Question

Water is pumped from a depth of 10 m and delivered through a pipe of cross section ${10^{ - 2}}{{\rm{m}}^2}$. If it delivered through a volume of ${10^{ - 1}}{{\rm{m}}^3}$ per second the power required will be:
(A) 10 kW
(B) 9.8 kW
(C) 15 kW
(D) 4.9 kW

Answer 1

In the given question, we can consider the water in the reservoir is at zero potential. When the water is pumped up from that depth, it attains some amount of potential energy. The amount of potential gained by the water is the amount of energy that needs to be supplied externally.

Complete step by step answer:
Given;
The water is pumped from a depth of h = 10 m.
The cross-section of the pipe, $A = {10^{ - 2}}{{\rm{m}}^2}$.
Rate of transfer of water, $r = {10^{ - 1}}{{\rm{m}}^3}$.
The potential energy obtained by the water when it is raised from that depth is;

P.E. = mgh\\\ = m \times 9.8 \times 10\\\ = \left( {98 \times m} \right){\rm{ J}} \end{array}$$ Also, the volume of water pumped up in time, t, is given as: $$\begin{array}{l} V = r \times t\\\ V = {10^{ - 1}} \times t\;{{\rm{m}}^3} \end{array}$$ So, the mass of the water pumped up is given by,h $$\begin{array}{l} m = \rho \times V\\\ m = {10^3}{\rm{ kg}}{{\rm{m}}^{ - 3}} \times {10^{ - 1}} \times {\rm{t }}{{\rm{m}}^3}\\\ m = 100 \times t{\rm{ kg}} \end{array}$$ Now, power can be expressed as energy consumed per unit time. So, the power required to pump up the water is, $$P = \dfrac{{P.E.}}{t}$$ On substituting the values in the above expression, we get, $$\begin{array}{l} P = \dfrac{{98 \times 100 \times t}}{t}\\\ P = 9800{\rm{ J}}{{\rm{s}}^{ - 1}}\\\ P = 9800{\rm{ W}}\\\ P = {\rm{9}}{\rm{.8 kW}} \end{array}$$ Therefore, the power required to pump up the water is 9.8 kW, and option (B) is correct. **Note:** In this question, we can tell that the power required to deliver the water from the reservoir to the height is the potential energy gained by the volume of water pumped out of the pipe per second. The power delivered to pump up the water from the reservoir is manifested as the stored potential energy of the water collected at the height (h = 10 m). The process is not spontaneous, but the supply of external power makes it an externally driven process.