Water is to be pumped through a vertical height difference of 12.5 m. The pump is driven by a wind turbine that has an efficiency of 50 % and an area swept by the blades of 100 m2. The average wind speed is 5.0 m s−1 and the air density is 1.2 kg m−3.What is the maximum mass of water that can be pumped every second?A. 3 kgB. 30 kgC. 60 kgD. 120 kg
Question
Water is to be pumped through a vertical height difference of 12.5 m. The pump is driven by a wind turbine that has an efficiency of 50 % and an area swept by the blades of 100 m2. The average wind speed is 5.0 m s−1 and the air density is 1.2 kg m−3.What is the maximum mass of water that can be pumped every second?A. 3 kgB. 30 kgC. 60 kgD. 120 kg
Solution
To solve this problem, we first need to calculate the power generated by the wind turbine. The power P of a wind turbine can be calculated using the formula:
P = 0.5 * ρ * A * v^3 * η
where: ρ is the air density (1.2 kg/m^3), A is the area swept by the blades (100 m^2), v is the wind speed (5.0 m/s), and η is the efficiency of the turbine (50% or 0.5).
Substituting these values into the formula gives:
P = 0.5 * 1.2 kg/m^3 * 100 m^2 * (5.0 m/s)^3 * 0.5 = 7500 W or 7500 J/s
Next, we need to calculate the energy required to pump the water. The energy E required to lift a mass m of water through a height h is given by the formula:
E = m * g * h
where: g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height difference (12.5 m).
We want to find the maximum mass of water that can be pumped every second, so we set E equal to P and solve for m:
7500 J/s = m * 9.8 m/s^2 * 12.5 m m = 7500 J/s / (9.8 m/s^2 * 12.5 m) = 60.2 kg
So, the maximum mass of water that can be pumped every second is approximately 60 kg, which corresponds to answer C.
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