Electrical Machines DC Motor New

7/30/2019 Electrical Machines DC Motor New

1/27

Experiment No: 3

Characteristics of DC Machines

1 Aim

1 To obtain

Open circuit and External characteristics of separately excited (S.E.) DC generator

External characteristics of shunt generator

2 To study the variation of speed of DC motor by

Armature voltage

Field current

2 Theory

2.1 Basic Theory of Electromagnetic Power Conversion:

Electric machines convert electrical energy to mechanical energy and vice-versa, as shown in Fig.1. Inthe motoring mode, the electric power is converted into mechanical power. All the machines have astationary part called the stator and a rotating part, called a rotor. They are separated by an air gapthus allowing the rotor to rotate freely on its shaft, supported by bearings. The stator is firmly fixedto a foundation to prevent it from turning. Both stator and rotor are made up of high permeabilityferromagnetic material and the length of the air gap is kept as small as possible ( of the order of0.5-1mm ) so that the ampere turns required to establish the flux crossing the airgap as shown in fig.2 is very small. If rotor and stator are perfectly cylindrical, the air gap is uniform and the magnetic

reluctance ( similar to resistance in electric circuits) in the path of flux lines crossing the air gap isuniform. Machines with such structures are called non-salient pole machines ( Fig.2 a). Sometimesthe machines are purposely designed to have saliency so that the magnetic reluctance is unequal alongvarious paths as shown in Fig.2(b). Such saliency results in what is called, the reluctance torquewhich may be the primary or a significant means of producing torque. You should also note that toreduce eddy-current losses, the stator and rotor are constructed by laminations of silicon steel ( thesesheets are insulated from each other by a layer of thin varnish). These laminations are stacked togetherperpendicular to the shaft axis. Slots may be cut into these laminations to place the conductors.There are two basic principles that govern the operation of electric machines :

A current carrying conductor placed in a magnetic field experiences a force.

A voltage is induced in the conductor moving in a magnetic field.

Consider a conductor of length L meters shown in fig.3 (a) which is carrying a current i amperesand is subjected to an externally established magnetic field of a uniform flux density B Wb/m2 inperpendicular to the conductor length. A force F is exerted on the conductor. The magnitude ofthis force is given by F = B I L Nm. The direction of the force is perpendicular to the directionsof both I and B. The direction of this force can be obtained by superimposing the flux lines due tothe conductor current as shown in Fig. 3 (b). The flux lines assist each other on the right side of theconductor and oppose each other on the left side as shown in Fig.3(c). Therefore, the force F acts from

1


2/27

higher concentration of flux lines to the lower concentration, i.e, from right to left.In the fig. 4, conductor of length L meters is moving to the right at a speed V meters/sec. The Bfield is uniform and is perpendicular, into the plane of paper. The magnitude of induced emf in Voltsis given by e = B L V. The polarity of the induced emf can be established as follows: Due to theconductor motion, the force on the charges within the conductor can be written as f= q(VB). Dueto orthogonality ofV and B, the force on the positive charge is upward. Thus, the upper end will havea positive potential with respect to the lower end.

Now, consider two bar magnets pivoted at their center on the same shaft. There will be a torqueproportional to the angular displacement of the bar magnets, which will act to align them. Thisphysical picture is useful in analysing the torque production in machines.Currents in the machine windings create magnetic flux in the air gap between stator and rotor. The fluxpath gets completed through the stator and rotor iron. This condition corresponds to the appearanceof magnetic poles on both the stator and the rotor, centered on their respective magnetic axes as shownin Fig.5. Torque is produced by the tendency of these two magnetic fields to align. The resulting torqueis proportional to the product of the amplitudes of the stator mmf (Fs) and rotor mmf (Fr), and sine ofthe angle sr measured from the axis of stator mmf wave to that of the rotor. Therefore the generalizedexpression for torque is given by

Te Fs Fr sin sr (1)

In a typical machine most of the flux produced by the stator and rotor windings crosses the air gapand links both windings. This flux is termed the mutual flux, directly analogous to the mutual ormagnetizing flux in a transformer. However, some of the flux produced by the rotor and stator windingsdoes not cross the air gap. This is analogous to the leakage flux in a transformer. Only the mutual fluxis of direct concern in torque production (though the leakage fluxes do affect machine performance).From Fig 5(b), we can also write Fs sin sr = Fsr sin r and Fr sin sr = Fsr sin s. Substitutingthese values in eqn.(1), the generalized expression for torque can also be written as

Te Fsr Fr sin r or Te Fs Fsr sin s (2)

A steady torque is developed only when both the fields are stationary with respect toeach other. This is the essential condition for the machine to develop steady torque .

2.2 DC Machine

The stator of a dc machine has projected poles ( salient poles) and a coil is wound on these polesas shown in Fig.6 . When excited by dc current air-gap flux distribution created by this winding issymmetric about the center line of the field poles. The field produced by the stator current isstationary with respect to stator. This axis is called the field axis or direct (d) axis .The rotor of dc machine has slots which contain a winding. This winding handles electric power forconversion to (or from) mechanical power at the shaft. In addition, there is a commutator affixed

to the rotor. The commutator on its outer surface contains copper segments, which are electricallyinsulated from each other by means of mica. The coils of the rotor ( armature) winding are connectedto these commutator segments. A set of stationary carbon brushes are placed on the rotatingcommutator(These brushes are fitted to the stator). It should be noted that the wear due tomechanical contact between the commutator and the brushes requires regular maintenance, which isthe main drawback of these machines.

2


3/27

2.2.1 Principle of Operation of dc Machines:

Consider a two pole dc machine as shown in Fig.7 (a). The voltage induced in the rotor coil rotatingin a uniform field established by the stator is alternating. The air gap flux distribution and the voltageinduced in the coil are shown in Fig.7 (b) and (c) respectively. Consider a coil a-b is placed on dia-metrically opposite slots of the rotor. The two ends of this coil are two commutator segments whichare rotating. Coil a is connected to segment Ca and coil b to segment Cb. Let B1 and B2 be the twocarbon brushes. These brushes are stationary. For counterclockwise motion of the rotor, the terminalunder north pole is positive with respect to the terminal under south pole. Therefore B1 is alwaysconnected to positive end of the coil and B2 to the negative end of the coil. Therefore even though thevoltage induced in the coil is alternating, the voltage at the brush terminals is unidirectional as shownin fig. 7(d). The rectified voltage across the brushes has a large ripple ( magnitude is not constant, it isvarying). In a actual machine a large number of turns are placed in several slots around the peripheryof the rotor. By connecting these in series through the commutator segments a reasonably constant dcvoltage is obtained. The magnitude of the voltage induced in the armature is proportional to ( )where is the airgap flux per pole and is the speed of rotation.Current reversal in a turn by commutator and brushes is shown in Fig. 8. In fig. 8(a) the end atouches the brush B1. The current flows from a to b. As this coil rotates, at a particular position it

gets short circuited ( refer fig.8b). The angle between the d-axis and this position is 90o

. As the coirotates from this position, end a now touches brush B2. The current now flows from b to a as shownin fig.8(c). Hence, the commutator and brushes rectify the alternating voltage induced inthe armature to dc. This combination is also known as mechanical rectifier. Though thecurrent flowing in the armature conductors is ac, the current flowing in/out of carbonbrushes is dc. The change over from positive to negative value takes place at a particularaxis. The angle between this axis and the d-axis is always 90o. Therefore, this axis isknown as q-axis or the quadrature axis. Since, the armature mmf (Fa) is along this axisthe angle sr is always 90

o.

It is often convenient to discuss a dc machine in terms of its equivalent circuit shown in Fig. 9(a

which shows the conversion between electrical and mechanical power or vice-versa. Field produced bythe stator is represented by current IF flowing in the coil. At steady state the equation relating thiscurrent and applied voltage to the field coil is Vf = IF RF, where RF is resistance of the field coil. Infigure 9(b) armature current Ia is entering the positive brush and produces electromagnetic torque torotate the rotor at a speed . Due to this rotation a voltage is induced in the armature.This voltage isproportional to the air gap field and the speed of rotation. If is the air gap flux, and is the speedof rotation, the expression for the voltage induced in the armature is given by E = K , where Kis a constant. If Ra is the armature resistance and Ia is the armature current flowing into the positivecarbon brush as shown in Fig.9(b) (the machine is operated as motor), the relationship between thesteady state voltage induced in the armature E and the armature terminal voltage Va is

Va = E+ IaRa or Ia =Va E

Ra(3)

In case Ia is leaving the positive brush as shown fig. 9(c) (machine is operated as generator; the rotoris being driven by the prime mover), the relationship between the steady state voltage induced in thearmature E and the armature terminal voltage Va is

Va = E IaRa or Ia =E Va

Ra(4)

The following points may be noted from these two equations:

3


4/27

During motoring operation, armature current flows in opposition to the voltage induced in thearmature (also known as speed voltage). Hence E for motoring operation is called as back emfdenoted by Eb.

when operated as generator, voltage induced in the armature is forcing the current to leave thepositive brush. Hence E for generator operation is called as induced emf and is denoted byE.

For motoring action: Multiplying equation 3 by Ia, we get

VaIa = I2aRa + EbIa

The term on the LHS in above equation is the power input to the armature, first term on RHS is thepower lost as heat in the armature resistance and last term is power developed in the armature. Thisdeveloped power should be equal to the mechanical output power (Te ) if friction is neglected.In case of generator action: Multiplying equation 4 by Ia, we get

EIa = I2aRa + VaIa

The term on LHS in the above equation is the power developed in the armature, first term on RHS isthe power lost as heat in the armature resistance and last term is power supplied to the load.Based on the method of excitation dc generators can be divided into two groups. Separately excitedgenerators and self-excited generators. In a separately excited (S.E.) machine, the field winding isconnected to a separate voltage source while, in a self excited generator field winding is connectedacross the armature terminals (provides its exciting current). In these machines the required fieldcurrent is a very small fraction of the rated current (could be around 2-3% of the current flowing inthe armature circuit at full load). Hence, the field winding has a large number of turns with a thinconductor.

2.3 DC Generator

2.3.1 Magnetization Characteristics of dc machine

The induced emf in the armature winding of a dc machine is directly proportional to flux and speed ofrotation. Let us assume that the field winding is connected to a variable dc source that is capable ofsupplying a desired field current. If the armature terminals are left open and the armature is rotatedat constant speed, then the induced emf in the armature is E = K1 , where K1 is a constant. In otherwords the induced emf is directly proportional to the airgap flux. Flux depends on the magneto-motiveforce ( MMF) provided by the current in the field winding. That is, = Kf IF, where Kf depends onthe operating flux density. Therefore induced emf can now be written as E = K1 Kf If.The magnetic circuit of a dc machine consists of both linear (airgap) and non-linear (magnetic materia

of the stator and rotor) parts. Hence, Kf changes ( it decreases as the magnetic circuit gets saturated)with the change in flux density in the machine. The relationship between E and If can be determinedby measuring the open circuit voltage ( voltage across armature terminals) at different values of If ata constant speed. This curve is known as open circuit characteristics (O.C.C). This variation is shownin Fig.10. Since E is an indirect measure of air gap flux ( at constant speed of rotation), the curve issimilar to the B H curve ( or Vs If ) of the magnetic material. For this reason, o.c.c. can also bereferred to as the magnetization curve. It should be noted that E does not start at zero when the fieldcurrent is zero but at some value (of the order of 8-10 V). This is due to residual magnetism.

4


5/27

2.3.2 Armature Reaction

When there is no current in the armature winding ( no-load condition), the flux produced by the fieldwinding is uniformly distributed over the pole faces as shown in Fig.11(A). Let us assume that thistwo pole machine is driven by a prime mover in the clockwise direction ( generator operation). Thedirection of the currents in the armature conductors under load is shown in Fig.11(B). The armatureflux distribution due to armature mmf is also shown in this figure. Since both fluxes exist at the sametime when the armature is loaded, the resultant flux gets distorted. It can be seen that the armatureflux opposes the flux in one half of the pole and aids in the other half. If the magnetic circuit isunsaturated the decrease in flux in one-half of the pole is accompanied by an equal increase in theflux in the other half. The net flux per pole, therefore, is the same under load as at no load. On theother hand, if the magnetic circuit is very close to saturation point under no-load, the increase in fluxis smaller than the decrease in flux. In this case there is a net reduction in the total flux.

2.3.3 Separately excited dc generator

As the name suggests, a separately excited dc generator requires an independent dc external source forthe field winding. The equivalent circuit representation under steady state condition is show in Fig.12The equations under steady-state are:Vf = If Rt Rt = Rf + Rext Va = E + IL Ra IL = IaWhen the field current is held constant and the armature is rotating at a constant speed, the inducedemf in an ideal generator is independent of the armature current, as shown by the dotted line inFig.13. As the load current IL increases, the terminal voltage decreases as indicated by solid line. Ithe armature reaction is neglected, decrease in Va should be linear and equal to the voltage drop acrossRa and carbon brushes. However, if the generator is operated at the knee point in the magnetizationcurve, the armature reaction causes a further drop in terminal voltage.

2.3.4 Self Excited dc generator

In this generator, the field winding is connected across the armature. Hence, the terminal voltage isalso the field voltage and the armature current is the sum of load and field currents. When the rotor ofthis machine is rotated, the residual flux in the field winding induces some voltage ( Er) in the armaturewinding as shown in Fig.14 . Since the field winding is connected across the armature, because of thisinduced emf a small current starts flowing in the field winding. If the field mmf due to this current aidsthe residual flux, total airgap flux increases. This increase in flux increases the induced emf which, inturn, increases the field current. This action is cumulative in nature.If the field winding is connected in such a way that the flux produced by the field current opposes theresidual flux, the generator fails to build up. This problem can be corrected by either reversing thedirection of rotation or interchanging the field winding connections across the armature. The valueof no-load voltage at the armature terminals depends on the field resistance ( point of intersection of

field resistance line with magnetization curve is the operating point). A decrease in the field resistancecauses the generator to build up faster to a higher voltage.The equivalent circuit representation under steady state condition is shown in Fig.15. The equationsthat govern the operation of a self excited generator under steady state are:Ia = IL + If Va = If(Rf + Rext) = ILRL = E IaRaUnder no load condition, the armature current is equal to the field current, which is a small fractionof load current. Therefore the terminal voltage under no-load condition is nearly equal to the inducedemf ( IaRa drop on no-load is very small). As the load current increases, the terminal voltage decreasesdue to the following reasons:

5


6/27

Ia Ra drop demagnetization effect of the armature reaction decrease in the field current due to the reduction in the terminal voltage ( field current can be keptconstant in separately excited generator, while in self excited generator this current falls with terminalvoltage).The variation of terminal voltage with load is shown in Fig.16.

2.4 DC Motor

One of the unique features of the DC motor which has helped to maintain its supremacy over otherelectric drive systems, is its ability to provide a smooth and wide range of speed control with relativeease. This is because the mmf (magneto-motive force) produced by field coil and that produced byarmature coil are always at quadrature, and hence they can be controlled independently. As shownin Fig.19, field mmf (Fs) is taken along the x-axis (also known as direct or d-axis) and the armaturemmf (Fa) is along the y-axis (quadrature or q-axis). Since this angle is 90

o, the torque expressionbecomes Te FsFa. Now, Fs and Fa are proportional to field current (IF) and armature current (Ia)respectively. Therefore Te IFIa. Since the angle between IF and Ia is always 90

o, the ratio (TeIa

)

is maximum. This is another important feature of the DC machine (in other machines, additionacontrol is required to achieve this feature). Generally, it is assumed that the magnetic circuit is linear.Therefore, is proportional to IF. Hence, the torque expression becomes Te = K Ia, where K isconstant.

2.4.1 Speed Control

The other basic equations governing the steady-state operation of the DC motor are:Va = IaRa + Eb, Eb = K, and IF =

VFRF

Alternate method of deriving the expression for torque is as follows:Multiplying the first equation by Ia we get, VaIa = I

2aRa + EbIa

The first term in the above equation is the power input to the armature, second term is the power lostas heat in the armature resistance and the last term is the power developed in the armature. Thispower should be equal to the mechanical power (Te ) if mechanical losses ( friction and windage ) areneglected.As we know

EbIa = (K)Ia = Tetherefore Te = KIa

Substituting for Ia in terms of torque and flux, the relationship between Te and is given by: = Va

K

TeRa(K)2

If the armature terminal voltage Va and airgap flux are held constant, the above equation can bewritten as:

= A + BTe,where A and B are constants. This is an equation of a straight line wherein, A is the y-axis interceptand (B) is the slope. The y-axis intercept represents the no-load speed which depends only on theterminal voltage and air gap flux. The variation of speed with torque is shown in Fig.20. Generally, thedrop in speed with increase in torque is small (it is desirable that the speed of rotation is independent ofload coupled to the motor shaft). A point to be noted is that in an actual machine, armature reactionhelps in maintaining the speed almost constant. Therefore ifVa and are held constant, the speed of a

6


7/27

separately excited dc motor will remain almost constant and it is independent of torque applied to theshaft. Hence in order to vary the speed of rotation over a wide range, the no-load speed (magnitude ofA) should be varied. This can be achieved by the following methods: By controlling the voltage applied to the armature terminals of the machine By controlling the flux produced by the field winding

2.4.2 Armature Voltage control

The schematic diagram for this control technique is shown in Fig.22. In this method, the field current(hence ) is held constant at its rated value and Va is varied. The speed-torque characteristics for thismethod are shown in Fig.23. These characteristics are drawn for various values of Va and fixed valueof . This method of speed control is used for speed below the rated value. In addition, the followingpoint may be noted:

* Since is held constant, the speed of rotation changes linearly with Va. Motor will draw aconstant armature current Ia from the source if it is driving a constant torque load (Te = KIa)Under this working condition, the power (P) drawn by the motor varies linearly with the speed.This mode of operation is known as constant flux or constant torque mode. The variation

of the various quantities with speed is shown in Fig.24.

2.4.3 Field Control

The schematic diagram for this control technique is shown in Fig.25. In this method, the armaturevoltage is held constant at its rated value and the field current is reduced. The speed of the motorchanges in inverse proportion to . The T characteristics for this method are shown in Fig.21(b)These characteristics are drawn for various values of and fixed value of Va. It should be noted thatthe reduction in speed with torque is higher compared to that in the previous method. This method ofspeed control is used for speed above the rated value. In addition, the following point may be noted:

* non-linear inverse speed control of motor speed. This method also changes the value of developedtorque for a given armature current. If the armature current is held constant at the rated value,the input power and therefore output power remains approximately constant (assuming thatfrictional and windage losses remain constant). Hence, this operating zone is known as eitherconstant hp (horse-power) or field weakening zone. Generally the maximum speed orotation is kept within 150% of the rated value.

2.4.4 Starting

From the basic equation governing the steady state operation of dc motor, we have the following:Ia =

(VaK)Ra

, and Te = KIa

The torque developed at starting is determined by the product of total flux and armature ampere-turns.This torque is utilized partly in overcoming friction, partly in accelerating the armature, and partly inaccelerating the load. The value of starting torque required from a motor will then depend very largelyon the load. At standstill there is no back emf, so that to circulate full load current in the armaturea very small voltage is required. Neglecting the effect of armature inductance, the voltage that mustbe applied to the armature at starting depends only on armature resistance.v Since, this resistance isvery small, a large current will flow if the rated voltage is directly applied to the armature (for thegiven machine whose parameters are given below, this current is approximately 1802 = 90A, while thefull load rated current is of the order of 10.5 A!). The starting current can be limited to a safe valueby the following methods:

7


8/27

including external resistance only in the armature circuit so that machine develops necessary startingtorque. As the motor speeds up, the back emf is generated and the current falls. Generally a largeresistance is necessary to limit the starting current. If this resistance is left in the circuit, the steadystate speed would be very low and in addition there would be waste of power in the resistance. It istherefore becomes necessary to cut out the whole of the resistance so that rated speed is obtained. applying a low voltage by using a variable dc power supply. This voltage is increased as the machine

accelerates.

The developed torque and the rate at which back emf is generated depends on the air gap flux. Inorder to have faster acceleration rated voltage is applied to the field winding while starting.

2.5 Variable Voltage DC Source

Both armature voltage control and field current control methods require a variable voltage dc source.If a fixed-voltage dc power supply is available, voltage applied to the armature or field circuits can bevaried by connecting a variable resistance in series with these circuits. However, this results in increasedlosses, heat and poor efficiency Nowadays, power electronic controllers are increasingly being used toobtain variable dc ( or ac) voltage supply from ac source. The advantages are smooth & flexiblecontrol, and high efficiency (one such example is elegant, light weight miniature size fan regulator

This regulator is mounted inside the switch board, while old fan regulators are mounted on the switchboard. Apart from larger size, the old regulators dissipate heat at low speed of operation).Fig.22 shows the output voltage waveform of a full wave bridge. Let the switches here are diodes. Ithe input voltage to the bridge is Vm sin t, then the average value of the output voltage is given by:

Vav =2Vm

From the output voltage waveform, it can be observed that the waveform is continuous and theinstantaneous value is always finite and positive. Hence the average value of the output voltage dependsonly on the peak value of the input voltage. An autotransformer is now required to vary the output dcvoltage by varying the peak input voltage.

The average value of the output voltage can also be reduced if the instantaneous value of the outputvoltage is made either zero or negative. This is possible by using power semiconductor devices otherthan diodes (e.g. thyristors) as the switches. This results in the reduction of size and cost, andimprovement in efficiency.

2.6 Operation of the single phase converter(rectifier)

The generic bridge converter is shown in fig. 25. The AC - DC converter will have diodes in place ofswitches S1 through S4 as in fig 26. In that case it is uncontrolled rectifier. If the switches are suchthat their conduction can be controlled (e.g. thyristors) then it is a controlled rectifier.

The following points must be remembered while analysing the circuits of the rectifiers.: If either ofthe two pairs of switches S1 S2 or S3 S4 conduct then terminal voltage across the motor is V = Vs(supply voltage). A switch can conduct only if anode to cathode voltage VAK is positive AND a gatepulse is applied to its gate. VAK is determined by the current flowing through the bridge just beforethe gate pulse is applied. So there are two possibilities:

Ifia > 0 and switches S3 and S4 are conducting, and a gate pulse is applied to switches S1 and S2Then voltage across S1 and S2 is Vs. Whether switches S1 and S2 will conduct; will be decidedsolely by supply voltage.Thus ia decides voltage across the switch after the gate pulse is appliedWe know that this current is decided by the load torque applied to the motor.

8


9/27

If ia = 0 at the instant of firing any of the pair of switches then the states of the switches aftergiving the firing pulse will be decided by VAK = Vs Eb. Here VAK is decided by the backemfand the Vs. We know that backemf (Eb) will be decided by the speed of the motor.

Thus the waveform of the current drawn by the motor and hence the voltage at the motor terminalswill be decided by the load torque and the speed at which motor is running.

The armature controller drive used in the lab has a semi converter topology. So two of the switchesused in the bridge are diodes (uncontrolled switches) as shown in fig.27 , while two of them are thyristors(controlled switches).Hence the name semi-controlled.Note that, T1 is fired at an angle t = and T2is fired at angle t = + .

2.6.1 Starting

Initial voltage applied across the motor terminals must be very less and hence the value of is keptvery near to 180o. is reduced gradually from 180o to the desired value thereafter.

2.6.2 Steady-State operation

The steady state waveform of the motor terminal voltage is shown in fig.28 During steady state the

voltage across motor terminals consists of 3 parts.

1 Duty interval: (Portion a to b of the voltage waveform) When 2 of the opposite switches inthe bridge conduct and the voltage across the motor terminals follows the input supply voltageVa = Vs

2 Free wheeling interval: (Portion b to c of the voltage waveform) When any two of the switchesin a single leg conduct and the voltage at the output of the motor is zero (e.g. say T1 and D2conduct). Due to the current already flowing through the thyristor T1, it conducts till the currentis reduced to zero. But the diode D1 opposite to T1 is reverse biased due to negative half of the

supply voltage and diode D2 starts conducting. Hence the motor terminals are shorted. Notethat T1 is not reverse biased as the current through T1 is not reduced to zero. Remember thisfreewheeling interval is absent in full converter topologies.

3 Zero current interval: (Portion c to d of the voltage waveform) When none of the switches conductthen voltage across the motor terminals is equal to back emf and armature current is zero.Va = Eb and ia = 0

Note that applied voltage to the armature is continuously pulsating. In order to reduce the pulsation inthe current drawn by the motor, (developed torque depends not on the applied voltage but on armaturecurrent and flux) an inductor may be connected in series with the armature. Since the current flowing

through the inductor is dc, the average voltage drop across it is zero at steady state. However, thecurrent becomes almost constant (property of an inductor- current can not change instantaneously)When the load at the motor shaft increases the motor will draw more current and hence the dutyinterval will increase. Also when the motor speed is changed by flux control then zero current intervalwill be affected. Thus we can observe these changes in the waveform.

9


10/27

Note to TAs/RAs: There is a dc machine which is cut open. Show the following to students: carbon brushes, commutator, stator poles, rotor coil, stator and rotor laminations rotate the rotor and show that carbon brushes are stationary while commutator is rotating.

2.7 Precaution

Always start the motor (prime mover) by applying a low input voltage ( Vin) to the

armature, else the power electronic controller may get damaged due to heavy inrushcurrent Also, apply the rated voltage to the field winding of the motor. In casethe drive has tripped, bring back the voltage control knob on the power controllerfeeding armature of the dc motor to zero position and then press the green button.

3 Procedure:

There are three machines mounted on the stand, out of which two of them are dc machines. Note thename plate ratings of these machines and use one of them as a prime mover (motor) and the other asa dc generator(You may have to justify your selection ).

1.5 kW DC machine: Ra = 2.04, RF = 415, Friction & windage loss at 1500 rpm = 53 W. 1.1 kW DC machine: Ra = 2.1, RF = 415, Friction & windage loss at 1500 rpm = 53 W.

3.1 O.C.C. and load characteristics of the DC Generator

3.1.1 For Separately Excited DC generator

A. Connect the circuit diagram as shown in Fig.17. Keep the control knob (which is used for applyingvariable dc voltage to the field winding of the generator) on the field regulator of generator atzero output voltage position. Put off all switches of the lamp load and open the main switchS connected between the load and the armature of the dc generator. Also, open switches S1, S2

and S3. These are on machine stand.B. Using the power electronic controller slowly increase the input to the prime-mover. Speed of the

set will increase. By controlling the input to the prime mover adjust the speed to the rated speedof the machine.

C. Using a multi-meter note the voltage due to residual magnetism.

D. Using the knob on the controller feeding the field winding of the generator, increase the inputvoltage in steps ( field current will also increase) and for each case determine the open circuitvoltage. Repeat this procedure till the open circuit voltage is 110% of the rated value.

E. Now decrease the applied voltage to the field of the generator till the generated voltage is equato its rated value.

F. Close the main switch S and load the generator in steps by switching ON the lamps and foreach case adjust the prime mover input such that the speed remains constant. Note down loadvoltage and current, field voltage and current of the generator, armature current and voltageof the motor. Repeat this procedure till the load current is equal to the rated current of thegenerator.

G. Put off all the lamps, open the main switch S, reduce the prime mover input and put off the ACsupply to the controller.

10


11/27

3.1.2 For Self Excited DC generator

A. Connect the circuit diagram as shown in Fig.18 (you need to connect the field winding across thearmature instead of connecting it to a separate supply).

B. Slowly increase the input to the prime-mover. Speed of the set will increase. Observe the voltmeterconnected across the terminals of the dc generator. Above a certain speed the voltmeter readingstarts increasing (If this does not happen reduce the prime-mover input and switch off the supply.

Interchange the field terminals of the generator and repeat the same procedure). By controllingthe input to the prime mover adjust the speed to the rated speed of the machine.

C. Close the main switch S. Load the generator in steps by switching ON the lamps and for eachcase adjust the prime mover input such that the speed remains constant. Note down the loadvoltage and current, field current of the generator, armature current and voltage of the motor.Repeat this procedure till the load current is equal to the rated current of the generator.

D. Put off all the lamps, open the main switch S, reduce the prime mover input and put off the ACsupply to the controller.

E. Incase in step-B the machine did self excite in the first attempt, interchange the field terminalsof the dc generator and repeat the step - B.

3.2 Speed Control of the DC Motor

3.2.1 Armature Voltage control:

A. Connect the circuit as shown in Fig.17. In this experiment the motor is loaded by loading thegenerator. Put off all switches of the lamp load and open the main switch S connected betweenthe load and the armature of the DC generator. Also, open switches S1, S2 and S3. These areon machine stand.

B. Switch on the AC supply to the power electronic controller supplying power to the field windingof the motor. Using the knob on the controller, apply the rated voltage to the field winding.

C. Switch on the AC supply to the power electronic controller supplying power to the armature othe motor. Using the knob on the controller, slowly increase the voltage to the armature tilthe rated value. Also, apply the rated voltage to the field winding of the generator (the outputterminals are at the rear side of the controller feeding the armature of the motor). Note downthe meter readings, speed and direction of rotation.

E. Close switch S and load the generator in steps till it reaches full load. For each load keep theinput voltage to the armature constant & note down all the meter readings and speed. You may

find that beyond a certain load, it is not possible to keep the armature input voltageconstant. Do not increase the load beyond this point. Switch off the load and open S.

F. Now apply 85% of the rated voltage to the armature and repeat the above step. Do not switchoff the supply.

3.2.2 Field Control:

A. Using the power electronic controller apply the rated voltage to the armature of the DC motor.

11


12/27

B. Using the power electronic controller suppling power to the field of the DC motor, reduce thefield current to 0.4 A.

C. Close S and load the generator in steps till full load. For each load keep the input voltage to thearmature and field current constant, and note down all the meter readings and speed. You mayfind that beyond a certain load, it is not possible to keep the armature input voltageconstant. Do not increase the load beyond this point. Switch off the load and open S.

D. Now reduce the field current to 0.38A with armature voltage unchanged at its rated value. Repeatthe above step.

E. Reduce the voltage applied to the armature to zero.

3.2.3 Reversal of Direction of Rotation

A. Without disturbing other parts of the circuit interchange the supply terminals to the armature omotor. Apply the rated voltage to the field winding of the motor, and slowly increase the voltageto the armature till the rated value. Note down the speed and direction of rotation.

B. Reduce the voltage applied to the armature of the motor to zero and put off the supply to boththe power electronic controllers.

4 Results

1. GENERATOR

* Plot the variation of open circuit voltage with field current

* Plot the variation of terminal voltage with load current for separately excited generator.

2. MOTOR

* Using the plot of efficiency vs output power of the generator, for each output determine theinput to the generator.

* Assuming 100% coupling efficiency, the above input power is the output of the motor. Know-ing the speed of the motor, determine the torque. Plot T- characteristics.

5 Questions

1. GENERATOR

* Of the two machines which one did you choose to operate as motor? Justify your answer.

* Assume that a given machine has the following name plate ratings:

220 V, 1.5 kW, 1500 rpm dc generator

What do these numbers imply?

* There are motors without any rotor windings ( e.g. stepper motor). Explain how the torqueis produced in these machines ( Hint: Go through very carefully the theory given in the firstpage).

12


13/27


14/27

Figure 1: Electrical machine as an energy converter

Figure 2: Structure of machines

14


15/27

Figure 3: Electric force on a current crrying conductor in a magnetic field

Figure 4: Conductor moving in a magnetic field

15


16/27

Figure 5: Simplified 2 pole machine and vector diagram of mmf waves

Figure 6: Schematic representation of a dc machine

16


17/27

Figure 7: Voltage rectification by commutators and brushes

17


18/27

Figure 8: Current reversal in a turn by commutators and brushes

Figure 9: Equivalent circuit of DC machine

18


19/27

Figure 10: Magnetization (no load) characteristic of a dc machine

Figure 11: Armature reaction in DC Generator

19


20/27

Figure 12: Equivalent circuit of a seperately excited dc generator

Figure 13: External characteristics of a seperately excited dc generator

20


21/27

Figure 14: Voltage buildup in a self excited generator

Figure 15: Equivalent Circuit of a self excited DC generator

21


22/27

Figure 16: External Characteristic of a Self Excited DC generator

Figure 17: Circuit diagram for seperately excited DC generator : for O.C.C. and Load test

22


23/27

Figure 18: Circuit diagram for Self Excited DC Generator

Figure 19: Separately Exicited DC Motor

Figure 20: T/omega characteristics of separately excited DC Motor

23


24/27

Figure 21: Schematic diagram for armature voltage control

Figure 22: T /omega characteristics for armature and field control

24


25/27

Figure 23: Variation of various quantities with speed

Figure 24: Schematic diagram for field control

Figure 25: Single phase AC-DC converter(rectifier)

25


26/27

Figure 26: Single phase diode bridge: uncontrolled rectifier

Figure 27: Single phase semi-converter

26


27/27

Figure 28: Armature voltage and armature current waveforms

Documents

Electrical Machines DC Motor New