Testing of Dc Motors

TESTING OF DC MOTORS

The various losses taking place in a d.c. motor and efficiency can be calculated carrying out testing of d.c. motors. There are different methods of testing d.c. motors.       These methods are broadly classified as ;i) Direct method of testing.ii) Indirect method of testing

1.1 Direct Method of Testing       In this method the d.c. motor which is to be tested is actually loaded and input and output are measured. The efficiency is given by

       Generally this method is employed to small motors. The motor is loaded by means of a brake applied to the water cooled pulleys.       The main draw back of this method is that the accuracy in determining the mechanical power output of the motor is limited. Alternately it is difficult to provide full load for the large capacity motor.1.2 Indirect Method of Testing       In these methods the motor is not loaded directly but the losses and efficiency at different loads can be estimated. Out of the different methods available for testing of d.c. motors, Swinburne's test and Hopkinson's test are commonly used in practice on shunt-motors. Since series motors can not be started without load, the no load tests cannot be performed on d.c. series motor.

HOPKINSON TEST

This test is called regenerative test or to back test which can be carried out on two identical d.c. machines mechanically coupled to each other and simultaneously tested. Thus the full load test can be carried out on two identical shunt machines without wasting their outputs. One of the machines is made to act as a motor while the other as a generator. The mechanical output obtained from the motor drives the generator whose electrical output supplies the greater part of input to the motor. The motor is connected to the supply mains only to components for losses since in absence of losses, the motor-generator set would have run without any external power supply. But due to losses, the generator output is not sufficient to drive the motor. Thus motor takes current from the supply to account for losses.       The Fig. 1 shows the connection diagram for Hopikinson's test. The two shunt machines are connected in parallel. One of the machines is then started as a motor. Here the stator connections are not shown for simplicity.

Fig. Hopkinson's test

       The switch S is kept open. The other machine which is coupled to first will act as load on first which is acting as motor. Thus second machine will act as a generator. The speed of motor is adjusted to normal value with the help of the field rheostat. The voltmeter reading is observed. The voltage of the generator is adjusted by its field rheostat so that voltmeter reading is zero. This will indicate that the generator voltage is having same magnitude and polarity of that of supply voltage. This will prevent heavy circulating current flowing in the local loop of armatures on closing the switch. Now switch S is closed. The two machines can be put into any load by adjusting their field rheostats. The generator current I2 can be adjusted to any value by increasing the excitation of generator or by reducing the excitation of motor. The various reading shown by different ammeters are noted for further calculations.       The input to the motor is nothing but the output of the generator and small power taken from supply. The mechanical output given by motor after supplying losses will in turn drive the generator.       Let        V = Supply VOLTAGER                    I1 = Current taken from the supply                    I2 = Current supplied by generator                    I3 = Exciting current of generator                    I4 = Exciting current of motor                    Ra = Resistance of armature of each machine                    η = Efficiency of both generator and motor.      Input to the motor = V (I1 + I2 )      Output of motor = η x input = V (I1 +I2 )      The output of motor will be given as input to the generator.      Input to generator = η V (I1 +I2 )      Output of generator = η x input = η .η V (I1 + I2 )                                               = η2 V (I1 + I2 )               ...............  (I)       The output of generator can also be given as,       Output of generator = V . I2                           .......... (II)       From equations (i) and (ii),         η2 V (I1 + I2 )  = I2 

         But the assumption of equal efficiencies of two machines is true in case of only large output machines where difference in armature currents of two machines is not large. Also the difference in excitation current required to circulate full load current in the armature will not effect the iron losses. But in case of small machines the difference between armature and field currents is large. So efficiencies can not be assumed to be same. Here the stray losses are assumed to be equal whereas armature and field copper losses are separately determined for estimating the efficiencies separately.       Armature copper loss in generator = (I2 + I3 )2 Ra

       Armature copper loss in motor = (I1 + I2 - I4 )2        Copper loss in field winding of motor = V I3

       Copper loss in field winding of motor = V I4

       But total losses in generator and motor are equal to the power supplied by the mains.       Power drawn from supply =  V I2

       The stary loss for both the machines = V I2 - {{I2 + I3 )2  Ra + (I1 + I2 - I4 )2 Ra +  V I3 + V I4 }                                                            = Ws (say)       Assuming that stray losses are equally divided between the two machines.       Stray loss for each machine = Ws /2       The machine which is acting as a motor        Total losses = (I1 + I2 - I4 )2 Ra + V I4 + Ws /2       Input to motor = (I1 + I2 )       Efficiency of motor, ηm = Output/Input = (Input -  Losses)/Input

       The machine which is acting as a generator,       Total losses = (I2 + I3)2  Ra + V I3 + Ws /2       Output of generator = V I2

       Efficiency of generator, ηg = Output/Input = Output/(Output + Losses)

1.1 Advantages   The various advantages of Hopkinson's test are,1. The power required for conducting the test is small compared to full load powers of the two machines.

2. Since the machines are operated at full load conditions, change in iron loss due to distortion in flux at full load will be included in the calculations.3. As the machines are tested under full load conditions, the temperature rise and quality of communication of the two machines can be observed.4. The test is economical as power required to conduct the test is very small which is just sufficient to meet the losses.5. There is no need for arranging any actual load. Similarly by changing the field currents of two machines, the load can be easily changed and a load test over complete range of load can be taken.

1.2 Disadvantages   The various disadvantages of Hopkinson's test are,1. There is difficulty in availability of two tdentical machines.2. The iron losses in the two machines can not be separated. The iron losses are different in both the machines because of different excitations.3. The machines are not loaded equally in case of small machines which may lead to difficulty in analysis.       This test is better suited in case of large machines.

RETARDATION TEST OR RUNNING DOWN TEST

This method is generally employed to shunt generators and shunt motors. From this method we can get stary losses. Thus if armature and shunt copper losses at any given load current are known then efficiency of a machine can be easily estimated.       The connections required for conducting this test are shown in the Fig. 1.       The machine whose test is to be taken is run at a speed which is slightly above its normal speed. The supply to the motor is cut off while the field is kept excited. The armature consequently slows down and its kinetic energy is used in supplying the rotational or stray losses which includes iron, friction and winding loss.       If I is the amount of inertia of the armature ans is the angular velocity.       Kinetic energy of armature = 0.5 Iω2 ...   Rotational losses, W = Rate of change of kinetic energy

Fig. 1 Retardation test        Angular velocity, ω = (2 πN)/60

       Thus if we if to find the rotational losses, the moment of inertia I and dN/dt must be known.       These quantities can be found as follows ;-1.1 Determination of dN/dt       The voltmeter  V1 which is connected across the armature will read the back e.m.f. of the motor. We know that back e.m.f. is proportional to speed so that voltmeter is calibrated to read the speed directly.       When motor is cut off from the supply, the speed decrease in speed is noted with the help of stop watch. A curve showing variation between time and speed which is obtained from voltmeter which is suitably calibrated is shown in the Fig. 2.

Fig. 2       At any point C corresponding to normal speed, a tangent AB is drawn. Then

       The value obtained from above can be substituted in the expression for W which can give the rotational looses.1.2 Determination of moment of inertia (I)Method (a) Using Flywheel       The armature supply is cut off and time required for definite change in speed is noted to draw the corresponding curve as we have drawn in previous case. This curve is drawn considering only armature of the machine. Now a flywheel with known moment of the inertia say is  I1 keyed onto the shaft and the same curve is drawn again. The slowing down time will be extended as combined moment of inertia of the two is increased.       For any given speed (dN/dt1) and (dN/dt2) are determined same as previous case. It can be seen that the losses in both the cases are almost same as addition of flywheel will not make much difference to the losses.       In the first case where flywheel is not there then,

       Adding the flywheel to the motor armature in second case we get,

Method (b) Without using Flywheel       In this method time is noted for the machine to slow down by say 5 % considering the armature alone. The a retarding torque either mechanical or electrical is applied. Preferably electrical retarding torque is applied and time required to slow down by 5% is noted again. The method by which electrical torque can be provided is shown in the Fig. 1 in which the switch S after disconnecting from the supply is thrown to terminals 1'2'. The machine then gets connected to a non-inductive load resistance RL. The power drawn by this resistance will acts as a retarding torque on the armature which will make it slow more quickly.       The additional loss in the resistance will be equal to product of ammeter reading and the average reading of the voltmeter (for a fall of 5% of voltmeter reading, the time is noted.) The ammeter reading is also changing so its average reading is taken. Thus the additional losses is I a

2

(Ra + R). Let t1 be the time when armature is considered alone and t2 be the time when armature is connected across a load resistance, V be average voltage across R and Ia be the average current and W' is additional retarding electrical torque supplied by motor.

        If dN i.e. change in speed is same in two cases then

       Here dN/dt1 is rate of change in speed without extra load whereas dN/dt2 is rate change in speed with extra electrical load which provides retarding torque.Example :A retardation test is made on a separately excited d.c. machine as a motor. The induced voltage falls from 240 V to 220 V in 25 seconds on opening the armature circuit and in 6 seconds on suddenly changing the armature connection from supply to a load resistance which takes average current of 10 A. Find the efficiency of the machine when running as a motor taking a current of 25 A on a supply of 250 V. The resistance of its armature is 0.3 Ω and that of its field winding is 200 Ω.Solution :               Iav = 10 A       Average voltage across load = (240+220) / 2 = 230       Power absorbed,   W' = Vav Iav = 230 x 10 = 2300 W       t1 = 25 sec       t2 = 6 sec

       Stary losses = 726.31 W       Motor in input current, IL = 25 A       Shunt current, Ish = V/Rsh = 250/200 = 1.25A       Armature current, Ia = IL - Ish = 25-1.25 = 23.75 A       Armature copper loss = Ia

2 Ra = (23.75)2 x 0.3 = 169.21 W       Shunt copper loss = Ish

2 Rsh = (1.25)2 x 200 = 312.5 W       Total losses = 169.21 + 312.5 + 726.31 = 1208.02 W       Input to the motor = V .I = 250 x 25 = 6250 W       Output of motor = Input - Losses = 6250 - 1208.02 = 5041.98 W%Efficiency of motor ηm = (Output/Input) x 100 = (5041.98/6250) x 100 = 80.67                                ηm  = 80.67 %

FIELD TEST

This is one of the methods of testing the d.c. series motors. Unlike shunt motors, the series motor can not be tested by the methods which area available for shunt motors as it is impossible to run the motor on no load. It may run at dangerously high speed on no load. In case of small series motors brake test may be employed.       The series motors are usually tested in pairs. The field test is applied to two similar series motors which are coupled mechanically. The connection diagram for the test is shown in the Fig. 1.

Fig. 1 Field test

       As shown in the Fig. 1 one machine is made to run as a motor while the other as a generator which is separately excited. The field of the two machines are connected in series so that both the machines are equally excited. This will make iron losses same for the two machines. The two machines are running at the same speed. The generator output is given to the variable resistance R.       The resistance R is changed until the current taken by motor reaches full load value. This will be indicated by ammeter A1. The other readings of different meters are then recorded.      Let      V = Supply voltage                 I1  = Current taken by motor                 I2 = Load current                 V2 = Terminal p.d. of generator                Ra, Rse = Armature and series field resistance of each machine      Power taken from supply = VI1 

      Output obtained from generator = V2  I2 

      Total losses in both the machines, WT  = VI1  - V2  I2 

      Armature copper and field losses, WCU  = ( Ra + 2 Rse ) I12   +   I22 Ra

      Total stray losses = WT  - WCU  

       Since the two machines are equally excited and are running at same speed the stray loses are equally divided.

For Motor ;       Input to motor = V1  I1 

       Total losses = Aramture Cu loss + Field Cu loss + Stray loss                         = I1

2 ( Ra + Rse) + Ws

       Output of motor = Input - Total losses = V1  I1  -  [ I12  ( Ra + Rse) + Ws ]

For Generator :       Efficiency of generator is of little importance because it is running under conditions of separate excitation. Still it can be found as follows.      Output of generator = V2  I2 

       Field Cu loss =  I12  Rse

       Armature Cu loss = I22  Ra

       Total losses = Armature Cu loss + Field Cu loss + Stray loss                        = I2

2  Ra +  I12  Rse + Ws

       Input to generator = Output + Total losses = V2  I2  + [ I22  Ra + I1

2  Rse + Ws ]

       The important point to be noted is that this is not regenerative method though the two machines are mechanically coupled because the generator output is not fed back to the motor as in case of Hopkinson's test but it is wasted in load resistance.

BRAKE TEST

Another method of testing the d.c. motor is brake test method. This is a direct method of testing the motor. In this method, the motor is put on the direct load by means of a belt and pulley arrangement. Bu adjusting the tension of belt, the load is adjusted to give the various values of currents. The load is finally adjusted to get full load current. The power developed gets wasted against the friction between belt and shaft. Due to the braking action of belt the test is called brake test.       The Fig. 1(a) shows the experimental setup for performing brake test on a d.c. shunt motor while the Fig. 1(b) shows the belt and pulley arrangement mounted on the shaft of the motor.

Fig. 1 Brake test       The tension in the belt can be adjusted using the handle. The tension in kg can be obtained from the spring balance readings.       Let   R = Radius of pulley inmeter               N = Speed in r.p.m.             W1 = Spring balance reading on tight side in kg             W2 = Spring balance reading on slack side in kg       So net pull on the belt due to friction at the pulley is the difference between the two spring balance readings.

       As radius R and speed N are known, the shaft torque developed can be obtained as,

       Hence the output power can be obtained as,

       Now let, V = Voltage applied in volts                     I = Total line current drawn in amps.

       Thus if the readings are taken on full load condition then the efficiency can be obtained as,

       Adjusting the load step by step till full load, number of readings can be obtained. The speed can be measured by tachometer. Thus all the motor characteristics can be plotted.1.1 Advantages   The advantages of brake test,1. Actual efficiency of the motor under working conditions can be found out.2. The method is simple and easy to perform.

3. Can be performed on any type of d.c. motor.

1.2 DisadvantagesThe disadvantages of brake test,1. Due to friction, heat generated and hence there is large dissipation of energy.2. Some type of cooling arrangement is necessary3. Convenient only for small machines due to limitations regarding heat dissipation arrangements.4. The power developed gets wasted hence method is expensive.5. The efficiency observed is on lower side.

 Example :       In a brake test conducted on a d.c. shunt motor the full load readings are observed as,       Tension on tight side = 9.1 kg       Tension on slack side = 0.8 kg       Total current = 10 A       Supply voltage = 100 V       Speed = 1320 r.p.m.       The radius of the pulley is 7.5 cm.       Calculate its full load efficiency.

Solution :       W1 = 9.1 kg,  W2 = 0.8 kg,         I = 10 A,          V = 110 V,     R = 7.5 cm       Tsh = (W1 - W2) 9.81 x R = (9.1 - 0.8) x 9.81 x 0.075             = 6.1067       Pout = Tsh x ω =  Tsh x (2πN/60)                     = (6.1067 x 2π x 1320/ 60) = 844.133 W       Pin = VI = 110 x 10 = 110 W       %η = (/) x 100 = (844.133/1100 ) x 100             =76.74 %

SWINBURNE’S TEST OR NO-LOAD TEST

This is indirect method of testing d.c. motors in which flux remains practically constant i.e. specially in case of shunt and compound motors. Without actually loading the motor the losses and hence efficiency at different loads can be found out.       The motor is run on no load at its rated voltage. At the starting some resistance is connected in series with the armature which is cut when motor attains sufficient speed.       Now the speed of the motor is adjusted to the rated speed with the help of shunt field rheostat as shown in the Fig. 1.

Fig. 1 Swinburnes's       The no load armature current Ia is measured by ammeter A1 whereas the shunt current is measured by ammeter A2.       If V is the supply voltage then motor input at no load will be,

       There will be Cu loss in the field winding which will be given as,

       Let Ra be the resistance of armature,

       Thus the stary losses which includes iron, friction and windage losses can be obtained as,Stary lossses = Input at no load - Field copper losses - No load armature copper losses

      In the field and armature windings there will be copper loss due to flow of current which will increase the temperature of the field and armature winding when the motor is loaded. This increase in temperature will affect their resistance.       Thus the new value of field resistance R'sh and that of armature R'a can be found by considering that rise in temperature as about 40 oC.       If α1 = Resistance temperature coefficient of copper at room temperature

       At room temperature the shunt field winding resistance will be,Rsh = V/ Ish

       Now shunt winding current, I'sh = V/R'sh

          Now if we want to find the efficiency of the motor at say 1/4 th full load. It can be calculated as follows,       Let      IF.L. = Full load current of motor                 WF = Field copper loss                  W = Stary losses       Load current at 1/4 th full load =   IF.L./4...     Motor input at 1/4 th full load = V x ( IF.L./4) watts.       Armature current at 1/4 th full load, I'a  = ( IF.L./4) - I'sh

       This is the efficiency of motor when the load on motor is 1/4 th of full load which can be found without loading the motor. The efficiencies at other loads can be calculated similarly.1.1 Advantages1. Since constant losses are known, the efficiency can be estimated at any load.2. The method is convenient and economical as less power is required for testing even a large motor i.e. only no load power is to be supplied.3. The motor is not required to be loaded i.e. only test to be carried out is the no load test.1.2 Disadvantages1. In this method, the iron losses are assumed to be constant which is not the true case a sthey change from no load to full load. Due to armature reaction at full load there will be distortion in flux which will increase the iron loass.2. The only test which is carried out is the no load test. Hence it is difficult to know whether there will be satisfactory communication at full load.3. We have assumed that there is rise in temperature of 40  oC at full load which can not be checked actually as we are not actually loading the motor.4. As it is a no load test it can not be performed on a series motor.