Answer to Question #259200 in Mechanics | Relativity for ساره

Question

Solar ponds are small artificial lakes of a few meters deep that are

used to store solar energy. The rise of heated (and thus less dense)

water to the surface is prevented by adding salt at the pond bottom.

In a typical salt gradient solar pond, the density of water increases in

the gradient zone, as shown in Fig. 1–52, and the density can be

expressed as

where ρ° is the density on the water surface, s is the vertical distance

measured downward from the top of the gradient zone (s 5 2z), and

H is the thickness of the gradient zone. For H 5 4 m, r0 5 1040 kg/m3,

and a thickness of 0.8 m for the surface zone, calculate the gage

pressure at the bottom of the gradient zone

Answer 1

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

P₁​=ρgh₁​=(1040  kg/m³)(9.81  m/s²)(0.8  m)(1 kN/1000  kg⋅m/s²​)=8.16 kPa

since 1 kN/m² = 1 kPa. Since s = −z, the differential change in hydrostatic pressure across a vertical distance of ds is given by

dP=ρg ds

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

P−P₁​=$\int^s_0$ ​ρg ds→P=P₁​+$\int^s_0$ ρ₀​$\sqrt{1+tan^2(\frac{\pi s}{4 H}​)g}ds$ ds

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

P=P₁​+ρ₀​g$\frac{4H}{\pi}$ ​sinh⁻¹(tan$\frac{\pi s}{4H}$ ​)

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

P₂​=8.16 kPa+(1040 kg/m³)(9.81 m/s²)$\frac{4(4m)}{\pi}$ ​sinh⁻¹(tan$\frac{4\pi}{4\cdot4}$ ​)(1 kN/1000 kg⋅m/s²​)=54.0 kPa (gage)

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