DC GENERATORS - A machine that converts mechanical energy to electrical energy DIAGRAM COUPLING PRIME MOVER (MECHANICAL) DC GENERATOR (ELECTRICAL) Prime Mover - A machine or equipment that drives the generator TYPES OF PRIME MOVER 1. Motor 2. Turbine COMMON SOURCES OF MECHANICAL ENERGY 1. Water/River – Hydroelectric 2. Steam – Thermal Plant (Geothermal, Coal Fired, Dentrothermal, Nuclear) 3. Wind - Windmill 4. Waves 5. Heat from the sun –Solar 6. Combustible Materials – Diesel/Gasoline, Gas Turbine FLUX CUTTING Flemmings Right Hand Rule - The thumb represents the direction of motion of the conductor The first finger represents field The second finger represents current FARADAY’S PRINCIPLE OF GENERATING EMF - According to Michael Faraday, a voltage can be induced in a conductor if the said conductor is moving across the magnetic field that results to cutting of magnetic flux. Basic Generator Operation An armature rotates through the magnetic field. At an initial position of zero degrees, the armature conductors are moving parallel to the magnetic field and not cutting through any magnetic lines of flux. No voltage is induced. The armature rotates from 0 to 90°. The conductors cut through more and more lines of flux, building up to a maximum induced voltage in the positive direction. The armature continues to rotate from 90 to 180°, cutting less lines of flux. The induced voltage decreases from a maximum value to zero. The armature continues to rotate from 180° to 270°. The conductors cut more and more lines of flux, but in the opposite direction. Voltage is induced in the negative direction building up to a maximum at 270°. The armature continues to rotate from 270 to 360°. Induced voltage decreases from a maximum negative value to zero. This completes one cycle. The armature will continue to rotate at constant speed. The cycle will continuously repeat as long as the armature rotates. THREE IMPORTANT PARAMETERS IN INDUCING EMF 1. Conductor 2. Speed 3. Magnetic Flux Basing on Faraday’s Principle π = π·ππ (πππ ππππ πππππ) Where; Note: Tesla = π = induced emf in volt β = flux density in Tesla π = length of conductor in m π£ = velocity in m/s πππππ π2 (SI Unit) ; Gauss = πππ₯π€πππ ππ2 (CGS Unit) 1 weber = 1 x 108 max 1 maxwell = 1 magnetic line of force Also: π= π½ (πππ ππππ πππππ) π Where; θ = flux in wb t = time in cutting the flux but: if we let N = no. of conductors in series then, π½ π = π΅( ) π Ex. A 30cm conductor placed to slide along a parallel metal bar spaced 30cm apart. The end of the railway is connected to a voltmeter. A uniform magnetic field density of 100 gauss passes perpendicular through the rail and conductor. If the conductor will move along the rail at a uniform velocity of 20cm in 0.1 sec, determine the voltage registered on the voltmeter. PARTS OF A BASIC D.C. GENERATOR 1. Yoke or Frame or Casing – protects the entire machine 2. Armature core – made up of laminated sheet of iron (that has very low reluctance). It carries the conductors that are responsible in cutting the flux. 3. Armature Winding – made up of copper wires or copper bars placed on the slots of the armature core. 4. Pole Core – made up of laminated sheet of iron bolted on the frame. 5. Field Winding – made up of copper wire usually smaller in size 6. Commutator – made up of copper brass insulated in between. It is used to rectify the A.C. signals to D.C. signals. Also called as mechanical rectifier. 7. Brush – made up of carbon compound used to transmit the current to the load without twisting the wire. COMMUTATOR ACTION IN RECTIFYING A.C. SIGNALS Flemmings Right Hand Rule - The thumb represents the direction of motion of the conductor The first finger represents field The second finger represents current Simple loop generator with split ring - The output voltage always has the same polarity The current is a pulsating current To produce a steady current, many loops and commutators around the axis of rotation are used TYPES OF WINDING 1. Lap Winding - the windings are connected in parallel 2. Wave Winding - the windings are connected in series PARALLEL PATHS - the path for the current generated - represented by “a” If lap winding, then π = ππ If wave winding, then π = ππ Where: m = the number of conductors used simplex: m = 1 duplex: m = 2 triplex: m = 3 etc… p = number of magnetic poles GENERATORS INDUCED VOLTAGE π¬= Where: πΊπ·ππ πππ E = voltage in volts S = speed in rpm P = number of poles Z = number of conductors φ = magnetic flux in weber a = number of parallel paths Example 1: A four pole D.C. generator with simplex lap winding has 48 slots and 4 conductors per slot. How many coil does it have? Example 2: A 900rpm, 6-pole generator has simplex lap winding. There are 300 conductors in the armature. The poles are 10 inches square and the flux density is 50,000 maxwells per square inch. Determine the induced emf. Example 3: A four pole D.C. generator with duplex wave winding has 48 slots and 4 conductors per slot. The flux per pole is 2.5x106 maxwells and it runsat 1500rpm. What is the induced emf? TYPES OF DC GENERATOR ACCORDING TO THE TYPE OF EXCITER USED 1. Separately-Excited Generator - the field winding is supplied from a separate source Circuit Diagram: + πΌπ πΌπ ππ π π πΌπΏ π πΏ π π A R M + πΈ − L O A D ππΏ − Where: ππ = Field Voltage πΌπ = Field Current π π = Field Resistance πΈ = Generated Voltage π π = Armature Resistance πΌπ = Armature Current ππΏ = Load Voltage πΌπΏ = Load Current π πΏ = Load Resistance Flux Analysis: π = π ∗ π΅ ∗ π°π Where: k = proportionality constant N = no. of turns (but if it is constant, then) π = π′ ∗ π°π Current Analysis: π°π = π°π³ Voltage Analysis: π¬ − π°ππΉπ − π°π³ πΉπ³ = π or π¬ = π°ππΉπ + π½π³ or π¬ = π°π(πΉπ + πΉπ³ ) ; π¬ = π°π³ (πΉπ + πΉπ³ ) Power Analysis: π·π = π¬ ∗ π°π and π·π³ = π½π³ π°π³ Where: Pg = Generated Power PL = Load Power Example 4: A DC Generator has no-load output voltage of 120V. Its armature circuit resistance is 0.95β¦ and its field coil is separately excited. If a load rated at 2kW, 115V is connected across the terminal, what power would be absorbed by the load? Example 5: If a no-load voltage of a separately excited DC Generator is at 110V at 1350rpm, what would be the voltage if the speed increases to 1600rpm? And if it is decreased to 1100rpm? Example 6: A separately-excited DC Generator has an effective armature resistance of 0.04β¦. it supplies a load current of 200A at 125V to a constant load when driven at 1000rpm. Calculate the current that the generator delivers if the speed is 800rpm. The field excitation remains unchanged and allow a brush voltage drop of 2V. 2. Self-Excited Generator - the source of the field circuit is from the generated voltage of the armature TYPES OF SELF-EXCITED GENERATOR A. Shunt Generator - the field circuit is in parallel with the armature. Circuit Diagram: + πΌπ π π A R M + πΌπ β πΌπΏ π π β π πΏ πΈ L O π πΏ A D − − Where: πΌπ β = Shunt current ππ β = Shunt Voltage Current Analysis: π°π = π°ππ + π°π³ Voltage Analysis: π¬ − π°ππΉπ − π°π³ πΉπ³ = π Or π¬ = π°ππΉπ + π½π³ Power Generated: π·π = π¬ ∗ π°π and π·π³ = π½π³ π°π³ Example 7: The emf induced in the armature at 400kW, 250V shunt generator is 258.866V when the terminal voltage and load current are at rated values and the shunt field current is 12A. The armature resistance including the brushes is 0.0055β¦. Find, (a) Terminal Voltage (b) Power Generated (c) Power Output Example 8: A 4-pole shunt generator has a simplex lap connected armature with 728 conductors and a flux per pole of 25πWb. The generator supplies two hundred fifty 110V, 75W incandescent lamps. The field and armature resistances are 110β¦ and 0.025β¦ respectively. Determine the speed of this generator. Example 9: A shunt generator supplies a load of 5500W at 110V through a pair of feeder conductors having a resistance of 0.02β¦ each. The armature and shunt field resistances are 0.15β¦ and 50β¦ respectively. Find the generated emf. B. Series Generator - the field winding is connected in series with the armature. Circuit Diagram: π π π + πΌπ πΌπ π π π πΌπΏ π πΏ A R M + πΈ − L O π πΏ A D − Where: πΌπ π = Series Field Current π π π = Series Field Resistance Current Analysis: π°π = π°ππ = π°π³ Voltage Analysis: π¬ − π°ππΉπ − π°πππΉππ − π°π³ πΉπ³ = π or π¬ = π°ππΉπ + π°πππΉππ + π°π³ πΉπ³ or π¬ = π°π (πΉπ + πΉππ + πΉπ³ ); π¬ = π°ππ (πΉπ + πΉππ + πΉπ³ ); π¬ = π°π³ (πΉπ + πΉππ + πΉπ³ ) Power Generated: π·π = π¬ ∗ π°π and π·π³ = π½π³ π°π³ Example 10: A series generator is used to supply a series lighting system that draws a total current of 4.5A. if the armature and the series field resistances are 0.1β¦ and 0.2β¦ respectively. Calculate the induced emf if the terminal voltage is 600V with a diverter resistance of 0.3β¦ connected across the field circuit. C. Compound Generator - a shunt generator with additional series field winding Types of Field According to the Connection of the Series Field with Respect to the Shunt Field 1. Commulative-Compounded - the flux produced by each field are in the same direction. ππ» = πππ + πππ 2. Differential-Compounded - for special reason, the flux produced by each field are at opposite direction. ππ» = πππ + πππ Degree of Compounding 1. Flat-Compound Generator - if the ππΉπΏ = πππΏ - if the transmission distance between the generator and load is short 2. Over-Compound Generator - if ππΉπΏ > πππΏ - if the transmission distance is long 3. Under-Compound Generator - if ππΉπΏ < πππΏ - if the transmission distance is medium distance Types of Compound Generator 1. Long-Shunt Compound Generator - the series field is connected in series with the armature Circuit Diagram: π π π + πΌπ πΌπ π π π A R M πΌπ β + πΈ π π β πΌπΏ π πΏ L O π πΏ A D Current Analysis: π°π = π°ππ = π°ππ + π°π³ Voltage Analysis: π¬ − π°ππΉπ − π°πππΉππ − π°π³ πΉπ³ = π or π¬ = π°ππΉπ + π°πππΉππ + π°π³ πΉπ³ or π¬ = π°π(πΉπ + πΉππ) + π½π³ ; π¬ = π°ππ(πΉπ + πΉππ) + π½π³ 2. Short-Shunt Compound Generator - the series field is in series with the load Circuit Diagram: π π π + πΌπ πΌπ π π π πΌπΏ πΌπ β A R M + π πΏ π π β πΈ − L O π πΏ A D − Current Analysis: π°ππ = π°π³ and π°π = π°ππ + π°π³ or π°π = π°ππ + π°ππ Voltage Analysis: π¬ − π°ππΉπ − π°πππΉππ − π°π³ πΉπ³ = π or π¬ = π°ππΉπ + π°πππΉππ + π°π³ πΉπ³ or π¬ = π°ππΉπ + π°ππ(πΉππ + πΉπ³ ); π¬ = π°ππΉπ + π°π³ (πΉππ + πΉπ³ ) GENERATORS EFFICIENCY - the ratio of the power output to that of the power input %π¬ππ = or π·π ∗ πππ% π·π %π¬ππ = π·π ∗ πππ% π·π + π·π³πΆπΊπΊπ¬πΊ Types of Power Losses 1. Electrical Losses - losses due to its windings and brush contact a) Copper Loss, Pcu a.1) Armature Loss, Pa ππ = πΌπ2 π π a.2) Shunt Field Loss, Psh ππ β = πΌπ β2 π π β a.3) Series Field Loss, Pse ππ π = πΌπ π 2 π π π a.4) Diverter Loss, Pd ππ = πΌπ2 π π a.5) Brush Contact Loss, Pb ππ = πΌπ 2 π π 2. Mechanical Losses - sometimes called as CONSTANT LOSS or STRAY POWER LOSS or ROTATIONAL LOSS - it is constant provided that the speed and flux are also constant a) Core Loss, Pco a.1) Eddy Current Loss, Pe ππ = ππ π 2 π 2 a.2) Hysteresis Loss, Ph πβ = πβ π(π½π)1.6 Where: βm = maximum flux density b) Friction and Windage Loss, Pfw - due to friction, ventilation bearing, vibration etc. c) Stray Load Loss, PSL - flux distortion - 1% of the output for machines 150kW and over Example 11: A 250kW, 230V compound generator is delivering 800A at 230V. the shunt field current is 12A. Armature resistance is 0.007β¦ and series field resistance is 0.002β¦. The stray power loss is 5500W and the generator is connected long shunt. Determine the full-load efficiency at rated voltage. Example 12: A short-shunt compound generator supplies a load of 50A at a terminal voltage of 250V. the armature, series field and shunt field resistances are 0.05β¦, 0.03β¦ and 250β¦ respectively. Calculate the over-all efficiency if the constant loss is 325W. Example 13: A 230V shunt generator has an armature resistance of 0.03β¦ and shunt field resistance of 110β¦. At rated load, the driving engine delivered 59.5Hp. If the stray power loss is 1550W calculate the current delivered to the load and the over-all efficiency. PARALLEL OPERATION OF D.C. GENERATORS - power plants generally have several small generators rather than one large unit capable of taking care of the maximum peak loads. Advantages Over Single Operation 1. Proper sharing of loads because every machine can be adjusted by its speed or field, meaning no machine is overloaded. 2. Higher efficiency because the power plant will operate according to its demand with respect to time. 3. Continuity of service with slight power interruption. 4. Proper maintenance is met because every unit can be shut down while others are operating. 5. Easy to add another machine for increase of load demand. Requirements for Parallel Operation 1. The same external characteristics 2. Terminal Voltage and Polarity must be the same Voltage Regulation - percent change from no-load to full-load voltage with respect to full-load voltage. %ππ = πππΏ − ππΉπΏ ∗ 100% ππΉπΏ Characteristic Triangle of DC Generator π πππΏ ππ ππΉπΏ 0 βπ βπΌ πΌπ πΌπΉπΏ πΌ By similar triangle: βπ½ π½π΅π³ − π½ππ³ = βπ° π°ππ³ or βπ½ %π½πΉ ∗ π½ππ³ = βπ° π°ππ³ Example 14: Two 220V generators operate in parallel. One machine has a terminal voltage of 270V at no-load and 220V at a full-load current of 35A. The other machine has a terminal voltage of 280V at no-load and 220V at a full-load current of 50A. Calculate the new bus voltage and power share of each machine if the total load current is 60A. Example 15: Two DC shunt generators in parallel supply together a 2000A load current. The machines have armature resistances of 0.04β¦ and 0.025β¦, field resistances of 25β¦ and 20β¦, and induced emf’s of 440V and 420V respectively. Find the terminal voltage and power share of each machine.
