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Note for Electrical Machines 2 - EM2 by Balachander K

• Electrical Machines 2 - EM2
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ELECTRICAL MACHINES 1 CIA-II ANSWER KEY Part B 15 a i) Determination of Efficiency The efficiency of DC machine like any other machine is determined by the ratio of output power to that of the input power. There are three methods of determining the efficiency of a machine. 1. 2. 3. Direct method Indirect method Regenerative method The 1st equation is giving an idea about the direct estimation of the efficiency. In this method the machine is fully loaded and the output is directly measured. This method of measurement is only applied for the small machines. The 2nd and 3rd equations are giving an idea about the indirect estimation of the efficiency. Indirect method is helpful of determining the efficiency of shunt wound generator and compound wound generators. In this method it is required to determine to determine the losses only. So, power supply is required to supply the losses only without loading the machine. For the regenerative method of determining efficiency, it is required to have two identical machines. One machine is used as motor and drives the other and the other is used as generator and feedback the power into the supply. Two machines are mechanically coupled. Therefore the losses can be determined because the internal power drawn is only to supply losses of the two machines. Except these testes, the insulation test and the test for making the commutation satisfactory is done while building up the machine. ii) Losses of a DC machines Copper Losses or Electrical Losses in DC Machine or Winding Loss The copper losses are the winding losses taking place during the current flowing through the winding. These losses occur due to the resistance in the winding. In DC machine, there are only two winding, armature and field winding. Thus copper losses categories in three parts; armature loss, field winding loss, and brush contact resistance loss. The copper losses are proportional to square of the current flowing through the winding. Armature Copper Loss in DC Machine Armature copper loss = Ia2Ra Where, Ia is armature current and Ra is armature resistance. These losses are about 30% of the total full load losses. Field Winding Copper Loss in DC Machine Field winding copper loss = If2Rf Where, If is field current and Rf is field resistance. These losses are about 25% theoretically, but practically it is constant.

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15) b) Hopkinsons Test Hopkinson's Test is another useful method of testing the efficiency of a DC machine. It is a full load test and it requires two identical machines which are coupled to each other. One of these two machines is operated as a generator to supply the mechanical power to the motor and the other is operated as a motor to drive the generator. For this process of back to back driving the motor and the generator, Hopkinson's test is also called back-to-back test or regenerative test.If there are no losses in the machine, then no external power supply would have needed. But due to the drop in the generator output voltage we need an extra voltage source to supply the proper input voltage to the motor. Hence, the power drawn from the external supply is therefore used to overcome the internal losses of the motor-generator set. Hopkinson’s test is also called regenerative test or back to back test or heat run test. Connection Diagram of Hopkinson's Test Here is a circuit connection for the Hopkinson's test shown in figure below. A motor and a generator, both identical, are coupled together. When the machine is started it is started as motor. The shunt field resistance of the machine is adjusted so that the motor can run at its rated speed. The generator voltage is now made equal to the supply voltage by adjusting the shunt field resistance connected across the generator. This equality of these two voltages of generator and supply is indicated by the voltmeter as it gives a zero reading at this point connected across the switch. The machine can run at rated speed and at desired load by varying the field currents of the motor and the generator.

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Calculation of Efficiency by Hopkinson's Test Let, V = supply voltage of the machines Then, Motor input = V(I1 + I2) I1 = The current from the generator I2 = The current from the external source And, Generator output = VI1..................(1) Let, both machines are operating at the same efficiency 'η'. Then, Output of motor = η x input = η x V(I1 + I2) Input to generator = Output of the motor = η X V(I1 + I2) Output of generator = η x input = η x [η x V(I1 + I2)] = η2 V(I1 + I2)..................(2) From equation 1 an 2 we get, VI1 = η2 V(I1 + I2) or I1 = η2 (I1 + I2) Now, in case of motor, armature copper loss in the motor = (I1 + I2 I4) Ra. Ra is the armature resistance of both motor and generator. I4 is the shunt field current of the motor. Shunt field copper loss in the motor will be = VI4 2 Next, in case of generator armature copper loss in generator = (I1 + I3)2Ra I3 is the shunt field current of the generator. Shunt field copper loss in the generator = VI3 Now, Power drawn from the external supply = VI2 Therefore, the stray losses in both machines will be W = VI2 - (I1 + I2 I4)2 Ra + VI4 + (I1 + I3)2 Ra + VI3 Let us assume that the stray losses will be same for both the machines. Then, Stray loss / machine = W/2 Efficiency of Generator Total losses in the generator, WG = (I1 + I3)2 Ra + VI3 + W/2 Generator output = VI1 Then, efficiency of the generator, Efficiency of Motor Total losses in the motor, WM = (I1 + I2 - I4)2 Ra + VI4 + W/2 Motor input = V(I1 + I2) Then, efficiency of the motor, Advantages of Hopkinson's Test The merits of this test are… 1. This test requires very small power compared to full-load power of the motor-generator coupled system. That is why it is economical. Large machines can be tested at rated load without much power consumption. 2. Temperature rise and commutation can be observed and maintained in the limit because this test is done under full load condition. 3. Change in iron loss due to flux distortion can be taken into account due to the advantage of its full load condition. 4. Efficiency at different loads can be determined. Disadvantages of Hopkinson's Test The demerits of this test are 1. 2. 3. It is difficult to find two identical machines needed for Hopkinson's test. Both machines cannot be loaded equally all the time. It is not possible to get separate iron losses for the two machines though they are different because of their excitations.