3-Phase Induction Motors

LessonAbout 1.5 h

Three-Phase Induction Motor is an asynchronous AC machine whose rotor rotates at a speed less than the synchronous speed of the rotating magnetic field established by the stator windings.

1. Slip

Synchronous Speed: speed of rotating stator magnetic field
Rotor Speed: actual rotational speed of motor shaft
Slip: relative speed difference normalized to synchronous speed

Slip and Rotor Speed

\[ s = \dfrac{N_s - N}{N_s} \]

\[ N_s = \dfrac{120f}{P} \]

\[ N = \dfrac{120f}{P}(1 - s) \]

Where: \(s\) = slip (per unit), \(N_s\) = synchronous speed (rpm), \(N\) = rotor speed (rpm), \(f\) = frequency of stator voltage (Hz), \(P\) = number of poles

EE Board April 1992

Example. The rotor speed of a six-pole, \(50\ \text{Hz}\) induction motor is \(960\ \text{rpm}\). Calculate the percent slip.

Solution.

\[ N_s = \dfrac{120(50)}{6} = 1000\ \text{rpm} \qquad\qquad s = \dfrac{1000 - 960}{1000} \times 100\% = 4.0\% \]

Answer: \(4.0\%\)

2. Rotor Voltage and Frequency

Rotor Frequency: frequency of induced rotor voltage proportional to slip
Rotor Induced Voltage: magnitude of rotor induced emf proportional to slip
Standstill Condition: maximum rotor voltage and line frequency occur when blocked (\(s = 1\))

Rotor Voltage and Frequency

\[ f_r = s f \]

\[ E_2 = s E_{\text{oc}} \]

Where: \(f_r\) = rotor frequency at slip \(s\) (Hz), \(f\) = stator supply frequency (Hz), \(s\) = slip (per unit), \(E_2\) = rotor induced voltage per phase at slip \(s\) (V), \(E_{\text{oc}}\) = standstill open-circuit rotor voltage per phase (V)

EE Board March 1998

Example. A \(50\ \text{hp}\) three-phase motor, rated \(220\ \text{V}\), \(75\%\) power factor has \(6\ \text{poles}\). The slip at full load is \(3\%\). What is the frequency of the rotor current at full load?

Solution.

\[ f_r = s f = (0.03)(60) = 1.8\ \text{Hz} \]

Answer: \(1.8\ \text{Hz}\)

3. Equivalent Circuit Per Phase of a Wound Rotor Induction Motor at Standstill

Transformer Action: blocked rotor (\(s=1\)) operates as a multi-phase transformer
Stator Parameters: stator winding resistance \(r_1\) and leakage reactance \(x_1\)
Rotor Parameters: standstill rotor resistance \(r_2\) and leakage reactance \(x_2\)
Excitation Branch: parallel magnetizing reactance \(X_m\) and core-loss resistance \(R_m\)

Standstill Rotor Circuit Impedance

\[ Z_{2,\text{standstill}} = (r_2 + R_x) + j x_2 \]

Where: \(Z_{2,\text{standstill}}\) = standstill rotor impedance per phase (\(\Omega\)), \(r_2\) = internal rotor resistance per phase (\(\Omega\)), \(R_x\) = external resistance per phase across slip rings (\(\Omega\)), \(x_2\) = standstill rotor leakage reactance per phase (\(\Omega\))

4. Equivalent Circuit Per Phase of Wound Rotor Induction Motor Referred to the Stator Side

Referred Parameters: rotor resistance and reactance multiplied by stator-to-rotor transformation ratio squared
Variable Load Resistance: mechanical load represented by variable rotor resistance \(R_2/s\)

4.1 Apparent Power Absorbed

Apparent Power Absorbed Per Phase

\[ S = \sqrt{P^2 + Q^2} \]

Where: \(S\) = apparent power absorbed per phase (VA), \(P\) = active power absorbed per phase (W), \(Q\) = reactive power absorbed per phase (var)

4.2 Reactive Power Absorbed

Reactive Power Absorbed Per Phase

\[ Q = \dfrac{E_p^2}{X_m} + I_1^2 (x_1 + x_2) \]

Where: \(Q\) = reactive power absorbed per phase (var), \(E_p\) = stator phase voltage (V), \(X_m\) = magnetizing reactance (\(\Omega\)), \(I_1\) = rotor current referred to stator (A), \(x_1\) = stator leakage reactance (\(\Omega\)), \(x_2\) = rotor leakage reactance referred to stator (\(\Omega\))

4.3 Apparent Power

Total Three-Phase Apparent Power

\[ S_{3\phi} = 3 S = \sqrt{3} V_L I_L \]

Where: \(S_{3\phi}\) = total three-phase apparent power (VA), \(S\) = per-phase apparent power (VA), \(V_L\) = line-to-line voltage (V), \(I_L\) = line current (A)

4.4 Power Factor

Motor Power Factor

\[ \text{pf} = \dfrac{P}{S} = \cos\theta \]

Where: \(\text{pf}\) = operating power factor, \(P\) = active power absorbed per phase (W), \(S\) = apparent power absorbed per phase (VA), \(\theta\) = phase angle between stator voltage and current (deg)

4.5 Line Current Drawn

Line Current Drawn

\[ I_p = \dfrac{S}{E_p} \qquad\qquad I_1 = \dfrac{E_p}{\sqrt{\left(r_1 + \frac{R_2}{s}\right)^2 + (x_1 + x_2)^2}} \]

Where: \(I_p\) = total stator line current per phase (A), \(I_1\) = rotor load current referred to stator (A), \(S\) = apparent power per phase (VA), \(E_p\) = stator phase voltage (V), \(r_1\) = stator resistance (\(\Omega\)), \(R_2\) = total rotor resistance referred to stator (\(\Omega\)), \(s\) = slip

4.6 Active Power Supplied to Rotor

Active Power Supplied to Rotor

\[ P_r = I_1^2 \left(\dfrac{R_2}{s}\right) \]

Where: \(P_r\) = active power transferred across air gap to rotor per phase (W), \(I_1\) = rotor current referred to stator (A), \(R_2\) = total rotor resistance referred to stator (\(\Omega\)), \(s\) = slip

4.7 Power Dissipated to the Rotor Circuit

Rotor Copper Loss

\[ P_{\text{cu2}} = I_1^2 R_2 = s P_r \]

Where: \(P_{\text{cu2}}\) = rotor circuit \(I^2R\) copper loss per phase (W), \(I_1\) = rotor current referred to stator (A), \(R_2\) = rotor resistance referred to stator (\(\Omega\)), \(s\) = slip, \(P_r\) = rotor input power per phase (W)

EE Board April 1993

Example. The rotor of a \(220\ \text{V}\), \(60\ \text{Hz}\), \(4\text{-pole}\) induction motor has a power input of \(80\ \text{kW}\). It is observed that the rotor emf makes \(90\ \text{cycles per minute}\). Calculate the rotor copper loss.

Solution.

\[ f_r = \dfrac{90}{60} = 1.5\ \text{Hz} \qquad\qquad s = \dfrac{1.5}{60} = 0.025 \qquad\qquad P_{\text{cu2}} = s P_r = (0.025)(80) = 2\ \text{kW} \]

Answer: \(2\ \text{kW}\)

4.8 Mechanical Power Developed

Mechanical Power Developed

\[ P_m = P_r - P_{\text{cu2}} = P_r (1 - s) \]

Where: \(P_m\) = gross mechanical power developed per phase (W), \(P_r\) = rotor input power per phase (W), \(P_{\text{cu2}}\) = rotor copper loss per phase (W), \(s\) = slip

EE Board April 1993

Solution.

\[ f_r = \dfrac{90}{60} = 1.5\ \text{Hz} \qquad\qquad s = \dfrac{1.5}{60} = 0.025 \qquad\qquad P_m = P_r (1 - s) = 80(1 - 0.025) = 78\ \text{kW} \]

Answer: \(78\ \text{kW}\)

4.9 Torque Developed

Electromagnetic Torque Developed

\[ T = \dfrac{9.55 P_m}{N} = \dfrac{9.55 P_r}{N_s} \]

Where: \(T\) = gross torque developed (\(\text{N}\cdot\text{m}\)), \(P_m\) = total mechanical power developed (\(3\phi\)) (W), \(N\) = rotor speed (rpm), \(P_r\) = total rotor power input (\(3\phi\)) (W), \(N_s\) = synchronous speed (rpm)

EE Board October 1990

Example. A \(10\ \text{hp}\), \(4\text{-pole}\), \(25\ \text{Hz}\), \(3\text{-phase}\) induction motor takes \(9100\ \text{W}\) from the line. Core loss is \(290\ \text{W}\); stator copper loss is \(568\ \text{W}\); rotor copper loss is \(445\ \text{W}\) and friction and windage loss is \(121\ \text{W}\). Determine the output torque.

Solution.

\[ P_r = 9100 - (568 + 290) = 8242\ \text{W} \qquad\qquad s = \dfrac{445}{8242} = 0.0539 \implies N = \dfrac{120(25)}{4}(1 - 0.0539) = 709.58\ \text{rpm} \]

\[ P_{\text{out}} = 9100 - (568 + 290 + 445 + 121) = 7676\ \text{W} \implies T_{\text{out}} = \dfrac{9.55(7676)}{709.58} = 103.3\ \text{N}\cdot\text{m} \]

Answer: \(103.3\ \text{N}\cdot\text{m}\)

4.10 Efficiency

Overall Efficiency

\[ \eta = \dfrac{P_{\text{out}}}{P_{\text{in}}} \times 100\% = \dfrac{P_m - P_{\text{fw}}}{P} \times 100\% \]

Where: \(\eta\) = overall motor efficiency (\%), \(P_{\text{out}}\) = useful shaft output power (W), \(P_{\text{in}}\) = total electrical input power (W), \(P_m\) = gross mechanical power developed (W), \(P_{\text{fw}}\) = friction and windage loss (W), \(P\) = active power input (W)

5. No-Load Test of Induction Motor

Unloaded Operation: motor runs uncoupled at near synchronous speed (\(s \approx 0\))
High Slip Resistance: rotor load branch \(R_2/s\) becomes an open circuit
Rotational Losses: total no-load input power represents core, friction, and windage losses

No-Load Test Equations

\[ S_{\text{NL}} = \sqrt{3} E_{\text{NL}} I_{\text{NL}} \qquad\qquad Q_{\text{NL}} = \sqrt{S_{\text{NL}}^2 - P_{\text{NL}}^2} \]

\[ X_m = \dfrac{(E_{\text{NL}}/\sqrt{3})^2}{Q_{\text{NL}}/3} \qquad\qquad R_m = \dfrac{(E_{\text{NL}}/\sqrt{3})^2}{P_{\text{NL}}/3} \]

\[ P_{\text{co}} + P_{\text{fw}} = P_{\text{NL}} - 3 I_{\text{NL}}^2 r_1 \]

Where: \(S_{\text{NL}}\) = no-load apparent power (VA), \(E_{\text{NL}}\) = line-to-line no-load voltage (V), \(I_{\text{NL}}\) = no-load line current (A), \(P_{\text{NL}}\) = 3-phase no-load power input (W), \(Q_{\text{NL}}\) = no-load reactive power (var), \(X_m\) = magnetizing reactance per phase (\(\Omega\)), \(R_m\) = no-load loss resistance (\(\Omega\)), \(P_{\text{co}}\) = core loss (W), \(P_{\text{fw}}\) = friction and windage loss (W), \(r_1\) = stator resistance per phase (\(\Omega\))

6. Locked Rotor Test of Induction Motor

Blocked Rotor: rotor is mechanically locked stationary (\(s = 1\))
Low Voltage Applied: reduced line voltage applied until rated stator current flows
Short-Circuit Impedance: yields total equivalent resistance \(R_{e1}\) and leakage reactance \(X_{e1}\)

Locked Rotor Test Equations

\[ S_{\text{LR}} = \sqrt{3} E_{\text{LR}} I_{\text{LR}} \qquad\qquad Q_{\text{LR}} = \sqrt{S_{\text{LR}}^2 - P_{\text{LR}}^2} \]

\[ R_{e1} = \dfrac{P_{\text{LR}}}{3 I_{\text{LR}}^2} \qquad\qquad X_{e1} = \dfrac{Q_{\text{LR}}}{3 I_{\text{LR}}^2} \]

Where: \(S_{\text{LR}}\) = locked-rotor apparent power (VA), \(E_{\text{LR}}\) = line-to-line locked-rotor voltage (V), \(I_{\text{LR}}\) = locked-rotor line current (A), \(P_{\text{LR}}\) = 3-phase locked-rotor power input (W), \(Q_{\text{LR}}\) = locked-rotor reactive power (var), \(R_{e1}\) = equivalent resistance per phase referred to stator (\(\Omega\)), \(X_{e1}\) = equivalent leakage reactance per phase referred to stator (\(\Omega\))

7. DC Stator Resistance Test of Induction Motor

DC Ohmic Measurement: direct current applied across two stator line terminals
AC Effective Resistance: skin effect multiplier (\(1.25\) to \(1.50\)) applied to obtain AC resistance

DC Stator Resistance Equations

\[ r_1 = \dfrac{1}{2} R_t \qquad\qquad (\text{for Wye}) \]

\[ r_1 = \dfrac{3}{2} R_t \qquad\qquad (\text{for Delta}) \]

Where: \(r_1\) = DC stator resistance per phase (\(\Omega\)), \(R_t\) = terminal DC resistance ohmmeter reading (\(\Omega\))

8. Power Flow Diagram

Electrical Input Power: total active power \(P_{\text{in}}\) drawn by stator from 3-phase line
Stator Losses: stator copper loss \(P_{\text{cus}}\) and core loss \(P_{\text{co}}\) subtracted before air gap
Rotor Power Input: air gap power \(P_r\) transferred to rotor via magnetic field
Rotor Losses: \(I^2R\) copper loss \(P_{\text{cur}} = s P_r\) dissipated in rotor conductors
Mechanical Power Developed: gross mechanical power \(P_m = (1-s)P_r\)
Useful Shaft Power: friction and windage loss \(P_{\text{fw}}\) subtracted to give net output \(P_{\text{out}}\)

Power Flow Equations

\[ P_{\text{in}} = P_r + P_{\text{cus}} + P_{\text{co}} \]

\[ P_r = P_m + P_{\text{cur}} \]

\[ P_m = P_{\text{out}} + P_{\text{fw}} \]

Where: \(P_{\text{in}}\) = stator input power (W), \(P_r\) = air gap power input to rotor (W), \(P_m\) = gross mechanical power developed (W), \(P_{\text{out}}\) = net shaft output power (W), \(P_{\text{cus}}\) = stator copper loss (W), \(P_{\text{co}}\) = stator core loss (W), \(P_{\text{cur}}\) = rotor copper loss (W), \(P_{\text{fw}}\) = friction and windage loss (W)

EE Board April 1993

Example. A 3-phase squirrel-cage induction motor is drawing \(45\ \text{kW}\) from a \(440\ \text{V}\), \(60\ \text{Hz}\) 3-phase line. If stator losses amount to \(900\ \text{W}\), determine total mechanical power developed. Assume a slip of \(3\%\).

Solution.

\[ P_r = P_{\text{in}} - P_{\text{stator losses}} = 45000 - 900 = 44100\ \text{W} \]

\[ P_m = P_r (1 - s) = 44100 (1 - 0.03) = 42777\ \text{W} = 42.78\ \text{kW} \]

Answer: \(42.78\ \text{kW}\)

9. Formula Summary

Formula Name	Formula	Variables
Per-Unit Slip	\(s = \dfrac{N_s - N}{N_s}\)	\(s\) = slip (per unit) \(N_s\) = synchronous speed (rpm) \(N\) = rotor speed (rpm)
Synchronous Speed	\(N_s = \dfrac{120f}{P}\)	\(N_s\) = synchronous speed (rpm) \(f\) = frequency (Hz) \(P\) = poles
Rotor Speed	\(N = \dfrac{120f}{P}(1 - s)\)	\(N\) = rotor speed (rpm) \(f\) = frequency (Hz) \(P\) = poles \(s\) = slip
Rotor Frequency	\(f_r = s f\)	\(f_r\) = rotor frequency (Hz) \(s\) = slip \(f\) = supply frequency (Hz)
Rotor Induced Voltage	\(E_2 = s E_{\text{oc}}\)	\(E_2\) = rotor phase voltage at slip \(s\) (V) \(s\) = slip \(E_{\text{oc}}\) = standstill open-circuit rotor voltage (V)
Standstill Rotor Impedance	\(Z_{2,\text{standstill}} = (r_2 + R_x) + j x_2\)	\(Z_{2,\text{standstill}}\) = standstill rotor impedance (\(\Omega\)) \(r_2\) = rotor resistance (\(\Omega\)) \(R_x\) = external resistance (\(\Omega\)) \(x_2\) = rotor reactance (\(\Omega\))
Apparent Power Absorbed	\(S = \sqrt{P^2 + Q^2}\)	\(S\) = apparent power per phase (VA) \(P\) = active power per phase (W) \(Q\) = reactive power per phase (var)
Reactive Power Absorbed	\(Q = \dfrac{E_p^2}{X_m} + I_1^2 (x_1 + x_2)\)	\(Q\) = reactive power per phase (var) \(E_p\) = phase voltage (V) \(X_m\) = magnetizing reactance (\(\Omega\)) \(I_1\) = rotor current (\(\Omega\)) \(x_1, x_2\) = leakage reactances (\(\Omega\))
Total Three-Phase Apparent Power	\(S_{3\phi} = \sqrt{3} V_L I_L\)	\(S_{3\phi}\) = 3-phase apparent power (VA) \(V_L\) = line voltage (V) \(I_L\) = line current (A)
Motor Power Factor	\(\text{pf} = \dfrac{P}{S}\)	\(\text{pf}\) = power factor \(P\) = active power per phase (W) \(S\) = apparent power per phase (VA)
Rotor Load Current	\(I_1 = \dfrac{E_p}{\sqrt{\left(r_1 + \dfrac{R_2}{s}\right)^2 + (x_1 + x_2)^2}}\)	\(I_1\) = rotor current referred to stator (A) \(E_p\) = phase voltage (V) \(r_1\) = stator resistance (\(\Omega\)) \(R_2\) = total rotor resistance (\(\Omega\)) \(s\) = slip \(x_1, x_2\) = reactances (\(\Omega\))
Active Power Supplied to Rotor	\(P_r = I_1^2 \left(\dfrac{R_2}{s}\right)\)	\(P_r\) = rotor power input per phase (W) \(I_1\) = rotor current (A) \(R_2\) = rotor resistance (\(\Omega\)) \(s\) = slip
Rotor Copper Loss	\(P_{\text{cu2}} = s P_r\)	\(P_{\text{cu2}}\) = rotor copper loss per phase (W) \(s\) = slip \(P_r\) = rotor power input per phase (W)
Mechanical Power Developed	\(P_m = P_r (1 - s)\)	\(P_m\) = mechanical power developed per phase (W) \(P_r\) = rotor power input per phase (W) \(s\) = slip
Electromagnetic Torque Developed	\(T = \dfrac{9.55 P_m}{N}\)	\(T\) = developed torque (\(\text{N}\cdot\text{m}\)) \(P_m\) = total mechanical power developed (W) \(N\) = rotor speed (rpm)
Overall Efficiency	\(\eta = \dfrac{P_{\text{out}}}{P_{\text{in}}} \times 100\%\)	\(\eta\) = efficiency (\%) \(P_{\text{out}}\) = shaft output power (W) \(P_{\text{in}}\) = total input power (W)
No-Load Magnetizing Reactance	\(X_m = \dfrac{(E_{\text{NL}}/\sqrt{3})^2}{Q_{\text{NL}}/3}\)	\(X_m\) = magnetizing reactance (\(\Omega\)) \(E_{\text{NL}}\) = no-load line voltage (V) \(Q_{\text{NL}}\) = 3-phase no-load var
Locked-Rotor Equivalent Resistance	\(R_{e1} = \dfrac{P_{\text{LR}}}{3 I_{\text{LR}}^2}\)	\(R_{e1}\) = equivalent resistance referred to stator (\(\Omega\)) \(P_{\text{LR}}\) = 3-phase locked-rotor power (W) \(I_{\text{LR}}\) = locked-rotor line current (A)
DC Stator Resistance (Wye)	\(r_1 = \dfrac{1}{2} R_t\)	\(r_1\) = DC phase resistance (\(\Omega\)) \(R_t\) = terminal resistance (\(\Omega\))
DC Stator Resistance (Delta)	\(r_1 = \dfrac{3}{2} R_t\)	\(r_1\) = DC phase resistance (\(\Omega\)) \(R_t\) = terminal resistance (\(\Omega\))
Rotor Power Balance	\(P_{\text{in}} = P_r + P_{\text{cus}} + P_{\text{co}}\)	\(P_{\text{in}}\) = stator input power (W) \(P_r\) = air gap power (W) \(P_{\text{cus}}\) = stator copper loss (W) \(P_{\text{co}}\) = stator core loss (W)
Shaft Output Power Balance	\(P_{\text{out}} = P_m - P_{\text{fw}}\)	\(P_{\text{out}}\) = net output power (W) \(P_m\) = gross mechanical power developed (W) \(P_{\text{fw}}\) = friction and windage loss (W)