Download Chapter One D.C. Generators and more Schemes and Mind Maps Engineering in PDF only on Docsity!
College of Engineering Mech. Eng. Dept. Prepared by Mohanad A. A. Alheety ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ
Chapter One
D.C. Generators
Alternating Current (AC) In alternating current the electric charges flow changes its direction periodically. AC is the most commonly used and most preferred electric power for household equipment office and buildings Alternating current can be identified in wave form called as sine wave Direct Current (DC) Unlike alternating current, the flow of current in direct current do not changes periodically. The current flows in a single direction in a steady voltage. The major uses of DC is to supply power for electrical devices and also to charge batteries. For example, mobile phone batteries, flashlights, flat-screen television, hybrid and electric vehicles.
College of Engineering Mech. Eng. Dept. Prepared by Mohanad A. A. Alheety ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ Generator Principle An electric generator is a machine that converts mechanical energy into electrical energy. An electric generator is based on the principle that whenever flux is cut by a conductor, an e.m.f. is induced which will cause a current to flow if the conductor circuit is closed. The direction of induced e.m.f. (and hence current) is given by Fleming’s right hand rule. Therefore, the essential components of a generator are: (a) A magnetic field (b) Conductor or a group of conductors (c) Motion of conductor w.r.t. magnetic field. Difference Between Alternating Current and Direct Current Alternating Current Direct Current AC can carry and safe to transfer longer distance even between two cities, and maintain the electric power. DC cannot travel for very longer distance. If does, it loses electric power. The rotating magnets cause the change in direction of electric flow. The steady magnetism makes the DC to flow in a single direction. The frequency of AC is depended upon the country. But, generally the frequency is 50Hz or 60Hz. DC has no frequency of zero frequency. In AC the flow of current changes it direction backwards periodically. It flows in single direction steadily. Electrons in AC keep changing its directions – backward and forward Electrons only move in one direction – that is forward.
College of Engineering Mech. Eng. Dept. Prepared by Mohanad A. A. Alheety ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ to the flux and, therefore, a low e.m.f. is generated as indicated by point 2. (iii) When the loop is in position no. 3, the coil sides (AB and CD) are at right angle to the flux and are, therefore, cutting the flux at a maximum rate. Hence at this instant, the generated e.m.f. is maximum as indicated by point 3 in Figure (iv) At position 4, the generated e.m.f. is less because the coil sides are cutting the flux at an angle. (v) At position 5, no magnetic lines are cut and hence induced e.m.f. is zero as indicated by point 5 in Fig. (vi) At position 6, the coil sides move under a pole of opposite polarity and hence the direction of generated e.m.f. is reversed. The maximum e.m.f. in this direction (i.e., reverse direction) will be when the loop is at position 7 and zero when at position 1. This cycle repeats with each revolution of the coil.
College of Engineering Mech. Eng. Dept. Prepared by Mohanad A. A. Alheety ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ Note that e.m.f. generated in the loop is alternating one. It is because any coil side, say AB has e.m.f. in one direction when under the influence of N-pole and in the other direction when under the influence of S-pole. If a load is connected across the ends of the loop, then alternating current will flow through the load. The alternating voltage generated in the loop can be converted into direct voltage by a device called commutator. We then have the d.c. generator. In fact, a commutator is a mechanical rectifier. Commutator If, somehow, connection of the coil side to the external load is reversed at the same instant the current in the coil side reverses, the current through the load will be direct current. This is what a commutator does. The figure shows a commutator having two segments C 1 and C 2. It consists of a cylindrical metal ring cut into two halves or segments C 1 and C 2 respectively separated by a thin sheet of mica. The commutator is mounted on but insulated from the rotor shaft. The ends of coil sides AB and CD are connected to the segments C 1 and C 2 respectively as shown. Two stationary carbon brushes rest on the commutator and lead current to the external load. With this arrangement, the commutator at all times connects the coil side under S-pole to the +ve brush and that under N-pole to the −ve brush. (i) In Fig below, the coil sides AB and CD are under N-pole and S-pole respectively. Note that segment C 1 connects the coil side AB to point P of the load resistance R and the segment C 2 connects the coil side CD to point Q of the load. Also note the direction of current through load. It is from Q to P.
College of Engineering Mech. Eng. Dept. Prepared by Mohanad A. A. Alheety ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ Construction of a DC Generator
- Yoke: The outer frame of a dc machine is called as yoke. It is made up of cast iron or steel. It not only provides mechanical strength to the whole assembly but also carries the magnetic flux produced by the field winding.
- Poles and pole shoes: Poles are joined to the yoke with the help of bolts or welding. They carry field winding and pole shoes are fastened to them. Pole shoes serve two purposes; (i) they support field coils and (ii) spread out the flux in air gap uniformly.
College of Engineering Mech. Eng. Dept. Prepared by Mohanad A. A. Alheety ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ
- Field winding: Each pole core has one or more field coils (windings) placed over it to produce a magnetic field. The copper wire is used for the construction of field or exciting coils. The coils are wound on the former and then placed around the pole core When direct current passes through the field winding, it magnetizes the poles, which in turns produces the flux. The field coils of all the poles are connected in series in such a way that when current flows through them, the adjacent poles attain opposite polarity.
- Armature core: Armature core is the rotor of the machine. It is cylindrical in shape with slots to carry armature winding. The armature is built up of thin laminated circular steel disks for reducing eddy current losses. The armature core of a DC generator or machine serves the following purposes.
- It houses the conductors in the slots.
- It provides an easy path for the magnetic flux
College of Engineering Mech. Eng. Dept. Prepared by Mohanad A. A. Alheety ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ Separately Excited D.C. Generators A d.c. generator whose field magnet winding is supplied from an independent external d.c. source (e.g., a battery etc.) is called a separately excited generator. Fig. below shows the connections of a separately excited generator. The voltage output depends upon the speed of rotation of armature and the field current The greater the speed and field current, greater is the generated e.m.f. It may be noted that separately excited d.c. generators are rarely used in practice. The d.c. generators are normally of self-excited type.
College of Engineering Mech. Eng. Dept. Prepared by Mohanad A. A. Alheety ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ Armature current, Ia = IL e.m.f generated, Eg = V + IaRa Electric power developed = EgIa Power delivered to load = 𝑉𝐼𝑎 Self-Excited D.C. Generators A d.c. generator whose field magnet winding is supplied current from the output of the generator itself is called a self-excited generator. There are three types of self-excited generators depending upon the manner in which the field winding is connected to the armature, namely; (i) Series generator; (ii) Shunt generator; (iii) Compound generator (i) Series generator In a series wound generator, the field winding is connected in series with armature winding so that whole armature current flows through the field winding as well as the load. Figure below shows the connections of a series wound generator. Since the field winding carries the whole of load current, it has a few turns of thick wire having low resistance. Series generators are rarely used except for special purposes e.g., as boosters. Armature current, Ia = Ise = IL = I e.m.f generated, Eg = V+I(Ra + Rse) Power developed in armature = EgIa Power delivered to load = VI or VIL
College of Engineering Mech. Eng. Dept. Prepared by Mohanad A. A. Alheety ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ (iii) Compound generator In a compound-wound generator, there are two sets of field windings on each pole—one is in series and the other in parallel with the armature. A compound wound generator may be: (a) Short Shunt in which only shunt field winding is in parallel with the armature winding (b) Long Shunt in which shunt field winding is in parallel with both series field and armature winding Short shunt Series field current, Ise = IL Shunt field current= 𝐼𝑠ℎ = V+𝐼𝑠𝑒𝑅𝑠𝑒 𝑅𝑠ℎ e.m.f generated, Eg = V+ IaRa + IseRse Power developed in armature = EgIa Power delivered to load = VIL Long shunt
College of Engineering Mech. Eng. Dept. Prepared by Mohanad A. A. Alheety ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ Series field current, Ise = Ia = IL + Ish Shunt field current, Ish = V/Rsh e.m.f generated, Eg = V + Ia(Ra + Rse) Power developed in armature = EgIa Power delivered to load = VIL Brush Contact Drop It is the voltage drop over the brush contact resistance when current flows. Obviously, its value will depend upon the amount of current flowing and the value of contact resistance. This drop is generally small. Ex1: A shunt generator delivers 450 A at 230 V and the resistance of the shunt field and armature are 50 Ω and 0.03 Ω respectively. Calculate the generated e.m.f. Solution: Shunt current Ish = 230 50
= 4. 6 A
Armature current Ia = I + Ish = 450 + 4. 6 = 454. 6 A Armature voltage drop IaRa = 454. 6. 0. 03 = 13. 6 V Eg = terminal voltage + armature drop = V + IaRa = 230 + 13. 6 = 243. 6 V
College of Engineering Mech. Eng. Dept. Prepared by Mohanad A. A. Alheety ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ Brush drop = 2. 1 = 2 V Eg = V + series drop + brush drop + IaRa = 220 + 9 + 2 + 1. 56 = 232. 56 V Ex4: In a long-shunt compound generator, the terminal voltage is 230 V when generator delivers 150 A. Determine (i) induced e.m.f. (ii) total power generated and. Given that shunt field, series field, divertor and armature resistance are 92 Ω, 0.015 Ω, 0.03 Ω and 0.032 Ω respectively. Solution: Ish = 230 / 92 = 2. 5 A Ia = 150 + 2. 5 = 152. 5 A Since series field resistance and divertor resistances are in parallel (thei r combined resistance is = 0. 03 × 0. 015 / 0. 045 = 0. 01 Ω Total armature circuit resistance is = 0. 032 + 0. 01 = 0. 042 Ω Voltage drop = 152. 5 × 0. 042 = 6. 4 V (𝐢) Voltage generated by armature Eg = 230 + 6. 4 = 236. 4 V (𝐢𝐢) Total power generated in armature EgIa = 236. 4 × 152. 5 = 36 , 051 W
College of Engineering Mech. Eng. Dept. Prepared by Mohanad A. A. Alheety ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ Generated E.M.F. or E.M.F. Equation of a Generator Let Φ = flux/pole in weber Z = total number of armature conductors = No. of slots. No. of conductors/slot P = No. of generator poles A = No. of parallel paths in armature N = armature rotation in revolutions per minute (r.p.m.) E = e.m.f. induced in any parallel path in armature Generated e.m.f. Eg = e.m.f. generated in any one of the parallel paths i.e. E. Average e.m.f. generated/conductor =𝑑𝛷/𝑑𝑡 v (∵ n = 1) Now, flux cut/conductor in one revolution d Φ = Φ P Wb No. of revolutions/second = N /60 ∴ Time for one revolution, dt = 60/ N second Hence, according to Faraday’s Laws of Electromagnetic Induction, E.M.F. generated/conductor = 𝑑𝛷 𝑑𝑡
𝛷 𝑃 𝑁 60 volt
Where, A=2 for a simplex wave-wound generator A=P for a simplex lap-wound generator
College of Engineering Mech. Eng. Dept. Prepared by Mohanad A. A. Alheety ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ Ex7: A 4-pole lap-connected armature of a d.c. shunt generator is required to supply the loads connected in parallel: (1) 5 kW Geyser at 250 V, and (2) 2.5 kW Lighting load also at 250 V. The Generator has an armature resistance of 0.2 ohm and a field resistance of 250 ohms. The armature has 120 conductors in the slots and runs at 1000 rpm. Allowing 1 V per brush for contact drops and neglecting friction, find Flux per pole. Solution: Geyser current = 5000 / 250 = 20 A Current for Lighting = 2500 / 250 = 10 A Total current = 30 A Field Current for Generator = 1 A (250v\250ohm) Hence, Armature Current = 31 A Armature resistance drop = 31 × 0. 2 = 6. 2 volts Generated e. m. f. = 250 + 6. 2 + 2 ( 2 brushes) = 258. 2 V, Eg =
Φ P N
Z
A
Φ × 1000 × 120
Φ = 129. 1 mWb
College of Engineering Mech. Eng. Dept. Prepared by Mohanad A. A. Alheety ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ Ex8: A 4-pole, d.c. shunt generator with a shunt field resistance of 100 Ω and an armature resistance of 1 Ω has 378 wave-connected conductors in its armature. The flux per pole is 0. Wb. If a load resistance of 10 Ω is connected across the armature terminals and the generator is driven at 1000 r.p.m., calculate the power absorbed by the load. Solution: Induced e.m.f. in the generator is Eg =
Φ P N
Z
A
Eg =
0. 02 × 4 × 1000
Load current =V/ Shunt current= V/ Armature current =
V
V
11V
V = Eg − armature drop V = 252 − 1 ×
V = 227 V
Load current = 227 / 10 = 22. 7 A Power absorbed = 227 × 22. 7 = 5135 W
