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12th National Certification Exam Energy Managers & Auditors OCTOBER 2011 Paper 3

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Jawaharlal Nehru Technological University (JNTUH), Hyderabad

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Paper 3 Set A Key Regn No: _________________ Name: ___________________ (To be written by the candidates) 12th NATIONAL CERTIFICATION EXAMINATION October, 2011 FOR ENERGY MANAGERS & ENERGY AUDITORS PAPER 3: Energy Efficiency in Electrical Utilities Date: 16.10.2011 Timings: 0930-1230 HRS Duration: 3 HRS Section II: SHORT DESCRIPTIVE QUESTIONS (i) (ii) S-1 Ans Max. Marks: 150 Marks: 8 x 5 = 40 Answer all Eight questions Each question carries Five marks Briefly explain transformer losses and how the total transformer losses at any load level can be computed. Transformer losses consist of two parts: No-load loss and Load loss 1. No-load loss (also called core loss) is the power consumed to sustain the magnetic field in the transformer's steel core. Core loss occurs whenever the transformer is energized; core loss does not vary with load. Core losses are caused by two factors: hysteresis and eddy current losses. Hysteresis loss is that energy lost by reversing the magnetic field in the core as the magnetizing AC rises and falls and reverses direction. Eddy current loss is a result of induced currents circulating in the core. 2. Load loss (also called copper loss) is associated with full-load current flow in the transformer windings. Copper loss is power lost in the primary and secondary windings of a transformer due to the ohmic resistance of the windings. Copper loss varies with the square of the load current. (P=I2R). For a given transformer, the manufacturer can supply values for no-load loss, PNO-LOAD, and load loss, PLOAD. The total transformer loss, PTOTAL, at any load level can then be calculated from: PTOTAL = PNO-LOAD+ (% Load/100)2 x PLOAD Where transformer loading is known, the actual transformers loss at given load can be computed as: S-2 No load loss 2 kVA Load full load loss Rated kVA A 15 kW, 3 phase, 415 V induction motor draws 25 A and 12 kW input power at 410 V. Calculate the Apparent and Reactive Power drawn by the motor at the operating load? Ans Apparent power = 1.7321 x 0.410 x 25 = 17.15 Kva Reactive power = sqrt (apparent power 2- active power2) Active power = 12 kW Therefore reactive power = sqrt (315.2-144) = sqrt (171.2) = 13.08 kVAr _______________________ Bureau of Energy Efficiency 1

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