The sewage outlet of a house constructed on a slope is $\mathbf{6}\mathbf{.59}\mathbf{m}$below street level. If the sewer is localid="1657605982734" $\mathbf{2}\mathbf{.16}\mathbf{m}$below street level, find the minimum pressure difference that must be created by the sewage pump to transfer waste of average density localid="1657605999636" $\mathbf{1000}\mathbf{.}\mathbf{00}\mathit{k}\mathit{g}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}$from outlet to sewer.

• The depth of the sewer outlet of the house below the street level is$h_0=6.59\mathrm{m}$

• The depth of the sewer below the street level is$h_s=2.16\mathrm{m}$

• The density of the waste is$\rho =1000\mathrm{kg}/{\mathrm{m}}^3$.

A fluid is a substance that can flow; it conforms to the boundaries of its container because it cannot withstand shearing stress. It can, however, exert a force perpendicular to its surface. If the force is uniform over a flat area, then the force is described in terms of pressure p as,

$\mathit{p}\mathbf{=}\frac{\mathbf{F}}{\mathbf{A}}$

We can find the minimum differential that must be created by the savage pump to transfer waste by taking the pressure difference between the sewer outlet and the sewer.

The pressure at a point in a fluid in static equilibrium is $p=p_0+\rho gh$

The minimum differential that must be created by the savage pump to transfer waste is equal to the pressure difference between the sewer outlet and sewer. The street is at atmospheric pressure.

So, the pressure at the sewer outlet below the street level is

$p^{\mathrm{\prime }}{p}_0+\rho g{h}_0$

The pressure at the sewer below the street level is

$p=p_0\mid \rho g{h}_s$

Hence, the pressure difference between sewer outlet and sewer is,

$\begin{array}{r}p-p^{\mathrm{\prime }}-p\\ =p_0+\rho g{h}_o-p_0+\rho g{h}_s\\ -\rho g{h}_0-\rho g{h}_s\\ =\rho g\left(h_0-h_s\right)\end{array}$

Substitute the values in the above equation.

$\begin{array}{r}=\left(1000\mathrm{kg}/{\mathrm{m}}^3\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)(6.59\mathrm{m}-2.16\mathrm{m})\\ =43414\mathrm{Pa}\\ \approx 4.34\times {10}^4\mathrm{Pa}\end{array}$

Therefore, the pressure difference between sewer outlet and sewer is $4.34\times {10}^4\mathrm{Pa}$.