ATTEMPT ALL QUESTIONS (EACH QUESTION 20 Marks, FULL MAKS = 60) Given v 1 = 100 sin(100πt+π/6) (i) Find the MS, period and the frequency of v 1 (ii) If v 2 =75sin(100πt-π/10) find V 1, V 2, 2V 1 -V 2 (phasor) & 2v 1 -v 2. (time function) 5% In Fig., given that 1 =15Ω, 2 =4 Ω, L=47.75mH, C=318.3µF and V=220V, 50Hz. Find (i) the impedance Z A and Z in polar form (ii) the voltages V 1,V 2 and current I in phasor format (iii) the overall active P and reactive Q. (iv) the capacitance of a capacitor connected in parallel with Z for improving the power factor to 1.0 (v) Draw the power triangle to illustrate your answers C I 1 Z A V 1 V 2 Z V 2 Fig. Z = Z A and Z in series L 15% -1-
An balanced three-phase delta load is connected to a 415V, (V =415 0 o ), 3-phase supply as shown in fig. a, Z 1 =Z 2 =Z 3 =10+10j. Find the ED line current I. 5% I 415 0 V Z 1 Z 3 Fig. 2a Z 2 A 3-phase, 3-wire, 380V, 50Hz, supply system connected to an unbalanced star-load as shown in Fig. b. Given that Z =5.66+5.66j, Z =8.66-5j and Z =13+7.5j, find (i) Displacement neutral voltage V ON at the star-point O (ii) Phase currents I the load (iii) Explain what will happen if a neutral wire is connected between supply neutral and point O. (iv) Draw diagram to show how 2-wattmeter method can be used to measure the load in Fig. b. Taken V N as reference. 15% I V N N V ON Z O Z I Z I Fig. 2b -2-
An auto-transformer is made by connecting up the primary and secondary windings of a 50kVA, 2000/200V, single phase transformer as shown in Fig., given that V A =200V and I 1 =250A. Determine (i) I C and I 2 (ii) Input and Output kva of the auto-transformer 7% -3-
Fig. 3 A 4-pole, 3-phase induction motor is fed from a 50Hz supply running at 0.05 slip, find Synchronous Speed, otor speed, otor frequency 6% (c) riefly describe the advantages and disadvantages of using (i) Salient pole 3-phase synchronous generator (ii) cylindrical rotor. 7% - END OF TEST - -4-
ATTEMPT ALL QUESTIONS (EACH QUESTION 20 Marks, FULL MAKS = 60) Given v 1 = 100 sin(100πt+π/6) Find the MS, period and the frequency of v 1 If v 2 = 75 sin(100πt-π/10) voltage source is connected in series with v 1, find the phasors V 1, V 2, V 1 +V 2 and v 1 + v 2. v 1 = 100 sin(100πt+π/6) MS of v 1 = 1/ 2 x 100 = 70.7 f = ω/ 2 π = 100 π/2 π=50hz T = 1/f = 1/50 = 0.02s V 1 = 1/ 2 x 100 π/6 V 2 = 1/ 2 x 75 -π/10 2V 1 -V 2 =2x70.7 π/6-53.03 π/10) =122.48+70.72j-50.44+16.39j=72.04+87.11j=113.03 50.4 o 2v 1 -v 2 = 2x113.03 sin(100πt+50.4 o )=160 sin(100πt +50.4 o ) In Fig., given that 1 =15Ω, 2 =4 Ω, L=47.75mH, C=318.3µF and V=220V, 50Hz (i) Find the impedance Z A and Z in polar form C 1 2 L (ii) Determine the voltage V 1, V 2 and the current I I in phasor format Z A Z (iii) Determine the overall active P and reactive Q. (iv) Find the capacitance of a capacitor connected in parallel with Z (Z A and Z in series) for V 1 V V 2 improving the power factor to 0.95. (v) Draw the power triangle to illustrate your Fig. answers X L = 2π 50 x 47.75 x10-3 = 15Ω Z A = 15-j10 = 18-33.7 o X C = 1/(2π50x10-6 318.3) = 10Ω Z = 4+j15 = 15.5 75.1 o Z = 15+15j+4-10j = 19 + 5j = 19.65 14.75 o I = 220 0 o / 19.65 14.75 o = 10.83-2.85j=11.20-14.75 o V 1 = IZ A = 11.20-14.75 o x 18-33.7 o = 201.6-48.4 o V 2 = IZ = 11.20-14.75 o x 15.5 75.1 o = 173.6 60.36 o S = V I * = 220 0 o (11.20-14.75 o )* = 220 0 o x11.20 14.75 o =2464 14.75 o S = 2382.3+j627.03, P = 2382.3 W, Q = 627.03 VA Power factor = cos [ tan-1(627.03/6382.3)]=0.967 For improving the p.f. to unity, Q = kvar rating of capacitor = 627.03 VA (Leading) Q = V 2 / X C X C = V 2 /Q =77.19 Ω 1/(2πfC)=1/(2π50C)=77.19 C=41.23µF An unbalanced three-phase delta load is connected to a 415V,, 3-phase supply as shown in fig. a, Z 1 =Z 2 =Z 3 =10+10j. Find the ED line current I. 415 0 V I Z 1 Z 3 4% 6% 1% 4% 6% 1% 5% I = 415 0 o /[10+10j]=29.35-45 o =20.8-20.8j I = 415 120 o /[10+10j]=29.35 75 o =7.6+28.3j Z 2 Fig. 2a I = I - I =13.17-49.1j=50.8-75 o -5-
A 3-phase, 3-wire, 380V, 50Hz, supply system connected to an unbalanced star-load as shown in Fig. b. Given that Z =5.66+5.66j, Z =8.66-5j and Z =13+7.5j, find Displacement neutral voltage V ON at the star-point O Phase currents to the load V N N I V ON Z O Z 8% 8% I Z (i) (ii) (iii) (iv) Fig. 2b Z =5.66+5.66j=8 45 o Z =8.66-5j = 10-30 o Z =13+7.5j = 10 30 o V N as reference, the phase voltages phasors are: V N = 380 / 3 0 o =219.4 0 o V N = 219.4-120 o V N = 219.4 120 o =0.0883-0.0883j=0.125-45 o =0.0866+0.05j=0.1 30 o =0.0577-0.033j=0.067-30 o y Millman s Theorem : V ON =[V N + V N + V N ] / [ + + ] V ON =[(219.4 0 o )(0.125-45 o )+(219.4-120 o )(0.1 30 o )+(219.4 120 o )(0.067-30 o )] /[(0.125-45 o )+ (0.1 30 o )+ (0.067-30 o )] V ON = 108.65 81.64j=135.9-36.9 o V O =V N V ON = 11.35+81.64j=138.07 36.25 o I = V O /Z =17.048-2.625j=17.25-8.75 o *Further calculation for reference only V O =V N V ON = -218.65-108.9j=244.26-153.53 o V O =V N V ON = -218.65+272.16j=349.11 128.8 o I = V O /Z =-13.49-20.36j=24.43-123.5 o I = V O /Z =-3.557+22.98j=23.26 98.8 o If a neutral wire is connected between supply neutral and point O. V O =V N V O =V N V O =V N I = V N /Z I = V N /Z I = V N /Z Neutral current will becomes I N = I + I + I () I M V + O A() V ON L V N I Positive Sequence I C() Z 1 Z 2 Z 3 4% An auto-transformer is made by connecting up the primary and secondary windings of a 50kVA, 2000/200V, single phase transformer as shown in Fig., given that V A =200V and I 1 =250A. Determine (i) I C and I 2 (ii) Input and Output kva of the auto-transformer V + M V L I 7% Fig. 3-6-
Given I A = I 1 =250A, V A I A = 200 x 250V = 50kVA V C = 2000 I C = 50000 / 2000 = 25A I 2 = I A + I C = 250 + 25 = 275A Input kva = (2000+200) x 250 = 550kVA Output kva = V C I 2 = 2000 x 275=550kVA A 4-pole, 3-phase induction motor is fed from a 50Hz supply running at 0.05 slip, find Synchronous Speed, otor speed, otor frequency 6% Slip = (N s N r ) / N s, f s = p N s Speed of rotating field = N s = Stator Frequency / Pole-pair = 50/2=25 rev/s = 1500 rpm S = 0.05 = (N s N r ) / N s N r = 1500(1-0.05)=1425 rpm Speed of rotor with respect to the rotating field = N s N r [rev/s] 1 revolution cut p pole pairs 1 sec p x (N s N r ) electrical cycles Since N s = Stator Frequency / Pole-pair Pole-pair = Stator Frequency / N s otor Frequency = Stator Frequency x (N s N r ) / N s = Stator Frequency x slip otor Frequency = Stator Frequency x slip f r = s f s = 0.05 x 50 = 2.5Hz riefly describe the advantages and disadvantages of using (i) Salient pole 3-phase synchronous (c) generator (ii) cylindrical rotor. Salient pole synchronous generator (c) Advantages: The air gap between the stator and the rotor can be adjusted so that the magnetic flux is sinusoidal in distribution. As a result the output waveform will also be sinusoidal in nature Disadvantages: The salient pole has a weak structure so that this machine is not suitable for high speed application such as the turbo-generator on air-plane. The salient pole generator is expensive Cylindrical rotor synchronous generator Advantages - The cylindrical rotor is cheaper than the salient pole rotor The cylindrical rotor is robotic in design, because it is symmetrical in shape, dynamically balance can be easily obtained. Hence it can be used for high speed application, say, coupled to turbo-engine such as the generator in an air-craft. Disadvantages - The air gap is uniform for the rotor, the generated voltage will have a polygonal waveform depending on the number of windings on each of the rotor slots. Though the shape of the polygon is adjusted to be nearly sinusoidal, the output waveform still defers from the sine wave and therefore the harmonic content of the cylindrical rotor generator is high compared with that of the salient pole design State any 2 advantages of using auto-transformer as compared to two-winding transformer It effects a saving in winding material (copper or aluminum), since the secondary winding is part of the primary current. Lower copper loss, therefore efficiency is higher than in the two winding transformer. Lower leakage reactances, lower exciting current. Variable output voltage can be obtained. - END OF TEST - 7% -7-