Unit 5 - Week 4. Week 4: Assignment. Course outline. Announcements Course Forum Progress Mentor

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1 14/12/2017 Electrical Machies - I - - Uit 5 - Week 4 X reviewer2@ptel.iitm.ac.i Courses» Electrical Machies - I Aoucemets Course Forum Progress Metor Uit 5 - Week 4 Course outlie How to access the portal Week 4: Assigmet. Week1 Week 2 Week 3 Week 4 Lecture 10 : Itroductio to Three Phase Trasformer Lecture 11:Costructio of Three Phase Trasformers Lecture 12: Three Phase Trasformers Coectios Quiz : Week 4: Assigmet Week 4 : Assigmet Solutio 1) A 3-phase trasformer bak cosistig of three sigle phase 10 poits trasformers is used to step-dow the voltage of a 3-phase, 11000V trasmissio lie. If the primary lie curret is 100A, calculate the secodary lie voltage for Y/Δ coectio. The turs ratio is 10. Neglect losses kv kv 63.5 kv 1.1 kv kv 2) Three sigle phase trasformers of ratig 50 kva, 2000V/1000V are coected i Δ / Δ. 10 poits Fid the curret i the primary widigs of the trasformer, whe supplyig a balaced three phase load of 40kW at 1000V ad 0.8 power factor laggig? A A 8.33 A A Feedback for week 4 Week 5 Week 6 Week 7 Week 8 Week A 3) Two idetical trasformers of ratig 15 kva, 2000/1000V, 50 Hz are coected i ope delta. Calculate the kva ratig of the ope delta bak whe HV side is used as primary? kva 30 kva kva 13 kva 10 poits Week 10 Week kva Week /4

14/12/2017 Electrical Machies - I - - Uit 5 - Week 4 4) Two idetical sigle phase trasformers are coected i ope delta (V-V) to supply a 10 poits three phase load of 180kW at 0.

Fid the real ad reactive power supplied by each of the trasformers?

2 14/12/2017 Electrical Machies - I - - Uit 5 - Week 4 4) Two idetical sigle phase trasformers are coected i ope delta (V-V) to supply a 10 poits three phase load of 180kW at 0.85 pf laggig ad sigle phase loads of 120kW at 0.9 pf laggig at 440V. The sigle phase loads are distributed equally amog the phases of the trasformers. Fid the real ad reactive power supplied by each of the trasformers? 118 kw ad 182 kw 156 kw ad 144 kw 192 kw ad 108 kw 101 kw ad 199 kw 101 kw ad 199 kw 5) Three sigle phase trasformers are coected i delta/zig-zag, to 10 poits form a three phase trasformer bak. The ratio of umber of turs of secodary to primary is 0.1. The secodary coils of the trasformer are split ito two parts of equal umber of turs. The coectio is show i figure. If the lie to lie voltage at the primary side is 1000V, what is the lie to lie voltage at the secodary side V 150 V 50 V 100V 150 V 6) A three phase step dow trasformer is coected to 3300V supply. The turs ratio per 10 poits phase is 12 ad it draws a curret of 5 A from the mais. Calculate the secodary lie voltage ad lie curret if trasformer is delta-star coected. 476V, 35A 275V, 60A 476V, 60A 275V, 35A 476V, 35A 7) A three phase step dow trasformer is coected to 3300V supply. The turs ratio per 10 poits phase is 12 ad it draws a curret of 5 A from the mais. Calculate the secodary lie voltage ad lie curret if trasformer is star- delta coected. 275V, 60A 159V, 60A 159V, 104A 2/4

3 14/12/2017 Electrical Machies - I - - Uit 5 - Week 4 275V, 104A 159V, 104A 8) A 50hp, 220 V, 3-phase iductio motor has efficiecy 0.9 ad power factor 0.8 a full load. 10 poits A 3300/220 V, delta-star trasformer, supplies it. What is the curret i the high voltage widig of the trasformer whe the motor is ruig at full load? 9A 5.2A 136A 235A 5.2A 9) From the mid poits of secodaries of a 4000/440 V trasformer, cotacts are 10 poits provided to provide lower voltage as show i figure. A curret of 50A is draw at upf from each of the two sets of low voltage termials abc ad ABC. What will be the curret i the high voltage widig? Igore magetizig curret of the trasformer. 14.3A 8.25A 75A 6A 8.25A 10) Three sigle phase trasformers are coected i delta. It supplies 100A per lie to a 3-10 poits phase 3-wire system. Oe of the uits of the trasformer develops a fault ad is removed. By how much does the capacity of the set of trasformers reduce for the same temperature rise? 33.33% 66.67% 42.3% 57.7% 3/4

4 14/12/2017 Electrical Machies - I - - Uit 5 - Week % Previous Page Ed A project of 2014 NPTEL - Privacy & Terms - Hoor Code - FAQs - I associatio with Fuded by Powered by 4/4

5 Assigmet - 4 : Solutio Q1.Solutio 100A 10:1 I L2 11,000V V ph1 V ph2 Figure 1: Phase voltage o the primary side, V ph1 = 11, = 6.35kV Secodary side phase voltage, V ph2 = = 0.635kV Secodary side lie voltage = Secodary side phase voltage Secodary side lie voltage, V LL2 = kv Primary side lie curret, I L1 = 100A Primary side lie curret = Primary side phase curret Primary side phase curret, = 100A Secodary side phase curret, I ph2 = = 1000A Secodary side lie curret, I L2 = = 1732A Q2. Solutio 3VL I L cos(φ) = 40, 000W IL 0.8 = 40, 000W I L = 40, = 28.86A The curret i the widigs of the trasformer secodary, I ph2 = = 16.66A Curret i the primary widigs of the trasformer, I ph1 = = 8.33A Assigmet No :3 oliecourses.ptel.ac.i Page 1 / 6

6 Q3.Solutio 7.5A 15A 2000V 7.5A 15A 1000V 2000V 7.5A 15A 1000V Figure 2: The ope-delta trasformer bak is show i Fig.2. I this cofiguratio each of the trasformer coil is coected across the lies. The lie to lie voltage to be applied at the HV side is 2000V. Iduced voltage i the LV coils will be 1000V.Hece, the lie-to-lie voltage at the LV side 1000V. Sice, the HV ad LV coils are i series with the lies, i order to avoid overload, the lie curret at the LV side must be limited to 15A, which automatically fixes the HV side lie curret to 7.5A. Hece, the kva hadled by the bak of two sigle phase trasformers. T otal kv A = = = 25.98kVA Assigmet No :3 oliecourses.ptel.ac.i Page 2 / 6

7 Q4.Solutio Sice, the loads o each trasformer is balaced; P P L1 = 180kW Q L1 = P L1 ta(φ L1 ) = 180 ta(cos 1 (0.85)) = kV AR P L2 = 120kW Q L2 = 120 ta(cos 1 (0.9)) = 58.11kV AR Total active power, P T ot = = 300kW Total reactive power, Q T ot = = 170.1kV AR [ ] Hece, power factor, P F = cos ta = 0.87 laggig 300 The curret i the secodary side of the trasformer is give by, I L2 = P T ot 3(440)(P F ) = 300k 3(440)(0.87) = 452A The lie currets are shifted by 30 o due to the differece betwee lie ad phase quatities ad further by φ due to the power factor of the load I a (-30- ) I b (90- ) I bph I c (-150- ) Figure 3: The real power supplied by each traformer ca be calculated as follows P T 1 = V ab I a cos(0 30 φ) = cos( ) P T 1 = kW P T 2 = V bc I bph cos( 120 (90 φ 180)) = cos( 120 ( )) P T 2 = kW Assigmet No :3 oliecourses.ptel.ac.i Page 3 / 6

8 Q5.Solutio Figure 4: The ratio of voltage iduced i coil a2a1 to coil A2A1 is give by, V a2a1 : V A2A1 = 1 : 20 Hece,V a2a1 = = 50V 20 The ratio of voltage iduced i coil a4a3 to coil A2A1 is give by, V a2a1 : V A2A1 = 1 : 20 Hece,V a4a3 = = 50V 20 V a4b4 Figure 5: Now, V a4a2 = 3 50 = 86.6V Hece, the lie voltage, V a4b4 = = 150V Assigmet No :3 oliecourses.ptel.ac.i Page 4 / 6

9 Q6. Solutio Delta-Star coectio Secodary phase voltage = = 275V Secodary lie voltage = Secodary phase voltage 3 = V Secodary lie curret = Secodary phase curret (sice, secoday is star) Secodary phase curret = (Primary phase curret x 12) Primary lie curret Primary phase curret = 3 Secodary lie curret = = A Q7. Solutio Star-Delta coectio Primary phase voltage = Primary lie voltage / 3 = = V Secodary Lie voltage = Secodary phase voltage = Primary phase voltage / 12 = V Secodary lie curret= 3 Secodary phase curret (sice, secoday is delta) Secodary phase curret= (Primary phase curret x 12) = 5 12 = 60A Secodary lie curret =60 3 = 103.9A Assigmet No :3 oliecourses.ptel.ac.i Page 5 / 6

10 Q8. Solutio Turs ratio of phase = / 3 = 26 Power draw by motor = Power output Efficiecy Power output of motor i watts = Power output i HP 746 Power draw by motor = = 41.4kW Power draw by motor = I 0.8 = I = = A Curret i LV widig Curret i HV side of trasformer = Turs ratio /26 = 5.23 A Q9. Solutio Power draw by both widigs ABC ad abc = = W Power draw from primary = Power draw from secodary Power draw from primary = 3 V pl I pl Primary Lie curret(i pl ) = I pl = 8.25 A Q10. Solutio Referig to solutio of questio 2, we ca see that the capacity of the bak reduces to 57.7% of the origial ratig This is because i ope delta the lie ca oly carry the phase curret Reductio i capacity of trasformers = = 42.3% Assigmet No :3 oliecourses.ptel.ac.i Page 6 / 6

ELE B7 Power Systems Engineering. Symmetrical Components

ELE B7 Power Systems Engineering. Symmetrical Components ELE B7 Power Systems Egieerig Symmetrical Compoets Aalysis of Ubalaced Systems Except for the balaced three-phase fault, faults result i a ubalaced system. The most commo types of faults are sigle liegroud