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What does the top pressure gauge read?
P= (in kPa)
♦ Relevant knowledgePressure is the ratio of force to the area. The pressure unit is the psi. Pressure can be determined using pressure measuring instruments like a simple pressure gauges, U-tube manometers and inverted U-tube manometers, and others.
This equation is the Bernoulli equation.
Convert the pressure to Pascal first. Next, create the Bernoulli equation. To calculate the pressure at top, add the Bernoulli equation’s values for pressure, density and velocities to the question.
The equation of Bernoulli is:
[katex]{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}[/katex]
Here, [katex]{P_1}[/katex]
The pressure is on
[katex]\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}[/katex]
The tube’s height at the bottom is:
[katex]{h_1} = 0{\rm{ m}}[/katex]
The equation of Bernoulli is:
[katex]{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}[/katex]
Substitute 0m for [katex]{h_1}[/katex]
[katex]\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}[/katex]
Substitute [katex]200 \times {10^3}{\rm{ Pa}}[/katex] for [katex]{P_1}[/katex], [katex]{900{\rm{ kg/}}{{\rm{m}}^3}}[/katex] for [katex]\rho[/katex], [katex]{2{\rm{ m/s}}}[/katex] for [katex]{v_1}[/katex], [katex]{9.8{\rm{ m/}}{{\rm{s}}^2}}[/katex] for [katex]{g}[/katex], [katex]{3{\rm{ m/s}}}[/katex] for [katex]{v_2}[/katex], [katex]10 m[/katex] for [katex]{h_2}[/katex]
[katex]\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}[/katex]Ans:
The top position pressure is [katex]{\rm{110 kPa}}[/katex].