Question 4.AE.8: Design a wind-power plant (Fig. E4.8.1) feeding Pout_transfo......
Design a wind-power plant (Fig. E4.8.1) feeding Pout_transformer=10 kW into the distribution system at a line-to-line voltage of V_{L-L system} =12.47 kV=V_{L-L\_secondary\ of\ transformer}. The wind-power plant consists of a Y–Δ three-phase (ideal) transformer connected between three-phase PWM inverter and power system, where the Y is the primary and Δ is the secondary of the ideal transformer (Fig. E4.8.2). The input voltage of the PWM inverter (see Chapter1, Fig.E1.5.1)isVDC=600 V,where the inverter delivers an output AC current I_{inverter} at about 18° lagging displacement power factor (consumer notation) cos φ≈0.95 lagging the primary line-neutral voltage of the Y – Δ transformer, and the inverter output line-to-line voltageis V_{L-L\_inverter}=240 V=V_{L-L\_primary of transformer}. The three-phase rectifier (see Chapter 1, Fig. E1.2.1) with one self-commuted PWM-operated switch (IGBT) is fed by a three-phase synchronous generator. The input line-to-line voltages of the three-phase rectifier is VL-L rectifier=480 V=VL-L_generator. Design the mechanical gear between wind turbine and generator (with p=2 poles at nS=3600 rpm, utilization factor C=1.3 kWmin/m3 , Drotor=0.2 m, the leakage reactance X_{sleakage}=2πfL_{sleakage} is 10% of the synchronous reactance X_S=1.5 pu, maximum flux density B_{max}=0.7 T, iron-stacking factor k_{fe}=0.95, stator winding factor k_S=0.8, rotor winding factor k_r=0.8, and torque angle δ=30°) so that a wind turbine can operate at n_{turbine}=30 rpm. Lastly, the wind turbine – using the famous Lanchester-Betz-Joukowsky limit [45,84–86] for the maximum energy efficiency of a wind turbine as a guideline – is to be designed for the rated wind velocity of v=10 m/s at an altitude of 1600 m and a coefficient of performance (actual efficiency) c_p=0.3. You may assume that all components (transformer, inverter, rectifier, generator, mechanical gear) except the wind turbine have each an efficiency of η=95%.
a) Based on the given transformer output power of P_{out\_transformer}=10 kW compute the output powers of inverter, rectifier, generator, gear, and wind turbine
b) For the circuit of Fig. E4.8.1 determine the transformation ratio (N_Y/N_△) of the Y–Δ transformer, where the Y is the primary (inverter side) and Δ the secondary (power system side).
c) What is the phase shift between \widetilde{\mathrm{V}}_{L-L \text { system }} \text { and } \widetilde{\mathrm{V}}_{L-L \_ \text {inverter }} \text { ? }
d) What is the phase shift between \widetilde{I}_{L-s y s t e m} \text { and } \widetilde{I}_{L-\text { inverter }} ?
e) Use sinusoidal PWM to determine for the circuit of Fig. E1.5.1 the inverter output current~Iinverter which lags V_{eL-N\_inverter} = 138.57 V about 18° at an inverter switching frequency of fsw=7.2 kHz. You may assume R_{system} =50 mΩ, X_{system} =0.1 Ω, R_{inverter}=10 mΩ,\ and\ X_{inverter}=0.377 Ω at f=60 Hz.
f) For the circuit of Fig. E1.2.1 and a duty cycle of δ=50%, compute the input current of the rectifier I_{rectifier}.
g) Design the synchronous generator for the above given data provided it operates at unity displacement power factor.
h) Design the mechanical gear.
i) Determine the radius of the wind turbine blades for the given conditions.
a)
\begin{aligned} & P_{ out\_transformer }=10 \mathrm{~kW}, \quad P_{out\_inverter }=10.53 \mathrm{~kW} \\ & P_{out\_rectifier }=11.08 \mathrm{~kW}, \quad P_{out\_generator }=11.67 \mathrm{~kW}, \\ & P_{out\_ear }=12.28 \mathrm{~kW}, \quad P_{out\_wind turbine }=12.92 \mathrm{~kW} \text {. } \end{aligned} (E4.8-1)
b) \left(\frac{N_Y}{N_{\Delta}}\right)=\frac{1}{89.9} (E4.8-2)
c) Note \widetilde{V}_{L-L \text { system }}=\tilde{V}_{L-L \Delta} \text { and } \tilde{V}_{L-L \text { inverter }}=\tilde{V}_{L-L Y}; ; therefore,
\tilde{V}_{L-L \Delta}=\widetilde{V}_{L-L Y} \angle+30^{\circ} . (E4.8-3)
\widetilde{I}_{L \text { system }}=\widetilde{I}_{\text {Linverter }} \angle+30^{\circ} (E4.8-4)
One concludes that the line-to-line voltages of the secondary Δ as well as the line currents fed into the power system are shifted by +30 degrees.
e) The computer program for the inverter is given in Chapter 1, resulting in an inverter output current of I_{out\_inverter}=25.33 A, and an inverter input current of I_{in\_inverter}=18.47 A. The inverter output phase voltage and output line current are shown in Fig. E4.8.3.
f) Computer program for rectifier is given in Chapter 1 resulting in a rectifier output current of I_{out\_rectifier}=18.47 A and a rectifier input current of I_{in\_rectifier}=14.04 A. The output voltage of the rectifier is depicted in Fig. E4.8.4.
g) Design of synchronous generator:
• Stator frequency f=\frac{n_s \cdot p}{120}=\frac{(3600)(2)}{120}=60 \mathrm{~Hz} (E4.8-5)
• Phase voltage V_{L-N}=480 \mathrm{~V} / 1.732=277.14 \mathrm{~V} (E4.8-6a)
Base voltage V_{\text {base }}=V_{L-N}=277.14 \mathrm{~V} (E4.8-6b)
• Base (rated) \text { current } I_{\mathrm{base}}=\frac{P_{out\_generator }}{3 \cdot V_{L-N}}=\frac{11670}{3(277.14)}=14.04 \mathrm{~A} (E4.8-7)
• Base impedance Z_{\text {base }}=\frac{V_{\text {base }}}{I_{\text {base }}}=\frac{277.14 \mathrm{~V}}{14.04 \mathrm{~V}}=19.74 \Omega (E4.8-8)
• Generator loss P_{\text {loss }}=P_{in\_gnerator }-P_{\text {out generator }}=(12280-11670) \mathrm{W}=610 \mathrm{~W} (E4.8-9)
• Stator resistance \mathrm{e} R_a=\frac{P_{\text {loss }}}{3 \cdot I_{\text {base }}^2}=1.03 \Omega (E4.8-10)
• Ideal core length l_i=\frac{P_{\text {out } \_ \text {generator }}}{D_{\text {rotor }}^2 \cdot C \cdot n_s}=\frac{11670}{\left(0.2^2\right)(1.3)(3600)}=0.0623 \mathrm{~m} (E4.8-11)
• Actual iron core length (with interlamination insulation)
l_{\mathrm{act}}=\frac{l_i}{k_{f f}}=\frac{0.0623}{0.95} \mathrm{~m}=0.0656 \mathrm{~m}, (E4.8-12)
• Stator synchronous reactance
X_S=X_S[\mathrm{pu}] \cdot Z_{\text {base }}=(1.5)(19.74)=29.61 \Omega (E4.8-13)
with X_d \approx X_S=X_{a d}+X_{a b}+\frac{3}{2} X_{a d 2} \text { one gets approximately } X_S=2 \cdot(2 \pi f) L_{a d} or for the self-inductance L_{aa} of one phase
L_{a d}=\frac{X_S}{4 \pi f}=0.0393 \mathrm{H} (E4.8-14)
and the leakage inductance of one stator (s) phase
L_{s \ell}=0.1\left(\frac{29.61}{2 \pi 60}\right)=0.0079 \mathrm{H} (E4.8-15)
• The area per pole is
\text { area }_p=\frac{2 \pi \cdot D_{\text {rotor }}}{2} \cdot \frac{l_i}{2}=\frac{2 \pi(0.2)(0.0623)}{4}=0.0196 \mathrm{~m}^2 . (E4.8-16)
• The maximum flux linked with one pole is
\Phi_{s_{-} \max }=B_{\max } \cdot \operatorname{area}_p=0.7(0.0196)=0.0137 \mathrm{~Wb} . (E4.8-17)
• The induced voltage in one phase of the stator winding becomes (generator operation)
E_{p h}=1.05 \cdot V_{L-N}=4.44 \cdot f \cdot N_s \cdot k_s \cdot \Phi_{s_{-} \max }=291.0 \mathrm{~V} (E4.8-18)
or the number of stator turns per phase is
\begin{aligned} N_s & =\frac{E_{p h}}{4.44 \cdot f \cdot k_s \cdot \Phi_{s_{-} \max }}=\frac{291}{4.44(60)(0.8)(0.0137)} \\ & =100 \text { turns. } \end{aligned} (E4.8-19)
The one-sided air-gap length can be obtained from
L_{a d}=\frac{\lambda_{a d}}{i_a}=\frac{N_s \cdot \Phi_{s \_\max }}{i_{\max }} (E4.8-20)
where
\Phi_{s_{-} \max }=\frac{\mu_0 \cdot \frac{4}{\pi} \cdot k_s \cdot N_s \cdot i_{\max } \cdot \text { area }_p}{p \cdot g} (E4.8-21)
Therefore,
L_{v i a}=\frac{\mu_o \cdot \frac{4}{\pi} \cdot k_s N_s^2 \cdot \operatorname{area}_p}{p \cdot g}or solved for the air-gap length
\begin{aligned}g=\frac{\mu_0 \cdot \frac{4}{\pi} \cdot k_s \cdot N_s^2 \cdot \operatorname{area}_p}{p \cdot L_{\mathrm{ua}}} & =\frac{\left(4 \pi \times 10^{-7}\right)\left(\frac{4}{\pi}\right)(0.8)(10000)(0.0196)}{2(0.0393)} \\& =0.0032 \mathrm{~m}=3.2 \mathrm{~mm} .\end{aligned} (E4.8-22)
• The torque of the generator is
(-T)=\frac{P_{out\_generator }}{\omega_s}=\frac{11670}{377}=30.94 \mathrm{Nm} (E4.8-23)
For a torque angle of δ=30° one gets with
T=-\frac{p \cdot \pi \cdot D_{\text {rotor }} \cdot l_{\text {act }} \cdot B_{\max } \cdot F_r}{4} \cdot \sin \delta (E4.8-24)
or the rotor mmf
\begin{aligned} F_r & =\frac{(-T) \cdot 4}{p \cdot \pi \cdot D_{\text {rotor }} \cdot l_{\text {act }} \cdot B_{\max } \cdot \sin \delta} \\ & =\frac{(30.94)(4)}{(2)(\pi)(0.2)(0.0656)(0.7)(0.5)}=4289.6 \mathrm{At} . \end{aligned} (E4.8-25)
• The ampere turns for the rotor are F_r=\frac{4}{\pi} \cdot k_r \cdot \frac{N_{r(\omega \omega t)}}{p} \cdot I_r or solved for the ampere turns
N_{r(\text { tot })} \cdot I_r=\frac{F_r \cdot p \cdot \pi}{4 \cdot k_r}=\frac{(4289.6)(2)(\pi)}{(4)(0.8)}=8422.4 \mathrm{At} . (E4.8-26)
• Selecting 16 rotor slots with 20 turns each, one obtains the field (rotor) current:
I_r=8422.4 \text { Aturns } /(16 \cdot 20 \text { turns })=26.3 \mathrm{~A} . (E4.8-27)
h) Gear ratio. The gear ratio is
\text { gear ratio }=\frac{n_{\text {generator }}}{n_{\text {windturbine }}}=\frac{3600}{30}=120 (E4.8-28)
i) Radius of (three) windturbine blades.The limiting efficiency for windturbines is according to Lanchester-Betz-Joukowsky [45,84–86] η_{Lanchester-Betz-Joukowsky}=(16/27)=0.59. The wind turbine design depends on the density of the air and the ambient temperature. Table E4.8.1 provides the air density as a function of altitude.
The wind turbine has to provide a rated output power of P_{out\_wind} turbine=12.92 kW at a rated wind velocity of v=10 m/s. The efficiency (coefficient of performance) is c_p=0.3, where c_p is the ratio between the power delivered by the rotor of the wind turbine to the power in the wind with velocity v striking the area A swept by the rotor.
According to [45] the output power of the wind turbine is
P_{ wind\_turbine }=\frac{1}{2} \cdot c_p \cdot v^3 \cdot A \cdot \rho \text {. } (E4.8-29)
The area swept by the rotor is
A=\frac{2 \cdot P_{\text {wind-turbine }}}{c_p \cdot v^3 \cdot \rho}=\frac{2(12920)}{0.3(1000)(1.05)}=82 \mathrm{~m}^2 (E4.8-30)
or the rotor blade radius is
R_{\text {blade }}=\sqrt{\frac{A}{\pi}}=\sqrt{\frac{82}{\pi}}=5.1 \mathrm{~m} \text {. } (E4.8-31)
The height of the tower can be assumed to be about 2 times the blade radius, that is, H_{tower}=10 m.
Table E4.8.1 Air Density as a Function of Altitude\begin{array}{l|l|l|l|l|l|l|l|l} \hline \text { Altitude }(\mathrm{m}) & -500 & 0 & 500 & 1000 & 1500 & 2000 & 2500 & 3000 \\ \text { Air (mass) density } \rho\left(\mathrm{kg} / \mathrm{m}^3\right) & 1.2854 & 1.2255 & 1.1677 & 1.112 & 1.0583 & 1.0067 & 0.957 & 0.9092 \\ \hline \text { Altitude }(\mathrm{m}) & 3500 & 4000 & 4500 & & & & & \\ \text { Air (mass) density } \rho\left(\mathrm{kg} / \mathrm{m}^3\right) & 0.8623 & 0.8191 & 0.7768 & & & & & \\ \hline \end{array}
