Question

In 2008 , the global average electricity consumption per head was \(3.0 \mathrm{MWh}\). The solar power striking Earth every day averages 168 watts per square meter. Considering that present technology for solar energy conversion is about \(10 \%\) efficient, from how many square meters of land must sunlight be collected in order to provide this power?

Short answer

First, we find the total energy consumption per head in kWh: \(3000 kWh\). Then, we convert it to energy per day in watts: \(\frac{3000 * 1000 \text{ Wh}}{86,400 \text{ seconds/day}}\). Next, we divide this by the \(0.1\) efficiency to find the required solar power. Finally, we divide the required solar power by the average solar power per square meter (\(168 \text{ W/m²}\)) to find the required area of land.

Step-by-step solution

Calculate total energy consumption per head

First, let's determine the total energy consumption per head in kWh. We are given that the average electricity consumption per head is \(3.0 MWh\), which can be converted to kWh by multiplying by 1000 (since 1 MWh = 1000 kWh): Total energy consumption per head (kWh) = \(3.0 MWh * 1000 = 3000 kWh\)

Convert energy consumption to energy per day

Next, we need to calculate the energy consumption per day, as the solar power is given in watts (energy per second). Since there are \(24 * 60 * 60 = 86,400\) seconds in a day, we can divide the total energy consumption by the number of seconds in a day to get the energy consumption per second (watts): Energy consumption per day (kWh/day) = \(\frac{3000 kWh}{365 \text{ days}}\) Energy consumption per day (W) = \(\frac{3000 * 1000 \text{ Wh}}{86,400 \text{ seconds/day}}\)

Find the required solar power

We are given that the efficiency of the solar energy conversion is \(10\%\) or \(0.1\). To find the total solar power required, we can divide the energy consumption per day (W) by the efficiency: Required solar power (W) = \(\frac{\text{Energy consumption per day (W)}}{0.1}\)

Calculate the required area of land

Finally, we are given that the average solar power striking Earth is 168 watts per square meter. We can now find the area of land required to collect this power by dividing the required solar power (W) by the average solar power per square meter: Required area of land (m²) = \(\frac{\text{Required solar power (W)}}{168 \text{ W/m²}}\) Now that we have the required area of land, we will substitute all values into the expressions to find the answer.

Key concepts

Electricity Consumption

Electricity consumption per capita is a measure of how much electrical energy the average person uses within a specific timeframe. In the exercise, the global average electricity consumption in 2008 was provided as 3.0 megawatt-hours (MWh) per person per year. To make sense of this measurement in daily terms, it was necessary to convert MWh into kilowatt-hours (kWh), which are more commonly used in everyday energy calculations.

Understanding how to translate annual consumption into daily and even hourly usage allows us to better relate to the figures, offering a clearer picture of our personal impact on energy demands and resource use. In our case, converting from a yearly total to a per-second rate is essential for comparing with solar power potential. This step helps us break down large numbers and conceptualize daily electricity use for individuals across the globe.

Solar Power Efficiency

Solar power efficiency refers to how effectively solar energy can be converted into usable electricity. Modern solar cells can convert approximately 10% of the sunlight that hits them into electrical energy. It's a crucial factor because not all the sunlight hitting the solar panels gets turned into electricity.

• Efficiency, which is represented as a decimal (e.g., 0.1 for 10%), directly influences how much surface area is necessary to capture a desired amount of energy.
• Efficiency improvements mean that less space would be needed for the same energy output.

In the exercise, this 10% efficiency means that for every 168 watts of solar power hitting a square meter of the Earth, only 16.8 watts is turned into electricity. This realization helps us understand why advancements in solar technology are critical—the more efficient, the better use of space and resources.

Energy Measurement Conversion

Converting energy measurements is pivotal in aligning various units to better comprehend real-world applications. In everyday calculations, we find ourselves switching between different energy units such as joules, watts, and kilowatt-hours.

• One megawatt-hour (MWh) equals 1,000 kilowatt-hours (kWh).
• Further conversions help align the annual personal electricity consumption with solar power potential calculated in watts.

The conversion from kWh to Wh supports the transition from yearly or daily energy figures to the necessary calculations in terms of watts, which solar power data is typically presented in. This ability to shift between units empowers us to interrelate energy consumption with production capabilities, making solar energy calculations accessible and practical.

Land Area Calculation

Calculating the land area needed for harvesting solar power involves dividing the total required solar power by the average solar power received per square meter, given the efficiency of conversion.

This step is critical to determine how much space would be needed to supply energy for one person as defined in the given problem.

• We know that the average solar energy available is 168 watts per square meter.
• Given this and an efficiency rate, we calculate the required solar output to support individual energy needs.

By adjusting these calculations, we can tailor our understanding of space and energy relation for various locales and property sizes. This knowledge is essential for site planning and for understanding how much land would be required to transition away from conventional power sources toward more sustainable solutions like solar energy.