You have installed a solar panel at your home that is connected to the electric grid of your local utility. Assume the system produces 8 kW from 9:00 AM to 6:00 PM, and you consume 2 kW during off-peak demand hours (9:00 PM - noon) and 10 kW during peak demand hours (noon - 9:00 PM). See underneath for the energy profile described. The User Purchase price: $0.105/ kWh On-peak sell price: $0.075/ kWh Off-peak sell price: $0.043/ kWh What is the economic value of the solar panel to the homeowner per day ($/day, with two decimal places)?
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The solar panel produces 8 kW from 9:00 AM to 6:00 PM, which is a duration of 9 hours. \[ \text{Total energy produced} = 8 \, \text{kW} \times 9 \, \text{hours} = 72 \, \text{kWh} \] Show more…
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The customer's house in Binghamton, NY uses 5750 kWh per year. The full-sun-hours per year in Binghamton is 1496. The system efficiency is 0.75. Use these inputs to calculate the system size and cost. Here is one formula to get you started: Solar electric size in kilowatts (kW) = yearly electric usage in kWh divided by (full-sun-hours multiplied by system efficiency). In formula form: PV size in kW = load in kWh Ă· (Full-sun-hours per year x system efficiency).
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Today, the best commercially available solar panels are about $18 \%$ efficient at converting the energy from sunlight into electricity. A typical home in the United States uses about $40 \mathrm{~kW} \cdot \mathrm{h}$ of electricity a day, and U.S. electricity costs currently average about $\$ 0.10$ per kW-h. (a) Assuming $8.0$ hours of useful sunlight per day with an incident solar energy of $0.5 \mathrm{~kW} \cdot \mathrm{m}^{-2}$, calculate the minimum panel area needed to provide for a typical home's electrical needs. Will a solar panel of this size fit on the roof of a typical single-family home? (b) If it costs $\$ 15000$ to install a solar panel of this size, how long will it take to pay for itself? Assume U.S. energy prices remain constant over this time.
The solar panels on the roof of a house measure $4.0 \mathrm{m}$ by $6.0 \mathrm{m} .$ Assume they convert $12 \%$ of the incident EM wave's energy to electric energy. (a) What average power do the panels supply when the incident intensity is $1.0 \mathrm{kW} / \mathrm{m}^{2}$ and the panels are perpendicular to the incident light? (b) What average power do the panels supply when the incident intensity is $0.80 \mathrm{kW} / \mathrm{m}^{2}$ and the light is incident at an angle of $60.0^{\circ}$ from the normal? (W) tutorial: solar collector) (c) Take the average daytime power requirement of a house to be about 2 kW. How do your answers to (a) and (b) compare? What are the implications for the use of solar panels?
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