SECTION B (Q4-Q6) Q5(a) Compare between distributed and centralised grids. [5] Q5(b) A homeowner is considering a rooftop installation of a solar PV system to offset a part of his grid energy cost. The total surface area of the rooftop is $$ 400 \mathrm{~m}^{2} $$. (i) Determine the system maximum installed capacity in kW if the only available [6] solar PV modules are 100W with dimensions of ( $$ 1.3 \mathrm{~m} imes 2.1 \mathrm{~m} $$ ). (ii) If the daily total solar irradiation for the site where the homeowner plans to [5] set up a solar PV system is $$ 4.5 \mathrm{kWh} / \mathbf{m}^{2} $$. Using this information and considering a conversion efficiency of $$ 21 \% $$, calculate the expected annual energy from the projected solar PV system. Q5(c) For a typical wind turbine: (i) What does the power coefficient, $$ C_{p} $$, indicate and what is its theoretical [5] maximum value? (ii) Explain why $$ C_{p} $$ is zero when $$ \lambda=0 $$ and $$ \lambda=12 $$ ? [4] Total [25]

Question

SECTION B (Q4-Q6) Q5(a) Compare between distributed and centralised grids. [5] Q5(b) A homeowner is considering a rooftop installation of a solar PV system to offset a part of his grid energy cost. The total surface area of the rooftop is $$ 400 \mathrm{~m}^{2} $$. (i) Determine the system maximum installed capacity in kW if the only available [6] solar PV modules are 100W with dimensions of ( $$ 1.3 \mathrm{~m} 	imes 2.1 \mathrm{~m} $$ ). (ii) If the daily total solar irradiation for the site where the homeowner plans to [5] set up a solar PV system is $$ 4.5 \mathrm{kWh} / \mathbf{m}^{2} $$. Using this information and considering a conversion efficiency of $$ 21 \% $$, calculate the expected annual energy from the projected solar PV system. Q5(c) For a typical wind turbine: (i) What does the power coefficient, $$ C_{p} $$, indicate and what is its theoretical [5] maximum value? (ii) Explain why $$ C_{p} $$ is zero when $$ \lambda=0 $$ and $$ \lambda=12 $$ ? [4] Total [25]

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### QUESTION 5(a)

**Compare between Distributed and Centralized Grids.**  
*[5 Marks]*

**Distributed Grids:**

1. **Generation Location:** Electricity is generated close to the point of consumption using multiple small-scale sources (e.g., rooftop solar panels, small wind turbines).
2. **Scalability:** Easily scalable by adding more local generation units as demand increases.
3. **Reliability:** Enhanced reliability and resilience since the failure of one source has minimal impact on the overall system.
4. **Transmission Losses:** Reduced transmission losses due to shorter distances between generation and consumption points.
5. **Environmental Impact:** Often utilizes renewable energy sources, leading to lower environmental footprints.

**Centralized Grids:**

1. **Generation Location:** Electricity is produced at large, centralized power plants (e.g., coal, nuclear, large hydroelectric) and transmitted over long distances to consumers.
2. **Economies of Scale:** Beneficial from economies of scale, typically resulting in lower cost per unit of electricity generated.
3. **Infrastructure:** Requires extensive transmission and distribution infrastructure to deliver electricity from central plants to end-users.
4. **Reliability:** Vulnerable to large-scale outages; a failure at a central plant can disrupt supply to a wide area.
5. **Environmental Impact:** Centralized plants, especially those using fossil fuels, can have significant environmental impacts, including higher greenhouse gas emissions.

**Comparison Summary:**

- **Flexibility:** Distributed grids offer greater flexibility and adaptability compared to centralized grids.
- **Cost:** Centralized grids may have lower generation costs due to scale but higher transmission costs, whereas distributed grids can reduce transmission losses but may have higher per-unit generation costs.
- **Resilience:** Distributed grids are generally more resilient to disruptions than centralized systems.

---

### QUESTION 5(b)

**A homeowner is considering a rooftop installation of a solar PV system to offset a part of his grid energy cost. The total surface area of the rooftop is $$ 400 \mathrm{~m}^{2} $$.**

#### (i) Determine the system maximum installed capacity in kW if the only available solar PV modules are 100 W with dimensions of $$ 1.3 \mathrm{~m} \times 2.1 \mathrm{~m} $$.
*[6 Marks]*

**Given:**
- **Total Rooftop Area:** 400 m²
- **Solar PV Module Rating:** 100 W
- **Module Dimensions:** 1.3 m × 2.1 m = 2.73 m² per module

**Calculation Steps:**

1. **Determine the number of modules that can fit on the rooftop:**
   $$
   \text{Number of Modules} = \frac{\text{Total Rooftop Area}}{\text{Area per Module}} = \frac{400\, \text{m}²}{2.73\, \text{m}²/\text{module}} \approx 146.45
   $$
   Since partial modules aren't feasible, the maximum number is **146 modules**.

2. **Calculate the total installed capacity:**
   $$
   \text{Total Capacity} = \text{Number of Modules} \times \text{Power per Module} = 146 \times 100\, \text{W} = 14,600\, \text{W} = 14.6\, \text{kW}
   $$

**Answer:**  
The maximum installed capacity is **14.6 kW**, achieved by installing 146 modules on the 400 m² rooftop.

#### (ii) If the daily total solar irradiation for the site where the homeowner plans to set up a solar PV system is $$ 4.5 \mathrm{kWh} / \mathrm{m}^{2} $$. Using this information and considering a conversion efficiency of $$ 21\% $$, calculate the expected annual energy from the projected solar PV system.
*[5 Marks]*

**Given:**
- **Daily Solar Irradiation:** 4.5 kWh/m²/day
- **Conversion Efficiency:** 21%
- **Total Rooftop Area:** 400 m²
- **Installed Capacity:** 14.6 kW (from part i)

**Approach:**
Calculate the energy produced by the entire rooftop area considering the solar irradiation and efficiency.

**Calculation Steps:**

1. **Total Daily Solar Energy Incident:**
   $$
   \text{Daily Solar Energy} = \text{Solar Irradiation} \times \text{Total Area} = 4.5\, \text{kWh/m}²/\text{day} \times 400\, \text{m}² = 1,800\, \text{kWh/day}
   $$

2. **Effective Energy Converted by PV System:**
   $$
   \text{Effective Daily Energy} = \text{Daily Solar Energy} \times \text{Efficiency} = 1,800\, \text{kWh/day} \times 0.21 = 378\, \text{kWh/day}
   $$

3. **Annual Energy Production:**
   $$
   \text{Annual Energy} = \text{Effective Daily Energy} \times 365\, \text{days} = 378\, \text{kWh/day} \times 365 = 137,970\, \text{kWh/year}
   $$

**Answer:**  
The expected annual energy from the solar PV system is **137,970 kWh**.

---

### QUESTION 5(c)

**For a typical wind turbine:**

#### (i) What does the power coefficient, $$ C_p $$, indicate and what is its theoretical maximum value?
*[5 Marks]*

**Power Coefficient ($$ C_p $$):**

- **Definition:** The power coefficient $$ C_p $$ represents the fraction of the kinetic energy in the wind that is converted into mechanical energy by the wind turbine.
- **Formula:**
  $$
  P = C_p \cdot \frac{1}{2} \rho A v^3
  $$
  where:
  - $$ P $$ = Power extracted
  - $$ \rho $$ = Air density
  - $$ A $$ = Swept area
  - $$ v $$ = Wind velocity

**Theoretical Maximum Value:**

- **Betz's Law:** Establishes that no wind turbine can capture more than 59.3% of the kinetic energy in wind.
- **Maximum $$ C_p $$:** $$ \frac{16}{27} \approx 0.593 $$ or **59.3%**

**Answer:**  
$$ C_p $$ indicates the efficiency with which a wind turbine converts the wind's kinetic energy into mechanical energy. Its theoretical maximum value, as defined by Betz's Law, is **59.3%**.

#### (ii) Explain why $$ C_p $$ is zero when $$ \lambda = 0 $$ and $$ \lambda = 12 $$.
*[4 Marks]*

**Definitions:**
- **Tip Speed Ratio ($$ \lambda $$):** The ratio of the tangential speed of the turbine blade tip to the wind speed.
  $$
  \lambda = \frac{\omega R}{v}
  $$
  where:
  - $$ \omega $$ = Angular velocity
  - $$ R $$ = Rotor radius
  - $$ v $$ = Wind speed

**Explanation:**

1. **When $$ \lambda = 0 $$:**
   - **Interpretation:** This implies that the rotor is not rotating ($$ \omega = 0 $$).
   - **Effect on $$ C_p $$:** Without rotation, the turbine does not extract any energy from the wind.
   - **Result:** $$ C_p = 0 $$

2. **When $$ \lambda = 12 $$:**
   - **Interpretation:** The rotor is rotating too fast relative to the wind speed.
   - **Effect on $$ C_p $$:** At very high tip speed ratios, the turbine blades cannot effectively capture the wind energy as the angle of attack becomes inefficient, leading to minimal energy extraction.
   - **Result:** $$ C_p = 0 $$

**Answer:**  
When $$ \lambda = 0 $$, the turbine is stationary and does not rotate, resulting in no energy extraction ($$ C_p = 0 $$). Conversely, when $$ \lambda = 12 $$, the turbine spins too rapidly relative to the wind speed, making energy capture inefficient, which also causes $$ C_p $$ to be zero.
### QUESTION 5(a) **Distributed vs. Centralized Grids** - **Distributed Grids:** - **Generation:** Local, using small sources like rooftop solar. - **Scalability:** Easily expandable. - **Reliability:** More resilient to failures. - **Transmission:** Less loss due to shorter distances. - **Environment:** Often uses renewables, reducing impact. - **Centralized Grids:** - **Generation:** Large-scale plants far from consumers. - **Economies of Scale:** Lower per-unit costs. - **Infrastructure:** Requires extensive transmission lines. - **Reliability:** Vulnerable to large outages. - **Environment:** Higher emissions from fossil fuels. **Summary:** Distributed grids offer flexibility and resilience, while centralized grids are more cost-effective but less resilient. --- ### QUESTION 5(b) **Solar PV System Installation** #### (i) Maximum Installed Capacity - **Modules:** 100 W each, 1.3 m × 2.1 m. - **Rooftop Area:** 400 m². - **Modules per Area:** 400 / (1.3 × 2.1) ≈ 146 modules. - **Total Capacity:** 146 × 100 W = 14.6 kW. **Answer:** Maximum installed capacity is **14.6 kW**. #### (ii) Annual Energy Production - **Daily Solar Irradiation:** 4.5 kWh/m²/day. - **Conversion Efficiency:** 21%. - **Annual Energy:** 4.5 × 400 × 21% × 365 ≈ 137,970 kWh/year. **Answer:** Expected annual energy is **137,970 kWh**. --- ### QUESTION 5(c) **Wind Turbine Power Coefficient ($$ C_p $$)** #### (i) Definition and Maximum Value - **$$ C_p $$:** Fraction of wind energy converted to mechanical energy. - **Theoretical Maximum:** 59.3% (as per Betz's Law). **Answer:** $$ C_p $$ measures the efficiency of energy conversion, with a maximum of **59.3%**. #### (ii) $$ C_p = 0 $$ at $$ \lambda = 0 $$ and $$ \lambda = 12 $$ - **$$ \lambda = 0 $$:** Turbine not rotating, no energy extraction. - **$$ \lambda = 12 $$:** Turbine spins too fast, inefficient energy capture. **Answer:** $$ C_p $$ is zero when the turbine is not rotating or spinning too fast, making energy extraction inefficient.

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