Chapter 13, Problem 67P

(a)

To determine

The speed of the water.

Answer to Problem 67P

  $v_i=9.28\times {10}^{-2}\text{m/s}$

Explanation of Solution

Given:

Temperature of water is $20°\text{C}$

Flow rate of water is $10.5{\text{cm}}^{\text{3}}\text{/s}$

Diameter of the tap is 1.20 cm

Formula used:

Let us represent the flow rate of the water as $I$ ,

Now we can relate the flow rate of water to the area of cross section of the circular tap $\left(A\right)$ and the speed of the water $\left(v_i\right)$ through the expression,

  $I=A{v}_i$$\left(1\right)$

Area of cross section of circular tap interms of its diameter $\left(d\right)$ is

  $\begin{array}{l}A=\pi r^2\\ A=\pi {\left(\frac{d}{2}\right)}^2\\ A=\frac{1}{4}\pi d^2\end{array}$

Substituting for $A$ in equation $\left(1\right)$ ,

  $\begin{array}{l}I=\frac{1}{4}\pi d^2{v}_i\\ \Rightarrow v_i=\frac{I}{\frac{1}{4}\pi d^2}\end{array}$$\left(2\right)$

Calculation:

Substituting the numerical values in equation $\left(2\right)$ ,

  $\begin{array}{c}{v}_i=\frac{\text{10}{\text{.5 cm}}^{\text{3}}\text{/s}}{\frac{\text{1}}{\text{4}}\text{π}{\left(\text{1}\text{.20 cm}\right)}^{\text{2}}}\\ v{i}_{}=9.28\times {10}^{-2}\text{m/s}\end{array}$

Conclusion:

The speed of the water is $9.28\times {10}^{-2}\text{m/s}$ .

(b)

To determine

The diameter of the stream at a point $7.50$ cm below the tap.

Answer to Problem 67P

  $d_f=3.31\times {10}^{-3}$ m.

Explanation of Solution

Given:

Temperature of water is $20°\text{C}$

Flow rate of water is $10.5{\text{cm}}^{\text{3}}\text{/s}$

Stream has circular cross section

Neglect any effects of drag forces acting on the water

Formula used:

Let us apply the continuity equation to the stream of water,

  $A_f{v}_f=A_i{v}_i$$\left(3\right)$

Where, $A_f$ is cross sectional area of the stream of water at $7.50$ cm below the tap,

  $v_f$ is speed of the stream of water at $7.50$ cm below the tap,

  $A_i$ Area of cross section of circular tap

  $v_i$ is initial speed of water

Interms of diameter, cross sectional areas can be written as,

  $A_f=\frac{1}{4}\pi d_f^2$ and $A_i=\frac{1}{4}\pi d_i^2$

Substituting these in equation $\left(3\right)$ ,

  $\frac{1}{4}\pi d_f^2{v}_f=\frac{1}{4}\pi d_i^2{v}_i$

  $\Rightarrow d_f^2{v}_f=d_i^2{v}_i$

  $\Rightarrow d_f^2=\frac{d_i^2{v}_i}{v_f}$

  $\Rightarrow d_f=d_i\sqrt{\frac{v_i}{v_f}}$$\left(4\right)$

Now correlate $v_f$ and $v_i$ to the distance fallen by water ( $\Delta h$ ), using a constant acceleration equation,

  $v_f^2=v_i^2+2g\Delta h$

  $\Rightarrow v_f=\sqrt{v_i^2+2g\Delta h}$$\left(5\right)$

  $g$ is acceleration due to gravity, $g=9.81$ m/s²

Calculation:

Substituting the numerical values in equation $\left(5\right)$ ,

  $v_f=\sqrt{{\left(\text{9}{\text{.28×10}}^{\text{-2}}\text{m/s}\right)}^{\text{2}}\text{+2}\left(\text{9}{\text{.81m/s}}^{\text{2}}\right)\left(\text{7}{\text{.50×10}}^{\text{-2}}\text{m}\right)}$

  $=1.22\text{m/s}$

Substituting the numerical values in equation $\left(4\right)$ ,

  $\begin{array}{l}{d}_f=\left(\text{1}{\text{.20×10}}^{\text{-2}}\text{m}\right)\sqrt{\frac{\text{9}{\text{.28×10}}^{\text{-2}}\text{m/s}}{\text{1}\text{.22m/s}}}\\ d_f\text{=}3.31\times {10}^{-3}\text{m}\end{array}$

Conclusion:

The diameter of the stream at a point $7.50$ cm below the tap is $3.31\times {10}^{-3}$ m.

(c)

To determine

The distance does the water have to fall before it becomes to turbulent and to see whether this match our everyday observations.

Answer to Problem 67P

  $\Delta d=0.76\text{m}$

Explanation of Solution

Given:

Temperature of water is $20°\text{C}$

Flow rate of water is $10.5{\text{cm}}^{\text{3}}\text{/s}$

Reynolds number $\left(N_R\right)$ for turbulent flow is above $2300$ .

Formula used:

Correlating the fall-distance to-turbulence $\left(\Delta d\right)$ to its initial speed $\left(v_i\right)$ and its speed $\left(v_t\right)$ when its flow becomes turbulent, using a constant-acceleration equation,

  $\begin{array}{l}{v}_t^2=v_i^2+2g\Delta d\\ \Rightarrow 2g\Delta d=v_t^2-v_i^2\end{array}$

  $\Rightarrow \Delta d=\frac{v_t^2-v_i^2}{2g}$$\left(6\right)$

Express Reynolds number $N_R$ for turbulent flow:

  $N_R=\frac{2r\rho v_t}{\eta }$$\left(7\right)$

Where $\eta$ is viscosity of water, $\eta =1.00\times {10}^{-3}\text{Pa}\cdot \text{s}$

The volume flow rate equals to $Av$ . Now express the volume flow rate at the speed $v_t$ ,

  $I=\pi r^2{v}_t$$\left(8\right)$

Eliminating $r$ from equations $\left(7\right)$ and $\left(8\right)$ , and solve for $v_t$ :

  $v_t=\frac{\pi N_R^2{\eta }^2}{4{\rho }^{2}I}$$\left(9\right)$

Calculation:

Substituting the numerical values in equation $\left(9\right)$ ,

  $\begin{array}{l}{v}_t=\frac{\text{π}{\left(\text{2300}\right)}^{\text{2}}{\left(\text{1}{\text{.00×10}}^{\text{-3}}\text{Pa}\text{.s}\right)}^{\text{2}}}{\text{4}{\left({\text{1000kg/m}}^{\text{3}}\right)}^2\left(\text{10}{\text{.5×10}}^{\text{-6}}{\text{m}}^{\text{3}}\text{/s}\right)}\\ =0.396\text{m/s}\end{array}$

Substituting the numerical values in equation $\left(6\right)$ ,

  $\begin{array}{l}\Delta d=\frac{{\left(\text{0}\text{.396m/s}\right)}^{\text{2}}\text{-}{\left(\text{9}{\text{.28×10}}^{\text{-2}}\text{m/s}\right)}^{\text{2}}}{\text{2}\left(\text{9}{\text{.81m/s}}^{\text{2}}\right)}\\ \Delta d=0.76\text{m}\end{array}$

This value is in reasonable agreement with everyday life.

Conclusion:

The distance does the water have to fall before it becomes to turbulent is $0.76\text{m}$ .

This value is in reasonable agreement with everyday life.