How Much Land Needed to Install a Solar Plant with a Capacity of 11.1 GW?

Question: Assuming that virtually unlimited flat land is available in a desolate desert, how many square kilometers would be required to install a solar plant with a capacity of 11.1 GW?

Answer: Here’s an estimate of the land area required for an 11.1 GW solar plant, along with some important considerations;

Assumptions:

  • Solar Panel Efficiency: We’ll use an average efficiency of 20% for this calculation.
  • Solar Panel Density: A typical large-scale solar farm has a panel density of around 1 MW per hectare (1 hectare = .01 square kilometers). This can vary depending on the specific panel type and layout.

Calculations:

  1. Total Land Area (Hectares):
  2. Total Capacity (MW) / Panel Density (MW/hectare) = Land Area (hectares).
  3. 11,100 MW / 1 MW/hectare = 11,100 hectares
  4. Total Land Area (Square Kilometers):
  5. Land Area (hectares) *.01 = Land Area (square kilometers)
  6. 11,110 hectares * .01 = 111 square kilometers.

Important Considerations:

  • Real-World Variations: The actual land area required could be larger or smaller depending on factors such as:
  • Panel Efficiency: Higher-efficiency panels would require less land.
  • Panel tilt and spacing: the angle and spacing of solar panels affect how much sunlight they capture.
  • Site Layout: The shape and topography of the land can influence the layout and efficiency of the solar farm.
  • Infrastructure: Space for roads, substations, and other infrastructure will also be needed.
  • Environmental impact: Even in a desolate desert, large-scale solar installations can have environmental impacts. It’s essential to consider potential effects on wildlife, vegetation, and soil.

Water Use: Solar panels themselves don’t use water for electricity generation, but cleaning them can require significant amounts of water. Water availability and management are crucial consideration, especially in a desert environment.

How Much Solar to Produce 611,800,000,000 kg of Hydrogen/Year?

Question:

Given: a recent breakthrough of researchers at the University of Adelaide has yielded an electrolysis process that is 100% efficient with a catalyst that prevents the anode and cathode from decay; the amount of hydrogen desired is 611,800,000,000 kg per year; the temperature of the seawater is 60 degrees C; the salinity of the water is 35 parts per thousand; the voltage used in the process is 1.87 V; the intensity of sunlight is 7.5 peak solar hours per day. Assuming the above parameters are correct, how much solar energy would be required during the time frame specified?

Answer:

Let’s calculate the solar energy required, given the assumption of 100% electrolysis efficiency:

  1. Energy for Electrolysis
  2. The theoretical energy calculation is 1.095 x 1017 Joules.
  3. Hours of sunlight per year: 7.5 hours/day * 365 days/year = 2737.5 hours/year.
  4. Seconds of sunlight per year: 2737.5 hours/year * 3600 seconds/hour = 9.86 x 106 seconds/year

Required Solar Power

  • Power (Watts) = Energy (Joules) / Time (seconds)
  • Power = 1.095 x 1017 J / 9.86 x 106 s = 1.11 x 1010 Watts = 11.1 Gigawatts (GW)

A solar power plant with a capacity of approximately 11.1 GW would be needed to generate enough energy to produce 611,800,000,000 kg of hydrogen per year, assuming 100% electrolysis efficiency and the given parameters.

Important notes:

  • 100% Efficiency: While the breakthrough claims 100% efficiency for the electrolysis process itself, it’s crucial to remember that there will still be losses in other parts of the system (e.g., capturing sunlight, converting electricity, etc.).
  • Real-World Conditions: Solar power output varies depending on weather, time of day, and location. The calculated power requirement assumes constant peak sunlight, which is not realistic.
  • Land Use: a 11.1 GW solar power plant would require a significant land area.
  • Other Factors: This calculation doesn’t account for energy storage, transportation of hydrogen, or potential maintenance requirements.

In conclusion, even with the remarkable breakthrough of 100% efficient electrolysis, producing such a massive amount of hydrogen would still necessitate an enormous solar energy infrastructure and careful consideration of real-world factors.

Required Number of Rooftops Equipped With Solar Panels to Produce 611,800,000,000 kg of Hydrogen

The first step in this estimation is access the Global Solar Atlas; click on the specific location and look for ‘specific photovoltaic power output’, or PVOUT. This estimates the annual productivity of solar panels for a given location, measured in kilowatt-hours generated per kilowatt of peak capacity (kWh/kWp). For purposes of these calculations, we’ll assume that all rooftops can generate 12,000 kWh of electricity per year.

To calculate the number of rooftops equipped with solar panels needed to produce 611,800,000,000 kg of hydrogen using electrolysis powered by solar energy:

  1. The amount of electricity required for this process can be calculated based on the energy needed to perform electrolysis. It takes about 39 kWh of electricity to produce one kilogram of hydrogen through electrolysis.
  2. To find out how much total electricity is need:
  3. Multiply the amount of hydrogen needed (611,800,000,000 kg) by the energy requirement per kilogram (39 kWh), which gives us the total electrical energy requirement in kWh.
  4. Next, we need to calculate how many rooftop solar panels are necessary:
  5. Assuming an average yield from a rooftop solar panel system is around 1465 kWh/year and each roof has capacity equivalent or more than that then it would take approximately 163,000 rooftops equipped with sufficient solar panels.

This calculation provides an estimate under ideal conditions and does not consider factors such as efficiency losses or variations in sunlight availability throughout the year at different locations where these roofs might exist. The total number of single-family residential dwellings in California’s Fresno, Imperial, Kern, Los Angeles, Orange, Riverside, San Bernardino, San Diego, San Joaquin, Santa Barbara and Ventura counties is 9,126,412. This is roughly 56 times the minimum number of rooftops needed to produce 611,800,000,000 kg of hydrogen.

Hydrogen Needed to Replace the California & Colorado Aqueducts

Southern California’s Water

Most of southern California’s water is imported. The California Aqueduct delivers up to 4.2 million acre-feet, the Colorado River Aqueduct 3,069.6 acre-feet, and the Los Angeles Aqueduct 275,000 acre-feet, for a combined total of 4,478,069.6 acre-feet of water per year. Since 1 acre-foot = 1233.5 m3, 4,478,069.6 acre-feet = 5,522,006,580.6 m3. To determine the mass of 5,522,006,580.6 m3 we’ll use the formula: mass = volume x density. Since the density of water is defined as 1000 kg/m3, 5,522,006,580.6 m3 x 1000 kg/m3 = 5,522,006,580,000 kg of water (mass).

Hydrogen to Replace the California and the Colorado River Aqueducts

The next task is to calculate how much hydrogen would be needed to create 5,522,006,580,000 kg of water. The first step is to understand the chemical formula for water, which is H2O. This means that one molecule of water contains two hydrogen atoms and one oxygen atom.

Next, we need to determine the molar mass of water, which is approximately 18.01528 grams per mole. Since we are given the mass of water in kg, we need to convert the mass to moles using molar mass.

First, we need to convert the mass of water to grams:

5,522,006,580,000 kg * 1000 = 5,522,006,580,000,000 g

Next, we calculate the number of moles of water:

5,522,006,580,000,000 g ÷ 18.01528 g/mol = approximately 3.07 x 1014 mol.

Since we need two molecules of hydrogen for every molecule of water, the number of moles of hydrogen required is double that of water: 3.097 x 1014 * 1.00784 g/mol = approximately 6.18 x 1014 g, or 611,800,000,000 kg of hydrogen.

This much hydrogen would be needed to replace the water from the California and Colorado Aqueducts.

California

There are several good reasons for selecting the Golden State. Unlike China, which lacks an ocean west of and adjacent to the Gobi Desert, California’s southern shore is a stone’s throw away from its own sparsely populated sun-drenched desert. In addition, the state features towering mountains, a vast southern metropolitan area with innumerable individual residential rooftops, and Death Valley, the lowest depression in North America, about 4.5 hours northeast of Los Angeles, 2.5 hours northwest of Las Vegas, and about 200 miles (321.9 km) as the crow flies from the ocean.

Death Valley Statistics

The stated depth of the Death Valley canal is based on this topographic map, which suggests an estimated average altitude at 2,500 feet (762 meters) along its possible 425 km route. The staggering volume of earth that would need to be excavated can be calculated by the formula Volume = length * width * depth. That works out to 74,154,000,000 cubic meters, or 74.154 km3. These dimensions assume the canal would be perfectly rectangular. In practice, the actual excavated volume would likely be significantly larger due to slope of the sides and other factors, but the numbers illustrate the magnitude of the project. Essentially, it’s a trench roughly equivalent to 10 times the volume of Mount Everest and about half as deep as the Grand Canyon. While the Colorado River formed the latter over approximately 35 million years, modern equipment could strive to complete the project in 12 years, the same time frame as China’s Three Gorges Dam and the $2 billion Teen-Tom Waterway. Death Valley would exceed both by many orders of magnitude. Teen-Tom required 100-million dump truck loads to move 229.3 million cubic meters of excavated earth. Extrapolating, Death Valley would require about 32.36 billion dump truck loads (7,388 per day, or about 308 per hour every 24 hours) and cost about $53.9 billion per year, or $647.2 billion over 12 years. However, this simplified calculation doesn’t include additional potential costs due to natural and bureaucratic barriers.

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