KULLIYYAH OF ENGINEERINGDEPARTMENT OF MECHATRONICS ENGINEERING

LAB REPORT

MCT 2119

EXPERIMENT 6A: DC GENERATOR

EXPERIMENT 6B: DC MOTOR

NAME MATRIC NO.

1. IZZUL FIKRI BIN MOHAMAD KHAFID 10160572. MUHAMMAD HASBI SHIDDIQ BIN SUPARMAN 10185853. IMADUDDIN BIN HAMZAH 1010765

GROUP : 1

PROGRAMME : MECHATRONICS ENGINEERING

DATE OF EXPERIMENT : 12/3/2012

DATE OF SUBMISSION : 19/3/2012

Experiment 6A :Dc Generator

Part A:Connection and Starting

Introduction

Dc generators are basics needs in our daily live. Electricity is generated by generator. From the name itself we can get to know the usage of the DC generator that is to convert mechanical energy to electricity energy. From part A, the mechanical energy influences the rate of voltage produce. Next, the experiment involved student in connecting the DC wound machine to the system and familiar with the equipment and machine in the laboratory. Moreover, we can use table Torque Vs Voltage produce to see the characteristics of the generator. The generator used is Shunt-Wound Machine. This type of machine runs practically constant speed, regardless of the load. It is the type generally used in commercial practice and is usually recommended where starting conditions are not usually severs. Speed of the shunt-wound machine may be regulated in two ways: first, by inserting resistance in series with the armature, thus decreasing speed: and second, by inserting resistance in the field circuit, the speed will vary with each change in load: in the latter, the speed is practically constant for any setting of the controller. This latter is the most generally used for adjustable-speed service, as in the case of machine tools.

Figure show the connection of shunt generator.

Objective

1. Connect and operate a D.C shunt wound machine as a self-excited shunt wound generator.

2. Deduce from the measurements that the magnitude of the generated voltage is determined by the speed and the armature or excite current.

3. Explain the function and purpose of the field regulator4. Familiarize students with the laboratory machine and equipment.

Instrument and Components

1. 1 D.C shunt wound machine

2. 1 Three-phase asynchronous motor with squirrel-cage rotor3. 1 Magnetic powder brake4. 1 Control unit for brake5. 2 Rubber coupling sleeves6. 2 Coupling guard7. 1 Shaft end guard8. Field regulator for D.C generators9. Load resistance10. Motor protection switch11. 3 Multimeter12. Set of connection cables

Procedures

IA t Rf RB

IE A1 q E1 E2 A2

1. Connect the generator with the system or controller as shown in the above figure .2. The asynchronous generator and wound-rotor machine are already been assemble

together.3. Next, ask the lab moderator to check the connections are properly assembled.4. Set the control unit:

(a) Speed n = 3000 rpm(b) Torque M = 1 Nm(c) Operating Mode N = Constant

(2) Range in Multimeter :

(a) Voltage V = 300 V(b) Armature Current IA = 1 A(c) Exciter Current IE = 0.3 A

Set the field generator to 0% and Load Resistance to 100% ~(1000Ω)

5. Operate the generator. Apply a load to the drive machine to obtain 2900 rpm (or at maximum rpm obtained. Use set/start value from the control unit and set it to fully clockwise. In our case the value starts at 2700 rpm.)

G

A

V

A

6. Record the value of voltage, armature current IA and exciter current IE into the table.(due to the value starts at 2700 rpm thus the table ends at 1700 rpm)

Results

This figure shows the result that obtained from the experiment. The power get by multiplying the voltage induced with current excite.

N (Rpm) VG (V) IE (mA) IA Power (W) VG * IE

2700 183 107.5 -0.28 19.672600 174 101.7 -0.26 17.692500 164.5 96 -0.24 15.792400 155.9 90.9 -0.23 14.172300 146.6 85.4 -0.22 12.522200 137 79.8 -0.2 10.932100 127.9 74.5 -0.19 9.532000 117.9 68.7 -0.17 8.11900 107.2 62.7 -0.16 6.721800 97.2 56.8 -0.15 5.561700 85.7 50.1 -0.13 4.29

Fig. 2

This figure shows the relationship between the voltages induced by the generator with the speed of rotation of the generator.

17001800

19002000

21002200

23002400

25002600

27000

20406080

100120140160180200

VG VS Rpm

VG VS Speed

Speed n , Rpm X 1000

Volta

ge ,

VG (V

)

Discussion

In the experiment 6 part A, firstly, we can get to know the generator producing electricity from the reading in fig. 2. The connection must be thoroughly checked because we need to reduce the possibilities the motor will damaged or burn. For this, we must check with the lab moderator to check the system connection. Next, from part A we can confirm that the generator was running properly without any problems encountered.

The working principle of the induction motor are when three phase voltage is applied to the stator winding, a rotating magnetic flux is established in the motor, that cuts the rotor conductors, induces voltage and generate current in the rotor. Due to nature of the process, the rotor current and the rotating magnetic flux are placed at 900

in space domain. The interaction between the rotating flux and the rotor current produces a torque that turns the motor into rotation.

The voltage induces are influence by the rotation of the generator. From the fig. 3, we can refer the voltage increases when the rotation per minute of the generator increases. From the fig. 3, the speed started at 2700 rpm instead of 3000 rpm. We can understand this problem by understanding slip of an induction machine. Slip induction is the difference between the synchronous speed and the rotor speed, expressed as a percent of synchronous speed.

s=(ns-n)/ns= slip, ns= synchronous speed, n = rotor speed

Synchronous Speed depends upon the frequency of the source and the number of poles on the stator. ns = 120f/p

ns = synchronous speed (r/min), f = frequency of the source (Hz), p = number of poles

The induction motor cannot achieve the synchronous speed as it has slip. The slip exists in induction motor because the stator and the rotor are having slight difference in the rotation. Thus, slip is occurred. From there, we can understand the phenomena the speed starts from 2700 rpm rather than 3000 rpm.

In this experiment, the speed (rpm) is manipulate and the voltage induced is the recorded value. In this experiment the torque is constant thus there no relation between torque and speed can be made in this experiment.

Next, we can see the uses of field regulator. The field regulator are the one that control the current flows from generator. The field regulator works by control the current flows out, and then it will control the electromagnetic field. The electromagnetic will control the strength of the magnetic field. As the magnetic field controlled thus we can control the current and voltage produced.

There are many application in from DC generator can be produced. We can see some example from the wind generator. The winds rotate the blade, then rotating the rotor. Thus, produce electricity. From this example, it clearly shows that the generator converting mechanical

power to electricity.

One of the problem that can occur is the objects can come from external sources or failure of internal components; they can pick up energy from the spinning rotor and do extensive damage. We can prevent by inspecting on a regular basis all internal parts that are prone to failure or can be dislodged. Inspection tests can be a combination of visual inspection along with ultrasonic or magnetic particle tests on rotating component.

Conclusion

All in all, DC generator give a lot of contribution as it become the basics to some other application around the globe. In this experiment, the voltages and current produce are directly proportional to the speed rotation of the armature and the field regulator.

Question

1. What is the effect of reduction in speed?

The speed influence the voltage induced by the armature. The higher the speed the higher the voltage induced and vice versa. The magnetic field produced by the stator will be reduced thus reducing the production of electromotive force which is the voltage induced.

2. What is the purpose of the field regulator?

The field regulator are the one that control the current flows from generator. The field regulator works by control the current flows out, and then it will control the electromagnetic field. The electromagnetic will control the strength of the magnetic field. As the magnetic field controlled thus we can control the current and voltage produced.

3. What function has the load resistance?

The load resistance also used to control the value of output or voltages induce. When the resistances increase, speed of rotation of the armature reduces thus producing low output of voltage. The load resistance will reduce the speed below its nominal speed.

Part B :(Load Characteristics)

Introduction

DC shunt-excited generator is a machine with the field winding in parallel with the armature terminals. This eliminates the need for an external source of excitation. The generator becomes self-excitation. Load characteristics can be determined by adjusted the load resistance to produce the generated voltage. Load characteristic curve is a relationship between the generated voltage and the armature current.

Objective

1. To determine the load characteristics of DC self-excited shunt wound generator.2. To determine the armature current when the load is varied.3. To calculate the delivered electrical power.4. To determine the relationship between generator voltage, armature current and power.

Instrument and Components

13. 1 D.C shunt wound machine 14. 1 Three-phase asynchronous motor with squirrel-cage rotor15. 1 Magnetic powder brake16. 1 Control unit for brake17. 2 Rubber coupling sleeves18. 2 Coupling guard19. 1 Shaft end guard20. Field regulator for D.C generators21. Load resistance22. Motor protection switch23. 3 Multimeter24. Set of connection cables

Procedure

IA

RB

A1

E1 E2

A2

1. The circuit is connected as shown in the above figure .2. The 3 phase asynchronous motor is connected. 3. The control unit is set as follow:

Speed (n) = 3000 rpm Torque (M) = 1 Nm Operating mode (n) = constant

Range on multimeters: Voltage = 300 V Armature current, IA = 1 A Exciter current, IE = 0.3 A

4. The generator is operated.5. The load resistance is adjusted to produce the generated voltage given in the table. The

armature current are measured and recorded in the table. 6. The delivered electrical power , P= I x V, is calculated.7. The load characteristic is drawn by the relationship between generator voltage and

armature current.

Results:

A

G

V

Table

VG (V) 175 150 125 100 75 50 25 0IA (A) 0.4 0.64 0.77 0.86 1.01 1.35 1.51 0.44Pout (W) 70 96 96.25 86 75.75 67.5 37.5 0

Sample calculation of delivered electrical power, Pout = IA x VG

Example:Pout= 0.4A x 175V = 70 W

Figure 1: shows the relationship between generator voltage and armature current

0.2 0.4 0.6 0.8 1 1.2 1.4 1.60

20406080

100120140160180200

VG (V) vs IA (A)

IA (A)

VG

(V)

Figure 2: shows the relationship between power output and armature current

0.2 0.4 0.6 0.8 1 1.2 1.4 1.60

20

40

60

80

100

120

Pout (W) vs IA (A)

Series2

IA (A)

Pout

(W)

Discussion:

In this experiment we manage to determine load characteristics by drawing the load characteristics curve. In the figure 1, when the voltage decreases by adjusting the load resistance, the armature current will decrease. The generator operating under constant speed and field excitation. The exciting current is controlled by potentiometer.

In figure 2, we can conclude that the power output is linearly rise to maximum when the armature current reach 0.77A and start to decrease when the current above 0.77A. It shows that 0.77 A is the limit of the armature current to get maximum of the power output.

Questions:

1. Delivered power depends on the armature current -The basic formula for power (Watts) is: P = V * I, where P is power (in Watts), V is generator terminal voltage (in Volts), and I is armature current (the current flowing in the stator of the generator). (For a three-phase generator the entire formula is P = V* I * (3^(0.5)) * PF, where, 3^(0.5) is the square root of three (a fixed value, 1.732, I think), and PF is the power factor of the generator (which is a number never greater than 1.0, and which we will presume to be 1.0 for the purposes of our discussion). Coincidentally, the terminal voltage of most synchronous generators is almost a fixed value, as well, and doesn't usually vary by more than approximately +/- 5.0%, which on an 11,0000 Volt generator is only about 550 Volts (out of 11,000).

So, since one of the terms of the three-phase power formula *is* a fixed value (the square

root of three), and we are presuming one of the terms (PF) to be fixed and equal to 1.0, and the generator terminal voltage is, for all intents and purposes, a fixed value, the way that a generator produces more power is to increase the number of amps flowing in the stator. The way that amps are increased in the generator stator is by providing more torque from the turbine into the generator; more torque equals more amps. Less torque equals less amps. (We are presuming that the prime mover is always producing at least sufficient torque to keep the generator rotor spinning at synchronous speed. When it doesn't, the generator actually becomes a motor and keeps spinning at synchronous speed and draws current from other generators on the grid. This is what's known as "reverse power" or, "motorizing the generator.

2. The shape of the load characteristic curve-the voltage at the terminals is equal to the induced voltage at no-load current condition. As the voltage generator decrease, current armature will increase. The voltage generator is inversely proportional to the armature current.

Conclusion

In conclusion the objectives of the experiment are well-achieved and proven where load characteristic curve are successfully obtained from the experimental and calculated results.

Experiment 6B :DC Motor

Part A: Connection and Starting

Introduction

Motor are significance in our daily lives. As we all know, there are many application that uses motor such as combustion engine. It converts from electrical energy to mechanical energy. Furthermore, motor are opposite of generator. Besides that, motor must have starter to start the motor so that the motor do not damage with high voltage input.

The figure shows that the current move into the motor to produce mechanical energy.

Objective

1. Connect and operate a D.C shunt wound machine as a D.C shunt wound motor with and without starter.

2. Measure the starting current and the armature voltages.3. Deduce that the starter reduces the starting current.4. To know to identify the specification of the motor machine.

Instrument and Components

1 D.C shunt wound motor machine 1 Magnetic powder brake 1 Control unit for brake 2 Rubber coupling sleeves 2 Coupling guard 1 Shaft end guard 1 Starter for D.C motors 1 Field regulator for D.C generators 1 Load resistance

2 Multimeter 1 Set of connection cables

Procedure

L+

A1

E1 E2A2

L- Figure 1: without starter

1. Check the DC motor working properly by consulting with the lab demonstrator.2. Assemble the connection as shown in the above figure 1.3. Set the control unit as follows:

(a) Speed n = 3000 Rpm(b) Torque m = 1 N.M(c) Operating mode n = constant

Range on Multimeter

(a) Voltage , V = 300V(b) Armature current , IA = 1 A(c) Exciter current , IE = 0.3 A

Set the field generator to 0 % and the load resistance to 100% (~1000 Ω)4. Operate the motor and set the torque on the control unit to M=0.3 Nm.

M

A

V

Measure the starting current and the armature voltage. Enter the measuredvalues into fig. 1

L+

RA

A1

E1 E2

A2 L-

Figure 2: with starter

5. Connect the machine with starter, according to circuit diagram in figure 2. Thesettings on the control unit remain unchanged. Set the starting resistor to100% (47 Ω) Measure the starting current and the armature voltage. Enter themeasured values into fig. 1

6. Make sure to record the reading quickly at the maximum value of current that used.

ResultsThe motor that had been used specification are shown.Type : SE 2662 – 3AV : 220 VI : 1.4 AVE : 220 VIE : 0.14 A

This table shows the motor current when starter used and starter at rest.

Starter R = 0 Ω R = 47 ΩLoad M = 0. 3 Nm M = 0.3 NmStarter Current IA (A) 5.5 1.5Armature Voltage IA (V)

266 242.8

Discussion

A

MV

In this experiment the student must learn the type of the motor and know how to read the specification that are used. From there, we can get to know the voltage, voltage excited, current and current exciter from the specification.

Next, to get the fig. 1, we need to find assemble the connection and run the experiment. The motor that we get for the experiment are used for high end performance. Thus, current value data recording must be quickly taken due to fluctuation of the ammeter really fast. We must take the maximum point that the pointer of the ammeter go. Then, we get the value of the starter current. From this, it will produce random error and the value will not precise.

When the starter in used, the starter current is lower than when the starter off. Thus it will help to prevent some damage to the armature winding due to high starting current. As the rotating armature of dc motor picks up speed, the starter resistance is gradually reduced so that the motor is able to attain.

At full speed the motor starts running normally, of course, without the help of starter. In other words, the starter offers resistance to armature current during starting of dc motor only. Under normal working condition of dc motor, the starter is electrically out of armature circuit of the motor. The starter protects the armature of dc motor from getting damaged.

The electromotive force (emf) induced in the armature winding during starting builds up from zero value to max value to restrict the armature current within the permissible value at full speed. As the speed of armature/motor build up, armature induced emf also starts building thus reducing the role resistance offered by the starter, hence requiring it to gradually reduce as the motor picks up full speed.

ConclusionIn conclusion, the starter current will help to maintain the motor from damages by

reducing the high starting current. Besides, the starter gives benefit to motor machine because it will produce the same voltage as non-starter motor as it will gradually increase to produce the same output voltage.

Question

What is the function of starter?

When the starter in used, the starter current is lower than when the starter off. Thus it will help to prevent some damage to the armature winding due to high starting current. As the rotating armature of dc motor picks up speed, the starter resistance is gradually reduced so that the motor is able to attain.

Part B: Load Characteristics

Introduction

Shunt motor gets its names from the way it is connected where the field coil is connected in parallel (shunt) with the armature.

As DC shunt motor’s speed does not change more than 12% between no-load to full-load, the shunt motor is considered as a constant speed motor. It is the type generally used in commercial practice and is usually recommended where starting conditions are not usually severs. Speed of the shunt-wound motors may be regulated in two ways: first, by inserting resistance in series with the armature, thus decreasing speed, and second, by inserting resistance in the field circuit, the speed will vary with each change in load, in the latter, the speeds are practically constant for any setting of the controller. This latter is the most generally used for adjustable-speed service, as in the case of machine tools. A shunt wound motor has a high-resistance field winding connected in parallel with the armature. It responds to increased load by trying to maintain its speed and this leads to an increase in armature current. This makes it unsuitable for widely-varying loads, which may lead to overheating.

Objectives

1. To connect and operate a D.C shunt wound machine as a shunt wound motor for recording the load characteristics.

2. To obtain the load characteristics curve based on the values obtained by measurement and by calculations.

3. To deduce from the load characteristics the highest efficiency of the motor at its nominal speed and approved the deduction through calculations.

4. To describe the response of the shunt wound motor under various load conditions.

Instrument and Components

1 D.C shunt wound motor machine 1 Magnetic powder brake 1 Control unit for brake 2 Rubber coupling sleeves 2 Coupling guard

1 Shaft end guard 1 Starter for D.C motors 1 Field regulator for D.C generators 1 Load resistance 2 Multimeter 1 Set of connection cables

Procedure

L+

A1

E1 E2

L- A2

1. The circuit is connected as shown in above figure .2. The control unit is set as follow:

Speed (n) = 3000 rpm Torque (M) = 1 Nm Operating mode (M) = constant

Range on multimeter: Voltage = 300 V Armature current, IA = 1 A Exciter current, IE = 0.3 A

3. The generator is operated with D.C power supply is set at 220 V.4. The control unit is used to vary the values of the torque based on the values given in the table. At the given values, the speed, armature current and exciter current are measured.

M

A A

V

5. The consumed electrical power, delivered mechanical power and efficiency of the motor is calculated at each torque’s varied values. The calculated values are entered into the table.

i. Consumed electrical power:

P1 = UA × Itotal

Itotal = IA + IE

ii. Delivered mechanical power:

P2 = M.N/9.55

iii. Efficiency of the motor: ƞ = P2/P1

6. The graph of load characteristics is plotted from the measured and calculated values obtained from the result.

N (rpm) vs M(Nm) IA (A) vs M (Nm) P2 (W) vs M (Nm) Ƞ vs M (Nm)

Results:

Table 1:

U (V) 220M (Nm) 0.3 0.5 0.7 0.8 0.9 1 1.1 1.2 1.3N (rpm) × 103 2.2 2.15 2.15 2.15 2.1 2.1 2.1 2.1 2.1IA (A) 0.36 0.53 0.7 0.78 0.87 0.94 1.04 1.13 1.23IE (A) 0.12 0.12 0.12 0.12 0.12 0.12 0.12 0.12 0.12Itot (A) 0.48 0.65 0.82 0.9 0.99 1.06 1.16 1.25 1.35

P1 (W)105.6 143 180.4 198 217.8 233.2 255.2 275 297

P2 (W)69.11 112.57 157.59 180.1 197.91 219.9 241.88 263.87

285.86

ƞ 0.65 0.79 0.87 0.91 0.91 0.94 0.95 0.96 0.96

Figure 1:

The graph below is obtained from the tabulated data from Table 1 which is showing the relationship between speed, N (rpm) and torque, M (Nm).

0.2 0.4 0.6 0.8 1 1.2 1.42.042.062.08

2.12.122.142.162.18

2.22.22

N (rpm) vs M (Nm)

M (Nm)

N (r

pm)

Figure 2:

The graph below is obtained from tabulated data in Table 1 where shows the relationship between armature current, IA (A) and torque, M (Nm).

0.2 0.4 0.6 0.8 1 1.2 1.40

0.2

0.4

0.6

0.8

1

1.2

1.4

IA (A) vs M (Nm)

M (Nm)

IA (A

)

Figure 3:

The graph below is obtained from calculated results based from the data obtained in Table 1 which shows the relation between mechanical output power, P2 (W) and torque, M (Nm) .

0.2 0.4 0.6 0.8 1 1.2 1.40

50

100

150

200

250

300

350

P2 (W) vs M (Nm)

M (Nm)

P2 (W

)

Sample calculation:

P2=M.N/9.55

=(0.9).(2.1)/9.55 =197.1 Watt

Figure 4:

The graph below is obtained from the calculated results which are based on data obtained in Table 1 where it shows the relation between efficiency, ƞ torque, M (Nm).

0.2 0.4 0.6 0.8 1 1.2 1.40

0.2

0.4

0.6

0.8

1

1.2

ƞ vs M (Nm)

M (Nm)

ƞ

Sample calculation

ƞ = P2/P1

=285.86/297 = 0.96

Discussion:

DC shunt motor’s speed is considered constant as its speed does not regulate beyond 12 % from no-load to full-load condition and vise-versa. From Figure 1 which shows the relationship between the motor speeds, N (rpm) as the torque, M (Nm) is varied. As observed from the graph of Figure 1, the speed of the motor is decreasing as the torque is increased and finally starts to equalize at 0.9 Nm of torque magnitude with speed of 2.1 × 103 rpm. This is due to the fact that when the motor switches from no-load to load condition; the motor will eventually slow down. The reduction in speed is proportional to the reduction of counter-electromotive force (CEMF). As the armature net voltage increases:

VA – CEMF = Net Voltage

As the net voltage increases, the armature current, IA (A) will increases and result in increasing torque, N (Nm). These properties can be observed from Figure 2, where from the graph it is shown that as the torque is increased the armature current also increases. However, the speed (referring to Figure 1) will eventually equalize and stop changing when the torque reaches a

level that is required to turn the larger load. The speed at which it equalize is known as nominal speed, referring to figure 1 the nominal speed is 2.1 ×103 rpm. Figure 3 shows the relationship between the mechanical output power of the motor (P2 in Watt) and the torque (M in Nm). The graph shows the relation almost linearly proportional between the increases of torque with the mechanical output torque. As referred to Figure 4 and Table 1, the motor is most efficient at its nominal speed, which is at 2.1 × 103 rpm with efficiency up to 0.96.

Conclusion

In conclusion the objectives of the experiment are well-achieved and proven where load characteristic curve are successfully obtained from the experimental and calculated results. Besides, it is also proven that the motor operate at the best efficiency when it reach its nominal speed which is at 2.1× 103 rpm with efficiency of 0.96.