11.1.1 Future Plans for Implementation of Renewable Energy Sources: Europe

Provided Germany reduces its energy consumption by 35% within the next 10 years then the cost/kWh would be reduced from 0.28 to 0.15 Euro, excluding value-added tax. Investments for conventional (natural gas) power plant would be minimized, and newly to be installed high-voltage AC and DC transmission lines could be reduced from 8500 km to 1750-5000 km. The energy efficiency is called in [3] a “sleeping giant”. To make the European Union (EU) countries less dependent upon imported natural gas and oil, the EU’s recommendation is to increase the renewable energy portfolio to 27% by 2030 [4]. This will be very challenging because in 2013 the total renewable energy consumption was in the EU 14% [5] and in Germany 11.5% [6]. The renewable energy portfolio can be increased by installing storage plants and reducing base-load power plants, otherwise Northern Europe's storms send wind turbines spinning and help strengthen a new phenomenon for the region - negative electricity prices [7]. In Germany 23.4% of the electricity generated originated in solar, hydro, and biomass plants in 2013 [8]. The first international off-shore grid is planned with the name “Offshore Interkonnektor” [9]: electricity between Central Europe and the Scandinavian countries Denmark and Sweden will be generated in the Baltic Sea connecting for the first time two national off-shore wind farms. The EU is determined to increase by 2020 the renewable energy percentage to 20% [5]. The EU parliament intends to accelerate climate protection by reducing electricity consumption: in 2030 EU’s energy consumption should be reduced by 40% and the CO₂ output could be reduced by the same percentage [10]. It is interesting to note that the United Kingdom plans to increase the nuclear and PV plant capacities, while France intends to scale back [11]. To make all plants work in unison time-synchronized measurement are employed to improve grid performance and reliability. Using time-synchronized measurements [12], one can improve local- and wide-area protection and control, state monitoring, power system modeling, and forensic event analysis.

11.1.2 Application Example 11.1 and its Solution: Efficient Residence in Germany where Photovoltaic (PV) Generation Exceeds Total Energy Consumption

A very good example for energy consumption reduction of a residential dwelling, a single-family house, is presented in Figure E11.1.1 illustrating a 220 m² house with traditional three levels (basement, 1^st and 2^nd floors) built in 2013 with insulated clay bricks combined with humidity-absorbing plaster [13], rated peak power PV plant of 6.48 kW_p @ US $3,600/kW, and triple-pane large windows [14].

The energy requirements of this building [15] are depicted in Figure E11.1.2, where the “energy consumption in house” (Endenergiebedarf) per square meter is 18 kWh/(m²-a) and the “primary energy consumption” (Primärenergiebedarf), which represents the energy consumption in house for heating and cooling as well as energy required to process and provide the energy consumption in house (e.g., energy required for natural gas drilling and pipeline operation) and is 46 kWh/(m²-a). These two values represent predicted values –based on calculations-- before the house was constructed. Note “a” means annual.

As of 2014 the energy savings regulations (Energieeinsparverordnung, EnEV) in Germany requires that the total energy consumption of newly constructed residences be reduced by 25%, and furnaces/boilers more than 30 years old must be replaced by more efficient ones.

To check the energy (E) generation and consumption of the house, daily readings of the following energy components were recorded and averaged for each week during one year from July 14, 2013 to July 14, 2014:

a) generated DC energy (EI_i) by PV system feeding two inverters (I) during week i,

b) supplied energy (EG_i) to the utility grid (G) by PV system during week i,

c) consumed energy (EHC_i) for heating (H) and cooling (C) during week i, and

d) consumed energy (ELA_i) by lighting (L) and appliances (A) during week i.

The floor-heating of the house maintained the bedroom temperatures at 17-20°C and all other rooms at 20-22°C. The residence is equipped with a 7.64 kW ground-water heat pump [16]. During summer the cooling of the house is via floor-cooling using ground-water (e.g., 8-11°C) maintaining room temperatures at less than 24°C. The above described generation (EI_i), energy fed-into grid (EG_i), energy requirements for both heating (including hot water heating during entire year for kitchen and baths) and cooling modes (EHC_i) as well as the energy supplied by the utility (ELA_i) are reflected in Figure E11.1.3 and Table E11.1.1.

Note that EI = 7·${\displaystyle \sum _{i=1}^{52}E{I}_i}$+ EI_7/14/2014 = 6577 kWh/year, EG = 7·${\displaystyle \sum _{i=1}^{52}E{G}_i}$+ EG_7/14/2014 = 6088 kWh/year, EHC = 7·${\displaystyle \sum _{i=1}^{52}EH{C}_i}$+ EHC_7/14/2014 = 3376 kWh/year, and ELA = 7·${\displaystyle \sum _{i=1}^{52}\mathit{EL}{A}_i}$+ ELA_7/14/2014 = 593 kWh/year.

The predicted value of 18 kWh/(m²-a) can be checked using some of the values (EHC_i) related to either Figure E11.1.3 or Table E11.1.1 as follows:

With 7·[${\displaystyle \sum _{i=1}^{52}EH{C}_i}$] + EHC_7/14/2014 + 7·[${\displaystyle \sum _{i=1}^{52}\mathit{EL}{A}_i}$] + ELA_7/14/2014] = (3376 + 593) kWh/year = 3969 kWh/year,

$\mathrm{energy}\mathrm{consumption}\mathrm{in}\mathrm{house}\mathrm{without}\mathrm{P}\mathrm{V}\mathrm{system}=\left\{\begin{array}{l}\mathrm{energy}\mathrm{input}\mathrm{or}\mathrm{energy}\mathrm{consumption}\mathrm{in}\mathrm{house}\\ \begin{array}{c}\mathrm{on}{14}^{\mathrm{t}\mathrm{h}}\mathrm{July}2014\left(E,H,C,+,E,L,A\right)–\mathrm{energy}\mathrm{input}\mathrm{or}\mathrm{energy}\\ \mathrm{consumption}\mathrm{in}\mathrm{house}\mathrm{on}{14}^{\mathrm{t}\mathrm{h}}\mathrm{July}2013\left(E,H,C,+,E,L,A\right)\end{array}\end{array}\right\}/\left(220{\mathrm{m}}^2\cdot \mathrm{a}\right)=\left\{5982\mathrm{k}\mathrm{W}\mathrm{h}-2013\mathrm{k}\mathrm{W}\mathrm{h}\right\}/220{\mathrm{m}}^2=3969\mathrm{k}\mathrm{W}\mathrm{h}/220{\mathrm{m}}^2=18.04\mathrm{k}\mathrm{W}\mathrm{h}/\left({\mathrm{m}}^2\cdot \mathrm{a}\right)\text{.}$

This measured value is about the same as predicted by the energy disclosure (18 kWh/(m²·a) of Figure E11.1.2 based on calculations of König [15]. If the generation of the PV system is taken into account via net metering —discussed later— then the energy coefficient of the house will be greatly reduced. Figure E11.1.3 (EI_i) depicts the generation of the PV system as a function of time from 14 July 2013 to 14 July 2014. The CO₂ = 15 (kg-force)/(m²·a) emission (Figure E11.1.2) will be greatly reduced as well, and will approach zero.

The total generated electric energy EI is for this time span 6577 kWh/year taking the inverter efficiency of 0.96 into account one obtains the generated energy available at the meter EI_meter = EI_out = 0.96·6577 kWh/year = 6313.9 kWh/year. Therefore, one obtains:

$\mathrm{energy}\mathrm{consumption}\mathrm{in}\mathrm{house}\mathrm{taking}\mathrm{P}\mathrm{V}\mathrm{generation}\mathrm{into}\mathrm{account}=\left\{\begin{array}{l}\mathrm{energy}\mathrm{consumption}\mathrm{on}14\mathrm{July}2014\left(,E,H,C,+,E,L,A\right)-\mathrm{energy}\mathrm{consumption}\\ \begin{array}{c}\mathrm{on}14\mathrm{July}2013\left(EHC+ELA\right)-\mathrm{generation}\mathrm{of}\mathrm{P}\mathrm{V}\mathrm{system}\left({EI}_{\mathrm{out}}\right)\end{array}\end{array}\right\}/\left(220{\mathrm{m}}^2\cdot \mathrm{a}\right)=\left(5982\mathrm{k}\mathrm{W}\mathrm{h}-2013\mathrm{k}\mathrm{W}\mathrm{h}-6313.9\mathrm{k}\mathrm{W}\mathrm{h}\right)/\left(220{\mathrm{m}}^2\cdot \mathrm{a}\right)=-10.66\mathrm{k}\mathrm{W}\mathrm{h}/\left({\mathrm{m}}^2\cdot \mathrm{a}\right)\text{.}$

This modified energy coefficient corresponds to that of the “passive house” construction [17], and it is negative which is even better. Because of utility policy, the PV system does not supply the water-to-water heat pump system with electricity.

The average coefficient of performance [1] of the heat pump is from April 2013 to November 2014 with the measured heat energy Q_H = 25,070 kWh as provided by the heat pump and (P_compressor + P_{water handlers}) = EHC = (P_compressor + P_{water handlers}) = 6,299 kWh

${\mathrm{COP}}_{\mathrm{H}}=Q_{\mathrm{H}}/\left(P_{\mathrm{compressor}}+P_{\mathrm{water}\mathrm{handlers}}\right)=25,070/6,299=3.98\text{,}$

corresponding to a seasonal energy efficiency ratio (SEER) of

$\mathrm{SEER}=3.413\cdot {\mathrm{COP}}_{\mathrm{H}}=13.58$

which corresponds to a low-efficiency heat pump. However, if the P_{water handlers} required for cooling is neglected in (E11.1-2a) during the summer season then the COP_H will be somewhat above 4. High-efficiency [1] heat pumps have CO_H = 5.57 or SEER = 19. This means there is still room for efficiency increase, requiring higher investment costs.

The feed-in compensation of the PV power fed-into the public utility system is 0.18 Euros/kWh = US$ 0.25/ kWh: of this amount a value-added tax of 19% must be paid in Germany. The feed-in arrangement is shown in Figure E11.1.4 where the supply ELA from utility is the all-time available electricity, EHC includes the high- and low-tariff electricity available from utility only at certain times during working days, nights, weekends and holidays.

The ELA is about 1.65 kWh/day and in addition about 0.6 kWh/day are supplied by the PV plant, that is, self-consumption (e.g., lights, appliances, outlets, electronic equipment) is about 2.25 kWh/day, which amounts to about 13% of the total generated PV power 0.96·EI = 6313.9 kWh/year. Comparing the graphs of Figure E11.1.3, one notes that energy consumption EHC is highest and energy generation (EI or EG) is lowest during winter. Correspondingly, during summer, energy consumption is lowest --even if cooling is employed-- and energy generation is highest. This oppositeness calls for long-term storage facilities such as hydro-pump [18,19] and compressed-air [20,21]. According to Figure E11.1.3 from October to February (EHC + ELA) is larger than {EG+(ELAR = 365days·0.6 kWh/day = 219 kWh/year)} requiring storage plant at the residence. New hydro pump-storage plants are planned in the Black Forest [22] and the Bavarian alpine regions of Germany [23, 24]. As expected in both cases recreationists oppose these new necessary storage plants.

Comparing the annual energy yield of the 6.15kW_p PV plant with one inverter in Boulder, Colorado [1] with the 6.48 kW_p PV plant with two inverters in Munich, Germany, one concludes that for the same array conditions

$\left(\mathrm{energy}\mathrm{yield}\mathrm{in}\mathrm{Munich}\right)/{\left(\mathrm{energy}\mathrm{yield}\mathrm{in}\mathrm{Boulder}\right)}_{\mathrm{measured}}=6242.1/7953=0.785\text{,}$

where the measured energy yield EI in Boulder [1] during one year (12/31/2008 to 12/31/2009) for the 6.15kW_p plant = 18370 kWh-10417 kWh = 7953kWh/year, and the measured energy yield EI in Munich (7/14/2013 to 7/14/2014) for an equivalent 6.15kW_p plant is (6.15 kW_p/6.48 kW_p)· 6577 kWh/year = 6242.1 kWh/year. The ratio (E11.1-3a) is approximately confirmed through calculations using National Renewable Energy (NREL) software [25] resulting in

$\left(\mathrm{energy}\mathrm{yield}\mathrm{in}\mathrm{Munich}\right)/{\left(\mathrm{energy}\mathrm{yield}\mathrm{in}\mathrm{Boulder}\right)}_{\mathrm{predicted}\mathrm{b}\mathrm{y}\mathrm{NREL}\mathrm{software}}\approx 0.65\text{.}$

The above significant difference (17.2%) may lie in the use of two inverters in the Munich plant minimizing cloud influence as compared to using one inverter in Boulder, whereby in the Munich 6.48 kWp residential PV plant two inverters each process approximately half the rated power. In addition the environmental conditions (temperature, snow coverage, clouds) might be somewhat different during 2008/2009 and 2013/2014 time frames. Inverters have an efficiency of about 96%. This was measured when there was no electrical consumption in the house during a sunny day (7/21/2013) in Munich when the generated power was EI_7/21/2013 = 37.57 kWh and the corresponding delivered power to the grid was EG_7/21/2013 = 36.14 kWh resulting in a total inverter efficiency of η_inverter = 36.14/37.57 ≈ 0.96.

11.1.3 Application Example 11.2 and its Solution: Energy Conservation within Residences

The appliances, e.g., LED (light emitting diode) lights [26], refrigerator, induction-type stove in a dwelling consume very little energy as compared to heating of the house. As for the induction-type stove, one must be aware that the induced energy in the bottom of pots is based on a frequency in the range of 20 to 50 kHz [27]. For control of the induction heat either burst-current (Figure E11.2.1) or phase-angle current (Figures E11.2.2a, b) control schemes are applied. Both methods cause power quality problems due to the generation of integer and non-integer current (sub) harmonics. Also, persons with heart pacemakers should keep some distance from the induction-type stove [28]. LED lights are equipped with peak rectifiers [29] generating all odd current harmonics when connected to a single-phase voltage including a dominant 3^rd current harmonic. The high efficient residence exhibits a greater current harmonic total distortion THD_i than low efficient houses equipped with incandescent light bulbs and resistive electric stoves representing linear loads.

Calculation of the output power P as a function of the duty ratio/cycle T_on/T and rated power P_rat:

$P=P_{\mathit{rat}}\frac{T_{\mathit{on}}}{T}\text{.}$

Calculation of the root-mean-squared (rms) voltage V_{t_rms} as a function of the duty ratio T_on/T and the rated voltage V_t__{rms_rat}:

$V_{t\_rms}=V_{t\_rms\_\mathit{rat}}\sqrt{\frac{T_{\mathit{on}}}{T}}\text{.}$

Figure E11.2.2a represents a simple circuit for phase-current control. The firing angle α can be generated by implementing the gating circuits of the thyristors which are modelled by self-commutating switches (MOSFETs) and diodes (D) in series, as described in [1, Chapter 5].

Calculation of the root-mean-squared (rms) voltage V_t__rms as a function of the ratio t_delay/T and the rated voltage V_t__{rms_rat} for phase-angle voltage control:

$V_{t\_rms}=V_{t\_rms\_\mathit{rat}}\sqrt{1-\frac{t_{\mathit{delay}}}{T/2}+\frac{1}{2\pi }sin\left(2\pi \frac{t_{\mathit{delay}}}{T/2}\right)}\text{.}$

The 7.64 kW groundwater (at about 10°C ground-water temperature during the entire year) heat pump provides the floor-heating system with warm water between 24°C and 30°C, and the hot water tank with 45-50°C, whereby the cold season outside temperatures varied mostly between -10°C and + 15°C. Air heat pumps appear not to be very efficient during very cold spells (less than 0°C outdoor temperature [30]). To further increase the heating efficiency of the house, heat exchangers can be employed where ventilated warm air heats the incoming fresh air [31]. This is especially important to avoid mold [32] in houses due to humidity generated by humans. During the warm season when the outside temperature varies between 25°C and 40°C the groundwater is circulated via heat exchangers through the piping within the floor and used for cooling, maintaining the room/house temperature below 24°C. The increase of the thermal efficiency of a house is important because about 20% of the total energy consumption in Germany is used for residences, and 40% of the total consumed energy is used for heating and cooling of all buildings – residential, commercial, and industrial [33]. This is the reason why the German Renewable Energy Act [34] (Erneuerbare-Energien-Gesetz, EEG) was designed to encourage cost reductions based on improved energy efficiency from economies of scale over time. The EU and German Requirements on Energy Efficiency of Residential Buildings – the Energy Conservation Act (Energieeinspargesetz, EnEG) complement the EEG providing Energy Savings Regulations (Energieeinsparverordnung, EnEV) [34]. To date residential consumers in Germany pay about additional 0.0624 Euro/kWh for the switch from conventional electric energy generation based on coal, gas and nuclear plants to renewable sources, the so-called energy turnaround (“Energiewende”) amounting to an electricity price of about 0.33 Euro/kWh including value-added tax. In 2013 renewable energy generation represents 24.5%, and it is anticipated that by 2025 renewable generation will be 45%.

Lesser sources of energy are the geothermal sources, which have water temperatures of about 93°C [35] and are mainly used for district heating. Co-generation (either electricity and heat or electricity and cooling) as well as tri-generation (electricity, heat, and cooling) [36] are part of the generation portfolio. Environmentally sound approaches are biomass [37] and non-recyclable trash power plants [38] in the range of 10 MW.

The deployment of PV and WP plants entails the use of rectifiers and inverters generating harmonics and unwanted electrical noise such as reverse-recovery currents [1]. The magnitude of the DC input voltage of inverters determines the displacement (power) factor angle of the output current: the larger the DC voltage the greater a leading (consumer system) displacement (power) factor angle is possible. To keep the DC voltage of an inverter as low as possible, a power factor of cos φ ≈ 1 is generally preferred. The distributed nature of the renewable sources makes harmonic compensation and elimination more difficult than in a conventional power system with central power stations, where fewer harmonic sources make voltage control through optimal placement of capacitors easier. The large conventional sources such as natural gas, coal and nuclear plants serve as frequency leaders, even in systems employing renewable distributed sources.

Recent publications [39–41] address the issue of supplying residences with DC power directly from PV and WP systems without DC/AC conversion, therefore avoiding the losses in transformers and rectifier/inverter systems. These losses can amount to about 8% at rated operation.

11.1.4 Tiered Energy Rates

To encourage conservation of energy in residences within the European countries progressive energy rates could be introduced. As a function of persons living in a residence, their energy consumption, and their income level “f(y)” the lowest tier of energy should pay “f(y) ·Euro/(unit of energy)”, the second tier should pay “2f(y) ·Euro/(unit of energy), the n^th tier should pay nf(y)·Euro/(unit of energy). This tiered structure of energy rates will encourage conservation so that less wealthy households have access to a minimum amount of energy while wealthy households will have to pay for their excessive energy consumption. Energy is a good which must be available to all populations, not only for the richest group. Of course such a tiered structure should be introduced in some form in all countries, not only in Europe.

11.1.5 Solutions to Power Quality Problems

Conservation of energy is taking place in Germany: the total electrical energy consumption in 2008 was 641 TWh while in 2013 it was 629 TWh [42], where T stands for tera (10¹²). The above tiered energy rates may hasten this conservation process. In a complementary manner the generated electricity originating in solar, wind, hydro, and biomass power plants increased in 2013 to (average) 23.4% [43–45]. Transients due to reverse-recovery currents and switching actions as well as (sub)harmonic currents and voltages caused by burst-current/voltage and phase-angle current/voltage controls can be eliminated by passive and active filters as discussed in Chapter 9. The design of highly efficient drives (e.g., pumping) relies on variable-speed drives causing harmonic currents and voltages as well. The Energy Conservation Act [34] stipulates that PV plants with ratings between 10 kW_p and 1000 kW_p only 90% of the total generated power can be fed into the grid. The solution to this restriction is the construction of short- and long-term storage plants discussed later which rely on rectifiers and inverters causing harmonics in the power network.

At present, the integration of renewable energy sources into the interconnected power system is the predominant mode of operation. This represents challenges for the engineer such as harmonic current/voltage control, power-factor compensation associated with fundamental voltage [46–48] and frequency control [49]. Before 1940 the interconnected power system was not generally in use. Major advantages of interconnection are the power exchange between neighboring power companies and the reduction of reserve capacity. The transmission of renewable energy from high-energy density generation areas (e.g., hydro plants, off-shore wind farms) to high-energy density consumption areas (e.g., cities, manufacturing facilities) is possible with AC and DC high-voltage transmission lines. The latter is in operation, for example, from the Northwest USA to the Los Angeles area [50] at a maximum transmission power capability of 2,000 MW. Interconnected systems are more prone to cascaded failure than stand-alone systems, as has been experienced in the past [51], if sufficient safety and reliability measures are not in place. In addition the susceptibility to geomagnetic disturbances is greater for wide-area interconnected grids [52–54] near igneous rock formations (e.g., Rocky Mountains). For this reason provisions for security, self-healing, reliability, and intentional islanding maintaining power system operation during major emergencies must be in place [55]. Thermal solar systems on roofs in residences [56] in the kW range or on industrial buildings in the 100 MW range are relied on for preheating warm water. Thermal systems in the MW range using mirrors [57] are mostly employed for large-scale applications.

11.2.2 Stand-Alone Mode

Stand-alone or micro grids in particular have the advantage of higher efficiency than interconnected systems because the transmission losses are 8-15% smaller. To date WP sources are not tolerated within city limits due to visual impact and acoustic noise [58]. Thermal [56,57] and PV sources are acceptable. However, they require back-up sources such as diesel generators [59] and long-term storage facilities [60] such as batteries (e.g., flow batteries) and short-term storage such as batteries, DC capacitor banks or flywheels [61]. In micro grids, inverters mostly supply the currents from renewable sources operating preferably at unity power factor/displacement factor to keep the DC input voltage of the inverters as low as possible. The frequency control becomes more difficult compared with an interconnected system due to the lack of a clear frequency leader [49].

11.2.3 Solutions to Power Quality Problems of Photovoltaic and Thermal Solar Systems

To accommodate wind, PV and thermal solar power plants located at distant locations from the consumers it is recommended to build additional AC and DC transmission lines. This calls for the application of fundamental, harmonic [46–50] and transient [62] power flow programs which are readily available. Although some engineers have the vision of a grid with completely renewable but intermittently operating sources, the question of the frequency leader within the power pool must be addressed. This can be done by either natural-gas fired, hydro or long-term storage plants, as will be discussed later.

11.2.4 Custom Power Devices

The most important components for renewable energy systems containing PV and variable-speed WP plants are rectifiers and inverters where the harmonic content and the displacement [1] and power factors can be controlled. Note the displacement factor refers to fundamental voltages and currents while the total power factor takes into account harmonics as addressed in preceding chapters.

11.2.4.1 Pulse-Width-Modulated (PWM) Rectifier

Rectifiers are an integral part for the conversion of AC to DC of most intermittently operating renewable sources such as PV, WP, and storage plants. Figure 11.1 shows one type of rectifier which is used within a three-phase power system. This rectifier is analyzed with PSpice [63] where the input output voltages can be determined as a function of the duty cycle [1, 64]. The PSpice program is listed in Table 11.1 and the computed results are given in Table 11.2. Figures 11.2a, b illustrate some of the results of Table 11.2.

Table 11.1

PSpice input program for PWM rectifier operation

*Controlled three-phase rectifier *input voltages VaN 1 0 sin(0 465 60 0 0 0) VbN 2 0 sin(0 465 60 0 0 -120) VcN 3 0 sin(0 465 60 0 0 -240) *switch gating signal of 3 kHz vg 15 12 pulse(0 50 10u 0n 0n +166.6u 333u) *diodes D1 5 9 ideal D2 10 7 ideal D3 6 9 ideal D4 10 5 ideal D5 7 9 ideal D6 10 6 ideal Df 10 12 ideal *input filter Lfa 1 5 90u Lfb 2 6 90u Lfc 3 7 90u Cfa 1 4 200u Cfb 2 4 200u Cfc 3 4 200u Rf1 4 10 10meg Cf2a 5 8 50u Cf2b 6 8 50u Cf2c 7 8 50u Rf2 8 10 10meg *switch

MOS 9 15 12 12 SMM *snubber resistors and capacitors Rsn 9 11 10 Csn 11 12 0.1u Rsnf 12 13 10 Csnf 13 10 0.1u *output filter and load resistor Ls 12 14 0.001 Cs 14 10 1000u Rload 14 10 8.9 *Model for MOSFET .model SMM NMOS(level = 3 gamma = 0 kappa = 0 + tox = 100n rs = 0 kp = 20.87u l = 2u w = 2.9 delta = 0 + eta = 0 theta = 0 vmax = 0 xj = 0 uo = 600 phi = 0.6 + vto = 0 rd = 0 cbd = 200n pb = 0.8 + mj = 0.5 cgso = 3.5n cgdo = 100p rg = 0 is = 10f) *diode model .model ideal d(is = 1p) *options for improvement of convergence .options abstol = 10u chgtol = 10p reltol = 0.1  + vntol = 100 m itl4 = 200 itl5 = 0 .tran 0.5u 350 m 300 m 0.5 m *plotting software .probe *fourier analysis .four 60 60 I(Rload) .end

Table 11.2

Dependency of the input voltage V_{aN max} of a three-phase rectifier for a given output voltage and current V_{DC load} and I_{DC load}, respectively, as a function of the duty cycle δ

δ = 0.05	IDC load = 45.05A	VDC load = 400.95 V	VaN max = 3000 V
δ = 0.25	IDC load = 45.29A	VDC load = 403.08 V	VaN max = 890 V
δ = 0.50	IDC load = 45.22A	VDC load = 402.46 V	VaN max = 465 V
δ = 0.75	IDC load = 45.40A	VDC load = 404.06 V	VaN max = 320 V
δ = 0.95	IDC load = 45.16A	VDC load = 401.92 V	VaN max = 256 V

Reverse-recovery currents [1]: When semiconductor switches (e.g., diodes, thyristors, MOSFETs, IGBTs) are experiencing a change of polarity of applied voltage (e.g., from forward biased to reverse biased, and vice versa) reverse-recovery currents occur due to the device (parasitic) capacitances. For the circuit of Figure 11.3a the input current i_in, and input and output voltages are computed, where R = 10 Ω, C = 100 μF, and firing angle α = 30°; all other data can be found in the PSpice program listed in Chapter 5 of [1]. In Figure 11.3b the input current i_in exhibits large negative current spikes. These large reverse-recovery currents of the thyristors can be reduced by introducing small inductances in the range of nH between T₁ and T₄, as well as between T₃ and T₂ of Figure 11.3a.

Experimental determination of reverse-recovery current: In the circuit of Figure 11.3c the input voltage at the gate of MOSFET Q₂ consists of two pulses, each having an amplitude of 15 V and a duration (pulse width) of 15 μs; the gate resistance is R_G = 100 Ω.

The time in between the two pulses is 15 μs as well. The rise and fall times t_r and t_f, respectively, of the two pulses one may assume to be t_r = t_f = 0.01 μs. This circuit is simulated using PSpice in Chapter 5 of [1] determining the magnitude A_{reverse-recovery magnitude} and the duration T_{reverse-recovery time} of the reverse-recovery current i_DQ1 through enhancement MOSFET Q₁.

Figure 11.3d shows the current i_DQ1 through MOSFET Q₁ during 200 μs and Figure 11.3e depicts the gating voltage V_in and the current i_DQ1 through MOSFET Q₁ from 90 μs to150 μs. Figure 11.3f shows the reverse-recovery magnitude A_{reverse-recovery magnitude} and the reverse-recovery time T_{reverse-recovery time}, and Figure 11.3g represents the current in the gate of Q₁ and that of Q₂. From these plots one obtains A_{reverse-recovery magnitude} = 12.5 A, and T_{reverse-recovery time} = 0.204 μs.

Differential-mode and common-mode electrical noise: Differential-mode noise is the most common one and can be removed or mitigated through passive or active filters discussed in Chapter 9. Common-mode noise occurs frequently in gating circuits for semiconductor switches and for internet connections as well as power supply leads for computers operating at low voltage levels (e.g., 3.3 V). Common-mode noise is caused through poor grounding conditions and is a function of the lengths of wires/cables. Differential-mode and common-mode noise is explained in Figures 11.4a–g [65] and the arrangement of a core-type filter with very high permeability μ is illustrated in Figures 11.5a, b.

The differential-mode current is (see Figure 11.4b) by definition

$i_{\mathrm{differential}}=\left(i_{\mathrm{s}}-i_{\mathrm{g}2}\right)/2\text{,}$

  (11-1a)

and the common-mode current is by definition

$i_{\mathrm{common}}=\left(i_{\mathrm{s}}+i_{\mathrm{g}2}\right)/2\text{.}$

  (11-1b)

By definition

$i_{\mathrm{s}\mathrm{d}}=-i_{\mathrm{g}2\mathrm{d}}\text{,}$

  (11-1c)

where i_sd is the differential-mode current of the signal current i_s and i_g2d is the differential-mode current of the ground current i_g2, and

$i_{\mathrm{s}\mathrm{c}}=i_{\mathrm{g}2\mathrm{c}}\text{,}$

  (11-1d)

where i_sc is the common-mode current of the signal current i_s and i_g2c is the common-mode current of the ground current i_g2.

With

$i_{\mathrm{g}3}=i_{\mathrm{s}}+i_{\mathrm{g}2}=\left(i_{\mathrm{differential}}+i_{\mathrm{common}}\right)+\left(-i_{\mathrm{differential}}+i_{\mathrm{common}}\right)=2i_{\mathrm{common}}=i_{\mathrm{g}1}$

  (11-1e)

one obtains due to the ideal ground condition i_g3 = i_g1. This shows that the common-mode current can only exist for poor grounding conditions.

A popular filter for the suppression of differential-mode noise is the low-loss capacitor. If the differential-mode noise current i_d through capacitor C is proportional to hωC and the signal current i_s through capacitor C is proportional to ωC then

$i_{\mathrm{d}}/i_{\mathrm{s}}=h\text{,}$

  (11-1f)

if h is very large (e.g., h → ∞) then the differential current i_d is short-circuited by the capacitor, or in other words for a finite v_s across the capacitor, the capacitor provides a short circuit for i_d.

In contrast, a good filter for the suppression of common-mode noise is a doughnut/core type high permeability ring as illustrated in Figures 11.5a, b. For high permeability (μ = μ_r μ_o → ∞) one obtains from Ampere’s law applied to Figure 11.5a, see Eq. (11-1e), for a finite flux density due to the finite voltage v_s, the finite core length ℓ, and with the number of turns N = 1:

$\left(i_{\mathrm{s}}+i_{\mathrm{g}2}\right)\cdot N=2i_{\mathrm{common}}\cdot N=H\cdot \mathrm{\ell }=\left(B/\mu \right)\cdot \mathrm{\ell }\to 0,\mathrm{or}{i}_{\mathrm{common}}\approx 0\text{.}$

  (11-1g)

11.2.4.2 Current-Controlled PWM Inverter

Similar to rectifiers, inverters (see for example Figures 11.6a, b) must be employed for some renewable energy sources and for storage plants to convert DC to AC. Table 11.3 lists the PSpice program on which the results of Table11.4 are based. This table illustrates the dependency of the input DC voltage V_DC of inverter as function of the output displacement (power) factor angle φ, that is, the angle between the fundamental of the phase current of inverter i_aN and the fundamental of the line-to-neutral voltage v_aN of power system as well as the modulation index m [1, 64]. Figures 11.7a, b illustrate some of the results of Table 11.4. According to IEEE Standard 519 [66] and IEC 61000-3-2 (2001-10) [67] a total harmonic distortion of the inverter output current I_THDi of about 3% --ignoring the switching ripple which can be mitigated by an output filter as indicated in Figure 11.6a— is acceptable. By decreasing the modulation index 0 < m < 1.00, say to m = 0.5 which increases V_DC, and by increasing the wave-shaping inductance L_w an almost ideal sinusoid for the output current can be achieved and reactive power can be supplied to the power system. However, this approach might not be feasible in most cases, resulting in low real power output of inverter.

Table 11.3

PSpice input program for PWM inverter operation at a gating signal voltage (vtrial) frequency of 5.76 kHz at lagging displacement (power) factor (generator notation, over-excited, delivering reactive power to system)

*Current-controlled pulse-width-  +modulated (PWM) voltage-source +inverter vsuppl 2 0 360 *switches msw1 2 11 10 10 qfet dsw1 10 2 diode msw5 2 21 20 20 qfet dsw5 20 2 diode msw3 2 31 30 30 qfet dsw3 30 2 diode msw4 10 41 0 0 qfet dsw4 0 10 diode msw2 20 51 0 0 qfet dsw2 0 20 diode msw6 30 61 0 0 qfet dsw6 0 30 diode *waveshaping inductors l_w1 10 15 1 m l_w2 20 25 1 m l_w3 30 35 1 m R_w1 15 16 10 m R_w2 25 26 10 m R_w3 35 36 10 m *current references in terms of voltages vref1 12 0 sin(0 56.6 60 0 0 0) vref2 22 0 sin(0 56.6 60 0 0 -120) vref3 32 0 sin(0 56.6 60 0 0 -240) eout1 13 0 15 16 100 eout2 23 0 25 26 100 eout3 33 0 35 36 100 *errorsignals produced as difference + between vref and eout rdiff1 12 13a 1 k rdiff2 22 23a 1 k rdiff3 32 33a 1 k cdiff1 12 13a 1u cdiff2 22 23a 1u cdiff3 32 33a 1u rdiff4 13a 13 1 k rdiff5 23a 23 1 k rdiff6 33a 33 1  ecin1 14 0 12 13a 2 ecin2 24 0 22 23a 2 ecin3 34 0 32 33a 2 vtria1 5 0 pulse(-10 10 0 86.5u + 86.5u 0.6u 173.6u) *gating signals for upper switches xgs1 14 5 11 10 comp xgs2 24 5 21 20 comp xgs3 34 5 31 30 comp *gating signals for lower switches egs4 41 0 poly(1) (11,10) 50 -1 egs5 51 0 poly(1) (21,20) 50 -1 egs6 61 0 poly(1) (31,30) 50 -1 *Filter is deleted because PSpice is limited to + 64 nodes

*lfi1 16 15b 45u *lfi2 26 25b 45u *lfi3 36 35b 45u *rfi1 15b 15c 0.01 *rfi2 25b 25c 0.01 *rfi3 35b 35c 0.01 *cfi1 15c 26 10.3u *cfi2 25c 36 10.3u *cfi3 35c 16 10.3u *parameters of power system with a line-to-  + line voltage of 340 V RM1 16 18 50 m LM1 18 19 265u Vout1 19 123 sin(0 196 60 0 0 -30) RM2 26 28 50 m LM2 28 29 265u Vout2 29 123 sin(0 196 60 0 0 -150) RM3 36 38 50 m LM3 38 39 265u Vout3 39 123 sin(0 196 60 0 0 -270) *subcircuit for comparator .subckt comp 1 2 9 10 rin 1 3 2.8 k r1 3 2 20meg e2 4 2 3 2 50 r2 4 5 1 k d1 5 6 zenerdiode1 d2 2 6 zenerdiode2 e3 7 2 5 2 1 r3 7 8 10 c3 8 2 10n r4 3 8 100 k e4 9 10 8 2 1 *models for zener diodes .model zenerdiode1 D (Is = 1p BV = 0.1) .model zenerdiode2 D (Is = 1p BV = 50) .ends comp *model for switch .model qfet nmos(level = 3 gamma = 0 kappa = 0  + tox = 100n rs = 42.69 m kp = 20.87u l = 2u + w = 2.9 delta = 0 eta = 0 theta = 0 vmax = 0   + xj = 0 uo = 600 phi = 0.6 + vto = 3.487 rd = 0.19 cbd = 200n   + pb = 0.8 mj = 0.5 cgso = 3.5n cgdo = 100p + rg = 1.2 is = 10f) *model for diodes .model diode d(is = 1p) *options to aid convergence .options abstol = 0.01 m chgtol = 0.01 m  + reltol = 50 m vntol = 1 m itl5 = 0 itl4 = 200 *transient analysis .tran 5u 350 m 300 m 5u *plotting of traces .probe *Fourier analysis .four 60 12 I(l_w1) .end

Table 11.4

Dependency of the input DC voltage V_DC of inverter as a function of the output displacement (power) factor angle φ (generator notation, lagging displacement power factor, where a negative φ corresponds to over-excited operation delivering reactive power to the system) is the angle between phase current of inverter I_{rms ph} and phase line-to-neutral voltage V_{rms l-n} of power system and m is the modulation index of inverter

φ = 90°	Vrms l-n = 139 V	Irms ph = 58.18 A	VDC = 375 V	ITHDi = 0.107%	m = 1.04
φ = 60°	Vrms l-n = 139 V	Irms ph = 54.17A	VDC = 368 V	ITHDi = 0.093%	m = 1.07
φ = 30°	Vrms l-n = 139 V	Irms ph = 50.44 A	VDC = 360 V	ITHDi = 0.131%	m = 1.09
φ = 0°	Vrms l-n = 139 V	Irms ph = 49.03 A	VDC = 388 V	ITHDi = 0.117%	m = 1.01
φ = -30°	Vrms l-n = 139 V	Irms ph = 49.74A	VDC = 410 V	ITHDi = 0.060%	m = 0.96
φ = -60°	Vrms l-n = 139 V	Irms ph = 52.06 A	VDC = 415 V	ITHDi = 0.085%	m = 0.95
φ = -90°	Vrms l-n = 139 V	Irms ph = 55.46 A	VDC = 415 V	ITHDi = 0.073%	m = 0.95

11.2.5 Operation at Maximum Power Point

PV arrays [68] as well as wind turbines [69] must be operated at the maximum power point to increase the yield of such plants. Many schemes for searching the maximum power point exist, however to operate at the maximum-power point without searching for such a point is very difficult. Searching for the maximum point entails a change in output of such sources causing frequency variations in the grid which might not be acceptable. PV power plants operate mostly intermittently due to changing insolation as is shown in Figure 11.8 recorded at the San Luis Valley of Colorado [70]. Figure 11.9 explains that for a PV power plant under certain conditions where two maxima may exist.

11.2.6 Safety Considerations

Renewable sources are mostly of smaller rating (100 MW or less) while for conventional power plants units from 100 MW to 600 MW or even 1200 MW are presently employed. The first one of the conventional plants is the output of a natural gas turbine, the second one that of a coal-fired plant, and the latter one that of a nuclear plant [71]. The reason for large ratings is efficiency -- the larger the unit the larger the efficiency. Single units have an efficiency of about 28 to 35%, while combined-cycle power plants -- where natural gas/oil turbines and steam turbines connected in series have efficiencies of about 60% [72]. In a combined-cycle plant the natural gas is combusted in the gas turbine and the high-temperature exhaust of the gas turbine heats steam for a steam turbine. Gas turbine power plants become important for distributed generation including renewable energy sources due to their fast cycle properties, that is, they can change their output quickly.

Due to the smaller rating of renewable sources many distributed plants must be operated on the grid, aggravating the control of the voltage associated with reactive power/load flow and frequency. In almost all cases the distributed source relies on the voltage of the power system, that is, the voltage cannot be changed by an individual renewable source of small rating and the current fed into the grid can only be controlled. In other words, as soon as the voltage of the grid is below or above certain limits the renewable source of small rating must be disconnected for safety reasons from the grid. No matter how good the solar panels on the roof are, they are not good enough to collect energy at night. But researchers at the Japan Aerospace Exploration Agency have hatched a plan to harvest the sun’s energy nearly 24 hours a day. The idea is to launch solar-energy collectors into geosynchronous orbit. The special satellites would beam microwave energy down to an array of tiny rectifying antennas that would convert microwaves into DC electricity [73]. Problems with this approach are the microwave beams in certain areas of the globe, which must be avoided by air planes, animals, and humans. In residential areas the glare of reflecting light from solar arrays might be a problem for some of the neighboring residences [74]. In particular in urban areas birds (e.g., pigeons) might find the gap between the solar arrays and the roof tiles an ideal nesting place in particular during cold seasons due to the warmth radiated from PV systems. These bird populations can cause damage on the structure of the building, the excrements of the birds cause a smell, parasitic bugs might be transmitted, and damage the wiring under the PV arrays [75].

11.2.7 Solutions to Power Quality Problems of Custom Power Devices with Switching Action

Any switching action causes transients due to parasitic capacitances of either a mechanical or semiconductor switch, resulting in the latter case in reverse-recovery currents. The suppression of these in thyristors, metal-oxide-field-effect transistors (MOSFETs), insulated-gate bipolar transistors (IGBTs) and others is discussed in [1] through the insertion of small inductors or through the minimization of parasitic capacitances. The differential-mode noise can be reduced by either passive or active filters (Chapter 9) and common-mode noise is minimized through reduction of grounding resistance through common-mode filters, e.g., ferromagnetic coil wrapped around a high-permeability core enclosing the positive and negative leads. Rectifiers and inverters can generate integer and non-integer or fractional harmonics as well as inter- and sub-harmonics. The displacement (fundamental power) factor of inverters can be made leading (consumer system of notation) by increasing the input DC voltage of the inverter. This calls for multilevel inverters where switches must be connected in series due to their limited maximum permissible reverse voltages. For the reduction of the switching harmonics a low modulation index m (e.g., 0.5 to 0.8) is recommended. Although many papers on maximum power point tracking for PV plants exist, no single approach is completely satisfactory. More work in this respect must be done, especially for PV plants with output power in the MW range: searching for the maximum power point induces frequency variations due to large varying output powers. Safety in distributed power systems with many sources can be improved through internet and power–line carrier control.

WP plants operate mostly intermittently due to changing wind conditions as is shown in Figure 11.10.

Some WP plants do not employ mechanical gears [1] and permanent-magnet synchronous generators (PMSG) with many poles (e.g., 40 or larger) [77, 78] are used resulting in pancake-type generators with a short axial length, large diameters and weights – they are called direct-drive plants. Some WP plant designs, however, employ mechanical gears because WP blades rotate at low speed (e.g., 0.2 rps) while generators must be designed –to reduce their weight --for high speed (3600 rpm corresponding to 60 rps). The changing speed conditions of WP plants wear out the mechanical gear and therefore, electrical gears can be employed based on pole-changing and series-turns techniques [79, 80]. When a sudden change of the pole numbers occurs through mechanical switching of poles, say from 8 to 4 poles, sudden transient torques stress the mechanical gear. To mitigate these stresses, pole-changing techniques combined with a change in the number of series turns of the generator by electric means, supplying rectifiers/inverters, are recommended resulting in electronic gears which also can be used for electric vehicles permitting a large gear ratio and therefore high efficiency at high speed.

It is possible to build on-shore 175 m tall [81] and off-shore 6 MW 100 m [82] WP plants. In Spain a 195 m tall chimney rose above the plain in Manzanares south of Madrid: the solar updraft tower [83], which had turbines at its base, rose from the middle of a 46, 000 square-meter greenhouse-like collection area; it generated up to 50 kW.

11.3.2 Basic Principle of (V·p)/f and (V/f·N) Control of Electronic Gear [79, 80]

The induced voltage E of an alternating current electric machine

$E=4.44\cdot f\cdot B_{\text{max}}\cdot N_{\mathit{rated}}\cdot 4\cdot R\cdot L/p$

is related to the rated maximum flux density B_max, the rated number of series turns per phase N_rated, the radius of the location of the stator winding R, the active (core) machine length L, the frequency of the voltages and currents f, and the pole number p. For rated induced voltage E_rated the frequency f and the maximum flux density assume rated values. If E < E_rated then the flux density is less than its rated value, and for E > E_rated the flux density will be above its rated value. The relation

$\frac{E}{f}=4.44\cdot B_{\text{max}}\cdot N_{\mathit{rated}}\cdot 4\cdot R\cdot L/p=\mathit{const}\text{.}$

must be maintained for all operating conditions where E $\le$ E_rated and f $\le$ f_rated. For f > f_rated the induced voltage can be at the most E = E_rated and the flux density will be less than its rated value, called flux-weakening operation. In this case the induced voltage E can be replaced in the above formula by the terminal voltage V_t. By changing the number of series turns per phase N and the pole number p of the machine flux weakening can be compensated and the machine can be operated under rated flux density under most operating conditions. This results in superior performance above rated speed (e.g., increased output power) and below rated speed (e.g., increased torque) of variable-speed drives.

It is well-known that the torque-speed characteristic of commonly available variable-speed electric machines employing field (flux) weakening may not match the load torque because the load torque may be proportional to the square of the speed, and that of a variable-speed electric machine is inversely proportional to the speed in the field-weakening region. This applies to variable-speed drives where the critical (maximum) speed is limited by the reduction of the developed torque, resulting in less than the rated output power. This mismatch can be mitigated by an electronic change in the number of poles p and/or a change in the number of series turns per phase N of the stator winding via application-specific-integrated circuit (ASIC). The change in the number of poles p is governed by

$\frac{E\cdot p}{f}=4.44\cdot B_{\text{max}}\cdot N_{\mathrm{rated}}\cdot 4\cdot R\cdot L\text{.}$

The change in the number of series turns per phase N is governed by

$\frac{E}{f\cdot N}=4.44\cdot B_{\text{max}}\cdot 4\cdot R\cdot L/p\text{.}$

The introduction of an additional degree of freedom --either by a change of p or N or both-- permits an extension of the rated flux density (B_max) region up to a multiple of the rated speed. Figure 11.11 illustrates in a qualitative manner the speed n_m, torque T (locus of maximum torque), and power P increase due to the change in the number of turns from N_rated to N_rated/2.

11.3.2.1 Starting and Motoring Operation

Some WP turbines must be brought up to speed by the generator acting as a motor. The overall speed-torque diagram of a 1 kW induction machine --which circuit data are listed below-- is shown in Figure 11.12 with 4 natural speed-torque characteristics and the characteristic which will be important for starting at V_{line-line-rated} and f_start < f_base (characteristics in quadrant I) where p₁ = 8 poles, n_base = 750 rpm, and T_rated = 2.8 Nm. In Figure 11.12 characteristic 1 pertains to f_base, N_rated^p1 and p₁; characteristic 2 is valid for f_base, N_rated^p2= N_rated and p₂; characteristic 3 is based on 1.5∙f_base, N_reduced = N_rated/2 and p₂; characteristic 4 is computed for 2∙f_base, N_reduced = N_rated/2 and p₂, and the starting characteristic is computed for f_start, which is less than f_base. The current drawn at zero speed is I_start > I_rated. As start-up occurs the starting current will decrease due to the frequency and the associated speed increase, and the machine in motor operation reaches at the base frequency f_base = 50 Hz the base speed. The transition from (natural) characteristic #1 to (natural) characteristic #2 occurs through pole switching from p₁ poles to p₂ poles [79, 80]. The transition from the natural characteristic #2 to the characteristic #3 occurs based on flux weakening (V/f) control, whereby the flux density is reduced to below its rated value. At the characteristic #3 the number of series turns $N_{\mathit{rated}}^{p2}=$N_rated will be reduced to N_reduced = N_rated/2 by a relay (having normally closed, NC, and normally open, NO, contacts) or electronic switches, thus effectively restoring the flux density to its rated (or larger) value and therefore increasing the torque and the output power of the motor. This increase in torque is about proportional to the ratio (N_rated/N_reduced). From characteristic #3 --at an increased (about rated) flux density and at, for example, 3 times the base speed -- to characteristic #4 (at about 4 times the base speed) flux weakening (V/f) control with N_reduced is performed. It is advisable to minimize the speed range between characteristics 2 and 3, as depicted in Figure 11.12. The frequency 1.5∙f_base has been chosen to clearly illustrate the effect of the reduction in the number of turns, but an optimization of the dynamics of the drive requires that 1.5∙f_base should be less, say 1.1∙f_base at reduced terminal voltage V_t.

The data of analyzed 3-phase induction machine of Figures 11.12 are P_rat = 1 kW, number of poles p = 4, f_rated = 50 Hz, V_L-N/V_L-L = 220/380 V, η = 0.735, cosφ = 0.705, B_max = 0.68 T, X_m = 130 Ω, X_s = 9.45 Ω, R_s = 7.55 Ω, X_r^'= 9.47 Ω, R_r^'= 6.65 Ω, stator bore diameter = 0.074 m, stator core length = 0.095 m.

11.3.2.3 Experimental Verification [84]

For testing, a 4-pole, 1 kW induction machine was purchased off the shelf and rewound to accommodate the 3 different configurations (e.g., Δ at p₁ = 8 poles, Y at N_rated and p₂ = 4 poles, and Y at N_rated/2 and p₂ = 4 poles) so that the number of poles and turns could be externally changed. This machine is about comparable to the machine data listed above except that the tested machine has a rated frequency of 60 Hz. The equivalent circuit of this three-phase induction machine at the rated phase voltage of 70 V during motoring operation is depicted in Figure 11.13. Table 11.5 lists the measured circuit parameters for all 3 configurations based on open-circuit and short-circuit tests at f = 60 Hz. Note that for all applications the terminal (line-to-line) voltages were below or equal to the rated value of about 120 V. This is important for electric car applications where for safety reasons the source voltage must be below a certain value (e.g., battery voltage of 300 V). A flux weakening compensation via increase of the battery voltage from 300 V to 600 V is not desirable for road vehicle applications.

Table 11.5

Measured equivalent circuit parameters for 8-pole and 4-pole configurations at f = 60 Hz

configuration	Rs [Ω]	Xs [Ω]	Xr′ [Ω]	Rr′ [Ω]	RFe [Ω]	Xm [Ω]
8-pole (Nrated)	3.13	0.84	0.84	0.077	12	3.49
4-pole (Nrated)	1.79	2.37	2.37	1.418	288	34.9
4-pole (Nrated/2)	0.56	0.67	0.67	0.389	63	8.66

Figure 11.14 depicts the measured starting torque (T_start) as a function of the starting current (I_start) for 10 Hz and 60 Hz operations. It is well known that induction machines cannot operate at a very low frequency (e.g., 3 Hz) applied to the stator because of the increase in the magnetizing current due to the decrease of the magnetizing reactance X_m.

In this application the lowest stator frequency that could be applied while maximizing the starting torque was 10 Hz. Unfortunately the inverter used could not deliver more than 15 A which is why the torque curve at 10 Hz is limited to 15 Nm. The 60 Hz operation was limited by the autotransformer to about 18 A. The current during starting can greatly exceed the rated current for a short time (few seconds) because the thermal time constant of an electric machine is in the range of a few hours [85].

The start-up current transient for motoring operation is shown in Figure 11.15. This oscillogram demonstrates the switchover from p₁ = 8 with N_rated to p₂ = 4 with N_rated, and from p₂ = 4 with N_rated to p₂ = 4 with N_rated/2. The latter switchover as shown in Figure 11.15 can be controlled, in other words, the torque can be increased as a function of the switching instant and the applied voltage and frequency. It is advisable to minimize the operating time of the p₂ pole, N_rated configuration. At the time of switching from p₂ = 4 poles at N_rated to p₂ = 4 poles at N_rated/2 operation the current spike increases when the p₂ = 4 and N_rated/2 operation is starting at low frequency. N_rated/2 operation (at low frequency) corresponds to low reactance and resistance circuit values of the machine. Therefore, maintaining the terminal voltage at the low impedances results in an increased input current.

11.3.2.6 Generator Operation

Figure 11.16 shows the output voltage of the motor (upper oscillogram) which is the input voltage to the rectifier, and the output voltage of the rectifier (lower oscillogram) measured at about the rated power of the induction machine of 1 kW. The rectifier input voltage is 152 V_rms, the rectifier output voltage is 202.5 V, and rectifier output current is 4.84 A.

11.3.3 Influence of Tower on Energy Production and Flicker

In horizontal WP plants the rotor blades rotate in the wind shadow of the tower and therefore the blade’s torque is periodically changed at low frequencies causing flicker due to changing (wind) shadowing [92, 93]. In addition, this flicker effect must be eliminated within the output voltage of the generator due to causing ill effects in humans such as nausea and epilepsy [94].

11.3.4 Constant-Speed Wind Power Plants

Constant-speed wind machines were used in the first generation of WP plants [95]; it turned out they are less efficient and prone to mechanical failure due to mechanical transient stresses caused by changing wind conditions. On-shore WP towers can now be 175 m high due to the availability of large cranes [81] which increases the efficiency and decreases fluctuation due to wind conditions.

11.3.5 Variable-Speed Wind Power Plants

These machines can handle changing wind speeds due to the combination of either permanent-magnet synchronous generators (PMSG) [77, 96–103], or doubly fed induction generators (DFIG) with transformer/rectifier/inverter combinations for rotor excitation [89] where the variable generator frequency of the rotor f_rot is transformed via AC-DC-AC conversion from the systems frequency of nominally f_syst (e.g., 60 Hz). The required power of the rotor AC-DC-AC conversion system is only about 8% of the stator output power and the speed of the rotor can be controlled n_rot = n_rat ± 30%.

11.3.6 Operation of Wind Farms

Wind farms --based on the horizontal drive --built in remote areas due to their visual impact [104, 105] and acoustic noise [106] represent a significant part (about 15% of total energy generated) of renewable energy development as compared to solar thermal and PV plants contributing about 5% [107,108] of renewable energy generated or 0.7% of total energy generated. Wind farms consist of a great number --a few hundred to thousands-- of individual plants in the 1 to 8 MW range resulting in outputs of a few hundred megawatts [109] provided the wind drives them at rated power. If properly located, managed and operated wind farms behave like a large plant similar to a conventional plant: this is especially true when wind farms are located on-shore near the ocean or on islands e.g., Ireland [110]. The construction of wind farms off-shore appears to be more complicated due to the rapid drop of the shelf in the southern and western shores [111] of the USA and due to local opposition [112] near the eastern shore. An attempt has been made to construct WP plants/farms in different parts of the USA in order to exploit their different wind conditions: for example in the northern and the southern parts of Colorado [113]. The operation of WP plants is not possible at low and high wind conditions. At low wind speeds there is not sufficient torque and at high wind conditions the blades must be protected by furling [114]. The optimal wind speed depends upon the design (e.g., constant, variable-speed) and output power rating of the plant and is in the range of 3.5-25 m/s [115]. It is best to operate wind plants at sea level due to the highest density of the air. WP plants built on very high mountain ridges are more costly: wind conditions are not as steady and the air is not as dense as in the plains [116] resulting in more difficult power quality problems through changing wind conditions.

The control of power plants in general — including WP facilities — is based on the recently developed principle of synchronized phasors (synchrophasors) [12, 117, 118]. Synchrophasors make real-time measurement of electrical quantities such as voltages, currents, various (reactive, real, apparent) powers, displacement power factor, (total) power factor including harmonics, transients, and frequencies from across the power system possible. Applications include wide-area control [119], system model validation [120], determining stability margins [121], maximizing stable system loading [122], islanding detection [123], system-wide disturbance recording, and visualization of dynamic system response. The basic system building blocks are Global Positioning System (GPS) satellite-synchronized clocks [124], phasor measurement units (PMUs) [125], a phasor data concentrator (PDC), communication equipment, and visualization software as outlined in [126] by Schweitzer Engineering Laboratories (SEL).

Siemens finalized the construction contract [127] of a wind farm with the Cape Wind project in New England, on Horseshoe Shoal, Massachusetts that could be a significant step to advance wind farm activity in the United States. The wind turbines are to be spaced 500 m to 900 m apart. Siemens will supply 130 offshore turbines, each 3.6 MW. One of the reasons the off-shore wind industry in the US has not matched its WP potential has been opposing groups' objections to "visual pollution" and harm to birds, just two of the arguments that have been used against such initiatives. Another reason is the sharp drop-off of the shelf in the southern part of the USA.

For the first time an off-shore interconnected system is planned [9] involving the Scandinavian (Denmark, Sweden) and Central European (Germany) countries within the Baltic Sea connecting two off-shore wind farms operated by different countries. This interconnection (“Interkonnektorenloesung”) will serve as a pilot project for Europe about 30 km northwest of the island of Rügen. Via a converter station, starting in 2018, two parallel 150 kV cables will connect a Danish wind farm of 600 MW with the Energie Baden Württemberg (EnBW) wind farm “Baltic 2” of 288 MW. Both wind farms lay 15 km apart. This technical concept may be extended and include additional connections. The varying generated electricity can be utilized either in Scandinavia or in Germany without the need to curtail WP generation. Until now when there are strong wind conditions some of the wind turbines must reduce generation in order to prevent that the generation will be larger than the consumption, leading to stability problems. Any connection to other consumers is a contribution to the optimization of the grid. During low wind conditions the electricity generated through hydropower in Scandinavian countries can be supplied to Germany.

Vertical axis WP plants have not been used in wind farms due to their lower output powers. However, such plants might be acceptable for operation near residences. One problem with such designs is the non-constant torque development [128] which can be mitigated through power electronic and storage components.

11.3.7 Control of Wind Power Systems

Constant-speed induction generators have lower power efficiency than variable-speed generators of the PMSG (Figures 11.17–11.22) and DFIG types [129] which have pitch-blade control. The advantages of variable–speed control include maximum power extraction and improvement in dynamic behavior of turbine. The PMSG and DFIG are the most widely employed variable-speed wind generator systems. Note that the power coefficient C_p(λ,β) of wind energy conversion systems (WECS) depends upon the tip-speed ratio λ and the blade pitch angle β [130]. The DFIGs are simpler and less expensive due to the smaller rating (about 8% of the rated stator output power) of the rotor converters (rectifier, inverter) but require regular maintenance of slip rings and carbon brushes. Any grid voltage fluctuation may produce oscillations in stator output power and can cause rotor torque pulsations. PMSG systems have a higher efficiency with reduced size and higher power density but require more complicated control circuitry to maintain converter voltages. In both cases a mechanical gear is not required: the pole number of the PMSG is large and the rotor converter frequency control of the DFIG can accommodate at least ± 30% speed deviation from rated speed. One of the important aspects of variable-speed WECS is to capture maximum energy from wind. Maximum power point tracking (MPPT) is important similar to PV systems. It can be achieved through an intelligent adaptive radial basis function neural network (RBFNN) controller [131] for PMSG wind systems. A superconducting magnetic energy storage system (SMES) is proposed next to the grid-side inverter [132]. Use of storage capacitors [131] are proposed as well to compensate the power fluctuation of the WECS. In [133] the stable operation of doubly fed induction generators (DFIGs) is analyzed when installed in weak and islanding/standalone power systems utilizing the swing equation [134], equal area criterion [134], and the 1.5 MW DFIG model of General Electric Company [135].

11.3.8 Maximum Power Extraction from Horizontal-Axis Wind Turbine

The Lanchester-Betz-Joukowsky [1] limit — and the maximum efficiency for horizontal axis wind turbines -- is discussed in [137–139]. Unfortunately, it does not apply to vertical-axis wind turbines addressed next.

11.3.9 Vertical-Axis Wind Turbines

This type of wind turbine [140–142] might be suitable for either urban or remote areas. The key to their use is the reduction of audible noise, visual acceptance limiting their output to less than 50 kW, the management of the non-constant torque, and the employment of electronic gears and electrical storage to mitigate their non-constant power output.

11.3.10 Solutions to Power Quality Problems of Wind Power Plants

The consequences (e.g., acoustic noise, height restrictions, and visual pollution) of operating vertical-axis WP plants within residential areas should be researched. With respect to horizontal-axis plants off-shore WP farms/parks appear to be costlier to construct and to operate. The on-shore approach should be therefore favored until the mechanical stresses on the wind turbine blades are better understood due to strong wind conditions in off-shore regions. Permanent-magnet synchronous generators (PMSG) require expensive rare earth material (e.g., neodymium) making them expensive. For this reason it is proposed to rely on inexpensive squirrel-cage induction generators where the power factor/magnetizing current can be provided and controlled by the PWM rectifier feeding the PWM inverter. PMSG and DFIG require no mechanical gear, however, are very heavy due to the large pole numbers. Research into electronic gears should be supported in order to operate generators with low pole numbers resulting in low weight at variable speed.

11.3.11 Short- and Long-Term Energy Storage Systems

Electric storage components can store electricity in DC form only [143]. For this reason AC-DC converters (rectifiers), DC–AC converters (inverters), and DC-DC converters (step-down and step-up) must be relied on. All types of converters are discussed in [1].

11.3.12 Role and Design of Short-Term and Long-Term Storage Plants [144]

Short-term storage devices such as conventional battery, super-capacitor, and flywheel can be put online within a few 60 Hz cycles but cannot provide energy for more than about 10 minutes; however, flow batteries and variable-speed hydro plants (with DFIGs) can change their load within a few 60 Hz cycles and are able to deliver power for days. Long-term storage plants such as constant speed (pump-)hydro storage and compressed air plants require a start-up time of about 6-10 minutes (due to synchronization), but can operate for several hours or even days. Two short-term storage plants will be analyzed below.

11.3.13 Storage Devices for Short-Term and Long-Term Plants

Conventional batteries, super-capacitors, ultra-capacitors, and flywheels serve as short-term storage devices, while compressed air, (constant and variable speed) hydro plants, and thermal storage are used as long-term storage facilities. Variable-speed drives [77, 79] are proposed for short- and long-term storage plants in order to eliminate mechanical gears.

11.3.14 Hydro-Storage or Pump-Storage Plants

Hydro pump-storage plants require upper and lower reservoirs at different altitudes, where the water can be recycled during a pumping-generation cycle. Of course a sufficiently large source of water replenishing the loss of water due to evaporation must be available at either the upper or lower reservoir. This type of storage is of the long-term type with a fairly quick response (e.g., seconds) if variable-speed generators, e.g., doubly fed induction generators (DFIGs) for pump-storage, hydro plants similar to WP plants [145] are employed.

11.3.14.1 Application Example 11.3: Design of a 250 MW Pumped-Storage, Hydro-Power Plant [1]

A pumped-storage, hydro-power plant (Figure E11.3.1) is to be designed [18, 19, 146–148] for a rated power P_rated = 250 MW and a rated energy capacity 1500 MWh per day. It consists of an upper and a lower reservoir with a water capacity C_H2O each. In addition there must be an emergency reserve for 625 MWh. Water evaporation must be taken into account and is 10% per year for each reservoir, and the precipitation per year is 20 inches. The maximum and minimum elevations of the upper reservoir are 1050 ft and 1000 ft, respectively. The maximum and minimum elevations for the lower reservoir are 200 ft and 150 ft, respectively. The water turbine is of the Francis type and it is coupled with a salient-pole synchronous machine with p = 24 poles, which can be used as a generator for generating electricity by releasing the water from the upper reservoir to the lower one, and as a motor for pumping the water from the lower reservoir to the upper one. A capacity factor of 100% can be assumed: in a real application the capacity factor may vary between 70-90%.

a) If the power efficiencies of the water turbine, the synchronous generator, and the Δ-Y transformer are η_turbine = 0.8, η_{synchronous machine} = η_{Y-Δ transformer} = 0.95, respectively, compute the required turbine input power P_turbine^required.

b) Provided the head of the water is H = 850 ft, the frictional losses between water and pipe amount to 15%, and the water flow measured in cubic feet per second is Q = 6000 cfs, compute the mechanical power available at the turbine input P_kW [146–148].

c) How does P_turbine^required compare with P_kW?

d) Compute the specific speed N_q [146–148]. Is the selection of the Francis turbine justified?

e) What other types of water turbines exist?

f) What is the amount of water the upper or lower reservoirs must hold to generate E = (1500 MWh + 625 MWh) = 2125 MWh per day during an 11.3 hour period?

g) Is the given precipitation per year sufficient to replace the evaporated water? If not, what is the required “rain-catch” area to replace the yearly water loss through evaporation?

h) The pumped-storage plant delivers the energy E = (1500 MWh + 625 MWh) = 2125 MWh per day for which customers pay $0.20/kWh due to peak-power generation. What is the payback period of this pumped storage plant if the construction price is $3000 per installed power capacity of 1 kW, the cost for pumping is $0.03/kWh, and the interest rate is 3%?

Solution to Application Example 11.3

a) Computation of the turbine input power P_turbine^required:

$P_{\mathrm{turbine}\_\mathrm{input}}^{\mathrm{required}}=\frac{P_{\mathrm{rated}}}{{\eta }_{\mathrm{turbine}}\cdot {\eta }_{\mathrm{synchronous}\_\mathrm{machine}}\cdot {\eta }_{\mathrm{\Delta }-\mathrm{Y}\_\mathrm{transformer}}}=346.25\mathrm{M}\mathrm{W}$

b) Computation of the mechanical power available at the turbine input P_kW:

$P_{\mathrm{kW}}=\frac{H\cdot Q\cdot W\cdot 0.746}{550}=\frac{722.5\cdot 6000\cdot 62.4\cdot 0.746}{550}=366.9\mathrm{kW}\text{,}$

where H is measured in ft, Q is measured in cubic feet per second (cfs), W is the weight of 1 cubic foot of water of 62.4 (lb-force)/(ft)³, 0.746 kW = 1 hp, and 550 is a constant converting ft∙(lb-force)/s to horse power (hp).

c) How does P_turbine^required compare with P_kW?

$P_{\mathrm{turbine}\_\mathrm{input}}^{\mathrm{required}}\le P_{\mathrm{kW}}\text{.}$

d) Computation of the specific speed. Is the selection of the Francis turbine justified?
Synchronous speed of generator $n_{\mathrm{s}}=\frac{120\cdot f}{p}=300\mathrm{rpm}$, Q = 6000·0.0283 = 169.8 m³/s, H_eff = 722.5·0.3048 = 220.22 m. Specific speed of a water turbine is now $N_{\mathrm{q}}=n_{\mathrm{s}}\frac{Q^{0.5}}{H_{\mathrm{eff}}^{0.75}}=68.38$, which means a Francis turbine is satisfactory.

e) What other types of water turbines exist?
Pelton wheel, Kaplan turbine, bulb-type turbine, mixed-flow turbine, propeller turbine, and Gorlov turbine.

f) The amount of water the upper or lower reservoirs must hold:
Q = 6000 cfs, therefore, the water required during one day is Capacity_H2O = 6000·0.0283·3600·11.3 = 6.908·10⁶ m³, this requires two (an upper and lower) reservoirs of 20 m depth, 588 m long and 588 m wide, each. As is well known the water inlets at the upper and lower reservoirs will be in the middle of the reservoir depths in order to prevent that sand and rocks enter the turbine and the pump. This is to say the 20 m depths are effective depths and the actual depths might be more than twice as large, say 50 m.

g) Required “rain-catch” area:
Capacity_{H2Oevaporation} = 6.908·10⁵ m³, Capacity_H2O + Capacity_{H2Oevaporation} = 6.908·10⁶ + 6.908·10⁵ = 7.599·10⁶ m³. For 20 inches of precipitation one gets per year Capacity_{H2Oprecipitation} = 20·0.0254 (588)² = 1.756·10⁵ m³. Note, Capacity_{H2Oevaporation} exceeds Capacity_{H2Oprecipitation} from this follows that a rain-catch area is required. At the time of commissioning the plant the additional rain-catch area is required to fill the upper reservoir; some run-off will be required after the commissioning of the plant. The additional water to be supplied initially by the rain-catch area is
Capacity_{H2O additional} = Capacity_H2O + Capacity_{H2Oevaporation} - Capacity_{H2O precipitation} = 7.599·10⁶ m³-1.756·10⁵ m³ = 7.42·10⁶ m³ requiring a rain-catch area of (area) _rain-catch = 7.42·10⁶/(20·0.0254) = 14.61·10⁶ m² = 14.61 km².

h) Payback period of pumped storage plant:
Earnings = $2125·10³·0.20·365y = 155.125·10⁶y, expenses = $750·10⁶+ (2125·10³·0.03·365y)/(0.95·0.95·0.80·0.85) = $750·10⁶ + $37.92y∙10⁶.
Payback period in y years without interest payments: y = 750·10⁶/(155.125·10⁶- 37.92·10⁶) = 6.4 years
Payback period with interest payments: 155.125y = 750(1.03)^y + 37.92y resulting in about y = 8 years.

11.3.15 Flywheel Storage

This type of storage can be constructed anywhere, and it is of the short-term type with a quick response time [149].

11.3.15.1 Application Example 11.4: Design of a 10 MWh Flywheel Short-Term Storage Power Plant [1]

Design a flywheel storage system which can provide for about 6 minutes a power of 100 MW, that is, energy of 10 MWh. The flywheel power plant consists (see Figures E11.4.1a, b) of a flywheel, mechanical gear, synchronous machine, inverter-rectifier set and a step-up transformer. The individual components of this plant must be designed as follows:

a) For the flywheel (made from steel) as shown in Figure E11.4.1a (h = 0.9 m, R_1o = 1.5 m, R_1i = 1.3 m, R_2o = 0.50 m, R_2i = 0.10 m, b = 0.2 m), compute the ratio of inertia (J) to the weight (W) of the flywheel, that is (J/W).
The axial moment of inertia of a flywheel with 4 spokes is

$J_{\mathrm{wheel}}=\frac{1}{2}\cdot \pi \cdot \gamma \cdot h\cdot \left(R_{1\mathrm{o}}^4-R_{1\mathrm{i}}^4\right)+\frac{4}{3}\cdot h\cdot b\cdot \gamma \cdot \left(R_{1\mathrm{i}}^3-R_{2\mathrm{o}}^3\right)+\frac{1}{2}\cdot \pi \cdot \gamma \cdot h\cdot \left(R_{2\mathrm{o}}^4-R_{2\mathrm{i}}^4\right)\left[{\mathrm{k}\mathrm{g}\mathrm{m}}^2\right]$

b) For the given values h = 0.9 m, R_1o = 1.5 m, R_1i = 1.3 m, R_2o = 0.50 m, R_2i = 0.10 m, and b = 0.2 m calculate for the wheel-type configuration the stored energy E_{stored rated} provided the flywheel rotates at n_{flywheel rated} = 21,210 rpm.

11.3.16 Battery Storage

Battery storage plants can be constructed anywhere, they have a quick response and are usually of the short-term type. However, if fluid-flow batteries are chosen then these plants can continually supply energy and act therefore as long-term batteries [150].

11.3.16.1 Application Example 11.5: Design a 10 MWh Battery Short-Term Storage Plant

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 (long-term) compressed-air power plant with a (short-term) battery plant for bridging the time from when the WP 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 $V_{\mathrm{L}-\mathrm{L}}^{\mathrm{i}}$ = 240 V, power system voltage $V_{\mathrm{L}-\mathrm{L}}^{\mathrm{s}}$= 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 $V_{\mathrm{L}-\mathrm{L}}^{\mathrm{i}}$= 240 V, power system voltage $V_{\mathrm{L}-\mathrm{L}}^{\mathrm{s}}$= 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(ωt-120°), 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 = 50 μF, ideal diodes D₁ to D₆ 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.

11.3.17 Compressed-Air Storage

This type of long-term storage has been available and operating for quite some time in Germany [21] and in the USA, Alabama [20].

11.3.17.1 Application Example 11.6: Design of a 100 MW Compressed-Air Storage Facility

Design a compressed-air energy storage (CAES) plant for P_{out_generator} = P_rated = 100 MW, V_L-L = 13,800 V, and cosφ = 1 which can deliver rated power for two hours. The overall block diagram of such a plant [20,21] is shown in Figure E11.6.1. It consists of a compressor, cooler, booster compressor, booster cooler, booster-compressor three-phase, 2-pole induction motor with a 1:2 mechanical gear to increase the compressor speed, underground air-storage reservoir, combustor fired either by natural gas or oil or coal, a modified (without compressor) gas turbine, and a three-phase, 2-pole synchronous generator/ motor.

Design data:

Air inlet temperature of compressor: 50 ° F ≡ 283.16 K at ambient pressure 1 atm ≡ 14.696 psi ≡ 101.325 kPa(scal).

Output pressure of compressor: 11 atm ≡ 161 psi ≡ 1114.5 kPa.

Output temperature of booster cooler: 120 ° F ≡ 322.05 K at 1000 psi ≡ 6895 kPa.

Output temperature of combustor: 1500 ° F ≡ 1089 K at 650 psi ≡ 4482 kPa.

Output temperature of gas turbine: 700 ° F ≡ 644 K at ambient pressure 1 atm ≡ 14.696 psi ≡ 101.325 kPa.

Generation operation: 2 h at 100 MW.

Re-charging (loading) operation of underground air storage reservoir: 8 h at 20 MW.

Capacity of air storage reservoir: 3.5 · 10⁶ ft³ ≡ 97.6 · 10³ m³ ≡ (46 m x 46 m x 46 m).

Start-up time: 6 minutes.

Turbine and motor/generator speed: n_{s mot/gen} = 3600 rpm

Compressor speed: n_{s comp} = 7000 rpm.

Cost per 1 kW installed power capacity: $2000.

a) Calculate the Carnot efficiency of the gas turbine η_{carnot gas turbine} = η_compressor ∙ η_turbine=$\frac{T_1-T_2}{T_1}$.

b) Note that the compressor of a gas turbine has an efficiency of η_compressor = 0.50. In this case the compressor is not needed because pressurized air is available from the underground reservoir. The absence of a compressor increases the overall efficiency of the CAES, if “free” WP is used for charging the reservoir. Provided the synchronous machine (motor/generator set) has an efficiency of η_generator = 0.9 calculate

b1) the efficiency of the turbine without the compressor η_compressor,

b2) P_{out turbine},

b3) P_{out combustor},

b4) For a heat rate of the combustor of 5500 BTU/kWh, determine the input power of the combustor required (during 1 hour) P_{in combustor}.

b5) The two air compressors (charging the reservoir during 8 hours) including the two coolers require an input power of 20 MW (whereby wind energy is not free) calculate the overall efficiency of the CAES (during 1 hour of operation)

${\eta }_{\mathrm{overall}}^{\mathrm{wind}\mathrm{energy}\mathrm{is}\mathrm{not}\mathrm{free}}=\frac{P_{\mathrm{out}\_\mathrm{generator}}}{P_{\mathrm{in}\mathrm{combustor}}+P_{2\_\mathrm{compressors}+2\_\mathrm{coolers}}}\text{.}$

b6) Compute the overall efficiency if wind energy is free, that is, ${\eta }_{\mathrm{overall}}^{\mathrm{wind}\mathrm{energy}\mathrm{is}\mathrm{free}}=\frac{P_{\mathrm{out}\_\mathrm{generator}}}{P_{\mathrm{in}\mathrm{combustor}}}\text{.}$

c) What is the overall construction cost of this CAES plant, if the construction price is $2000 per installed power capacity of 1 kW?

d) The CAES plant delivers the energy E = 200 MWh per day (during 2 hours) for which customers pay $0.25/kWh due to peak-power generation. What is the payback period of this CAES plant, if the cost for pumping (loading, recharging) is neglected (free wind power), the cost of 1 cubic foot of natural gas is $0.005, and the interest rate is 3%?

e) Repeat the analysis for a heat rate of 4000 BTU/kWh at a gas turbine output temperature of 600 K.
Hint: For the calculation of the Carnot efficiency of the gas turbine and of the compressors you may use the software available on the Internet address:
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/adiab.html#c3

Solution to Application Example 11.6

a) The Carnot efficiency of the gas turbine is

${\eta }_{\mathrm{carnot}\mathrm{gas}\mathrm{turbine}}={\eta }_{\mathrm{compressor}}\cdot {\eta }_{\mathrm{turbine}}=\frac{T_1-T_2}{T_1}=\frac{1089\mathrm{K}-644\mathrm{K}}{1089\mathrm{K}}=0.41\text{.}$

b1) The efficiency of the gas turbine without the compressor η_compressor is ${\eta }_{\mathrm{turbine}}=\frac{{\eta }_{\mathrm{carnot}\mathrm{gas}\mathrm{turbine}}}{{\eta }_{\mathrm{compressor}}}=\frac{0.41}{0.5}=0.82\text{.}$

b2) $P_{\mathrm{out}\mathrm{turbine}}=\frac{P_{\mathrm{generator}}}{{\eta }_{\mathrm{generator}}}=\frac{100\mathrm{M}\mathrm{W}}{0.9}=111\mathrm{M}\mathrm{W}$

b3) $P_{\mathrm{out}\mathrm{turbine}}=\frac{P_{\mathrm{out}\mathrm{turbine}}}{{\eta }_{\mathrm{turbine}}}=\frac{111\mathrm{M}\mathrm{W}}{0.82}=135.37\mathrm{M}\mathrm{W}$

b4) For a heat rate of the combustor of 5500 BTU/kWh, determine the input power of the combustor required (during 1 hour) P_{in combustor}. Figure E11.6.2 shows the output powers and the associated efficiencies of generator, gas turbine without compressor, and combustor.
The heat rate = 5500 BTU/kWh indicates the thermal input energy to the combustor required to generate 1 kWh of electrical energy, that is, 5500 BTU = 1.612 kWh of thermal energy is required to generate 1 kWh of electrical output energy.
The input energy delivered within 1 hour to the combustor is identical to the input power of the combustor $P_{\mathrm{in}\mathrm{combustor}}=P_{\mathrm{out}\_\mathrm{generator}}\left(\frac{1.612\mathrm{k}\mathrm{W}\mathrm{h}}{1\mathrm{k}\mathrm{W}\mathrm{h}}\right)=100\mathrm{M}\mathrm{W}\frac{1.612\mathrm{k}\mathrm{W}\mathrm{h}}{1\mathrm{k}\mathrm{W}\mathrm{h}}=161.2\mathrm{M}\mathrm{W}\text{.}$
The efficiency of the combustor is ${\eta }_{\mathrm{combustor}}=\frac{135.37}{161.2.}=0.84\text{.}$

b5) Two air compressors (charging the reservoir during 8 hours) including the two coolers require an input power of 20 MW (whereby wind energy is not free) calculate the overall efficiency of the CAES (during 1 hour of operation).
The energy required during an 8 hour period for the 2 compressors and 2 coolers is
E_{2 compressors + 2 coolers} = 20 MW ∙ 8 h = 160 MWh. If we want to deliver 100 MW during 1 h then the power required is P_{2 compressors + 2 coolers} = 80 MW. The overall efficiency is

${\eta }_{\mathrm{overall}}^{\mathrm{wind}\mathrm{energy}\mathrm{is}\mathrm{not}\mathrm{free}}=\frac{P_{\mathrm{out}\_\mathrm{generator}}}{P_{\mathrm{in}\mathrm{combustor}}+P_{2\_\mathrm{compressors}+2\_\mathrm{coolers}}}=\frac{100\mathrm{M}\mathrm{W}}{\left(161.2+80\right)\mathrm{M}\mathrm{W}}=0.415\text{.}$

b6) If wind energy is free, then ${\eta }_{\mathrm{overall}}^{\mathrm{wind}\mathrm{energy}\mathrm{is}\mathrm{free}}=\frac{P_{\mathrm{out}\_\mathrm{generator}}}{P_{\mathrm{in}\mathrm{combustor}}}=\frac{100\mathrm{M}\mathrm{W}}{161.2\mathrm{M}\mathrm{W}}=0.62\text{.}$

c) The construction cost of this CAES plant is $200 M.

d) Earnings per day by discharging the CAES
Earning = 200 000 kWh∙$0.25 = $50 k/day.
Natural gas cost: 5500 BTU = 1.6117·10³ Wh or 1 Wh = 3.413 BTU resulting in 1 MWh = 3.413·10⁶ BTU.
The fuel energy delivered to the combustor is E_{natural gas} = 161.2 MWh∙2·3.413·10⁶ BTU/MWh = 1100·10⁶ BTU. 1(ft)³ of natural gas contains 1021 BTU, and the volume of natural gas required is ${\mathrm{V}\mathrm{o}\mathrm{l}}_{\mathrm{natural}\mathrm{gas}}=\frac{1100\cdot {10}^6}{1021}{\left(\mathrm{ft}\right)}^3=1.08\cdot {10}^6{\left(\mathrm{ft}\right)}^3$ within a 2 h period or per day. The fuel costs are then Cost_fuel = 1.08·10⁶∙$0.005 = $5400 for 2 h or per day.
Payback period:

$(1.03)^y100000kW∙$2000/kW + $5400·365∙y = $50000·365∙y or (1.03)^y200 = 16.28∙y resulting in approximately y = 30 years.

e) Repetition of the analysis for a heat rate of 4000 BTU/kWh at a gas turbine output temperature of 600 K.
The Carnot efficiency of the gas turbine is determined from Figure E11.6.3.
The Carnot efficiency of the gas turbine is
${\eta }_{\mathrm{carnot}\mathrm{gas}\mathrm{turbine}}={\eta }_{\mathrm{compressor}}\cdot {\eta }_{\mathrm{turbine}}=\frac{T_1-T_2}{T_1}=\frac{1089K-600K}{1089K}=0.449$, where the turbine power is 488974.29 J/3600 s = 135.83 MW.
The efficiency of the gas turbine without the compressor η_compressor is

${\eta }_{\mathit{turbine}}=\frac{{\eta }_{\mathit{carnot}\mathit{gas}\mathit{turbine}}}{{\eta }_{\mathit{compressor}}}=\frac{0.449}{0.5}=0.898\text{.}$

$P_{\mathit{out}\mathit{turbine}}=\frac{P_{\mathit{generator}}}{{\eta }_{\mathit{generator}}}=\frac{100MW}{0.9}=111.11MW$

$P_{\mathit{out}\mathit{combustor}}=\frac{P_{\mathit{out}\mathit{turbine}}}{{\eta }_{\mathit{turbine}}}=\frac{111.11MW}{0.898}=123.73MW$

For a heat rate of the combustor of 4000 BTU/kWh at a gas turbine output temperature of 600 K, determine the input power of the combustor required (during 1 hour) P_{in combustor}.
The heat rate = 4000 BTU/kWh indicates the thermal input energy to the combustor required to generate 1 kWh of electrical energy, that is, 4000 BTU = 1.1723 kWh of thermal energy is required to generate 1 kWh of electrical energy.
The input energy delivered within one hour to the combustor is identical to the input power of the combustor P_{in combustor}=$P_{\mathit{out}\_\mathit{generator}}\left(\frac{1.1723kWh}{1kWh}\right)=100MW\left(\frac{1.1723kWh}{1kWh}\right)=117.23MW\text{.}$
The efficiency of the combustor is η_combustor=$\frac{123.73}{117.23}$> 1.0 which is not possible. This design is not feasible.