Question 1 |
A separately excited DC motor rated 400 \mathrm{~V}, 15A, 1500 RPM drives a constant torque load at rated speed operating from 400 \mathrm{~V} DC supply drawing rated current. The armature resistance is 1.2 \Omega. If the supply voltage drops by 10 \% with field current unaltered then the resultant speed of the motor in RPM is ___ (Round off to the nearest integer).
1343 | |
2541 | |
1145 | |
1852 |
Question 1 Explanation:
Seperately excited motor :
Back emf, \mathrm{E}_{\mathrm{b}_{1}}=400-15 \times 1.2=382 \mathrm{~V}
Given: \quad \tau= constant
\because \quad \tau \propto \phi I_{a}
where \phi \rightarrow constant
\therefore \mathrm{I}_{\mathrm{a}} \rightarrow consant
If supply voltage is drops by 10 \%, then new value of supply voltage =0.9 \times 400=360 \mathrm{~V}
Now, back emf, \mathrm{E}_{\mathrm{b}_{2}}=360-15 \times 1.2=342 \mathrm{~V}
We have,
\begin{aligned} \mathrm{E}_{\mathrm{b}} & =\phi \omega_{\mathrm{m}} \\ \therefore \quad \frac{\mathrm{E}_{\mathrm{b}_{2}}}{\mathrm{E}_{\mathrm{b}_{1}}} & =\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}} \\ \Rightarrow \quad \frac{342}{382} & =\frac{\mathrm{N}_{2}}{1500} \\ \Rightarrow \quad \mathrm{N}_{2} & =342.93 \mathrm{rpm} \\ &\approx 1343 \mathrm{rpm} \end{aligned}
Back emf, \mathrm{E}_{\mathrm{b}_{1}}=400-15 \times 1.2=382 \mathrm{~V}
Given: \quad \tau= constant
\because \quad \tau \propto \phi I_{a}
where \phi \rightarrow constant
\therefore \mathrm{I}_{\mathrm{a}} \rightarrow consant
If supply voltage is drops by 10 \%, then new value of supply voltage =0.9 \times 400=360 \mathrm{~V}
Now, back emf, \mathrm{E}_{\mathrm{b}_{2}}=360-15 \times 1.2=342 \mathrm{~V}
We have,
\begin{aligned} \mathrm{E}_{\mathrm{b}} & =\phi \omega_{\mathrm{m}} \\ \therefore \quad \frac{\mathrm{E}_{\mathrm{b}_{2}}}{\mathrm{E}_{\mathrm{b}_{1}}} & =\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}} \\ \Rightarrow \quad \frac{342}{382} & =\frac{\mathrm{N}_{2}}{1500} \\ \Rightarrow \quad \mathrm{N}_{2} & =342.93 \mathrm{rpm} \\ &\approx 1343 \mathrm{rpm} \end{aligned}
Question 2 |
A 280 V, separately excited DC motor with armature resistance of 1\Omega and constant
field excitation drives a load. The load torque is proportional to the speed. The
motor draws a current of 30 A when running at a speed of 1000 rpm. Neglect
frictional losses in the motor. The speed, in rpm, at which the motor will run, if an
additional resistance of value 10\Omega is connected in series with the armature, is
_______. (round off to nearest integer)
482 | |
254 | |
365 | |
625 |
Question 2 Explanation:
Back emf,
E_{b1}=280-30 \times 1 =250V
When additional 10\Omega resistance connected in series with armature.
E_{b2}=280-(10+1) I_{a_2} =280-11I_{a_2}
Given: \tau \propto N
We know, \tau \propto \phi I_a \text{ (Here, }\phi \rightarrow constant)
So,
\begin{aligned} \tau &\propto I_a \\ \therefore \; N &\propto I_a \\ or \; \frac{N_2}{N_1}&=\frac{I_{a_2}}{I_{a_1}}\\ I_{a_2}&=N_2 \times \frac{30}{1000}\\ \therefore \; E_{b_2}&=280-\left ( 11 \times \frac{30}{1000} \right )N_2\\ &\text{We also have,}\\ E_b &\propto\phi \omega \\ \therefore \frac{E_{b_2}}{E_{b_1}}&=\frac{N_2}{N_1}\\ &\frac{280-\left ( 11 \times \frac{30}{1000} \right )N_2}{250}=\frac{N_2}{1000}\\ \Rightarrow N_2&=482.758rpm \end{aligned}
E_{b1}=280-30 \times 1 =250V
When additional 10\Omega resistance connected in series with armature.
E_{b2}=280-(10+1) I_{a_2} =280-11I_{a_2}
Given: \tau \propto N
We know, \tau \propto \phi I_a \text{ (Here, }\phi \rightarrow constant)
So,
\begin{aligned} \tau &\propto I_a \\ \therefore \; N &\propto I_a \\ or \; \frac{N_2}{N_1}&=\frac{I_{a_2}}{I_{a_1}}\\ I_{a_2}&=N_2 \times \frac{30}{1000}\\ \therefore \; E_{b_2}&=280-\left ( 11 \times \frac{30}{1000} \right )N_2\\ &\text{We also have,}\\ E_b &\propto\phi \omega \\ \therefore \frac{E_{b_2}}{E_{b_1}}&=\frac{N_2}{N_1}\\ &\frac{280-\left ( 11 \times \frac{30}{1000} \right )N_2}{250}=\frac{N_2}{1000}\\ \Rightarrow N_2&=482.758rpm \end{aligned}
Question 3 |
A belt-driven \text{DC}
shunt generator running at 300\: \text{RPM}
delivers 100 \:\text{kW}
to a 200 \:V\: \text{DC}
grid. It continues to run as a motor when the belt breaks, taking 10 \:\text{kW}
from the \text{DC}
grid. The armature resistance is 0.025 \Omega, field resistance is 50\:\Omega, and brush drop is 2 \:V. Ignoring armature reaction, the speed of the motor is _____ \text{RPM}. (Round off to 2 decimal places.)
275.18 | |
254.12 | |
362.24 | |
148.44 |
Question 3 Explanation:
\begin{aligned} I_{L} &=\frac{100 \times 10^{3}}{200}=500 \mathrm{~A} \\ I_{S h} &=\frac{200}{50}=4 \mathrm{~A} \\ I_{a} &=504 \mathrm{~A} \\ E_{q} &=V+I_{a} R_{a}+\text { Brush drop } \\ &=200+504(0.025)+2 \mathrm{~V} \\ &=214.6 \mathrm{~V}\\ \text{In motoring case:} V I&=10 \mathrm{~kW}, V=200 \mathrm{~V}\\ \therefore \qquad \qquad I &=\frac{10000}{200}=50 A \\ I_{f} &=4 \mathrm{~A}, I_{a}=I_{L}-I_{f} \\ &=46 \mathrm{~A} \\ E_{b} &=V-I_{a} R_{a}-\text { Brush Drop } \\ &=200-46(0.025)-2=196.85 \mathrm{~V} \\ \frac{N_{m}}{N_{g}} &=\frac{E_{b}}{E_{g}} \\ N_{m} &=\frac{E_{b}}{E_{g}} \times N_{g} \\ \therefore \qquad \qquad N_{m} &=\frac{196.85}{214.6} \times 300=275.18 \mathrm{rpm} \end{aligned}
Question 4 |
Consider a permanent magnet dc (PMDC) motor which is initially at rest. At t=0, a
dc voltage of 5 V is applied to the motor. Its speed monotonically increases from 0 rad/s
to 6.32 rad/s in 0.5 s and finally settles to 10 rad/s. Assuming that the armature
inductance of the motor is negligible, the transfer function for the motor is
\frac{10}{0.5s+1} | |
\frac{2}{0.5s+1} | |
\frac{2}{s+0.5} | |
\frac{10}{s+0.5} |
Question 4 Explanation:
\begin{aligned}
\text{Input} &= 5 V \\ \Rightarrow \; R(s)&=\frac{5}{s} \\ \frac{C(s)}{R(s)}&=T.F.=\frac{K}{1+Ts} \\ C(s)&=\frac{5K}{s(1+Ts)} \\ \lim_{s\rightarrow 0}&=5K=10\\ \Rightarrow \; K&=2\\ \text{Steady state speed}&=10 \text{ rad/sec.(Given)} \\ C(s)&=\frac{2}{1+0.5s}\times \frac{5}{s} \\ \Rightarrow \; \lim_{s\rightarrow 0}sC(s)&=10
\end{aligned}
Question 5 |
A 250 V dc shunt motor has an armature resistance of 0.2\Omega and a field resistance of
100\Omega. When the motor is operated on no-load at rated voltage. It draws an armature
current of 5 A and runs at 1200 rpm. When a load is coupled to the motor, it draws
total line current of 50 A at rated voltage, with a 5% reduction in the air-gap flux due
to armature reaction. Voltage drop across the brushes can be taken as 1 V per brush
under all operating conditions. The speed of the motor, in rpm, under this loaded
condition, is closest to:
1200 | |
1000 | |
1220 | |
900 |
Question 5 Explanation:
no load current 5 A
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\begin{aligned}
B.R.D&=\text{1 V per brush}\\ \text{loaded, }I_{L}&=50 A \\R_{sh}&=100 \Omega\\ I_{sh}&=\frac{250}{100}=2.5\: A \\I_{a0}&=2.5\: A\\I_{aL}&=47.5\: A \\V&=E_{b}+I_{a}R_{a}+B.R.D\\ E_{b\, no\, load}&=V-I_{a0}R_{a}-B.R.D \\&=250-2.5(0.2)-1\times 2 \\&=247.5\, Volts \\E_{b\, load}&=250-47.5(0.2)-1\times 2 \\&=238.5\, volts\\ \frac{N_{2}}{N_{1}}&=\frac{E_{b_{2}}}{E_{b_{1}}}\times \frac{\phi_{1} }{\phi _{2}} \\ \frac{N_{2}}{1200}&=\frac{238.5}{247.5}\times \frac{\phi _{1}}{0.95\phi_{1}}\\ N_{2}&=1217.22\, rpm
\end{aligned}
There are 5 questions to complete.