Assessment of Safety Issues in Electrical Machines
It is important for a machine in satisfying its intended purpose to comply with all the set directives. This is fostered by determining the operation limits of the machine, the defined safety measures and how to apply them and assessing user instructions and technical files before operating the machines. In assessment of the machines, the following issues need to be checked:
- Dangerous parts of the machine.
The machine should be assessed whether dangerous parts such as rotating shafts, electrical terminals or heat emission points are accessible by the user. If in the assessment they are found to be accessible then guards should be accurately positioned and securely held in place to prevent reach to such areas. However, for purposes of maintenance of the machine, they could be disabled but only after the machine becomes safe probably by being off or having cooled.
- Protection against specified hazards
It is important to first establish possible hazards that could result from operation of the machine. Such may be fire, explosion or discharge of hazardous material. Protection mechanisms for such hazards can then be set in place for example on proper disposal of the hazardous material, access to fire extinguishers, emergency stops and fire alarms.
- Controls for machine starting, stopping or making a significant change in operation.
The operator should be well familiar with the starting procedure or the sequence clearly shown on the machine. Any control button should be clearly marked and well differentiated from the other in the purpose intended for example green for start, yellow for stall and red for stop. Key things to establish are:
- Whether a reset facility is provided and the ability of the machine to restart automatically after stopping.
- Whether the emergency stops are correctly located and functioning.
- Whether the machine stops safely or the stopping procedure.
- Whether the operator has view all around the machine from the point of operation
- Whether there is a delayed start and a working warning system.
- Isolation of energy sources.
Isolation refers to removal or disconnection of an energy source to prevent unintended restoration of energy through start-up of installations or release of stored energy. Energy isolating devices include:
- Electrical circuit breakers
- Line valve
- Disconnect switch
- Block
The energy isolating devices should be in good working condition to ensure safety. Lock-out devices prevent the unintended energizing of energy sources on installations. They should be able to meet minimum safety requirements with the following characteristics;
- Visually identified instantly as a safety device well marked for example “DANGER: LOCK-OUT”
- Avoid inadvertent energisation at the point of attachment.
- Unique locking mechanism for each lock-out device.
- Tamper proof.
- Lighting and ventilation
Ambient lighting of enough intensity is appropriate around the machine. The operator should be able to appropriately see essential parts of the machine. If ambient lighting lacks or not enough then lighting should be provided with adequate lighting. The code of practice for the electricity (wiring) regulation specifies a vertical illumination level average of 120 luminous intensity for proper operation of the machine. Suitable ventilation for free flow of air is needed to avoid the development of high ambient air temperatures exceeding the permissible temperature around the machine.
The code of practice for the electricity (wiring) regulation recommends a clearance space of at least 900 mm for full width and in front of meters and of all low voltage control panels including switch gear with a rating in excess of 100 A. However, a high voltage electrical equipment should have a minimum clearance space of 1400 mm. For accessibility behind or by the side of the machine, a clearance space of at least 600 mm is recommended. The clearance space should be enough for proper operation of draw-out type equipment and opening of the hinged panels to at least 90 degrees.
Constraints related to environmental and sustainability limitations
Energy losses to the environment in form of heat during conversion of electrical energy to mechanical energy and vice versa limits machine charging. The rise in temperature leads to faster aging of insulation and reduced operation time.
Electrical machines impact the environment negatively by generating magnetic fields, noise and vibration. The machines that use brushes produce graphite powder fumes and products resulting from heating of insulation material cause air pollution.
Sustainable development of electrical machines should therefore ensure: pollution is minimized as much as possible and reduced consumption of conventional energy sources by preventing energy losses.
LOAD TEST ON DC SHUNT MOTOR
Aim: To obtain the performance characteristics by using load test.
Load test is a simple method of testing a dc machine and involves measurement of motor output, speed and efficiency at varied load conditions. A dc shunt motor has its field winding in parallel with armature. The field windings produce a magnetomotive force by having relatively large number of turns and higher resistance.
Rotation of dc shunt motor occurs as a result of torque developed in the armature upon connection of the terminals to the supply. Back EMF is induced in the armature during rotation as it cuts the magnetic field. Fleming’s right hand rule and Lenz law explains the direction of the induced EMF where it tends to oppose the applied voltage.
Equation describing the motor is V = +
Where = and
= is zero during motor starting, hence armature has to be increased gradually.
Power developed in the armature = = Tω
Where ω = and T = φ
Where φ = flux produced per pole, = armature current and is torque proportionality constant.
Losses are due to friction and windage, eddy currents and hysteresis. Stable speed is achieved when the torque developed and the resisting torques (load torque) is at equilibrium.
increases with increase in mechanical load, hence increase in the voltage drop () and decrease of the back EMF ();
= V – = φ ω
Where is back EMF constant.
If flux (φ) is constant, speed (ω) decreases as decreases. is very small hence is a small value. The speed therefore slightly decreases as increases.
Armature current vs torque characteristics
At constant flux, an increase in armature current leads to corresponding increase in torque developed. However, flux slightly falls with increased armature current in real situation as a result of armature reaction.
Load Test on DC Shunt Motor
Armature current vs induced EMF
The induced EMF slightly decreases with increase in armature current.
Increasing the load leads to an increase in armature current since the flux is constant. The decrease in speed is small because the voltage drop due to armature resistance is negligible.
Name plate details:
Apparatus
No. |
Equipment |
Range |
Type |
Qty. |
1. |
Ammeter |
0 – 5 A 0 – 2 A |
MC MC |
1No 1No |
2. |
Voltmeter |
0 – 250 V |
MC |
1No |
3. |
Rheostats |
100/5 A 400/1.7 A |
Wire wound Wire wound |
1No 1No |
4. |
Tachometer |
0-2000 rpm |
Digital |
1No |
5. |
Connecting wires |
LS |
Circuit diagram
Fig. 1: Circuit diagram for load test of DC shunt motor, Lab manual.
Procedure
- Connections were made as shown in Fig. 1.
- DPST switch was closed and starter resistance gradually removed after checking the no load condition.
- The motor was then brought to its rated speed by adjusting the field rheostat.
- Ammeter, voltmeter, tachometer and spring balance readings were noted under no load condition.
- The load was then added to the motor gradually and for each load, voltmeter, ammeter, spring balance readings and speed of the motor were noted.
- The motor was then brought to no load condition and field rheostat to minimum position then DPST switch opened.
Results
S.No. |
Volts |
Amps |
(Kg) |
(Kg) |
– (Kg) |
rpm |
T (Nm) |
(W) |
(W) |
? (%) |
1 |
225 |
0.94 |
0.05 |
0.58 |
0.53 |
1548 |
0.1950 |
0.316 |
0.2115 |
14.9 |
2 |
225 |
0.09 |
0.165 |
0.07 |
0.095 |
1544 |
0.0350 |
0.0057 |
0.0203 |
27.9 |
3 |
225 |
0.92 |
0.185 |
0.07 |
0.115 |
1542 |
0.0423 |
0.0068 |
0.207 |
3.3 |
4 |
225 |
0.96 |
0.255 |
0.105 |
0.15 |
1540 |
0.0552 |
0.0089 |
0.216 |
4.12 |
5 |
225 |
0.97 |
0.21 |
0.145 |
0.065 |
1432 |
0.0239 |
0.0039 |
0.2183 |
1.78 |
A graph of armature current against speed and torque against speed were drawn representing the data collected during the experiment.
From the graph, it is possible to deduce that armature current varies inversely with speed.
Torque against speed
From the graph, torque varies inversely with speed.
Operation simulation and results verification of dc shunt motor using Matlab Simulink.
Field winding of the shunt motor is connected across the same voltage supply as the armature hence the same voltage drop. Total current drawn by the motor is given by the addition of field current.
The dynamics for the dc shunt motor can be modelled by the following equations;
= + + ω ………………………………………………… (1)
From equation 1, = – – * ω + …………………………………. (2)
= + …………………………………………………………….. (3)
From equation 3, = – + …………………………………………………. (4)
Motor rotates to move the load either by rotation or translation. For purposes of representation, an equivalent rotational system is convenient.
The fundamental torque equation illustrated by Fig. 3 is;
= + b ω + J = ………………………………………………… (5)
From equation 5, = – ω + – ………………………………………… (6)
From equations 1 and 3, the following representation can be made;
= + ………………………………………………… (7)
At steady state, inductances are zero hence currents are given by:
= and = ………………………………………………… (8)
Substituting equations 8 in torque equation;
= gives
= ………………………………………………………… (9)
On further substitution;
= – – …………………………………………. (10)
Applying a constant armature voltage of 50 V and step input at the field, and taking Ra = 0.61 ?, La = 0.013 H, Rf = 230 ?, Lf = 125 H, Kt = 1, Kb = 1, J =1, b = 0.1, r = 0.05m and Tl = 392.4 N. The following graphs were obtained at 10 ms snapshot.
LOAD TEST ON COMPOUND GENERATOR.
AIM: To conduct load test on compound generator and determine performance characteristics.
Introduction
A dc compound generator equivalent circuit is made up of both series and shunt field wildings. Connection of the shunt and series fields is possible in two ways; short shunt and long shunt. Two types of compound generators exist; cumulative compound generator and differential compound generator.
Performance Characteristics of DC Shunt Motor
Flux produced by both shunt field windings and series field windings add up and the net flux increases as load on generator gets to increase. At higher series field strength, terminal voltage increases with increase in load current leading to over-compounding. When full-load terminal voltage and no-load terminal voltage are at equilibrium, the generator is flat-compounded. Where series field is weak, terminal voltage gets to decrease with increase in load current hence under-compounded.
Flux produced by both series field windings and shunt field windings opposes each other. There is decrease in the air gap net flux and the generated electromotive force (EMF) decreases with increased load. Graph between and gives external characteristics of the motor while that between E and gives internal characteristics of the motor.
Apparatus
S.No |
Equipment |
Range |
Type |
Qty. |
1. |
Ammeter |
0 – 5 A 0 – 2 A |
MC MC |
1No. 1No. |
2. |
Voltmeter |
0 – 250 V |
MC |
1No. |
3. |
Rheostats |
100/5 A 400/1.7 A |
Wire wound Wire wound |
1No. 1No. |
4. |
Tachometer |
0 – 2000 rpm |
Digital |
1No. |
5. |
Connecting wires |
LS |
Procedure
- Connections were as per the circuit diagrams shown above.
- The machine was run at rated speed and the rated voltage was obtained by varying field excitation.
- The switch was then closed for the load to be connected across the generator.
- The load was increased step by step at an interval of 0.125 KW and all meter readings noted.
Results
No. |
(Amps) |
(Volts) |
(Amps) |
= + () |
1 |
0 |
225 |
0.34 |
496.25 |
2 |
0.32 |
224 |
0.36 |
538 |
3 |
0.4 |
224 |
0.37 |
583 |
4 |
0.42 |
224 |
0.38 |
633 |
5 |
0.44 |
224 |
0.39 |
633 |
The unsaturated core of the compound motor has very high reluctance relative to the airgap. There is a linear increase in flux at first but when saturation is attained there is a great increase in core reluctance hence flux increases slowly with increase in magnetomotive force.
LOAD TEST ON SINGLE PHASE INDUCTION MOTOR.
AIM: To conduct direct load test on a single phase induction motor and determine the performance characteristics.
Single phase induction motors are machines that are simple in construction and do not self-start. It constitutes a squirrel cage rotor and a single phase stator winding. The motor’s starting torque is zero and therefore rotation does not occur with connection of stator winding to single phase ac supply. However, when provided for an auxiliary winding on the stator the motor can quickly attain final speed. This behavior is based on the principle of double revolving theory.
The single phase induction motor has a rotor made up of three phase winding with three leads coming out through slip rings and brushes. The leads are short-circuited during running of the motor. Introduction of resistances in the circuit happen via slip rings during starting to boost starting torque. The stator winding creates a rotating field that cuts the shorted rotor conductors leading to induction of current in the rotor. The currents that have been induced produce their own field that rotates at a speed equal to that of stator-produced field. With the interaction of the two fields, torque is developed. Normally the rotor tends to run at a speed that is almost synchronous but slightly lower thus helping in generation of torque since relative speed between the two fields is not zero.
Apparatus
S.No |
Equipment |
Range |
Type |
Qty. |
1. |
Ammeter |
0 – 5 A 0 – 2 A |
MC MC |
1No. 1No. |
2. |
Voltmeter |
0 – 250 V |
MC |
1No. |
3. |
Rheostats |
100/5 A 400/1.7 A |
Wire wound Wire wound |
1No. 1No. |
4. |
Tachometer |
0 – 2000 rpm |
Digital |
1No. |
5. |
Connecting wires |
LS |
Procedure
- The apparatus were set up as shown in the circuit diagram above.
- The supply was switched on at no load condition.
- The rotor voltage was applied to the motor using the variac and readings from ammeter and wattmeter noted.
- The load was varied in suitable steps and all meter readings noted until full load condition.
Results
S.No |
Volts |
Amps |
(Kg) |
(Kg) |
– (Kg) |
rpm |
T (Nm) |
(W) |
(W) |
? |
1 |
233 |
2.8 |
0 |
0 |
0 |
1486 |
0 |
0 |
0.652 |
0 |
2 |
233 |
2.84 |
2 |
1 |
1 |
1472 |
0.357 |
0.056 |
0.661 |
0.0856 |
3 |
233 |
2.9 |
3 |
1 |
2 |
1460 |
0.736 |
0.112 |
0.675 |
0.1647 |
4 |
233 |
2.98 |
4 |
1.5 |
2.5 |
1448 |
0.919 |
0.139 |
0.694 |
0.2008 |
5 |
233 |
3.14 |
5.5 |
1.7 |
3.8 |
1432 |
1.397 |
0.219 |
0.732 |
0.2865 |
Results analysis
As speed increases torque decreases so torque varies inversely with speed.
Simulation of single phase induction motor on Simulink.
Single phase induction motor model.
Stator magnetic flux
= +
= +
Rotor magnetic flux
= +
= +
Stator currents
= ( – )
= ( – )
Rotor currents
= ( – )
= ( – )
Motor torque
= p
Dynamic equation
= ()
Upon simulation of the developed model above, the following graphs were obtained.
The torque-speed curve of the induction motor can be divided into three regions representing braking, motoring and regenerative braking. From the torque-speed relationship, torque can be seen to be positive whilst speed is negative. Plugging or braking is achieved when power converted is negative and airgap power positive. Motoring on the hand has torque and motion in the same direction whereas in generating mode torque is positive while speed is negative.
AIM: To simulate the operation and control of separately excited dc motor using PI controller in Matlab Simulink.
Introduction
A separately excited dc motor constitutes field circuit and armature circuit energized by separate dc sources. A permanent magnet is established in the field circuit with constant flux. The armature circuit is found at the rotor and is made up of rotor winding and commutator segments.
Referring to Fig. 10, for energy balance summation of the torques must be zero.
– – – = 0 …………………………………………………………. (1)
By considering output position of shaft and substituting in equation (1);
– – – = 0 ………………………………………….. (2)
– – – = 0 ………………………………………………….. (3)
Solving equation (3) with Laplace transforms yields;
– – – = 0
– = + ……………………………………………. (4)
Equation (4) can be made to relate as follows;
= ……………………………………………………. (5)
Making angular speed subject of the formula;
= ……………………………………………………………. (6)
Sum of field voltage is;
= + …………………………………………………………….. (7)
Taking Laplace transforms and making subject of the formula;
= +
= ……………………………………………………………………… (8)
Summation of armature voltage;
= + +
= + I(s) + ……………………………………………….. (9)
Separating the armature current and field current;
=
=
Substituting in equation (4) gives;
– =
Transfer function with input armature voltage as input and motor angular speed as output:
=
Applying step input at armature and field voltages, then setting proportional gain () and integral gain () as 9.46 and 20.94 respectively, the following graphs were obtained.
References
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Hindmarsh, J. and Renfrew, A., 1996. Electrical machines and drive systems. Elsevier.
Chapman, S., 2005. Electric machinery fundamentals. Tata McGraw-Hill Education.
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Wagner, C.F., 1939. Self-excitation of induction motors. Electrical Engineering, 58(2), pp.47-51
Pillay, P. and Krishnan, R., 1989. Modeling, simulation, and analysis of permanent-magnet motor drives. I. The permanent-magnet synchronous motor drive. IEEE Transactions on industry applications, 25(2), pp.265-273
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