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Q93GP

Expert-verifiedFound in: Page 260

Book edition
7th

Author(s)
Douglas C. Giancoli

Pages
978 pages

ISBN
978-0321625922

**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 \(11878.25\;{\rm{N/}}{{\rm{m}}^2}\).

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 \({d_{\rm{2}}} = 1.2\;{\rm{cm}}\).

**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.**

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.

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

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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