Chapter 5 - Wind Energy

Question 1

What is the wind pattern on earth dominated by?

Answer 1

• Convection due to different temperatures of earth surface.
• Coriolis-force due to earth rotation
• Local high and low pressure zones.

Question 2

How is the Coriolis-force described mathematically, and what consequences for the wind pattern has it?

Answer 2

F = 2m(v x omega)

North-bound winds on northern hemisphere tend to bend eastward (North-East winds), and south-bound winds westward (South-West, passat/trade winds in the tropical cell). At the southern hemisphere this is the other way around. At the equator, only the uprising winds experience the Coriolis force.

Question 3

How much of the solar irradiation does the global energy stored in wind correspond to?

Answer 3

About 2%.

Question 4

Draw a schematic of the Earth showing the different cells

Answer 4

At equator: intertropical conversion zone, low pressure zone.
At 0-30 deg: tropical cell
At 30-60 deg: moderate cell
At 60-90 deg: polar cell

High pressure zones between tropical cell and moderate cell, and at the polar caps.

Question 5

What is the troposphere?

Answer 5

The part of the atmosphere that contains most of the mass. Extends about 20km at the equator, only about 8 km at the polar cap due to centrifugal forces.

Question 6

For wind energy, what is the most important parameter?

Answer 6

The long term average wind velocity, since this determines the power density in the wind.

Question 7

What is the power density of moving air?

Answer 7

P/A = 1/A * dEkin/dt = 1/2 rho * v^3

Question 8

Calculate the power density of moving air.

Answer 8

See own calculation in notes.

Question 9

How is the variation of the wind power in height?

Answer 9

It is not so pronounced. It can be approximated with an empirical formula:

v(h) = v(10m) * (h/10 m)^g,

where g is a factor that varies depending on the surface structure (≈ 0.16 for flat surfaces, 0.04 for cities with highrises).

Question 10

What are typical wind velocities in Europe?

Answer 10

0 - 30 m/s

Question 11

In which wind speed interval are the commercial operation of wind mills?

Answer 11

4 - 25 m/s

Question 12

What important connection can be made from the continuity equations an Bernoullis principle, when talking about wind power?

Answer 12

That v_1*A_1 = v_2 * A_2

Question 13

How is the extracted power given?

Answer 13

P_R = 1/2 * dm/dt * (v_1^2 - v_2^2)

Question 14

What is the Betz’ efficiency? How to we obtain a maximum efficiency?

Answer 14

It shows the efficiency of wind power extraction. It has a maximum when v_2 = 1/3 v_1, and then the efficieny is 16/27 ≈ 59%.

It is given:
C_p = P_R / P_1 = 1/2 (1 + v2/v1)(1 - (v2_v1)^2)

Question 15

Calculate the Betz’ efficiency.

Answer 15

See calculations in notes.

Question 16

What are drag-type rotors?

Answer 16

Rotors using the drag force, F_d = c_w * 1/2 * rho * A * v^2, as the driving force for the rotor. The drag coefficient c_w depends on the shape of the body.

In a drag-type rotor we want to increase c_w as much as possible. Usually it can be operated either by blocking half of the revolution for wind (to avoid drag forces in the opposite direction), or designing the blades so that the drag force in one direction is significantly higher than in the other.

Question 17

What is the efficiency of a drag-type rotor.

Answer 17

It is given as c_p = c_w ( 1 - u/v1)^2(u/v1).

It has a maximum when u, the rotation speed, is 1/3 v_1. When u = 0 or u = v_1, the efficiency drops to 0. The max efficiency of a drag type rotor is less than half of the Betz’ efficiency.

Question 18

What are lift-type rotors? Draw a schematic of the lift-force.

Answer 18

Lift-type rotors uses the lift force to generate power. The lift comes from the Bernoulli principle, that we have high speeds of wind above the blade profile, making the air pressure lower there, and thus creating lift.

Lift force is defined: F_L = c_L * 1/2 * rho * A_|| * v1^2

See page 7 for schematic.

Question 19

What is the angle of attack?

Answer 19

The angle of attack is the angle between the effective wind direction (due to the motion of the blades) and the incoming wind direction.

Question 20

How does the lift and drag coefficient evolve for optimized blade profiles as a function of angle of attack? Draw a plot.

Answer 20

The drag coefficient increases from 0.01 to 0.1 for angle of attack between 0 and 10 deg. The lift coefficient increases from 0.5 to 1 between 0 and 10 deg. The lift coefficient can be approximated with 5*alpha * π/180.

See page 7 for plot.

Question 21

What is pitching?

Answer 21

Pitching is the changing of the blade angles, to change the angle of attack. This can be used to optimize the efficiencies for a given wind speed.

Question 22

What is the glide number?

Answer 22

The glide number is the ratio between the lift force and the drag force. For optimized profiles, this can be around 100.

Question 23

Draw a schematic of the forces on a blade profile.

Answer 23

See pages 8 and 9.

Question 24

What is the speed ratio?

Answer 24

The speed ratio is the ratio between the rotation speed and the wind speed.

Question 25

What is the tip speed ratio?

Answer 25

It is the ratio of the rotation speed of the tip of the rotor to the wind speed.

lambda_S = omega R / v1

Question 26

How can one classify rotor types depending on the tip speed ratio?

Answer 26

For tip speed ratios larger than 3, we call them fast rotors, and for tip speed ratios less than 3 we call them slow rotors.

Question 27

What is the power for a lift type rotor with Z blades? What are important notes to make about this?

Answer 27

P = 1/2 * omega * v1^2 * b * Z * ∫ sqrt(1+lambda(r)^2) * c_A(r) * r dr

It is proportional to omega and v1^2.

Question 28

Calculate the power generated from a lift type rotor.

Answer 28

See page 10.

Question 29

Why does the blade profile has to be optimized?

Answer 29

Because that the rotation speed of the blade increases outwards on the blade. This means that the value and direction of the angle of attack changes, and the profile shape has to be optimized accordingly.

Question 30

What can be said about the rotation speed in regards to extracting power from the wind generator?

Answer 30

Too slow rotation, not enough power is extracted from the wind to reach the optimum wind speed (1/3 v1). For tip speed ratios approaching 0, v2 -> v1.

Too fast rotation, and too much power is extracted from the wind to reach the optimum speed. For tip speed rations approaching ∞, v2 -> 0.

The optimum number of blades must therefore be decreased for increasing tip speed ratios.

Question 31

What kind of losses do we have with wind engines?

Answer 31

Profile losses: caused by finite drag coefficient.
Tip losses: caused by the pressure difference before and behind the rotor.
Wake losses: caused by non-linear movement of air behind rotor.

Question 32

For a fast rotor, and r/R > 0.1, how does the breadth of an optimal blade profile vary with r (distance from center) and lambda_S (tip speed ratio).

Answer 32

b(r) decreases like 1/r and 1/lambda^2

Question 33

What is the profile efficiency?

Answer 33

The profile efficiency is the ratio between power of a real and ideal blade profile.

eta = 1 - 3/2 * r/R lambda_S/G

G = glide number.

This is the frictional losses. The tip losses are harder to estimate, and the tip efficiency is given empirically:

eta_tip = 1 - 1.84/(Z * lambda_S)

Question 34

What is the Schmitz efficiency?

Answer 34

The Schmitz efficiency includes the wake losses. This approaches asymptotically the Betz’ efficiency for tip speed ratios approaching ∞. For tip speed ratios -> 0, the efficiency -> 0.

Question 35

What efficiencies can lift-type rotors be run with?

Answer 35

Almost 50%.