91Ó°ÊÓ

The power of a process is the amount of some type of energy converted into a different type divided by the time interval\({\rm{\Delta t}}\)in which the process occurred:

\({\rm{P}} = \frac{{{\rm{\Delta E}}}}{{{\rm{\Delta t}}}}\) ….. (1)

The SI unit of power is the watt.

As you know, one watt is equal to one joule per second. Therefore,

\(1{\rm{ }}W = 1{\rm{ }}{J \mathord{\left/ {\vphantom {J s}} \right. \ } s}\)

The power of any electric circuit is defined by the formula

\({\rm{P}} = {\rm{I\Delta V}}\) ….. (2)

Here,\({\rm{I}}\)is the current and\({\rm{I\Delta V}}\)is the potential difference.

The gravitational potential energy is given by

\({{\rm{U}}_{\rm{g}}} = {\rm{mgh}}\) ….. (3)

Here,\({\rm{m}}\)is the mass of the particle,\({\rm{g}}\)is the acceleration due to gravity and\({\rm{h}}\)is the height of the particle.

The maximum current produced by the generators, \({\rm{I}} = {\rm{8}}{\rm{.00 \times 1}}{{\rm{0}}^{\rm{3}}}{\rm{\;A}}\).

The voltage produced by the generators is,

\(\begin{align}{\rm{\Delta V}} &= \left( {{\rm{250 kV}}} \right)\left( {\frac{{{\rm{1000\;V}}}}{{{\rm{1 kV}}}}} \right)\\ &= {\rm{250 \times 1}}{{\rm{0}}^{\rm{3}}}{\rm{\;V}}\end{align}\).

The kinetic energy of the water is constant.

The change in height of the water,\({\rm{\Delta y}} = {\rm{160\;m}}\).

The efficiency of the energy conversion process, \({\rm{\eta }} = {\rm{85}}{\rm{.0\% }}\).

The density of water,

\(\rho = 1000{\rm{ }}{{kg} \mathord{\left/ {\vphantom {{kg} {{m^3}}}}\right. \ {m^3}}}\).

The output power of the generator is,

\({\rm{P}} = {\rm{I\Delta V}}\)

Substituting the known values in the above equation.

\(\begin{align}{\rm{P}} &= \left( {{\rm{8}}{\rm{.00 \times 1}}{{\rm{0}}^{\rm{3}}}{\rm{\;A}}} \right)\left( {{\rm{250 \times 1}}{{\rm{0}}^{\rm{3}}}{\rm{\;V}}} \right)\\ &= {\rm{2}}{\rm{.00 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{\;W}}\end{align}\)

Therefore, the output power is given by \({\rm{2}}{\rm{.00 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{\;W}}\).

The gravitational potential energy of the water is converted into electrical energy with an efficiency of \({\rm{85}}{\rm{.0\% }}\).

Therefore, the electrical energy is,

\({\rm{E}} = \left( {{\rm{0}}{\rm{.85}}} \right){{\rm{U}}_{\rm{g}}}\)

Here, \({\rm{E}}\) is the electrical energy and \({{\rm{U}}_{\rm{g}}}\) he gravitational potential energy.

Substitute \({\rm{P\Delta t}}\) for \({\rm{E}}\) in the above equation.

\(\begin{align}P\Delta t &= \left( {0.85} \right)mg\Delta y\\\frac{m}{{\Delta t}} &= \frac{P}{{\left( {0.85} \right)g\Delta y}}\end{align}\)

Substituting the known numerical values in the above equation.

\(\begin{align}\frac{m}{{\Delta t}} &= \frac{{2.00 \times {{10}^9}{\rm{ }}{J \mathord{\left/ {\vphantom {J s}} \right. \ }s}}}{{(0.85)\left( {9.80{\rm{ }}{m \mathord{\left/ {\vphantom {m {{s^2}}}} \right. \ }{{s^2}}}} \right)(160{\rm{ }}m)}}\\ &= 1.50 \times {10^6}{\rm{ }}{{kg} \mathord{\left/ {\vphantom {{kg} s}}\right. \ }s}\end{align}\)

The volume occupied by the water is found in terms of its mass and density from the basic definition of density:

\(\begin{align}\frac{V}{{\Delta t}} &= \frac{{{\raise0.7ex\hbox{$m$} \! \mathord{\left/ {\vphantom {m {\Delta t}}}\right.\ }\!\lower0.7ex\hbox{${\Delta t}$}}}}{\rho }\\ &= \frac{{1.50 \times {{10}^6}^6{\rm{ }}{{kg} \mathord{\left/ {\vphantom {{kg} s}}\right. \ }s}}}{{{{1000}^6}{\rm{ }}{{kg} \mathord{\left/ {\vphantom {{kg} {{m^3}}}} \right. \ }{{m^3}}}}}\\ &= 1.50 \times {10^3}{\rm{ }}{{{m^3}} \mathord{\left/ {\vphantom {{{m^3}} s}} \right. \ } s}\end{align}\)

Therefore, the required solution is \(1.50 \times {10^3}{\rm{ }}{{{m^3}} \mathord{\left/ {\vphantom {{{m^3}} s}} \right. \ }s}\).