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Physics Principles with Applications
Found in: Page 260
Physics Principles with Applications

Physics Principles with Applications

Book edition 7th
Author(s) Douglas C. Giancoli
Pages 978 pages
ISBN 978-0321625922

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Short Answer

A drinking fountain shoots water about 12 cm up in the air from a nozzle of diameter 0.60 cm (Fig. 10–57). The pump at the base of the unit (1.1 m below the nozzle) pushes water into a 1.2-cm-diameter supply pipe that goes up to the nozzle. What gauge pressure does the pump have to provide? Ignore the viscosity; your answer will therefore be an underestimate.

The gauge pressure is \(11878.25\;{\rm{N/}}{{\rm{m}}^2}\).

See the step by step solution

Step by Step Solution

Step 1: Given Data

The height of the fountain water is \(h = 12\;{\rm{cm}}\).

The diameter of the nozzle is \({d_1} = 0.60\;{\rm{cm}}\).

The distance below the nozzle is \(d = 1.1\;{\rm{m}}\).

The diameter of the pipe is \({d_{\rm{2}}} = 1.2\;{\rm{cm}}\).

Step 2: Understanding the Bernoulli’s equation

In this problem, the Bernoulli’s equation will be utilized for determining the gauge pressure. Initially, the speed of the water at exit of the nozzle will be evaluated.

Step 3: Calculating the velocity at the nozzle

The relation of velocity is given by,

\(\begin{array}{l}{v_1} = \sqrt {2gh} \\{v_1} = \sqrt {2\left( {9.80\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {0.12\;{\rm{m}}} \right)} \\{v_1} = 1.53\;{\rm{m/s}}\end{array}\)

Here, \(g\) is the gravitational acceleration.

The relation from equation of continuity is given by,

\(\begin{array}{c}{A_1}{v_1} = {A_2}{v_2}\\{v_2} = \left( {\frac{{{A_1}}}{{{A_2}}}} \right){v_1}\\{v_2} = \left( {\frac{{\frac{1}{2}\pi {{\left( {{d_1}} \right)}^2}}}{{\frac{1}{2}\pi {{\left( {{d_2}} \right)}^2}}}} \right){v_1}\\{v_2} = {v_1}{\left( {\frac{{{d_1}}}{{{d_2}}}} \right)^2}\end{array}\)

Here, \({A_1}\) and \({A_2}\) are the area at the inlet and outlet of the nozzle.

On plugging the values in the above relation.

\(\begin{array}{l}{v_2} = \left( {1.53\;{\rm{m/s}}} \right){\left( {\frac{{0.60\;{\rm{cm}}}}{{1.2\;{\rm{cm}}}}} \right)^2}\\{v_2} = 0.38\;{\rm{m/s}}\end{array}\)

Step 4: Calculating the gauge pressure

The relation from Bernoulli’s equation is given by,

\({P_{\rm{g}}} = \rho \left( {\frac{1}{2}\left( {{v_1}^2 - {v_2}^2} \right) + gd} \right)\)

Here, \(\rho \) is the density of water.

On plugging the values in the above relation.

\(\begin{array}{l}{P_{\rm{g}}} = \left( {1000\;{\rm{kg/}}{{\rm{m}}^3}} \right)\left[ {\frac{1}{2}\left( {{{\left( {1.53\;{\rm{m/s}}} \right)}^2} - {{\left( {0.38\;{\rm{m/s}}} \right)}^2}} \right) + \left( {9.80\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {1.1\;{\rm{m}}} \right)} \right]\\{P_{\rm{g}}} = 11878.25\;{\rm{N/}}{{\rm{m}}^2}\end{array}\)

Thus, the gauge pressure is \(11878.25\;{\rm{N/}}{{\rm{m}}^2}\).

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