Question

A 30 year study by the National Renewable Energy Lab \(^{112}\) indicates that in San Diego, a typical year delivers an annual average of \(5.0 \mathrm{kWh} / \mathrm{m}^{2} /\) day of insolation for a flat panel facing straight up. Convert this to \(\mathrm{W} / \mathrm{m}^{2}\).

Short answer

Answer: The annual average insolation energy in San Diego is approximately 0.2778 W/m².

Step-by-step solution

Given values and units

We are given that the average annual insolation energy in San Diego is \(5.0 \mathrm{kWh} / \mathrm{m}^{2}\) per day. Our goal is to convert this value into units of \(\mathrm{W} / \mathrm{m}^{2}\).

Conversion factor connection between kWh and Wh

First, we need to convert the energy value from kilowatt-hours (kWh) to watt-hours (Wh) since our final unit should be in Watts. There are 1000 Watts in a kilowatt, so we can multiply the given energy value by 1000:$$5.0 \mathrm{kWh} / \mathrm{m}^{2} \times 1000 = 5000 \mathrm{Wh}/ \mathrm{m}^{2}$$ The insolation energy is now in watt-hours per square meter per day.

Conversion factor connection between hours and seconds

Now, we need to convert the time unit from hours to seconds. There are 3600 seconds in an hour and 24 hours in a day. Hence, we need to divide the energy value by the total number of seconds in a day:$$5000 \mathrm{Wh}/ \mathrm{m}^{2} \times (1/ (24 \, \mathrm{hours} \times 3600 \, \mathrm{seconds})) = \frac{5000}{86400} \frac{\mathrm{Wh}}{\mathrm{m}^{2}}$$

Conversion factor connection between Wh and W

Finally, we convert watt-hours (Wh) to watts (W) by dividing by the number of seconds per hour (3600). A watt is the rate of energy consumption or generation, and one watt-hour means one watt of energy is used or generated per hour.$$ \frac{5000}{86400} \frac{\mathrm{Wh}}{\mathrm{m}^{2}} \times (1/3600) \frac{\mathrm{W}}{\mathrm{Wh}} = \frac{5}{18} \frac{\mathrm{W}}{\mathrm{m}^{2}}$$

Final answer

The annual average insolation energy in San Diego, in units of \(\mathrm{W}/\mathrm{m}^{2}\), is:$$\frac{5}{18} \mathrm{W} / \mathrm{m}^{2} \approx 0.2778 \mathrm{W} / \mathrm{m}^{2}$$

Key concepts

Kilowatt-hour to Watt Conversion

Understanding how to convert energy units from kilowatt-hours (kWh) to watts (W) is crucial when dealing with various energy systems, including solar panels. A kilowatt-hour is a measure of energy, equivalent to the power consumption of 1000 watts for one hour. Conversely, a watt is a unit of power, which refers to the rate of energy use or generation per second.

When you want to convert kilowatt-hours to watts, you're essentially calculating the rate at which energy was consumed or generated. When working with solar energy measurements, we often start with kilowatt-hours because that's how energy output is typically recorded over a period, like a day. However, to understand the instantaneous power available at any moment — such as what an appliance might draw — we need to express the energy in watts.

Here's a simplified example: If a solar panel generates 5 kWh over the course of an hour, this means it has an output of 5000 watts for that hour. But since we want an instantaneous rate, not an hourly total, we need to account for the number of seconds in an hour (3600). So the conversion becomes a two-step process: first, multiplying kilowatt-hours by 1000 to get watt-hours, and then dividing by the number of seconds in an hour to arrive at watts.

Energy Unit Conversion

Energy unit conversions are a staple for understanding different energy systems and for comparing various forms of energy output or consumption. The exercise we're examining requires converting a daily average energy value into an instantaneous power value. It's important to grasp the distinction: the energy unit watt-hour (Wh) measures the total energy used over time, while the power unit watt (W) measures the rate of energy use per second at any given instant.

To transition between these units, one must take the amount of time into account, which in this case is represented in hours or seconds. When we deal with watt-hours and want to find watts, we divide the energy by the time in seconds during which the energy was used. In essence, we convert a total amount of energy into an energy rate.

Remember that while the math may look simple, it's easy to get tangled up if you lose track of your units or fail to convert time correctly. Always double-check your work to ensure that you've converted all your units consistently and correctly. Following the correct conversion factors is key to getting to the accurate measurement of power in watts.

Solar Energy Mathematics

Solar energy mathematics involves a range of calculations, including those relating to insolation, which is the amount of solar energy received on a given surface area in a given time. Calculating insolation is important for determining the potential solar power generation of a location.

In the example provided, the insolation value given in kilowatt-hours per square meter per day represents an average energy received during a typical day. However, for practical applications like sizing solar panels or understanding their output at any given moment, we need to know the average power, which requires converting this number into watts per square meter.

The process involves converting kilowatt-hours to watt-hours, then accounting for the number of seconds in each day to transition from a total daily energy value to an average power value. Such conversions are vital for engineers and technicians in sizing solar installations and for homeowners considering solar energy as they inform decisions about the number and type of solar panels needed to meet energy requirements.

Overall, solar energy mathematics is the backbone of designing and implementing efficient solar power systems. It ensures that solar panels are not undersized, leading to insufficient power, or oversized, which could result in unnecessary costs.