Problem 125 - Chemistry An Atoms First Approach

Question

The best solar panels currently available are about \(15 \%\) efficient in converting sunlight to electricity. A typical home will use about \(40 .\) kWh of electricity per day \((1 \mathrm{kWh}=1\) kilowatt hour; \(1 \mathrm{kW}=1000 \mathrm{J} / \mathrm{s}\) ). Assuming 8.0 hours of useful sunlight per day, calculate the minimum solar panel surface area necessary to provide all of a typical home's electricity. (See Exercise 124 for the energy rate supplied by the sun.)

Step-by-Step Solution

Answer

The minimum solar panel surface area necessary to provide all of a typical home's electricity is \(33.33 \, m^2\).

Step 1: Calculate the total energy required per day.

To find the energy consumption per day in Joules, we first need to convert the energy consumption from kilowatt-hours (kWh) to Joules. We'll use the following conversion factors: 1 kWh = 1 kW * 1 hour, and 1 kW = 1000 J/s. Total energy required per day (in Joules) = 40 kWh/day * 1 kW / 1 kWh * 3600 s/hour Total energy required per day = 144,000 kJ/day

Step 2: Calculate the energy provided by sunlight per day.

Next, we need to find out how much energy the sun provides per day. To do this, we use the given energy rate of 1.00 kW/m² (1.00 kJ/s/m²), multiplied by the number of seconds of useful sunlight per day (8 hours * 3600 seconds/hour). Energy provided by sunlight per day (per m²) = 1.00 kJ/s/m² * 8 hours/day * 3600 s/hour Energy provided by sunlight per day (per m²) = 28,800 kJ/day/m²

Step 3: Calculate the energy output of the solar panels per day.

Now, we need to calculate how much energy the solar panels can generate per day, considering their efficiency. The efficiency is given as 15%, which means that 15% of the sunlight energy is converted into electricity. Energy output from the solar panels per day (per m²) = energy provided by sunlight per day (per m²) * efficiency Energy output from the solar panels per day (per m²) = 28,800 kJ/day/m² * 0.15 Energy output from the solar panels per day (per m²) = 4,320 kJ/day/m²

Step 4: Calculate the minimum solar panel surface area required.

Finally, we need to find the minimum surface area of solar panels required to provide all the energy needed for a typical home. To do this, we divide the total energy required per day by the energy output from the solar panels per day (per m²). Minimum solar panel surface area required = Total energy required per day / Energy output from the solar panels per day (per m²) Minimum solar panel surface area required = 144,000 kJ/day / 4,320 kJ/day/m² Minimum solar panel surface area required = 33.33 m² Thus, the minimum solar panel surface area necessary to provide all of a typical home's electricity is 33.33 m².

Key Concepts

Efficiency of Solar Panels

Efficiency in solar panels is a measure of how much sunlight that strikes a panel is converted into usable electricity. Current top-of-the-line solar panels typically reach about 15% efficiency. This means that out of all the solar energy striking the panel, only 15% is turned into electrical energy that you can use around the house.

This efficiency can be influenced by several factors:

• Quality of materials: Higher quality materials tend to be more efficient.
• Angle and position: Panels installed at the optimal angle according to the sun’s path typically produce more electricity.
• Temperature: Most panels perform better in cooler conditions than in hot environments.

Efficiency is a fundamental concept, because even a small increase can significantly decrease the surface area required for powering a household. It's crucial to understand that not all sunlight's energy can be turned into electricity due to physical limits and technical constraints.

Energy Conversion

Energy conversion is what occurs when solar panels transform sunlight into electricity. The process is executed through photovoltaic (PV) cells that are embedded within the solar panel. Here's how it works step-by-step:

• Photon absorption: When sunlight, which is composed of photons, hits the solar panel, these photons are absorbed by the PV cells.
• Electron excitation: This energy within the photons excites electrons in the PV cells, freeing them to flow as an electric current.
• Electrical output: This current is captured and transferred as electrical energy.

This conversion process turns sunlight into direct current (DC). An inverter then changes this into alternating current (AC), which is used by most household appliances. A notable aspect here is how efficiency ties into conversion; not all the solar energy striking a panel will be captured and converted by the PV cells due to inherent energy losses at each conversion stage.

Surface Area Calculation

Calculating the surface area of solar panels necessary to meet energy needs starts with understanding how much energy is consumed by the household and how much energy a single square meter of solar panels can produce. In this problem, a home consumes 144,000 kJ of electricity daily.

Using the provided efficiency and sunlight energy data, we determine the energy output of the solar panels per day per square meter:

• Total energy provided by the sun per day per m² = 28,800 kJ
• Efficiency of solar panels = 15%
• Energy from solar panels per day per m² = 4,320 kJ

The formula for calculating the minimum surface area required is:\[\text{Minimum Surface Area} = \frac{\text{Total Energy Required per Day}}{\text{Energy Output per m² per Day}}\]Substituting the values, you find:\[\text{Minimum Surface Area} = \frac{144,000 \text{ kJ/day}}{4,320 \text{ kJ/day/m²}} = 33.33 \text{ m²}\]This calculation shows that to supply all of a home's energy through solar power, at least 33.33 m² of solar panels at 15% efficiency are needed. Proper calculations ensure that all power is met without underestimating or overestimating the resources required.