Problem 11.17: Design of a 50 MW Flywheel Storage Power Plant

Design a flywheel storage system which can provide for about 6 minutes (= 0.1 h) a power of 50 MW, that is, energy of 5 MWh. The flywheel power plant consists (see Figure P11.17.1) 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) Possible flywheel configurations
For the flywheels (made from steel) based on a spherical (R_1o = 1.5 m), cylindrical (R_1o = 1.5 m, h = 0.9 m), and wheel-type (with 4 spokes) configurations as shown in Figure P11.17.2 (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). You may assume that the flywheel has magnetic bearings; see internet address: http://www.skf.com/portal/skf_rev/home
The axial moment of inertia of a rotating sphere is:

$J_{\mathrm{sphere}}=\frac{8}{15}\cdot \pi \cdot \gamma \cdot R_{1\mathrm{o}}^5\left[{\mathrm{k}\mathrm{g}\mathrm{m}}^2\right]$

where the mass density of iron/steel is

${\gamma }_{\mathrm{steel}}=7.8\left[{\mathrm{kg}/\mathrm{dm}}^3\right]$

The axial moment of inertia of a rotating cylinder is:

$J_{\mathrm{cylinder}}=\frac{1}{2}\cdot \pi \cdot \gamma \cdot h\cdot R_{1\mathrm{o}}^4\left[{\mathrm{k}\mathrm{g}\mathrm{m}}^2\right]$

The axial moment of inertia of a rotating wheel-type configuration with 4 spokes is

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

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} = 15,000 rpm.

Problem 11.18: Steady-State Characteristics of a Battery Powered by Solar Array/Panel with Inherent Peak-Power Tracking

Figure P11.18.1 shows the circuit diagram of a battery directly connected to a solar generator (array) and Figure P11.18.2 depicts the 12 V lead-acid battery and solar-power plant characteristics. The operating points (intersections of battery and solar array characteristics) can be made rather close to the maximum power points of the solar power plant (see reference 33 of of Chapter 3 [1]). Note that a 12 V lead-acid battery has a rated voltage of V_BAT = E₀ = 12.6 V. A voltage of V_BAT = E₀ = 11.7 V if almost empty, and a voltage of V_BAT = V_BAToc + V_BATover = 14 V if fully charged, that is, the voltage drops across the electrochemical capacitance C_B and the nonlinear capacitance C_b (V_BATover) of battery cells are included (see Chapter 3 of [1]).

For the characteristics of the solar panel of Figure P11.18.2 the loci of the knees change as a function of the insolation E. In order to obtain peak-power tracking the voltages V_BAT, V_L1, and V_L2 of Figure P11.18.1 must be adjusted as a function of the insolation.

a) Augment the circuit of Figure P11.18.1 with step-down, step up DC converters as described in Chapter 5 of [1] so that the battery characteristic of Figure P11.18.2 can be shifted in a parallel manner to either lower or to higher voltages resulting in either an increase or decrease of the battery voltage V_BAT of Figure P11.18.1.

b) The slope of the battery characteristic can be changed by altering the voltages V_L1 and V_L2 by step-down, step-up DC converters as well. Draw these converters in the circuit as requested in Part a).

Problem 11.19: Design of a 200 MW Compressed-Air Storage Facility

Design a compressed-air energy storage (CAES) plant for P_{out_generator} = P_rated = 200 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 P11.19.1. 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: 570 ° F ≡ 524.40 K at ambient pressure 7.921 atm ≡ 116.40 psi ≡ 802.57 kPa.

Generation operation: 2 h at 200 MW.

Re-charging (loading) operation of underground air storage reservoir: 8 h at 40 MW.

Capacity of air storage reservoir: 7.0 · 10⁶ ft³ ≡ 195.2 · 10³ m³ ≡ (57.78 m x 57.78 m x 57.78 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: $3000.

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 estimated efficiency of η_compressor = 0.63. 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 40 MW (whereby wind energy is not free) calculate the overall efficiency of the CAES (during 1 hour of operation) ${\eta }_{\mathrm{overall}}^{\mathrm{wind}\mathrm{energy}\mathrm{is}\mathrm{not}\mathrm{free}}=\frac{P_{\mathrm{out}\_\mathrm{generator}}}{P_{\mathrm{in}\mathrm{combustor}}+P_{2\_\mathrm{compressors}+2\_\mathrm{coolers}}}$.

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

c) What is the overall construction cost of this CAES plant, if the construction price is $3000 per installed power capacity of 1 kW?

d) The CAES plant delivers the energy E = 400 MWh per day (during 2 hours) for which customers pay $0.27/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 $0.007, and the interest rate is 2%?

e) Repeat the analysis for a heat rate of 4000 BTU/kWh at a gas turbine output temperature of 424 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

Problem 11.20: Wind Power Plant of 8 MW Produces Hydrogen Based on Electrolysis and its Application to Electric Cars with Fuel Cells

A well-known way of splitting water (H₂O) is by electrolysis. A container (trough) contains pure (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 1 kcal: 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.

a) The energy of a P_out = 8 MW wind-power plant can be used to generate hydrogen during 8 hours of operation per day during 365 days per year. The construction cost of the 8 MW plant, and the generation of pure water including the electrolysis equipment is $4000/kW installed power capacity. How much hydrogen energy (expressed in MWh) can be obtained during one year taking into account the above-described losses?

b) Calculate the payback period provided one kg-force of hydrogen (weight) can be sold for $4.00. For your payback period calculation you may neglect interest for borrowing the money for the construction cost.

c) How many “equivalent-gasoline gallons” of hydrogen can be produced during one year?

d) Provided a car owner (on average) travels 15,000 miles per year with an average “mileage” of 40 miles/(gallon “equivalent gasoline”), how many car owners can get the “equivalent gasoline gallons of hydrogen” from the 8 MW plant?

Problem 11.21: Energy Efficiency of 80 kW PEM Fuel Cell as Used for Electric Automobiles

Pertinent data of a PEM fuel cell are as follows:

Output power: P_rat = 80,000 W

Output current: I_rat = 364 A

DC voltage range: V_rat = 220-500 V

Operating lifetime: T_life = 1,500 h

Hydrogen volume: V = 1,233 standard liters per minute (SLPM)

Weight of PEM fuel cell: W = 83 kg-force

Emission: 58 liters of water per hour

a) The energy efficiency is defined by η_energy = (rated electrical energy output)/ (hydrogen energy input). Calculate the energy efficiency of the fuel cell. Note that the nominal energy density of hydrogen is 28 kWh/(kg-force), significantly larger than that of gasoline. This makes hydrogen a desirable fuel for automobiles.

b) Find the specific energy density (energy per unit of weight) of this fuel cell expressed in Wh/(kg-force).

c) How does this specific energy density compare with that of a nickel-metal hydrideenergy density compare with that of a nickel-metal hydride battery?

Problem 11.22: Frequency/Load Control with Droop Characteristics of an Interconnected Power System consisting of Conventional and Renewable Plants and Broken into Two Areas each Having one Generator

Section 11.4.5 illustrates the instability of a two area network when both networks have identical droop characteristics and stability when both have different droop characteristics.

In this problem it will be shown that the stability also depends upon the time constants (τ_G1, τ_CH1, τ_G2, τ_CH2) of the two plants as shown in Figure P11.22.1 which shows the block diagram of two generators interconnected by a transmission tie line [49].

a) List the ordinary differential equations and the algebraic equations of the block diagram of Figure P11.22.1

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) =$\frac{0.8}{s}$ pu and run the program with zero step-load changes ΔP_L1 = 0, ΔP_L2 = 0 (for 10 s) in order to establish the equilibrium condition. After 10 s impose step-load changes $\mathrm{\Delta }{P}_{\mathrm{L}1}\left(\mathrm{s}\right)=\frac{\mathrm{\Delta }{P}_{\mathrm{L}1}}{s}\mathrm{p}\mathrm{u}=\frac{0.2}{s}$ pu, and after 30 s impose $\mathrm{\Delta }{P}_{\mathrm{L}2}\left(\mathrm{s}\right)=\frac{\mathrm{\Delta }{P}_{\mathrm{L}2}}{s}\mathrm{p}\mathrm{u}=\frac{-0.2}{s}$ pu to find the transient response Δω₁(t) = Δω₂(t) = Δω(t) for a total of 50 s. Repeat part b) for R₁ = 0.5 pu, (e.g., wind-power plant), and R₂ = 0.01 pu (e.g., coal-fired plant).

c) Compute the transient of the angular frequency Δω(t) for R1 = 0.5; 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 = 0.5; and Lr2 = 0.8.

d) Compute the transient of the angular frequency Δω(t) for
R1 = 0.5; d1 = 0.8; M1 = 4.5; Tg1 = 0.4; Tch1 = 0.5; Lr1 = 0.8; DPL1[t_]:=If[t < 10,0,0.2]; R2 = 0.02; d2 = 1.0; M2 = 6; Tg2 = 0.4; Tch2 = 0.5; and Lr2 = 0.8.

Problem 11.23: Complementary Generation of Long-Term and Short-Term Storage Facilities Together with Two Renewable Sources, where a Natural-Gas Fired Plant is the Frequency Leader

In Section 11.4.6 the complementary operational frequency/load behavior of a natural-gas fired plant as frequency leader and two long-term storage Plants together with two renewable sources is investigated.

The variation of parameters of a control circuit is important because it may result in instability. In this problem compute for block diagram of Figures E11.11.2a, b, c the frequency change Δω(t) for doubling all T_chj for j = 1 to 6 time constants, that is,T_ch1 = 1.80s to T_ch6 = 0.2 s. The Mathematica program of Table E11.11.1 can be modified to solve this stability/instability problem. Note each time constant occurs two times in this program.