Sharjah Maritime Academy Electrical Machines Student Workbook Fall 2024/2025 PDF

Shape Description automatically generated **SHARJAH MARITIME ACADEMY** **Department of Marine Engineering Technology** **Electrical Machines** **MET 323** **Student Workbook**  **Fall 2024/2025** **CADET'S PERSON...

Shape Description automatically generated **SHARJAH MARITIME ACADEMY** **Department of Marine Engineering Technology** **Electrical Machines** **MET 323** **Student Workbook**  **Fall 2024/2025** **CADET'S PERSONAL DATA** **Student Name** **Register Number** ------------------ -------------------------- ----------------------- -- **Department** SHARJAH MARITIME ACADEMY **Student Signature** ***OBJECTIVES OF THE PRACTICAL TRAINING*** *The purpose of the practical training is to train the cades intentionally and effectively to cultivate the necessary attribute and ability to be a competent ship\'s engineer.* *The immediate objectives would be set up as follows.* 1. *To cultivate such attributes of the cadets as the adaptability, the discipline, the sense of responsibility, the determination, the endurance, the spirit of cooperation that are indispensable elements for ship\'s engineer.* 2. *To develop the practical knowledge of proficiency of cadets through practical experience which makes integration of their theoretical study, to a desired standard based on the syllabus for the coarse practical CLOs.* ***Bases of course selection:*** *This course is selected to achieve the following:* 1. *The syllabus of the practical training phase.* 2. *The course of an engineering officer in charge of a watch according to (STCW Regulations).* 3. *To match the engineering equipment on board ships.* **[Table of Content]** [Coils in the A.C. circuit 7](#coils-in-the-a.c.-circuit) [DC Generator Characteristics (no load and loaded) **Error! Bookmark not defined.**](#_Toc175573282) [Operating DC Compound Machines as a DC Motor 26](#dc-motor-characteristics) [Test Motor Insulation using Megger 13](#test-motor-insulation-using-megger) [Transformer 34](#transformer) [Three-Phase Squirrel Cage Induction Motor 40](#three-phase-squirrel-cage-induction-motor) [Three-Phase slip ring (wound rotor) Induction Motor 46](#three-phase-slip-ring-wound-rotor-induction-motor) [Operating Synchronous Machines as a synchronous Motor 52](#operating-synchronous-machines-as-a-synchronous-motor) ------------------ ------------------------------- -------------- ---------------- **Department** Marine Engineering Technology **Semester** Fall 2024/2025 **Course Title** Electrical Machines **Code** MET 323 ------------------ ------------------------------- -------------- ---------------- **[PRACTICAL GUIDE]** **Experiment** **Marks** **CLO's** ------------------- ----------------------------------------------------------- ------------ ------ **Experiment** **Actual** **1** **Coils in the A.C. circuit** CLO3 **2** **DC Generator Characteristics** CLO3 **3** **DC Motor Characteristics** CLO3 **4** **Test Motor Insulation using Megger** CLO3 **5** **Transformer** CLO3 **6** **Three-Phase Squirrel Cage Induction Motor** CLO3 **7** **Three-Phase slip ring Induction Motor** CLO3 **8** **Operating Synchronous Machines as a synchronous Motor** CLO3 **Total Marks** **Lecturer Name** Eng: Ibtihal Ahmed **Experiment** **Experiment** **Mark** **CLO's** ------------------- ----------------------------------------------------------- ---------- ----------- **1** **Coils in the A.C. circuit** 3 CLO3 **2** **DC Generator Characteristics** 4 CLO3 **3** **DC Motor Characteristics** 3 CLO3 **4** **Test Motor Insulation using Megger** 2 CLO3 **5** **Transformer** 3 CLO3 **6** **Three-Phase Squirrel Cage Induction Motor** 3 CLO3 **7** **Three-Phase slip ring Induction Motor** 4 CLO3 **8** **Operating Synchronous Machines as a synchronous Motor** 4 CLO3 **Total Marks** **26** **Lecturer Name** Eng. Ibtihal Ahmed **Signature**  **SHARJAH MARITIME ACADEMY** **[Laboratory Experiment]** ------------------------------ ------------------- ------------------------------- ---------------- **SHARJAH MARITIME ACADEMY** Fall 2024/2025 **Semester** Marine Engineering Technology **Department** Electrical Machine/ MET323 **Course/Code** Eng. Ibtihal Ahmed **Lecturer** 1 hour 50 mins **Duration** 12:30 -- 14:20 **Time** 3 **Grade** 04/09/2024 **Date** 5 **No. of papers** 11-11 **Room No.** ------------------------------ ------------------- ------------------------------- ---------------- **[LAB assessment Experiment 1]** -------------- ------------------------------ ------------ ---------- **Question** **Marks** **CLOs** **Available** **Actual** 1 **1.5** **CLO3** 2 **1.5** **CLO3** **Total** **3** **Lecturer** **Name:** Eng. Ibtihal Ahmed **Sign:** **Date:** -------------- ------------------------------ ------------ ---------- **[Experiment 1]** **[Coils in the A.C. circuit]** =========================================== **[Objectives]** - Study the behavior and characteristics of inductive coils when subjected to alternating current. **[Instruments & Equipment:]** 1. Electrical circuit board 2. Coils 3. Oscilloscope **[Theory]** A current flowing through conductive material (for example, copper wire) creates a magnetic field whose lines of force you can imagine as arranged in concentric circles around the center of the wire. Coils react to changes of the current by increasing or decreasing their magnetic fields. The reaction is always in the opposite direction to the cause: - If the current grows, the coil develops a mutual induction UL in the opposite direction of the external voltage. - If the current in the circuit decreases, mutual induction in the coil creates a voltage intending to maintain the current flow. The ability of a coil to react to current changes depends on their inductance L. The higher inductance L, the greater the influence of the coil by mutual induction in the current circuit. Inductance L results from the coil properties: winding arrangement, number of windings, diameter (or cross-sectional area), length and material of the wire. A decisive factor is whether the coil is filled with a core whose material supports the magnetic flux. \ [\$\$\\mathbf{L =}\\frac{\\mathbf{(N}\^{\\mathbf{2}}\\mathbf{\\mu}\_{\\mathbf{0}}\\mathbf{\\mu}\_{\\mathbf{r}}\\mathbf{A)}}{\\mathbf{l}}\$\$]{.math.display}\ [*L*]{.math.inline}: Inductance; unit: Henry \[H\] [*N*]{.math.inline}: Number of windings \[no unit\] [*A*]{.math.inline}: Cross-sectional area \[m2\] [*l*]{.math.inline}: Length \[m\] [*μ*~0~]{.math.inline}: Magnetic field constant 1.257×10^-6^ \[Vs/Am\] [*μ*~*r*~]{.math.inline}: Permeability number \[no unit\] - Without a core, area and length information refer to the coil itself. In addition, permeability number [*μ*~*r*~]{.math.inline} can be omitted in the formula. - With core, you need to enter cross section and length of the core as well as permeability number [*μ*~*r*~]{.math.inline} for the respective material into the equation. The magnitude of inductive reactance XL is a function of frequency f and inductance L. The following applies: \ [**X**L **=** **ωL** **=** **2**πfL]{.math.display}\ **[Procedure]** 1. Connect the circuit as shown:  2. Record the voltages on R and L using the oscilloscope. (1 Mark) **Coil 1:** L = 100mH (device in plastic housing) **Coil 2:** Transformer coil N = 900 and upper half of the iron core inserted. - To measure the U~L~ value, connect the probe to the L and ground then interchange L and R places to measure U~R.~ - For coil N=900, first measure all voltage values for inductance then measure U~R~ and make sure to reduce the amplitude to zero before removing L. **Q1** --------- **1.5** +--------+--------+--------+--------+--------+--------+--------+--------+ | [**f** | **1** | **2** | **3** | **4** | **6** | **8** | | | ]{.mat | | | | | | | | | h | | | | | | | | |.inlin | | | | | | | | | e}**\[ | | | | | | | | | KHz\]* | | | | | | | | | * | | | | | | | | +========+========+========+========+========+========+========+========+ | **U~L~ | **N=90 | 1.88 | 3.6 | 5.12 | 6.36 | 8.16 | 9.36 | | ** | 0** | | | | | | | | | | | | | | | | | **~\[V | | | | | | | | | PP\]~* | | | | | | | | | * | | | | | | | | +--------+--------+--------+--------+--------+--------+--------+--------+ | | **100m | 6.4 | 9.2 | 10.4 | 11.2 | 11.7 | 11.9 | | | H** | | | | | | | +--------+--------+--------+--------+--------+--------+--------+--------+ | **U~R~ | **N=90 | 11.7 | 11.2 | 10.5 | 9.84 | 8.32 | 7.12 | | ** | 0** | | | | | | | | | | | | | | | | | **~\[V | | | | | | | | | PP\]~* | | | | | | | | | * | | | | | | | | +--------+--------+--------+--------+--------+--------+--------+--------+ | | **100m | 9.28 | 7.12 | 5.52 | 4.4 | 3.12 | 2.48 | | | H** | | | | | | | +--------+--------+--------+--------+--------+--------+--------+--------+ | **I~L~ | **N=90 | 11.7 | 11.2 | 10.5 | 9.84 | 8.32 | 7.12 | | =U~R~/ | 0** | | | | | | | | R** | | | | | | | | | | | | | | | | | | **~\[m | | | | | | | | | APP\]~ | | | | | | | | | ** | | | | | | | | +--------+--------+--------+--------+--------+--------+--------+--------+ | | **100m | 9.28 | 7.12 | 5.52 | 4.4 | 3.12 | 2.48 | | | H** | | | | | | | +--------+--------+--------+--------+--------+--------+--------+--------+ | **X~L~ | **N=90 | 0.16 | 0.32 | 0.49 | 0.65 | 0.98 | 1.31 | | ** | 0** | | | | | | | | | | | | | | | | | **\[kΩ | | | | | | | | | \]** | | | | | | | | +--------+--------+--------+--------+--------+--------+--------+--------+ | | **100m | 0.68 | 1.29 | 1.88 | 2.45 | 3.75 | 4.79 | | | H** | | | | | | | +--------+--------+--------+--------+--------+--------+--------+--------+ 3. Plot the arithmetic values for reactance X~L~ for Coil 1 & 2. (0.5 Mark) Comment on the graph (0.25 Mark) **Q2** --------- **1.5** XL increases as the frequency increases due to direct relation Xl=2πfL. XL for 100mH is higher than N=900mH. 4. Arithmetically verify value XL at f (6kHz) established by measurement for coil L = 100mH and compare to the value form measurement. (0.25 Mark) \ [*X*~*l*~ = 2πfL = 2*π* × 6*k* × 100*m* = 3.76Ko*hm*]{.math.display}\ The arithmetic value is close to value from measurement which is 3.75 Kohm 5. Verify the nominal value of coil L = 100mH arithmetically. To do so, use your readings at 4kHz and compare to the actual value. (0.25 Mark) \ [\$\$L = \\frac{X\_{l}}{2\\pi f} = \\frac{2.45k}{2\\pi \\times 4k} = 0.097\\ H = 97mH\$\$]{.math.display}\ The arithmetic value is close to actual value 100mH 6. Calculate the unknown inductance L of the transformer coil (N = 900) used in the measurement from the reading XL = f(3kHz). (0.25 Mark) \ [\$\$L = \\frac{X\_{l}}{2\\pi f} = \\frac{0.49k}{2\\pi \\times 3k} = 0.026 = 26\\ mH\$\$]{.math.display}\ 7. What is inductance L of the transformer coil (N = 900) without core? Establish the value by measurement. Use measuring frequency f = 8kHz. (0.25 Mark) At 8kHz without core: U~L~= 4.08 V, U~R~= 11.2 V, I~R~= 11.2 mA \ [\$\$X\_{l} = \\frac{U\_{L}}{I\_{L}} = \\frac{4.08}{11.2m} = 0.43\\ \\text{ko}hm\$\$]{.math.display}\ \ [\$\$L = \\frac{X\_{l}}{2\\pi f} = \\frac{0.43k}{2\\pi \\times 8k} = 0.0085 = 8.5\\ mH\$\$]{.math.display}\ 8. Apply again for N=300, What is inductance L of the transformer coil (N = 300) without core? Establish the value by measurement. Use measuring frequency f = 8kHz. (0.25 Mark) At 8kHz without core: U~L~= 0.484 V, U~R~= 12 V, I~R~= 12 mA \ [\$\$X\_{l} = \\frac{U\_{L}}{I\_{L}} = \\frac{0.48}{11.2m} = 0.04\\ \\text{ko}hm\$\$]{.math.display}\ \ [\$\$L = \\frac{X\_{l}}{2\\pi f} = \\frac{0.040k}{2\\pi \\times 8k} = 0.79\\ mH\$\$]{.math.display}\ 9. How does the core affect the inductance value? And from experiment what does another parameter affect the inductance value (0.25 Mark) The core material is often a ferromagnetic material, adding core to inductance turn increase its inductance. The core enhances the magnetic flux generated by the current flowing through the coil, leading to a higher inductance value. Another parameter affecting the inductance value is the number of turns, as the number of turns increases the inductance increases.  **SHARJAH MARITIME ACADEMY** **[Laboratory Experiment]** ------------------------------ ------------------- ------------------------------- ---------------- **SHARJAH MARITIME ACADEMY** Fall 2024/2025 **Semester** Marine Engineering Technology **Department** Electrical Machines/ MET323 **Course/Code** Eng. Ibtihal Ahmed **Lecturer** 1 hour 50 mins **Duration** **Time** 2 **Grade** **Day & Date** 6 **No. of papers** 11-03 **Room No.** ------------------------------ ------------------- ------------------------------- ---------------- **[LAB assessment experiment 2]** -------------- ------------------------------ ------------ ---------- **Question** **Marks** **CLOs** **Available** **Actual** 1 **2** **CLO3** **Total** **2** **Lecturer** **Name:** Eng. Ibtihal Ahmed **Sign:** **Date:** -------------- ------------------------------ ------------ ---------- **[Experiment 2]** **[Test Motor Insulation using Megger]** ==================================================== **[Objectives]** - The objective of a Megger test experiment is to evaluate the insulation resistance of electrical equipment, cables, or other components, ensuring their safety and reliability in operation. **[Instruments & Equipment:]** 1. Motor 2. Megger **[Theory:]** The theory behind the Megger test for a motor is centered on measuring the insulation resistance of the motor\'s windings to ensure they are properly insulated from the motor\'s casing and each other. Proper insulation is crucial to prevent electrical leakage, short circuits, and potential motor failures. The Megger test, also known as an insulation resistance test, uses a high DC voltage to measure the resistance offered by the insulation. Over time, insulation can degrade due to factors like heat, moisture, mechanical stress, and contamination. The Megger test measures the resistance of this insulation. The working principle of a Megger, or insulation resistance tester, is based on the application of a high direct current (DC) voltage to an electrical component and the measurement of the resulting current that flows due to the resistance of the insulation. This resistance is then used to evaluate the condition of the insulation. Usually, there are two types of meggers: Digital megger and Electrical Megger meter. In digital meter, we use a battery to provide the supply, whereas in Electrical megger is provided with a hand-driven DC generator as source. Insulation resistance readings should be considered relative. They can be quite different for one motor or machine tested three days in a row, yet not mean bad insulation. What really matters is the trend in readings over a time period, showing lessening resistance and warning of coming problems. periodic testing is, therefore, your best approach to preventive maintenance of electrical equipment, using record cards. Whether you test monthly, twice a year, or once a year depends upon the type, location, and importance of the equipment. **Types of tests using Megger test:** 1. **Continuity testing:** Tests the resistance between two points. If there is low resistance, the two points are electrically connected. If there is higher resistance, the circuit is open. 2. **Winding to winding insulation resistance test** To check the insulation resistance between different phases of the motor windings to ensure that they are insulated from each other. The resistance should be high and roughly equal across all phase pairs. Significant differences could indicate insulation breakdown between phases. 3. **Winding to body insulation resistance test** Measures the resistance between the motor windings and the motor frame (ground) to ensure that the windings are properly insulated from the ground. A high resistance value indicates good insulation. Low resistance suggests potential grounding issues or insulation degradation. **[Procedure: ]** **Continuity test:** 1. In this test either Multimeter or Megger can be used. 2. Using Megger tester, set the position on continuity. Connect the two terminals of Megger to the motor terminal and record the resistance values: ---------------------------------------------------- **Continuity test** **Resistance** ----------------------------------- ---------------- \ 41.5 [*U*~1~\_*U*~2~]{.math.display}\ \ 41.5 [*V*~1~\_*V*~2~]{.math.display}\ \ 41.5 [*W*~1~\_*W*~2~]{.math.display}\ ---------------------------------------------------- 3. Draw the winding connection to Megger tester  4. Comment on the output: The resistance between two points for each coil is equal and the resistance is low means the coil points are electrically connected. Resistance value is same for the coils. **Winding to winding IR test:** 5. First select the voltage of Megger that will be applied to the motor. As the motor operation voltage is 400, select the voltage of 500 V. 6. Connect the Megger leads between two phase terminals (U-V, V-W, W-U) 7. Apply the voltage and measure the resistance by pressing the button and record the IR value. ---------------------------------------------------- **Winding to winding IR test** **Resistance** ----------------------------------- ---------------- \ 22 M [*U*~1~\_*V*~1~]{.math.display}\ \ 24 M [*U*~2~\_*V*~2~]{.math.display}\ \ 10 M [*V*~1~\_*W*~1~]{.math.display}\ \ 10.5 M [*V*~2~\_*W*~2~]{.math.display}\ \ 13.5 M [*W*~1~\_*U*~1~]{.math.display}\ \ 14 M [*W*~2~\_*U*~2~]{.math.display}\ ---------------------------------------------------- 8. Draw the winding connection to Megger tester 9. Comment on the output. The insulation resistance between the winding \> 1 M ohm, which means that the windings of phases are insulated from each other 10. What does it indicate if the resistance value is low, less than 1M? Getting low resistance between phases indicates poor insulation in the motor. This can suggest that the insulation material has degraded. Low insulation resistance values are a sign that the equipment may be unsafe for operation and could fail under normal working conditions, requiring maintenance or replacement. **Winding to body IR test:** 11. Connect the Megger leads between motor body and single phase (U, V, W). 12. Apply the voltage and measure the resistance by pressing the button and record the IR value. **Winding to body IR test** **Resistance** ---------------------------------- ---------------- [*U*~1~]{.math.inline}to ground 2.54 M [*U*~2~]{.math.inline}to ground 2.54 M [*V*~1~]{.math.inline}to ground 2.56 M [*V*~2~]{.math.inline}to ground 2.54 M [*W*~1~]{.math.inline}to ground 2.54 M [*W*~2~]{.math.inline}to ground 2.56 M 13. Comment on the output. The insulation resistance between the winding and ground is \> 1 M ohm, which means good insulation between the winding and the ground. This suggests that the insulation is in good condition and is effectively preventing current from leaking to the ground. 14. What does it indicate if the resistance value is low, less than 1M? Getting low resistance between winding and ground indicates poor insulation in the motor. This can suggest that the insulation material has degraded. Low insulation resistance values are a sign that the equipment may be unsafe for operation and could fail under normal working conditions, requiring maintenance or replacement. 15. What is the difference between a Megger test and a standard multimeter insulation resistance test? A Megger test applies a high DC voltage typically ranging from 250V to 1000V to measure insulation resistance to be used for equipment such as motors, transformers, A standard multimeter typically uses a much lower voltage often around 3V to 12V when measuring resistance to be used in low voltage circuit measurement.  **SHARJAH MARITIME ACADEMY** **[Laboratory Experiment]** ------------------------------ ------------------- ------------------------------- ---------------- **SHARJAH MARITIME ACADEMY** Fall 2024/2025 **Semester** Marine Engineering Technology **Department** Electrical Machine/ MET323 **Course/Code** Eng. Ibtihal Ahmed **Lecturer** 1 hour 50 mins **Duration** **Time** 4 **Grade** **Date** 7 **No. of papers** 11-03 **Room No.** ------------------------------ ------------------- ------------------------------- ---------------- **[LAB assessment Experiment 3]** -------------- ------------------------------ ------------ ---------- **Question** **Marks** **CLOs** **Available** **Actual** 1 **0.5** **CLO3** 2 **2.5** **CLO3** 3 **1** **CLO3** **Total** **4** **Lecturer** **Name:** Eng. Ibtihal Ahmed **Sign:** **Date:** -------------- ------------------------------ ------------ ---------- **[Experiment 3]** **[DC Generator Characteristics]** ============================================== **[Objectives]** - To understand connections of the DC Generator. - To determine V-I characteristics of the DC Generator (at load and no-load). - To understand the effect of load on characteristics of generator in case of loading. **[Instruments & Equipment:]** 1. DC Generator. 2. Dynamometer. 3. DC Supply. 4. Switching, protection and measurement units (modules). 5. Computer + Cassy software. 6. Resistive Loads. ### [Theory:] ### A DC machine is an electro-mechanical energy conversion device. When it converts mechanical power ([**ω**]{.math.inline}T) into DC electrical power (EI), it is known as a DC generator. {#a-dc-machine-is-an-electro-mechanical-energy-conversion-device.-when-it-converts-mechanical-power-mathbfomegat-into-dc-electrical-power-ei-it-is-known-as-a-dc-generator.} ### Generators produce electrical power based on the principle of Faraday's law of electromagnetic induction. This law states that when a conductor moves in a magnetic field it cuts magnetic lines of force, which induces an electromagnetic force (EMF) in the conductor. The magnitude of this induced emf depends upon the rate of change of flux linkage with the conductor. This emf will cause a current to flow if the conductor circuit is closed. ### Diagram of a circular structure with arrows and lines Description automatically generated ### The magnetic characteristic, also known as the no-load saturation characteristic, depicts the connection between the generated emf when the generator is operating with no load and the field current at a constant speed. This curve essentially resembles the magnetization curve and is the same across various generator types. The field current is gradually increased, and the corresponding terminal voltage is recorded. ### ###  ### ### According to the emf equation, the generated emf should be directly proportional to the field flux. Nonetheless, even with no field current applied, some emf is generated. This initial induced emf is a result of residual magnetism remaining in the field poles. Because of this residual magnetism, a small initial emf is induced in the armature. ### ### ### ### ### ### ### The equations that govern the operation of a generator under steady state: ### \ [**V**~**t**~**=E**~**a**~**−I**~**a**~**R**~**a**~]{.math.display}\ ### ### At no-load, the voltage across the terminals is maximum and is equal to generated emf. As the load increases gradually, the load current IL increases but the terminal voltage decreases. The decrease in voltage is because of the following reasons 1. ### The increase in [**I**~**a**~**R**~**a**~]{.math.inline} drop {#the-increase-in-mathbfi_mathbfamathbfr_mathbfa-drop} 2. ### The demagnetization effect of the armature reaction 3. ### The decrease in the field current due to the drop in the induced emf ### ### ### ### ### ### ### ### ### ### ### ### ### ### ### ### ### ### ### ### [Procedure:] 1. Draw the block diagram connection of DC generator. What type of DC generator connected according to the excitation method? **Q1** ---------- **0.50**  **Q2** --------- **2.5** 2. At Speed **n**=2000 RPM At Speed **n**=1800 --------------------------------- --------------------------- --------------------------------- --------------------------- **Excitation Current I~E~ (A)** **Output voltage Vo (V)** **Excitation Current I~E~ (A)** **Output voltage Vo (V)** **0** 7.1 **0** 5.7 **0.05** 33.6 **0.05** 37.8 **0.1** 68.6 **0.1** 74.6 **0.15** 102.3 **0.15** 112 **0.2** 132.6 **0.2** 147.2 3. Draw V-I curve based on readings in table 1. (**Show axis title, units, legend, and chart title**). Comment on the obtained V-I characteristics. (0.5 Mark) 4. 5. Set the speed to 2000 RPM, excitation current value to 0.1A, adjust the resistive load to the values shown in the table and record the reading. (1 Mark) At Speed **n**=2000 RPM, **I~E~** =0.1A and with resistive load ----------------------------------------------------------------- ------ ------ ------ 79.5 78.5 76.9 6.1 7.7 10 13.9 17.6 23.4 ### ### ### ### ### \ [*V* = *E* − *I* ⋅ *R*~*a*~]{.math.display}\ 6. Using the rated data for DC generator, find the generator efficiency and compare it to values from experiment. ----------------------------------------- **Nominal Data** **Value** ----------------------------- ----------- \ 200V [*V*~*E*~]{.math.display}\ \ 0.3A [*I*~*E*~]{.math.display}\ \ 2.7 [*T*]{.math.display}\ \ 220 [*V*~*o*~]{.math.display}\ \ 3 [*I*~*o*~]{.math.display}\ \ 2300 [*N*]{.math.display}\ ----------------------------------------- **Q3** -------- **1** \ [\$\$efficiency\\ = \\frac{P\_{\\text{out}}}{P\_{\\text{in}}}\$\$]{.math.display}\ \ [*P*~*o*~ = 660*W*  → idle *V* = 220  *I* = 3]{.math.display}\ \ [\$\$P\_{m} = T \\times \\omega = T \\times \\frac{2\\text{πN}}{60} = 2.7 \\times \\left( \\frac{2\\pi \\times 2300}{60} \\right) = 650W\$\$]{.math.display}\ \ [*P*~*e*~ = *V*~*e*~*I*~*e*~ = 200 × 0.3 = 60*W*]{.math.display}\ \ [\$\$\\mathbf{\\eta}\_{\\mathbf{\\text{ideal}}}\\mathbf{=}\\frac{\\mathbf{P}\_{\\mathbf{o}}}{\\mathbf{P}\_{\\mathbf{\\text{in}}}}\\mathbf{=}\\frac{\\mathbf{660}}{\\mathbf{650}\\mathbf{.}\\mathbf{3}\\mathbf{+}\\mathbf{60}}\\mathbf{=}\\mathbf{92}\\mathbf{.}\\mathbf{9}\\mathbf{\\%}\$\$]{.math.display}\ Shape Description automatically generated **SHARJAH MARITIME ACADEMY** **[Laboratory Experiment]** ------------------------------ ------------------- ------------------------------- ---------------- **SHARJAH MARITIME ACADEMY** Fall 2024/2025 **Semester** Marine Engineering Technology **Department** Electrical Machine/ MET323 **Course/Code** Eng. Ibtihal Ahmed **Lecturer** 1 hour 50 mins **Duration** 12:30 -- 14:20 **Time** 4 **Grade** 09/10/2024 **Date** 8 **No. of papers** 11-03 **Room No.** ------------------------------ ------------------- ------------------------------- ---------------- **[LAB assessment Experiment 4]** -------------- ------------------------------ ------------ ---------- **Question** **Marks** **CLOs** **Available** **Actual** 1 **1** **CLO3** 2 **1** **CLO3** 3 **1** **CLO3** **Total** **3** **Lecturer** **Name:** Eng. Ibtihal Ahmed **Sign:** **Date:** -------------- ------------------------------ ------------ ---------- **[Experiment 4]** **[DC Motor Characteristics]** ========================================== **[Objectives]** - To understand and demonstrate the operation principles of DC motor. - To determine characteristics of as DC Motor **[Instruments & Equipment:]** 1. DC Compound Machines. 2. Dynamometer. 3. DC Power supply. 4. Switching, protection and measurement units (modules). 5. Computer + Cassy software. **[Theory:]** A DC motor is defined as a class of electrical motors that convert direct current electrical energy into mechanical energy. As a typical DC machine, the DC motor consists of three main parts namely -- magnetic field system, armature, commutator, and brush-gear. When the field coil of the DC motor is powered, a magnetic field is generated in the air gap. This magnetic field aligns with the radii of the armature. It enters the armature from the North pole side of the field coil and exits from the South pole side of the field coil. The conductors situated on the opposite pole experience a force of equal magnitude but in the opposite direction. The interaction of these two opposing forces results in a torque that induces the rotation of the motor armature. ### ### ###  ### ###  ### The characteristic of torque and armature current is a straight line from the origin. The shaft torque is always less than the gross torque. This is because of stray losses. ### ### **T-I~a~ curve** As the armature rotates, each coil on the armature experiences a change in the flux passing through its plane. Therefore, an electromotive force (emf) is induced in each coil. In accordance with Faraday's law of induction, the induced emf must oppose the current entering the armature. In other words, the induced emf opposes the applied voltage. For this reason, we commonly refer to the induced emf in a motor as the back emf or counter emf of the motor. The back emf of the motor is represented by: The back emf and flux remain consistent during regular operations, resulting in a constant motor speed relative to the armature current. But in practice due to the armature reaction effect, the distribution of air-gap flux gets distorted. Thus, reducing the resultant flux. Suppose if the load on the motor increases, the armature current Ia increases, thereby increasing the drop Ia Ra. This causes a small drop in speed because at constant Φ there will be very little change in the difference (V - Ia Ra) since the armature resistance Ra of a dc motor is kept very small. ###  **N-I~a~ curve** ### The torque speed characteristics are similar to speed and armature current characteristics, it can be seen that the speed decreases as the load torque increases. ### ### A graph of a speed-torque Description automatically generated **N-T curve** ### [Procedure:] 1. Draw the connection diagram of DC motor.  **Q1** -------- **1** 2. --------------------------------- --------- --------- --------- --------- --------- ------------ No load 0.1 N.m 0.3 N.m 0.5 N.m 0.7 N.m 11\. 1 N.m 1706 1685 1642 1610 1581 1539 200 200 200 200 200 200 0.466 0.54 0.7 0.86 1.05 1.29 0.3 0.3 0.3 0.3 0.3 0.3 \ 66.1 65.9 65.6 65.6 65.5 65.5 [**P**~**E**~]{.math.display}\ \ 94 109.2 1422 173 210 260 [**P**~elec~]{.math.display}\ \ 3.7 17.5 51.5 84.7 116.7 161.6 [**P**~mech~]{.math.display}\ 2 10 25 35 42 50 --------------------------------- --------- --------- --------- --------- --------- ------------ 3. Using below equations and readings from Table 1 verify the power result: **Q2** -------- **1** \ [**P**~elec~**=U**~**M**~**I**~**M**~ ]{.math.display}\ \ [\$\$\\mathbf{P}\_{\\mathbf{\\text{mech}}}\\mathbf{=}\\frac{\\mathbf{2}\\mathbf{\\text{πTN}}}{\\mathbf{60}}\$\$]{.math.display}\ ----------------------------------------------------------------- 0.5 N.m 0.7 N.m 11\. 1 N.m -------------------------------- --------- --------- ------------ 60 60 60 \ 172 210 258 [**P**~elec~]{.math.display}\ \ 84.3 115.9 161.1 [**P**~mech~]{.math.display}\ 36.3 42.9 50.6 ----------------------------------------------------------------- 4. Using the nominal values for the machine find the theoretical efficiency of the motor ----------------------------------------- **Nominal Data** **Value** ----------------------------- ----------- \ 200V [*V*~*E*~]{.math.display}\ \ 0.24A [*I*~*E*~]{.math.display}\ \ 3.5 [*T*]{.math.display}\ \ 220 [*V*~*M*~]{.math.display}\ \ 4.2 [*I*~*M*~]{.math.display}\ \ 2040 [*N*]{.math.display}\ ----------------------------------------- \ [*P*~in~ = *P*~ele~ + *P*~*e*~]{.math.display}\ \ [\$\$P\_{o} = P\_{m} = T \\times \\omega = T \\times \\frac{2\\text{πN}}{60} = 3.5 \\times \\left( \\frac{2\\pi \\times 2040}{60} \\right) = 747.7W\$\$]{.math.display}\ \ [*P*~*e*~ = *V*~*e*~*I*~*e*~ = 200 × 0.24 = 48*W*]{.math.display}\ \ [*P*~ele~ = *V*~*M*~*I*~*M*~ = 220 × 4.2 = 924*W*]{.math.display}\ \ [\$\$\\mathbf{\\eta}\_{\\mathbf{\\text{ideal}}}\\mathbf{=}\\frac{\\mathbf{P}\_{\\mathbf{o}}}{\\mathbf{P}\_{\\mathbf{\\text{in}}}}\\mathbf{=}\\frac{\\mathbf{747}\\mathbf{.}\\mathbf{7}}{\\mathbf{924}\\mathbf{+}\\mathbf{48}}\\mathbf{=}\\mathbf{76}\\mathbf{.}\\mathbf{9}\\mathbf{\\%}\$\$]{.math.display}\ 5. Draw curves based on readings in table 1. (**Show axis title, units, legend, and chart title**) - Speed with Armature current curve. - Torque with Armature current curve. - Speed with torque curve. **Q3** -------- **1**  Comment on the obtained charateristics: N -- I characteristics: As the load increases, the armature current increases, and this causes the speed to decrease. T- I characteristics: As the load increases, the armature current increases, leading to a proportional increase in torque. N- T charateristics: The torque speed characteristics are similar to speed and armature current characteristics, it can be seen that the speed decreases as the load torque increases.  **SHARJAH MARITIME ACADEMY** **[Laboratory Experiment]** ------------------------------ ------------------- ------------------------------- ---------------- **SHARJAH MARITIME ACADEMY** Fall 2024/2025 **Semester** Marine Engineering Technology **Department** Electrical Machine/ MET323 **Course/Code** Eng. Ibtihal Ahmed **Lecturer** 1 hour 50 mins **Duration** **Time** 3 **Grade** **Date** 6 **No. of papers** 11-11 **Room No.** ------------------------------ ------------------- ------------------------------- ---------------- **[Lab assessment experiment 5]** -------------- ------------------------------ ------------ ---------- **Question** **Marks** **CLOs** **Available** **Actual** 1 **1** **CLO3** 2 **1** **CLO3** 3 **1** **CLO3** **Total** **3** **Lecturer** **Name:** Eng. Ibtihal Ahmed **Sign:** **Date:** -------------- ------------------------------ ------------ ---------- **[Experiment 5]** **[Transformer]** ============================= **[Objectives]** - Understand the basic operation of a transformer, including the step-up and step-down of voltages based on the turn's ratio between the primary and secondary windings. **[Instruments & Equipment:]** 1. Circuit board 2. transformer 3. Resistor **[Theory:]** Transformers consist of two or more magnetically coupled windings (coils). The primary winding (primary coil) draws electric energy from an A.C. voltage generator. The energy is converted into a changing magnetic field and directed and/or transferred to the output side of the transformer. The secondary coil (secondary winding) reconverts the magnetic flux into electric energy and makes it available to a consumer. The purpose of transformers is to adjust the electric conditions of an A.C. voltage source (primary side) to the requirements of a consumer (secondary side). A diagram of a transformer core Description automatically generated  **Transformer** Ideal transformers are often used to describe the properties of a transformer: - Ideal transformers experience no loss. They have an efficiency [*η*]{.math.inline} = 1. - In off-load mode (= no load resistance), the ideal transformer draws no active energy. - The entire magnetic flux of the primary coil also pushes through to the secondary coil; outside of the core there are no components of the magnetic field (no lines of force in the space around the coils); thus, magnetic coupling between the windings is 100%. - The waveforms of input and output A.C. voltage is identical. In other words: The signal shape on the transformer output shows no distortions. The ratio between the number of the windings (N) of the primary side and those of the secondary side of the transformer is called transmission ratio a. \ [\$\$a = \\frac{N\_{1}}{N\_{2}} = \\frac{U\_{1}}{U\_{2}} = \\frac{I\_{2}}{I\_{1}}\$\$]{.math.display}\ **[Procedure:]** 1. Connect the circuit as shown in the below Figure: First, choose the secondary coil with [*N*~2~ = 900]{.math.inline} windings. Set a sinewave A.C. voltage [*U*~rms~= 3*V*]{.math.inline}, [*f* = 50]{.math.inline}Hz, on the 3-phase source for [*U*~1~]{.math.inline}. 2. Measure the secondary voltage for [*N*~2~ = 900]{.math.inline} without load. Then replace the secondary coil by [*N*~2~ = 300]{.math.inline} and repeat the voltage measurement. (0.5 Mark) 3. For current measurement, keep the two multimeters in mA. **Q1** -------- **1** --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- \ \ [*U*~1~]{.math.inline}\[V\] [*U*~2~]{.math.inline}\[V\] a (from Voltage) [*I*~1~]{.math.inline}\[mA\] [*I*~2~]{.math.inline}\[mA\] a (from current) [*N*~1~]{.math.display}\ [*N*~2~]{.math.display}\ --------------------------- --------------------------- ------------------------------ ------------------------------ ------------------ ------------------------------- ------------------------------- ------------------ 900 900 3 2.91 1.03 60.5 59.2 0.97 300 0.947 3.16 40.2 117.2 2.91 --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- 4. What is transmission ratio [*a*]{.math.inline} if you base your calculation on the winding numbers of the coils? (0.25 Mark) \ [\$\$a = \\frac{N\_{1}}{N\_{2}} = \\frac{900}{900} = 1\\ \$\$]{.math.display}\ \ [\$\$a = \\frac{N\_{1}}{N\_{2}} = \\frac{900}{300} = 3\\ \$\$]{.math.display}\ 5. What type of transformer does this represent for each case? (0.25 Mark) 1. Isolating transformer , N1=N2 2. Step down, N2\N1 9. What setting of input voltage U1 is required so that you can measure a voltage in the off-load mode of [*U*~2~= 6.5*V*]{.math.inline}rms, [*f* = 100]{.math.inline}Hz, on the output of the transformer? Set the input to the value you're getting and record the output. (0.25 Mark)  \ [\$\$a = \\frac{N\_{1}}{N\_{2}} = \\frac{U\_{1}}{U\_{2}}\$\$]{.math.display}\ \ [\$\$\\frac{900}{300} = \\frac{U\_{1}}{6.5}\$\$]{.math.display}\ \ [\$\$U\_{1} = \\frac{300\*6.5}{900} = 2.167\\ V\$\$]{.math.display}\ Output form the circuit \ [*U*2 = 6.04*V*]{.math.display}\ 10. Connect the circuit as shown in the below Figure: First, choose the secondary coil with [*N*~2~ = 900]{.math.inline} windings. Set a sinewave A.C. voltage [*U*~rms~= 3*V*]{.math.inline}, [*f* = 50]{.math.inline}Hz, on the 3-phase source for [*U*~1~]{.math.inline}. A diagram of a circuit Description automatically generated **Q3** --------- **0.5** --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- \ \ a \ [*U*~1~]{.math.inline}\[V\] [*U*~2~]{.math.inline}\[V\] [*I*~1~]{.math.inline}\[mA\] [*I*~2~]{.math.inline}\[mA\] \ \ [*N*~1~]{.math.display}\ [*N*~2~]{.math.display}\ [*R*~*L*~]{.math.display}\ [*R*~1~ = *U*~1~*s*]{.math.inline} increases. Since the rotor current is almost constant, the torque developed by the motor increases with the increase in the effective resistance [*R*/*s*]{.math.inline}. Thus, the torque developed by the motor keeps increasing with the decrease in the slip. When the slip falls below a certain value called the **breakdown slip**, the hypothetical resistance becomes the dominating factor. The torque developed by the motor is now proportional to the slip [*s*]{.math.inline}. As the slip decreases, so does the torque developed. At no load, the slip is almost zero, the hypothetical rotor resistance is nearly infinite, the rotor current is approximately zero, and the torque developed is virtually zero. A diagram of a motor Description automatically generated  A diagram of a phase induction motor Description automatically generated ### ### ### [Procedure:] 1. Draw the connection diagram of 3Ø squirrel cage induction motor. **Q1** --------- **0.5** ### ### ### ### ### ### ### ### ### ### 2. Measure the required readings and record in the table. (1 Mark) **Q2** --------- **1.5** ### ----------------------------------------------------------------------------------------- **Loads** **7** -------------------------- ------- ---------- ------- --------- ------- --------- ------- **0** **0.22** **1** **1.5** **2** **2.5** **3** \ [**⌀**]{.math.display}\ ----------------------------------------------------------------------------------------- ### Table 1 ### 3. Use readings in table 1 and equation to verify the required values in Table2. (0.5 Mark) **2** **2.5** **3** -- ------- --------- ------- ### ### ### ### ### ### ### ### ### ### ### ### ### Table 2 ### ### 4. Draw curves based on readings. (**Show axis title, units, legend, and chart title**) - Torque T- Speed N curve based on readings in question 1 with efficiency values shown in each point. - Torque T-Slip S curve based on readings in question 1 with current readings shown in each point. **Q3** -------- **1** ### ### ###  Comment: ### ### Shape Description automatically generated **SHARJAH MARITIME ACADEMY** **[Laboratory Experiment]** ------------------------------ ------------------- ------------------------------- ---------------- **SHARJAH MARITIME ACADEMY** Fall 2024/2025 **Semester** Marine Engineering Technology **Department** Electrical Machine/ MET323 **Course/Code** Eng. Ibtihal Ahmed **Lecturer** 1 hour 50 mins **Duration** **Time** 4 **Grade** **Date** 6 **No. of papers** 11-03 **Room No.** ------------------------------ ------------------- ------------------------------- ---------------- **[Lab assessment experiment 7]** -------------- ------------------------------ ------------ ---------- **Question** **Marks** **CLOs** **Available** **Actual** 1 **0.5** **CLO3** 2 **1** **CLO3** 3 **0.5** **CLO3** 4 **1** **CLO3** 5 **1** **CLO3** **Total** **4** **Lecturer** **Name:** Eng. Ibtihal Ahmed **Sign:** **Date:** -------------- ------------------------------ ------------ ---------- **[Experiment 7]** **[Three-Phase slip ring (wound rotor) Induction Motor]** ===================================================================== ### ### [Objective] - - - - ### ### ### [Instruments & Equipment] - 3Ø slip ring (wound rotor) Induction Motor. - Dynamometer. - 3Ø Power supply. - Switching, protection and measurement units (modules). - Computer + Cassy software. - Rotor Starter Resistance ### [Theory:] An induction motor is an electrical device that converts electrical energy into mechanical energy. It is most widely used for industrial applications due to its self-starting attribute. Slip ring induction motor is one of the types of 3-phase induction motor and is a wound rotor motor type. Squirrel cage motor produces a low starting torque which is not suitable for some applications. In this case, slip ring induction motor is used for high starting torque. When a 3-phase supply is connected to the stator winding, it produces a rotating magnetic field in the air gab between the stator and rotor. According to Faraday's law of electromagnetic induction, an electromotive force is induced in the rotor bars. Because of the rotor parts are short circuit by end ring, this emf generates a current to flow through the rotor bars. According to Lorentz's law, when a current carrying conductor is placed in a magnetic field, it will experience force. These collective forces make the rotor turn. The three-phase windings on the rotor are internally connected to form an internal neutral connection. The other three ends are connected to the sliprings. With the brushes riding on the sliprings, we can add external resistances in the rotor circuit. In this way the total resistance in the rotor circuit can be controlled. By controlling the resistance in the rotor circuit, we are, in fact, controlling the torque developed by the motor. This way slip ring induction motors are able to produce high torque even as they are starting. The below graph shows higher starting torque produced by slip ring motors in comparison to squirrel cage motors. A diagram of a motion curve Description automatically generated ### [Procedure:] ### Draw the connection diagram of the slip ring induction motor. ### **Q1** --------- **0.5** ### ### ### ### ### ### ### ### ### ### Measure the required values and record the results in the table. **Q2** -------- **1** ---------------------------------------------------------------------------------------- **Load** -------------------------------- ------- --------- --------- --------- ------- --------- **0** **0.2** **0.4** **0.8** **1** **1.2** \ [**P**~ele~]{.math.display}\ \ [**P**~mech~]{.math.display}\ ---------------------------------------------------------------------------------------- 3. Set the rotor starter to level 6 before starting the experiment. 4. Allow the machine to ramp up without a load then change the resistance. Note that: the starter resistance (level 6: resistance value 0 Ω, level 1: highest **Q3** --------- **0.5** resistance value). (0.25 Mark) **Rotor Starter Resistance** **Speed** ------------------------------ ----------- **6** **5** **4** **3** **2** **1** 5. What is the relation between the starter resistance and speed? (0.25 Mark) 6. Now apply a load for each resistor setting and find the corresponding speed. (1 Mark) **Q4** -------- **1** +-----------------+-----------------+-----------------+-----------------+ | **Rotor Starter | **6** | **4** | **3** | | Resistance | | | | | (Level)** | | | | +=================+=================+=================+=================+ | **0** | | | | +-----------------+-----------------+-----------------+-----------------+ | **0.2** | | | | +-----------------+-----------------+-----------------+-----------------+ | **0.4** | | | | +-----------------+-----------------+-----------------+-----------------+ | **0.8** | | | | +-----------------+-----------------+-----------------+-----------------+ | **1** | | | | +-----------------+-----------------+-----------------+-----------------+ | **1.2** | | | | +-----------------+-----------------+-----------------+-----------------+ 7. Draw T-S curve based on readings. **Q5** -------- **1** 8. ###  **SHARJAH MARITIME ACADEMY** **[Laboratory Experiment]** ------------------------------ ------------------- ------------------------------- ---------------- **SHARJAH MARITIME ACADEMY** Spring 2023/2024 **Semester** Marine Engineering Technology **Department** Electrical Machines/ MET323 **Course/Code** Eng. Ibtihal Ahmed **Lecturer** 1 hour 50 mins **Duration** **Time** 4 **Grade** **Day & Date** 6 **No. of papers** 11-03 **Room No.** ------------------------------ ------------------- ------------------------------- ---------------- **[LAB assessment experiment 8]** -------------- ------------------------------ ------------ ---------- **Question** **Marks** **CLOs** **Available** **Actual** 1 **0.5** **CLO3** 2 **1** **CLO3** 3 **1** **CLO3** 4 **1.5** **CLO3** **Total** **4** **Lecturer** **Name:** Eng. Ibtihal Ahmed **Sign:** **Date:** -------------- ------------------------------ ------------ ---------- **[Experiment 8]** **[Operating Synchronous Machines as a synchronous Motor]** ======================================================================= **[Objectives]** - To study and understand the working principle of a synchronous motor, including how it starts, operates, and maintains synchronous speed. - To measure and analyze the performance characteristics of the synchronous motor, such as speed, torque, power factor, efficiency under different load conditions. **[Instruments & Equipment:]** 1. Synchronous machine. 2. Dynamometer. 3. DC Power supply. 4. Switching, protection and measurement units (modules). **[Theory:]** A synchronous motor, as the name suggests, runs under steady-state conditions at a fixed speed called the **synchronous speed** irrespective of the load acting on it. The synchronous speed depends only upon (a) the frequency of the applied voltage and (b) the number of poles in the machine. The synchronous speed in revolutions per minute (rpm) at which the flux revolves around the periphery of the airgap: \ [\$\$\\mathbf{N}\_{\\mathbf{s}}\\mathbf{=}\\frac{\\mathbf{120}\\mathbf{f}}{\\mathbf{P}}\$\$]{.math.display}\ where [**f**]{.math.inline} is the frequency of the three-phase power source and [**P**]{.math.inline} is the number of poles in the motor. The stator is wound for the similar number of poles as that of rotor and fed with three phase AC supply. The 3 phase AC supply produces rotating magnetic field in the stator. The rotor winding is fed with DC supply which magnetizes the rotor and thus the stationary magnetic field develops on the rotor. If the motor does not get initial rotation, the starting speed will be very low, and the rotor will not be able to start (not inherently self-starting). To make synchronous motor self-starting, a squirrel cage arrangement is fitted across its' poles. At the start, the rotor is not connected to the DC power supply and the rooting magnetic field induces electricity in squirrel cage bars and rotor starts rotating as an induction motor. When the rotor reaches its maximum speed the rotor is connected to DC supply.  A diagram of a machine Description automatically generated ### [Procedure:] 1. **Q1** --------- **0.5** 2. 3. **Q2** -------- **1** --------------------------------- --------- --------- --------- --------- --------- --------- ------- No load 0.1 N.m 0.3 N.m 0.5 N.m 0.7 N.m 0.9 N.m 1 N.m \ [**I**~**M**~]{.math.display}\ \ [**V**~**M**~]{.math.display}\ \ [**P**~**m**~]{.math.display}\ \ [**P**~el~]{.math.display}\ \ [**η**]{.math.display}\ --------------------------------- --------- --------- --------- --------- --------- --------- ------- 4. Draw Torque with speed curve. **Q3** -------- **1**  A graph paper with a grid Description automatically generated Comment on Graph: Comment on the synchronous motor operation compared to induction motors. **Q4** --------- **1.5** Why the DC supply is required in Synchronous motor. Why isn\'t DC applied to the rotor field coil right away instead of waiting until the rotor is up to speed?