Chapter 6, Problem 93GP

To determine

The volume of water to be pumped from lower to the upper reservoir

Answer to Problem 93GP

Solution: $1.74\times {10}^5{\text{m}}^{\text{3}}$

Explanation of Solution

The energy of the water required to pump it to the given height is equal to the gravitational potential energy of water at the same height.

Given: Time,

  $\begin{array}{l}t=1\text{hr}\\ t=3600\text{s}\end{array}$

Power,

  $\begin{array}{l}P=180\text{MW}\\ P=180\times {10}^6\text{W}\end{array}$

The height of the upper reservoir with respect to the lower one is $h=380\text{m}$ .

The density of water is $d=1.0\times {10}^3\frac{\text{kg}}{{\text{m}}^{\text{3}}}$ .

Formula used

Energy and power are related as $E=Pt$ , P is the power and tis the time.

Gravitational potential energy is given as,

  $\begin{array}{l}W=mgh\\ W=Vdgh\end{array}$

Here, V is volume and dis density.

Both E and W are numerically equal, $E=W$ .

Calculations

  $\begin{array}{l}E=W\\ Pt=Vdgh\\ V=\frac{Pt}{dgh}\\ V=\frac{180\times {10}^6\times 3600}{1.0\times {10}^3\times 9.8\times 380}\\ V=1.74\times {10}^5{\text{m}}^{\text{3}}\end{array}$

Conclusion: The approximate water volume is $1.74\times {10}^5{\text{m}}^{\text{3}}$ .