Question

Calculate the time for which a motor pump of $10HP$ and efficiency $80\%$ must be switched on, so as to pump $20m^3$ of water through a vertical height of $20m$. [Density of water = $\raisebox{1ex}{$1000kg$}\!\left/ \!\raisebox{-1ex}{$m^3$}\right.$, $g=\raisebox{1ex}{$10m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$, $1HP=750W$]

Answer 1

Step 1: Given data,

Efficiency=$80\%$

Power=$10HP$

Height=$20m$

Volume of water=$20m^3$

Density of water = $\raisebox{1ex}{$1000kg$}\!\left/ \!\raisebox{-1ex}{$m^3$}\right.$

Acceleration due to gravity $g=\raisebox{1ex}{$10m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$

Horse power in terms of watts is $1HP=750W$

Step 2: To find mass of water,

Let ‘m’ mass of water can be lifted.

Formula used: $$\begin{gathered} Density=\frac{mass}{volume} \\ Mass=density\times volume \end{gathered}$$

$Massofwater=1000\times 20=20000kg$

Step 3: To convert horsepower into watts:

$1HP=750W$

$10HP=7500W$

Step 4: To find the power that is efficiently used:

P' =$80\%of7500$

$=\frac{80}{100}\times 7500=6000watts$

Step 5: To find force and work done,

Formula used : $force=weight=mg$

$force=20000\times 10=200000N$

Work done in terms of force and displacement,

Formula used : $Workdone=force\times displacement$

$$\begin{gathered} W=200000\times 20 \\ =4000000J \end{gathered}$$

Step 6: To find time:

Formula used : $power=\frac{work}{time}$

$6000=\frac{4000000}{time}$

$time=\frac{4000000}{6000}$

$=666.67s$

Thus, the water can be lifted for $666.67s$.