11.3.18 Hydrogen Generation and Storage

A well-known way of splitting water (H₂O) is by electrolysis [151]. A container (trough) contains distilled water and two electrodes -- the anode (A) and cathode (C) consisting of either platinum or nickel -- connected to a DC current source. Pure H₂O is a poor conductor; the addition of H₂SO₄ to the water makes the solution conducting. Although the solution of H₂O and H₂SO₄ heats up - due to the losses caused by the current flowing from anode to cathode- at the anode oxygen O₂ and at the cathode hydrogen H₂ is accumulated. The efficiency of splitting H₂O based on electrolysis is about η_electrolysis = 80%, that is, 20% of the energy is converted to heat within the electrolysis process. The energy density of hydrogen is about 2.28 times that of gasoline (E_gasoline = 12.3 kWh/kg-force), that is, E_H2 = 28 kWh/kg-force. {Note 1 liter (dm³) of water has a mass m of 1 kg at 4° Celsius, and exerts a (force or) weight of 1 kg-force on a scale}. The additional energy loss due to the production of distilled water can be deduced from the definition of unit 1 kcal: The energy of 1 kcal is required to heat 1 kg-force of H₂O by 1 ° C (this is approximately true only because the energy required depends upon the temperature range. It is exactly true heating pure water from 14.5°C to 15.5°C). Note that 1 kcal = 4.186 kWs.

11.3.19 Fuel Cells

A fuel cell takes fuel (e.g., hydrogen and oxygen) as input and produces electricity as output: it will continue to produce output as long as raw material (e.g., fuel) is supplied. This is the difference between a fuel cell and an ordinary battery, except flow batteries. While both rely on electrochemistry a fuel cell is not consumed as an ordinary battery when it produces electricity: it transforms the chemical energy stored in the fuel into electrical energy [1,152].

11.3.20 Molten Salt Storage

Solar power generation [153–157] around the clock is possible where a solar concentrator array, consisting of thousands of mirrors on the ground and a tower supporting at the focal point of the mirrors a salt container, see Figure 11.23. Figure 11.24 illustrates construction details. The great advantage of this approach is that solar heat can be stored in the molten salt. SolarReserve [153] has developed, constructed and sold solar power systems that have output powers from 100MW to 200MW for a single molten salt power tower. For example, their Crescent Dunes project has 110MW output power that can store and generate energy for 10 hours of full-load output power by focusing mirrors onto millions of gallons of molten salt which flows through a heat exchanger (receiver) producing steam and driving a steam turbine, allowing the plant to provide electricity for 24 hours a day.

The solar energy is generated by a massive circular array of, for example, 10,347 heliostats, each measuring at Crescent Dunes 37 feet wide and 34 feet tall. The heliostat field encircles a concrete solar power tower which is 540 feet high with a 100-foot high receiver on top. Molten salt flows through the receiver, but it is not held there. The molten salt is held in the storage tanks, each tank holding 3.6 million gallons (or 70 million pounds-force) of molten salt. When the heliostats focus the sunlight onto the receiver the salt is heated to over 1,000 degrees Fahrenheit. When it is needed such as at night or at peak times, the heat is released by passing the molten salt through a heat exchanger and steam generator that drives a turbine to produce electricity. The cooled salt is then recirculated to the receiver for re-heating.

The salt used is a mixture of sodium and potassium nitrate (the same as that used in fertilizers), which is inexpensive, reliable, and environmentally friendly. It will be mixed on site with no additives. Apart from a few unique components such as the high heat flux hardware in the tower, the system uses existing technologies such as turbines and steam generators. The system was proven over a four-year period in the 1990s at a 10 MW demonstration project near Barstow in California.

Other solar systems also use salt as storage, but they use synthetic oil in the steam generation. Using salt for both means the system is more efficient, since it can produce steam at higher temperatures and can harvest three times as much energy for the same amount of salt, as compared to the oil approach. The system can be air cooled, thus avoiding criticisms about water use, but its height, at 640 feet (with a maintenance crane on top), could spark other (e.g., environmental) criticisms.

11.3.21 Solutions to Power Quality Problems of Molten Salt Storage

The molten salt approach appears to be best suited for long-term energy storage requiring not necessarily cooling water as discussed above. The disadvantages are the great number of heliostats and the tall tower with the heavy salt container. This approach deserves more research effort as it is suitable for desert areas. Unfortunately such a location requires long transmission lines to transmit the generated power to urban areas.

11.4.1 Energy Efficiency and Reliability Increases Based on Distributed Generation of Interconnected Power System and Islanding Operation

Prior to 1940 there were a limited number of interconnected power systems and the servicing of the load was elementary since the systems were primarily radial circuits (Figure 11.25a) and many power systems were operated in islanding mode. In more recent years, the numerous advantages of interconnection of many power systems were recognized within the three US power grids [1] (Western, Eastern and Texan generation systems [ERCOT]), see Figure 11.25b for example. Many loop circuits with many load/generation buses and high levels of power exchange between neighboring companies exist. The latter point relates closely to interconnection advantages.

With no addition of generation capacity, it is possible to increase generation capability, efficiency and reliability through interconnection. For example, in the case of an outage of a generating unit, power may be purchased from a neighboring company. The cost savings realized from lower installed capacity usually far outweigh the cost of the transmission circuits required to access neighboring companies. Steady-state power flow is discussed in Chapter 7. While on the one hand the energy efficiency of an interconnected system is increased by more fully loading existing generation plants, it is on the other hand decreased by transporting energy via transmission lines over longer distances. Thus on average the transmission loss within interconnected systems is about 8%.

11.4.2 Distributed Power Plants

The generation mix (e.g., coal, natural gas, nuclear, hydro plants) of existing power systems will change in the future to mainly natural-gas fired plants, distributed renewable generation facilities [158] and storage plants. In addition a transition from the interconnected system to an islanding mode of operation must be possible [55] to increase reliability. This means that any islanding system must have a frequency leading plant --called “frequency leader”-- in addition to renewable plants and storage plants. Renewable and storage plants cannot be frequency leaders because of their intermittent and limited output powers, respectively.

11.4.3 Review of Current Methods and Issues of Present-Day Frequency and Voltage Control

Present-day frequency/load and voltage control of interconnected systems take place at the transmission level and are based on load sharing and demand-side management. Load sharing relies on drooping characteristics [49], see Figures 4a, b, c of [144] or Figures 11.26 and E11.9.1 where natural gas, coal, nuclear and hydro plants or those with spinning reserves supply the additional load demand. If this additional load cannot be provided by the interconnected plants then demand-side management (e.g., load shedding) will set in and some of the less important loads will be disconnected. This method of frequency/load control (when no spinning reserve is available) cannot be employed if renewable sources operate at peak power exploiting the renewable sources to the fullest in order to displace as much fuel as possible. This approach works well as long as renewable energy represents a small fraction of the generation capacity, i.e. the highly-variable renewable characteristic on the left side of Figure 11.26 is significantly smaller than the plant with spinning reserve characteristic on the right. As renewable penetrations increase at either the distribution or transmission levels, control problems may result. In present-day interconnected systems a frequency variation between f_max = 59 and f_min = 61 Hz, that is Δf =$\pm$ 1.67% is acceptable [159].

11.4.4 Control or Frequency/Load Control of an Isolated Power Plant with one Generator only

Figure 11.27 illustrates the block diagram of governor, prime mover (steam turbine) and rotating mass & load of a turbo generator set [49].

11.4.5 Load/Frequency Control with Droop Characteristics of an Interconnected Power System Broken into Two Areas each Having one Generator

Figure 11.28 shows the block diagram of two generators interconnected by a transmission tie line [49].

Solution to Application Example 11.10

Differential and algebraic equations

System # 1: ${\varepsilon }_{11}=\mathrm{load}{\mathrm{r}\mathrm{e}\mathrm{f}}_1-\frac{{\mathrm{\Delta }\omega }_1}{R_1}$, ${\mathrm{\Delta }P}_{\mathrm{valve}\_1}+{\tau }_{\mathrm{G}1}\frac{d\left({\mathrm{\Delta }P}_{\mathrm{valve}\_1}\right)}{dt}={\varepsilon }_{11}$, ${\mathrm{\Delta }P}_{\mathrm{mech}\_1}+{\tau }_{\mathrm{C}\mathrm{H}1}\frac{\mathit{d}\left({\mathrm{\Delta }P}_{\mathrm{mech}\_1}\right)}{dt}={\mathrm{\Delta }P}_{\mathrm{valve}\_1}$, ${\varepsilon }_{21}={\mathrm{\Delta }P}_{\mathrm{mech}\_1}-{\mathrm{\Delta }P}_{\mathrm{L}1}-{\mathrm{\Delta }P}_{\mathrm{tie}}$, ${\mathrm{\Delta }\omega }_1{D}_1+M_1\frac{\mathit{d}\left({\mathrm{\Delta }\omega }_1\right)}{dt}={\varepsilon }_{21}$.

Coupling (tie, transmission) network: $\frac{1}{\mathit{T}}\frac{\mathit{d}\left({\mathrm{\Delta }P}_{\mathrm{tie}}\right)}{\mathit{d}t}={\varepsilon }_3$, where ${\varepsilon }_3={\mathrm{\Delta }\omega }_1-{\mathrm{\Delta }\omega }_2$.

System # 2: ${\mathit{\varepsilon }}_{12}=\mathrm{load}{\mathrm{r}\mathrm{e}\mathrm{f}}_2-\frac{{\mathrm{\Delta }\omega }_2}{R_2}$, ${\mathrm{\Delta }P}_{\mathrm{valve}\_2}+{\tau }_{\mathrm{G}2}\frac{d\left({\mathrm{\Delta }P}_{\mathrm{valve}\_2}\right)}{dt}={\varepsilon }_{12}$, ${\mathrm{\Delta }P}_{\mathrm{mech}\_2}+{\tau }_{\mathrm{C}\mathrm{H}2}\frac{\mathit{d}\left({\mathrm{\Delta }P}_{\mathrm{mech}\_2}\right)}{dt}=\mathrm{\Delta }{P}_{\mathrm{valve}\_2}$, ${\varepsilon }_{22}={\mathrm{\Delta }P}_{\mathrm{mech}\_2}-{\mathrm{\Delta }P}_{\mathrm{L}2}+{\mathrm{\Delta }P}_{\mathrm{tie}}$, ${\mathrm{\Delta }\omega }_2{D}_2+M_2\frac{d\left({\mathrm{\Delta }\mathit{\omega }}_2\right)}{dt}={\varepsilon }_{22}$.

b) The results based on Mathematica are shown in Figures E11.10.1a, b. Figure E11.10.1a illustrates the occurrence of instability for similar (R₁ = 0.01 pu, R₂ = 0.02 pu) droop characteristics and Figure E11.10.1b stability for dissimilar (R₁ = 0.5 pu, R₂ = 0.01 pu) droop characteristics. The Mathematica input program is listed in Table E11.10.1.

Table E11.10.1

Mathematica input program for two interconnected systems

R1 = 0.01; d1 = 0.8; M1 = 4.5; Tg1 = 0.01; Tch1 = 0.5; Lr1 = 0.8; DPL1[t_]:=If [t < 10, 0, 0.2]; R2 = 0.02; d2 = 1.0; M2 = 6; Tg2 = 0.02; Tch2 = 0.75; Lr2 = 0.8; DPL2[t_]:=If [t < 30, 0, -0.2]; Xtie = 0.2; Tie = 377/Xtie; ic1 = Dw1[0]= = 0; ic2 = DPmech1[0]= = 0; ic3 = DPvalve1[0]= = 0; ic4 = DPtie[0]= = 0; ic5 = Dw2[0]= = 0; ic6 = DPmech2[0]= = 0;

ic7 = DPvalve2[0]= = 0; E11[t_]:=Lr1-Dw1[t]/R1; E12[t_]:=DPmech1[t]-DPL1[t]-DPtie[t]; E3[t_]:=Dw1[t]-Dw2[t]; E22[t_]:=Lr2-Dw2[t]/R2; E21[t_]:=DPmech2[t]-DPL2[t] + DPtie[t]; eqn1 = Dw1'[t]= =(1/M1)*(E12[t]-d1*Dw1[t]); eqn2 = DPmech1'[t]= =(1/Tch1)*(DPvalve1[t]-DPmech1[t]); eqn3 = DPvalve1'[t]= =(1/Tg1)*(E11[t]-DPvalve1[t]); eqn4 = DPtie'[t]= = Tie*E3[t]; eqn5 = Dw2'[t]= =(1/M2)*(E21[t]-d2*Dw2[t]); eqn6 = DPmech2'[t]= =(1/Tch2)*(DPvalve2[t]-DPmech2[t]); eqn7 = DPvalve2'[t]= =(1/Tg2)*(E22[t]-DPvalve2[t]); sol = NDSolve[{eqn1,eqn2,eqn3,eqn4,eqn5,eqn6,eqn7, ic1,ic2,ic3,ic4,ic5,ic6,ic7},{Dw1[t], DPmech1[t],DPvalve1[t],DPtie[t],Dw2[t], DPmech2[t],DPvalve2[t]},{t,0,50},MaxSteps- > 100000]; Plot[Dw1[t]/.sol,{t,0,50},PlotRange- > All,AxesLabel - > {"t[s]","Dw1[t][pu]"}]

Figures 11.28 and E11.10.1a, b lead to the conclusion that one power plant must be the so-called frequency leader having a small-sloped droop characteristic which can accommodate overloads. Note that Δω₁(t) = Δω₂(t) = Δω(t) are identical, which is due to the transmission line transfer function (T/s) acting as an integrator where at steady-state the perturbation at the receiving end of the line is zero.

11.4.6 Complementary Operation of Renewable Plants with Short-Term and Long-Term Storage Plants Analyzed with either Mathematica or Matlab [64]

The stability of a smart/micro grid consisting of natural gas-fired power plant serving as frequency leader, a long-term storage power plant, and two intermittently operating plants (e.g., PV and WP plants) with associated short-term storage plants is the objective of this section. In order to achieve stability for a given smart/micro grid structure the following constraints must be satisfied:

1) instability and frequency variations are minimized through appropriate switching (in and out) of the short-term storage plants; in particular the time instant of switching is important.

2) transmission line parameters are optimized;

3) time constants of the governors and the valves must be within feasible regions;

4) droop characteristics of the individual plants must satisfy certain constraints.

11.4.6.1 Application Example 11.11: Operation of Natural-Gas Fired and Long-Term Storage Plants with Two Renewable Sources with Complementary Short-Term Storage Plants

Figure E11.11.1 illustrates the sharing (increase) of the additional load among a short-term storage plant (e.g., R₂ = 0.01 pu), a long-term-storage plant (e.g., R₃ = 10 pu), and a PV plant (e.g., R₁ = 10 pu) causing a frequency decrease. One obtains a stable frequency control when the short-term storage plant compensates the intermittent power output of the PV plant and the plant with spinning reserve (natural-gas fired plant) is replaced by a long-term storage plant. The long-term storage plant is connected all the time to the power system and serves therefore as frequency leader. The PV and the short-term storage plants may operate intermittently only.

In Figure 11.26 the spinning reserve plant can be replaced by a long-term storage plant (e.g., pump-hydro or compressed air facility), as shown in Figures E11.11.2a, b, c. The intermittently operating PV and WP plants are complemented by short-term storage plants (e.g., battery-fed inverter). During times of high power demand, the storage plants supply power to maintain the power balance between generation and the served loads. At low power demand, the renewable sources will supply the storage plants. The continuous load increase of conventional peak-power plants (which can provide additional load through spinning reserve) are replaced by putting short-term (located next to renewable plants) and long-term storage plants [1] online, reducing fossil-fuel generation and contributing to renewable portfolio standards. Renewable sources are operated at their peak-power point e.g., the slopes of the droop characteristic R₃ and R₅ of Figure E11.11.2b are large while those of R₁ and R₂ are relatively small to permit the increase of their output power upon demand. Thus the PV and WP plants cannot participate in frequency/load control. In Figure E11.11.2c all droop characteristics have a relatively small slope and all participating plants can output increased power and participate in frequency/load control.

In the block diagram of Figure E11.11.2a the PV plant, the governor, and prime mover represent the solar array and inverter/rectifier while in the short-term storage plant associated with the PV plant the governor and prime mover represent the storage device (e.g., battery, super-capacitor, flywheel) with inverter/rectifier. Similar considerations apply to the WP plant and its associated short-term storage plant.

Data for natural-gas fired plant (system #1): Angular frequency change (Δω₁) per change in generator output power (ΔP₁), that is, R₁= $\frac{{\mathrm{\Delta }\omega }_1}{{\mathrm{\Delta }P}_1}\mathrm{p}\mathrm{u}=$ 0.01 pu, the frequency-dependent load change (ΔP_L1|_frequ) per angular frequency change (Δω₁), that is D₁= $\frac{{\left.{\mathrm{\Delta }P}_{\mathrm{L}1}\right|}_{\mathrm{frequ}}}{{\mathrm{\Delta }\omega }_1}$ = 0.8 pu, step-load change ${\mathrm{\Delta }P}_{\mathrm{L}1}\left(s\right)=\frac{{\mathrm{\Delta }P}_{\mathrm{L}1}}{s}\mathrm{p}\mathrm{u}=\frac{0.1}{s}$ pu, angular momentum of gas turbine and generator set M₁ = 4.5, base apparent power S_base = 500 MVA, governor time constant T_G1 = 0.3 s, valve charging time constant T_CH1 = 0.9 s, and load ref₁(s) = 0.8 pu.

Data for long-term storage plant (system #2): Angular frequency change (Δω₂) per change in generator output power (ΔP₂), that is R₂= $\frac{{\mathrm{\Delta }\omega }_2}{{\mathrm{\Delta }P}_2}\mathrm{p}\mathrm{u}=$ 0.1 pu (e.g., hydro-power plant), the frequency-dependent load change (ΔP_L2|_frequ) per angular frequency change (Δω₂), that is D₂= $\frac{{\left.{\mathrm{\Delta }P}_{\mathrm{L}2}\right|}_{\mathrm{frequ}}}{{\mathrm{\Delta }\omega }_2}$ = 1.0 pu, step-load change ${\mathrm{\Delta }P}_{\mathrm{L}2}\left(s\right)=\frac{{\mathrm{\Delta }P}_{\mathrm{L}2}}{s}\mathrm{p}\mathrm{u}=\frac{-0.2}{s}$ pu, angular momentum of hydro turbine and generator set M₂ = 6, base apparent power S_base = 500 MVA, governor time constant T_G2 = 0.2 s, valve charging time constant T_CH2 = 0.2 s, and load ref₂(s) = 0.5 pu.

Data for tie line: T = $\frac{377}{X_{\mathrm{tie}}}$ with X_tie = 0.2 pu.

Data for PV plant (system #3): Angular frequency change (Δω₁) per change in inverter output power (ΔP₃), that is R₁= $\frac{{\mathrm{\Delta }\omega }_1}{{\mathrm{\Delta }P}_3}\mathrm{p}\mathrm{u}=$ 0.3 pu, governor time constant T_G3 = 0.1 s, equivalent valve time constant T_CH3 = 0.1 s, and load ref₃(s) = 0.01 pu.

Data for short-term storage plant associated with PV plant (system #4): Angular frequency change (Δω₁) per change in generator output power (ΔP₄), that is R₄= $\frac{{\mathrm{\Delta }\omega }_1}{{\mathrm{\Delta }P}_4}\mathrm{p}\mathrm{u}=$ 0.5 pu, governor time constant T_G4 = 0.2 s, equivalent valve time constant T_CH4 = 0.1 s, and load ref₄(s) = 0.01 pu.

Data for wind power (WP) plant (system #5): Angular frequency change (Δω₂) per change in generator output power (ΔP₅), that is R₅= $\frac{{\mathrm{\Delta }\omega }_1}{{\mathrm{\Delta }P}_5}\mathrm{p}\mathrm{u}=$ 0.7 pu, governor time constant T_G5 = 0.1 s, equivalent valve time constant T_CH5 = 0.1 s, and load ref₅(s) = 0.01 pu.

Data for short-term storage plant associated with WP plant (system #6): Angular frequency change (Δω₂) per change in generator output power (ΔP₆), that is R₆ = $\frac{{\mathrm{\Delta }\omega }_2}{{\mathrm{\Delta }P}_6}\mathrm{p}\mathrm{u}=$ 0.5 pu, governor time constant T_G6 = 0.2 s, equivalent valve time constant T_CH6 = 0.1 s, and load ref₆(s) = 0.01 pu.

a) List the ordinary differential equations and the algebraic equations of the block diagram of Figure E11.11.2a.

b) Use either Mathematica or Matlab to establish steady-state conditions by imposing a step function for load ref₁(s)= $\frac{0.8}{s}$ pu, load ref₂(s) = $\frac{0.5}{s}$ pu, load ref₃(s)= $\frac{0.01}{s}$ pu, load ref₄(s) = $\frac{0.01}{s}$ pu, load ref₅(s)= $\frac{0.01}{s}$ pu, load ref₆(s) = $\frac{0.01}{s}$ pu, and run the program with a zero step-load changes ΔP_L1 = 0, ΔP_L2 = 0 for 200 s. Save the steady-state values for all variables at 200 s. Plot the calculated angular frequency response.

c) Initialize the parameters with the steady-state values as obtained in Part b). After 300 s impose step-load change ${\mathrm{\Delta }P}_{\mathrm{L}1}\left(s\right)=\frac{\mathrm{\Delta }{P}_{\mathrm{L}1}}{s}\mathrm{p}\mathrm{u}=\frac{0.1}{s}$ pu, and after 400 s impose $\mathrm{\Delta }{P}_{\mathrm{L}2}\left(s\right)=\frac{\mathrm{\Delta }{P}_{\mathrm{L}2}}{s}\mathrm{p}\mathrm{u}=\frac{-0.1}{s}$ pu. Thereafter, for load ref₃(s), load ref₄ (s), load ref₃(s), DPstorage₄(s), DPstorage₆(s) and load ref₄ (s):

Lr3[t_]:=If [t < 600, 0, 0.06];

Lr4[t_]:=If [t < 600.1, 0,- 0.6];

Lr3[t_]:=If [t < 1200,If[t < 1120,If[t < 1000,If[t < 940,If[t < 910,0,0.03],0.09],0.05],0.03],0.0];

Lr5[t_]:=If [t < 1200,If[t < 1120,If[t < 1000,If[t < 940,If[t < 910,0,0.3],0.9],0.5],0.3],0.0];

DPstorage4[t_]:=If[t < 1200.2,If[t < 1120.2,If[t < 1000.2,If[t < 940.2,If[t < 910.2,0,0.15],0.45],0.25],0.15],0.0];

DPstorage6[t_]:=If[t < 1200.2,If[t < 1120.2,If[t < 1000.2,If[t < 940.2,If[t < 910.2,0,0.15],0.45],0.25],0.15],0.0];

Lr4[t_]:=If [t < 700, 0,- 0.01]; Plot the given WP plant load reference (Lr3[t]) and calculated the transient response Δω(t) for a total of 1500 s.

d) Initialize the parameters with the steady-state values as obtained in Part b). After 300 s impose step-load change $\mathrm{\Delta }{P}_{\mathrm{L}1}\left(\mathit{s}\right)=\frac{\mathrm{\Delta }{P}_{\mathrm{L}1}}{s}\mathrm{p}\mathrm{u}=\frac{0.1}{s}$ pu, and after 400 s impose $\mathrm{\Delta }{P}_{\mathrm{L}2}\left(s\right)=\frac{\mathrm{\Delta }{P}_{\mathrm{L}2}}{s}\mathrm{p}\mathrm{u}=\frac{-0.1}{s}$ pu. Thereafter:

Lr3[t_]:=If [t < 600, 0, 0.6];

Lr4[t_]:=If [t < 600.1, 0,- 0.6];

Lr3[t_]:=If [t < 1200,If[t < 1120,If[t < 1000,If[t < 940,If[t < 910,0,0.03],0.09],0.05],0.03],0.0];

Lr5[t_]:=If [t < 1200,If[t < 1120,If[t < 1000,If[t < 940,If[t < 910,0,0.3],0.9],0.5],0.3],0.0];

DPstorage4[t_]:=If[t < 1200.2,If[t < 1120.2,If[t < 1000.2,If[t < 940.2,If[t < 910.2,0,0.01],0.01],0.01],0,0.01],0.0];

DPstorage6[t_]:=If[t < 1200.2,If[t < 1120.2,If[t < 1000.2,If[t < 940.2,If[t < 910.2,0,0.30],0.90],0.50],0.30],0.0];

Lr4[t_]:=If [t < 700, 0,- 0.3]; calculate and plot the transient response Δω(t) for a total of 1500 s.

e) Initialize the parameters with the steady-state values as obtained in Part b). After 300 s impose step-load change $\mathrm{\Delta }{P}_{\mathrm{L}1}\left(s\right)=\frac{\mathrm{\Delta }{P}_{\mathrm{L}1}}{s}\mathrm{p}\mathrm{u}=\frac{0.1}{s}$ pu, and after 400 s impose $\mathrm{\Delta }{P}_{\mathrm{L}2}\left(s\right)=\frac{\mathrm{\Delta }{P}_{\mathrm{L}2}}{s}\mathrm{p}\mathrm{u}=\frac{-0.1}{s}$ pu. Thereafter:

Lr3[t_]:=If [t < 600, 0, 0.6];

Lr4[t_]:=If [t < 600.1, 0,- 0.6];

Lr3[t_]:=If [t < 1200,If[t < 1120,If[t < 1000,If[t < 940,If[t < 910,0,0.03],0.09],0.05],0.03],0.0];

Lr5[t_]:=If [t < 1200,If[t < 1120,If[t < 1000,If[t < 940,If[t < 910,0,0.3],0.9],0.5],0.3],0.0];

DPstorage4[t_]:=If[t < 1205.2,If[t < 1125.2,If[t < 1005.2,If[t < 950.2,If[t < 920.2,0,0.30],0.90],0.50],0.30],0.0];

DPstorage6[t_]:=If[t < 1200.2,If[t < 1120.2,If[t < 1000.2,If[t < 940.2,If[t < 910.2,0,0.01],0.01],0.01],0,0.01],0.0];

Lr4[t_]:=If [t < 700, 0,- 0.3]; calculate and plot the transient response Δω(t) for a total of 1500 s.

f) Initialize the parameters with the steady-state values as obtained in Part b). After 300 s impose step-load change $\mathrm{\Delta }{P}_{\mathrm{L}1}\left(s\right)=\frac{\mathrm{\Delta }{P}_{\mathrm{L}1}}{s}\mathrm{p}\mathrm{u}=\frac{0.1}{s}$ pu, and after 400 s impose $\mathrm{\Delta }{P}_{\mathrm{L}2}\left(s\right)=\frac{\mathrm{\Delta }{P}_{\mathrm{L}2}}{s}\mathrm{p}\mathrm{u}=\frac{-0.1}{s}$ pu. Thereafter:

Lr3[t_]:=If [t < 600, 0, 0.6];

Lr4[t_]:=If [t < 600.1, 0,- 0.6];

Lr3[t_]:=If[t < 1200,If[t < 1120,If[t < 1000,If[t < 940,If[t < 910,0,0.03],0.09],0.05],0.03],0.0];

Lr5[t_]:=If [t < 1200,If[t < 1120,If[t < 1000,If[t < 940,If[t < 910,0,0.3],0.9],0.5],0.3],0.0];

DPstorage4[t_]:=If[t < 1230,If[t < 1150,If[t < 1030,If[t < 970,If[t < 940,0,0.030],0.090],0.050],0.030],0.0];

DPstorage6[t_]:=If[t < 1200.2,If[t < 1120.2,If[t < 1000.2,If[t < 940.2,If[t < 910.2,0,0.01],0.01],0.01],0,0.01],0.0];

Lr4[t_]:=If [t < 700, 0,- 0.3]; calculate and plot the transient response Δω(t) for a total of 1500 s.

Solution to Application Example 11.11

a) Differential and algebraic equations
System # 1: ${\varepsilon }_{11}=\mathrm{load}{\mathrm{r}\mathrm{e}\mathrm{f}}_1-\frac{\mathrm{\Delta }{\omega }_1}{R_1}$, $\mathrm{\Delta }{P}_{\mathrm{valve}\_1}+{\tau }_{G1}\frac{d\left(\mathrm{\Delta }{P}_{\mathrm{valve}\_1}\right)}{dt}={\varepsilon }_{11}$, $\mathrm{\Delta }{P}_{\mathrm{mech}\_1}+{\tau }_{\mathrm{C}\mathrm{H}1}\frac{d\left(\mathrm{\Delta }{P}_{\mathrm{mech}\_1}\right)}{dt}=\mathrm{\Delta }{P}_{\mathrm{valve}\_1}$, ${\varepsilon }_{12}=\mathrm{\Delta }{P}_{\mathrm{mech}\_1}-\mathrm{\Delta }{P}_{\mathrm{L}1}-\mathrm{\Delta }{P}_{\mathrm{tie}}$+ $\mathrm{\Delta }{P}_{\mathrm{mech}\_3}+\mathrm{\Delta }{P}_{\mathrm{mech}\_4}$−ΔP_{storage_4}, $\mathrm{\Delta }{\omega }_1{D}_1+M_1\frac{d\left(\mathrm{\Delta }{\omega }_1\right)}{dt}={\varepsilon }_{12}$.
Coupling (tie, transmission) network: $\frac{1}{T}\frac{d\left(\mathrm{\Delta }{P}_{\mathrm{tie}}\right)}{dt}={\varepsilon }_3$, where ${\varepsilon }_3=\mathrm{\Delta }{\omega }_1-\mathrm{\Delta }{\omega }_2$.
System # 2: ${\varepsilon }_{22}=\mathrm{load}{\mathrm{r}\mathrm{e}\mathrm{f}}_2-\frac{\mathrm{\Delta }{\omega }_2}{R_2}$, $\mathrm{\Delta }{P}_{\mathrm{valve}\_2}+{\tau }_{\mathrm{G}2}\frac{d\left(\mathrm{\Delta }{P}_{\mathrm{valve}\_2}\right)}{dt}={\varepsilon }_{22}$, $\mathrm{\Delta }{P}_{\mathrm{mech}\_2}+{\tau }_{\mathrm{C}\mathrm{H}2}\frac{d\left(\mathrm{\Delta }{P}_{\mathrm{mech}\_2}\right)}{dt}=\mathrm{\Delta }{P}_{\mathrm{valve}\_2}$, ${\varepsilon }_{21}=\mathrm{\Delta }{P}_{\mathrm{mech}\_2}-\mathrm{\Delta }{P}_{\mathrm{L}2}+\mathrm{\Delta }{P}_{\mathrm{tie}}$+ $\mathrm{\Delta }{P}_{\mathrm{mech}\_5}+\mathrm{\Delta }{P}_{\mathrm{mech}\_6}$− ΔP_{storage_6}, $\mathrm{\Delta }{\omega }_2{D}_2+M_2\frac{d\left(\mathrm{\Delta }{\omega }_2\right)}{dt}={\varepsilon }_{21}$.
System # 3: ${\varepsilon }_{33}=\mathrm{load}{\mathrm{r}\mathrm{e}\mathrm{f}}_3-\frac{\mathrm{\Delta }{\omega }_1}{R_3}$, $\mathrm{\Delta }{P}_{\mathrm{valve}\_3}+{\tau }_{\mathrm{G}3}\frac{d\left(\mathrm{\Delta }{P}_{\mathrm{valve}\_3}\right)}{dt}={\varepsilon }_{33}$, $\mathrm{\Delta }{P}_{\mathrm{mech}\_3}+{\tau }_{\mathrm{C}\mathrm{H}3}\frac{d\left(\mathrm{\Delta }{P}_{\mathrm{mech}\_3}\right)}{dt}=\mathrm{\Delta }{P}_{\mathrm{valve}\_3}$.
System # 4: ${\varepsilon }_{44}=\mathrm{load}{\mathrm{r}\mathrm{e}\mathrm{f}}_4-\frac{\mathrm{\Delta }{\omega }_2}{R_4}$, $\mathrm{\Delta }{P}_{\mathrm{valve}\_4}+{\tau }_{\mathrm{G}4}\frac{d\left(\mathrm{\Delta }{P}_{\mathrm{valve}\_4}\right)}{dt}={\varepsilon }_{44}$, $\mathrm{\Delta }{P}_{\mathrm{mech}\_4}+{\tau }_{CH4}\frac{d\left(\mathrm{\Delta }{P}_{\mathrm{mech}\_4}\right)}{dt}=\mathrm{\Delta }{P}_{\mathrm{valve}\_4}$.
System # 5: ${\varepsilon }_{55}=\mathrm{load}{\mathrm{r}\mathrm{e}\mathrm{f}}_5-\frac{\mathrm{\Delta }{\omega }_2}{R_5}$, $\mathrm{\Delta }{P}_{\mathrm{valve}\_5}+{\tau }_{\mathrm{G}5}\frac{d\left(\mathrm{\Delta }{P}_{\mathrm{valve}\_5}\right)}{dt}={\varepsilon }_{55}$, $\mathrm{\Delta }{P}_{\mathrm{mech}\_5}+{\tau }_{\mathrm{C}\mathrm{H}5}\frac{d\left(\mathrm{\Delta }{P}_{\mathrm{mech}\_5}\right)}{dt}=\mathrm{\Delta }{P}_{\mathrm{valve}\_5}$.
System # 6: ${\mathit{\varepsilon }}_{66}=\mathrm{load}{\mathrm{r}\mathrm{e}\mathrm{f}}_6-\frac{\mathrm{\Delta }{\omega }_2}{R_6}$, $\mathrm{\Delta }{P}_{\mathrm{valve}\_6}+{\tau }_{\mathrm{G}6}\frac{d\left(\mathrm{\Delta }{P}_{\mathrm{valve}\_6}\right)}{dt}={\varepsilon }_{66}$, $\mathrm{\Delta }{P}_{\mathrm{mech}\_6}+{\tau }_{\mathrm{C}\mathrm{H}6}\frac{d\left(\mathrm{\Delta }{P}_{\mathrm{mech}\_6}\right)}{dt}=\mathrm{\Delta }{P}_{\mathrm{valve}\_6}$.
The frequency variation of a power system should be within 59-61 Hz, that is, a frequency band of $\pm$ 1.66 Hz. Table E11.11.1 lists the Mathematica program for Figure E11.11.2a on which the following figures are based. Figure E11.11.3a illustrates the establishment of steady-state conditions at time t = 200 s for given systems parameters. Figure E11.11.3b illustrates the given WP plant load reference (Lr5[t]). Figure E11.11.3c shows the transient response Δω(t) if storage plants 4 and 6 absorb the additional energy generated by WP plant with 0.2 s delay, and Figure E11.11.3d depicts the transient response Δω(t) if only storage plant 6 absorbs the additional energy generated by WP plant with 0.2 s delay. If the change of the PW plant is absorbed by the short-term storage plant 4 associated with the PV plant then Figure E11.11.3e is obtained. Figure E11.11.3f shows the transient response Δω(t) if no storage plant absorbs the additional energy generated by WP plant. The influence of the transmission line, switching instants, and storage plants increasing Δω(t) can clearly be observed.

Table E11.11.1

Mathematica input program for stability analysis of Figures E11.11.2, E11.11.3

R1 = 0.01; d1 = 0.8; M1 = 4.5; Tg1 = 0.3; Tch1 = 0.9; Lr1 = 0.8; DPL1[t_]:=If [t < 210, 0, 0.1]; R2 = 0.1; d2 = 1.0; M2 = 6; Tg2 = 0.2; Tch2 = 0.2; Lr2 = 0.5; DPL2[t_]:=If [t < 400, 0, -0.1]; Xtie = 0.2; Tie = 377/Xtie; R3 = 0.3; Tg3 = 0.1; Tch3 = 0.1;	DPvalve4b = (DPvalve4[t]/.sol1[[1]]/.t- > 200); DPmech5b = (DPmech5[t]/.sol1[[1]]/.t- > 200); DPvalve5b = (DPvalve5[t]/.sol1[[1]]/.t- > 200); DPmech6b = (DPmech6[t]/.sol1[[1]]/.t- > 200); DPvalve6b = (DPvalve6[t]/.sol1[[1]]/.t- > 200); ic16 = Dw1[200] = Dw1b; ic17 = DPmech1[200] = DPmech1b; ic18 = DPvalve1[200] = DPvalve1b; ic19 = DPtie[200] = DPtie1b; ic20 = Dw2[200] = Dw2b; ic21 = DPmech2[200] = DPmech2b; ic22 = DPvalve2[200] = DPvalve2b; ic23 = DPmech3[200] = DPmech3b; ic24 = DPvalve3[200] = DPvalve3b; ic25 = DPmech4[200] = DPmech4b; ic26 = DPvalve4[200] = DPvalve4b; ic27 = DPmech5[200] = DPmech5b; ic28 = DPvalve5[200] = DPvalve5b; ic29 = DPmech6[200] = DPmech6b;
Lr3[t_]:=If [t < 600, 0, 0.06]; R4 = 0.5; Tg4 = 0.2; Tch4 = 0.1; Lr4[t_]:=If [t < 600.1, 0,- 0.6]; R5 = 0.7; Tg5 = 0.1; Tch5 = 0.1; Lr5[t]:=If [t < 600.2, 0, 0.3]; R6 = 0.5; Tg6 = 0.2; Tch6 = 0.1; Lr6 = 0.01; ic1 = Dw1[0]= = 0; ic2 = DPmech1[0]= = 0; ic3 = DPvalve1[0]= = 0; ic4 = DPtie[0]= = 0; ic5 = Dw2[0]= = 0; ic6 = DPmech2[0]= = 0; ic7 = DPvalve2[0]= = 0; ic8 = DPmech3[0]= = 0; ic9 = DPvalve3[0]= = 0; ic10 = DPmech4[0]= = 0; ic11 = DPvalve4[0]= = 0; ic12 = DPmech5[0]= = 0; ic13 = DPvalve5[0]= = 0; ic14 = DPmech6[0]= = 0; ic15 = DPvalve6[0]= = 0; E11[t_]:=Lr1-Dw1[t]/R1; E12[t_]:=DPmech1[t]-DPL1[t]-DPtie[t] + DPmech3[t] + DPmech4[t]; E3[t_]:=Dw1[t]-Dw2[t]; E22[t_]:=Lr2-Dw2[t]/R2; E21[t_]:=DPmech2[t]-DPL2[t] + DPtie[t] + DPmech5[t] + DPmech6[t]; E33[t_]:=Lr3[t]-Dw1[t]/R3; E44[t_]:=Lr4[t]-Dw1[t]/R4; E55[t_]:=Lr5[t]-Dw2[t]/R5; E66[t_]:=Lr6-Dw2[t]/R6; eqn1 = Dw1'[t]= =(1/M1)*(E12[t]-d1*Dw1[t]); eqn2 = DPmech1'[t]= =(1/Tch1)*(DPvalve1[t]-DPmech1[t]); eqn3 = DPvalve1'[t]= =(1/Tg1)*(E11[t]-DPvalve1[t]);	ic30 = DPvalve6[200] = DPvalve6b; R1 = 0.01; d1 = 0.8; M1 = 4.5; Tg1 = 0.3; Tch1 = 0.9; Lr1 = 0.8; DPL1[t_]:=If [t < 300, 0, 0.1]; R2 = 0.1; d2 = 1.0; M2 = 6; Tg2 = 0.2; Tch2 = 0.2; Lr2 = 0.5; DPL2[t_]:=If [t < 400, 0, -0.1]; Xtie = 0.2; Tie = 377/Xtie; R3 = 0.3; Tg3 = 0.1; Tch3 = 0.1; Lr3[t_]:=If [t < 1200,If[t < 1120,If[t < 1000,If[t < 940,If[t < 910,0,0.03],0.09],0.05],0.03],0.0]; DPstorage4[t_]:=If [t < 1200.2,If[t < 1120.2,If[t < 1000.2,If[t < 940.2,If[t < 910.2,0,0.15],0.45],0.25],0.15],0.0]; DPstorage6[t_]:=If [t < 1200.2,If[t < 1120.2,If[t < 1000.2,If[t < 940.2,If[t < 910.2,0,0.15],0.45],0.25],0.15],0.0]; R4 = 0.5; Tg4 = 0.2; Tch4 = 0.1; Lr4[t_]:=If [t < 700, 0,- 0.01]; R5 = 0.7; Tg5 = 0.1; Tch5 = 0.1; Lr5[t]:=If [t < 1200,If[t < 1120,If[t < 1000,If[t < 940,If[t < 910,0,0.3],0.9],0.5],0.3],0.0]; R6 = 0.5; Tg6 = 0.2; Tch6 = 0.1; Lr6 = 0.01; E11[t_]:=Lr1-Dw1[t]/R1; E12[t_]:=DPmech1[t]-DPL1[t]-DPtie[t] + DPmech3[t] + DPmech4[t]-DPstorage4[t];
eqn4 = DPtie'[t]= = Tie*E3[t]; eqn5 = Dw2'[t]= =(1/M2)*(E21[t]-d2*Dw2[t]); eqn6 = DPmech2'[t]= =(1/Tch2)*(DPvalve2[t]-DPmech2[t]); eqn7 = DPvalve2'[t]= =(1/Tg2)*(E22[t]-DPvalve2[t]); eqn8 = DPmech3'[t]= =(1/Tch3)*(DPvalve3 [t]-DPmech3[t]); eqn9 = DPvalve3'[t]= =(1/Tg3)*(E33[t]-DPvalve3[t]); eqn10 = DPmech4'[t]= =(1/Tch4)*(DPvalve4[t]-DPmech4[t]); eqn11 = DPvalve4'[t]= =(1/Tg4)*(E44[t]-DPvalve4[t]); eqn12 = DPmech5'[t]= =(1/Tch5)*(DPvalve5 [t]-DPmech5[t]); eqn13 = DPvalve5'[t]= =(1/Tg5)*(E55[t]-DPvalve5[t]); eqn14 = DPmech6'[t]= =(1/Tch6)*(DPvalve6[t]-DPmech6[t]); eqn15 = DPvalve6'[t]= =(1/Tg6)*(E66[t]-DPvalve6[t]); sol1 = NDSolve[{eqn1,eqn2,eqn3,eqn4,eqn5,eqn6,eqn7,eqn8,eqn9,eqn10,eqn11,eqn12,eqn13,eqn14,eqn15,ic1,ic2,ic3,ic4,ic5,ic6,ic7,ic8,ic9,ic10,ic11,ic12,ic13,ic14,ic15},{Dw1[t],DPmech1[t],DPvalve1[t],DPtie[t],Dw2[t],DPmech2[t],DPvalve2[t],DPmech3[t],DPvalve3[t],DPmech4[t],DPvalve4[t],DPmech5[t],DPvalve5[t],DPmech6[t],DPvalve6[t]},{t,0,200},MaxSteps- > 100000000]; Plot[Dw1[t]/.sol1[[1]],{t,0,200},PlotRange- > All,AxesLabel- > {"t[s]","Dw1[t][pu]"}] Dw1b = (Dw1[t]/.sol1[[1]]/.t- > 200) DPmech1b = (DPmech1[t]/.sol1[[1]]/.t- > 200); DPvalve1b = (DPvalve1[t]/.sol1[[1]]/.t- > 200); DPtie1b = (DPtie[t]/.sol1[[1]]/.t- > 200); Dw2b = (Dw2[t]/.sol1[[1]]/.t- > 200)	E3[t_]:=Dw1[t]-Dw2[t]; E22[t_]:=Lr2-Dw2[t]/R2; E21[t_]:=DPmech2[t]-DPL2[t] + DPtie[t] + DPmech5[t] + DPmech6[t]-DPstorage6[t]; E33[t_]:=Lr3[t]-Dw1[t]/R3; E44[t_]:=Lr4[t]-Dw1[t]/R4; E55[t_]:=Lr5[t]-Dw2[t]/R5; E66[t_]:=Lr6-Dw2[t]/R6; eqn1 = Dw1'[t]= =(1/M1)*(E12[t]-d1*Dw1[t]); eqn2 = DPmech1'[t]= =(1/Tch1)*(DPvalve1[t]-DPmech1[t]); eqn3 = DPvalve1'[t]= =(1/Tg1)*(E11[t]-DPvalve1[t]); eqn4 = DPtie'[t]= = Tie*E3[t]; eqn5 = Dw2'[t]= =(1/M2)*(E21[t]-d2*Dw2[t]); eqn6 = DPmech2'[t]= =(1/Tch2)*(DPvalve2[t]-DPmech2[t]); eqn7 = DPvalve2'[t]= =(1/Tg2)*(E22[t]-DPvalve2[t]); eqn8 = DPmech3'[t]= =(1/Tch3)*(DPvalve3 [t]-DPmech3[t]); eqn9 = DPvalve3'[t]= =(1/Tg3)*(E33[t]-DPvalve3[t]); eqn10 = DPmech4'[t]= =(1/Tch4)*(DPvalve4[t]-DPmech4[t]); eqn11 = DPvalve4'[t]= =(1/Tg4)*(E44[t]-DPvalve4[t]); eqn12 = DPmech5'[t]= =(1/Tch5)*(DPvalve5 [t]-DPmech5[t]); eqn13 = DPvalve5'[t]= =(1/Tg5)*(E55[t]-DPvalve5[t]); eqn14 = DPmech6'[t]= =(1/Tch6)*(DPvalve6[t]-DPmech6[t]); eqn15 = DPvalve6'[t]= =(1/Tg6)*(E66[t]-DPvalve6[t]); sol2 = NDSolve[{eqn1,eqn2,eqn3,eqn4,eqn5,eqn6,eqn7,eqn8,eqn9,eqn10,eqn11,eqn12,eqn13,eqn14,eqn15,ic1,ic2,ic3,ic4,ic5,ic6,ic7,ic8,ic9,ic10,ic11,ic12,ic13,ic14,ic15},{Dw1[t],DPmech1[t],DPvalve1[t],DPtie[t],Dw2[t],DPmech2[t],DPvalve2[t],DPmech3[t],DPvalve3[t],DPmech4[t],DPvalve4[t],
DPmech2b = (DPmech2[t]/.sol1[[1]]/.t- > 200); DPvalve2b = (DPvalve2[t]/.sol1[[1]]/.t- > 200); DPmech3b = (DPmech3[t]/.sol1[[1]]/.t- > 200); DPvalve3b = (DPvalve3[t]/.sol1[[1]]/.t- > 200); DPmech4b = (DPmech4[t]/.sol1[[1]]/.t- > 200);	DPmech5[t],DPvalve5[t],DPmech6[t],DPvalve6[t]},{t,200,1500},MaxSteps- > 100000000]; Plot[Evaluate[Lr5[t]],{t,200,1500},PlotRange- > All,AxesLabel- > {"t (s)","Lr5[t] [pu]"}] Plot[Dw1[t]/.sol2[[1]],{t,200,1500},PlotRange- > All,AxesLabel- > {"t[s]","Dw1[t][pu]"}]

b) The results based on Mathematica are shown in Figures E11.11.3. The Mathematica input program is listed in Table E11.11.1.

11.4.7 Power Quality Issues Due to Renewable Sources

In addition to power-load balance issues, distributed renewable sources lead to high system impedances at the distribution level because these plants cannot deliver additional transient currents when faults occur due to their operation at peak power [68]. Even with acceptable current harmonics, this high system impedance may result in unacceptably high voltage harmonics, single-time events (e.g., spikes due to network switching and synchronization) and non-periodic but repetitive (e.g., flicker) events, and contribute to power quality problems [160–162,46–48,50].

The frequency/load control in distributed grids with a majority of renewable sources (e.g., renewable generation larger than 50%) appears to be the most difficult part of a smart grid and requires extensive research. In addition the transfer of power via AC and DC lines --associated with harmonic and power factor control--must be studied: most necessary software programs, e.g., symmetric (based on single phase) and asymmetric power (based on three phases) flow programs for this effort are available for integer harmonics, but are not available for sub and non-integer harmonics.

11.4.8 Control of Voltage and Reactive Power

Today frequency and voltage control predominantly occurs at the AC transmission level. In distributed systems with renewable sources these control functions will take place both at the transmission and distribution levels. As the intermittency of renewable sources makes frequency and voltage control more difficult, short- and long-term storage plants for electric energy must be relied on to compensate the energy flow [163]. This involves an increase of switching actions which may impair the quality of energy delivered to the consumer.

11.4.9 Frequency Variations Due to Charging and Discharging of Storage Plants of Grid with Renewable Sources

The objectives of operation of the grid with renewable sources and storage plants where a natural-gas-fired plant is the frequency leader are:

1) Modeling of an AC power system with renewable sources and storage plants and its frequency variation due to switching actions, if energy is transmitted from renewable sources to load centers 800 km away with and without energy storage plants [49];

2) Determination of length of AC transmission line sections to ensure steady-state and dynamic stability;

3) Selection of time constants of conventional, renewable, and storage plants;

4) Operation of WP and PV plants and charging and discharging of energy storage plants must be such that the transmission line power changes only gradually. Frequency control within acceptable limits (49-51) Hz [159] can be achieved by compensating/complementing the output of PW and PV plants with those of storage plants.

11.4.9.1 Simplified Grid

Figure 11.29 illustrates a simplified AC grid [49] with predominantly renewable energy sources: WP farm in the North of Germany and associated short-term storage plant to compensate for the intermittent output of wind farm. Natural-gas and long-term storage plants are used for peak-power generation near the two load systems. In the South there is a PV farm with associated storage plant. Figures 11.30a, b depict the frequency control for either storage-plant operation or renewable-source operation. In Figure 11.30a the WP and PV plants have no output power and the short-term storage PV and WP plants together with the natural-gas fired plant control though appropriate droop characteristics the frequency of the AC power system. In Figure 11.30b the PV and WP plants operate at their maximum output power through peak-power tracking together with the natural-gas fired and long-term storage plants while the PV and WP storage plants have zero output. There are three modes of operation:

Mode #1: renewable sources (RS) and storage plants supply power to two load systems.

Mode #2: RS charge storage plants [144].

Mode #3: RS supply energy to two load systems.

If the transmitted power throughput across the AC transmission line is not changing slowly enough, frequency stability problems set in due to discontinuous AC power transmission and associated frequency perturbations. For example, a three-phase, 735 kV (line-to-line), 50 Hz, 800 km line has a reactance of 0.55 Ω/km [173] or X_L = 440 Ω resulting at base apparent power S = 1000 MVA with the base impedance of Z_base = (735)²/1000 = 540 Ω in a per-unit line reactance of X_Lpu = (440/540) = 0.815pu. With a transformer reactance of X_Tpu = 0.08pu the total line reactance is X_total = X_Lpu + 2X_Tpu = (0.815 + 0.16)pu = 0.98 pu @ 50 Hz. Transient analysis shows that this long AC transmission line will have to be subdivided into about five sections each having a per-unit reactance of X_section = X_total/5 = 0.98/5pu = 0.196 pu, and at the terminals of each section the voltage will have to be controlled to guarantee stability of transmission. The differential and algebraic equation system of Figure 11.29 is listed below. The AC transmission/tie line acts as an integrating component [144] enforcing Δω(s) = Δω₁(s) = Δω₂(s). The differential and algebraic equations are solved with Mathematica software; s is the Laplace operator.

System #1:

${\varepsilon }_{11}=\mathrm{load}{\mathrm{r}\mathrm{e}\mathrm{f}}_1-\mathrm{\Delta }{\omega }_1/R_1\text{,}$

  (11-6)

$\mathrm{\Delta }{P}_{\mathrm{valve}1}+T_{\mathrm{G}1}\cdot d\left(\mathrm{\Delta }{P}_{\mathrm{valve}1}\right)/dt={\varepsilon }_{11}\text{,}$

  (11-7)

$\mathrm{\Delta }{P}_{\mathrm{mech}1}+T_{\mathrm{C}\mathrm{H}1}\cdot d\left(\mathrm{\Delta }{P}_{\mathrm{mech}1}\right)/dt=\mathrm{\Delta }{P}_{\mathrm{valve}1}\text{,}$

  (11-8)

${\varepsilon }_{12}=\mathrm{\Delta }{P}_{\mathrm{mech}1}-\mathrm{\Delta }{P}_{\mathrm{L}1}-\mathrm{\Delta }{P}_{\mathrm{tie}}+\mathrm{\Delta }{P}_{\mathrm{mech}3}+\mathrm{\Delta }{P}_{\mathrm{mech}4}-\mathrm{\Delta }{P}_{\mathrm{storage}4}\text{,}$

  (11-9)

$\mathrm{\Delta }{\omega }_1{D}_1+M_1\cdot d\left(\mathrm{\Delta }{\omega }_1\right)/dt={\varepsilon }_{12}\text{.}$

  (11-10)

Coupling (tie, transmission) network:

$\left(1/T\right)\cdot d\left(\mathrm{\Delta }{P}_{\mathrm{tie}}\right)/dt={\varepsilon }_3\text{,}$

  (11-11)

${\varepsilon }_3=\mathrm{\Delta }{\omega }_1-\mathrm{\Delta }{\omega }_2\text{.}$

  (11-12)

System #2:

${\varepsilon }_{22}=\mathrm{load}{\mathrm{r}\mathrm{e}\mathrm{f}}_2-\mathrm{\Delta }{\omega }_2/R_2\text{,}$

  (11-13)

$\mathrm{\Delta }{P}_{\mathrm{valve}2}+T_{\mathrm{G}2}\cdot d\left(\mathrm{\Delta }{P}_{\mathrm{valve}2}\right)/dt={\varepsilon }_{22}\text{,}$

  (11-14)

$\mathrm{\Delta }{P}_{\mathrm{mech}2}+T_{\mathrm{C}\mathrm{H}2}\cdot d\left(\mathrm{\Delta }{P}_{\mathrm{mech}2}\right)/dt=\mathrm{\Delta }{P}_{\mathrm{valve}2}\text{,}$

  (11-15)

${\varepsilon }_{21}=\mathrm{\Delta }{P}_{\mathrm{mech}2}-\mathrm{\Delta }{P}_{\mathrm{L}2}+\mathrm{\Delta }{P}_{\mathrm{tie}}+\mathrm{\Delta }{P}_{\mathrm{mech}5}+\mathrm{\Delta }{P}_{\mathrm{mech}6}-\mathrm{\Delta }{P}_{\mathrm{storage}6}\text{,}$

  (11-16)

$\mathrm{\Delta }{\omega }_2{D}_2+M_2\cdot d\left(\mathrm{\Delta }{\omega }_2\right)/dt={\varepsilon }_{21}\text{.}$

  (11-17)

System #3:

${\varepsilon }_{33}=\mathrm{load}{\mathrm{r}\mathrm{e}\mathrm{f}}_3-\mathrm{\Delta }{\omega }_1/R_3\text{,}$

  (11-18)

$\mathrm{\Delta }{P}_{\mathrm{valve}3}+T_{\mathrm{G}3}\cdot d\left(\mathrm{\Delta }{P}_{\mathrm{valve}3}\right)/dt={\varepsilon }_{33}\text{,}$

  (11-19)

$\mathrm{\Delta }{P}_{\mathrm{mech}3}+T_{\mathrm{C}\mathrm{H}3}\cdot d\left(\mathrm{\Delta }{P}_{\mathrm{mech}3}\right)/dt=\mathrm{\Delta }{P}_{\mathrm{valve}3}\text{.}$

  (11-20)

System #4:

${\varepsilon }_{44}=\mathrm{load}{\mathrm{r}\mathrm{e}\mathrm{f}}_4-\mathrm{\Delta }{\omega }_1/R_4\text{,}$

  (11-21)

$\mathrm{\Delta }{P}_{\mathrm{valve}4}+T_{\mathrm{G}4}\cdot \mathit{d}\left(\mathrm{\Delta }{P}_{\mathrm{valve}4}\right)/dt={\varepsilon }_{44}\text{,}$

  (11-22)

$\mathrm{\Delta }{P}_{\mathrm{mech}4}+T_{\mathrm{C}\mathrm{H}4}\cdot d\left(\mathrm{\Delta }{P}_{\mathrm{mech}4}\right)/dt=\mathrm{\Delta }{P}_{\mathrm{valve}4}\text{.}$

  (11-23)

System #5:

${\varepsilon }_{55}=\mathrm{load}{\mathrm{r}\mathrm{e}\mathrm{f}}_5-\mathrm{\Delta }{\omega }_2/R_5\text{,}$

  (11-24)

$\mathrm{\Delta }{P}_{\mathrm{valve}5}+T_{\mathrm{G}5}\cdot d\left(\mathrm{\Delta }{P}_{\mathrm{valve}5}\right)/dt={\varepsilon }_{55}\text{,}$

  (11-25)

$\mathrm{\Delta }{P}_{\mathrm{mech}5}+T_{\mathrm{C}\mathrm{H}5}\cdot d\left(\mathrm{\Delta }{P}_{\mathrm{mech}5}\right)/dt=\mathrm{\Delta }{P}_{\mathrm{valve}5}\text{.}$

  (11-26)

System #6:

${\varepsilon }_{66}=\mathrm{load}{\mathrm{r}\mathrm{e}\mathrm{f}}_6-\mathrm{\Delta }{\omega }_2/R_6\text{,}$

  (11-27)

$\mathrm{\Delta }{P}_{\mathrm{valve}6}+T_{\mathrm{G}6}\cdot d\left(\mathrm{\Delta }{P}_{\mathrm{valve}6}\right)/dt={\varepsilon }_{66}\text{,}$

  (11-28)

$\mathrm{\Delta }{P}_{\mathrm{mech}6}+T_{\mathrm{C}\mathrm{H}6}\cdot d\left(\mathrm{\Delta }{P}_{\mathrm{mech}6}\right)/dt=\mathrm{\Delta }{P}_{\mathrm{valve}6}\text{.}$

  (11-29)

11.4.9.3 Non-Constant Power Output of Renewable Sources

Mode #1: Figures 11.31a-d show that at X_section = 0.196 pu at large-step outputs of WP plant and associated storage plant permit a stable AC energy transmission to the two load systems in the South, defined by M₁, M₂, D₁, D₂, ΔP_L1(s), ΔP_L2(s) [49]. However, the angular frequency variations due to switching are large (0.25pu). At X_{section_max} = 0.33 pu instability sets in as indicated in Figure 11.32 where load ref₅(s) = ΔP_storage6(s) = 0. Reduced time constant T_CH1 = 1.4 s results in the stability as demonstrated in Figure 11.33. For medium-step outputs of WP illustrated in Figure 11.34a and associated storage plant the frequency variation due to switching is still large (0.07 pu), as indicated in Figures 11.34b, c. Best performance with lowest angular frequency variation due to switching (0.01 pu) is obtained with small-step outputs of WP (as shown in Figure 11.35a) and associated storage plant, as depicted in Figures 11.35b, c; however, instability results for X_section = 0.33 pu as indicated in Figure 11.36.

Mode #2: If the WP plant large-step output (load ref₃(s) ≠ 0, ΔP_storage4(s) = load ref₃(s)) charges the associated short-term storage plant ΔP_storage4(s) large (0.05 pu) angular frequency variations of Figure 11.37a are obtained while the transmission tie line power (neglecting switching transients) is about constant as depicted in Figure 11.37b. For small-step charging the angular frequency variation is smaller: 0.02 pu, see Figure 11.38a and the tie line power as given in Figure 11.38b is about constant and shows smaller transient excursions during switching. The spikes in the Δω(t) = Δω₁(t) = Δω₂(t) are mitigated by smoothing the output of the WP plant based on small-step operation. The AC transmission of energy from the South PV farm to the two load circuits in the North is similar as discussed above.

According to the newspaper [169] “Financial Times Deutschland”, DC transmission lines are planned [170] to avoid stability problems associated with long AC lines and result in smaller line voltage drops (no reactive drop). DC lines could be mounted on existing towers/pylons presently used for AC transmission lines to reduce costs and avoid protests by citizens. Although DC lines solve the voltage stability problem associated with AC lines, DC lines still require short-term and long-term storage to control the frequency at the AC load side by enforcing power balance. DC lines are known for their voltage distortions, thus harmonic filters would also be required as discussed in Chapter 9. Frequency control can be achieved by employing short-and long-term storage plants [1,144] complementing output of renewable sources by small step-wise control of the AC transmission-line power. Stable AC transmission results by appropriate power management: wave shaping of power inputs/outputs, appropriate transmission line segmentation, and selection of power plant and storage plant time constants.

11.4.10 Reduced Life Time of Storage Plants

Renewable sources as well as all types of storage plants are operated intermittently, and therefore the winding structures of generators are thermally exposed to changing temperatures through changing loss and heating conditions: this causes the winding to contract and expand depending upon the changing temperature resulting in mechanical degradation (see Chapter 6) of the insulating material of electrical components such as generators (Chapter 4). The rated lifetime of about 40 years is thus reduced by a factor of two resulting in long outages (e.g., half a year) of plants.

11.4.11 Solutions to Power Quality Problems Due to Renewable Plants with Energy Storage Systems

Frequently charging and discharging of storage plants impact the lifetime of the insulation material due to mechanical stresses and temperature changes. In addition the transients due to rapid frequency changes induce mechanical (e.g., flywheel) and electrical (e.g., overvoltage) stresses in the equipment. To operate storage and renewable sources in a complementary stable manner the droop characteristics must be different where the small-sloped characteristic is associated with the frequency leader. In addition the control of storage and renewable sources must have sufficient gains and small enough time constants as required by fast mechanical [174] and electronic valving and switching, respectively. Parameter variations are an important subject in the field of power quality because these may cause instability of the grid.

11.5.1 Solutions to Power Quality Problems of High-Voltage AC Transmission Lines

The problem with buried AC transmission lines is that at voltages larger than 20 kV at a certain length ferroresonance sets in due to the large cable capacitances. This is the reason why at high voltages, overhead lines are installed which have small network capacitances. Gas-filled cables – avoiding ferroresonance-- have a reduced capacitance and can be buried. However, gas-filled cables are very expensive. DC transmission lines do not experience ferroresonance, and can be either overhead or underground lines. The application of rectifiers and inverters complicate their application at large transmission power ratings, say 2,000 MW.

11.8.1 Net Metering

Most electricity meters are bi-directional and can measure current flowing in two directions [1]. This permits one to easily bank excess electricity from your solar panels for future credit. Net metering only requires one power meter, while feed-in tariffs require two. Unlike feed-in tariffs and power purchase agreements, the credits one accumulates through net metering are always at full retail value. Net metering was first adopted by utilities in Idaho in 1980. Since the Energy Policy Act of 2005, every public electric utility in USA is required to offer net metering to their customers.

11.8.2 Virtual Net Metering

Virtual net metering is basically net metering (with one electric meter) shared between several people. This enables homeowners that are unsuited for solar for one reason or the other, to participate in community-owned solar farms (also known as solar gardens). There are about one dozen virtual net metering systems operating in the U.S. at the time of writing.

11.8.3 Feed-In Tariff

Feed-in tariff requires one extra power meter in order to measure outflow of electricity from your home independently. This enables electricity consumption and electricity generation to be priced separately as is done in Germany by Stadtwerke München (SWM) and most other utilities. Feed-in tariff schemes are typically based on a 15-20 yearlong contract where prices are pre-defined above retail with a tariff that is subject to changes, either by regular degression or by legislative changes, which effectively reduces your earnings over time. For every kWh one generates one gets paid. Only six states across the U.S. currently have some form of feed-in tariff scheme as of today: California, Florida, Vermont, Oregon, Maine, and Hawaii.

A PV power plant consists of solar array, peak (maximum)-power tracker [1], a step-up/step-down DC-to-DC converter, a deep-cycle battery for part f) only, a single-phase inverter, single-phase transformer, and a residence requires a maximum inverter AC output power of P_inv^max = 5.91 kW as shown in Figure P11.1.1. Note, the maximum inverter output AC power has been specified because the entire power must pass through the inverter for all operating modes as is explained below. In addition, inverters cannot be overloaded even for a short time due to the low heat capacity of the semiconductor switches.

Three operating modes will be investigated:

1) in part f) the operating mode #1 is a stand-alone configuration (546 kWh are consumed per month in Munich, Germany)

2) in part g) the operating mode #2 is a configuration, where the entire energy (546 kWh) is consumed by the residence and the utility system is used as storage device only,

3) in part h) the operating mode #3 is a configuration, where 60 kWh are consumed by the residence and 486 kWh are sold to the utility. The consumption in the residence -- disregarding heating energy -- is low due to LED lighting, high efficiency appliances (e.g., refrigerator, induction-type stove), and electronic equipment

a) The power efficiencies of the maximum power tracker, the step-up/step-down DC to DC converter, the battery, and the inverter are 97% each, while that of the transformer is about 1.00. What maximum power P_max^solararray must be generated by the solar plant (array), provided during daytime E_{month_day} = 60 kWh will be delivered via the inverter to the residence (without storing this energy in battery), and sufficient energy will be stored in the battery so that the battery energy of E_{month_night} = 486 kWh can be delivered during nighttime by the battery via the inverter to the residence: that is a total of E_month = E_{month_day} + E_{month_night} = 546 kWh can be delivered to the residence during one month?

b) For a commercially available solar panel the V-I characteristic of Figure 3.23 of [1] was measured at an insolation of Q_s = 0.9 kW/m². Plot the power curve of this solar panel: P_panel = f(I_panel).

c) At which point of the power curve P_panel = f(I_panel) would you operate assuming Q_s = 0.9 kW/m² is constant? What values for power, voltage and current correspond to this point?

d) How many solar panels would you have to connect in series (N_s) in order to achieve a DC output voltage of $V_{\mathrm{D}\mathrm{C}}^{\text{max}}=$ 480 V of the solar plant (array)? How many solar panels would you have to connect in parallel (N_p) in order to generate the inverter output power P_inv^max = 5.91 kW?

e) How much would be the purchase price of this solar power plant, if 1 kW installed output capacity of the inverter (this includes the purchase costs of solar cells + peak-power tracker + DC-to-DC converter + inverter) costs US $3,500 (without utility rebates and state/federal government tax-related subsidies)?

f) Operating mode #1
What is the payback period (in years, without taking into account interest payments) of this solar plant if you use in your residence 546 kWh per month at an avoided cost of US $0.37/kWh (includes service fees and tax)? (You may assume that this solar plant can generate every month 546 kWh and there is no need to buy electricity from the utility: 60 kWh per month will be used in the residence during daytime and during nighttime 486 kWh per month will be supplied via inverter from the battery to the residence). However, there is a need for the use of a 20 kWh deep-cycle battery as a storage element so that electricity will be available during hours after sunset. This battery must be replaced every four years at a cost of US $4,000.

g) Operating mode #2
What is the payback period (in years, without taking into account interest payments) of this solar plant if you use in your residence 60 kWh per month at an avoided cost of US $0.37 per kWh? You may assume that this solar plant can generate every month 546 kWh and every month the solar plant feeds 486 kWh into the power system of the utility company which reimburses you US $0.22 per kWh (so-called “net metering”, see reference 27 of Chapter 1 [1]). In this case there is no need for a battery as a storage element because during the hours after sunset the electricity can be supplied by the utility: 486 kWh at US $0.22 per kWh. There is a connection charge of US $7.00 per month.

h) Operating mode #3
What is the payback period (in years, without taking into account interest payments) of this solar plant if you use in your residence 60 kWh per month at an avoided cost of US $0.37 per kWh? You may assume that this solar plant can generate every month 546 kWh of which every month the solar plant feeds 486 kWh per month into the power system of the utility company which reimburses you US $0.22 per kWh. There is a connection charge of US $7.00 per month.

i) Which power plant configuration (e.g., f, g or h) is more cost effective (e.g., has the shortest payback period)?

j) What is the total surface of the solar panels provided the efficiency of solar cells is 15% at Q_s = 0.9 kW/m²?

k) Instead of obtaining tax rebates and state/federal government subsidies the owner of a PV power plant obtains a higher price (feed-in tariff, see Chapter 1 [1]) for the electricity delivered to the utility: Provided 546 kWh are fed into the utility grid at a reimbursement cost of US $0.22/kWh and the utility supplies 60 kWh to the residence at a cost of US $0.37/kWh, what is the payback period if the entire plant generating 5.91 kW_AC (there are no batteries required for storage) costs US $24.355 k. You may neglect interest payments, and there is a connection charge of $7.00 per month.

l) Repeat part k) taking into account interest payments of 2.85%.

Problem 11.2: Calculation of Energy Consumption in Residence per Living Space (m²) and per Year (a)

A residence with 300 m² finished living space requires 12,000 kWh of energy (input/consumption in house) during one year. Determine the energy consumption in house per square meter and per year.

Problem 11.3: Photovoltaic (PV) Energy Yield as a Function of Latitude

Calculate based on the National Renewable Laboratory (NREL) software available at http://rredc.nrel.gov/solar/calculators/PVWATTS/version1/ the electric energy generation for a PV residential plant of 6.48 kW installed output power capacity when either located at a latitude (array tilt) 40.02 degrees (e.g., Boulder Colorado, cost of electricity is US$ 0.16/kWh) and at latitude (array tilt) 48.13 degrees (e.g., Munich Germany cost of electricity is 0.26 Euros/kWh = US$ 0.36/kWh) in the northern hemisphere for an array azimuth of 180 degrees with an overall DC to AC derating (electrical efficiency of all components) factor of 0.77 and an array type 20 degree tilt in Boulder and an array tilt of 48.13 degrees in Munich.

Problem 11.4: Burst-Current/Voltage Control for Induction-Type Stove/Cooker

An induction-type stove has a rated power of P_rat = 800 W with (T_on/T) = 0.8 at 50 Hz find the output heating power of the induction-stove coil. Normally burst-current control relies on the frequency range 20-50 kHz, but in this problem 50 Hz are chosen for simplicity.

Problem 11.5: Burst-Current/Voltage Control for Induction Stove/Cooker

An induction-type stove has a rated voltage of V_{t_rms_rat} = 240 V at f = 50 Hz with (T_on/T) = 0.8 find the input voltage of the induction-stove coil. Normally burst-current control relies on the frequency range 20-50 kHz, but in this problem 50 Hz are chosen for simplicity.

Problem 11.6: Current Harmonics Injected into Power System by Burst-Current Control

It can be assumed that for burst-voltage control, current and voltage are in phase because of resistive load R_load, and both have sinusoidal wave shapes. Assume a burst period of T = T_on + T_off, where T_on/T = 0.5 at f = 50 Hz for sinusoidal terminal voltage with V_{t_rms_rat} = 240 V. Normally burst-current control relies on the frequency range 20-50 kHz, but in this problem 50 Hz are chosen for simplicity.

a) What are the firing angles of the two thyristors of Figure E11.2.2a for T_on/T = 0.5?

b) Write a PSpice program for circuit of Figure E11.2.2a and compute the current through R_load = 2.2 Ω of Figure E11.2.2a.

c) Determine the Fourier coefficients of the current through R_load up to the 21^th harmonic.

d) Plot the current through R_load together with the terminal voltage V(10)-V(0).

e) What types of current harmonics are generated in I(R_load):

(1) sub-harmonics,

(2) inter-harmonics/fractional harmonics/non-integer harmonics,

(3) integer harmonics?

Problem 11.7: Phase-Angle Current Control for Induction Stove/Cooker

An induction-type stove has a rated voltage of V_{t_rms_rat} = 240 V at f = 50 Hz with (t_delay/T/2) = 0.4 find the input voltage of the induction-stove coil. Phase-angle current control relies sometimes on a higher frequency than 50 Hz, but in this problem 50 Hz is chosen for simplicity.

Problem 11.8: Current Harmonics Injected into Power System by Phase-Angle Current Control

It can be assumed that for phase-voltage control the load is resistive as indicated in Figure E11.2.2a and the voltage has sinusoidal wave shape. Compute the phase current through load resistor R_load = 2.2 Ω as shown in Figure E11.2.2a for a sinusoidal terminal voltage with rms value V_{t_rms} = 240 V and a current phase angle delay time t_delay = 3 ms at f = 50 Hz. Phase-current control relies sometimes on a high frequencies (20-50 kHz), but in this problem 50 Hz are chosen for simplicity.

a) What are the firing angles of the two thyristors of Figure E11.2.2a for t_delay = 3 ms?

b) Write a PSpice program for circuit of Figure E11.2.2a and compute the current through R_load of Figure E11.2.2a.

c) Determine the Fourier coefficients of the current through R_load up to the 21^th harmonic.

d) Plot the current through R_load together with the terminal voltage V(10)-V(0).

e) What types of current harmonics are generated in I(R_load):

(1) sub-harmonics,

(2) inter-harmonics/fractional harmonics/non-integer harmonics,

(3) integer harmonics?

Problem 11.9: PWM Rectifier Output Voltage V_{DC load} = V(R_load) = f(V_rms) at Constant Duty-Cycle δ for Variable Input Phase Voltage V_rms

Section 11.2.4.1 computes for a pulse-width-modulated (PWM) three-phase rectifier the dependency of the input voltage V_{aN max} at a constant output voltage V_{DC load} and constant output current I_{DC load} provided the duty cycle δ is variable.

In this problem compute and plot for the rectifying circuit of Figure 11.1 the input phase currents I(V_aN), I(V_bN), I(V_cN), and output voltage V_{DC load} = V(R_load) as a function of the input phase voltages V_aNmax, V_bNmax, V_cNmax = √2V_rms = (233 V, 465 V, 930 V) at a constant duty cycle δ = 0.50 from 300 ms to 325 ms. The PSpice program of Table 11.1 can be used as a starting point with switch gating signal frequency = 20 kHz.

Problem 11.10: PWM Rectifier Output Voltage V_{DC load} = f(δ) for Constant Input Voltage V_rms = 329 V

Section 11.2.4.1 computes for a pulse-width-modulated (PWM) three-phase rectifier the dependency of the input voltage V_{AN max} at a constant output voltage V_{DC load} and constant output current I_{DC load} provided the duty cycle δ is variable.

In this problem compute for the rectifying circuit of Figure 11.1 the output voltage V_{DC load} as a function of the duty cycle δ at constant input voltage V_rms = 329 V, where the input sinusoidal phase-voltage amplitude is V_{aN max} = 465 V. The PSpice program of Table 11.1 can be used as a starting point.

Problem 11.11: Current-Controlled PWM Inverter Output Voltage V_rms = f(V_DC) for Given Modulation Index m

Section 11.2.4.2 computes for a current-controlled pulse-width-modulated (PWM) three-phase inverter the dependency of the input voltage V_DC at constant output voltage V_{rms l-n} as a function of the displacement (power factor) angle φ and the modulation index m.

In this problem compute the output currents I_{rms ph} and the modulation index m of the inverter circuit of Figure 11.6a, b for the output voltages V_rms = (240/√3) V = 139 V, (380/√3) V = 220 V, (480/√3) V = 277 V as a function of the input DC voltage V_DC = 700 V for leading, lagging and unity displacement (power) factors (generator notation) with displacement power factor angle φ. The PSpice program of Table 11.3 can be used as a starting point with switch gating signal voltage (vtrial) frequency of 17.28 kHz.

Problem 11.12: Current-Controlled PWM Inverter Output Voltage V_rms = f(m) for Given Input Voltage V_DC

Section 11.2.4.2 computes for a current-controlled pulse-width-modulated (PWM) three-phase inverter the dependency of the input voltage V_DC at constant output voltage V_{rms l-n} as a function of the displacement (power factor) angle φ and the modulation index m.

In this problem compute for the inverter circuit of Figure 11.6a, b the modulation index m for constant input DC voltage V_DC = 400 V and variable output inverter phase voltages V_{rms ln} (240/√3 = 139 V, 380/√3 = 220 V, 480/√3 = 277 V, 600/√3 = 346 V, 2400/√3 = 1386 V) at unity power factor. The PSpice program of Table 11.3 can be used as a starting point. The frequency of the gating signal voltage is 5.76 kHz.

Problem 11.13: Differential-Mode Electrical Noise Filter

The differential electrical noise of f_10kHz = 10 kHz with the amplitude of ${\tilde{V}}_{10\mathit{kHz}}=1V\angle 0°$ across the terminals of a V_DC = 12 V power supply for a radio must be attenuated to V_noise = 10 m V. Figure P11.13.1 shows the circuit of the battery voltage of V_DC = 12 V including the ${\tilde{V}}_{10\mathit{kHz}}=1V\angle 0°$ as well as the differential-mode filter with impedance Z_filter = R_filter + 1/(j2π·f_10kHz·C_filter). You may assume R_filter = 0.1Ω, Z_battery ≈ R_battery = 0.5 Ω, and R_radio = 1 Ω.

a) Draw the AC circuit of Figure P11.13.1.

b) Compute the required capacitance C_filter of the filter.

Problem 11.14: Common-Mode Electrical Noise Filter

The internet connection of a computer is from the modem 20 m long, and the transmission is frequently interrupted due to common-mode noise. Design a suitable filter and specify its components to reduce the maximum value of the common-mode current to i_common = 10 μA.

Problem 11.15: Maximum Power Point of a PV Array of a P_out = 8 MW_AC Photovoltaic (PV) Power Plant

A PV power plant consists of solar panels, peak-power tracker, step-down/step-up DC-to-DC converter(s) (20 connected in parallel), three-phase inverter(s) (20 connected in parallel), Y-Δ three-phase transformer, and a three-phase power system absorbing the rated (nominal) output power P_{out inverter} of the 20 inverters at rated operation and rated (maximum) insolation Q_s = 0.9 kW/m². The block diagram of this plant is shown in Figure P11.15.1. The power efficiencies of the peak-power tracker, step-down/step-up DC-to-DC converters, inverters, and transformer are η = 95% each.

a) Using the output characteristics of the solar array of Figure P11.15.2a determine at an insolation of Q_s = 0.9 kW/m² the output (peak) power of the solar array $P_{\mathrm{solararray}}^{\mathrm{Q}\mathrm{s}=0.9\mathrm{kW}/{\mathrm{m}}^2}$, and the resulting output power of the 20 parallel-connected inverters ${\mathit{P}}_{\mathrm{outinverter}}^{\mathrm{Q}\mathrm{s}=0.9\mathrm{kW}/{\mathrm{m}}^2}$.

b) If the line-to-line voltage of the power system -which is identical to the output line-to-line voltage of the Δ-secondary of the Y-Δ transformer (t)- is $V_{\mathrm{L}-\mathrm{L}}^{\mathrm{t}}$= 13.2 kV determine the input line-to-line voltage of the Y-primary of the transformer -which is identical with the output line-to-line voltage of the 20 (parallel-connected) inverters (i) $V_{\mathrm{L}-\mathrm{L}}^{\mathrm{i}}$- provided the DC input voltage of one inverter is V_DCⁱ= 2 kV and the modulation index m = 0.8, see modulation index m for inverters in [1, Chapter 5]. What is the turns ratio of the transformer N_p/N_s where the primary (p) is the Y-winding and the secondary (s) is the Δ-winding of the Y-Δ transformer, see 3-phase transformer connections in [1, Chapter 8]?

c) Repeat the analysis of part a) at an insolation of Q_s = 0.6 kW/m², that is, determine the output (peak) power of the solar array $P_{\mathrm{solar}\mathrm{array}}^{\mathrm{Q}\mathrm{s}=0.6\mathrm{kW}/{\mathrm{m}}^2}$ and the resulting output power of the 20 (parallel-connected) inverters $P_{\mathrm{out}\mathrm{inverter}}^{\mathrm{Q}\mathrm{s}=0.6\mathrm{kW}/{\mathrm{m}}^2}$.

d) The output characteristic of the solar array changes due to shadowing, see references 33 and 36 of [1, Chapter 3], caused by clouds, snow, dirt and leaves even though there are bypass diodes mitigating this effect. Repeat the analysis of part a) for the output characteristic of Figure P11.15.2b, that is, determine the output (peak) power of the solar array $P_{\mathrm{solar}\mathrm{array}}^{\mathrm{Q}\mathrm{s}=0.9\mathrm{kW}/{\mathrm{m}}^2\mathrm{with}\mathrm{shadowing}}$ and the resulting output power of the 20 parallel-connected inverters $P_{\mathrm{out}\mathrm{inverter}}^{\mathrm{Q}\mathrm{s}=0.9\mathrm{kW}/{\mathrm{m}}^2\mathrm{with}\mathrm{shadowing}}$.

e) The solar array consists of 670,000 solar panels each having an area of (0.8x1.60) m². What is the total area of all solar panels (Area_total)?

f) What is the payback period (in years) if 1 kW installed output power capacity of the 8 MW PV plant costs $2,500 provided the average wholesale price of 1 kWh is $0.22? Note: The fuel costs are zero and the operational costs are negligible; for the payback period calculation you may assume on the average a 6 h operation of the plant per day. No rebates and tax breaks are available, and no interest must be paid.

Problem 11.16: Design of a 500 MW Pumped-Storage Hydro-Power Plant

A pumped-storage hydro-power plant (Figure P11.16.1) is to be designed (see references [14–18] of Chapter 12 of [1]) for a rated power P_rated = 500 MW and a rated energy capacity 3000 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 50 MWh. Water evaporation must be taken into account and is 10% per year for each reservoir, and the precipitation per year is 40 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 = 12000 cfs, compute the mechanical power available at the turbine input P_kW (see references [146–148].)

c) How does P_turbine^required compare with P_kW?

d) Compute the specific speed N_q (see references [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 = (3000 MWh + 50 MWh) = 3050 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 = (3000 MWh + 50 MWh) = 3050 MWh per day for which customers pay $0.27/kWh due to peak-power generation. What is the payback period of this pumped storage plant if the construction price is $4000 per installed power capacity of 1 kW, the cost for pumping is $0.06/kWh, and the interest rate is $2%?