Question

A pump is used to lift $500\mathrm{kg}$ of water from a depth of $80\mathrm{m}$ in $10\mathrm{s}$. Calculate the power rating of the pump if its efficiency is $40\%$

• A. $100\mathrm{kW}$
• B. $40\mathrm{kW}$
• C. $16\mathrm{kW}$
• D. None of the above

Answer 1

The correct option is A

$100\mathrm{kW}$

Step 1: Calculate the force exerted by the pump

Given, mass of the water lifted is $500\mathrm{kg}$

We know that $\mathrm{F}=\mathrm{m}g$, where $\mathrm{F}$ is the force exerted, $\mathrm{m}$ is the mass of the object and $g$ is the acceleration due to gravity. Assuming $g=10{\mathrm{ms}}^{-2}$,

$\begin{array}{rcl}\mathrm{F}& =& 500\times 10\\ \therefore \mathrm{F}& =& 5000\mathrm{N}\end{array}$

There the pump exerts $5000\mathrm{N}$ of force.

Step 2: Calculate the work done by the pump

Given, water is moved $80\mathrm{m}$ high.

The formula for calculating work done is,

$\mathrm{W}=\mathrm{F}·\mathrm{s}$

Where, $\mathrm{W}$ is the work done, $\mathrm{F}$ is the force exerted and $\mathrm{s}$ is the displacement of the object.

Thus, work done is,

$$\begin{gathered} \mathrm{W}=5000·80 \\ \therefore \mathrm{W}=400000=4\times {10}^5\mathrm{J} \end{gathered}$$

Therefore, the pump does $4\times {10}^5\mathrm{J}$ of work.

Step 3: Calculate the power of the pump

Given, the pump does the work in $10\mathrm{s}$

The formula for power ($\mathrm{P}$) is given as,

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

Where, $t$ is the time taken to do the work.

Thus,

$\mathrm{P}=\frac{4\times {10}^5}{10}=40\mathrm{kW}$

Step 4: Calculate power rating

Given, efficiency$=40\%=0.4$

We know that,

$\mathrm{Power}\mathrm{rating}=\frac{\mathrm{Useful}\mathrm{power}}{\mathrm{efficieny}}$

Thus, the power rating of the given pump is,

$\mathrm{Power}\mathrm{rating}=\frac{40}{0.4}=100\mathrm{kW}$

Hence, option A is correct.