Question

Water flows steadily from an open tank as shown in the figure. (Figure 1) The elevation of point 1 is 10.0m , and the elevation of points 2 and 3 is 2.00 m . The cross-sectional area at point 2 is 4.80x10^-2m ; at point 3, where the water is discharged, it is 1.60?10^?2m. The cross-sectional area of the tank is very large compared with the cross-sectional area of the pipe. Part A Assuming that Bernoulli's equation applies, compute the volume of water DeltaV that flows across the exit of the pipe in 1.00 s . In other words, find the discharge rate \Delta V/Delta t. Express your answer numerically in cubic meters per second.

Answer 1

Concepts and reason

The concepts required to solve the given questions is the speed of efflux and the continuity equation.

Initially, calculate the discharge area. Later, use the continuity equation. Finally, by using the continuity equation calculate the gauge pressure at point 2.

Fundamentals

The expression for the speed of efflux is as follows:

$v = \sqrt {2gh}$

Here, g is the acceleration due to gravity and h is the height.

The height of the efflux is as follows:

$h = {y_1} - {y_2}$

Here, ${y_1}$ is the elevation of the point 1 and ${y_2}$ is the elevation at point 2.

Therefore, the speed of efflux is as follows:

$v = \sqrt {2g\left( {{y_1} - {y_2}} \right)}$

The expression for the discharge charge is as follows:

${Q_3} = {v_3}{A_3}$

Here, ${v_3}$ is the speed of efflux and ${A_3}$ is the area at point 3, where the water is discharged.

The expression for the equation of continuity is as follows:

${A_2}{v_2} = {A_3}{v_3}$

Here, ${A_2}$ is the area at point 2 and ${v_2}$ is the speed of efflux.

The expression to calculate the gauge pressure at point 2 is as follows:

${P_2} = \frac{1}{2}\rho \left( {v_3^2 - v_2^2} \right)$

Here, $\rho$ is the volume charge density.

Substitute $9.8{\rm{ m/}}{{\rm{s}}^2}$ for g, 10.0 m for ${y_1}$ , and 2.00 m for ${y_2}$ in the equation $v = \sqrt {2g\left( {{y_1} - {y_2}} \right)}$ .

$\begin{array}{c}\\v = \sqrt {2\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {10.0{\rm{ m}} - 2.00{\rm{ m}}} \right)} \\\\ = 12.52{\rm{ m/s}}\\\end{array}$

Now, calculate the discharge charge using the equation ${Q_3} = {v_3}{A_3}$ .

Substitute 12.52 m/s for ${v_3}$ and $0.0160{\rm{ }}{{\rm{m}}^2}$ for ${A_3}$ in the equation ${Q_3} = {v_3}{A_3}$ .

$\begin{array}{c}\\{Q_3} = \left( {12.52{\rm{ m/s}}} \right)\left( {0.0160{\rm{ }}{{\rm{m}}^2}} \right)\\\\ = 0.20{\rm{ }}{{\rm{m}}^3}{\rm{/s}}\\\end{array}$

Rearrange the equation ${P_2} = \frac{1}{2}\rho \left( {v_3^2 - v_2^2} \right)$ as follows:

$\begin{array}{c}\\{P_2} = \frac{1}{2}\rho v_3^2\left( {1 - \frac{{v_2^2}}{{v_3^2}}} \right)\\\\ = \frac{1}{2}\rho v_3^2\left( {1 - {{\left( {\frac{{{A_3}}}{{{A_2}}}} \right)}^2}} \right)\\\\ = \frac{1}{2}\rho \left( {2g\left( {{y_1} - {y_3}} \right)} \right)\left( {1 - {{\left( {\frac{{{A_3}}}{{{A_2}}}} \right)}^2}} \right)\\\\ = \rho \left( {g\left( {{y_1} - {y_3}} \right)} \right)\left( {1 - {{\left( {\frac{{{A_3}}}{{{A_2}}}} \right)}^2}} \right)\\\end{array}$

Now, substitute $1000{\rm{ kg/}}{{\rm{m}}^3}$ for $\rho$ , $9.8{\rm{ m/}}{{\rm{s}}^2}$ for g, 10.0 m for ${y_1}$ , 2.00 m for ${y_3}$ , $0.0160{\rm{ }}{{\rm{m}}^2}$ for ${A_3}$ , and $0.0480{\rm{ }}{{\rm{m}}^2}$ for ${A_2}$ in the equation ${P_2} = \rho \left( {g\left( {{y_1} - {y_3}} \right)} \right)\left( {1 - {{\left( {\frac{{{A_3}}}{{{A_2}}}} \right)}^2}} \right)$ .

$\begin{array}{c}\\{P_2} = \left( {1000{\rm{ kg/}}{{\rm{m}}^3}} \right)\left( {9.8{\rm{ m/}}{{\rm{s}}^2}\left( {10.0{\rm{ m}} - 2.00{\rm{ m}}} \right)} \right)\left( {1 - {{\left( {\frac{{0.0160{\rm{ }}{{\rm{m}}^2}}}{{0.0480{\rm{ }}{{\rm{m}}^2}}}} \right)}^2}} \right)\\\\ = \left( {1000{\rm{ kg/}}{{\rm{m}}^3}} \right)\left( {9.8{\rm{ m/}}{{\rm{s}}^2}\left( {8.00{\rm{ m}}} \right)} \right)\left( {1 - {{\left( {\frac{1}{3}} \right)}^2}} \right)\\\\ = 6.97 \times {10^4}{\rm{ Pa}}\\\end{array}$

Ans:

The gauge pressure at point 2 is equal to $6.97 \times {10^4}{\rm{ Pa}}$ .