Problem 87 - Chemistry The Central Science In Si Units

Question

The total rate at which power is used by humans worldwide is approximately 15 TW (terawatts). The solar flux averaged over the sunlit half of Earth is \(680 \mathrm{~W} / \mathrm{m}^{2}\) (assuming no clouds). The area of Earth's disc as seen from the Sun is \(1.28 \times 10^{14} \mathrm{~m}^{2}\). The surface area of Earth is approximately 197,000,000 square miles. How much of Earth's surface would we need to cover with solar energy collectors to power the planet for use by all humans? Assume that the solar energy collectors can convert only \(10 \%\) of the available sunlight into useful power.

Step-by-Step Solution

Answer

We need to cover approximately 41,279 square miles with solar collectors.

Step 1: Calculate Incident Solar Power

First, calculate the total power incident on the Earth from the Sun. This can be found by multiplying the solar flux by the area of Earth's disc as seen from the Sun:\[ P_{incident} = 680 \, \text{W/m}^2 \times 1.28 \times 10^{14} \, \text{m}^2 \]This gives the total solar power incident on Earth.

Step 2: Determine Solar Power Converted to Electricity

Since the solar energy collectors convert only 10% of the incident solar energy into useful power, calculate the power that can be converted using the formula:\[ P_{convertible} = 0.10 \times P_{incident} \]This is the power that could be realistically converted to electricity by the collectors.

Step 3: Calculate Required Collector Area

Determine how much area needs to be covered with collectors to meet the world's power demand of 15 TW. Rearrange the formula for power to solve for the area:\[ \text{Collector Area} = \frac{15 \, \text{TW}}{P_{convertible}} \]This is the area required in square meters to meet the power needs of humans.

Step 4: Convert Collector Area to Square Miles

Convert the area in square meters to square miles, knowing that \(1 \, \text{mile}^2 \approx 2.59 \times 10^6 \, \text{m}^2 \). Use the formula:\[ \text{Area in miles}^2 = \frac{\text{Collector Area}}{2.59 \times 10^6} \]This gives the required area in square miles to be covered by solar collectors.

Key Concepts

Energy Conversion Efficiency

The concept of energy conversion efficiency is crucial in understanding how solar energy collectors work. Efficiency in this context refers to the ability of a solar energy system to convert incident sunlight into usable electrical energy. For solar panels, this is typically expressed as a percentage.
In the exercise at hand, the efficiency is given as 10%, which means that only 10% of the sunlight reaching the surface of the solar collectors is transformed into electricity.
For example:

• If a solar panel receives 1000 watts of sunlight, it would convert 100 watts into electricity given a 10% efficiency.

Several factors affect this efficiency rate:

• Material Type: Different types of photovoltaic materials (like silicon or thin-film technology) have varying efficiencies.
• Temperature: As temperature increases, efficiency typically decreases.
• Angle and Position: The angle at which sunlight strikes the panel can significantly change its effective power conversion.
• Shading and Dirt: Any obstruction on the panel, such as leaves or dust, reduces efficiency.

Maximizing efficiency often involves technological advances and optimizing placement and maintenance. Understanding these elements can help design more effective solar energy systems.

Solar Power Calculation

Calculating solar power involves determining how much sunlight can be potentially captured and used. In the given exercise, the calculation starts with identifying the solar flux, which represents the power per unit area received from the Sun.
Firstly, the incident solar power on Earth is determined by multiplying the solar flux (\(680 \, \text{W/m}^2\)) with the area of Earth's disc as seen from the Sun (\(1.28 \times 10^{14} \, \text{m}^2\)). This calculation provides the total solar power reaching Earth:

• \[ P_{incident} = 680 \, \text{W/m}^2 \times 1.28 \times 10^{14} \, \text{m}^2 = 8.704 \times 10^{16} \, \text{W} \]

After finding the incident power, we then focus on how much of this power becomes usable electricity, given the panel's efficiency. The exercise simplifies this by offering a standard 10% conversion efficiency. Hence:

• \[ P_{convertible} = 0.10 \times 8.704 \times 10^{16} \, \text{W} = 8.704 \times 10^{15} \, \text{W} \]

This step showcases the practical conversion potential of solar systems under current technology constraints. Even though a gigantic amount of energy hits the Earth, only a fraction is converted into usable power, highlighting the importance of improving solar technologies.

Global Energy Demand

Understanding global energy demand is key to realizing the potential and limitations of solar power installations. The current exercise assumes a global energy use rate of 15 terawatts (TW). This total represents all the power consumed by human activities across the world.
Meeting this demand with solar power necessitates a large-scale deployment of solar collectors. Once the potential convertible solar power is calculated, the next step is to figure out what surface area of Earth needs to be covered by solar panels.
The required collector area in square meters is calculated using:

• \[ \text{Collector Area} = \frac{15 \, \text{TW}}{8.704 \times 10^{15} \, \text{W}} \]

Converting this number into square miles for easier interpretation provides a concrete visualization of solar energy infrastructure needs.
This exercise underscores the massive scale required for solar panels to satisfy global power demands. It also emphasizes the broader theme of integrating various renewable energy sources and advancing technology to bridge the gap between current capabilities and future energy needs.