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Wind Energy Conversion Systems Questions and Answers

1.

A horizontal axis drag type wind mill with square blades and a horizontal axis lift type wind mill with airfoil section blades having same rotor size are installed at a height of 10 m above the ground. The average wind speed is 25 km h−1. The maximum power coefficient for drag type and lift type wind mills is 0.148 and 0.593, respectively. If the maximum power extracted by drag type wind mill is 5 kW, the corresponding power extracted by lift type wind mill, in kW is

Exam Year(s)

GATE AG - 2012,

Answer: D

The lift type wind mill extracts about 20.03 kW.

Power extracted is \(P = C_{p} \cdot \tfrac{1}{2}\rho A V^{3}\). Both machines have the same rotor size, height and wind, so the available wind power \(\tfrac{1}{2}\rho A V^{3}\) is the same and the extracted power scales with the power coefficient alone:

\[ \begin{aligned} P_{lift} &= P_{drag} \frac{C_{p,lift}}{C_{p,drag}} \\ &= 5 \times \frac{0.593}{0.148} \\ &= 20.03\ \text{kW} \end{aligned} \]

The wind speed and height are not needed. 0.593 is the Betz limit, the theoretical maximum, which is why the airfoil (lift) machine can extract about four times what the drag machine can. The smaller options do not follow from the ratio.

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2.

A small horizontal-axis wind turbine (HAWT) is being designed to deliver 2 kW shaft power at a steady wind speed of 7 m.s−1. The HAWT operates with an aerodynamic power coefficient of 0.36, drivetrain efficiency of 0.90, and air density of 1.225 kg.m−3. Neglecting other losses, the required rotor diameter is X m. When the wind speed drops to 6 m.s−1, the shaft power is Y kW. Identify the nearest correct option for X and Y combination

(Take π = 3.14)

Exam Year(s)

GATE AG - 2026,

Answer: A

The nearest combination is X = 6.12 m and Y = 1.28 kW.

Shaft power is the power in the wind through the swept area, times the power coefficient and the drivetrain efficiency:

\[ \begin{aligned} P &= \tfrac{1}{2}\rho A V^{3} C_{p}\eta \\ 2000 &= 0.5(1.225)A(343)(0.36)(0.90) \\ A &= \frac{2000}{68.07} = 29.38\ \text{m}^{2} \\ D &= \sqrt{\frac{4A}{\pi}} = \sqrt{37.43} = 6.12\ \text{m} \end{aligned} \]

With the same rotor, power varies as the cube of wind speed:

\[ \begin{aligned} Y &= 2\left(\frac{6}{7}\right)^{3} \\ &= 2 \times 0.630 = 1.26\ \text{kW} \end{aligned} \]

That is 1.26 kW, and the option set offers 1.28 alongside the correct diameter, so it is the nearest correct combination, as the question asks. X = 3.06 halves the diameter (it is the radius), X = 6.12 with Y = 2.00 ignores the fall in wind speed, and X = 6.75 does not come from the power equation.

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3.

A grid-connected induction generator is used to generate electricity from a wind turbine. Which one of the following statements is true while the generator is feeding power to the grid?

Exam Year(s)

GATE XE - 2026,

Answer: C

While feeding power to the grid, the slip of the induction generator is less than zero.

Slip is \(s = (N_{s}-N_{r})/N_{s}\). An induction machine acts as a generator only when the rotor is driven faster than the synchronous speed set by the grid frequency, so \(N_{r} \gt N_{s}\) and the slip is negative, typically a few per cent.

Zero slip means the rotor turns at synchronous speed and no power is converted. Slip between 0 and 1 is motoring, slip of 1 is a stalled rotor, and slip greater than 1 means the rotor turns against the field, braking (plugging) rather than generating.

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