Air moving at 13.0 m/s in a steady wind encounters a windmill of diameter 2.30 m and having an efficiency of 29.0%. The energy generated by the windmill is used to pump water from a well 31.0 m deep into a tank 2.30 m above the ground. At what rate in liters per minute can water be pumped into the tank?
To find the rate at which water can be pumped into the tank, we need to calculate the power generated by the windmill and then convert it to the rate of water pumping.
First, let's calculate the power generated by the windmill using the formula:
Power = 0.5 * density * area * velocity^3 * efficiency
Given:
Density of air (ρ) = 1.2 kg/m^3 (typical value)
Diameter of windmill (d) = 2.30 m
Radius of windmill (r) = 2.30 / 2 = 1.15 m
Efficiency of windmill = 29.0% = 0.29
Velocity of air (v) = 13.0 m/s
Area of windmill (A) = π * r^2
Density (ρ) = 1.2 kg/m^3
Plugging in the values:
Area of windmill (A) = π * (1.15)^2
Density (ρ) = 1.2 kg/m^3
Velocity (v) = 13.0 m/s
Efficiency of windmill (η) = 0.29
Power = 0.5 * 1.2 * π * (1.15)^2 * 13.0^3 * 0.29
Now we have the power generated by the windmill.
To convert this power to the rate of water pumping, we need to account for the energy required to pump water against gravity.
The energy required to lift water is given by:
Energy = m * g * h
Given:
Height of well (h) = 31.0 m
Height of tank (h_tank) = 2.30 m
Acceleration due to gravity (g) = 9.8 m/s^2
Now, let's calculate the energy required to lift the water:
Energy = m * g * (h + h_tank)
Rate of water pumping (R) = Power / Energy
Finally, we need to convert the rate of water pumping to liters per minute.
Conversion:
1 liter = 0.001 m^3
1 minute = 60 seconds
Rate of water pumping (R_liters_per_minute) = R * (1/0.001) * (1/60)
Now you can substitute the values into the equations and calculate the rate at which water can be pumped into the tank.