Question

Solar energy stored in large bodies of water, called solar pounds, is being used to generate electricity. If such a solar power plant has an efficiency of 3 percent and a net power output of \(180 \mathrm{kW}\), determine the average value of the required solar energy collection rate, in \(\mathrm{Btu} / \mathrm{h}\).

Short answer

Answer: The average value of the required solar energy collection rate is approximately 20,460,006.67 Btu/h.

Step-by-step solution

Convert net power output to Btu/h

We are given the net power output in kW, but we need to determine the solar energy collection rate in Btu/h. To do this, we will use the following conversion factors: 1 kW = 3412.14 Btu/h So, we multiply the given net power output by the conversion factor to obtain the output in Btu/h. \(180 \mathrm{kW} \times 3412.14 \frac{\mathrm{Btu}}{\mathrm{h \cdot kW}}\)

Calculate output in Btu/h

Multiply the net power output by the conversion factor: \(180 \mathrm{kW} \times 3412.14 \frac{\mathrm{Btu}}{\mathrm{h \cdot kW}} = 613800.2 \mathrm{Btu/h}\) So, the net power output in Btu/h is 613800.2 Btu/h.

Use the efficiency to find the required solar energy collection rate

We are given the efficiency of the solar power plant as 3 percent (0.03). Efficiency is defined as the ratio of the useful power output to the input power, so we have: Efficiency = \(\frac{\mathrm{Output \ Power}}{\mathrm{Input \ Power}}\) Now, we can rearrange the formula to find the input power (required solar energy collection rate): Input Power = \(\frac{\mathrm{Output \ Power}}{\mathrm{Efficiency}}\) So, we can plug in the given values: Input Power = \(\frac{613800.2 \mathrm{Btu/h}}{0.03}\)

Calculate the required solar energy collection rate

Divide the output power in Btu/h by the efficiency: Input Power = \(\frac{613800.2 \mathrm{Btu/h}}{0.03} = 20460006.67 \mathrm{Btu/h}\) Thus, the average value of the required solar energy collection rate is approximately 20,460,006.67 Btu/h.

Key concepts

Solar Power Plant Efficiency

Solar power plant efficiency is a measure of how effectively a plant converts solar energy into electrical power. It's an important factor because it directly affects the amount of solar energy that needs to be collected to meet energy production goals. In the exercise, with an efficiency of 3%, the plant is only converting 3% of the solar energy it collects into usable electricity. This low efficiency rate means that the plant must collect a large amount of solar energy to produce a relatively small amount of electrical power.

The efficiency of a solar power plant can be impacted by numerous factors, including the technology used in the solar panels, the configuration and design of the system, and external factors like weather conditions. For instance, newer solar panels with higher conversion efficiencies can produce more power from the same amount of sunlight compared to older modules. Improving efficiency is key in making solar energy a more viable and economical power source.

Energy Conversion Factors

Energy conversion factors are essential when working with different units of energy, as they allow for the proper conversion from one to another. In our example, we see the conversion of kilowatts (kW) to British thermal units per hour (Btu/h), which is crucial for calculating the required solar energy collection rate in the units requested. The conversion factor used is 1 kW = 3412.14 Btu/h.

Understanding these conversion factors is vital for engineers and scientists in analyzing energy systems and for making meaningful comparisons. Being fluent with these values facilitates clear communication of power output or energy consumption figures across various disciplines and regions that may use different units of measurement. Always remember to use the correct conversion factor, as using an incorrect one could result in significant calculation errors.

Thermodynamics Problem Solving

Thermodynamics problem solving involves understanding the principles that govern energy interactions and conversions. In our exercise, we apply a basic thermodynamics principle that relates the output power to the input power via the efficiency of the system. This relationship is expressed by the formula: Efficiency = Output Power / Input Power. By rearranging this formula, we can solve for the unknown variable, which in this case is the input power, or the solar energy collection rate required for the power plant to function.

Practicing thermodynamics problem solving prepares students to tackle real-world challenges, where they can apply these principles to optimize energy systems for efficiency and sustainability. It's crucial to think systematically and follow logical steps when tackling such problems, ensuring all units are consistent and all known values are accurately applied to find the unknown variables. In the improvement of exercises, highlighting these systematic approaches encourages a deeper understanding of the principles at play.