Water is pumped at 250m^3/min from a lake into a tank 65.0 m above the lake.
(a) What power, in kW, must be delivered by the pump? (b) What horsepower rating
does this pump motor have? (c) What is the increase in potential energy of the water
each minute?
To answer these questions, we need to use the principles of work, power, and potential energy.
(a) To determine the power required by the pump, we can use the formula:
Power = Work / Time
Work can be calculated using the formula:
Work = Force * Distance
In this case, the force acting on the water is the weight of the water being lifted, given by:
Force = Mass * Gravity
The mass of the water can be calculated using the formula:
Mass = Volume * Density
Given that water has a density of 1000 kg/m³, and the volume flow rate is 250 m³/min, we can calculate the mass of the water per minute:
Mass = 250 m³/min * 1000 kg/m³ = 250,000 kg/min
The distance the water is lifted is 65.0 m, and the acceleration due to gravity is approximately 9.8 m/s².
Substituting these values into the formulas, we have:
Force = Mass * Gravity = 250,000 kg/min * 9.8 m/s² = 2,450,000 N
Work = Force * Distance = 2,450,000 N * 65.0 m = 159,250,000 J
Power = Work / Time = 159,250,000 J / 60 s/min = 2,654,167 W = 2654.167 kW
Therefore, the power required by the pump is approximately 2654.167 kW.
(b) To convert the power from kilowatts to horsepower, we can use the conversion factor:
1 horsepower = 0.7457 kW
Substituting the value of the power in kilowatts, we have:
Power (in horsepower) = 2654.167 kW * (1 horsepower / 0.7457 kW) ≈ 3560.75 horsepower
Therefore, the horsepower rating of the pump motor is approximately 3560.75 horsepower.
(c) The increase in potential energy of the water can be calculated using the formula:
Potential Energy = Mass * Gravity * Height
Substituting the values, we have:
Potential Energy = 250,000 kg/min * 9.8 m/s² * 65.0 m = 160,250,000 J
Therefore, the increase in potential energy of the water each minute is approximately 160,250,000 J.