To draw water from a well, a pump with a power of 3675 W is used. The well depth is 30 m. Calculate the
volume of water taken in 24 h. The density of water is 1000 kg/m3.
Problem data:
-
Power of pump:
\( \mathscr{P} \) = 3675 W;
- Depth of well: h = 30 m;
- Pump operating time interval: Δ t = 24 h;
- Density of water: ρ = 1000 kg/m3;
- Acceleration due to gravity: g = 9.8 m/s2.
Problem diagram:
We choose a Reference Level (RL) on the surface of the well (Figure 1).
Solution::
First, we convert the interval of time given in hours (h) to seconds used in the
International System of Units (SI).
\[
\begin{gather}
\Delta t=24\;\mathrm{\cancel h}\times\;\frac{60\;\mathrm{\cancel{min}}}{1\;\mathrm{\cancel h}}\times\frac{60\;\mathrm s}{1\;\mathrm{\cancel{min}}}=86400\;\mathrm s
\end{gather}
\]
Power is given by energy variation per unit of time.
\[
\begin{gather}
\bbox[background:#99CCFF; padding:10px; border-radius:10px;]
{\mathscr{P} =\frac{\Delta E}{\Delta t}} \tag{I}
\end{gather}
\]
The change of total energy will be given by the change of potential energy.
\[
\begin{gather}
\bbox[background:#99CCFF; padding:10px; border-radius:10px;]
{\Delta E=U_f-U_i} \tag{II}
\end{gather}
\]
the potential energy is given by
\[
\begin{gather}
\bbox[background:#99CCFF; padding:10px; border-radius:10px;]
{U=mgh} \tag{III}
\end{gather}
\]
substituting equation (III) into equation (II), the change in potential energy when a mass of water
is taken from the bottom of the well (reference level, h0 = 0) until the surface will be
\[
\begin{gather}
\Delta E=mgh-mgh_0\\[5pt]
\Delta E=mgh-mg\times 0\\[5pt]
\Delta E=mgh \tag{IV}
\end{gather}
\]
substituting the equation (IV) into equation (I)
\[
\begin{gather}
\mathscr{P} =\frac{mgh}{\Delta t} \\[5pt]
m=\frac{\mathscr{P} \Delta t}{gh} \\[5pt]
m=\frac{(3675\;\mathrm W)\times(86400\;\mathrm s)}{\left(9.8\;\frac{\mathrm m}{\mathrm{s^2}}\right)\times(30\;\mathrm m)} \\[5pt]
m=1080000\;\text{kg}
\end{gather}
\]
The density is given by
\[
\begin{gather}
\bbox[background:#99CCFF; padding:10px; border-radius:10px;]
{\rho =\frac{m}{V}}
\end{gather}
\]
The volume of water removed will be
\[
\begin{gather}
V=\frac{m}{\mu} \\[5pt]
V=\frac{1080000\;\mathrm{\cancel{kg}}}{1000\;\mathrm{\frac{\cancel{{kg}}}{m^3}}}
\end{gather}
\]
\[
\begin{gather}
\bbox[background:#FFCCCC; padding:10px; border-radius:10px;]
{V=1080\;\mathrm m^3}
\end{gather}
\]
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