Question

Figure P14.43 on page 442 shows a stream of water in steady flow from a kitchen faucet. At the faucet, the diameter of the stream is $\mathbf{0}\mathbf{.}\mathbf{960}\mathit{c}\mathit{m}$. The stream fills a $\mathbf{125}\mathbf{-}\mathit{c}{\mathit{m}}^{\mathbf{3}}$ container in $\mathbf{16}\mathbf{.}\mathbf{3}\mathit{s}$. Find the diameter of the stream $\mathbf{13}\mathbf{.}\mathbf{0}\mathit{c}\mathit{m}$ below the opening of the faucet

Short answer

The diameter of the stream $13.0cm$ below the opening of the faucet is $d_2=0.247cm$

Step-by-step solution

Given data

$$\begin{gathered} ∆V=125c{m}^3 \\ ∆t=16.3s \end{gathered}$$

At the faucet, the diameter of the stream is $d_1=0.96cm$

Definition of Bernoulli’s equation

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

$P+\rho gy+\frac{\text{1}}{\text{2}}\rho v^2=\text{constant}$

Determining the diameter of the stream 13.0 cm  below the opening of the faucet

The volume flow rate is given by:

$$\begin{gathered} \frac{∆V}{∆t}=\frac{125c{m}^3}{16.3s} \\ =7.67c{m}^3/s \\ =A{v}_1 \end{gathered}$$

Where $d_1=0.96cm$, and $A={\mathrm{\pi r}}^2=0.724{\mathrm{cm}}^2$. The speed at the top of the falling column is

$$\begin{gathered} v_1=\frac{{\displaystyle \raisebox{1ex}{$∆V$}\!\left/ \!\raisebox{-1ex}{$∆t$}\right.}}{A} \\ =\frac{7.67c{m}^3/s}{0.724c{m}^3} \\ =10.6cm/s \end{gathered}$$

Take point 2 at 13.0cm below:

$$\begin{gathered} P_1+\rho g{y}_1+\frac{1}{2}\rho v_1^2=P_2+\rho g{y}_2+\frac{1}{2}\rho v_2^2 \\ {P}_0+(1000kh/m^3)(9.8m/s^2)0.130m+\frac{1}{2}(1000kg/m^3){(0.106m/s)}^2=P_0+0+\frac{1}{2}(1000kg/m^3)v_2^2 \end{gathered}$$

Solving for the velocity gives

$$\begin{gathered} v_2=\sqrt{2gh+v_1^2} \\ {v}_2=\sqrt{2\left(9.8{\text{m/s}}^{\text{2}}\right)\left(0.130\text{m}\right)+{\left(0.106\text{m/s}\right)}^2} \\ {v}_2=1.60\text{m/s} \end{gathered}$$

The volume flow rate is constant:

$$\begin{gathered} 7.67{\text{cm}}^{\text{3}}\text{/s}=\pi \frac{{d_2}^2}{4}\times 160\text{cm/s} \\ {d}_2=0.247\text{cm} \end{gathered}$$

So the diameter of the stream $13.0cm$ below the opening of the faucet is $d_2=0.247cm$