At \(4 \mathrm{~km}\) down, we expect the \(\Delta T\) to be about \(100^{\circ} \mathrm{C}\), and each cubic meter of rock would contain about \(250 \mathrm{MJ}\) of thermal energy. If \(50 \%\) of the maximum theoretical efficiency were achieved from an ambient environment at \(288 \mathrm{~K}\), how much rock thickness would have to be depleted in a year to satisfy a \(1 \mathrm{~km}^{2}\) campus whose output electricity demand is \(20 \mathrm{MW}\) ?

To find out the yearly energy demand for the campus, multiply the power demand by the number of hours in a year. Yearly demand in energy (E_demand) can be calculated as: E_demand = Power_demand × hours_in_a_year Where Power_demand = 20 MW and hours_in_a_year = 24 hours × 365 days

Given that 50% of the maximum theoretical efficiency is achieved, we need to convert it into a decimal form in order to calculate the actual usable energy from the extracted thermal energy. Efficiency (η) = 50% = 0.50

To determine the total amount of energy that needs to be extracted from the rocks, we need to calculate the energy required yearly with the efficiency taken into consideration. Total energy required (E_required) = E_demand ÷ η

Given that each cubic meter of rock contains 250 MJ of thermal energy, we want to find out how many cubic meters of rocks are needed to extract the total required energy. Number of cubic meters (N_cubic_meters) = E_required ÷ Energy_per_cubic_meter Where Energy_per_cubic_meter = 250 MJ

Since we need to provide energy to a campus with an area of 1 km², we need to convert the number of cubic meters of rock needed to the amount of rock needed per square meter. Volume_per_square_meter (V_rock_sqm) = N_cubic_meters ÷ Area Area = 1 km² = 1,000,000 m²

Finally, we can find the thickness of the rock to be depleted in a year (T_rock) using the volume of rock needed per square meter. T_rock = V_rock_sqm Now we can compute the values: Step 1: E_demand = 20 MW × (24 hours/day × 365 days/year) = 20 × 10^6 W × 8760 hours = 175,200,000 MJ Step 3: E_required = E_demand ÷ η = 175,200,000 MJ ÷ 0.50 = 350,400,000 MJ Step 4: N_cubic_meters = E_required ÷ Energy_per_cubic_meter = 350,400,000 MJ ÷ 250 MJ = 1,401,600 cubic meters Step 5: V_rock_sqm = N_cubic_meters ÷ Area = 1,401,600 cubic meters ÷ 1,000,000 m² = 1.4016 m³/m² Step 6: T_rock = V_rock_sqm = 1.4016 m Therefore, to satisfy the electricity demand of a 1 km² campus with a 20 MW output, a thickness of 1.4016 meters of rock per square meter needs to be depleted in a year.