Question 1 |

A delta-connected, 3.7 kW, 400 V(line), three-phase, 4-pole, 50-Hz squirrel-cage induction motor has the following equivalent circuit parameters per phase referred to the stator:

R_1=5.39\Omega,R_2=5.72\Omega,X_1=X_2=8.22\Omega.

Neglect shunt branch in the equivalent circuit. The starting line current in amperes (round off to two decimal places) when it is connected to a 100 V (line), 10 Hz, three-phase AC source is ______

R_1=5.39\Omega,R_2=5.72\Omega,X_1=X_2=8.22\Omega.

Neglect shunt branch in the equivalent circuit. The starting line current in amperes (round off to two decimal places) when it is connected to a 100 V (line), 10 Hz, three-phase AC source is ______

8.25 | |

14.95 | |

25.45 | |

32.85 |

Question 1 Explanation:

\begin{aligned} I_L &=\sqrt{3}I_{ph} \\ \text{At} \;\; f &= 10Hz, \\ X_1=X_2 &=8.22 \times \frac{10}{50}=1.644\Omega \\ I_{ph} &= \frac{V_{ph}}{\sqrt{(R_1+R_2)^2+(X_1+X_2)^2}} \\ &=\frac{100}{\sqrt{(5.39+5.72)^2+(1.644+1.644)^2}} \\ &= \frac{100}{11.586}=8.63\\ I_L&= 14.95A \end{aligned}

Question 2 |

The parameter of an equivalent circuit of a three-phase induction motor affected by reducing the rms value of the supply voltage at the rated frequency is

rotor resistance | |

rotor leakage reactance | |

magnetizing reactance | |

stator resistance |

Question 3 |

A star-connected, 12.5 kW, 208 V (line), 3-phase, 60 Hz squirrel cage induction motor has
following equivalent circuit parameters per phase referred to the stator: R_{1}=0.3\Omega ,R_{2}=0.3\Omega ,X_{1}=0.41\Omega ,X_{2}=0.41\Omega. Neglect shunt branch in the equivalent circuit. The starting current (in Ampere) for this motor when connected to an 80 V (line), 20 Hz, 3-phase AC source is __________.

50 | |

60 | |

70 | |

80 |

Question 3 Explanation:

Equivalent circuit representation during starting

-according to question

Total resistance to stator =0.6\Omega

Total reactance reference to stator =0.82\Omega

Reactance for 60Hz \;\rightarrow \;0.82\Omega

Reactance for 60Hz \;\rightarrow ?

\because \;\; X_e\propto F

Reactance for 20Hz\;\Rightarrow \; \frac{20}{60} \times 0.82

\therefore \;\; X for 20 Hz frequency =0.273\Omega

\therefore \;\; Starting current

\Rightarrow \;\; I_{st}=\frac{80\sqrt{3}}{\sqrt{(0.6)^2+(0.273)^2}}=70.05A

-according to question

Total resistance to stator =0.6\Omega

Total reactance reference to stator =0.82\Omega

Reactance for 60Hz \;\rightarrow \;0.82\Omega

Reactance for 60Hz \;\rightarrow ?

\because \;\; X_e\propto F

Reactance for 20Hz\;\Rightarrow \; \frac{20}{60} \times 0.82

\therefore \;\; X for 20 Hz frequency =0.273\Omega

\therefore \;\; Starting current

\Rightarrow \;\; I_{st}=\frac{80\sqrt{3}}{\sqrt{(0.6)^2+(0.273)^2}}=70.05A

Question 4 |

A 3-phase, 4-pole, 400 V, 50 Hz squirrel-cage induction motor is operating at a slip of 0.02. The
speed of the rotor flux in mechanical rad/sec, sensed by a stationary observer, is closest to

1500 | |

1470 | |

157 | |

154 |

Question 4 Explanation:

\begin{aligned} N_s &= \frac{120 \times 50}{4}=1500\; \text{rpm}\\ s &=0.02 \\ N&=N_s(1-s) \\ &=1500(1-0.02) \\ &=1470 \; \text{rpm} \end{aligned}

As it was specified rotor rmf with respect to stationary part/sector.

Rotor rmf w.r.t. stator is always at N_s.

\therefore \; Speed of rotor flux in mechanical radian/sec.

=\frac{2 \times 3.14 \times 1500}{60}=157 \; rad/sec

As it was specified rotor rmf with respect to stationary part/sector.

Rotor rmf w.r.t. stator is always at N_s.

\therefore \; Speed of rotor flux in mechanical radian/sec.

=\frac{2 \times 3.14 \times 1500}{60}=157 \; rad/sec

Question 5 |

A 4 pole induction machine is working as an induction generator. The generator supply
frequency is 60 Hz. The rotor current frequency is 5 Hz. The mechanical speed of the rotor in
RPM is

1350 | |

1650 | |

1950 | |

2250 |

Question 5 Explanation:

Induction Generator Mode:

\begin{aligned} N & \gt N_s \;\;\text{i.e.(s is negative)} \\ \because \;\; P &=4,\; f=60Hz \\ N_s&=1800 \; \text{rpm} \end{aligned}

Rotor frequency given = 5Hz \Rightarrow (s \times f)

\begin{aligned} \because \;\; f &=60Hz \\ \therefore \;\; s &=\frac{5}{60}=0.0833 \end{aligned}

\because It is induction generator slip will be -0.0833.

\begin{aligned} s &=\frac{1800-N}{1800}=-0.0833 \\ N &=1950 \; \text{rpm} \end{aligned}

\begin{aligned} N & \gt N_s \;\;\text{i.e.(s is negative)} \\ \because \;\; P &=4,\; f=60Hz \\ N_s&=1800 \; \text{rpm} \end{aligned}

Rotor frequency given = 5Hz \Rightarrow (s \times f)

\begin{aligned} \because \;\; f &=60Hz \\ \therefore \;\; s &=\frac{5}{60}=0.0833 \end{aligned}

\because It is induction generator slip will be -0.0833.

\begin{aligned} s &=\frac{1800-N}{1800}=-0.0833 \\ N &=1950 \; \text{rpm} \end{aligned}

Question 6 |

The starting line current of a 415 V, 3-phase, delta connected induction motor is 120 A, when the rated voltage is applied to its stator winding. The starting line current at a reduced voltage of 110 V,
in ampere, is _________.

11.4 | |

22.5 | |

31.8 | |

44.2 |

Question 6 Explanation:

415V, 3-phase, \Delta connected induction motor

(I_{st})_{line}=120A at rated voltage.

at, V=110V, i.e. reduced voltage

\begin{aligned} I_{st} &=x(I_{st})_{rated} \\ \text{where}\;\; x&= \frac{V_{reduced}}{V_{rated}}\\ &=\frac{110}{415} \\ (I_{st})_{at \; 110V} &= \left ( \frac{110}{415} \right )\times 120\\ &=31.807A \end{aligned}

(I_{st})_{line}=120A at rated voltage.

at, V=110V, i.e. reduced voltage

\begin{aligned} I_{st} &=x(I_{st})_{rated} \\ \text{where}\;\; x&= \frac{V_{reduced}}{V_{rated}}\\ &=\frac{110}{415} \\ (I_{st})_{at \; 110V} &= \left ( \frac{110}{415} \right )\times 120\\ &=31.807A \end{aligned}

Question 7 |

In a constant V/f induction motor drive, the slip at the maximum torque

is directly proportional to the synchronous speed. | |

remains constant with respect to the synchronous speed. | |

has an inverse relation with the synchronous speed. | |

has no relation with the synchronous speed. |

Question 7 Explanation:

\begin{aligned} f_0 &=\text{nominal frequency} \\ \omega _s &=\left ( \frac{f}{f_0} \right )\omega _{s0}\;\;...(i) \\ S_{max,T}&=\left ( \frac{f}{f_0} \right ) \left ( \frac{R_2}{X_{20}^1} \right )\;\;...(ii) \\ &\text{From (i) and (ii)} \\ S_{max,T} &= \left ( \frac{\omega _{s0}}{\omega _s} \right )\left ( \frac{R_2}{X_{20}^1} \right )\\ S_{max,T} &\propto \frac{1}{\omega _s} \end{aligned}

Question 8 |

The figure shows the per-phase equivalent circuit of a two-pole three-phase induction motor operating at 50 Hz. The "air-gap" voltage, V_{g} across the magnetizing inductance, is 210 V rms, and the slip, s, is 0.05. The torque (in Nm) produced by the motor is ________.

200.45 | |

305.25 | |

401.64 | |

455.65 |

Question 8 Explanation:

\begin{aligned} i &=\frac{210}{1+j0.22} \\ &=205.09\angle -12.40^{\circ} \\ \text{Active power} &= (205.09)^2 \times 1=P_g\\ P_m &= (1-0.05) \times P_g \\ P_m &= 39958.81 \; \text{Watt}\\ \text{Torque} &= \frac{39958.81 \times 60}{2 \pi \times 2850} \times 3\\ &= 133.88 \times 3=401.64 N-m \end{aligned}

Question 9 |

A 3-phase 50 Hz square wave (6-step) VSI feeds a 3-phase, 4 pole induction motor. The VSI line voltage has a dominant 5^{th} harmonic component. If the operating slip of the motor with respect to fundamental component voltage is 0.04, the slip of the motor with respect to 5^{th} harmonic component of voltage is _________.

1.20 | |

1.82 | |

2.14 | |

6.23 |

Question 9 Explanation:

N_S=\frac{120f}{P}=1500

Slip due to 5th harmmonic

S_5=\frac{N_S+\frac{N_s}{5}}{N_S}=\frac{6}{5}=1.2

Slip due to 5th harmmonic

S_5=\frac{N_S+\frac{N_s}{5}}{N_S}=\frac{6}{5}=1.2

Question 10 |

The torque speed characteristics of motor (T_M) and load (T_L) for two cases are
shown in the figures (a) and (b). The load torque is equal to motor torque at
points P, Q, R and S.

The stable operating points are

The stable operating points are

P and R | |

P and S | |

Q and R | |

Q and S |

Question 10 Explanation:

S=\frac{N_s-N_r}{N_s}

For given system,

slip=(2-s)(speed =negative)

Below point P, T_M-T_l \lt 0

\Rightarrow speed decreases

\therefore S increases

\therefore slip=(2-s) decreases

Above point P, T_m-T_l \gt 0

\Rightarrow speed increases

\therefore S decreases and (2-s) increasaes.

Thus, system is stable near point P.

POINT R:

Below point R: T_M-T_l \lt 0\;\Rightarrow \; speed decreases

Above point R: T_M-T_l \gt 0\;\Rightarrow \; speed increases

Thus, system can't regain stability at point R.

Therefore, R is unstable point.

POINT S:

Below point S: T_M-T_l \gt 0\;\Rightarrow \; speed increases

Above point S: T_M-T_l \lt 0\;\Rightarrow \; speed decreases

Thus, system can regain stability near point S.

There are 10 questions to complete.