GATE 2024 IN Q.62: solution and answer

Official answer: 17.00 to 17.30Source: Official answer key, GATE 2024 IN (IISc Bengaluru)

1. Neglecting stator impedance, the torque-slip relation near the maximum-torque point is $$\frac{T}{T_{max}}=\frac{2}{\frac{s}{s_{maxT}}+\frac{s_{maxT}}{s}}$$

2. At starting, $s=1$, so $$\frac{T_{st}}{T_{max}}=\frac{2s_{maxT}}{1+s_{maxT}^2}$$

3. Given $T_{st}=1.0T_{fl}$ and $T_{max}=3.0T_{fl}$, so $\frac{T_{st}}{T_{max}}=\frac{1}{3}$.

4. Substituting: $\frac{2s_{maxT}}{1+s_{maxT}^2}=\frac{1}{3}\Rightarrow 6s_{maxT}=1+s_{maxT}^2\Rightarrow s_{maxT}^2-6s_{maxT}+1=0$

5. Solving the quadratic: $$s_{maxT}=\frac{6\pm \sqrt{36-4}}{2}=3\pm 2\sqrt{2}$$

6. Since slip at maximum torque must satisfy $0<s_{maxT}<1$ (starting torque at $s=1$ is only 100% of full load torque, well below the 300% peak, so the peak lies at a smaller slip than the starting slip), the valid root is $$s_{maxT}=3-2\sqrt{2}\approx 0.17157$$

7. Expressed as a percentage: $s_{maxT}\approx 17.16\%$.