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Q5.15:
A pump on the ground floor of a building can pump up water to fill a tank of volume (30 m^{3}) in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump ?

Solution :

Step 1: Given Variables & Physical Constants

  • Volume of tank ($V$) = $30 \, m^3$
  • Time taken ($t$) = $15 \, min = 15 \times 60 = 900 \, s$
  • Height of tank ($h$) = $40 \, m$
  • Efficiency ($\eta$) = $30\% = 0.3$
  • Density of water ($\rho$) = $10^3 \, kg/m^3$
  • Acceleration due to gravity ($g$) = $9.8 \, m/s^2$

Step 2: Calculating Mass of Water

The mass ($m$) of the water pumped is determined by the volume and the density of water [By definition: $m = \rho V$].

$m = 10^3 \, kg/m^3 \times 30 \, m^3 = 30,000 \, kg$

Step 3: Calculating Useful Work Done (Output Energy)

The work done ($W_{out}$) against gravity to raise the water to height $h$ is defined by potential energy [Per the Work-Energy Theorem]:

$W_{out} = mgh$

$W_{out} = 30,000 \, kg \times 9.8 \, m/s^2 \times 40 \, m$

$W_{out} = 11,760,000 \, J = 1.176 \times 10^7 \, J$

Step 4: Calculating Useful Power Output

Power is the rate at which work is performed ($P = W/t$):

$P_{out} = \frac{1.176 \times 10^7 \, J}{900 \, s}$

$P_{out} = 13,066.67 \, W$

Step 5: Calculating Input Electric Power

Efficiency is the ratio of useful power output to total power input [Per the Law of Conservation of Energy]:

$\eta = \frac{P_{out}}{P_{in}} \implies P_{in} = \frac{P_{out}}{\eta}$

$P_{in} = \frac{13,066.67 \, W}{0.3}$

$P_{in} = 43,555.57 \, W \approx 43.56 \, kW$

Visual Representation of Power Conversion

Input Power Pump (30%) Output Work

Final Solution: The total electric power consumed by the pump is $43.56 \, kW$.


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