Egypt Engineering Consultant Electrical study materials
Egypt Engineering Consultant Electrical study resources

Egypt Engineering Consultant Electrical: Renewable Energy and Energy Efficiency

Egypt Engineering Consultant Electrical · Work through your exam, one topic at a time.

All guide sections

Renewable Energy and Energy Efficiency

What this domain tests

Renewable Energy and Energy Efficiency tests PV sizing, inverter selection, performance ratio, power-factor correction, motor efficiency, and Egyptian interconnection rules. Egypt’s renewable-energy framework includes Law 203/2014, NREA projects, net metering, and competitive tenders, while efficiency work focuses on motors, transformers, lighting, and reactive compensation.

Core rules and concepts

  • PV energy yield: E = kWp × GHI × PR, where GHI is global horizontal irradiance in kWh/m²/day and PR is performance ratio, typically 0.75–0.85 after temperature, soiling, mismatch, and inverter losses.
  • Inverter DC/AC ratio is often 1.1–1.3. Oversizing the array increases yield but may clip at peak irradiance; undersizing wastes inverter capacity.
  • Power-factor correction: Qc = P × (tan φ1 − tan φ2). Correcting to 0.95 avoids utility penalties and frees transformer capacity.
  • Use high-efficiency motors (IE3 or better), variable-speed drives for variable-torque loads, and proper cable sizing to reduce losses.
  • Net metering and feed-in arrangements require utility approval, bidirectional metering, protection settings, and compliance with EgyptERA rules.

Worked example

A 20 kWp PV array receives 5.8 kWh/m²/day and has PR = 0.82. Daily energy = 20 × 5.8 × 0.82 = 95.1 kWh/day. Annual energy is about 34.7 MWh before availability losses. If the inverter is rated 17 kW AC, the DC/AC ratio is 1.18, which is reasonable but will clip occasionally at peak irradiance. For a 100 kW load at pf 0.70 corrected to 0.95: tan φ1 = 1.020, tan φ2 = 0.329, so Qc = 100 × (1.020 − 0.329) = 69 kVAr. Select standard capacitor steps and check harmonics before installation.

Traps that cost marks

Treating kWp as actual AC output; using irradiance in W/m² without converting to kWh/m²/day; ignoring temperature derating and soiling; correcting power factor without checking resonance or harmonics; and assuming net metering is automatic without utility approval. Also, a high-efficiency motor saves energy only if it is correctly sized and controlled.

Exam-focused summary

Show the energy or reactive-power equation, state the assumptions, and confirm the interconnection or power-quality check.