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Chapter 10: Q53P (page 260)

Water at a gauge pressure of \({\bf{3}}{\bf{.8}}\;{\bf{atm}}\) at street level flows into an office building at a speed of \({\bf{0}}{\bf{.78}}\;{\bf{m/s}}\) through a pipe \({\bf{5}}{\bf{.0}}\;{\bf{cm}}\)in diameter. The pipe tapers down to \({\bf{2}}{\bf{.8}}\;{\bf{cm}}\) in diameter by the top floor, \({\bf{16}}\;{\bf{m}}\) above (Fig. 10鈥�53), where the faucet has been left open. Calculate the flow velocity and the gauge pressure in the pipe on the top floor. Assume no branch pipes and ignore viscosity.

Figure 10-53

Short Answer

Expert verified

The flow velocity of water at the top floor is \(2.5\;{\rm{m/s}}\) and the gauge pressure on the top floor is \(2.22\;{\rm{atm}}\).

Step by step solution

01

Understanding equation of continuity and Bernoulli’s principle

The product of area and velocity of fluid in a pipe always remains constant as stated by equation of continuity.

The Bernoulli鈥檚 principle gives the information that the point where velocity of fluid is high, pressure at that point will be low.

02

Given data

The Gauge pressure of water at street level is \({P_{{\rm{street}}}} = 3.8\;{\rm{atm}} = 385035\;{\rm{Pa}}\).

The speed of water flow at street level is \({v_{{\rm{street}}}} = 0.78\;{\rm{m/s}}\).

The diameter of pipe at street level is \({D_{{\rm{street}}}} = 5.0\;{\rm{cm}}\).

The diameter of pipe at top floor is \({D_{{\rm{top}}}} = 2.8\;{\rm{cm}}\).

The height of top floor is \({y_{{\rm{top}}}} = 16\;{\rm{m}}\).

03

Step 3: Calculating the velocity of fluid

According to Equation of continuity, we get:

\(\begin{array}{c}{A_{{\rm{street}}}}{v_{{\rm{street}}}} = {A_{{\rm{top}}}}{v_{{\rm{top}}}}\\\pi {\left( {\frac{{{D_{{\rm{street}}}}}}{2}} \right)^2}{v_{{\rm{street}}}} = \pi {\left( {\frac{{{D_{{\rm{top}}}}}}{2}} \right)^2}{v_{{\rm{top}}}}\\{v_{{\rm{top}}}} = \left( {0.78\;{\rm{m/s}}} \right){\left( {\frac{{5.00\;{\rm{cm}}}}{{2.8\;{\rm{cm}}}}} \right)^2}\\ \approx 2.5\;{\rm{m/s}}\end{array}\)

04

Step 4: Calculating the pressure on top floor

The expression for the Bernoulli's equation is given as,

\(\begin{array}{c}{P_{{\rm{street}}}} + \frac{1}{2}\rho v_{{\rm{street}}}^2 + \rho g{y_{{\rm{street}}}} = {P_{{\rm{top}}}} + \frac{1}{2}\rho v_{{\rm{top}}}^2 + \rho g{y_{{\rm{top}}}}\\{P_{{\rm{top}}}} = {P_{{\rm{street}}}} + \frac{1}{2}\rho v_{{\rm{street}}}^2 + \rho g{y_{{\rm{street}}}} - \frac{1}{2}\rho v_{{\rm{top}}}^2 - \rho g{y_{{\rm{top}}}}\\{P_{{\rm{top}}}} = {P_{{\rm{street}}}} + \frac{1}{2}\rho \left( {v_{{\rm{street}}}^2 - v_{{\rm{top}}}^2} \right) + \rho g\left( {{y_{{\rm{street}}}} - {y_{{\rm{top}}}}} \right)\end{array}\)

Substitute the values in the above equation.

\(\begin{array}{c}{P_{{\rm{top}}}} = 385035\;{\rm{Pa}} + \frac{1}{2}\left( {1000\;{\rm{kg/}}{{\rm{m}}^3}} \right)\left( {{{\left( {{\rm{0}}{\rm{.78}}\;{\rm{m/s}}} \right)}^2} - {{\left( {2.5\;{\rm{m/s}}} \right)}^2}} \right) + \left( {1000\;{\rm{kg/}}{{\rm{m}}^3}} \right)\left( {9.{\rm{8}}\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {0\;{\rm{m}} - 16\;{\rm{m}}} \right)\\ = 385035\;{\rm{Pa}} + \left( {500\;{\rm{kg/}}{{\rm{m}}^3}} \right)\left( { - 5.64\;{{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^2}} \right)\left( {\frac{{{\rm{1}}\;{\rm{Pa}}}}{{{\rm{1}}\;{\rm{kg/m}} \cdot {{\rm{s}}^{\rm{2}}}}}} \right) + \left( {9800\;{\rm{kg/}}{{\rm{m}}^{\rm{2}}} \cdot {{\rm{s}}^2}} \right)\left( { - 16\;{\rm{m}}} \right)\\ = 385035\;{\rm{Pa}} - 2820\;{\rm{Pa}} - 156800\;{\rm{Pa}}\\ = 225415\;{\rm{Pa}}\end{array}\)

The Pressure at top level in atm can be calculated as,

\(\begin{array}{c}{P_{{\rm{top}}}} = \left( {224515\;{\rm{Pa}}} \right)\left( {\frac{{1\;{\rm{atm}}}}{{1.01 \times {{10}^5}\;{\rm{Pa}}}}} \right)\\ = 2.2\;{\rm{atm}}\end{array}\)

Therefore, the speed of fluid at top level is \(2.5\;{\rm{m/s}}\) and the pressure of fluid at top level is \(2.2\;{\rm{atm}}\).

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