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Chapter 2: Problem 12

Determine the pressure at the bottom of an open 5 -m-deep tank in which a chemical process is taking place that causes the density of the liquid in the tank to vary as $$\rho=\rho_{\text {surf }} \sqrt{1+\sin ^{2}\left(\frac{h}{h_{\text {bot }}} \frac{\pi}{2}\right)}$$ where \(h\) is the distance from the free surface and \(\rho_{\text {surf }}=1700 \mathrm{kg} / \mathrm{m}^{3}\).

Short Answer

Expert verified
The pressure at the bottom of the tank is represented by the equation \( P = g \int_0^{5} 1700 \sqrt{1+\sin^{2}\left(\frac{h}{5} \frac{\pi}{2}\right)} \, dh \). Solving this integral equation will provide the precise value.

Step by step solution

01

Understand the Given Information

Here the density of the liquid in the tank is given by the equation \(\rho=\rho_{\text {surf }} \sqrt{1+\sin^{2}\left(\frac{h}{h_{\text {bot }}} \frac{\pi}{2}\right)}\) where \(h\) is the distance from the free surface and \(\rho_{\text {surf }}=1700 \mathrm{kg} / \mathrm{m}^{3}.\) This equation indicates that the density of the liquid changes as the distance from the surface changes, with the highest density at the bottom and the lowest at the top.
02

Apply the Integral of the Density over Depth

To find the pressure at the bottom, calculate the integral of the density from the surface (\(h=0\)) to the bottom of the tank (\(h=h_{\text{bot}}.\)) This gives the equation for pressure as \( P = g \int_0^{h_{\text {bot }}} \rho \, dh \). Here, \(g\) is the acceleration due to gravity, assumed as \(9.81 m/s^2\). We already know that the depth of the tank is 5 m.
03

Substitute the Values and Solve

Substitute the given values and solve the equation. \( P = g \int_0^{5} 1700 \sqrt{1+\sin^{2}\left(\frac{h}{5} \frac{\pi}{2}\right)} \, dh \). Solving this integral using appropriate methods will give the pressure at the bottom of the tank.

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