.9) = + j0.9i I = j909 I = I = 496A p = co I = leading

4 EE 004 Final Solution : Thi wa a hr exam 9) A 460V, 60Hz Y connected -phae 4 pole induction motor ha the ollowing parameter: =0.065Ω, =0.0Ω, X =0.04Ω, X =0.059Ω, X m =.74Ω The motor operate at 770 rpm with.9kw rotational loe. a) Find the motor upply line current, power actor and input power. 0 mark b) Find the motor torque developed, output power in horepower ( Hp = 746W) and eiciency. 0 mark jxm V + jx I = o we need input impedance: Zin = = jx+ which in turn Z in + j ( X + jxm) 0 e 0 60 n depend on lip. n 800 nm 0 = = = rpm, = = = p 4 n Putting the number into the expreion or input impedance: Zin =.09 + j0.54ω V The motor i Y connected: LL V 65 V = = 65.6V o I = = = 00 j94a Zin.09 + j0.54 For Y connection line current equal phae current: I = I = 0A p = co I = 0.905lagging P= V I coθ = 58.6kW L LL L There are a ew option to ind torque: I jx m τ = with I = I O ω + j( Xm + X ) Pgap τ = with Pgap = Pin I O ω jx m jx P + gap τ = with Pgap = I ω + j( X + Xm ) For the above option, I = 0A and P gap = 54.8kW, which give τ = 8Nm π Pout = τωm Prot = = 49kW = 00hp 60 Pout 49 η = = = 9.9% P 58.6 in

5 EE 004 Final Solution : Thi wa a hr exam 0) A 40V, 50Hz, 4 pole ingle phae induction motor ha the ollowing parameter and operate at 440 rpm. =.4Ω, =.75Ω, X =.0Ω, X =.0Ω, X m =50.0Ω. a) Calculate the input impedance and power actor 0 mark b) Calculate the orward, backward and net torque 0 mark Zin = + jx+ ( ZF + ZB ) jx m + jx jx m jx + ZF = ZB = + j ( X + Xm) + j( X + Xm) 0 50 n nm 60 Now, or 50 Hz upply, n = = 500rpm = = = 0.04 =.96 4 n 500 Thi give Z = j8.7ω Z =.77 + j.99ω = 4.+ j.ω To ind torque: F B Z in V 40 and I = = = 9.04A Z 6.5 in X p = co tan = I I F ( F) F ( B) net F ω ω τ = Z τ = Z τ = τ + τ B Subtituting give: τ = 5.Nm τ = 0.46Nm τ = 4.85Nm F F net

6 EE 004 Final Solution : Thi wa a hr exam. A 6-4 Switched reluctance motor ha an aligned inductance o 6mH and an unaligned inductance o 8mH. The motor i operated to obtain maximum average torque, with phae current o 5A. a) Calculate the turn on and turn o angle or the current. 4 mark b) Calculate the average torque mark c) Calculate the peak to peak torque ripple mark d) Due to a poition meaurement error, the current i witched on and o 5 degree ater than deired value. What i the magnitude o the dicontinuity in the torque waveorm? 4 mark Thi quetion aume that an S motor can be modelled a a linear ytem, with torque a unction o inductance. With the number o tator pole, p S, = 6 and number o rotor pole, p, = 4: θ = θ = + p ps p ps θ = 7.5 θ = 7.5 τ ave = ˆ τ in pθd θ θ θ LI ˆ p where ˆ τ = and ˆ Lmax Lmin L = = 4mH. 60. τ [ co 4 ] 0.0 ave = θ = 0.65Nm π 4 Peak to peak ripple can be ound rom the dierence between maximum and minimum torque. Maximum torque = ˆ τ = 0.Nm, while minimum torque occur at either θ or θ. θ θ ( ) τmin = ˆ τ in 4θ = 0.Nm Thereore, pk-pk torque ripple = 0.Nm I the current witche at -.5 and -.5, there will be a dicontinuity when one phae witche o and the next phae witche on: τ = ˆin τ. 4 = 0.5Nm τ on o ( ) ( ) = ˆin τ.5 4 = 0.04Nm δτ = = 0.9Nm