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Calculate the power required to pump 5 m³ of water per minute to a reservoir 30 m above the water level if the efficiency is 80%.

Physics

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Answer

30656.25 W\text{30656.25 W}

Step-by-step solution

4 steps
  1. Identify the given values and constants.

    • Volume of water, V=5 m3V = 5 \text{ m}^3
    • Time, t=1 minute=60 secondst = 1 \text{ minute} = 60 \text{ seconds}
    • Height, h=30 mh = 30 \text{ m}
    • Efficiency, η=80%=0.80\eta = 80\% = 0.80
    • Density of water, ρ=1000 kg/m3\rho = 1000 \text{ kg/m}^3 (standard value)
    • Acceleration due to gravity, g=9.81 m/s2g = 9.81 \text{ m/s}^2 (standard value)
  2. Calculate the mass of water pumped per minute.

    The mass (mm) of water can be found using its density (ρ\rho) and volume (VV):

    m=ρ×Vm = \rho \times V m=1000 kg/m3×5 m3m = 1000 \text{ kg/m}^3 \times 5 \text{ m}^3 m=5000 kgm = 5000 \text{ kg}
  3. Calculate the useful power (power output) required to lift the water.

    Useful power (PoutP_{out}) is the rate at which potential energy is gained by the water.

    Pout=mghtP_{out} = \frac{mgh}{t} Pout=(5000 kg)×(9.81 m/s2)×(30 m)60 sP_{out} = \frac{(5000 \text{ kg}) \times (9.81 \text{ m/s}^2) \times (30 \text{ m})}{60 \text{ s}} Pout=1471500 J60 sP_{out} = \frac{1471500 \text{ J}}{60 \text{ s}} Pout=24525 WP_{out} = 24525 \text{ W}
  4. Calculate the actual power required (power input) by the pump, considering its efficiency.

    Efficiency (η\eta) is defined as the ratio of useful power output to the total power input (PinP_{in}):

    η=PoutPin\eta = \frac{P_{out}}{P_{in}}

    Rearranging the formula to find PinP_{in}:

    Pin=PoutηP_{in} = \frac{P_{out}}{\eta} Pin=24525 W0.80P_{in} = \frac{24525 \text{ W}}{0.80} Pin=30656.25 WP_{in} = 30656.25 \text{ W}

    The power required to pump the water is 30656.25 W\text{30656.25 W}.

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