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Assignment 3

Assignment 3. Page 222, Problems: 10-18, 10-23, 10-25, 10-30, 10-31 Due Date: Tuesday 15 th Feb, 2011 Quiz No.2 Next Week. Cooling methods.

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Assignment 3

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  1. Assignment 3 • Page 222, Problems: 10-18, 10-23, 10-25, • 10-30, 10-31 • Due Date: Tuesday 15th Feb, 2011 • Quiz No.2 Next Week Electro Mechanical System

  2. Cooling methods • To prevent rapid deterioration of the insulating materials inside a transformer, adequate cooling of the windings and core must be provided. • Indoor transformers below 200 kVA can be directly cooled by the natural flow of the surrounding air. • Distribution transformers above 200 kVA are usually immersed in mineral oil and enclosed in a steel tank. • Oil carries the heat away to the tank, where it is dissipated by radiation and convection to the outside air. • Oil is a much better insulator than air is; consequently, it is invariably used on high-voltage transformers. Electro Mechanical System

  3. Cooling methods • For transformers in the megawatt range, cooling may be effected by an oil-water heat exchanger. • Hot oil drawn from the transformer tank is pumped to a heat exchanger where it flows through pipes that are in contact with cool water. • Such a heat exchanger is very efficient. But also very costly, because the water itself has to be continuously cooled and re-circulated. • The type of transformer cooling is designated by the following symbols: • AA–dry-type, self-cooled • AFA–dry-type, forced-air cooled • OA–oil-immersed, self-cooled • OA/FA–oil-immersed, self-cooled/forced air cooled • AO/FA/FOA –oil immersed. self-cooled/forced air cooled/forced-air, forced-oil cooled Electro Mechanical System

  4. Cooling methods Three-phase, type FOA, transformer rated 1300 MVA, 24.5 kV/345 kV, 60 Hz Three-phase, type OA/FA/FOA transformer rated 36/48/60 MVA, 225 kV/26.4 kV, 60 Hz Electro Mechanical System

  5. Simpifying the equivalent cct • Equivalent circuit gives far more detail than required • At no load I2 is zero and so is I1 because T is ideal transformer • Only exciting current IO flows in R1 and Xf1. These impedences are so small that voltage drop is negligibly small. • We can easily neglect these four impedences R1 and Xf1 and R2 and Xf2 Electro Mechanical System

  6. Simpifying the equivalent cct • At full load IP is at least 20 times larger than IO, so we can neglect IO • simplified circuit can be used if the load is only 10% of rated capacity • It can be more simplified by shifting impedances • Summing up all the impedances Electro Mechanical System

  7. Simpifying the equivalent cct • Transformers above 500kVA posses leakage reactance XP which is 5 times higher than RP • We can even neglecting RP • Circuit further simplified with only XP Electro Mechanical System

  8. Voltage Regulation • Important attribute of a transformer • Constantly held primary voltage • Impact of secondary voltage due to changing loads • Where: • ENL = secondary voltage at no-load [V] • EFL = secondary voltage at full-load [V] • Voltage regulation depends upon the power factor of load Electro Mechanical System

  9. Example • A single-phase transformer rated at 3 MVA, 69 kV/4160 V with an internal impedance of 127 ohms as seen on the primary. Calculate • The rated primary and secondary currents • The voltage regulation from no-load to full load for 2000kW resistive load, knowing that primary voltage is fixed to 69kV • The primary and secondary currents if secondary is accidently short circuited • Rated primary current • Inp = Sn/Enp = 3000,000/69000 = 43.5 A • Rated secondary current • Ins = Sn/Ens = 3000,000/4160 = 721 A • above 500 kVA, so we neglecting resistance • ZP = XP = 127Ω Electro Mechanical System

  10. Example Approximate secondary side impedance with 2000kW load Z = E2S / P = 41602/2000000 = 8.65Ω Load impedance shifting to primary a2Z = (69/4.16)2 x 8.65 = 2380Ω Constant primary voltage 69kV results in no-load secondary voltage 4160V The Voltage regulation is excellent Electro Mechanical System

  11. Example • If secondary is accidently short circuited • aES = 0 • IP = EP/XP = 69000/127 = 543 A • Corresponding IS on secondary side: • IS = aIP =(69/4.16)x543 = 9006 A • Short circuit current in both primary and secondary are 12.5 times greater than rated value. • I2R losses are therefore 12.52 or 156 times greater then normal • Circuit breaker or fuse must open immediately to protect the transformer Electro Mechanical System

  12. Measuring Impedances • We can measure using two tests (open and short circuit) the actual values of Xm, Rm, Xpand Rp for a given transformer • Voltages, currents, and real powers are measured with Open-circuit test. It is performed with secondary opened. • Rated voltage applied at primary winding • Io, voltage Ep and active power Pm are measured • Es is also measured • Active power absorbed by core = Pm • Apparent power absorbed by core Sm = EpIo • Active power absorbed by the core • Core loss resistance Rm = Ep2/Pm • Magnetizing reactance Xm = Ep2/Qm • Turn ratio a = N1/N2 = Ep/Es Electro Mechanical System

  13. Measuring Impedances • During short-circuit test, secondary winding is short-circuited. • Voltage Eg must be lower than normal (Usually less than 5% of rated voltage) is applied to the primary. • Primary voltage is adjusted until current applied to primary is equal to the rated value. • Due to low voltage very little current is required for excitation and can be ignored. • Primary side impedance is Zp = Esc/Isc • Transformer resistance Rp = Psc/Isc2 • Leakage Reactance Home Work: Page 213 Example 10-7 Electro Mechanical System

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