![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 = Armature Cu loss + Field Cu loss + Stray loss
= I12 ( Ra + Rse) + Ws
Output of motor = Input - Total losses = V1 I1 - [ I12 ( Ra + Rse) + Ws ]](https://image.slidesharecdn.com/testingofdcmotor-170316173614/85/Testing-of-dc-motor-30-320.jpg)
![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
= I22 Ra + I12 Rse + Ws
Input to generator = Output + Total losses = V2 I2 + [ I22 Ra + I12 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.](https://image.slidesharecdn.com/testingofdcmotor-170316173614/85/Testing-of-dc-motor-31-320.jpg)
![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)](https://image.slidesharecdn.com/testingofdcmotor-170316173614/85/Testing-of-dc-motor-36-320.jpg)
Testing of dc motor
- 1. Losses In A DC Generator And DC Motor • A dc generator converts mechanical power into electrical power and a dc motorconverts electrical power into mechanical power. Thus, for a dc generator, input power is in the form of mechanical and the output power is in the form of electrical. • On the other hand, for a dc motor, input power is in the form of electrical and output power is in the form of mechanical. In a practical machine, whole of the input power can not be converted into output power as some power is lost in the process. • This causes the efficiency of the machine to be reduced. Efficiency is the ratio of output power to the input power. Thus, in order to design rotating dc machines with higher efficiency, it is important to study the losses occurring in them. • Various losses in a rotating DC machine (DC generator or DC motor) can be characterized as follows:
- 3. Losses In A Rotating DC Machine • Copper losses – Armature Cu loss – Field Cu loss – Loss due to brush contact resistance • Iron Losses – Hysteresis loss – Eddy current loss • Mechanical losses – Friction loss – Windage loss • The above tree categorizes various types of losses that occur in a dc generator or a dc motor. Each of these is explained in details below.
- 4. COPPER LOSSES • These losses occur in armature and field copper windings. Copper losses consist of Armature copper loss, Field copper loss and loss due to brush contact resistance. • Armature copper loss = Ia2Ra (where, Ia = Armature current and Ra = Armature resistance) This loss contributes about 30 to 40% to full load losses. The armature copper loss is variable and depends upon the amount of loading of the machine. • Field copper loss = If2Rf (where, If = field current and Rf = field resistance) In the case of a shunt wounded field, field copper loss is practically constant. It contributes about 20 to 30% to full load losses. • Brush contact resistance also contributes to the copper losses. Generally, this loss is included into armature copper loss.
- 5. IRON LOSSES (CORE LOSSES) • As the armature core is made of iron and it rotates in a magnetic field, a small current gets induced in the core itself too. Due to this current, eddy current loss and hysteresis loss occur in the armature iron core. Iron losses are also called as Core losses or magnetic losses. • Hysteresis loss is due to the reversal of magnetization of the armature core. When the core passes under one pair of poles, it undergoes one complete cycle of magnetic reversal. The frequency of magnetic reversal if given by, f=P.N/120 (where, P = no. of poles and N = Speed in rpm) The loss depends upon the volume and grade of the iron, frequency of magnetic reversals and value of flux density. • Hysteresis loss is given by, Steinmetz formula: Wh=ηBmax1.6fV (watts) where, η = Steinmetz hysteresis constant V = volume of the core in m3
- 6. • Eddy current loss: When the armature core rotates in the magnetic field, an emf is also induced in the core (just like it induces in armature conductors), according to the Faraday's law of electromagnetic induction. • Though this induced emf is small, it causes a large current to flow in the body due to the low resistance of the core. This current is known as eddy current. The power loss due to this current is known as eddy current loss. • Eddy current loss Pe = Ke Bmax f2 t2 V where Ke = Constant depending upon the electrical resistance of core and system of units used Bmax = Maximum flux density in Wb/m2 f = Frequency of magnetic reversals in Hz t = Thickness of lamination in m V = Volume of core in m3 It may be noted that eddy current loss depends upon the square of lamination thickness. For this reason, lamination thickness should be kept as small as possible.
- 7. Mechanical Losses • Mechanical losses consist of the losses due to friction in bearings and commutator. Air friction loss of rotating armature also contributes to these. These losses are about 10 to 20% of full load losses. • These losses are due to friction and windage. (i) friction loss e.g., bearing friction, brush friction etc. (ii) windage loss i.e., air friction of rotating armature. • These losses depend upon the speed of the machine. But for a given speed, they are practically constant. Note. Iron losses and mechanical losses together are called stray losses. Stray Losses • In addition to the losses stated above, there may be small losses present which are called as stray losses or miscellaneous losses. These losses are difficult to account. They are usually due to inaccuracies in the designing and modeling of the machine. Most of the times, stray losses are assumed to be 1% of the full load.
- 8. Power Flow Diagram • The most convenient method to understand these losses in a dc generator or a dc motor is using the power flow diagram. • The diagram visualizes the amount of power that has been lost in various types of losses and the amount of power which has been actually converted into the output. Following are the typical power flow diagrams for a dc generator and a dc motor.
- 10. Efficiency of D.C. Machines: Generator:
- 16. • Thus the efficiency increases with increase in load current, reaches a maximum value when load current equals the value given by eqn. (27) and then starts decreasing, • Efficiency curve: The efficiency of a machine is different at different values of power output. As the output increases, the efficiency increases till it reaches a maximum value. • As the output is further increased, the efficiency starts decreasing. A graph of efficiency vs. output is called efficiency curve. A typical efficiency curve is shown in Fig. 96. • The machines are so designed as to give maximum efficiency at or near the rated output of the machine. Since the generators operate at a constant terminal voltage V, the efficiency curve of a generator can be drawn between efficiency and load current I:
- 17. BRAKE TEST ON DC SHUNT MOTOR • 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. Brake 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.
- 20. 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 in meter 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,
- 23. Swinburne Test of DC Machine • This method is an indirect method of testing a dc machine. It is named after Sir James Swinburne. Swinburne's test is the most commonly used and simplest method of testing of shunt and compound wound dc machines which have constant flux. • In this test the efficiency of the machine at any load is pre-determined. We can run the machine as a motor or as a generator. In this method of testing no load losses are measured separately and eventually we can determine the efficiency. • The circuit connection for Swinburne's test is shown in figure below. The speed of the machine is adjusted to the rated speed with the help of the shunt regulator R as shown in fig.
- 25. Calculation of Efficiency: Let, I0 is the no load current ( it can be measured by ammeter A1 ) Ish is the shunt field current ( it can be measured by ammeter A2 ) Then, no load armature current = (I0 - Ish) Also let, V is the supply voltage. Therefore, No load power input = VI0 watts. In Swinburne's test no load power input is only required to supply the losses. The losses occur in the machine mainly are: Iron losses in the core Friction and windings losses Armature copper loss. Since the no load mechanical output of the machine is zero in Swinburne's test, the no load input power is only used to supply the losses. The value of armature copper loss = (I0 - Ish)2 Ra Here, Ra is the armature resistance. Now, no to get the constant losses we have to subtract the armature copper loss from the no load power input. Then, Constant losses WC = VI0 -(I0 - Ish)2 Ra After calculating the no load constant losses now we can determine the efficiency at any load. Let, I is the load current at which we have to calculate the efficiency of the machine. Then, armature current (Ia) will be (I - Ish), when the machine is motoring. And Ia = (I + Ish), when the machine is generating.
- 26. Calculation of Efficiency When the Machine is Motoring on Load: Power input = VI Armature copper loss, PCU = I2 Ra = (I - Ish)2Ra Constant losses, WC = VI0 -(I0 - Ish)2 Ra Total losses = PCU + WC Efficiency of the motor:∴ Calculation of Efficiency When the Machine is Generating on Load: Power input = VI Armature copper loss, PCU = I2 Ra = (I + Ish)2 Ra Constant losses, WC = VI0 - (I0 - Ish)2 Ra Total losses = PCU + WC ∴ Efficiency of the generator:
- 27. Advantages of Swinburne's Test: The main advantages of this test are : 1.This test is very convenient and economical as it is required very less power from supply to perform the test. 2.Since constant losses are known, efficiency of Swinburne's test can be pre- determined at any load. Disadvantages of Swinburne's Test The main disadvantages of this test are : •Iron loss is neglected though there is change in iron loss from no load to full load due to armature reaction. •We cannot be sure about the satisfactory commutation on loaded condition because the test is done on no-load. •We can’t measure the temperature rise when the machine is loaded. Power losses can vary with the temperature. •In DC series motors, the Swinburne’s test cannot be done to find its efficiency as it is a no load test.
- 28. Field’s Test for Series Motor • 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.
- 29. • 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
- 31. 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 = I22 Ra + I12 Rse + Ws Input to generator = Output + Total losses = V2 I2 + [ I22 Ra + I12 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.
- 32. HOPKINSON 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.
- 33. Connection Diagram of Hopkinson's Test:
- 34. • 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.
- 36. 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)
- 37. • Now, in case of motor, • armature copper loss in the motor = (I1 + I2 - I4)2 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 • 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
- 38. 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,
- 39. 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.
- 40. Disadvantages of Hopkinson's Test The demerits of this test are 1. It is difficult to find two identical machines needed for Hopkinson's test. 2. Both machines cannot be loaded equally all the time. 3. It is not possible to get separate iron losses for the two machines though they are different because of their excitations. 4. It is difficult to operate the machines at rated speed because field currents vary widely.