Wind Energy Conversion Systems Questions and Answers

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.

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.

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.

Answer: B

Maximum power extraction from the wind occurs when the fractional wind speed decrease at the turbine is 1/3. Treating the turbine as an actuator disc, the power coefficient is proportional to a(1 - a)^2 (with a as the fractional speed decrease); differentiating and setting it to zero gives a = 1/3, and the corresponding maximum fraction of the wind power extracted is 16/27, about 59 per cent, the Betz limit. At a = 1/2 the coefficient is lower, 1/2 x 1/4 = 0.125 against 4/27 = 0.148, and at a = 1 the wind is stopped and no power is extracted.

Answer: C

The coefficient of performance of a wind turbine at maximum power output is given by the Betz limit, 16/27, which is about 0.593. It follows from momentum theory: the maximum power is extracted when the wind speed behind the rotor falls to one third of the speed ahead of it, and no turbine can capture more than 59.3 percent of the kinetic energy of the wind passing through it. The values 13/18, 5/8 and 11/17 are not the Betz limit.

Answer: B

The power obtained from a horizontal wind turbine at maximum efficiency is proportional to the square of its diameter, option B. The power in the wind passing through the rotor is proportional to the swept area, and the area is proportional to the square of the diameter.

\[ P = \tfrac{1}{2}\,\rho A v^{3} C_{p} \]

\[ A = \frac{\pi D^{2}}{4} \Rightarrow P \propto D^{2} \]

Option A, a linear dependence, and option C, the power of three halves, do not follow from the formula. Option D, the square root, is the inverse relation. At maximum efficiency the coefficient is the Betz limit, a constant, and so does not change the dependence on diameter.

Answer: B

Wind turbines start generating at the cut-in speed of about 4 m/s, reach rated power at about 12 to 15 m/s and are shut down at high speed. So the most suitable range of wind speeds is 4 to 12 m/s. The range of 1 to 5 m/s is too low for useful power, and 10 to 20 and 20 to 50 m/s include speeds at which turbines are stopped to protect them. The answer is 4 - 12 m/s.

Answer: C

The power increases 27 times. The power in the wind is proportional to the cube of the wind speed, because the kinetic energy of air passing in a second rises with the square of the speed and the mass flow rate rises with the speed itself.

\[ P \propto v^{3} \]

\[ P_2 / P_1 = 3^{3} = 27 \]

The figures 3 and 9 come from using the first or second power, and 81 from the fourth power. The answer is option C.

Answer: D

The power output of the wind mill is about 7.74 kW, option (D). The power in the wind passing through the vane area is half the product of the air density, the area and the cube of the wind speed, and the conversion efficiency gives the output.

\[ P_{wind} = \tfrac{1}{2} \rho A v^{3} \]

\[ P_{wind} = 0.5(1.29)(30)(10)^{3} \]

\[ P_{wind} = 19350\ \mathrm{W} \]

\[ P_{out} = 0.4 \times 19350 = 7740\ \mathrm{W} \]

\[ P_{out} \approx 7.74\ \mathrm{kW} \]

The other options do not follow from this formula, for example option (C) is about half of the power in the wind.

Answer: B

The power in the wind is proportional to the cube of the wind speed, so the power output of an ideal wind mill is also proportional to \(v^{3}\). When the speed doubles from 4 m/s to 8 m/s:

\[ \frac{P_{2}}{P_{1}} = \left(\frac{8}{4}\right)^{3} \\ = 2^{3} = 8 \]

The output increases by a factor of 8, option (B). A factor of 2 would apply if power were proportional to speed, and a factor of 4 if it were proportional to the square of the speed, but the kinetic energy flux has the cube.