Load Tests and Simulations of Electrical Machines: A Report

This report presents a comprehensive analysis of various electrical machines, including DC shunt motors, compound generators, and single-phase induction motors. The report begins with a discussion of health, safety, and risk assessment issues related to the operation and grounding of electrical machines, as well as constraints related to environmental and sustainability limitations. The core of the report focuses on load tests conducted on each type of machine, detailing the aim, introduction, theory, apparatus, procedure, results, and analysis of results. Additionally, the report includes operation simulations and results verification of a DC shunt motor and a single-phase induction motor using MATLAB Simulink. The simulations provide a deeper understanding of the machines' behavior under different conditions. The report concludes with references and provides a valuable resource for students studying electrical machines and their performance characteristics.

Electrical machines 1
ELECTRICAL MACHINES
By (Name)
Course
Professor’s name
University name
City, State
Date of submission

Electrical machines 2
Table of Contents
Health, safety and risk assessment issues related to operations and grounding of electrical
machines and generators...............................................................................................................3
Constraints related to environmental and sustainability limitations............................................6
LOAD TEST ON DC SHUNT MOTOR.....................................................................................6
Aim:.............................................................................................................................................6
Introduction..................................................................................................................................6
Theory..........................................................................................................................................7
Armature current vs speed characteristics................................................................................7
Armature current vs torque characteristics...............................................................................8
Apparatus.....................................................................................................................................8
Circuit diagram............................................................................................................................9
Procedure.....................................................................................................................................9
Results........................................................................................................................................10
Analysis of results..................................................................................................................10
Operation simulation and results verification of dc shunt motor using Matlab Simulink.. .11
Results........................................................................................................................................14
LOAD TEST ON COMPOUND GENERATOR......................................................................14
AIM:...........................................................................................................................................14
Introduction................................................................................................................................15
Theory........................................................................................................................................15
Apparatus...................................................................................................................................16
Procedure...................................................................................................................................17
Results........................................................................................................................................17
Results analysis......................................................................................................................18

Electrical machines 3
LOAD TEST ON SINGLE PHASE INDUCTION MOTOR..................................................18
AIM:...........................................................................................................................................18
Introduction................................................................................................................................18
Theory........................................................................................................................................19
Apparatus...................................................................................................................................19
Circuit diagram..........................................................................................................................20
Procedure...................................................................................................................................20
Results........................................................................................................................................20
Results analysis..........................................................................................................................21
Simulation of single phase induction motor on Simulink........................................................22
Single phase induction motor model......................................................................................22
Stator magnetic flux...............................................................................................................22
Rotor magnetic flux................................................................................................................23
Rotor currents.........................................................................................................................24
Dynamic equation...................................................................................................................25
RESULTS..................................................................................................................................26
SIMULATION OF SEPARATELY EXCITED DC MOTOR................................................27
AIM:...........................................................................................................................................27
Introduction................................................................................................................................27
Mathematical modelling of SEDC.........................................................................................27
Results........................................................................................................................................30
References.....................................................................................................................................31

Electrical machines 4
Health, safety and risk assessment issues related to operations and grounding of electrical
machines and generators
Introduction
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:
1. 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.
2. 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.
3. 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

Electrical machines 5
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.
4. 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;

Electrical machines 6
 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.
5. 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.
6. Working space
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.

Electrical machines 7
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.
Introduction
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.
Theory

Electrical machines 8
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 = Eb + I a Ra
Where Eb= ( φZN
60 )( P
A ) and
I a = (V −Eb )
Ra
Eb is zero during motor starting, hence armature has to be increased gradually.
Power developed in the armature = Eb I a = Tω
Where ω = 2 πN
60 and T = k tφ I a
Where φ = flux produced per pole, I a = armature current and k t 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.
Armature current vs speed characteristics
I a increases with increase in mechanical load, hence increase in the voltage drop ( I a Ra)
and decrease of the back EMF (Eb);
Eb = V - I a Ra = k eφ ω

Electrical machines 9
Where k e is back EMF constant.
If flux (φ) is constant, speed (ω) decreases as Eb decreases. Ra is very small hence I a Ra is
a small value. The speed therefore slightly decreases as I a 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.
Armature current vs induced EMF
The induced EMF slightly decreases with increase in armature current.
Torque vs speed characteristics
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 Wire wound 1No

Electrical machines 10
400/1.7 A Wire wound 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
a) Connections were made as shown in Fig. 1.
b) DPST switch was closed and starter resistance gradually removed after checking the no
load condition.
c) The motor was then brought to its rated speed by adjusting the field rheostat.
d) Ammeter, voltmeter, tachometer and spring balance readings were noted under no load
condition.

Electrical machines 11
e) 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.
f) The motor was then brought to no load condition and field rheostat to minimum position
then DPST switch opened.
Results
S.No. Volts Amps S1
(Kg)
S2
(Kg)
S1- S2
(Kg)
rpm T
(Nm)
Po
(W)
Pi (W) Ƞ (%)
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
Analysis of results.
A graph of armature current against speed and torque against speed were drawn representing the
data collected during the experiment.

Electrical machines 12
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.

Electrical machines 13
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, I f and armature current I a as shown in Fig. 2.
Fig. 2: Equivalent circuit of dc shunt motor.
The dynamics for the dc shunt motor can be modelled by the following equations;
V a = Ra I a + La
d ia
dt + Kb If ω ………………………………………………… (1)
From equation 1, d ia
dt = - Ra
La
I a - Kb
La
If * ω + 1
La
V a …………………………………. (2)
V f = Rf I f + Lf
d if
dt …………………………………………………………….. (3)
From equation 3, d if
dt = - Rf
Lf
I f + 1
Lf
V f …………………………………………………. (4)

Electrical machines 14
Motor rotates to move the load either by rotation or translation. For purposes of representation,
an equivalent rotational system is convenient.
Fig. 3: Dynamic of motor – load system.
The fundamental torque equation illustrated by Fig. 3 is;
T m = T l + b ω + J dω
dt = Kb I a If ………………………………………………… (5)
From equation 5, dω
dt = - b
J ω + Lf
J I a I f - Tl
J ………………………………………… (6)
From equations 1 and 3, the following representation can be made;
[ia
if ]=
[−Ra
La
−ωKb
La
0 −Rf
Lf
] [ Ia
If ] +
[ V a
La
V a
Lf
] ………………………………………………… (7)
At steady state, inductances are zero hence currents are given by:
ia = V ¿−K t∗ω∗if
Ra
and if = V ¿
Rf
………………………………………………… (8)
Substituting equations 8 in torque equation;
T m = Kt Ia If gives

Electrical machines 15
T m = Kt
Ra Rf (1− Kt ω
Rf )V ¿
2 ………………………………………………………… (9)
On further substitution;
T l = Kt
Ra Rf (1− Kt ω
Rf )V ¿
2 - bm ω - Jm ωs …………………………………………. (10)
Based on the above model equations, Simulink block models representing the equations was
built as shown in Fig. 4.
Fig. 4: DC shunt motor Simulink model.
Results
Applying a constant armature voltage of 50 V and step input at the field, and taking Ra = 0.61 Ω,
La = 0.013 H, Rf = 230 Ω, 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.

Electrical machines 16
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.
Theory
Cumulative compound generator.
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.

Electrical machines 17
Differentially compounded generator.
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 V l and I l gives external characteristics of the motor while that
between E and I a gives internal characteristics of the motor.
Name plate details
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
Circuit diagrams

Electrical machines 18
Fig. 5: Circuit diagram for cumulative compound generator.
Fig. 6: Circuit diagram for differential compound generator.
Procedure
1. Connections were as per the circuit diagrams shown above.
2. The machine was run at rated speed and the rated voltage was obtained by varying field
excitation.
3. The switch was then closed for the load to be connected across the generator.

Electrical machines 19
4. The load was increased step by step at an interval of 0.125 KW and all meter readings
noted.
Results
No. I l (Amps) V l (Volts) I f (Amps) Eg = V l + I a(rn +r sc)
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
Results analysis
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.

Electrical machines 20
LOAD TEST ON SINGLE PHASE INDUCTION MOTOR.
AIM: To conduct direct load test on a single phase induction motor and determine the
performance characteristics.
Introduction
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.
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.
Name plate details.

Electrical machines 21
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
Circuit diagram.

Electrical machines 22
Fig. 7: Circuit diagram for single phase induction motor.
Procedure
1. The apparatus were set up as shown in the circuit diagram above.
2. The supply was switched on at no load condition.
3. The rotor voltage was applied to the motor using the variac and readings from ammeter
and wattmeter noted.
4. The load was varied in suitable steps and all meter readings noted until full load
condition.
Results
S.No Volts Amps S1
(Kg)
S2
(Kg)
S1- S2
(Kg)
rpm T
(Nm)
Po
(W)
Pi (W) Ƞ
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

Electrical machines 23
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.

Electrical machines 24
Simulation of single phase induction motor on Simulink.
Single phase induction motor model.
Fig. 8: Equivalent circuits: (a) Main winding circuit (b) Auxiliary winding circuit
Mathematical equations for modelling of the single phase induction motor are based on the
equivalent circuits in Fig. 8 above.
Stator magnetic flux
ψs ⍺ = Ls ⍺ I s ⍺ + Lm ⍺ I r ⍺
ψsβ = Lsβ Isβ + Lmβ I rβ

Electrical machines 25
Rotor magnetic flux
ψr ⍺ = Lm ⍺ I s ⍺ + Lr ⍺ I r ⍺
ψrβ = Lmβ Isβ + Lrβ Irβ
Stator currents
I s ⍺ = 1
Ls ⍺
(ψs ⍺ - Lm ⍺ Ir ⍺)

Electrical machines 26
I sβ = 1
Lsβ
(ψsβ - Lmβ Irβ)
Rotor currents
I r ⍺ = 1
Ls ⍺
(ψs ⍺ - Lm ⍺ I r ⍺)
I rβ = 1
Lrβ
(ψrβ - Lmβ I rβ)

Electrical machines 27
Motor torque
M m = p ( N1
N2
ψrβ I r ⍺− N2
N1
ψra I rβ )
Dynamic equation
dω
dt = 1
J ( M m−M load)

Electrical machines 28
Fig. 9: Complete induction motor model.
RESULTS
Upon simulation of the developed model above, the following graphs were obtained.

Electrical machines 29
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.
SIMULATION OF SEPARATELY EXCITED DC MOTOR.
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.
The equivalent circuit is as shown in Fig. 10 below.

Electrical machines 30
Fig. 10: Equivalent representation of SEDC electromechanical components.
Mathematical modelling of SEDC
Referring to Fig. 10, for energy balance summation of the torques must be zero.
T e - T ⍺ - T ω - T emf = 0 …………………………………………………………. (1)
By considering output position of shaft and substituting in equation (1);
Kt ia if - T load - Jm ( d2 θ
dt2 ) - bm ( dθ
dt ) = 0 ………………………………………….. (2)
Kt ia if - T load - Jm ( dω
dt ) - bm ω = 0 ………………………………………………….. (3)
Solving equation (3) with Laplace transforms yields;
Kt Ia ( s)I f (s) - T load - Jm s2 θ (s ) - bm θ (s) = 0
Kt Ia ( s)I f (s) - T load = ¿ + bm ¿ ω ( s) ………………………………………….... (4)
Equation (4) can be made to relate as follows;
ω (s)
Kt Ia (s) I f (s)−T load
= [ 1
(J m s +bm ) ] ……………………………………………………. (5)
Making angular speed subject of the formula;

Electrical machines 31
ω ( s ) = [ Kt Ia (s )I f (s)−Tlo ad ]
(J m s+bm) ……………………………………………………………. (6)
Sum of field voltage is;
V f = Rf if + Lf ( dif (t )
dt ) …………………………………………………………….. (7)
Taking Laplace transforms and making if subject of the formula;
V f = Rf if (t) + Lf s if
if = V f
(R¿¿ f +Lf (s)) ¿ ……………………………………………………………………… (8)
Summation of armature voltage;
V ¿ = Ra ia (t) + La ( dia (t )
dt ) + Kb if
dθ(t)
dt
V ¿ = Ra I a (s ) + La sI(s) + Kb If sθ (s) ……………………………………………….. (9)
Separating the armature current and field current;
I a (s ) = [ 1
La s + Ra ] [ V a ( s ) −Kb I f ω(s) ]
If (s) = V i nf
(s)
Lf s + Rf
Substituting I a in equation (4) gives;
Kt If [ 1
La s + Ra ] [ V ¿ ( s ) −Kb If ω( s) ] - T load = Jm s2 r

Electrical machines 32
Transfer function with input armature voltage as input and motor angular speed as output:
ω(s)
V a ( s ) =
Kt If
Ra Rf b
( La J
Ra b ) s2 +
( La
Ra
+ J
b ) s+ ( 1+ (K b V f )2
Ra Rf
2 b )
The derived mathematical models were used to develop the SEDC Simulink model shown in Fig.
11.
Fig. 11: Block diagram of PI control in SEDC motor

Electrical machines 33
Fig. 12: SEDC subsystem expanded.
Results
Applying step input at armature and field voltages, then setting proportional gain ( K p) and
integral gain (Ki) as 9.46 and 20.94 respectively, the following graphs were obtained.