9884Lecture 3.pptx on dc generator it's

AnshitDhiman 10 views 21 slides Oct 04, 2024
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Lecture # 3 DC GENERATOR CHARACTERISTICS 1

DC G ENERATOR C HARACTERISTICS o Generated voltage E g can be calculated by: Therefore, the generated voltage E g depends on two factors : Flux  Speed  It is evident that generated voltage E g varies proportionally with flux and speed at which the armature is turning. So as  or  increases, so does E g . E g = = ZP ϕ ω N P ϕ ω 2πa πa = k a ϕω ZP NP ; k a = 2πa = πa 2

DC G ENERATOR C HARACTERISTICS o If the flux is held constant and the speed is changing , the new E g : o If the speed is held constant , and the flux is varied , we get: o If both speed and flux are changed , the ratio becomes as follows: E g2 ϕ 2 = E ϕ g 1 1 g 2 ω 1 ω 2 E = E g 1 E g1 ϕ 1 ω 1 g 2 ϕ 1 ϕ 2 E = E g 1 E g2 ϕ 2 ω 2 = E g 2 ϕ 1 ω 1 ϕ 2 ω 2 = E g 1 E g2 E k a ϕ ω 2 ω 2 3 = = k ϕ ω ω g 1 a 1 1

DC G ENERATOR C HARACTERISTICS (a) (b) Example 1: A generator rotating at 200 rad/s develops 220 V. If the flux remains constant and the speed drops to 185 rad/s, (a) what will be the new generated voltage? (b) By how much must the flux be changed to restore the generated voltage to 220 V at the new slower speed? Sol. E g1 E g2 ω 2 = ω 1 g 2 ω 1 g 1 200 ω 2 185 E = E = × 220 = 203.5 𝑉 E g2 E g 1 ϕ 2 ω 2 = ϕ ω 1 1 E g2 ω 2 ϕ 2 = E g 1 1 220 × 200 ϕ 1 = × ϕ 1 ω 220 × 185 ϕ 2 = 1.081ϕ 1 Thus 𝛟 2 must increase by 8.1% of 𝛟 1 4

E QUIVALENT C IRCUIT OF A DC G ENERATOR o The field circuit is used to generate the flux in a magnetic core as described in Chapter 1. R f is the total resistance in field circuit and I f is current in the field. The armature circuit has two parts: Symbol for the dc voltage generated by rotating armature, E g . R a is the total resistance of the a r matu r e w in d ing and I a repre s e nts the armature current. I L represents load current and V t is the output (terminal) voltage of the generator. P i : Mechanical power is fed into the generator through the rotor shaft. P o : Electrical Power is supplied to the load via the armature terminals.  St a tor P o Rotor 5 Field circuit Armature circuit + V f _

6 S EPARATELY E XCITED G ENERATOR Unpopular type because it requires an external dc voltage source to set up a magnetic field (flux) in the field winding. A rheostat is connected to the field winding to vary the field current I f from its normal values to lower values; thus, changing the flux,  . Varying field current varies the generated voltage E g then output voltage V t Separately Excited generator E g = k a ϕ ω f ϕ α I α can be controlled: 1 R f Note: R f represents the total field resistance. V t = E g - R a I a V f R heos t at V f I f = R f

N O -L OAD M AGNETIZATION C URVE Magnetization curve is used to identify the internal characteristics of a particular generator. Magnetization curve is a plot of E g vs. I f at a given speed  . A generator is normally operated in the saturation region (region 3 to 4) . If I f increases, flux increases as well as generated voltage E g . The curve shown is due to the nonlinear behavior of a ferromagnetic materials. o Note: DC generator operating region No-load magnetization curve 7 E g = k a ϕ ω ∝ ϕ ∝ I f

N O -L OAD M AGNETIZATION C URVE o At no-load , I a = I L = 0 so no internal voltage drop across R a . Therefore, V t = E g . Because of residual magnetism , a small generated voltage E g_residual is present although I f = 0 . ( Point 1) As I f increases, the flux, E g increases linearly until Point 3 . Then, the core becomes saturated and any further increase in I f will have only a small affect on E g . As I f decreases, we get the case of hysteresis as described before. E g_residual V t = E g - R a I a = E g @ no-load I a = I L = No-load magnetization curve 8

9 V OLTAGE R EGULATION When the generator is connected to a load, I L = I a since the armature is in series with the load. Because I a flows in the armature, there is a voltage drop within the armature winding, I a R a . Because of the internal voltage drop, the terminal voltage ( V t ) is less than the generated voltage ( E g ): Separately Excited DC Generator R f f I R f V  f V f V t = E g - R a I a

Separately Excited DC Generator R f 10 V OLTAGE R EGULATION When the generator is connected to a load I L = I a , V t = E g - R a I a At no-load ( I a = 0 ), V t = E g . Voltage regulation (V.R.) is a measure of the change of the terminal voltage as I L increases. V.R. can be calculated at the rated conditions in kW or hp, and can be found by: g NL FL w h er e V V % V.R.  V NL  V FL x 100  E V f

V OLTAGE R EGULATION x 100 V N L  V F L % V.R.  x 100 V FL 138  120 % V.R.  120 % V.R.  15 % Example 2: A 5-KW, 120-V generator is under test. When the load is removed, the terminal voltage is found by measurement to be 138 V. Calculate the voltage regulation. Sol. 11

G ENERATOR E FFICIENCY o The losses in a DC machine (generator or motor) may be divided into three classes: Mechanical losses Iron or core losses Copper losses o All these losses appear as heat and thus raise the temperature of the machine. They also lower the efficiency of the machine. Where the efficiency: and the mechanical input power : %  P o x 100 P i P i  P o  P losses DC Generator Input Power P 12 i Output Power P o P losses Losses Power

13 G ENERATOR E FFICIENCY Mechanical power losses Friction losses : occurs because of the moving parts. The energy lost is in the form of heat. In DC machines, there is a brush frictions constant at any given speed, bearing friction . Windage losses : Resistance between the rotating armature and the air inside the machine. These losses depend upon the speed of the machine. But for a given speed , they are practically constant . Iron or Core losses : These losses occur in the armature of a d.c. machine and are due to the rotation of armature in the magnetic field of the poles. They are of two types 1) hysteresis loss 2) eddy current loss . Note : Iron losses and mechanical losses together are called stray losses .

G ENERATOR E FFICIENCY 14 Hysteresis loss Hysteresis loss occurs in the armature of the d.c. machine since any given part of the armature is subjected to magnetic field reversals as it passes under successive poles. The figure shows an armature rotating in two-pole machine, consider a small piece ab of the armature. When the piece ab is under N-pole , the magnetic lines pass from a to b . Half a revolution later, the same piece of iron is under S-pole and magnetic lines pass from b to a so that magnetism in the iron is reversed. In order to reverse continuously the molecular magnets in the armature core, some amount of power must be spent which is called hysteresis loss.

G ENERATOR E FFICIENCY 15 Eddy current loss In addition to the voltages induced in the armature conductors , there are also voltages induced in the armature core . These voltages produce circulating currents in the armature core as shown in the Figure below. These are called eddy currents and power loss due to their flow is called eddy current loss . The eddy current loss appears as heat which raises the temperature of the machine and lowers its efficiency.

G ENERATOR E FFICIENCY 16 Eddy current loss Core resistance can be greatly increased by constructing the core of thin, round. If a continuous solid iron core is used, the resistance to eddy current path will be small due to large cross-sectional area of the core. Consequently, the magnitude of eddy current and hence eddy current loss will be large. The magnitude of eddy current can be reduced by making core resistance as high as practical. The iron sheets called laminations . The laminations are insulated from each other with a coating of varnish . The insulating coating has a high resistance , so very little current flows from one lamination to the other. Also, because each lamination is very thin, the resistance to current flowing through the width of a lamination is also quite large. Thus laminating a core increases the core resistance which decreases the eddy current and hence the eddy current loss.

17 G ENERATOR E FFICIENCY Copper Losses If there is a current flowing through the resistance, then there is an I 2 R power loss dissipated in the form of heat and called a copper loss . There are copper losses in the field circuits I f 2 R f & I s R s and in the armature 2 circuit I a 2 R a . Note : There is also brush contact loss due to brush contact resistance (i.e., resistance between the surface of brush and surface of the commutator). This loss is generally included in armature copper loss . Stray Losses Iron losses and mechanical losses together are called stray losses .

G ENERATOR E FFICIENCY 18 x 100 P o P o  P l os s e s  %   x 100 P i %   P o Copper Losses P cu = P cua + P cuf + P cus Mechanical Input Power P converted = E g I a P i P i = P st + P conv Stray Power Losses (mechanical and core losses) P st P i  P o  P st  P cu P i  P o  P losses Electrical Output Power P o = V t I L P o = P conv - P cu

G ENERATOR E FFICIENCY 19 Example 3: A 1-KW, 125-V generator requires 2 hp to supply rated output. Calculate the generator efficiency. Sol. P i = (2 hp) (746 W/hp) = 1492 W P o = 1 KW %η = (P o /P i ) x 100 %η = (1x10 3 / 1492) x 100 = 67%

G ENERATOR E FFICIENCY P i = P o + P st + P cu P i = 2000 + 150 + 280 = 2430 W %η = (P o /P i ) x 100 %η = (2000/2430) x 100 = 82.3% The input power of the generator is the output power of the motor. Usually, the motor power is in hp. P out_motor = P in_generator = (2430 W)(1 hp/746 W) = 3.26 hp So l . P o i P st P cu P P co n v Example 4: A 2-KW, 220-V generator has constant stray power losses of 150 W. The copper losses at rated load are calculated to be 280 W. If it is driven by motor, find the power required from the motor and the efficiency of the generator, both at rated load. 20

21 S HUNT DC G ENERATOR Self-Excited DC Generator means the generator does not require an external source to set up a magnetic field (flux). The types of self-excited dc generators are: Shunt dc generator. Series dc generator. Shunt DC Generator 3. Compound (shunt and series) dc generator. Shunt DC Generator It is called shunt-generator because the field winding shunts (in parallel) with the armature. The residual magnetism in the core is responsible for the generation of the magnetic field; thus, a small voltage will be generated.  Field W inding A r m ature Winding L oad E g ∝ ϕ ω
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