EMC611S Electrical Machines 214 Notes - Three-Phase Induction Motors

EMC611S Electrical Machines 214: Three-Phase Induction Motors Tutorial Questions

General Information

• Tutorial Date: 28-05-2020 (10H30 – 12H30), 29-05-2020 (10H30 – 12H30 & 14H00 – 16H00)

• Lectures Covered:
    - Lecture 1: Design, construction, and principle of operation (Refer to Tutorial dated 26-05-2020)
    - Lecture 2: The equivalent circuit (Refer to Tutorial dated 26-05-2020)
    - Lecture 3: Induction motor tests

Lecture 3: Induction Motor Tests

Test Data

• Motor Specifications: 460-V, 60-Hz, Delta-connected, three-phase induction motor

• No-load test data:
    - Power input: 380 W
    - Line current: 1.15 A at rated voltage

• Blocked-rotor test data:
    - Power input: 14.7 W
    - Line current: 2.1 A at line voltage: 21 V

• Friction-and-windage loss: 21 W

• Winding resistance between any two lines: 1.2 Ω

Questions

Question 3.1: Determine the equivalent circuit parameters of the motor

• Part (a):
    1. D.C. Resistance Test:
       - Winding resistance between any two lines: $R_{dc} = 1.2 \, ext{Ω}$
       - Define equivalent resistance:
         $R_{ph} = rac{R_{dc}}{2} = rac{1.2}{2} = 0.6 \, ext{Ω}$
    2. From the No-load Test:
       - Power input = 380 W; Line current = 1.15 A
       - Friction-and-windage loss = 21 W
       - Equivalent single-phase measured power:
         $W_{oc} = rac{3W_{oc}}{3} = rac{380}{3} = 126.667 \, ext{W}$
       - Equivalent single-phase power loss:
         $P_{F+W} = rac{3P_{F+W}}{3} = rac{21}{3} = 7 \, ext{W}$
       - Core-loss resistance, $R_c$:
         - Core loss is given by:
           $P_{oc} = W_{oc} - P_{F+W} = 126.667 ext{W} - 7 ext{W} = 119.667 ext{W}$
         - $R_c = rac{V_{oc}^2}{P_{oc}} = rac{460^2}{119.667} <br>ightarrow 1768.24 \, ext{Ω}$
         - Approximated to: $R_c ≈ 1768 \, ext{Ω}$

Question 3.2: Draw exact per-phase equivalent circuit diagram

• Provide a diagram incorporating all parameters mentioned in Part (a) results.

Lecture 4: Power and Torque

Two Questions

Question 4.1: Determine developed torque in a two-pole, 50-Hz induction motor

• Motor output: 30 kW at speed 2950 rpm

• Part (a): Developed Torque $au_d$ Calculation:
    1. Assumptions:
       - Negligible windage and friction losses in rotor
       - $au_d ext{ must equal load torque}$
    2. Torque formula:
       - $au_d = rac{P_d}{ ext{ω}<em>r} ightarrow au_d = rac{P</em>{out} imes 60}{2 ext{π}N_r}$
       - Substituting known values:
         $au_d = rac{30,000 imes 60}{2 imes ext{π} imes 2950} = 97.1 \, ext{Nm}$

Question 4.2: Operating speed and power output when torque is trebled

• Part (b):
    1. Calculate existing synchronous speed:
       - s = rac{N_s - N_r}{N_s} imes 100 ext{%}
       - For existing speed 2950 rpm, calculate:
         - N_s = 3000 ext{ rpm}, ext{ slip} = 1.667 ext{%}
    2. New Torque Calculation:
       - Trebled torque: $au_{out(new)} = 3 imes 97.1 = 291.3 \, ext{Nm}$
    3. New Speed Calculation:
       - New slip: $s(new) = 3 imes s(old)$
         - s(new) = 5.001 ext{%}
       - New motor speed: $N_r(new) = N_s(1-s(new)) = 3000(1 - 0.05001) = 2849.97 ext{ rpm}$

Efficiency Calculation: Six-Pole Induction Motor

• Parameters: 230-V, 60-Hz, Rs = 0.5 Ω, R' = 0.25 Ω, Xs = j0.75 Ω, X' = j0.5 Ω, Rm = j100 Ω, Rc = 500 Ω.

• Load and losses:
    - Considering power-flow circuits, set up both equivalent and power-flow diagrams.

Lecture 5: Induction Motor Starting

Starting Conditions

Question 5.1: Maximum permissible full-load current rating

• Given:
    - D.O.L. Starting current = 6 imes I_{FL}
    - Auto-transformer has a 60% tapping.

• Solution Steps::
    1. Establish autotransformer starting current equality with allowable maximum:
       - $I_{L-AutoT} = 120 ext{ A}$
    2. Apply equivalent current relations to derive the max full-load current, transforming through respective calculations to yield:
       - Maximum permissible full-load current approximately: 56 A

Question 5.2: Calculation of rotor currents with slip rings shorted

• Analyze rotor current at start and during continued operation with specified loads and resistances. Include charts for clarity on current behaviors during various operational states.

Calculation Summary

• The comprehensive workings of the above sections should constitute all relevant figures and diagrams, including those summarizing the per-phase equivalent circuit results with full load and starting conditions metric calculations to enhance clarity and understanding of electrical machine operations within practical parameters.