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      <text>27.3.4.2 mass of water in a reservoir</text>
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      <text><![CDATA[<h4>Mass of water in a reservoir</h4>
<p>Let us return to the reservoir example. Imagine that a reservoir is rectangular in profile and that the width along the dam (i.e. measured in the \(x\) direction) is 100 m. Imagine also that the length of the reservoir (measured away from the dam i.e. in the \(y\) direction) is 400 m. The depth of the reservoir is given by \(40 - y/10\) m i.e. the reservoir is 40 m deep along the dam and the depth reduces to zero at the end away from the dam. 0.1inThe density of the water can be approximated by \(\rho(z) = a - b\times z\) where \(a = 998\) kg m\(^{-3}\) and \(b = 0.05\) kg m\(^{-4}.\) I.e. at the surface (\(z=0\)) the water has density \(998\) kg m\(^{-3}\) (corresponding to a temperature of \(20^{\circ}\)C) while \(40\) m down i.e. \(z = -40\), the water has a density of \(1000\) kg m\(^{-3}\) (corresponding to the lower temperature of \(4^{\circ}\)C).</p>

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<img src="@@PLUGINFILE@@/Fig363_6.svg" alt="A diagram on 3D axis. The points (100,0,0) and (0,0,-40) make up a back shaded rectangle, and it slopes up to (0,400,0)." width="328" height="175">
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<p>The mass of water in the reservoir is given by the integral of the function \(\rho(z) = a - b \times z\). For each value of \(x\) and \(y\), the limits on \(z\) will be from \(y/10-40\) (bottom) to \(0\) (top). Limits on \(y\) will be \(0\) to \(400\) m while the limits of \(x\) will be \(0\) to \(100\) m. The mass of water is therefore given by the integral 
\(\displaystyle M = \displaystyle\int_0^{100} \displaystyle\int_0^{400} \displaystyle\int_{y/10-40}^0 (a - bz) \;\mathrm{d}z \,\mathrm{d}y\,\mathrm{d}x\) which can be evaluated as follows:

\[\begin{align*} 
M &=\displaystyle\int_0^{100} \displaystyle\int_0^{400} \displaystyle\int_{y/10-40}^0 (a - bz) \;\mathrm{d}z\,\mathrm{d}y\,\mathrm{d}x \\ 
&= \displaystyle\int_0^{100} \displaystyle\int_0^{400} \left[ az - \frac{b}{2}z^2 \right] _{y/10-40}^0 \;\mathrm{d}y\,\mathrm{d}x\\ 
& = \displaystyle\int_0^{100} \displaystyle\int_0^{400} \left[0 - a(y/10-40) + \frac{b}{2} (y/10-40)^2 \right] \;\mathrm{d}y\,\mathrm{d}x \\ 
&= \displaystyle\int_0^{100} \displaystyle\int_0^{400} \left[40a - \frac{a}{10}y + \frac{b}{200} y^2 - 4 b y + 800 b\right] \;\mathrm{d}y\,\mathrm{d}x \\ 
&=\displaystyle\int_0^{100} \left[40ay - \frac{a}{20}y^2 + \frac{b}{600} y^3 - 2 b y^2 + 800 b y\right]_0^{400} \;\mathrm{d}x \\ 
&= \displaystyle\int_0^{100} \left[16 000 a - 8 000 a + \frac{320000}{3}b - 320 000 b + 320 000 b\right] \;\mathrm{d}x\\
&=\displaystyle\int_0^{100} \left[8 000 a + \frac{320000}{3}b \right] \;\mathrm{d}x \\
&=8 \times 10^5 a + \frac{3.2}{3} \times 10^7 b = 7.984 \times 10^8 + \frac{0.16}{3}\times 10^7 = 7.989 \times 10^8\ {\rm kg}\end{align*}\] So the mass of water in the reservoir is \(7.989 \times 10^8\) kg.<br>Notes :</p>
<ol>
<li>In practice, the profile of the reservoir would not be rectangular and the depth would not vary so smoothly.</li>
<li>The variation of the density of water with height is only a minor factor so it would only be taken into account when a very exact answer was required. Assuming that the water had a uniform density of \(\rho = 998\) kg m\(^{-3}\) would give a total mass of \(7.984 \times 10^8\) kg while assuming a uniform density of \(\rho = 1 000\) kg m\(^{-3}\) gives a total mass of \(8 \times 10^8\) kg.</li>
</ol>]]></text>
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