1. Testing of DC MachinesBy, Rohini Haridas Assistant Professor, Dept of Electrical Engineering, SSGM College of Engineering,Shegaon

2. Objectives of Testing When m/c is manufactured in industry as per particular design, it is necessary that it should undergo testing Testing is required to determine whether the performance of m/c is as per the designed data or not? If not how much variations are there in the actual values and calculated values, the difference is so recorded are within the permissible limits or not? 3. Continued.If the variations in the results such as regulation, iron losses, copper losses, no load current, no load speed etc. are not within the permissible limits, the reasons for either higher values or lower values should be investigated and manufacturing process modify accordingly. From testing we can also determine the quality of materials used. 4. Types of Testing Direct Method of Testing Indirect Method of Testing Regenerative Method of Testing 5. Direct Method of Testing m/c under test can be loaded directly and efficiency is given by Efficiency= output/input the machine can be loaded to its full load hence temp. rise at full load can be actually determined. As the output is measured by mechanical system which can not be so accurate and difficult to load large machines 6. Brake TestPrecaution: While performing this test with series machines care should be taken that brake applied is tight failing Which the motor will attain dangerously high speed and get damaged 7. Let , V= supply voltage measured by voltmeter V I = Input current measured by ammeter A W1 and W2 = spring balance reading in Kg N = speed of armature in rpm r = radius of pulley The net force acting on pulley is (W1 W 2) Kg9 .81(W 1 W 2 ) NewtonOR 8. Therefore, the torque developed by the motor T = (W1 W 2) * r kg-m OR T = 9.81* (W1 W 2) * r N-m Output of the Motor = T * w = 9.81* (W1 W 2) * r *2 N watt 60..............(1)Input to motor = V I................................ (2) Efficiency of Motor= output/input,2N * 9 .81 * (W 1 W 2 ) * r = 60 * VI 9. Numerical Pb1) In brake test the effective load on the pulley was 38.1Kg.The effective diameter of the pulley is 63.5cm and speed is 12 rps.The motor took 49Amp at 220 V. Calculate the o/p power and efficiency at this load. Given: (W1-W2)=38.1Kg d=63.5cm thus, r=(63.5/2)*10-2m N=12 rps=12*60 rpm I=49 Amp V=220V 10. Solution:The effective torque in Newton isT = 9.81* (W1 W 2) * r N-mT=9.81*38.1*0.3175 T=118.66 N-m Input to motor=V*I = 220*49 = 10,780 watt Output of the Motor = T*w =118.66*(2*pi*N)/60 watt =8947.416watt Efficiency=output/input = 8947.416/10,780 = 83% 11. Pb 2) In a brake test on dc shunt motor, the load on one side of brake band was 35Kg and the other side was 5Kg. The motor was running at 1300rpm, its input being 70 Amp at 420 V dc. The pulley diameter is 1m. Determine torque, output of motor and efficiency of motor. Given W1=35Kg W2=5Kg N=1300 rpm I=70Amp V=420V d= 1m therefore r=0.5m T=? o/p=? Efficiency=? 12. Solution:The effective torque in Newton isT = 9.81* (W1 W 2) * r N-mT=9.81*(35-5)*0.5 T=147.15N-m Input to motor=V*I = 420*70 = 29400 watt Output of the Motor = T*w =147.15*(2*pi*N)/60 watt =20032.36watt Efficiency=output/input = 20032.36/29400 = 68.13% 13. Indirect method of Testing In this method , the m/c under test is not directly loaded for determining its efficiency but its performance characteristics is determined by using the data obtained in no load test performed on the m/c. 14. Swinburne's test In this test, the machine under test is run as a motor although it may be generator At no load, we apply the rated voltage across its terminals and adjust its field current to run the motor at its rated speed Under this condition its 1. No load line current Io 2. Field current Ish 3. Rated voltage VL are recorded From which either constant losses or stray losses are computed. 15. Swinburne's test 16. No load arm current of the motor is Iao = Io Ish Constant losses (Pc)=No load input No load arm copper loss= VL * Io Iao2 Ra Extra: Therefore, stray losses = input on no load shunt field copper loss armature copper loss 22=V * Io Ish Rsh Iao Ra 17. From the detail plate of m/c, its full load current is known. So ,let full load line current is IL Then, input to motor on full load = VLILwattsIts armature current on full load, I a = I L Ish Full load armature copper loss = Ia 2 Ra= ( I L Ish ) 2 Ra Efficiency of the motor on full load = (input-losses)/inputVL I L Pc ( I L Ish) 2 Ra = VL I L 18. If machine under the test is generator having IL its full load output current and VL is the load voltage or terminal voltageThen, output of generator on full load = VLILwattsIts armature current on full load, I a = I L + Ish Full load armature copper loss = Ia 2 Ra= ( I L + Ish ) 2 Ra Efficiency of the generator on full load = output/(output+losses)VL I L = VL I L + Pc + ( I L + I sh ) 2 Ra