Sharjah Maritime Academy Laboratory Experiment PDF

Shape Description automatically generated **SHARJAH MARITIME ACADEMY** **[Laboratory Experiment]** ------------------------------ ------------------- ------------------------------- ---------------- **SHARJAH MARITIME ACADEMY** Fall 202...

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** 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*~ = 2*πfL* = 2*π* × 6*k* × 100*m* = 3.76*Kohm*]{.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\\ kohm\$\$]{.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\\ kohm\$\$]{.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. Shape Description automatically generated **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 A diagram of a circuit Description automatically generated 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. **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 [*U*~1~\_*V*~1~]{.math.display}\ \ 24 [*U*~2~\_*V*~2~]{.math.display}\ \ 10 [*V*~1~\_*W*~1~]{.math.display}\ \ 10.5 [*V*~2~\_*W*~2~]{.math.display}\ \ 13.5 [*W*~1~\_*U*~1~]{.math.display}\ \ 14 [*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 [*U*~2~]{.math.inline}to ground 2.54 [*V*~1~]{.math.inline}to ground 2.56 [*V*~2~]{.math.inline}to ground 2.54 [*W*~1~]{.math.inline}to ground 2.54 [*W*~2~]{.math.inline}to ground 2.56 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. 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** 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. ###  ### 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\\pi N}{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{650.3 + 60}}\\mathbf{= 92.9\\%}\$\$]{.math.display}\  **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. ### ### ### A metal object with a wire Description automatically generated with medium confidence  ### ### A diagram of a magnet Description automatically generated ### 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. ### ###  **N-T curve** ### [Procedure:] 1. Draw the connection diagram of DC motor. A diagram of a circuit Description automatically generated **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 1659 1623 1600 1565 200.8 200.5 201.1 200.5 0.559 0.73 0.88 1.068 0.3 0.3 0.3 0.3 \ 66.8 66.7 66.7 66.4 [**P**~**E**~]{.math.display}\ \ 112.2 146.6 177.4 214 [**P**~elec~]{.math.display}\ \ 17.4 50.3 83.3 114 [**P**~mech~]{.math.display}\ 9.7 23.6 34 40 --------------------------------- --------- --------- --------- --------- --------- ------------ 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 --------------------------------- --------- --------- ------------ \ [**P**~**E**~]{.math.display}\ \ [**P**~elec~]{.math.display}\ \ [**P**~mech~]{.math.display}\ ------------------------------------------------------------------ 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\\pi N}{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{747.7}}{\\mathbf{924 + 48}}\\mathbf{= 76.9\\%}\$\$]{.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**  A graph with a line Description automatically generated  Comment on the obtained charateristics: