A water pipe having a $\mathbf{2}\mathbf{.}\mathbf{5}\mathit{c}\mathit{m}$inside diameter carries water into the basement of a house at a speed of $\mathbf{0}\mathbf{.}\mathbf{90}\mathit{m}\mathbf{/}\mathit{s}$and a pressure of $\mathbf{170}\mathit{k}\mathit{P}\mathit{a}$.

(a) If the pipe tapers to $\mathbf{1}\mathbf{.}\mathbf{2}\mathit{c}\mathit{m}$and rises to the second floor $\mathbf{7}\mathbf{.}\mathbf{6}\mathit{m}$above the input point, what is the speed and

(b) If the pipe tapers to $\mathbf{1}\mathbf{.}\mathbf{2}\mathit{c}\mathit{m}$and rises to the second floor $\mathbf{7}\mathbf{.}\mathbf{6}\mathit{m}$above the input point, what is the water pressure at the second floor?

1. The speed of the water at the basement level, $v_b=0.90m/s$.

2. The diameter of the pipe at the basement level, $d_b=2.5cm=2.5\times {10}^{-2}m$.

3. The pressure at the basement level, $p_b=170kPa=170\times {10}^{3}Pa$.

4. The diameter of the pipe at the second floor, $d_s=1.2cm=1.2\times {10}^{-2}m$.

5. The height of the second floor, $h_s=7.6m$.

Determine the speed of the water at the second floor using the equation of continuity. Then, using Bernoulli’s equation, find the pressure of water on the second floor. According to Bernoulli’s equation, as the speed of a moving fluid increases, the pressure within the fluid decreases.

Formulae are as follows:

$pv+\frac{1}{2}\rho g^{2}h=cons\tan t$

$Av=cons\tan t$

Where, $p$is pressure, $v$is velocity,$h$is height, $g$is the acceleration due to gravity,$h$is height, $A$is the area, and $\rho$is density.

The water flow through the pipe obeys the equation of continuity,

$A_b{V}_b=A_s{V}_s$

Hence,

$$\begin{gathered} v_s=\frac{A_b{V}_b}{A_s} \\ =\frac{{d^2}_b{V}_b}{\left({d^2}_s\right)} \\ =\frac{{\left(2.5\times {10}^{-2}m\right)}^2\times 0.90m/s}{{\left(1.2\times {10}^{-2}m\right)}^2} \\ =3.9m/s \end{gathered}$$

Hence, the speed of the water at the second floor is$3.9m/s$.

The water flow obeys Bernoulli’s principle as well,

$p_{b}v+\frac{1}{2}\rho g{h^2}_b+p_b={\rho }_{s}v+\frac{1}{2}\rho g{h^2}_s$

ρ=1000kg/m3Considering, $h_b=0m$, the density of water,$\rho =1000kg/m^3$,

ρ=1000kg/m3$$\begin{gathered} p_{s}v=p_{b}v+\frac{1}{2}(p{g^2}_b+{h^2}_s)h+({}_b-{}_s) \\ =170\times {10}^{3}Pa+\frac{1}{2}\left(1000kg/m^3\right)\left({\left(0.90m/s\right)}^2-{\left(3.9m/s\right)}^2\right)+\left(1000kg/m^3\right)\left(9.8m/s^2\right)\left(0-7.6m\right) \\ =8.8\times {10}^{4}Pa \end{gathered}$$

ρ=1000kg/m3Hence, the pressure of water on the second floor is$8.8\times {10}^{4}Pa$