17.2 Renewable Energy Systems & Geothermal Power Plants

17.2 Renewable Energy Systems & Geothermal Power Plants

With renewable energy expansion accelerating under Republic Act 9513 (the Philippines Renewable Energy Act of 2008), renewable energy conversion systems form a substantial part of the modern PRC REE Board Examination. This section presents a technical examination of geothermal power generation, solar photovoltaic (PV) engineering, wind turbine aerodynamics, and battery energy storage systems (BESS).

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1. Geothermal Energy Conversion Systems

The Philippines ranks among the world's top producers of geothermal power (with major fields in MakBan, Tiwi, Leyte, and Palinpinon). Geothermal energy extracts thermal energy stored in sub-surface rock and water reservoirs to generate electricity.

Geothermal Conversion Cycles

                      GEOTHERMAL CONVERSION CYCLE SCHEMATICS

  1. DRY STEAM PLANT:      Reservoir Steam ===> Turbine ===> Condenser ===> Reinjection

  2. SINGLE-FLASH PLANT:   High-P Hot Water ==> Flash Tank ==[Steam]==> Turbine
                                                    || [Brine]
                                                    v
                                             Reinjection Well

  3. BINARY CYCLE (ORC):   Geo Fluid ===> Heat Exchanger ===> Reinjection Well
                                               | (Isobutane Loop)
                                               v
                                         Organic Turbine ===> Condenser

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2. Solar Photovoltaic (PV) Engineering & Performance

Solar photovoltaic systems convert solar irradiance directly into direct current (DC) electricity via the photovoltaic effect in semiconductor p-n junctions.

Electrical Characteristics of Solar PV Cells

A solar cell's non-linear $I\text{--}V$ characteristic curve under solar irradiance $G$ (in $\text{W/m}^2$) is governed by the single-diode model equation:

$$I = I_L - I_0 \left[ \exp\left(\frac{q (V + I R_s)}{n k T}\right) - 1 \right] - \frac{V + I R_s}{R_{sh}}$$

Key Standard Test Condition (STC: $1000\ \text{W/m}^2$, cell temperature $25^\circ\text{C}$, Air Mass 1.5 spectrum) parameters are:

• $V_{\text{oc}}$ (Open-Circuit Voltage): Voltage across terminals when $I = 0$.
• $I_{\text{sc}}$ (Short-Circuit Current): Current flowing when terminals are shorted ($V = 0$). Directly proportional to solar irradiance $G$.
• $P_{\text{mp}}$ (Maximum Power Point): Peak power operating point where the product $V \times I$ is maximized ($P_{\text{mp}} = V_{\text{mp}} \times I_{\text{mp}}$).

Fill Factor (FF) & Module Efficiency ($\eta$)

The Fill Factor (FF) quantifies the squareness of the $I\text{--}V$ curve:

$$\text{FF} = \frac{P_{\text{mp}}}{V_{\text{oc}} \cdot I_{\text{sc}}} = \frac{V_{\text{mp}} \cdot I_{\text{mp}}}{V_{\text{oc}} \cdot I_{\text{sc}}}$$

Typical silicon PV modules have Fill Factors between $0.70$ and $0.85$ ($70\%\text{--}85\%$).

Module conversion efficiency $\eta_{\text{pv}}$ under STC irradiance $G = 1000\ \text{W/m}^2$ for gross surface area $A$ (in $\text{m}^2$) is:

$$\eta_{\text{pv}} = \frac{P_{\text{mp}}}{G \cdot A} \times 100\% = \frac{V_{\text{oc}} \cdot I_{\text{sc}} \cdot \text{FF}}{G \cdot A} \times 100\%$$

Maximum Power Point Tracking (MPPT) & Inverters

Because solar irradiance and ambient temperatures fluctuate continuously, grid-tied PV inverters employ MPPT algorithms (such as Perturb & Observe or Incremental Conductance) to dynamically adjust DC terminal operating voltage $V_{\text{mp}}$, extracting peak available DC power before converting it into $60\ \text{Hz}$ AC electricity.

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3. Wind Energy Conversion Systems (WECS)

Wind turbines extract kinetic energy from moving air masses and convert it into mechanical torque, which drives an electric generator.

Available Power in Wind & Betz's Law

The kinetic energy of air mass $m$ moving at velocity $v$ is $E_k = \frac{1}{2} m v^2$. The mass flow rate through a rotor swept area $A = \pi R^2$ (where $R$ is blade length in meters) is $\dot{m} = \rho A v$. Thus, total theoretical power $P_{\text{wind}}$ available in an unperturbed wind stream is:

$$P_{\text{wind}} = \frac{1}{2} \rho A v^3 \quad (\text{Watts, W})$$

where:

• $\rho$ = air density ($1.225\ \text{kg/m}^3$ at standard sea-level conditions)
• $A$ = rotor swept area = $\pi R^2 = \frac{\pi D^2}{4}$ ($\text{m}^2$)
• $v$ = free-stream wind speed (m/s)

Betz Limit ($C_p$ Max)

According to Betz's Law, fluid momentum mechanics dictates that a wind turbine rotor cannot extract more than $\frac{16}{27}$ ($59.26\%$) of the kinetic energy in the wind stream. The ratio of extracted shaft power to total wind power is the Power Coefficient ($C_p$):

$$C_p = \frac{P_{\text{rotor}}}{P_{\text{wind}}} \le C_{p,\text{Betz}} = \frac{16}{27} \approx 0.5926$$

Accounting for aerodynamic drag, tip-loss, gearbox efficiency $\eta_m$, and generator efficiency $\eta_g$, practical wind turbine electrical power output is:

$$P_e = \frac{1}{2} \rho A v^3 \cdot C_p \cdot \eta_m \cdot \eta_g$$

                  WIND TURBINE POWER CURVE REGIONS

   Power Output (kW)
      ^                                  Cut-Out Speed (e.g. 25 m/s)
      |                       Rated Power | (Turbine Feathered/Stops)
   P_rated +---------------+--------------|
      |                   /               | 
      |                  /                | 
      |                 / (Region 2)      | 
      |  (Region 1)    /                  | 
    0 +---------------o-------------------|------------------> Wind Speed (m/s)
                    Cut-In Speed
                    (e.g. 3-4 m/s)

Wind Turbine Operational Regions

• Cut-In Wind Speed ($v_{\text{in}} \approx 3\text{--}4\ \text{m/s}$): Minimum wind speed at which the turbine begins producing net electrical power.
• Rated Wind Speed ($v_{\text{rated}} \approx 11\text{--}14\ \text{m/s}$): Wind speed at which the turbine reaches its full rated electrical generator capacity ($P_{\text{rated}}$).
• Cut-Out Wind Speed ($v_{\text{out}} \approx 25\ \text{m/s}$): Maximum safe operating wind speed. Blades are feathered (pitched out of wind) and mechanical brakes are applied to protect structural components from storm damage.

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4. Battery Energy Storage Systems (BESS)

Grid-scale Battery Energy Storage Systems (BESS) play an essential role in stabilizing utility grids with high penetrations of intermittent solar and wind generation.

• Primary Applications: Frequency Regulation (fast primary response within milliseconds), Ramp Rate Control, Peak Shaving, and Load Shifting.
• Round-Trip Efficiency (RTE): BESS efficiency measures net energy delivered during discharge relative to energy absorbed during charging: $$\text{RTE} = \frac{E_{\text{discharge}}}{E_{\text{charge}}} \times 100\%$$ Lithium-ion BESS facilities achieve high round-trip efficiencies of $85\%$ to $92\%$.

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Solved Board Exam Examples

Example 1: Solar PV Array Sizing & Peak Output

Problem: A commercial solar PV system consists of 120 crystalline silicon modules. Each module has STC parameters: $V_{\text{oc}} = 45.0\ \text{V}$, $I_{\text{sc}} = 10.2\ \text{A}$, $V_{\text{mp}} = 37.5\ \text{V}$, and $I_{\text{mp}} = 9.6\ \text{A}$. Each module has dimensions $1.7\ \text{m} \times 1.0\ \text{m}$. Calculate: (a) Fill Factor (FF) of a single module, (b) STC maximum peak power rating per module, (c) Total array STC peak power output in $\text{kW}_p$, and (d) Module conversion efficiency $\eta_{\text{pv}}$.

Solution:

1. Calculate module peak power output $P_{\text{mp}}$: $$P_{\text{mp}} = V_{\text{mp}} \times I_{\text{mp}} = 37.5\ \text{V} \times 9.6\ \text{A} = 360\ \text{W}_p$$
2. Calculate Fill Factor (FF): $$\text{FF} = \frac{P_{\text{mp}}}{V_{\text{oc}} \cdot I_{\text{sc}}} = \frac{360}{45.0 \times 10.2} = \frac{360}{459} = 0.7843 \quad (78.43\%)$$
3. Compute total array peak output power: $$P_{\text{array}} = 120 \times 360\ \text{W}_p = 43,200\ \text{W}_p = 43.2\ \text{kW}_p$$
4. Calculate module conversion efficiency under $G = 1000\ \text{W/m}^2$ for module area $A = 1.7\ \text{m}^2$: $$\eta_{\text{pv}} = \frac{P_{\text{mp}}}{G \cdot A} \times 100\% = \frac{360\ \text{W}}{1000\ \text{W/m}^2 \times 1.7\ \text{m}^2} \times 100\% = \frac{360}{1700} \times 100\% = 21.18\%$$

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Example 2: Wind Turbine Electrical Power Output

Problem: A utility-scale wind turbine features a rotor diameter of $90\ \text{m}$ operating in standard air density $\rho = 1.225\ \text{kg/m}^3$ under a steady wind speed of $12\ \text{m/s}$. The aerodynamic power coefficient is $C_p = 0.44$, gearbox efficiency is $96\%$, and generator efficiency is $95\%$. Determine: (a) Total wind power in the swept air stream, and (b) Net electrical power output in kW.

Solution:

1. Calculate swept rotor area $A$: $$A = \frac{\pi D^2}{4} = \frac{\pi \times (90)^2}{4} = 6361.73\ \text{m}^2$$
2. Compute total available power in unperturbed wind stream $P_{\text{wind}}$: $$P_{\text{wind}} = \frac{1}{2} \rho A v^3 = 0.5 \times 1.225 \times 6361.73 \times (12)^3$$ $$P_{\text{wind}} = 0.5 \times 1.225 \times 6361.73 \times 1728 = 6,733,359\ \text{W} = 6733.36\ \text{kW}$$
3. Calculate net electrical power output $P_e$: $$P_e = P_{\text{wind}} \times C_p \times \eta_m \times \eta_g = 6733.36\ \text{kW} \times 0.44 \times 0.96 \times 0.95$$ $$P_e = 6733.36 \times 0.40128 = 2701.96\ \text{kW} \approx 2.70\ \text{MW}$$