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
There are 5 questions to complete.