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13th National Certification Exam Energy Managers & Auditors SEPTEMBER 2012 Paper 3

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Click Here & Upgrade PDF Complete Expanded Features Unlimited Pages Documents Paper 3 Set A Key 13th NATIONAL CERTIFICATION EXAMINATION September, 2012 FOR ENERGY MANAGERS & ENERGY AUDITORS PAPER 3: Energy Efficiency in Electrical Utilities Date: 16.9.2012 Max. Marks: 150 Section I: (i) (ii) 1 Timings: 0930-1230 HRS OBJECTIVE TYPE In a 22 kV feeder, if the voltage is raised from 22 kV to 66 kV for the same loading conditions, the voltage drop in the same feeder system would be lowered to b) 1/3 d) unpredictable value b) 80 to 85 % c) 60 to 70% d) 50%- 60% What is the reduction in distribution loss if the current flowing through the distribution line is reduced by 10%? a)10% 4 c) 1/9 Normally, the efficiency of distribution transformer at full load varies anywhere between a) 96 to 99 % 3 Marks: 50 x 1 = 50 Answer all 50 questions Each question carries one mark a) 1/2 2 Duration: 3 HRS b) 90% c) 19% d) 81% Power factor is the ratio of a) kVAr/kW c) kW/ (kW 2+kVAr2)1/2 5 b) (kW 2+kVAr2)1/2/kW d) kVAr/ (kW 2+kVAr2)1/2 The electricity bill shows an average power factor of 0.72 with an average kW demand of 627. How much kVAr is required to improve the power factor to 0.95? (Given Data: tan 1 = 0.964, tan 2 = 0.329) a) 398 6 b) 144 c) 95 d) 627 Where transformer loading is known, the actual transformer loss at a given load can be computed as: a) No Load Loss+ (Actual kVA Load/rated kVA) X Load Loss 2 b) No Load Loss+ (Actual kVA Load/rated kVA) X Load Loss 2 c) No Load Loss+ (Actual kVA Load/rated kVA) X Load Loss 2 d ) [No Load Loss+{ (Actual kVA Load/rated kVA) X Load Loss}] _______________________ Bureau of Energy Efficiency 1

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