GATE AG 2025 • Q57 NAT • 2 Marks

Section 3: Farm Power

Sources of Power

Subtopic: Sources of power on the farm – human, animal, mechanical, electrical, wind, solar and biomass

Question Statement:

A horizontal-axis wind turbine (24 m diameter blades), average wind velocity 6.0 m/s, average air density 1.23 kg/m³, is used for pumping irrigation water. Cp = 0.3, transmission efficiency 90%, pump efficiency 60%. Find the pump discharge (L/s) for a total head of 20 m. (Rounded to 2 decimal places)

Official Answer Key:

49.00 to 50.00

Step-by-Step Solution & Explanation

• **Concept & Wind Power Formula:** - Blade diameter $D = 24\text{ m} \implies \text{Swept area } A = \frac{\pi}{4} D^2 = \frac{\pi}{4} (24)^2 = 452.389\text{ m}^2$. - Wind speed $v = 6.0\text{ m/s}$, air density $\rho_{air} = 1.23\text{ kg/m}^3$. - Total wind power: $P_{wind} = \frac{1}{2} \rho_{air} A v^3 = 0.5 \times 1.23 \times 452.389 \times (6)^3 = 60,104.4\text{ W}$. • **Power Train Efficiencies:** - Power coefficient $C_p = 0.30 \implies P_{rotor} = 0.30 \times 60,104.4 = 18,031.3\text{ W}$. - Transmission efficiency $\eta_t = 0.90 \implies P_{pump\_shaft} = 18,031.3 \times 0.90 = 16,228.2\text{ W}$. - Pump efficiency $\eta_p = 0.60 \implies \text{Water power } WP = 16,228.2 \times 0.60 = 9,736.9\text{ W}$. • **Hydraulic Output & Discharge:** $WP = \rho_w g Q H \implies 9,736.9 = 1000 \times 9.81 \times Q \times 20 = 196,200 \cdot Q$ $Q = \frac{9,736.9}{196,200} = 0.049627\text{ m}^3/\text{s} = 49.63\text{ L/s}$ • **Conclusion:** Pump discharge $= 49.63\text{ L/s}$ (Official range: 49.00 to 50.00).

Practice This in Real CBT Exam Interface

Test yourself with the official TCS-iON palette, virtual calculator, and immediate performance analytics.

Launch CBT Mock Test