Chapter 13, Problem 64P

(a)

To determine

To Show:The speed of the root beer leaving the spigot is approximately $\sqrt{2gh}$ .

Explanation of Solution

Given:

Height of the root beer is $h$

Calculation:

Let us apply the Bernoulli’s equation to the beer at the top of the keg and at the spigot,

  $P_1+{\rho }_{beer}g{h}_1+\frac{1}{2}{\rho }_{beer}{v}_1^2=P_2+{\rho }_{beer}g{h}_2+\frac{1}{2}{\rho }_{beer}{v}_2^2$$\left(1\right)$

Where, $P_1$ and $P_2$ are the pressure at the top of the keg and at the spigot respectively,

  ${\rho }_{beer}$ is density of the beer,

  $g$ is acceleration due to gravity,

  $h_1$ and $h_2$ are height of the keg and spigot respectively

  $v_1$ is speed of the root beer at the top of the keg,

  $v_2$ is speed of the root beer leaving the spigot,

Since, $v_1\approx 0$ , $h_2=0$ , $P_1=P_2=P_{atm}$ and $h_1=h$ , equation $\left(1\right)$ becomes,

  $\begin{array}{l}gh=\frac{1}{2}{v}_2^2\\ \Rightarrow v_2^2=2gh\\ \Rightarrow v_2=\sqrt{2gh}\end{array}$

Conclusion:

Thus, the speed of the root beer leaving the spigot is approximately $\sqrt{2gh}$ .

(b)

To determine

To Show:If

  $A_2\ll A_1$ , the rate of change of the height $h$ of the root beer is given by $\frac{dh}{dt}=-\left(\frac{A_2}{A_1}\right){\left(2gh\right)}^{\frac{1}{2}}$ .

Explanation of Solution

Given:

Height of keg is $H$

Cross sectional area of keg is $A_1$

Area of spigot opening is $A_2$

Calculation:

Let, $v_1$ be the speed of the root beer at the top of the keg, and $v_2$ be the speed of the root beer leaving the spigot.

Let us relate the speeds $v_1$ and $v_2$ using the continuity equation, that is

  $A_1{v}_1=A_2{v}_2$

Substituting $-\frac{dh}{dt}$ for $v_1$ and $\sqrt{2gh}$ For

  $v_2$ ,

  $-\frac{dh}{dt}{A}_1=A_2\sqrt{2gh}$

  $\Rightarrow \frac{dh}{dt}=-\frac{A_2\sqrt{2gh}}{A_1}$

Conclusion:

Thus, the speed of the root beer leaving the spigot is approximately $\sqrt{2gh}$ .

(c)

To determine

To Find: $h$ as a function of time.

Answer to Problem 64P

  $h={\left(\sqrt{H}+\frac{t{A}_2\sqrt{2g}}{2A_1}\right)}^2$

Explanation of Solution

Given:

  $h=H$ at $t=0$

Calculation:

Consider the equation,

  $\frac{dh}{dt}=-\frac{A_2\sqrt{2gh}}{A_1}$

Separating the variables,

  $\frac{\raisebox{1ex}{$-A_1$}\!\left/ \!\raisebox{-1ex}{$A_2$}\right.}{\sqrt{2g}}\frac{dh}{\sqrt{h}}=dt$

Taking the integral from the limits, $H$ to $h$ and $0$ to $t$ ,

  $\frac{\raisebox{1ex}{$-A_1$}\!\left/ \!\raisebox{-1ex}{$A_2$}\right.}{\sqrt{2g}}{\displaystyle \underset{H}{\overset{h}{\int }}\frac{dh}{\sqrt{h}}}={\displaystyle \underset{0}{\overset{t}{\int }}dt}$

  $\Rightarrow \frac{\raisebox{1ex}{$-2A_1$}\!\left/ \!\raisebox{-1ex}{$A_2$}\right.}{\sqrt{2g}}\left[\sqrt{H}-\sqrt{h}\right]=t$

  $\Rightarrow \sqrt{H}-\sqrt{h}=-\frac{t{A}_2\sqrt{2g}}{2A_1}$

  $\Rightarrow \sqrt{h}=\sqrt{H}+\frac{t{A}_2\sqrt{2g}}{2A_1}$

  $\Rightarrow h={\left(\sqrt{H}+\frac{t{A}_2\sqrt{2g}}{2A_1}\right)}^2$

Conclusion:

  $h$ as a function of time is $h={\left(\sqrt{H}+\frac{t{A}_2\sqrt{2g}}{2A_1}\right)}^2$

(d)

To determine

To Find:The total time needed to drain the keg.

Answer to Problem 64P

  $\text{1h 46 min}$

Explanation of Solution

Given:

  $H=2.0$ m

  $A_1=0.8$ m²

  $A_2=1\times {10}^{-4}{A}_1$

Formula used:

Consider the time dependent expression for $h$ ,

  $h={\left(\sqrt{H}+\frac{t{A}_2\sqrt{2g}}{2A_1}\right)}^2$

Solving this expression for the time to drain $\left(t\text{'}\right)$ ,

  $t\text{'}=\frac{A_1}{A_2}\sqrt{\frac{2H}{g}}$$\left(2\right)$

Calculation:

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

  $\begin{array}{l}t\text{'}=\frac{A_1}{1\times {10}^{-4}{A}_1}\sqrt{\frac{\text{2}\left(\text{2}\text{.00}\text{m}\right)}{\text{9}\text{.8}{\text{m/s}}^{\text{2}}}}\\ \text{=}6.39\times {10}^3\text{s}\\ \text{= 1h 46 min}\end{array}$

Conclusion:

The total time needed to drain the keg is $\text{1h 46 min}$ .