Question

A 3-phase induction motor drives a blower where load torque is directly proportional to speed squared. If the motor operates at 1450 rpm, the maximum current in terms of rated current will be nearly
1. 2.2
2. 3.4
3. 4.6
4. 6.8

Answer 1

Correct Answer - Option 1 : 2.2

\({T_{em}} \propto \frac{{I_2^2{R_2}}}{s}\)

Load torque, \({T_L} \propto N_r^2 \propto {N_s}{\left( {1 - s} \right)^2} \propto {\left( {1 - s} \right)^2}\)

\(\frac{{I_2^2{R_2}}}{s} \propto {\left( {1 - s} \right)^2}\)

\( \Rightarrow {I_2} \propto \sqrt s \left( {1 - s} \right)\)

By neglecting the stator impedance and the no-load current,

\({I_1} = {I_2} \propto \sqrt s \left( {1 - s} \right)\)

For slip at maximum current, \(\frac{{d{I_1}}}{{ds}} = 0\)

\( \Rightarrow \frac{d}{{ds}}\left( {\sqrt s \left( {1 - s} \right)} \right) = 0\)

\( \Rightarrow \frac{1}{2}\sqrt s = \frac{3}{2}s\sqrt s \)

\( \Rightarrow s = \frac{1}{3}\)

Slip at maximum current, s_max = 1/3

Slip at full load, \({s_{fl}} = \frac{{1500 - 1450}}{{1500}} = \frac{1}{{30}}\)

\(\frac{{{I_{max}}}}{{{I_{fl}}}} = \frac{{\sqrt {{s_{max}}} \left( {1 - {s_{max}}} \right)}}{{\sqrt {{s_{fl}}} \left( {1 - {s_{fl}}} \right)}}\)

\( = \frac{{\sqrt {\frac{1}{3}} \left( {1 - \frac{1}{3}} \right)}}{{\sqrt {\frac{1}{{30}}} \left( {1 - \frac{1}{{30}}} \right)}} = 2.2\)