Question 1.10: Hydrostatic Pressure in a Solar Pond with Variable Density S...

The variation of density of saline water in the gradient zone of a solar pond with depth is given. The gage pressure at the bottom of the gradient zone is to be determined.
Assumptions The density in the surface zone of the pond is constant.
Properties The density of brine on the surface is given to be 1040 kg/m³.
Analysis We label the top and the bottom of the gradient zone as 1 and 2, respectively. Noting that the density of the surface zone is constant, the gage pressure at the bottom of the surface zone (which is the top of the gradient zone) is

P_{1}=\rho g h_{1}=\left(1040 kg / m ^{3}\right)\left(9.81 m / s ^{2}\right)(0.8 m )\left(\frac{1 kN }{1000 kg \cdot m / s ^{2}}\right)=8.16 kPa

since 1 kN/m² = 1 kPa. The differential change in hydrostatic pressure across a vertical distance of dz is given by

d P=\rho g d z

Integrating from the top of the gradient zone (point 1 where z = 0) to any location z in the gradient zone (no subscript) gives

P-P_{1}=\int_{0}^{z} \rho g d z \quad \rightarrow \quad P=P_{1}+\int_{0}^{z} \rho_{0} \sqrt{1+\tan ^{2}\left(\frac{\pi}{4} \frac{z}{H}\right)} g d z

Performing the integration gives the variation of gage pressure in the gradient zone to be

P=P_{1}+\rho_{0} g \frac{4 H}{\pi} \sinh ^{-1}\left(\tan \frac{\pi}{4} \frac{z}{H}\right)

Then the pressure at the bottom of the gradient zone (z = H = 4 m) becomes

\begin{aligned}P_{2}=& 8.16 kPa +\left(1040 kg / m ^{3}\right)\left(9.81 m / s ^{2}\right) \frac{4(4 m )}{\pi} \\& \times \sinh ^{-1}\left(\tan \frac{\pi}{4} \frac{4}{4}\right)\left(\frac{1 kN }{1000 kg \cdot m / s ^{2}}\right) \\=& 5 4 . 0 kPa (\text { gage })\end{aligned}

Discussion The variation of gage pressure in the gradient zone with depth is plotted in Fig. 1–56. The dashed line indicates the hydrostatic pressure for the case of constant density at 1040 kg/m³ and is given for reference. Note that the variation of pressure with depth is not linear when density varies with depth.