Question

The best solar panels currently available are about 19\(\%\) efficient in converting sunlight to electricity. A typical home will use about \(40 .\) kWh of electricity per day \((1 \mathrm{kWh}=1 \text { kilowatt }\) 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 132 for the energy rate supplied by the sun.)

Short answer

The minimum solar panel surface area needed to provide all of a typical home's electricity is approximately \(18.93 m^2\).

Step-by-step solution

Find the total energy needed per day

First, we need to calculate the total energy in kWh that a typical home requires per day. We have been given that a home uses 40 kWh of electricity per day. Total energy needed per day = 40 kWh

Convert the energy from kWh to Joules

We need to convert the total energy needed per day from kWh to Joules to match the units in Exercise 132, which provide the energy rate supplied by the sun. We know that 1kWh = 3.6*10^6 J. Total energy needed per day in Joules = 40 kWh × 3.6 × 10^6 J/kWh = 1.44 × 10^8 J

Calculate the energy required per second

Next, we need to calculate how much energy is required per second, since the energy rate supplied by the sun is in watts (Joules/second). We have been given that there are 8 hours of useful sunlight per day, so we will divide the total energy needed per day by the number of seconds in 8 hours. Seconds in 8 hours = 8 hours × 3600 seconds/hour = 28800 seconds Energy required per second = (1.44 × 10^8 J) / (28800 seconds) = 5 × 10^3 J/s or 5 kW

Find the energy rate provided by the sun

We need to find the energy rate supplied by the sun, which is given in Exercise 132. In that exercise, we are told that the sun provides an energy rate of 1390 W/m^2. Energy rate provided by the sun = 1390 W/m^2

Calculate the solar panel energy conversion

We have been given that the best solar panels currently available are 19% efficient in converting sunlight to electricity. Therefore, we need to calculate how much of the energy rate provided by the sun will be converted to electricity. Solar panel energy conversion = Energy rate provided by the sun × Efficiency of solar panels = 1390 W/m^2 × 0.19 = 264.1 W/m^2

Calculate the minimum solar panel surface area

Finally, we need to calculate the surface area of the solar panel required to provide the necessary energy per second (5 kW). To do this, we will divide the energy required per second by the solar panel energy conversion. Minimum solar panel surface area = Energy required per second / Solar panel energy conversion = (5 × 10^3 W) / (264.1 W/m^2) ≈ 18.93 m^2 The minimum solar panel surface area needed to provide all of a typical home's electricity is approximately 18.93 square meters.

Key concepts

Energy Conversion

Energy conversion is a fundamental concept in understanding how solar panels work. When sunlight hits a solar panel, it is transformed into electricity, which can be used to power homes and other electrical devices. This process is not 100% efficient, meaning not all sunlight is converted into usable energy. In the scenario described, the solar panels have an efficiency rate of 19%. This implies that only 19% of the solar energy captured by the panels is turned into electricity.
To understand this better, consider that the sun provides a constant energy rate of 1390 watts per square meter. With a 19% efficiency rate, the solar panels convert approximately 264.1 watts ( 1390 W/m² × 0.19 = 264.1 W/m² ) of that energy per square meter. The rest of the energy is either reflected away, absorbed by the materials, or lost as heat.
This energy conversion is crucial in determining how much surface area of solar panels is needed to meet the energy demands of a home. For instance, if a home requires 5 kW of power, knowing the efficiency of the panels helps calculate the necessary area to meet this demand.

Joules to Kilowatt Hours Conversion

Joules and kilowatt-hours (kWh) are units of energy, but they are used in different contexts. In scientific calculations like solar energy analysis, it is common to see energy expressed in joules. However, electricity billing and consumption tend to be in kilowatt-hours. Understanding how to convert between these units is essential.
One kilowatt-hour is the amount of energy consumed when a 1,000-watt appliance runs for an hour. In joules, this equates to 3.6 million joules (1 kWh = 3.6 x 10^6 J). When dealing with large quantities of energy, such as a home's daily electricity use, it is often easier to work with kilowatt-hours because they deal with smaller numbers.

• To convert kWh to joules, you multiply by 3.6 million. For example, 40 kWh (a household's daily usage) converts to 144 million joules (40 kWh x 3.6 x 10^6 J/kWh).
• The reverse operation (joules to kWh) involves dividing by 3.6 million, useful when needing to understand energy use or costs in a more familiar kWh format.

These conversions allow homeowners and energy analysts to efficiently calculate and communicate energy usage data.

Solar Energy Calculation

Calculating solar energy involves determining how much sunlight a location receives and how it is converted into power by solar panels. A key part of this process is figuring out how much energy is needed daily and matching that with what solar panels can provide based on their efficiency and available sunlight hours.
Consider a typical household that consumes 40 kWh per day, with the solar panels getting about 8 hours of useful sunlight. First, convert the daily consumption into joules (totaling 144 million joules per day). Then, calculate the energy requirement per second over the sunlight hours. For 8 hours, that's 28,800 seconds (8 x 3600), necessitating approximately 5,000 joules per second (or 5 kW, as 1 watt = 1 joule/second).
Following this, determine how much energy solar panels produce per square meter. If the solar panels convert energy at 19% efficiency from sunlight that delivers 1390 watts/m², output is about 264.1 watts/m². To find the surface area necessary to meet demand, divide the required power (5 kW) by the panel output (264.1 W/m²), resulting in a minimum area of roughly 18.93 m².
This calculation illustrates how solar panel surface area depends on both panel efficiency and the amount of sunlight received. It underscores the importance of optimized energy arrangements for efficient renewable energy production. By understanding these calculations, homeowners can make informed decisions about solar installations to fully power their homes.