Question

A tube-well pumps out 2400 kg of water per minute. If water is coming out with a velocity of 3 ms^–1, what is power of the pump? How much work is done, if the pump runs for 10 hours?

Answer 1

Step 1: Given data

Mass of water pumped is, $m=2400kg$

Velocity of water is, $v=3m/s$

Time taken to pump the water is $t=1min=60s$

Step 2: Calculating kinetic energy

Kinetic energy is given as,

$KE=\frac{1}{2}m{v}^2$

where, $m$ is mass and $v$ is velocity

Substituting the known values, we get,

$$\begin{gathered} KE=\frac{1}{2}\times 2400kg\times {\left(3m/s\right)}^2 \\ KE=10800J \end{gathered}$$

Step 3: Calculating power

According to the work-energy theorem, work done equals the change in kinetic energy. Therefore,

$W=KE=10800J$

Power is given as,

$P=\frac{W}{t}$

where, $W$ is work done and $t$ is time

Substituting the known values, we get,

$$\begin{gathered} P=\frac{10800J}{60s} \\ P=180W \end{gathered}$$

Step 4: Calculating work done when time is 10 h

Substituting $t=10h=36\times {10}^{3}s$ in the equation for power, we get

$$\begin{gathered} P=\frac{W}{t} \\ 180Watt=\frac{W}{36\times {10}^{3}s} \\ W=180\times 36\times {10}^3 \\ W=6480000 \\ W=6.48\times {10}^{6}J \end{gathered}$$

Therefore, power of the pump is $180W$ and work done by the pump when it runs for $10h$ is $6.48\times {10}^{6}J$.