lecture-7_examples1.pptx
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The document describes a dc battery charging circuit using a half-wave rectifier connected to a 230V 50Hz source through an 8 ohm resistor. It asks to (a) find the average charging current, (b) find the power supplied to the battery and dissipated in the resistor, (c) calculate the supply power factor, (d) find the charging time if battery capacity is 1000Wh, and (e) find the rectifier efficiency and peak inverse voltage of the diode.
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امتحان جهد عالي 2016
This document contains a 3-part exam on high voltage engineering. Part 1 contains 3 questions, the first on measuring HVAC peak value and performing an accelerated aging test on a bushing. The second concerns generating HVDC using a Greinacher cascade circuit and a basic rectifier circuit. The third addresses impulse voltage generation and designing a 5-stage Marx generator. Part 2 includes questions on earthing systems, surge arrestors, and analyzing an RLC circuit with DC supply. Part 3 concerns calculating flash protection boundaries for a busbar with short circuit current. The exam tests knowledge of high voltage concepts and applications.
امتحان جهد عالي 2015-2016
This document contains a 3-part exam on high voltage engineering. Part 1 contains 3 questions, the first on measuring HVAC peak value and performing an accelerated aging test on a bushing. The second concerns generating HVDC using a Greinacher cascade circuit and a basic rectifier circuit. The third addresses impulse voltage generation and designing a 5-stage Marx generator. Part 2 contains 3 questions, two on RC and RLC circuits applied to DC sources, and one on flash protection boundaries for live busbars. Part 3 requests calculations regarding voltage regulation, ripple, and output of a Cockcroft-Walton circuit, as well as the design of a DC voltage doubling circuit.
Single phase transformer
The document describes tests conducted on a single-phase transformer to determine its efficiency and regulation. An open circuit test was conducted to measure no-load losses. A short circuit test was used to determine copper losses and develop an equivalent circuit model. Efficiency was calculated at various load levels and power factors based on losses from the two tests. Regulation was also calculated using the short circuit test results. Plots of efficiency versus load and tables of efficiency and regulation values are presented.
EMEC-II, unit 1
This document provides information about synchronous machines. It discusses:
- Synchronous generators are used to generate electrical power from steam, gas, or hydraulic turbines. They are the primary source of power generation.
- Synchronous machines can operate as generators or motors. Large synchronous motors are commonly used for constant speed industrial drives.
- The document describes the construction, types, operation, and testing of synchronous machines. It provides equations to calculate parameters like voltage, frequency, reactance, and regulation from test data.
- Parallel operation and synchronization of generators is discussed. Concepts like the infinite bus and power-angle characteristics are introduced.
chapter4 DC to AC Converter.ppt
This document summarizes inverters and their operation. It begins with an introduction that defines inverters as devices that convert DC to AC power by switching the DC input voltage in a predetermined sequence. It then discusses the basic principles of inverters including single-phase half-bridge and full-bridge inverter circuits. Fourier series analysis is introduced as a tool to analyze the output waveforms of inverters in terms of harmonic components. The document concludes with a discussion of total harmonic distortion as a measure of output waveform quality.
Solved problems to_chapter_09
1) The document provides solved problems related to power electronics circuits involving thyristors, inverters, and commutation circuits.
2) Problem 9.1 calculates the capacitance needed for a series RLC load to achieve zero current switching for SCRs in a single-phase bridge inverter.
3) Problem 9.2 checks if a self-commutated SCR circuit will reliably commutate when triggered from zero voltage, and calculates voltages and current derivative at the time of commutation.
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This document discusses the design of a three-phase current source inverter. It describes the main components of the current source inverter including a chopper circuit, inverter switching arrangement, and control circuit. Shift registers are used to generate six pulse signals with 60 degree phase shifts that are fed to the thyristors to produce the three-phase output. The current source inverter provides advantages over voltage source inverters like short circuit protection and simpler control circuits.
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lecture-7_examples1.pptx
- 1. Example1 A dc battery of constant emf E is being charged through a resistor using half- wave diode rectifier. For source voltage of 230 V, 50 Hz and for R =8Ω, E =150 V, (a) Find the value of average charging current, (b) Find the power supplied to battery and that dissipated in the resistor, (c) Calculate the supply pf, (d) Find the charging time in case battery capacity is 1000 Wh and (e) Find rectifier efficiency and PIV of the diode. Dr. Oday A. Ahmed
- 2. the value of average charging current Dr. Oday A. Ahmed
- 3. the power supplied to battery the power dissipated in the resistor Dr. Oday A. Ahmed
- 4. Calculate the supply pf, Find the charging time in case battery capacity is 1000 Wh (Power delivered to battery) ×(charging time in hours)= Battery capacity in Wh Hence, charging time is Dr. Oday A. Ahmed
- 5. rectifier efficiency PIV of the diode 𝑃𝑜𝑤𝑒𝑟 𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑒𝑑 𝑡𝑜 𝑏𝑎𝑡𝑡𝑒𝑟𝑦 𝑇𝑜𝑡𝑎𝑙 𝑖𝑛𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟 What is the PIV of the single-phase diode Rectifier Supply R-Load Dr. Oday A. Ahmed
- 6. Dr. Oday A. Ahmed A single-phase 230V: 1kW heater is connected across I-phase 230V, 50Hz supply through an SCR. For firing angle delays of 45° and 90°, calculate the power absorbed in the heater element. 230V, 1kW Heater Example2
- 7. Dr. Oday A. Ahmed Solution Heater resistance: For α=450, RMS output voltage: 𝑉 𝑜𝑟 = 2.230 2 𝜋 𝜋 − 𝜋 4 + 1 2 𝑠𝑖𝑛900 = 155.071 V Power absorbed by heater element for α=450
- 8. Dr. Oday A. Ahmed Power absorbed by heater element for α=450 𝑉 𝑜𝑟 2 𝑅 = 155.071 230 2 × 1000 = 454.57 𝑤𝑎𝑡𝑡 For α=900, RMS output voltage: 𝑉 𝑜𝑟 = 2. 230 2 𝜋 𝜋 − 𝜋 2 + 1 2 𝑠𝑖𝑛1800 = 115 V Power absorbed by heater element for α=900 𝑉 𝑜𝑟 2 𝑅 = 115 230 2 × 1000 = 250 𝑤𝑎𝑡𝑡
- 9. Dr. Oday A. Ahmed In a single-phase full-wave diode bridge rectifier, the diodes have a reverse recovery time of 40 µs. For an AC input voltage of 230 V, determine the effect of reverse recovery time on the average output voltage for a supply frequency of (a) 50 Hz and (b) 2.5 kHz. Solution D 1 and D2 will not be off at ωt =π but will continue to conduct until t=( π/ω )+trr Example3
- 10. Dr. Oday A. Ahmed With zero reverse recovery time, average output voltage,
- 11. Dr. Oday A. Ahmed For f= 50 Hz and trr = 40 µs, the reduction in the average output voltage, Percentage reduction in average outpace voltage
- 12. Dr. Oday A. Ahmed (a) For f= 2.5k Hz, the reduction in the average output voltage, Percentage reduction in average output voltage
- 13. Dr. Oday A. Ahmed Example 4 A separately-excited DC motor, shown below, drives a rated load torque of 85N.M at 1200rpm. The field circuit resistance is 200Ω and the armature resistance is 0.2Ω. The field winding connected to single-phase 400V source fed via single-phase full controlled AC-DC converter with zero degree firing angle. The armature circuit is also fed through another full converter from the same single phase 400V. With magnetic saturation neglected, the motor constant is 0.8Volte-sec/Amp-rad. For ripple free armature and field currents, determine: a. Rated armature current b. Firing angle delay of armature converter at rated load. c. Speed regulation at full load.
- 14. Dr. Oday A. Ahmed Some Points before solving this Example Variable-Speed DC Motor Drive Methods
- 15. Dr. Oday A. Ahmed basic DC motor types
- 16. Under steady-state operation, time derivatives is zero. Assuming the motor is saturated For field circuit, Vf = If Rf Eg = Kv ω if The armature circuit Va = Ia Ra + Eg Va = Ia Ra + Kv ω If Kv is the motor voltage constant (in V/A-rad/s) The motor back emf, which is also known as speed voltage, is expressed :
- 17. Steady-state Torque and Speed The motor speed can be easily derived : If Ra is a small value (which is usual), or when the motor is slightly loaded, i.e, Ia is small That is if the field current is kept constant, the motor speed depends only on the supply voltage. The developed torque is : The required power is : Td = Kt If Ia = B ω + TL Pd = Td ω
- 18. Dr. Oday A. Ahmed
- 19. Dr. Oday A. Ahmed
- 20. Dr. Oday A. Ahmed
- 21. Dr. Oday A. Ahmed It is a two quadrant drive
- 22. Dr. Oday A. Ahmed either converter 1 operates to supply a positive armature voltage Vo, or converter 2 operates to supply negative armature voltage –Vo. Converter 1 provides operation in first and fourth quadrants, and converter 2 operation in second and third quadrants. It is four quadrant drive and provides four modes of operation: forward motoring, forward braking (regeneration), reverse motoring, and reverse breaking (regeneration).
- 23. Dr. Oday A. Ahmed
- 24. Dr. Oday A. Ahmed Return to Example For ripple free armature
- 25. Dr. Oday A. Ahmed
- 26. Dr. Oday A. Ahmed a. Rated armature current Electromagnetic torque Te K has the unit Volte-sec/Amp-rad and equal to 0.8 rated armature current is
- 27. Dr. Oday A. Ahmed a. Firing angle delay of armature converter at rated load. For separately-excited DC motor:
- 28. Dr. Oday A. Ahmed a. Speed regulation at full load. At the same firing angle of 57.630 motor emf at no load, no-load speed is equal to the speed regulation is:
- 29. Dr. Oday A. Ahmed
- 30. Dr. Oday A. Ahmed
- 31. Dr. Oday A. Ahmed