15

Design Examples for Wind Turbine and Solar Panel Energy Systems

Wind turbine and solar panel energy are clean and renewable. In recent years, their applications have attracted much worldwide attention. Therefore, the design of wind turbine and solar panel energy systems is a very popular research area.

15.1  Introduction

We first introduce some units used to measure large values of power. They are grouped by order of magnitude as follows:

•  kW—kilowatt (10³ W)

•  MW—megawatt (10⁶ W)

•  GW—gigawatt (10⁹ W)

•  TW—terawatt (10¹² W)

•  PW—petawatt (10¹⁵ W)

•  EW—exawatt (10¹⁸ W)

•  ZW—zettawatt (10²¹ W)

•  YW—yottawatt (10²⁴ W)

The relationship between the watt and joule is: 1 joule = 1 watt × 1 second.

The sun radiates 3.8 × 10²⁰ MW into space. Earth receives 174 petawatts (PW) of incoming solar radiation (insolation) at the upper atmosphere, as shown in Figure 15.1 [1,2]. Approximately 30% of the power from the sun is reflected back to outer space, while the rest is absorbed by air, clouds, oceans, and land masses. The spectrum of solar light at the surface of Earth is mostly spread across the visible and near-infrared ranges with a small part in the near-ultraviolet range.

Earth’s land surface, oceans, and atmosphere absorb solar radiation, and their temperature rises. Warm air contains evaporated water from the oceans, causing atmospheric circulation or convection. When the air reaches a certain altitude where the temperature is low enough, water vapor condenses into clouds, which rain onto Earth’s surface, completing the water cycle. The latent heat of water condensation produces atmospheric phenomena such as wind and cyclones.

FIGURE 15.1
About half the incoming solar energy reaches Earth’s surface. (From en.wikipedia.org/wiki/Solar_energy.)

The total solar energy per year absorbed by Earth’s atmosphere, oceans, and land masses is approximately 3,850,000 exajoules (EJ) as shown in Table 15.1 [1]. Depending on the geographical location, the closer a location is to the equator, the more solar energy is available there. The whole world electricity energy (56.7 EJ) is much less than one ten-thousandth of the available solar energy (3, 850, 000 EJ). This indicates that if we can effectively use the solar and wind energy, the energy requirement is not a problem.

TABLE 15.1
Yearly Solar Fluxes and Human Energy Consumption

Solar	3,850,000 EJ
Wind	2,250 EJ
Biomass	3,000 EJ
Primary energy use	487 EJ
Electricity	56.7 EJ

Source: en.wikipedia.org/wiki/Solar_energy.

15.2  Wind Turbine Energy Systems

The wind turbine is one of the most promising energy sources, and it has attracted much attraction in recent decades and penetrated utility systems deeply compared to other renewable sources [2, 3, 4, 5 and 6]. Unfortunately, the output voltage and frequency of wind turbines are unstable as the wind speed is variable. These turbines are installed onshore or offshore, or sometimes as a wind farm where large numbers of turbines are installed and connected together.

Large-scale flowing air is called wind. Because of the sun, the wind always exists. Wind energy is from the sun and is a renewable energy. Figure 15.2 shows wind production. Atmospheric air circulates as in a boiler. The air becomes light at the equator, and heavy at the two poles. The wind flows day and night.

Wind energy has been used over the world from the ancient times; as an example, the windmill has been used for water pumping and grain grinding for over 3000 years. The wind turbine is modern equipment to convert the wind’s dynamic energy into electrical energy. Two types of wind turbines are used: vertical axis and horizontal axis turbines [2, 3, 4, 5 and 6].

15.2.1  Technical Features

The wind speed is uncertain and is presented as the Weibull probability density function (pdf) in Equation (15.1) or Rayliegh pdf with k = 2 and Figures 15.3 and 15.4 [2].

$f(\upsilon )=\frac{k}{c}{\left(\frac{\upsilon }{c}\right)}^{k-1}{\text{e}}^{-{\left(\frac{\upsilon }{c}\right)}^2}$

(15.1)

FIGURE 15.2
Wind production.

FIGURE 15.3
The Weibull probability density function (pdf).

In the above equation, k is the shape parameter, v is the wind speed, and c is the scale parameter, which can be 4, 6, 8, and so on depending on the particular location; for example, c ≈ 6 at Singapore (see Figure 15.4).

The mass of air flows m in Figure 15.5 through a wind turbine is

$m=A_1\upsilon =A{\upsilon }_b=A_2{\upsilon }_d$

(15.2)

The power extracted by the turbine is

$P=\frac{1}{2}m({\upsilon }^2-{\upsilon }_d^2)$

(15.3)

FIGURE 15.4
Rayleigh pdf.

FIGURE 15.5
The mass of air flows through a wind turbine.

Since m = ρAv_b and

${\upsilon }_b\approx \frac{1}{2}(\upsilon +{\upsilon }_d)$

(15.4)

$P=\frac{1}{2}m({\upsilon }^2-{\upsilon }_d^2)$

(15.5)

or

$P=0.5\text{ρ}A{\upsilon }^3\left[\frac{1}{2}(1+\text{λ)}(1-{\text{λ}}^{\text{2}}\text{)}\right]$

(15.6)

where ρ is the air density, A is the rotor area, v_b is the wind speed through the turbine, v is the blow-in wind speed, and v_d is the blow-out wind speed. To describe the power produced by the wind turbine, we use Betz’s law:

$P=0.5\text{ρπ}{R}^2{\upsilon }^3{C}_p$

(15.7)

where R is the radius of the windmill (or the length of the blade), and C_p is the power coefficient:

$C_p(\text{λ)=0}\text{.5(1+λ)(1}-{\text{ρλ}}^{\text{2}})$

(15.8)

with $\lambda =\frac{v_d}{v}$ being the wind speed deduction ratio.

The effect of temperature and pressure of air density is described by the gas state equation,

$PV=nRT\Rightarrow \frac{n}{V}=\frac{P}{RT}$

(15.9)

where n is the mass of air in mols; V is the volume of air in m³; P is the pressure in atm; R is the idea gas constant = 8.2056 ×10⁻⁵ in m³.atm.K⁻¹.mol⁻¹; and T is the absolute temperature in K.

The air density is given by

$\text{ρ=(}n/V)\cdot M=\frac{P}{RT}\cdot M$

where M is the molecular weight of air in kg/mol, 0.02897. Substituting M and R, we obtain

$\text{ρ=353}\frac{P}{T}$

(15.10)

The unit of ρ is kg/m³.

The atmospheric pressure is dependent on altitude. It is 1 atm at the sites at mean sea level. At sites above sea level, pressure is less; it can be shown as

$\frac{dP}{dh}=-\text{ρ}\cdot g[(N/m^2)/m]$

(15.11)

where h is the height above mean sea level, g is the gravitational acceleration constant (9.806 m/s²), and 1 atm = 1.01325×105 N/m². Therefore,

$P=e^{-\frac{0.0341}{T}h}$

(15.12)

The wind speed at a location varies with height h. The relation is

$\frac{\upsilon }{{\upsilon }_0}={\left(\frac{h}{h_0}\right)}^{\text{α}}$

(15.13)

where v is the wind speed at the height h, v₀ is the wind speed at the height h₀(usually h₀ = 10 m), and α is the friction coefficient (see Table 15.2).

Usually, the wind turbine works in a certain range as shown in Figure 15.6 [4]. The wind speed change causes the output voltage and frequency to vary.

15.2.2  Design Example for Wind Turbine Power System

Figure 15.7 shows the block diagram of a wind turbine power system, which consists of a wind turbine, converters, and control subsystems.

TABLE 15.2
Friction Coefficient α for Various Terrains

Terrain Characteristics	Friction Coefficient α
Smooth hard ground, calm water	0.10
Tall grass on level ground	0.15
High crops, hedges, and shrubs	0.20
Wooded countryside, many trees	0.25
Small town with trees and shrubs	0.30
Large city with tall buildings	0.40

FIGURE 15.6
The wind speed range (power versus wind speed). (From Johnson, G. L. 1985. Wind Energy Systems. New Jersey: Prentice-Hall. With permission.)

FIGURE 15.7
Block diagram of the wind turbine power system.

15.2.2.1  Design Example for Wind Turbine

The wind turbine feeds power to a 3-phase, 11 kV, 50 Hz grid through a wound-rotor induction generator operating with slip-power control. The configurations of the system are as follows [3, 4, 5 and 6]:

Induction generator:

Three-phase, 11 kV, 50 Hz, 4-pole, delta-connected

Per-phase magnetizing inductance referred to the stator = 7 H

Per-phase rotor winding resistance referred to the stator = 30 Ω

Stator to rotor turns ratio = 3:1

Gear box data: Power efficiency = 85%; speed ratio = 65

Site:

Altitude, 500 m above mean sea level

Average temperature, 30°C ≈ 303°K

Friction coefficient a of terrain is 0.15

Wind speed 7.79 m/s at a height of 10 m above ground.

Wind turbine:

Horizontal-axis, 2-blade

Diameter 50 m (R = 25 m)

Tower height 70 m

Efficiency = 45% at tip-speed ratio of 5.5.

Questions are to determine the following:

1.  Slip of the generator

2.  Mechanical power converted to electrical form

3.  The magnitude, phase, and frequency of the phase voltage injected into the rotor by taking the stator terminal voltage as the reference phasor

4.  Real and reactive power supplied by the rotor side converter

5.  Real power supplied to the grid by assuming no losses in the converters and in the stator winding

Solution:

1.  Slip of the generator

$s=\frac{{\text{ω}}_s-{\text{ω}}_m}{{\text{ω}}_s};{\text{ω}}_s=\frac{4\text{π}f}{P}=\frac{4\text{π}\ast \text{50}}{4}=157.1rad/\text{s}$

Shaft speed w_m is directly decided by the wind speed. The wind speed at height of 10 m above the ground is given. Average wind speed is calculated at the turbine midpoint, that is, at the top of the tower

$\frac{\upsilon }{{\upsilon }_0}={\left(\frac{h}{h_0}\right)}^{\text{α}}\Rightarrow \frac{\upsilon }{7.79}={\left(\frac{70}{10}\right)}^{0.15}\Rightarrow \upsilon =10.43m/\text{s}$

Tip-speed ratio (TSR) = 5.5 = tip speed/wind speed

${\text{ω}}_t\cdot \frac{D}{2}=\upsilon \cdot TSR\Rightarrow {\text{ω}}_t=10.43*5.5*\frac{2}{50}=2.295rad/\text{s}$

Speed conversion through gear box:

${\text{ω}}_m=65{\text{ω}}_t=65*2.295=149.175rad/\text{s}$

Therefore, the slip of the generator:

$s=\frac{{\text{ω}}_s-{\text{ω}}_m}{{\text{ω}}_s}=\frac{157.1-149.175}{157.1}=0.05$

2.  Mechanical power converted to electrical form:

$p_m={\text{η}}_g{P}_t={\text{η}}_g{\text{η}}_t{P}_w={\text{η}}_g{\text{η}}_t\left(\frac{1}{2}\text{ρA}{\upsilon }^{\text{3}}\right)$

Hence

$p_m=\frac{1}{2}{\text{η}}_g{\text{η}}_t=\left(\frac{353}{T}{e}^{-\frac{0.0341h}{T}}\right)\frac{\text{π}{D}^2}{4}{\upsilon }^3$

where T = 303 K; h = 500 +70 = 570 m; v = wind speed at turbine midpoint = 1.043 m/s; D = 50 m; η_g = efficiency of gearbox = 0.85; η_t = efficiency of turbine = 0.45. Substituting the data into the formula,

p_m = 465540 W = 465.54kW

3.  Rotor injected voltage

Delta-connected stator winding: V_phase = V_line = 11000 V

We choose it as the reference phasor: V₁ = 11000∠0° V

Since there are no stator losses,

$P_1=P_g=\frac{P_m}{1-s}=\frac{-465540}{1-0.05}=-490042W$

And

$P_1=3V_1{I}_1\cos {\Phi }_1\Rightarrow I_1=\frac{-490042}{3\ast 11000\ast 1.0}=-14.85\text{A}$

$I_m=V_1/j{X}_m=\frac{11000}{2\text{π*50*7}}=-j5\text{A}$

$I_2=I_1-I_m=-14.85+j5\text{A}$

$V_1=\frac{V_2}{s}+I_2\frac{R_2}{s}\Rightarrow$

$\frac{V_2}{0.05}=V-(-14.85+j5)\frac{30}{0.05}=19910-j3000$

V₂=0.05(19910 − j3000) = 995.5−j150=1006.7∠–8.57^°V

The rotor injected voltage is calculated using the turns ratio as

$V_{1’}=\frac{1}{3}V{}_2=335.6\angle -{8.57}^{\text{O}}V$

The rotor injected voltage should have a frequency given by

f₂=sf₁=0.05*50=2.5Hz

4.  Complex power absorbed by the rotor converter:

$S_r=3V_r{I}_r^{*}=3V_2{I}_2^{*}=3*1006.7\angle -{8.57}^{\text{O}}*(-14.85-j5)$

S_r=3*1006.7∠–8.57^O *–15.67∠18.6^O=–47325∠–10.04^O

S_r=P_r +jQ_r=–46600 + j8250

P_r=–46.6kW and Q_r=8.25kVAr

Real power and reactive power injected into the rotor winding by the rotor side converter are 46.6 kW and -8.25 kVAr.

5.  Real power drawn from the grid:

P_grid = P₁ - P_r = −490042 + 46600 = −443442 W

This is also equal to the sum of the power converted to mechanical form (P_m) and the rotor current loss.

15.2.2.2  Design Example for Converters

The wind speed is unstable and changes from time to time in a certain speed range. The output voltage and frequency of the double-feed induction from time to time generator (DFIG) change by about ±20%. In order to transfer the unstable electrical energy generated from DFIG to the grid, we design our converter system as follows [3, 4, 5 and 6].

We assume the output voltage (e.g., line-to-line rms 11,000 V) and frequency (e.g., 50 Hz) of the double-feed induction generator (DFIG) change about ±20%, and the grid voltage (e.g., line-to-line rms 11,000 V) and frequency (e.g., 50 Hz) are very stable with ±1% variation. The converter’s system design includes three parts: AC/DC rectifier, DC/DC converter, and DC/ AC inverter, as shown in Figure 15.7. The AC/DC rectifier is an uncontrolled diode full-bridge rectifier. Its output is an unstable DC voltage of about 14.86 kV ±20%. The DC/DC converter is a boost type with closed-loop PI control. Its output voltage is 20 kV ±1% and is very stable. The DC/AC inverter is a VSI. Its output is a three-phase, 50 Hz, 11 kV (line-to-line rms).

15.2.2.3  Simulation Results

The simulation diagram is shown in Figure 15.8. The simulation results are shown in Figure 15.9. When the input voltage and frequency changed by 20%, both output voltage and frequency of the system remained stable.

FIGURE 15.8
Simulation diagram of the wind turbine power system. (a) V_in = 8.8 kV (line-to-line, rms)/40 Hz (b) V_in = 11 kV (line-to-line, rms)/50 Hz (c) V_in = 13.2 kV (line-to-line, rms)/60 Hz

FIGURE 15.9
Simulation results of the wind turbine power system.

15.3  Solar Panel Energy Systems

The sun offered sunlight and heat (with chemical effects) to Earth over millions of years, and this will continue for millions of years. The tremendous energy from the sun is thousands of times higher than the current total energy consumption of the world.

15.3.1  Technical Features

The sunlight changes from time to time. If the rated voltage of a solar panel is 186 V with a current of about 13 A, during a day it varies from 186 - 20% to 186 + 20%, that is, from 148.8 to 223.2 V. In order to convert this energy into the grid, we have to design appropriate power electronic circuits. The objectives are as follows:

1.  To convert the unstable DC voltage to a stable DC voltage

2.  To invert the stable DC voltage into 3 Φ AC voltage

3.  To link the solar panel system to the main grid of 400 V/50 Hz/3 Φ

According to the above, the technical features are set as follows:

1.  To match the grid data, we need an inverter to provide its output of 400 V/50 Hz/3 Φ with 1% variation.

2.  To provide the inverter with output of 400 V/50 Hz/3 Φ, we need to offer a DC link voltage 700 V with 1% variation.

3.  Since the input voltage is 186 V with ±20% variation, we need a high-voltage-transfer-gain DC/DC converter. The positive output super-lift Luo converter is selected.

4.  To keep the link voltage at 700 V with 1% variation, we need a closed-loop control for the DC/DC converter.

We will briefly introduce each block of the system before beginning the system design.

15.3.2  P/O Super-Lift Luo Converter

The super-lift Luo converter is very good at high-voltage transformation and thus was used in the solar panel energy system. The positive output super-lift Luo converter [4, 5 and 6] is shown in Figure 15.10. It consists of a switch S, an inductor L, two capacitors C₁ and C₂, two diodes D₁ and D₂, and a resistive load R. The input voltage is V_in and output voltage V_O, the switch frequency is f, the period T = 1/f, and the switch-on duty cycle is k. To avoid the parasitic effect, k is (0.1 - 0.9).

When switch S is on, the source voltage V_in charges the capacitor C₁ to V_in, and current flows through the inductor L. The inductor current increases by

$\Delta I_L=\frac{V_{in}}{L}kT$

(15.14)

When the switch S is off, the inductor current decreases with the applied voltage (V_O - 2 V_in). Therefore, the inductor current decrement is

$\Delta I_L=\frac{V_O-2V_{in}}{L}(1-k)T$

(15.15)

FIGURE 15.10
Positive output super-lift Luo converter.

FIGURE 15.11
The voltage transfer gain M versus. the duty cycle k.

In the steady state, the inductor current increment must equal its decrement. Therefore, we obtain the voltage transfer gain M is

$M=\frac{V_{\text{O}}}{V_{in}}=\frac{2-k}{1-k}$

(15.16)

This voltage transfer gain is much higher than that of the boost converter and positive output Luo converter. When k is very small, the voltage transfer gain M ≈ 2. When k = 0.5, the output voltage V_O is equal to 3 × V_in. The voltage transfer gain M versus the duty cycle k is shown in Figure 15.11.

In our system, V_in = 186 V and V_O required for the DC/AC inverter is 700 V. The voltage transfer gain M requested is 3.76; thus, the duty cycle k = 0.638. Since the input voltage varies from 148.8 to 223.2 V, the voltage transfer gain M and the duty cycle k change: M = 3.136–4.704, and k = 0.532–0.73. These values are very good for the given variation range.

15.3.3  Closed-Loop Control

The input voltage from the solar panel varies in the range of 148.8 to 223.2 V. In order to obtain a stable output voltage, we have to design a closed-loop control for the positive output super-lift Luo converter. To this end, a proportional plus integral (PI) controller is used for outer voltage loop control, and a proportional (P) controller for inner current loop control. The control block diagram is shown in Figure 15.12.

The output PWM signal is used to control the duty cycle k for the positive output super-lift Luo converter. The switching frequency is usually chosen in the range of 50–500 kHz. Since this is an automatic control, no k value need be preset.

FIGURE 15.12
Double closed-loop controller.

15.3.4  PWM Inverter

The pulse width modulation technique is a popular method to implement DC/AC inversion technology. The pulse-width-modulated (PWM) voltage source inverter (VSI) introduced in Chapter 3 is used for this design. The three-phase full-bridge VSI is shown in Figure 15.13.

The triangular and modulating signals are shown in Figure 15.14.

There are two important modulation ratios for the PWM technique. We define the amplitude modulation ratio m_a as

$m_a=\frac{V_{in-m}}{V_{tri-m}}$

(15.17)

where V_in-m is the amplitude of the control (sine) waveform and V_tri-m is the amplitude of the triangle waveform. Usually, for nondistorted inversion the amplitude modulation ratio m_a is selected to be smaller than 1.0. We also define the frequency modulation ratio m_f as

$m_f=\frac{f_{tri-m}}{f_{in-m}}$

(15.18)

FIGURE 15.13
Three-phase full-bridge VSI.

FIGURE 15.14
The triangular and modulating signals.

where f_in-m is the frequency of the control (sine) waveform, and f_tri-m is the frequency of the triangle waveform. Usually, for nondistorted inversion, the frequency modulation ratio m_f is selected to be greater than 21. The AC output voltage and current (each phase) are shown in Figure 15.15.

In order to produce the three-phase AC voltage to synchronize to the main grid voltage, we take the grid signals as the control signal.

15.3.5  System Design

After all the blocks are prepared, we can install our system. The block diagram is shown in Figure 15.16. The solar panel yields an input voltage of 186 V ± 20%. The DC/DC converter is the positive output super-lift Luo converter with double closed-loop control. Its output voltage is the DC link voltage with 700 V ±1%. Since the DC link voltage is quite stable, there is no need for any closed-loop control for the DC/AC voltage source inverter.

FIGURE 15.15
The AC output voltage and current (each phase).

FIGURE 15.16
Block diagram of the solar panel power system.

Considering the synchronization, we use the grid voltage as the control signal of the VSI. Its output is a three-phase, 50 Hz, 400 V (line-to-line rms).

15.3.6  Simulation Results

The simulation diagram is shown in Figure 15.17 and the simulation results in Figure 15.18. Figure 15.18a shows the input voltage V_in is 186 - 20% = 148.8 V.

FIGURE 15.17
Simulation diagram of the solar panel power system. (a) V_in = 148.8 V, V_dc2 = 700 V and V_O = 400.155 V/50 Hz/3 Φ. (b) V_in = 186 V, V_dc2 = 700 V and V_O = 399.906 V/50 Hz/3 Φ. (c) V_in = 223.2 V, V_dc2 = 700 V and V_O = 400.321 V/50 Hz/3 Φ.

FIGURE 15.18
Simulation results of the wind turbine power system.

After the double closed-loop control, the output voltage of the P/O super-lift Luo converter V_dc2 is 700 V. We then obtain V_O = 400.155 V/50 Hz/3 Φ after the DC/AC inverter. Figure 15.18b shows the input voltage V_in = 186 V, V_dc2 = 700 V, and V_O = 399.906 V/50 Hz/3 Φ. Figure 15.18c shows the input voltage V_in = 186 V + 20% = 223.2 V, V_dc2 = 700 V, and V_O = 400.321 V/50 Hz/3 Φ. In all cases, when the input voltage varies, the output voltage remains stable. The requirements of the application are thus satisfied.

References

1.  en.wikipedia.org/wiki/Solar_energy

2.  Masters, G. M. 2005. Renewable and Efficient Electric Power Systems. New York: John Wiley & Sons.

3.  Ackermann, T. 2005. Wind Power in Power Systems. New York: John Wiley & Sons.

4.  Johnson, G. L. 1985. Wind Energy Systems. New Jersey: Prentice-Hall.

5.  Luo, F. L. and Ye, H. 2004. Advanced DC/DC Converters. Boca Raton, FL: CRC Press.

6.  Luo F. L. 2012. Lecture Notes on Renewable Energy Systems. NTU Course EE4504.