DC Machine: Construction and their Applications, Study notes of Electrical Engineering

Download DC Machine: Construction and their Applications and more Study notes Electrical Engineering in PDF only on Docsity! Electrical Machines –I Third Semester Govt. Polytechnic Sector-26 Panchkula Electrical Engg. Dept. DC Machine: Construction and their Applications The DC machine can be classified into two types namely DC motors as well as DC generators. Most of the DC machines are equivalent to AC machines because they include AC currents as well as AC voltages in them. The output of the DC machine is DC output because they convert AC voltage to DC voltage. The conversion of this mechanism is known as the commutator, thus these machines are also named as commutating machines. DC machine is most frequently used for a motor. The main benefits of this machine include torque regulation as well as easy speed. The applications of the DC machine are limited to trains, mills, and mines. As examples, underground subway cars, as well as trolleys, may utilize DC motors. In the past, automobiles were designed with DC dynamos for charging their batteries. DC Machine A DC machine is an electromechanical energy alteration device. The working principle of a DC machine is when electric current flows through a coil within a magnetic field, and then the magnetic force generates a torque which rotates the dc motor. The DC machines are classified into two types such as DC generator as well as DC motor. The main function can be built with the annealed steel laminations for reducing the power drop because of the eddy currents. Pole Shoe Pole shoe in DC machine is an extensive part as well as enlarge the region of the pole. Because of this region, flux can be spread out within the air-gap as well as extra flux can be passed through the air space toward armature. The materials used to build pole shoe is cast iron otherwise cast steed, and also used annealed steel lamination to reduce the loss of power because of eddy currents. Field Windings In this, the windings are wounded in the region of pole core & named as field coil. Whenever current is supplied through field winding then it electromagnetics the poles which generate required flux. The material used for field windings is copper. Armature Core Armature core includes the huge number of slots within its edge. Armature conductor is located in these slots. It provides the low-reluctance path toward the flux generated with field winding. The materials used in this core are permeability low- reluctance materials like iron otherwise cast. The lamination is used to decrease the loss because of the eddy current. Armature Winding The armature winding can be formed by interconnecting the armature conductor. Whenever an armature winding is turned with the help of prime mover then the voltage, as well as magnetic flux, gets induced within it. This winding is allied to an exterior circuit. The materials used for this winding are conducting material like copper. Commutator The main function of the commutator in the DC machine is to collect the current from the armature conductor as well as supplies the current to the load using brushes. And also provides uni-directional torque for DC-motor. The commutator can be built with a huge number of segments in the edge form of hard drawn copper. The Segments in the commutator are protected from thin mica layer. Brushes Brushes in the DC machine gather the current from commutator and supplies it to exterior load. Brushes wear with time to inspect frequently. The materials used in brushes are graphite otherwise carbon which is in rectangular form. Types of DC Machines The excitation of the DC machine is classified into two types namely separate excitation, as well as self-excitation. In separate excitation type of dc machine, the field coils are activated with a separate DC source. In self-excitation type of dc machine, the flow of current throughout the field-winding is supplied with the machine. The principal kinds of DC machine are classified into four types which include the following. twists of a huge cross-sectional region, as well as the shunt windings, include several fine wire twists. The connection of the compound machine can be done in two ways. If the shunt-field is allied in parallel by the armature only, then the machine can be named as the ‘short shunt compound machine’ & if the shunt-field is allied in parallel by both the armature as well as series field, then the machine is named as the ‘long shunt compound machine’. Characteristics of DC motors Generally, three characteristic curves are considered important for DC motors which are, (i) Torque vs. armature current, (ii) Speed vs. armature current and (iii) Speed vs. torque. These are explained below for each type of DC motor. These characteristics are determined by keeping the following two relations in mind. Ta 𝖺 ɸ.Ia and N 𝖺 Eb/ɸ For a DC motor, magnitude of the back emf is given by the same emf equation of a dc generator i.e. Eb = PɸNZ / 60A. For a machine, P, Z and A are constant, therefore, N 𝖺 Eb/ɸ Characteristics of DC series motors Torque vs. armature current (Ta-Ia) This characteristic is also known as electrical characteristic. We know that torque is directly proportional to the product of armature current and field flux, Ta 𝖺 ɸ.Ia. In DC series motors, field winding is connected in series with the armature, i.e. Ia = If. Therefore, before magnetic saturation of the field, flux ɸ is directly proportional to Ia. Hence, before magnetic saturation Ta α Ia2. Therefore, the Ta-Ia curve is parabola for smaller values of Ia. After magnetic saturation of the field poles, flux ɸ is independent of armature current Ia. Therefore, the torque varies proportionally to Ia only, T 𝖺 Ia.Therefore, after magnetic saturation, Ta-Ia curve becomes a straight line. The shaft torque (Tsh) is less than armature torque (Ta) due to stray losses. Hence, the curve Tsh vs Ia lies slightly lower. In DC series motors, (prior to magnetic saturation) torque increases as the square of armature current, these motors are used where high starting torque is required. Speed vs. armature current (N-Ia) We know the relation, N 𝖺 Eb/ɸ For small load current (and hence for small armature current) change in back emf Eb is small and it may be neglected. Hence, for small currents speed is inversely proportional to ɸ. As we know, flux is directly proportional to Ia, speed is inversely proportional to Ia. Therefore, when armature current is very small the speed becomes dangerously high. That is why a series motor should never be started without some mechanical load. But, at heavy loads, armature current Ia is large. And hence, speed is low which results in decreased back emf Eb. Due to decreased Eb, more armature current is allowed. Speed vs. torque (N-Ta) This characteristic is also called as mechanical characteristic. From the above two characteristics of DC series motor, it can be found that when speed is high, torque is low and vice versa. Characteristics of DC shunt motors Torque vs. armature current (Ta-Ia) In case of DC shunt motors, we can assume the field flux ɸ to be constant. Though at heavy loads, ɸ decreases in a small amount due to increased armature reaction. As we are neglecting the change in the flux ɸ, we can say that torque is proportional to armature current. Hence, the Ta-Ia characteristic for a dc shunt motor will be a straight line through the origin. Since heavy starting load needs heavy starting current, shunt motor Series Cumulative compound Differential compound Rated speed z2 yz Rated speed Series a Shunt Torque Differential compound Cumulative compound Armature current (Ia) Armature Current (Ia) Characteristics of DC compound motor The frame is used as an outer protecting cover that is used to protect against environmental conditions. The frame also acts as an outer periphery such that the inner parts can be easily housed. The stable state section of the equipment is stator on which the stator winding is enclosed. The rotor is the moving part that either move in clockwise or anti- clockwise depending upon thrust impelled on it. The bearings provide proper friction for the rotor to run smoothly. A fan is employed to remove the unwanted heat that gained during the running of the rotor. It is expelled out through the ventilation that is provided behind the machine. A shaft is provided to deliver the mechanical output as the rotor rotates. The slips rings are employed for a normal Ac machine where rotating armature stationary field winding is employed. In this situation, the slip rings allow the input alternating current to change continuously in the coils . Constructional detail : Shell type • Windings are wrapped around the center leg of a laminated core. Core type • Windings are wrapped around two sides of a laminated square core. Sectional view of transformers Note: High voltage conductors are smaller cross section conductors than the low voltage coils Shell type • The HV and LV windings are split into no. of sections • Where HV winding lies between two LV windings • In sandwich coils leakage can be controlled Fig: Sandwich windings Cut view of transformer Yep t terege epee terre wee eerie High-voltage Low-voltage Terminal Terminal Winding ali {| gE Tank Radiating tubes Transformer with conservator and breather Conservator . cs po Transformer Tank ve oil Breather —____ i Silica Ge Ideal Transformers • Zero leakage flux: -Fluxes produced by the primary and secondary currents are confined within the core • The windings have no resistance: - Induced voltages equal applied voltages • The core has infinite permeability - Reluctance of the core is zero - Negligible current is required to establish magnetic flux • Loss-less magnetic core - No hysteresis or eddy currents Ideal transformer V1 – supply voltage ; I1- noload input current ; V2- output voltgae; I2- output current Im- magnetising current; E1-self induced emf ; E2- mutually induced emf EMF equation of a transformer • Worked out on board / • Refer pdf file: emf-equation-of-tranformer Transformer on load Fig. a: Ideal transformer on load Fig. b: Main flux and leakage flux in a transformer Phasor diagram of transformer with UPF load Phasor diagram of transformer with lagging p.f load Equivalent circuit parameters referred to primary and secondary sides respectively Contd., • The effect of circuit parameters shouldn’t be changed while transferring the parameters from one side to another side • It can be proved that a resistance of R2 in sec. is equivalent to R2/k2 will be denoted as R2’(ie. Equivalent sec. resistance w.r.t primary) which would have caused the same loss as R2 in secondary, 2 2 2 2 1 2' 2 2 2 2 ' 2 2 1 k R R =       = = R I I RIRI Transferring secondary parameters to primary side Exact equivalent circuit referred to primary Approximate equivalent circuit • Since the noload current is 1% of the full load current, the nolad circuit can be neglected Transformer Tests Electrical Machines •The performance of a transformer can be calculated on the basis of equivalent circuit •The four main parameters of equivalent circuit are: - R01 as referred to primary (or secondary R02) - the equivalent leakage reactance X01 as referred to primary (or secondary X02) - Magnetising susceptance B0 ( or reactance X0) - core loss conductance G0 (or resistance R0) •The above constants can be easily determined by two tests - Oper circuit test (O.C test / No load test) - Short circuit test (S.C test/Impedance test) •These tests are economical and convenient - these tests furnish the result without actually loading the transformer In Open Circuit Test the transformer’s secondary winding is open-circuited, and its primary winding is connected to a full-rated line voltage. • Usually conducted on H.V side • To find (i) No load loss or core loss (ii) No load current Io which is helpful in finding Go(or Ro ) and Bo (or Xo ) 2 0 2 00 2 0 oc 00 2 0oc 0 0 o000 22 000m 00wc 00 0 000 B esusceptanc Exciting & V W G econductanc Exciting ;GV W Y ;YVI sinI I cosI I cos cosloss Core GY V I -IIIor Ior IV W IVW w oc oc −= == == == = = ==      Open-circuit Test 0 0 0 0 0 0 0 0 V I B V I G I V X I V R w w   = = = = Transformer Voltage Regulation and Efficiency Electrical Machines The output voltage of a transformer varies with the load even if the input voltage remains constant. This is because a real transformer has series impedance within it. Full load Voltage Regulation is a quantity that compares the output voltage at no load with the output voltage at full load, defined by this equation: %100down Regulation %100up Regulation , ,, , ,,  − =  − = nlS flSnlS flS flSnlS V VV V VV ( ) ( ) %100 / down Regulation %100 / up Regulation V V k noloadAt , , , , p s x V VkV x V VkV nlS flSP flS flSP − = − = = Ideal transformer, VR = 0%. Voltage regulation of a transformer voltageload-no voltageload-fullvoltageload-no regulationVoltage − =       = 1 2 12 N N VV p s p s N N V V =recall Secondary voltage on no-load V2 is a secondary terminal voltage on full load       −      = 1 2 1 2 1 2 1 regulationVoltage N N V V N N V Substitute we have Transformer Phasor Diagram 11/1/2023 Electrical Machines © Aamir Hasan Khan To determine the voltage regulation of a transformer, it is necessary understand the voltage drops within it. Transformer Phasor Diagram 11/1/2023 Electrical Machines © Aamir Hasan Khan With a leading power factor, VS is higher than the referred VP so VR < 0 Transformer Phasor Diagram Electrical Machines For lagging loads, the vertical components of Req and Xeq will partially cancel each other. Due to that, the angle of VP/a will be very small, hence we can assume that VP/k is horizontal. Therefore the approximation will be as follows: Formula: voltage regulation leadingfor '-' and laggingfor '' V sincos V V regulation % luesprimary va of In terms leadingfor '-' and laggingfor '' V sincos V V regulation % valuessecondary of In terms 1 10111011 1 ' 21 20 20222022 20 220 +  = − = +  = − = where XIRIV where XIRIV   Condition for maximum efficiency Cu loss» = 17Roy or 1 Ron =W,, Ironloss = Hysteresis loss + Eddy current loss = W, + W,=W; Considering primary side, Primary input = V,/, cos >, _ Vii, cos@, ~ losses _ Vi, cos — IPR —W; NiSat = eds e) eee Vil, 08, Spo V, cos'g, Vif, cos @, Differentiating both sides with respect to /,, we get Ga gee! wR ogy eye hinol Mya te. dl, V, cos d VI? cos 0, For 7 to be maximum, a = 0. Hence, the above equation becomes ay Sere = 5 or W,=1? or 17 V, cos , Vip cos >, v= ty Roy 2 Ree or Cu loss = tron loss Contd., The load at which the two losses are equal = All day efficiency hours) 24 ( kWhin Input kWhin output in wattsinput in wattsput out efficiency commercialordinary day forall = =  •All day efficiency is always less than the commercial efficiency Advantages *Less space *Weight Less *Cost is Less *Transported easily *Core will be smaller size *More efficient *Structure, switchgear and installation of single three phase unit is simpler Principal of Operation *The three cores are arrange at 120° from each other. Only primary windings are shown on the cores for simplicity. *The primaries are connected to the three phase supply. *The three fluxes is also zero at any instant. *Star-Star connection *Delta-Delta connection *Star-Delta connection *Delta-Star connection *Open Delta or V connection *Scott connection or T-T connection 3 Single Phase Transformers Coed PHIM Coeh «= #8 Primary Configuration | Secondary Configuration Deta(iest) /\ | Deta(vest) /\ Deta (Mest) | Sta (Wye) Y Stay) | Deta(Mest) /\ Star(Wye) | Stare) neon 4 Data les A Interconnected n A Sta (We) Star Connection Delta Connection om *The Transformers connected in parallel must have same polarity so that the resultant voltage around the local loop is zero. With improper polarities there are chances of dead short circuit. *The relative phase displacements on the secondary sides of the three phase transformers to be connected in parallel must be zero. The transformers with same phase group can be connected in parallel *As the phase shift between the secondary voltages of a star/delta and delta/star transformers is 30°, They cannot be connected in parallel. *But transformers with +30° and -30° phase shift can be connected in parallel by reversing phase sequence of one of them *The voltage ratio of the two transformers must be same. This prevents no load circulating current when the transformers are in parallel on primary and secondary sides. *As the leakage impedance is less, with a small voltage difference no load circulating current is high resulting in large I2R losses.