Question

The water supply of a building is fed through a main pipe $6.00cm$ in diameter. A $2.00cm$ diameter faucet tap, located $2.00m$ above the main pipe, is observed to fill a $25.0L$ container in $30.0s$. (a) what is the speed at which the water leaves the faucet? (b) What is the gauge pressure in the $6.00cm$ main pipe? Assume the faucet is the only “leak” in the building.

Short answer

(a) The speed at which the water leaves the faucet is $v=2.65m/s$

(b) The gauge pressure in the $6.00cm$ main pipe is$P_{gauge}=2.31\times {10}^{4}Pa$.

Step-by-step solution

A concept:

The flow rate (volume flux) through a pipe that varies in cross-sectional area is constant; that is equivalent to stating that the product of the cross-sectional area A and the speed v at any point is a constant. This result is expressed in the equation of continuity for fluids:

$A_1{V}_1-A_2{V}_2-cons\tan t$

The sum of the pressure, kinetic energy per unit volume, and gravitational potential energy per unit volume have the same value at all points along a streamline for an ideal fluid. This result is summarized in Bernoulli’s equation:

$P+\rho gy+\frac{1}{2}\rho v^2=cons\tan t$

Here, P is the pressure, $\rho$ is the density, g is the gravity, and y is the distance.

Known data:

Consider the known data as below.

Density of water, $\rho =1000kg/m^3$

Acceleration due to gravity, $g=9.8m/s^2$

The height,$y_2-y_1$

Volume, $V=25.0L$

Diameter, $d_1=6.00cm$

Diameter, $d_2=2.00cm$

The speed at which the water leaves the faucet:

The flow rate, $Av$, as given may be expressed as follows:

Flow rate $=\frac{25.0liters}{30.0s}=0.833liters/s$

$=833c{m}^3/s$

The area of the faucet tap is $\mathrm{\pi }{\mathrm{cm}}^2$, so you can find the velocity as,

$$\begin{gathered} V_2=\frac{flowrate}{A} \\ =\frac{833c{m}^3/s}{\mathrm{\pi }{\mathrm{cm}}^2} \\ =265cm/s \\ =2.65m/s \end{gathered}$$

(b) The gauge pressure in the 6.00 cm  main pipe:

Choose point 1 to be in the entrance pipe and point 2 to be at the faucet tap.

$A_1{V}_1=A_2{V}_2=cons\tan t$

The speed of the water in the larger cylinder is,

$V_1=\frac{A_2}{A_1}{V}_2$

data-custom-editor="chemistry" $$\begin{gathered} \left(\frac{{\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4^{{\mathrm{d}}_2^2}$}\right.}}{{\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4^{{\mathrm{d}}_1^2}$}\right.}}\right)V_2 \\ {\left(\frac{d_2}{d_1}\right)}^2{V}_2 \end{gathered}$$

Substitute known values in the above equation.

$$\begin{gathered} V_1={\left(\frac{2.00cm}{6.00cm}\right)}^2\times 2.65m/s \\ =0.29m/s \end{gathered}$$

Bernoulli’s equation is:

$$\begin{gathered} P_1+\rho g{y}_1+\frac{1}{2}\rho v_1^2=P_2+\rho g{y}_2+\frac{1}{2}\rho v_2^2 \\ {P}_1-P_2=\frac{1}{2}\rho (v_2^2-v_1^2)+\rho g(y_2-y_1) \end{gathered}$$

By substituting known values gives,

$$\begin{gathered} P_1-P_2=(1000kg/m^3){(2.65m/s)}^2-{(0.295m/s)}^2+(1000kg/m^3)(9.8m/s^2)(2.00m) \\ =2.31\times {10}^{4}Pa \\ =P_{gauge} \end{gathered}$$