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 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; π = induced emf in volt β = flux density in Tesla π = length of conductor in m π£ = velocity in m/s Note: Tesla = πππππ π2 ; Gauss = πππ₯π€πππ ππ2 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. Given: πππ₯ π = 30ππ; π½ = 100 πππ’π π ( 2 ) ; π = 20ππ; π‘ = 0.1 π ππ ππ From the formula: π = π·ππ Substitute the given, take note π£ = π π‘ π = (πππ π΄ππ ππ ππ ) (ππ ππ) ( ) π ππ π. π πππ π = πππ, πππ Converting to SI units, which is π΄ππ πππ πππππ π ππ π = πππ, πππ π΄ππ π πΎππππ ( ) πππ πππππ π΄ππ π = π. πππ π½ ππ π ππ½ 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 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: E = voltage in volts S = speed in rpm πΊπ·ππ πππ 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? Given πππππ’ππ‘πππ ππ¦ππ ππ π€ππππππ = π ππππππ₯ πππ; π πππ‘π = 48; =4 π πππ‘ Take note, for every coil there 2 conductors, so ππππ = ππ πππππ (π ππππ ππππππ π ππππ )( ) ππππ π ππππ ππππππ ππππ = ππ πππππ Example 2: 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? Given πππππ’ππ‘πππ π = 4; ππ¦ππ ππ π€ππππππ = ππ’ππππ₯ π€ππ£π; π πππ‘π = 48; = 4; π = 2.5π₯106 πππ₯; π = 1500πππ π πππ‘ Using the formula, πΊπ·ππ π¬= πππ Substituting values with π = 2π, and converting φ to weber to conform to SI units π ππππ ππππππ π πΎππππ (ππππ)(π)(ππ πππππ) ( ) (π. πππππ π΄ππ) ( π π΄ππ) ππππ ππππ π¬= ππ(π)(π) π¬ = πππ π½ 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 + L O A D ππΏ πΈ − − 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 c.2 Important terms in armature winding design a. Pole pitch is the distance between two adjacent poles. It is also defined as the number of armature slots per pole. ππππ πππ‘πβ = b. Winding Pitches ππ. ππ πππππ‘π’ππ π πππ‘π π b. Commutator Pitch [Yc] is defined as the distance between two commutator segments to which the two ends. πΉππ π ππππππ₯ πΏππ π€ππππππ, πΉππ π πππππ₯ πππ£π π€ππππππ, π¦π = 2 − 1 = 1 π¦π = 12 − 1 = 11 d. EMF Equation of DC Generator Let, Φ = Flux produced by each pole in weber (Wb) P = number of poles in the DC generator. N = speed of the armature conductor in rpm. Z = total numbers of conductor a = number of parallel paths According to Faraday’s second law of electromagnetic induction, the magnitude of the induced emf in each armature conductor is π= b.1 Back Pitch [Yb] is measured in terms of coil sides. It is always odd number, since it is the difference between odd and even number. π¦π = 14 − 1 = 13 b.2 Front Pitch [Yf] is defined as the distance measured between two coil sides which are connected to some commutator segment. πΉππ πΏππ π€ππππππ, πΉππ πππ£π π€ππππππ, π¦π = 14 − 3 = 11 π¦π = 27 − 14 = 13 b.3 Winding Pitch [Y] is defined as the distance between the starts of two consecutive coils measure in terms of coil sides. πΉππ πΏππ π€ππππππ, πΉππ πππ£π π€ππππππ, pg. 5 π = π¦π − π¦π π = π¦π + π¦π π∅ π‘ππ‘ππ πππ’π₯ = ππ‘ π‘πππ For one revolution of the conductor, total flux produced by all the poles π‘ππ‘ππ πππ’π₯ = ∅π And, Time taken to complete one revolution π‘πππ = Then, the induced emf of one conductor is π= 60 π ∅π ∅ππ = 60 60 π Let us suppose there are Z total numbers of conductor in a generator, and arranged in such a manner that all parallel paths are always in series. Then, Z/a= number of conductors connected in series We know that induced emf in each path is same across the line Therefore, 1st edition MARADAZA2017 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? Given πππππ’ππ‘πππ ππ¦ππ ππ π€ππππππ = π ππππππ₯ πππ; π πππ‘π = 48; =4 π πππ‘ Take note, for every coil there 2 conductors, so ππππ = ππ πππππ (π ππππ ππππππ π ππππ )( ) ππππ π ππππ ππππππ ππππ = ππ πππππ Example 2: 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? Given πππππ’ππ‘πππ π = 4; ππ¦ππ ππ π€ππππππ = ππ’ππππ₯ π€ππ£π; π πππ‘π = 48; = 4; π = 2.5π₯106 πππ₯; π = 1500πππ π πππ‘ Using the formula, πΊπ·ππ π¬= πππ Substituting values with π = 2π, and converting φ to weber to conform to SI units π ππππ ππππππ π πΎππππ (ππππ)(π)(ππ πππππ) ( ) (π. πππππ π΄ππ) ( π π΄ππ) ππππ ππππ π¬= ππ(π)(π) π¬ = πππ π½ 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 + L O A D ππΏ πΈ − − 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 1: 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? Take Note: No-load output voltage = Generated Voltage, E Given: E=120V; Ra=0.95β¦; Rated Power=2000W; Rated Voltage=115V; PL=? Using the formula, π¬ = π°ππΉπ + π½π³ V=115 cannot be used for VL because this is just the stated rated voltage/ideal voltage but in reality it will change. The only thing constant is the load resistance RL, therefore instead of using VL, its equivalent formula will be used which is ππΏ = πΌπΏ π πΏ . In solving for π πΏ , the formula for power will be used which is ππΏ = ππΏ2 , π πΏ so π½ππ³ ππππ πΉπ³ = = = π. ππππβ¦ π·π³ ππππ Going back to the formula, π¬ = π°ππΉπ + π½π³ π¬ = π°ππΉπ + π°π³ πΉπ³ But current is the same, therefore π¬ = π°π (πΉπ + πΉπ³ ) Substituting values, πππ = π°π (π. ππ + π. ππππ) π°π = ππ. ππ π¨ But take note πΌπ = πΌπΏ . Also, in solving for ππΏ , we cannot use the formula P=VI because of the voltage, so the formula used will be π·π³ = π°ππ³ πΉπ³ π·π³ = (ππ. πππ )(π. ππππ) π·π³ = π, πππ. π πΎ Example 2: 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? Given πΈ1 = 110π @ π1 = 1350 πππ; πΈ2 =? @ π 2 = 1600 πππ; πΈ3 =? @ π3 = 1100 πππ For this problem, since the generator will be the same even if the speed is changed, the following ratio and proportion formula can be used, π¬π πΊπ = π¬π πΊπ Solving for πΈ2 πππ ππππ = π¬π ππππ π¬π = πππ. ππ π½ Solving for πΈ3 πππ ππππ = π¬π ππππ π¬π = ππ. ππ π½ 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 πΌπ β π π β + A R M πΌπΏ π πΏ L O A D ππΏ πΈ − − 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 1: 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 Solution: Since the problem stated specifically that the generator voltage and current are at rated values, then we can directly use ππΏ = 250 π πππ ππΏ = 400 ππ For question a, terminal voltage means the voltage at the load side, therefore π»πππππππ π½ππππππ = π½π³ = πππ π½ For question b, π·π = π¬ ∗ π°π For Ia, π°π = π°π³ + π°ππ π°π = π°π = π·π³ + π°ππ π½π³ πππ, πππ + ππ πππ π°π = π, πππ π¨ So, π·π = (πππ. πππ)(ππππ) π·π = πππ, πππ. πππ πΎ For question c, Power output means the power at the load side, therefore π·ππππ πΆπππππ = π·π³ = πππ, πππ πΎ Example 2: 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. Solution: To determine the speed, the only formula that can be used is, πΊπ·ππ π¬= πππ Solving for E, π¬ = π°ππΉπ + π½π³ But, π°π = π°π³ + π°ππ π·π³ π½ππ + π½π³ πΉππ πππ(ππ) πππ π°π = + πππ πππ π°π = π°π = πππ. ππ So, π¬ = (πππ. ππ)(π. πππ) + πππ π¬ = πππ. ππ Therefore, solving for the speed with π = ππ πππ. ππ = πΊ(π)(πππ)(π. πππ) ππ(π)(π) πΊ = πππ. ππ πππ 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 L O A D ππΏ πΈ − − 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: 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. Solution: Diverter resistance is just a resistor in parallel with the series field resistance. Rx π π π + πΌπ π πΌπ π π πΌπΏ π π π πΏ + A R M L O A D ππΏ πΈ − − Solving for the total resistance between the parallel connection of the series field resistance and the diverter resistance, Rx, (πΉππ )(πΉπ ) πΉπ = πΉππ + πΉπ πΉπ = (π. π)(π. π) π. π + π. π πΉπ = π. ππ From this, the equivalent circuit becomes, π π₯ + πΌπ₯ πΌπ π π πΌπΏ π πΏ A R M + L O A D ππΏ πΈ − − From this, the circuit is just the same as an ordinary series generator. Take note that the induced emf in question is the generated voltage, E, and recall also that terminal voltage is the load voltage VL. From the problem, it was given that the load draws current of 4.5A, therefore, the load current is 4.5A. From this, solving for the generated voltage π¬ = π°ππΉπ + π°πππΉππ + π°π³ πΉπ³ Instead of using πΌπΏ π πΏ , we will use the given load voltage of 600V. Also, π π π will now become π π . Lastly, since current is the same for series generator, π¬ = π°π³ (πΉπ + πΉπΏ ) + π½π³ π¬ = (π. π)(π. π + π. ππ) + πππ π¬ = πππ π½ C. Compound Generator - a shunt generator with additional series field winding 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 πΌπΏ πΌπ β π πΏ π π β + A R M L O A D ππΏ πΈ − − 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 π β π πΏ L O A D ππΏ πΈ − − 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 = π·π ∗ πππ% π·π or %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 1: 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. Given: πΌπΏ = 800π΄; ππΏ = 230π; πΌπ β = 12π΄; π π = 0.007β¦; π π π = 0.002β¦; ππππβ = 5500; %πππ =? Solving first electrical losses due to all the resistance of the windings starting with the shunt field winding, but, since the shunt field resistance is not given, then the following formula will be used π·ππ = π½ππ π°ππ But, the shunt field resistance is in parallel with the load, therefore ππ β = ππΏ π·ππ = π½π³ π°ππ π·ππ = (πππ)(ππ) π·ππ = ππππ πΎ Next is the series field winding, since the series field winding resistance is given, the following formula can be used, π·ππ = π°πππ πΉππ But, π°ππ = π°π³ + π°ππ π°ππ = πππ + ππ π°ππ = πππ So, π·ππ = (πππ)π (π. πππ) π·ππ = π, πππ. πππ πΎ Last is for the armature winding, since the armature resistance is given, the following formula can be used, π·π = π°ππ πΉπ But, armature is in series with the series field, therefore π·π = π°πππ πΉπ π·π = (πππ)π (π. πππ) π·π = π, πππ. πππ πΎ Solving for the efficiency of the generator, %π¬ππ = %π¬ππ = π·π ∗ πππ% π·π + π·π³πΆπΊπΊπ¬πΊ (πππ)(πππ) ∗ πππ% (πππ)(πππ) + ππππ + ππππ + ππππ. πππ + ππππ. πππ %π¬ππ = ππ. ππ% Example 2: 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. Given: πΌπΏ = 50π΄; ππΏ = 250π; π π = 0.05β¦; π π π = 0.03β¦; π π β = 250β¦; %πππ =? Solving first electrical losses due to all the resistance of the windings starting with the series field winding since it is just in series with the load, π·ππ = π°πππ πΉππ Since it is in series with the load, then πΌπ π = πΌπΏ π·ππ = π°ππ³ πΉππ π·ππ = (ππ)π (π. ππ) π·ππ = ππ πΎ Next is the shunt field winding, π·ππ = π°πππ πΉππ π½ππ π π·ππ = ( ) πΉππ πΉππ π·ππ = ( π·ππ = [ π½π³ + π½ππ π ) πΉππ πΉππ π½π³ + (π°ππ )(πΉππ ) π ] πΉππ πΉππ π π½π³ + (π°ππ )(πΉππ ) π·ππ = [ ] πΉππ πΉππ π πππ + (ππ)(π. ππ) π·ππ = [ ] (πππ) πππ π·ππ = πππ. π πΎ Last is for the armature winding, π·π = π°ππ πΉπ But π°π = π°ππ + π°ππ π°π = π°π = π½π³ + (π°ππ )(πΉππ ) + π°ππ πΉππ πππ + (ππ)(π. ππ) + ππ πππ π°π = ππ. πππ π¨ so, π·π = (ππ. πππ)π (π. ππ) π·π = πππ. ππ πΎ Therefore, solving for the efficiency of the generator, %π¬ππ = %π¬ππ = π·π ∗ πππ% π·π + π·π³πΆπΊπΊπ¬πΊ (ππ)(πππ) ∗ πππ% (ππ)(πππ) + πππ + ππ + πππ. π + πππ. ππ %π¬ππ = ππ. ππ% DC MOTORS - A machine that converts electrical energy to mechanical energy Speed Characteristics of DC Motors πΊ = ππ π¬π π Where: S = Speed in rpm kb = proportionality constant Eb = Back emf or Counter emf = as the armature rotates, it also induces its own voltage Φ = flux per pole Torque Characteristics of DC Motors Torque – turning moment of rotating body Consider a shaft, π πΉ π» = π ∗ π (π΅. π) Where: T = Torque in Newton-meter F = Force in Newton r = radius in meters HORSEPOWER DEVELOPED IN A ROTATIONAL BODY π―π· = ππ πΊπ» πππππ Where: HP = mechanical power in horsepower S = speed in rpm T = torque in N-m Note: π π―π· = πππ πΎ From this, a variation of a formula for torque can be developed π» = ππ ππ°π Where: T = torque kt = proportionality constant φ = flux per pole πΌπ = armature current TYPES OF DC MOTORS A. Shunt Motor + πΌπ πΌπ π π πΌπ β ππ π π β + πΈπ A R M − − Where: Vs = Supply Voltage Is = Supply Current Current Analysis: π°π = π°ππ + π°π Voltage Analysis: π½π − π°ππΉπ − π¬π = π or π½π = π°ππΉπ + π¬π Power Analysis: π·π = π½ππ°π and π·π = π¬ππ°π Where: Ps = Supply Power in Watts Pd = Power Developed in the armature Example: A 50Hp, 550V shunt motor has an armature resistance of 0.36β¦ when operating at rated load speed while the armature takes 75A. What resistance should be inserted in the armature circuit to obtain 20% reduction in flux and when developing 70% of the rated torque? Given: π = 50π»π; ππ = 550π; π π = 0.36; πΌπ1 = 75π΄; π π =?; π2 = 0.8π1 ; π2 = 0.7π1 This problem has two scenarios where the first scenario is the standard shunt motor and the second scenario has an added resistance to the armature. First, recall the characteristics of Torque π» = ππ ππ°π But, π»π = π. ππ»π ππ ππ π°ππ = π. π(ππ ππ π°ππ ) But, constant remains constant as long as it’s the same motor and also from the problem π2 = 0.8π1 , thus ππ (π. πππ )π°ππ = π. π(ππ ππ π°ππ ) π°ππ = π. π (π° ) π. π ππ π. π (ππ) π. π = ππ. ππππ¨ π°π π = π°ππ With the resistance inserted, the circuit becomes like this, + πΌπ πΌπ2 π π πΌπ β ππ π π β + πΈπ2 A R M − − Using the formula, π½π = π°ππΉπ + π¬π But this time, since there is a resistance in series with the armature, the formula becomes, π½π = π°ππ (πΉπ + πΉπ) + π¬ππ This formula now needs πΈπ2 to be solved, recall the characteristics of motor speed, πΊ = ππ π¬π π Since the motor will still be the same even if there is an added resistance, and the problem did not state any changes to the motor speed after adding the resistance, it is safe to assume that π2 = π1. From this, we can have a ratio and proportion approach, π¬π πΊπ = ππ π π π π¬ππ πΊπ = ππ π π Since it is just the same motor, constant will stay the same. Add in also the scenario for flux which is π2 = 0.8π1 , the formula now becomes π¬ππ ππ π¬ππ πΊπ = ππ π. π(ππ ) πΊπ = ππ πΊπ π¬ππ = πΊπ π¬ππ π. π But since π2 = π1 , then π= π¬ππ π¬ππ π. π π¬ππ = π. ππ¬ππ Solving first for πΈπ1 , π½π = π°ππ πΉπ + π¬ππ πππ = (ππ)(π. ππ) + π¬ππ π¬ππ = ππππ½ Solving for πΈπ2 , π¬ππ = π. ππ¬ππ π¬ππ = π. π(πππ) π¬ππ = πππ. π Solving for π π π½π = π°ππ (πΉπ + πΉπ) + π¬ππ πππ = (ππ. πππ)(π. ππ + πΉπ) + πππ. π πΉπ = π. πππβ¦ B. Series Motor π π π + πΌπ πΌπ π πΌπ π π ππ + πΈπ − A R M − Current Analysis: π°π = π°ππ = π°π Voltage Analysis: π½π − π°πππΉππ − π°ππΉπ − π¬π = π or π½π = π°πππΉππ + π°ππΉπ + π¬π or π½π = π°π(πΉππ + πΉπ) + π¬π; π½π = π°ππ(πΉππ + πΉπ) + π¬π; π½π = π°π(πΉππ + πΉπ) + π¬π Power Analysis: π·π = π½ππ°π and π·π = π¬ππ°π Example: A 400V series motor working with an unsaturated field is taking 60A and running at 840RPM. If the armature and series field resistances are 0.04β¦ and 0.06β¦ respectively, at what speed will the motor run when developing half-full load torque? Given: ππ = 400π; πΌπ = 60π΄; π1 = 840 πππ; π π = 0.04β¦; π π π = 0.06β¦; π2 =?; π2 = 0.5π1 Recall the torque characteristics, π» = ππ ππ°π But, π»π = π. ππ»π ππ ππ π°ππ = π. π(ππ ππ π°ππ ) The problem did not state that there are changes in flux but flux is directly proportional to the field winding current and with the change in torque there will be changes in the armature current. Considering the motor connected in series, a change in armature current will also mean a change in field winding current, therefore a change will also occur in the field winding flux. Recall, π = π′ π°π For series motor, πΌπ = πΌπ π, but also πΌπ π = πΌπ, thus π = π′ π°π Substituting flux to the formula from torque, ππ (π′ π°ππ )π°ππ = π. π(ππ (π′ π°ππ )π°ππ ) Constant remains the same, thus (π°ππ )π = π. π(π°ππ )π π°ππ = √π. π(π°ππ )π π°ππ = (√π. π)(π°ππ ) π°ππ = (√π. π)(ππ) π°ππ = ππ. ππππ¨ Since the problems asks for the motor speed after the change in torque, the same ratio and proportion can be used therefore, π¬π πΊπ = ππ π π π π¬ππ πΊπ = ππ π π Solving first for πΈπ1 , π½π = π°ππ (πΉππ + πΉπ) + π¬ππ πππ = (ππ)(π. ππ + π. ππ) + π¬ππ π¬ππ = ππππ½ Solving for πΈπ2 , π½π = π°ππ (πΉππ + πΉπ) + π¬ππ πππ = (ππ. πππ)(π. ππ + π. ππ) + π¬ππ π¬ππ = πππ. πππππ½ Going back to the ratio and proportion π¬π πΊπ = ππ π π π π¬ππ πΊπ = ππ ππ π¬ππ π′ π°ππ π¬π πΊπ = ππ ′ π π π°ππ πΊπ = ππ πππ πππ ππ = πππ. ππππ πΊπ ππ. πππ πΊπ = π, πππ. ππ πΉπ·π΄ C. Compound Motor 1. Long-Shunt Compound Motor π π π + πΌπ πΌπ π πΌπ π π πΌπ β ππ π π β + πΈπ − A R M − Current Analysis: π°ππ = π°π so π°π = π°ππ + π°ππ Or π°π = π°ππ + π°π Voltage Analysis: π½π − π°πππΉππ − π°ππΉπ − π¬π = π or π½π = π°πππΉππ + π°ππΉπ + π¬π or π½π = π°ππ(πΉππ + πΉπ) + π¬π; π½π = π°π(πΉππ + πΉπ) + π¬π Power Analysis: π·π = π½ππ°π and π·π = π¬ππ°π 2. Short-Shunt Compound Motor π π π + πΌπ πΌπ π πΌπ π π πΌπ β ππ π π β + πΈπ A R M − − Current Analysis: π°π = π°ππ = π°ππ + π°π Voltage Analysis: π½π − π°πππΉππ − π°ππΉπ − π¬π = π or π½π = π°πππΉππ + π°ππΉπ + π¬π or π½π = π°ππΉππ + π°ππΉπ + π¬π Power Analysis: π·π = π½ππ°π and π·π = π¬ππ°π Example 5: A 550V long-shunt compound motor has an armature resistance of 0.815β¦ and series field resistance of 0.15β¦. the full-load speed is 1900rpm when the armature current is 22A. At what speed will the motor operate at noload if the armature current drops to 3A with a corresponding drop in flux to 90% of the full-load value? Given: ππ = 550π; π π = 0.815β¦; π π π = 0.15β¦; π1 = 1900πππ; πΌπ1 = 22π΄; π2 =?; πΌπ2 = 3π΄; π2 = 0.9π1 For this problem, the same ratio and proportion will be used, π¬π πΊπ = ππ π π π π¬ππ πΊπ = ππ π π Solving first for πΈπ1 , since this problem is a long-shunt compound motor, then π½π = π°ππ (πΉππ + πΉπ) + π¬ππ πππ = (ππ)(π. ππ + π. πππ) + π¬ππ π¬ππ = πππ. ππ π½ Solving for πΈπ2 , π½π = π°ππ (πΉππ + πΉπ) + π¬ππ πππ = (π)(π. ππ + π. πππ) + π¬ππ π¬ππ = πππ. πππ π½ Back to the ratio and proportion formula, π¬π πΊπ = ππ π π π π¬ππ πΊπ = ππ π π π¬π πΊπ = ππ π π π π¬ππ πΊπ = ππ π. π(ππ ) πΊπ π¬ππ = πΊπ π¬ππ π. π ππππ πππ. ππ = πππ. πππ πΊπ π. π πΊπ = π, πππ. ππ πππ MOTOR STARTING - at starting, the armature of the motor draws high current due to zero back emf. To limit the current, a rheostat is inserted/ connected in series with armature circuit. Example: A shunt motor inserted with a starting resistor π π + πΌπ πΌπ π π πΌπ β ππ π π β + πΈπ − A R M − Current Analysis: π°π = Where: π½π πΉπ + πΉπ Rs = Starting Resistance Note: After the motor developed its rated speed/torque or the motor starts to rotate, the rheostat must set back to zero/minimum value so the motor characteristics is not affected Example 1: A shunt motor is running at the speed of 720RPM and draws 50A from 500V source. The armature and field resistance are 0.4 & 250β¦ respectively. Determine; a. starting resistance to maintain the torque b. ohmic value of the rheostat to reduce the speed to 540RPM. Solution: a. to maintain the torques means to maintain the armature current, thus the given 50A must be maintained. Therefore, ππ = πππ π. π + πΉπ πΉπ = π. πβ¦ b. for the speed to drop to 540rpm, the starting resistor is still connected but the resistance is adjusted, with this solving for π π should follow the same ratio and proportion formula. Solving first for πΈπ1 during when the circuit has no resistance connected, πππ − (ππ)(π. π) − π¬ππ = π π¬ππ = πππ Solving for πΈπ2 , using the ratio and proportion formula, π¬ππ ππ π¬ππ πΊπ = ππ π π πΊπ = ππ Since the problem did not state that there changes in flux or torque, it is safe to assume that they are the same, therefore the resulting formula will be, πΊπ π¬ππ = πΊπ π¬ππ πππ πππ = πππ π¬ππ π¬ππ = πππ Solving for π π , π½π − π°ππ (πΉπ + πΉπ) − π¬ππ = π Since it was assumed that the torque is the same, therefore the armature current will still be the same with the starting resistance, thus πππ − (ππ)(π. π + πΉπ) − πππ = π πΉπ = π. πβ¦
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