Openai/6920e3bb-4978-800c-8ec1-0136167c6c1a

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gpt-5-mini

User: Calculate the size of solar panel that would be required to power a 100 W pedestrian light system in Ireland.

Calculate the size of solar panel that would be required to power a 100 W pedestrian light system in Ireland. It would have to operate 24/7. Use figures for solar irradiance in Ireland from Met Eireann and figures for panel efficiency

Assistant: Good question.

Good question. To size a solar array to power a 100 W pedestrian light system 24/7 in Ireland, we need to make a number of assumptions, then estimate the total energy demand, compare to typical Irish solar resource, and account for inefficiencies. Here’s a worked‑through calculation.

1. Estimate the energy demand

• A 100 W (watts) light means 0.1 kW.
• Running 24 hours/day, the energy use per day = 0.1 kW × 24 h = 2.4 kWh/day.
• Per year: 2.4 kWh/day × 365 = 876 kWh/year.

2. Estimate the solar resource in Ireland

We need an estimate of how much solar energy you can harvest per square metre in Ireland.

Some relevant figures:

1. According to Carbon Markets, Ireland receives about 900–1,200 kWh/m²/year of solar irradiance. Carbon Markets
2. SEAI-related documentation (from a local energy plan) uses a value of ~962 kWh/m²/year for annual solar insolation. Sustainable Energy Authority of Ireland
3. Met Éireann gives sunshine duration, e.g., average ~3¼ to ~3¾ hours of sunshine per day over the year. Met Éireann
4. Typical panel efficiency in Ireland: modern solar panels in Ireland are often in the ~14%–22% efficiency range. Going Solar

3. Calculate how much electricity a given area of panel can generate per year

Pick a representative solar irradiance and panel efficiency. Let’s choose:

• Solar resource (insolation) = 1,000 kWh/m²/year (in the mid-range of 900–1,200).
• Panel efficiency = 20% (a reasonable modern panel).

Energy hitting 1 m² panel per year = 1,000 kWh/m² × 0.20 = 200 kWh/m²/year (theoretical electrical output, before losses).

But real-world systems always have losses (inverter, wiring, shade, dirt, mismatch). A typical derating might be ~ 20% (this is just an assumption; real derating depends on setup). So:

• Derated output ~ 200 × (1 – 0.20) = 160 kWh/m²/year.

4. Calculate the required area of solar panels

We need to generate 876 kWh/year from the panels.

If each square metre gives ~160 kWh/year, then:

Required panel area=876 kWh/year160 kWh/m2 ⁣/year≈5.48 m2\text{Required panel area} = \frac{876\ \text{kWh/year}}{160\ \text{kWh/m}^2\!/year} \approx 5.48\ \text{m}^2Required panel area=160 kWh/m2/year876 kWh/year​≈5.48 m2

5. Convert area to a more practical panel size / capacity

If you have panels that are, say, 400 W each, typical panel area might be ~2 m² (varies by panel). Let’s do a rough check:

• 400 W panel, assuming standard test conditions, means 0.4 kW peak (kWp).
• If you need ~876 kWh/year, how many kWp do you need?
• Using a rule of thumb: in Ireland, 1 kWp produces ~900 kWh/year (varies) according to some sources. Sustainable Energy Authority of Ireland
• So required kWp = 876 / 900 ≈ 0.974 kWp, or about 1 kWp.

If using 400 W panels, you’d need:

1 000 W400 W/panel=2.5 panels.\frac{1\,000 \text{W}}{400 \text{W/panel}} = 2.5\ \text{panels}.400 W/panel1000 W​=2.5 panels. So, practically, 3 × 400 W panels (or equivalent) would cover it, giving some margin.

6. Consideration of storing / smoothing (batteries) or grid tie

• Since this is a 24/7 load, you’ll need storage (battery) or be grid-tied (if allowed) to supply power at night.
• The battery size will depend on how many nights of “autonomy” you want, and how much of the solar energy you can divert to charging.

7. Sensitivity / Margin

A few caveats & margins to include:

• In winter, solar irradiance is much lower (days are shorter, more cloud) → your output will be seasonal.
• Panel orientation, tilt, and shading will significantly affect real output.
• Panel degradation over time.
• Losses might be more than 20% depending on system design.

Given these, it's wise to over‑specify slightly (e.g., design for ~1.1 kWp rather than exactly 1 kWp) if budget allows, especially for a small, constant-load system like a light that must run 24/7.

Conclusion

• You need about 5.5 m² of solar panels (at ~20% efficiency) to generate ~876 kWh/year, which powers a 100 W light running 24 h/day.
• That corresponds to roughly 1 kWp of installed PV capacity in Irish conditions.
• Using typical high-power panels (~400 W), you'd likely install 3 panels.

If you like, I can run a more detailed simulation (with monthly irradiance, storage sizing, worst-case winter months) — do you want me to do that?