A 20 H.P., 3-phase, 6-pole, 50 Hz, 400 V induction motor runs at 960 rpm on full load. If it takes 120 A on direct starting, find the ratio of starting torque to full load torque with a star-delta starter. Full load efficiency and p.f. are 90% and 0.85.

Question

Accepted Answer

[latex]Output power = 20 H.P. = 20 \times 735.5 = 14710 W [/latex]

No. of poles, [latex]P = 6[/latex]

Supply voltage (line value), [latex]V_{L} = 400 V[/latex]

Supply frequency, [latex]f = 50 Hz[/latex]

Short circuit current, [latex]I_{sc} = 120 A [/latex]

Efficiency, [latex]\eta = 90\% = 0.9[/latex]

Power factor, [latex]\cos \phi = 0.85 [/latex]

Rotor speed, [latex]N = 960 rpm [/latex]

Full load current, [latex]I_{fl} = \frac{14170}{\sqrt{3} \times 400 \times 0.9 \times 0.85} = 27.75 A[/latex]

Synchronous speed, [latex]N_{s} = \frac{120f}{p} = \frac{120 \times 50}{6} = 1000 r.p.m[/latex]

Full-load slip, [latex]S_{fl} = \frac{N_{2} - N}{N_{2}} = \frac{1000 - 960}{1000} = 0.04[/latex]

Ratio of starting torque to full load torque,

[latex]\frac{T_{st}}{T_{fl}} = \frac{1}{3}\left\lgroup \frac{I_{sc}}{I_{fl}}\right\rgroup^{2} \times S \[0.5cm] \quad \; \, = \frac{1}{3}\left(\frac{120}{27.75} \right)^{2} \times 0.04 = \mathbf{0.2493} [/latex]

Question 10.5: A 20 H.P., 3-phase, 6-pole, 50 Hz, 400 V induction motor run...

Related Answered Questions