Question 1 |

A shunt-connected DC motor operates at its rated terminal voltage. Its no-load speed is 200 radian/second. At its rated torque of 500 Nm, its speed is 180 radian/second. The motor is used to directly drive a load whose load torque T _{L} depends on its rotational speed \omega _{r} (in radian/second), such that T _{L}=2.78 \times \omega _{r}. Neglecting rotational losses, the steady-state speed (in radian/second) of the motor, when it drives this load, is _______.

120 | |

180 | |

240 | |

285 |

Question 1 Explanation:

T_L=2.78 \times \omega

After neglecting rotational losses,

(T_{sh}=T_a)=T_L

At rated load,

\begin{aligned} T_a&=500Nm=2.78 \times \omega \\ \omega _r&=179.85 \\ \omega _r&\simeq 180 \text{ rad/sec} \end{aligned}

After neglecting rotational losses,

(T_{sh}=T_a)=T_L

At rated load,

\begin{aligned} T_a&=500Nm=2.78 \times \omega \\ \omega _r&=179.85 \\ \omega _r&\simeq 180 \text{ rad/sec} \end{aligned}

Question 2 |

The separately excited dc motor in the figure below has a rated armature current
of 20 A and a rated armature voltage of 150 V. An ideal chopper switching
at 5 kHz is used to control the armature voltage. If L_a=0.1mH, R_a=1\Omega,
neglecting armature reaction, the duty ratio of the chopper to obtain 50% of the
rated torque at the rated speed and the rated field current is

0.4 | |

0.5 | |

0.6 | |

0.7 |

Question 2 Explanation:

\begin{aligned} V_a &=E_b+I_aR_a \\ 150&=E_b+20 \times 1 \\ E_b&= 130V\\ \text{For half} &\text{ the rated torque,} \\ I_a&=10A \\ V_a&=130+10 \times 1=140 \\ D&=\frac{140}{200}=0.7 \end{aligned}

Question 3 |

A 3-phase squirrel cage induction motor supplied from a balanced 3-phase source drives a mechanical load. The torque-speed characteristics of the motor(solid curve) and of the load(dotted curve) are shown. Of the two equilibrium points A and B, which of the following options correctly describes the stability of A and B ?

A is stable, B is unstable | |

A is unstable, B is stable | |

Both are stable | |

Both are unstable |

Question 3 Explanation:

Let torque developed by the motor =T_m

Load torque =T_L

Accelerating torque,T_a=T_m-T_L

At point A,

T_m=T_L

\Rightarrow \;\;T_a=T_m-T_L=0

So, no accelerating , hence the motor rums at speed N_A.

When a smal disturbance decreases speed to N_c

At point C,

T_m \gt T_L

\Rightarrow \;\;T_a \gt 0

So, motor will accelerate and comes to point A. When the disturbance increases speed to N_D.

At point D,

T_m \lt T_L

\Rightarrow \;\;T_a \lt 0

Motor will decelerate and comes at point A

So, motor is stable at point A.

When the motor is operating at point B.,

T_m = T_L

\Rightarrow \;\;T_a= 0

If a small disturbance decreases speed to N_E.

T_m \lt T_L

\Rightarrow \;\;T_a \lt 0

Motor will decelerate and motor speed will keep on decreasing.

If the disturbance increases speed to N_F.

At point F,

T_m \gt T_L

\Rightarrow \;\;T_a \gt 0

Motor will accelerate and motor speed will keep on increasing.

So, motor is unstable at point B.

Question 4 |

A single phase fully controlled converter bridge is used for electrical braking of a
separately excited dc motor. The dc motor load is represented by an equivalent
circuit as shown in the figure.

Assume that the load inductance is sufficient to ensure continuous and ripple free load current. The firing angle of the bridge for a load current of I_0=10 A will be

Assume that the load inductance is sufficient to ensure continuous and ripple free load current. The firing angle of the bridge for a load current of I_0=10 A will be

44^{\circ} | |

51^{\circ} | |

129^{\circ} | |

136^{\circ} |

Question 4 Explanation:

Average output voltage of the converter,

V_0=\frac{2V_m}{\pi}\cos \alpha

Load current =I_0=10A

Back emf =E_b=150V

Armature resistance=R_a=2\Omega

Applying KVL,

\begin{aligned} &V_0-2I_0+150=0 \\ \Rightarrow \;& V_0=-150+2 \times 10\\ &=-130V\\ \Rightarrow \; & \frac{2V_m}{\pi} \cos \alpha =-130\\ \Rightarrow \; &\frac{2 \times \sqrt{2} \times 230}{\pi}\cos \alpha=-130 \\ \Rightarrow \;&\alpha =129^{\circ} \end{aligned}

V_0=\frac{2V_m}{\pi}\cos \alpha

Load current =I_0=10A

Back emf =E_b=150V

Armature resistance=R_a=2\Omega

Applying KVL,

\begin{aligned} &V_0-2I_0+150=0 \\ \Rightarrow \;& V_0=-150+2 \times 10\\ &=-130V\\ \Rightarrow \; & \frac{2V_m}{\pi} \cos \alpha =-130\\ \Rightarrow \; &\frac{2 \times \sqrt{2} \times 230}{\pi}\cos \alpha=-130 \\ \Rightarrow \;&\alpha =129^{\circ} \end{aligned}

Question 5 |

A three-phase, 440 V, 50 Hz ac mains fed thyristor bridge is feeding a 440 V dc,
15 kW, 1500 rpm separately excited dc motor with a ripple free continuos current
in the dc link under all operating conditions, Neglecting the losses, the power
factor of the ac mains at half the rated speed is

0.354 | |

0.372 | |

0.9 | |

0.955 |

Question 5 Explanation:

For a separately excited dc motor

back emf =E_a=V_0-I_aR_a

Since, losses are neglected R_a can be neglected

So,

\begin{aligned} E_a&=V_0 \\ V_0&=E_a =k_a\phi N \;\;...(i)\\ V_0&\propto N \end{aligned}

At rated voltage V_0=440V and N=1500rpm, so, at half the rated speed \left ( \frac{N}{2} =750 \;rpm\right ) output voltage of the bridge (V_0) is 220V.

if I_a is the average value of armature current rms value of supply current will be

I_s=I_a\sqrt{\frac{2}{3}}

Power delivered to the motor

P_0=V_0I_a

Input VA to the thyristor bridge

S_{in}=\sqrt{3}V_sI_s

Input power factor

=\frac{P_0}{S_{in}}=\frac{V_0I_a}{\sqrt{3}V_sI_s}=\frac{220 \times I_a}{\sqrt{3} \times 440 \times I_a\sqrt{\frac{2}{3}}}=0.345

Question 6 |

A solar cell of 350 V is feeding power to an ac supply of 440 V, 50 Hz through a
3-phase fully controlled bridge converter. A large inductance is connected in the
dc circuit to maintain the dc current at 20 A. If the solar cell resistance is 0.5 \Omega, then each thyristor will be reverse biased for a period of

125^{\circ} | |

120^{\circ} | |

60^{\circ} | |

55^{\circ} |

Question 6 Explanation:

Solar cell emf = 350V

DC current I_{dc}=20A

Solar cell resistance

\begin{aligned} R_{cell}&=0.5\Omega \\ V_0&=\text{Voltage across inverter} \\ &=-(E-I_{dc}R_{cell}) \\ &=-(350-20 \times 0.5)=-340V \end{aligned}

The bridge acts as inverter,

Output voltage of 3-\phi fully controlled bridge

\begin{aligned} V_0&=\frac{3V_{ml} \cos \alpha }{\pi}\\ -340&=\frac{3V_{ml} \cos \alpha }{\pi}\\ -340&=\frac{3 \times 440\sqrt{2} \cos \alpha }{\pi}\\ \Rightarrow \; \alpha &=125^{\circ} \end{aligned}

Therefore, each thyristor will be reverse biased for a period of 55^{\circ}.

DC current I_{dc}=20A

Solar cell resistance

\begin{aligned} R_{cell}&=0.5\Omega \\ V_0&=\text{Voltage across inverter} \\ &=-(E-I_{dc}R_{cell}) \\ &=-(350-20 \times 0.5)=-340V \end{aligned}

The bridge acts as inverter,

Output voltage of 3-\phi fully controlled bridge

\begin{aligned} V_0&=\frac{3V_{ml} \cos \alpha }{\pi}\\ -340&=\frac{3V_{ml} \cos \alpha }{\pi}\\ -340&=\frac{3 \times 440\sqrt{2} \cos \alpha }{\pi}\\ \Rightarrow \; \alpha &=125^{\circ} \end{aligned}

Therefore, each thyristor will be reverse biased for a period of 55^{\circ}.

Question 7 |

The speed of a 3-phase, 440 V, 50 Hz induction motor is to be controlled over a
wide range from zero speed to 1.5 time the rated speed using a 3-phase voltage
source inverter. It is desired to keep the flux in the machine constant in the
constant torque region by controlling the terminal voltage as the frequency
changes. The inverter output voltage vs frequency characteristic should be

A | |

B | |

C | |

D |

Question 7 Explanation:

For a 3-\phi induction motor, stator voltage per phase is given by

\begin{aligned} V_1 &=\sqrt{2}\pi f_1 N_{pn_1}\phi K_{\omega 1} \\ \phi &\propto \frac{V_1}{f_1} \end{aligned}

To keep the \phi constant, constant V/f control is applied and voltage (V_1) varies linearly with frequency (f_1). For frequency above 50 Hz, such control is not possible because voltage can not be increased above rated voltage. Therefore, for frequency above 50 Hz, voltage is kept constant, During this control induction motor is said to be working in field weakling mode.

\begin{aligned} V_1 &=\sqrt{2}\pi f_1 N_{pn_1}\phi K_{\omega 1} \\ \phi &\propto \frac{V_1}{f_1} \end{aligned}

To keep the \phi constant, constant V/f control is applied and voltage (V_1) varies linearly with frequency (f_1). For frequency above 50 Hz, such control is not possible because voltage can not be increased above rated voltage. Therefore, for frequency above 50 Hz, voltage is kept constant, During this control induction motor is said to be working in field weakling mode.

Question 8 |

An electric motor, developing a starting torque of 15 Nm, starts with a load
torque of 7 Nm on its shaft. If the acceleration at start is 2 rad/sec^{2} , the moment of inertia of the system must be (neglecting viscous and coulomb friction)

0.25 kgm^{2} | |

0.25 Nm^{2} | |

4 kgm^{2} | |

4 Nm^{2} |

Question 8 Explanation:

T_s= Starting torque developed by the motor =15 N-m

T_L= Load torque =7 N-m

T_a= Accelerating torque =T_s-T_l= 15-7=8 N-m

\alpha =Acceleration = 2 \; rad/sec^2

T_a=I\alpha

I= Moment of inertia =\frac{T_a}{\alpha }=\frac{8}{2}=4kgm^2

T_L= Load torque =7 N-m

T_a= Accelerating torque =T_s-T_l= 15-7=8 N-m

\alpha =Acceleration = 2 \; rad/sec^2

T_a=I\alpha

I= Moment of inertia =\frac{T_a}{\alpha }=\frac{8}{2}=4kgm^2

Question 9 |

A variable speed drive rated for 1500 rpm, 40 Nm is reversing under no load. Figure shows the reversing torque and the speed during the transient. The moment of inertia of the drive is

0.048 kg m^{2} | |

0.064 kg m^{2} | |

0.096 kg m^{2} | |

0.128 kg m^{2} |

Question 9 Explanation:

Speed changes from -1500 rpm to 500 rpm in 0.5 sec

So angular acceleration,

\begin{aligned} \alpha &=\frac{500-(-1500)}{0.5}\times \frac{2 \pi}{60}\\ &=418.88 \;rad/sec^2\\ \text{Torque}&=T=20N-m\\ T&=I\alpha \\ &\text{Moment of inertia}\\ I&=\frac{T}{\alpha }=\frac{20}{418.88}=0.048kgm^2 \end{aligned}

So angular acceleration,

\begin{aligned} \alpha &=\frac{500-(-1500)}{0.5}\times \frac{2 \pi}{60}\\ &=418.88 \;rad/sec^2\\ \text{Torque}&=T=20N-m\\ T&=I\alpha \\ &\text{Moment of inertia}\\ I&=\frac{T}{\alpha }=\frac{20}{418.88}=0.048kgm^2 \end{aligned}

Question 10 |

A single-phase half-controlled rectifier is driving a separately excited dc motor.
The dc motor has a back emf constant of 0.5 V/rpm. The armature current is 5A without any ripple. The armature resistance is 2\Omega. The converter is working
from a 230V, single-phase ac source with a firing angle of 30^{\circ}. Under this
operating condition, the speed of the motor will be

339 rpm | |

359 rpm | |

366 rpm | |

386 rpm |

Question 10 Explanation:

\begin{aligned} \text{Back emf }&=E_a=k\phi N\\ E_a&=k_bN \end{aligned}

where,

k_b= Back-emf constant =0.25 V/rpm

Average output voltage of 1-\phi half controlled rectifier =V

\begin{aligned} V &=\frac{V_m}{\pi} (1+ \cos \alpha ) \\ &= \frac{230\sqrt{2}}{2 \pi} (1+ \cos 30^{\circ})\\ \Rightarrow \; V &=96.6V \\ E_a&=V-I_aR_a \\ &= 96.6-5 \times 2=86.6V\\ \text{Speed}&=N=\frac{E_a}{k_b}=\frac{86.6}{0.25}=346.4V \end{aligned}

There are 10 questions to complete.