91Ó°ÊÓ

The law of conservation of energy states that energy can neither be created nor destroyed but can be converted from one form to another. The total energy of a closed system remains constant.

The initial speed (at the top) of the water is\({v_1} = 0\).

The distance covered by falling water is\({h_1} = 88\;{\rm{m}}\).

The flow rate of water is \(Q = 680\;{\rm{kg/s}}\).

The potential energy at the level of the turbine is zero\(\left( {{h_2} = 0} \right)\). Use the law of conservation of energy at the top of the dam and near the turbine blades where the water strikes.

\(\begin{aligned}K{E_{\rm{i}}} + P{E_{\rm{i}}} &= K{E_{\rm{f}}} + P{E_{\rm{f}}}\\\frac{1}{2}mv_1^2 + mg{h_1} &= \frac{1}{2}mv_2^2 + mg{h_2}\\0 + mg{h_1} &= \frac{1}{2}mv_2^2 + 0\\mg{h_1} &= \frac{1}{2}mv_2^2\end{aligned}\) … (i)

Equation (i) implies that the potential energy at the top is equal to the kinetic energy at the bottom. Rearrange equation (i) for the speed of the water at the bottom.

\(\begin{aligned}{v_2} &= \sqrt {2g{h_1}} \\ &= \sqrt {2\left( {{\rm{9}}{\rm{.8}}\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\left( {88\;{\rm{m}}} \right)} \\ &= \sqrt {1724.8\;{{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^{\rm{2}}}} \\ &\approx 42\;{\rm{m/s}}\end{aligned}\)

Thus, the speed of the water just before falling on the blades of the turbine is \(42\;{\rm{m/s}}\).

Power is directly proportional to the work done on an object. It is defined as the rate of doing work or the work done per unit time.

At the bottom, the potential energy of water is zero. The energy of the water at the level of the turbine blades is only kinetic energy.

The rate of transfer of energy to the turbine blades is the power generated by the water.

\(\begin{aligned}P &= 55\% \;of\;\frac{{KE}}{t}\\ &= \frac{{55}}{{100}} \times \frac{{\left( {\frac{1}{2}m{v^2}} \right)}}{t}\\ &= \frac{{0.55\left( {680\;{\rm{kg}}} \right){{\left( {42\;{\rm{m/s}}} \right)}^2}}}{{2 \times 1\;{\rm{s}}}}\\ &= 3.3 \times {10^5}\;{\rm{W}}\end{aligned}\)

Thus, the power transferred to the turbine blades is \(3.3 \times {10^5}\;{\rm{W}}\).