Question 11.AE.5: Relying solely on wind/solar energy is problematic because i......
Relying solely on wind/solar energy is problematic because it may not be available when needed, for example, a wind farm could lose as much as 60 MW within a minute. There are several scenarios of how the power change of 60 MW per minute can be mitigated through complementary, albeit more expensive power sources: one is the combination of a (longterm) compressed-air power plant with a (short-term) battery plant for bridging the time from when theWP plant output decreases (60 MW per minute) to when either a long-term compressed-air storage (CAES) plant [20,21] or a pump-storage hydro plant [18,19,146] can take over. A CAES plant requires a start-up time of about 6 minutes. To bridge this 6-minute gap for a 100 MW compressed-air power plant, a battery plant is proposed to provide up to 100 MW during a 6-minute interval amounting to a required energy storage of 10 MWh. Inverters fed from a battery can deliver power within a few 60 Hz cycles to the power system, replacing the lost power of 60 MW per minute almost instantaneously. This combination of CAES plant and battery storage plant as bridging energy sources can be employed for peak-power operation as well as for improving power quality by preventing brown/blackouts. Figure E11.5.1a depicts the block diagram of such a battery storage plant consisting of wind turbine mechanical gear, synchronous generator, 3-phase transformer, 3-phase rectifier, battery bank, three-phase inverter, 3-phase transformer, and power system.
Recommended voltages are: output line-to-line voltage of generator V_{\mathrm{L}-\mathrm{L}}^{\mathrm{g}}==677 V, input DC voltage of inverter V_{bat}=V_{DC}=600 V, output line-to-line voltage of inverter V_{\mathrm{L}-\mathrm{L}}^{\mathrm{g}}=V_{\mathrm{L}-\mathrm{L}}^{\mathrm{i}} = 240 V, power system voltage V_{\mathrm{L}-\mathrm{L}}^s = 480 V, the 2 transformers are of the Δ-Y type connected to the generator and of the Y-Δ type connected to the power system.
a) Draw a detailed circuit diagram depicting transformers, rectifier, capacitor, and inverter in more detail as compared to Figure E11.5.1b and specify the powers, voltages and currents, provided 10 rectifier/battery bank/inverter combinations are connected in parallel: this will improve the overall efficiency of the plant, because some of the components can be disconnected, if WP generation is low. The efficiency for each and every component is about η=0.96.
b) Determine the specifications of the Y-Δ three-phase transformer between inverter and power system.
Design of PWM (pulse-width-modulated) three-phase current controlled voltage–source inverter feeding power into the utility system via Y- Δ transformer
The inverter circuit of Figures 11.6a, b is to be analyzed with PSpice, where the DC voltage is V_{DC}=600 V, output voltage of inverter V_{\mathrm{L}-\mathrm{L}^2}^{\mathrm{i}}=240 V, power system voltage V_{L-L}^s=480 V, and the switching frequency of the IGBTs is f_{switch}=7.2 kHz. The DC input capacitance is C_{filter}=1000 μF (not shown in Figure 11.6a), the components connected at the output of the inverter bridge are L_w=1 mH, R_w=10 mΩ (wave-shaping inductor), L_f=45 μH, C_F=10.3 μF, R_F=10 mΩ (output filter); transformer and power system inductance and resistance per phase are L_N =265 μH, and R_N =50 mΩ, respectively, the power system (phase) voltages referred to the secondary (Y) side of the Y-Δ transformer are v_{aN}(t)=196Vsinωt, v_{bN}(t)=196Vsin(ωt120°), and v_{cN}(t)=196Vsin(ωt-240°).
c) Calculate with PSpice (you may ignore the output filter because of the 64-node limit of the PSpice student version program) and plot output phase voltage and phase current of the PWM inverter provided the output voltages are sinusoidal/co-sinusoidal as given above by the power system voltages.
d) Subject the output current of the inverter to a Fourier analysis.
Design of a controlled three-phase rectifier
Figure E11.5.1c shows the three-phase rectifier of Figure 11.1 with six diodes and one self-commutated switch, an insulated-gate bipolar transistor (IGBT).
For the load resistance R_{load}=3.19 Ω, the nominal input phase voltages of the rectifier can be assumed to be v_a=738Vsinωt, v_b=738Vsin(ωt-120°), and v_c=738Vsin(ωt-240°) resulting at a duty cycle of δ=50% in a DC output voltage of 600 V. The IGBT is gated with a switching frequency of 3 kHz at δ=50%. C_{fic}=200 μF, L_{fa}=90 μH, C_{f2c}¼50 μF, ideal diodes D_1\ to\ D_6\ and\ D_{fw},\ v_{switch}=100 V magnitude, R_{snf}=R_{sn}=10 Ω, C_{snf}=C_{sn}=0.1 μF, C_s=1000 μF, L_s=1 mH, and V_{load}=600 V.
e) Perform a PSpice analysis and plot output voltage of the rectifier v_{load}\ (t)≈V_{DC}, rectifier output current I_r, rectifier input phase current i_r(t), and the rectifier input phase voltage v_{ph}^r (t).
f) Subject the rectifier input current i_r(t) to a Fourier analysis.
g) Determine the specifications of the three-phase transformer between generator and rectifier.
h) Determine the overall costs of this power plant if a specific cost of $4000/kW installed output capacity is assumed. Note that a coal-fired plant has a specific cost of $2000/ kW installed output capacity.
a) The circuit diagram depicting transformers, rectifier, capacitor, and inverter in more detail as compared to Figure E11.5.1a is shown in Figure E11.5.1b with a rated output power of P_s=1.0 MW. For the generation of 100 MW 100 such plants must be connected in parallel. Assuming a unity power factor the output powers of the various components are as follows. Inverter-side transformer output power: P_s=1.0 MW, PWM inverter output power: P_t^i = 1:04MW, battery bank output power: P_i=1.08 MW, rectifier output power P_c =1.13 MW, generator-side transformer output power: P_r = 1.17MW, generator output power: P_t^g = 1.22MW, gear output power: P_g=1.3 MW, wind turbine output power: P_{gear}=1.35 MW as listed in Figure E11.5.1b.
b) The input voltage of inverter of V_{DC}=600 V results with a modulation index of m=0.655 in the output phase voltage of the inverter
V_{\mathrm{ph}}^{\mathrm{i}}=\frac{m \cdot V_{\mathrm{DC}}}{2 \sqrt{2}}=\frac{0.655 \cdot 600}{2 \sqrt{2}}=139 \mathrm{~V}, \text { or } V_{\mathrm{L}-\mathrm{L}}^{\mathrm{i}}=\sqrt{3} V_{\mathrm{ph}}^{\mathrm{i}}=240 \mathrm{~V}. The transformation ratio of the inverter-side Y-Δ transformer is \frac{N_{\mathrm{s}}^{\Delta}}{N_{\mathrm{s}}^{\mathrm{Y}}}=\frac{V_{\mathrm{L}-\mathrm{L}}^{\mathrm{s}}}{V_{\mathrm{ph}}^{\mathrm{i}}}=\frac{480}{139}=3.45 resulting in a current fed into the utility system of I_{\mathrm{s}}=\frac{P_{\mathrm{s}}^{\mathrm{s}}}{\sqrt{3} V_{\mathrm{L}-\mathrm{L}}^{\mathrm{s}}}=\frac{1 \mathrm{MW}}{\sqrt{3} \cdot 480}=1,202.8 \mathrm{~A} The output current of one inverter is I_{\mathrm{i}}=\frac{P_{\mathrm{t}}^{\mathrm{s}}}{3 \cdot V_{\mathrm{ph}}^{\mathrm{i}} \cdot 10}=\frac{1.04 \mathrm{MW}}{3 \cdot 139 \cdot 10}=249 \mathrm{~A}.
c) The PSpice input list of Table 11.3 yields for the given output inverter voltage the current wave shape of Figure E11.5.1d. Note that the amplitude of the inverter current is √2 . 249 A=352 A and the amplitude of the output lineto-line voltage of the inverter is √3 . 139 V=240 V. The inverter output current ii(t) leads the inverter line-to-line voltage v_{\mathrm{L}-\mathrm{L}}^{\mathrm{s}}(\mathrm{t}) by 60 degrees based on generator operation.
d) Fourier analysis of inverter output current: all current harmonics are negligibly small due to the small modulation index and relatively large wave-shaping inductance L_w.
e) Figure E11.5.1e depicts the input and output voltages and currents of the rectifier.
f) Dominant harmonic amplitudes including the fundamental are:
1st: 105.4 A; 5th: 17.7 A; 7th: 11.2 A; 36th: 12.8 A; 37th: 23.0 A; 43rd: 17.3 A; 45th: 18.2 A; 49th: 84.2 A; 51st: 75.7 A; 55th: 10 A. Note, to get rms values these values must be divided by √2. That is, the fundamental rms value is 74.54 A.
g) The rectifier output current is I_{\mathrm{r}}=\frac{P_{\mathrm{c}}}{V_{\mathrm{DC}} \cdot 10}=\frac{1.13 \mathrm{MW}}{600 \cdot 10}=188 \mathrm{~A} and results with V_{DC}=600 V in a load resistor of R_{load}=3.19 Ω. The PSpice simulation requires for a duty cycle of 50% with a load resistor R_{load}=3.19 Ω an input AC voltage of V_{ph}^r = 522V or an amplitude of 738 V. With the given generator voltage of V_{L-L}^g=677 V one obtains the generator-side transformer ratio of \frac{N_{\mathrm{g}}^{\Delta}}{N_{\mathrm{g}}^{\mathrm{Y}}}=\frac{677}{522}=1.3.
If the rectifier and the generator were matched with respect to their voltages one does not need a generator-side transformer. The input current to the rectifier is approximately [note there are harmonics as indicated in part f]
I_{\mathrm{t}}=\frac{P_{\mathrm{r}}}{3 \cdot V_{\mathrm{ph}}^{\mathrm{r}} \cdot 10}=\frac{1.17 \mathrm{MW}}{3 \cdot 522 \cdot 10}=74.7 \mathrm{~A}, this compares well with the rms value of the fundamental of part f), that is, 105.4/1.414=74.54 A. The input current to the generator-side transformer is I_{\mathrm{g}}=\frac{P_{\mathrm{t}}}{\sqrt{3} V_{\mathrm{L}-\mathrm{L}}^{\mathrm{g}}}=\frac{1.22 \mathrm{MW}}{\sqrt{3 \cdot 677}}=1040.5 \mathrm{~A}. That is, the generator output power is at unity power factor P_{\mathrm{t}}^{\mathrm{g}}=\sqrt{3} \cdot V_{\mathrm{L}-\mathrm{L}}^{\mathrm{g}} \cdot I_{\mathrm{g}}=\sqrt{3} \cdot 677 \cdot 1040.5=1.22 \mathrm{MW}, which confirms the value as calculated in part a).
h) The cost of the 100 MW battery plant is $400\cdot 10^6.
