Chapter 9, Problem 117P

A section of pipe with an internal diameter of 10.0 cm tapers to an inner diameter of 6.00 cm as it rises through a height of 1.70 m at an angle of 60.0° with respect to the horizontal. The pipe carries water and its higher end is open to air. (a) If the speed of the water at the lower point is 15.0 cm/s, what are the pressure at the lower end and the speed of the water as it exits the pipe? (b) If the higher end of the pipe is 0.300 m above ground, at what horizontal distance from the pipe out let does the water land?

To determine

The pressure at the lower end and the speed of the water as it exits the pipe.

Answer to Problem 117P

The pressure at the lower end is $118\text{ kPa}$ and the speed of the water as it exits the pipe is $41.7\text{ cm/s}$ .

Explanation of Solution

Take point 1 to be the lower end and the point 2 to be upper end of the pipe.

Write the continuity equation of fluids.

  $A_1{v}_1=A_2{v}_2$        (I)

Here, $A_1$ is the area through which the liquid moves at point 1 and $v_1$ is the speed of the liquid at point 1, $A_2$ is the area through which the liquid moves at point 2 and $v_2$ is the speed of the liquid at point 2.

Rewrite the above equation for $v_2$ .

  $v_2=\frac{A_1}{A_2}{v}_1$        (II)

Write the equation for $A_1$ .

  $A_1=\frac{\pi d_1^2}{4}$

Here, $d_1$ is the dimeter of the pipe at point 1.

Write the equation for $A_2$ .

  $A_2=\frac{\pi d_2^2}{4}$

Here, $d_2$ is the dimeter of the pipe at point 2.

Put the above two equations in equation (II).

  $\begin{array}{c}{v}_2=\frac{\frac{\pi d_1^2}{4}}{\frac{\pi d_2^2}{4}}\\ =\frac{d_1^2}{d_2^2}\\ ={\left(\frac{d_1}{d_2}\right)}^2\end{array}$        (III)

Write the Bernoulli’s equation.

  $P_1+\frac{1}{2}\rho v_1^2+\rho g{y}_1=P_2+\frac{1}{2}\rho v_2^2+\rho g{y}_2$

Here, $P_1$ is the pressure at the point 1, $\rho$ is the density of the liquid, $g$ is the acceleration due to gravity, $y_1$ is the height of the point 1 above the reference position, $P_2$ is the pressure at the point 2 and $y_2$ is the height of the point 2 above the reference position.

Substitute $P_{\text{atm}}$ for $P_1$ in the above equation and rewrite it for $P_2$ .

  $\begin{array}{c}{P}_{\text{atm}}+\frac{1}{2}\rho v_1^2+\rho g{y}_1=P_2+\frac{1}{2}\rho v_2^2+\rho g{y}_2\\ P_2=P_{\text{atm}}+\frac{1}{2}\rho \left(v_1^2-v_2^2\right)+\rho g\left(y_1-y_2\right)\\ =P_{\text{atm}}+\rho \left[\frac{1}{2}\left(v_1^2-v_2^2\right)+g\left(y_1-y_2\right)\right]\end{array}$        (IV)

Here, $P_{\text{atm}}$ is the atmospheric pressure.

Conclusion:

The atmospheric pressure is $101.3\text{ kPa}$ , value of $g$ is $9.80{\text{ m/s}}^2$ and the density of water is $1000{\text{ kg/m}}^3$ .

Substitute $10\text{ cm}$ for $d_1$ , $6.00\text{ cm}$ for $d_2$ and $15.0\text{ cm/s}$ for $v_1$ in equation (III) to find $v_2$ .

  $\begin{array}{c}{v}_2=\frac{10\text{ cm}}{6.00\text{ cm}}\left(15.0\text{ cm/s}\right)\\ =41.7\text{ cm/s}\end{array}$

Substitute $101.3\text{ kPa}$ for $P_{\text{atm}}$ , $1000{\text{ kg/m}}^3$ for $\rho$ , $15.0\text{ cm/s}$ for $v_1$ , $41.7\text{ cm/s}$ for $v_2$ , $9.80{\text{ m/s}}^2$ for $g$ and $1.70\text{ m}$ for $\left(y_1-y_2\right)$ in equation (IV) to find $P_2$ .

  $\begin{array}{c}{P}_2=101.3\text{ kPa}+\left(1000{\text{ kg/m}}^3\right)\left[\frac{1}{2}\left({\left(15.0\text{ cm/s}\frac{1\text{ m}}{100\text{ cm}}\right)}^2-{\left(41.7\text{ cm/s}\frac{1\text{ m}}{100\text{ cm}}\right)}^2\right)+\left(9.80{\text{ m/s}}^2\right)\left(1.70\text{ m}\right)\right]\\ =118\text{ kPa}\end{array}$

Therefore, the pressure at the lower end is $118\text{ kPa}$ and the speed of the water as it exits the pipe is $41.7\text{ cm/s}$ .

To determine

The horizontal distance from the pipe outlet where the water lands.

Answer to Problem 117P

The horizontal distance from the pipe outlet where the water lands is $5.98\text{ cm}$ .

Explanation of Solution

Write the equation for the horizontal distance from the pipe outlet where the water lands.

  $\Delta x=v_x\Delta t$        (V)

Here, $\Delta x$ is the horizontal distance from the pipe outlet where the water lands, $v_x$ is the velocity of water in horizontal direction and $\Delta t$ is the time taken.

Write the equation for $v_x$ .

  $v_x=v_2\cos \theta$

Here, $\theta$ is the angle made by the pipe with the horizontal.

Put the above equation in equation (V).

  $\Delta x=v_2\cos \theta \Delta t$        (VI)

Write the equation for the vertical distance through which the water falls.

  $\Delta y=v_y\Delta t+\frac{1}{2}{a}_y{\left(\Delta t\right)}^2$

Here, $\Delta y$ is the vertical distance through which the water falls, $v_y$ is the $y$ component of the velocity and $a_y$ is the $y$ component of the acceleration.

The value of $a_y$ will be the negative of the acceleration due to gravity since the upward direction is taken to be $+y$ direction.

Substitute $-g$ for $a_y$ in the above equation.

  $\begin{array}{c}\Delta y=v_y\Delta t+\frac{1}{2}\left(-g\right){\left(\Delta t\right)}^2\\ =v_y\Delta t-\frac{1}{2}g{\left(\Delta t\right)}^2\end{array}$        (VII)

Write the equation for $v_y$ .

  $v_y=v_2\sin \theta$

Put the above equation in equation (VII).

  $\begin{array}{c}\Delta y=v_2\sin \theta \Delta t-\frac{1}{2}g{\left(\Delta t\right)}^2\\ 2\Delta y=2v_2\sin \theta \Delta t-g{\left(\Delta t\right)}^2\\ g{\left(\Delta t\right)}^2-2v_2\sin \theta \Delta t+2\Delta y=0\end{array}$

The above equation is a quadratic equation in $\Delta t$ .

Write the equation for the root of the above quadratic equation.

  $\begin{array}{c}\Delta t=\frac{-\left(-2v_2\sin \theta \right)\pm \sqrt{{\left(-2v_{2}sin\theta \right)}^2-4\left(g\right)\left(2\Delta y\right)}}{2g}\\ =\frac{2v_2\sin \theta \pm \sqrt{4v_2^{2}si{n}^2\theta -8g\Delta y}}{2g}\end{array}$        (VIII)

Conclusion:

Substitute $41.7\text{ cm/s}$ for $v_2$ , $60.0°$ for $\theta$ , $9.80{\text{ m/s}}^2$ for $g$ and $-0.300\text{ m}$ for $\Delta y$ in equation (VIII) to find $\Delta t$ .

  $\begin{array}{c}\Delta t=\frac{2\left(41.7\text{ cm/s}\frac{1\text{ m}}{100\text{ cm}}\right)\sin 60.0°\pm \sqrt{4{\left(41.7\text{ cm/s}\frac{1\text{ m}}{100\text{ cm}}\right)}^{2}si{n}^{2}60.0°-8\left(9.80{\text{ m/s}}^2\right)\left(-0.300\text{ m}\right)}}{2\left(9.80{\text{ m/s}}^2\right)}\\ =0.287\text{ s or }-0.213\text{ s}\end{array}$

Time cannot be negative.

Substitute $41.7\text{ cm/s}$ for $v_2$ , $60.0°$ for $\theta$ and $0.287\text{ s}$ for $\Delta t$ in equation (VI) to find $\Delta x$ .

  $\begin{array}{c}\Delta x=\left(41.7\text{ cm/s}\right)\cos 60.0°\left(0.287\text{ s}\right)\\ =5.98\text{ cm}\end{array}$

Therefore, the horizontal distance from the pipe outlet where the water lands is $5.98\text{ cm}$ .