Question

What does the top pressure gauge read?

P= (in kPa)

Answer 1

Concepts and reason

The concept required to solve this problem is the Bernoulli’s equation.

Initially, convert the pressure into Pascal. Then, write the Bernoulli’s equation. Finally, substitute the values of the pressure, density, velocities, and the height given in the question into Bernoulli’s equation to calculate the pressure at the top.

Fundamentals

The expression of the Bernoulli’s equation is,

${P_1} + \frac{1}{2}\rho {v_1}^2 + \rho g{h_1} = {P_2} + \frac{1}{2}\rho {v_2}^2 + \rho g{h_2}$

Here, ${P_1}$ is the pressure at the bottom position, $\rho$ is the density, ${v_1}$ is the speed of the liquid at bottom, g is the acceleration due to gravity, ${h_1}$ is the height at the bottom, ${P_2}$ is the pressure at the top position, ${v_2}$ is the speed of the liquid at top, and ${h_2}$ is the height at the top.

The pressure is,

$\begin{array}{c}\\{P_1} = 200{\rm{ kPa}}\left( {\frac{{{{10}^3}{\rm{ Pa}}}}{{1{\rm{ kPa}}}}} \right)\\\\ = 200 \times {10^3}{\rm{ Pa}}\\\end{array}$

The height of the tube at the bottom position is,

${h_1} = 0{\rm{ m}}$

The expression of the Bernoulli’s equation is,

${P_1} + \frac{1}{2}\rho {v_1}^2 + \rho g{h_1} = {P_2} + \frac{1}{2}\rho {v_2}^2 + \rho g{h_2}$

Substitute 0 m for ${h_1}$ and rearrange the expression for ${P_2}$ .

$\begin{array}{c}\\{P_1} + \frac{1}{2}\rho {v_1}^2 + \rho g\left( 0 \right) = {P_2} + \frac{1}{2}\rho {v_2}^2 + \rho g{h_2}\\\\{P_1} + \frac{1}{2}\rho {v_1}^2 = {P_2} + \frac{1}{2}\rho {v_2}^2 + \rho g{h_2}\\\\{P_2} = {P_1} + \frac{1}{2}\rho \left( {{v_1}^2 - {v_2}^2} \right) - \rho g{h_2}\\\end{array}$

Substitute $200 \times {10^3}{\rm{ Pa}}$ for ${P_1}$ , $900{\rm{ kg/}}{{\rm{m}}^3}$ for $\rho$ , 2 m/s for ${v_1}$ , $9.8{\rm{ m/}}{{\rm{s}}^2}$ for g, 3.0 m/s for ${v_2}$ , 10 m for ${h_2}$ .

$\begin{array}{c}\\{P_2} = 200 \times {10^3}{\rm{ Pa}} + \frac{1}{2}\left( {900{\rm{ kg/}}{{\rm{m}}^3}} \right)\left( {{{\left( {2{\rm{ m/s}}} \right)}^2} - {{\left( {3{\rm{ m/s}}} \right)}^2}} \right) - \left( {900{\rm{ kg/}}{{\rm{m}}^3}} \right)\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {10{\rm{ m}}} \right)\\\\ = 200 \times {10^3}{\rm{ Pa}} - 2250{\rm{ kg/m}} \cdot {{\rm{s}}^2} - 88200{\rm{ kg/m}} \cdot {{\rm{s}}^2}\\\\ = 1.10 \times {10^5}{\rm{ Pa}}\\\\{P_2}{\rm{ = 110 kPa}}\\\end{array}$

Ans:

The pressure at the top position is ${\rm{110 kPa}}$ .

Answer 2

110 kPa