Q41GATE 2006MCQ

A 3-phase, 10 kW, 400 V, 4-pole, 50Hz, star connected induction motor draws 20 A on full load. Its no load and blocked rotor test data are given below. Neglecting copper loss in no load test and core loss in blocked rotor test, estimate motor's full load efficiency

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📖 Explanation

No-load losses, $P_1=1002W=1.002kW$ (does not depends on load) Copper loss at 15A $P_{cu}=762W$ Full load current =20A So, copper loss on full load

\begin{aligned} P_{cu,fl} &= (P_{cu}) \left ( \frac{20}{15} \right )^2\\ P_{cu,fl} &= 762 \times \left ( \frac{20}{15} \right )^2=1.354 kW \\ & \text{Total losses,}\\ P-1 +P_{cu,fl} &=1.002+1.354 \\ &= 2.356kW \end{aligned}

Power available at shaft or output power on full load $P_0=10kW$ Input power=Output power +losses $=10+2.356=12.356kW$

\begin{aligned} \% \text{efficiency} &=\frac{\text{Output power}}{Input Power}\times 100 \\ &= \frac{10}{12.356}\times 100\approx 81\% \end{aligned}

Q43GATE 2005MCQ

For an induction motor, operation at a slip $s$ , the ration of gross power output to air gap power is equal to

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📖 Explanation

Let, $P_g=$ Air gap power $P_{cu}=$ Rotor copper loss $=sP_g$ $P_m=$ Gross power output =Air gap power - Rotor copper loss $P_m=P_g-sP_g=P_g(1-s)$ $\Rightarrow \;\;\frac{P_m}{P_g}=(1-s)$

Q44GATE 2005MCQ

On the torque/speed curve of the induction motor shown in the figure, four points of operation are marked as W, X, Y and Z. Which one of them represents the operation at a slip greater than 1 ?

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📖 Explanation

In braking mode, the motor runs in opposite direction to rotating field i.e. $N_f$ (rotor speed) is negative, $s=\frac{N_s -(-N_r)}{N_s}=\frac{N_s +N_r}{N_s} \gt 1$

Q45GATE 2005MCQ

Determine the correctness or otherwise of the following assertion[A] and the reason[R]. Assertion [A] : Under V/f control of induction motor, the maximum value of the developed torque remains constant over a wide range of speed in the subsynchronous region. Reason [R] : The magnetic flux is maintained almost constant at the rated value by keeping the ration V/f constant over the considered speed range.

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📖 Explanation

The resultant air-gap flux per pole is given by $\phi _r=\frac{1}{4.44k_{\omega _1}N_{pn_1}}\left ( \frac{V}{f} \right )$ $\phi _r\propto \left ( \frac{V}{f} \right )$ Therefore, in order to avoid saturation in stator and rotor cores which would cause sharp increase in magnetization current, the flux $f_p$ must be kept constant. To achieve this, $\frac{V}{f}$ ratio must be kept constant. Let, $V_0=$ Nominal voltage $f_0=$ Nominal frequency $X'_{20}=$ Nominal rotor standstill reactance reffered to stator then at any frequency f. Then at any frequency f,

\begin{aligned} V &=\left ( \frac{f}{f_0} \right ) V_0\\ X'_2 &=\left ( \frac{f}{f_0} \right ) X'_{20} \\ \omega _s &=\left ( \frac{f}{f_0} \right )\omega _{20} \\ T_{max}&= \frac{3}{\left ( \frac{f}{f_0} \right )\omega _{s_0}}\left[ \frac{0.5 \left ( \frac{f}{f_0} \right )^2 V_0^2}{\left ( \frac{f}{f_0} \right )X'_{20}} \right]\$\\ &= \frac{3}{\omega _{s_0}}\frac{0.5V_0^2}{X'_{20}}\\ &=\text{constant, independent of f} \end{aligned}

Q46GATE 2005MCQ

Under no load condition, if the applied voltage to an induction motor is reduced from the rated voltage to half the rated value,

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📖 Explanation

Let no-load current $=I_0$ Equivalent-circuit model of induction motor at no-load is drawn. If voltage is reduced, no-load current ( $I_0$ ) reduces. Slip is extremely low (of the order 0.001) at no-load, even if voltage is reduced, there is negligible change in the speed.

Q47GATE 2005MCQ

A three-phase cage induction motor is started by direct-on-line (DOL) switching at the rated voltage. If the starting current drawn is 6 times the full load current, and the full load slip is 4%, then ratio of the starting developed torque to the full load torque is approximately equal to

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📖 Explanation

Torque in induction motor is given by,

\begin{aligned} T &=\frac{3}{\omega _s}(I'_2)\frac{R'_2}{s} \\ & \text{Full load torque}\\ T_{fl} &= \frac{2}{\omega _s}(I'_{2fl})^2 \frac{R'_2}{S_{fl}}\\ & \text{At starting,}\\ s &= 1\\ & \text{Starting torque}\\ T_{st}&= \frac{3}{\omega _s}(I'_{2st})^2 R'_2\\ \frac{T_{st}}{T_{fl}}&=\left ( \frac{I'_{2st}}{I'_{2fl}} \right )^2 S_{fl} \\ &= 6^2 \times 0.04=1.44 \end{aligned}

Q48GATE 2004MCQ

The direction of rotation of a 3-phase induction motor is clockwise when it is supplied with 3-phase sinusoidal voltage having phase sequence A-B-C. For counter clockwise rotation of the motor, the phase sequence of the power supply should be

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Q49GATE 2004MCQ

The synchronous speed for the seventh space harmonic mmf wave of a 3-phase, 8-pole, 50 Hz induction machine is

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📖 Explanation

Synchronous speed $N_s=\frac{120f}{P}=\frac{120 \times 50}{8}$ $=750 \; \text{rpm}$ mmf wave contain space harmonics e.g. fifth and seventh harmonics which correspond to poles five and seven time that of the fundamental. Since the space-phase difference between fundamental poles of the winding phase is $(0^{\circ},120^{\circ},240^{\circ})$ , this difference is $(0^{\circ},240^{\circ},120^{\circ})$ for fifth harmonic poles and $(0^{\circ},120^{\circ},240^{\circ})$ for the seventh. Hence, the fifth harmonic poles rotate backwards with synchronous speed $N_s/5$ and the seventh harmonic poles rotate forward at $\frac{N_s}{7}=\frac{750}{7}=107.14 \; \text{rpm}$

Q50GATE 2004MCQ

A 400 V, 15 kW, 4-pole, 50Hz, Y-connected induction motor has full load slip of 4%. The output torque of the machine at full load is

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📖 Explanation

\begin{aligned} N_s(\text{Syn. speed}) &= \frac{120f}{P}\\ &= \frac{120 \times 50}{4}=1500\; \text{rpm}\\ \text{Torque developed} &= T=\frac{P}{\omega _s(1-s)}\\ &= \frac{15 \times 10^3 \times 60}{2 \pi \times 1440}\\ &= 99.47Nm \end{aligned}

Q51GATE 2003MCQ

An ac induction motor is used for a speed control application. It is driven from an inverter with a constant V/f control. The motor name-plate details are as follows (no. of poles = 2) V:415 V Ph:3 f:50 Hz N:2850 rpm The motor runs with the inverter output frequency set at 40 Hz, and with half the rated slip. The running speed of the motor is

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📖 Explanation

$N_{S1}=3000 \; \text{rpm}$ $N_1=2850 \; \text{rpm}$ $S_{FL}=\frac{3000-2850}{3000}=0.05$ where, by (V/f) control, $N_{S2}=3000\left ( \frac{40}{50} \right )=2400 \; \text{rpm}$ (synchronous speed at 40 Hz) $N_2=$ New running speed of motor $N_{S2}\left ( 1-\frac{S_{FL}}{2} \right )=2400\left ( 1-\frac{0.05}{2} \right )$ $=2340 \; \text{rpm}$

Q52GATE 2003MCQ

No-load test on a 3-phase induction motor was conducted at different supply voltage and a plot of input power versus voltage was drawn. This curve was extrapolated to intersect the y-axis. The intersection point yields

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📖 Explanation

Power input at no-load $(P_0)$ provided losses only as the shaft output is zero. These losses comprise $P_i$ (Iron/core loss) and $P_{wf}$ (windage and friction loss). As the voltage is reduced below the rated value, the core-loss decreases as the square of voltage. Since the slip does not increase significantly, the windage and friction loss remains almost constant. The plot of $P_0$ versus V is extrapolated at V=0 which gives, $P_{wf}$ as $P_{i}=0$ art zero voltage.

Q53GATE 2003MCQ

A 3-phase induction motor is driving a constant torque load at rated voltage and frequency. If both voltage and frequency are halved, following statements relate to the new condition if stator resistance, leakage reactance and core loss are ignored 1. The difference between synchronous speed and actual speed remains same 2. The airgap flux remains same 3. The stator current remains same 4. The p.u. slip remains same Among the above, correct statements are

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📖 Explanation

\begin{aligned} V &\propto f\phi \\ &\text{\in induction machine}\\ V_2&= \frac{V_1}{2}\\ \text{and}\;\; f_2 &=\frac{f_1}{2} \\ \frac{V_1}{V_2} &=\frac{f_1\phi _1}{f_2\phi _2} \\ \phi _2&= \left ( \frac{V_2}{V_1} \right )\left ( \frac{f_1}{f_2} \right )\phi _1\\ &= \left ( \frac{V_1/2}{V_1} \right )\left ( \frac{f_1}{f_1/2} \right )\phi _1\\ &= \phi _1 \end{aligned}

So, air-gap flux remains same. $(X_2')_1=$ nominal rotor stanstill reactance referred to stator (at frequency $f_1$ )

\begin{aligned} (X_2')_2 &= \left ( \frac{f_2}{f_1} \right )(X_2')_1\\ &= \left ( \frac{f_1/2}{f_1} \right )(X_2')_1\\ &= \frac{(X_2')_1}{2}\\ (\omega _s)_2 &= \left ( \frac{f_2}{f_1} \right ) (\omega _s)_1\\ &= \left ( \frac{f_1/2}{f_1} \right ) (\omega _s)_1\\ &=\frac{(\omega _s)_1}{2} \end{aligned}

Torque developed in induction motor $T=\frac{3}{\omega _s} \frac{V^2R'_2/s}{(R'_2/s)^2 +{X'_2}^2}$ For normal steady state oprating region, slip is very low $\frac{R'_2}{s} \gt \gt X'_2$ $T=\frac{3}{\omega _s} \frac{V^2R'_2/s}{(R'_2/s)^2}=\frac{3}{\omega _s} \frac{sV^2}{R'_2}$ As the load torque is constant

\begin{aligned} T_1 &=T_2 \\ \frac{s_1V_1^2}{(\omega _s)_1R'_2}&= \frac{s_2V_2^2}{(\omega _s)_2R'_2}\\ \Rightarrow \;\; s_2 &=\frac{(\omega _s)_2}{(\omega _s)_1} \times \left ( \frac{V_1}{V_2} \right )^2 s_1\\ &=\frac{(\omega _s)_1/2}{(\omega _s)_1} \times \left ( \frac{V_1}{V_1/2} \right )^2 s_1 \\ s_2 &= \frac{1}{2} \times 4 \times s_1=2s_1 \end{aligned}

Difference between synchronous speed and actual speed

\begin{aligned} \omega _s -\omega _r &= s\omega _s\\ (\omega _s)_1 -(\omega _r)_2 &= s_2(\omega _s)_2\\ &=(2s_1) \times \frac{(\omega _s)_1}{2} =s_1(\omega _s)_1 \end{aligned}

Therefore, the difference between synchronous speed and actual speed remain same. Stator current

\begin{aligned} |I_s| &= \frac{V}{\sqrt{(R'_2/s)^{2+}{X'}_2^2}} \\ \text{as}\;\; R'_2/s & \gt \gt X'_2 \\ |I_s|&=\frac{sV}{R'_2} \\ |I_{s1}|&= s_1\frac{V_1}{R'_2}\\ |I_{s2}|&=s_2\frac{V_2}{R'_2} \\ ||I_{s2}&=(2s_1)\times \frac{V_1/2}{R'_2}=\frac{s_1V_1}{R'_2} \\ ||I_{s1}&=|I_{s2}| \end{aligned}

Therefore, stator current remains same.

Q54GATE 2002MCQ

The rotor of a three phase, 5 kW, 400V, 50 Hz, slip ring induction motor is wound for 6 poles while its stator is wound for 4 poles. The approximate average no load steady state speed when this motor is connected to 400V, 50 Hz supply is

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Q55GATE 2002MCQ

If a 400V, 50 Hz, star connected, 3 phase squirrel cage induction motor is operated from a 400 V, 75 Hz supply, the torque that the motor can now provide while drawing rated current from the supply?

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📖 Explanation

$T=3 I^2\frac{R_2}{s}\frac{60}{2 \pi N_s}$ As the motor is drawing rated current from the supply.

\begin{aligned} T &=\propto \frac{1}{N_s} \\ T & \propto \frac{1}{\text{frequency}} \;\;\left[ \because \; N_s=\frac{120f}{P} \right] \end{aligned}

Hence as frequency increases, torque decreases.