How high should the pressure head be if water is to come from a faucet at a speed of \({\bf{9}}{\bf{.2\;m/}}{{\bf{s}}^{\bf{2}}}\) Ignore viscosity.

Whenever a liquid flows with a specific flow rate through a pipe/tube, then the flow parameters like velocity, pressure, and many more, can obtain with the help of Bernoulli's equation.

Bernoulli's equation consists of energy in the form of pressure, kinetic, and datum heads.

The speed of water is \({v_1} = 9.2\;{\rm{m/s}}\).

According to Bernoulli’s equation, as the pressure head and faucet are at atmospheric pressure then the equation can be given by,

\(\begin{array}{c}\left( {{P_0} + \frac{1}{2}\rho v_1^2 + \rho g{y_1}} \right) = \left( {{P_0} + \frac{1}{2}\rho v_2^2 + \rho g{y_2}} \right)\\\left( {\frac{1}{2}\rho v_1^2 + \rho g{y_1}} \right) = \left( {\frac{1}{2}\rho v_2^2 + \rho g{y_2}} \right)\\\left( {\frac{1}{2}v_1^2 + g{y_1}} \right) = \left( {\frac{1}{2}v_2^2 + g{y_2}} \right)\end{array}\)

Here, \({P_0}\) is the atmospheric pressure, \(\rho \) is the density, \(g\) is the gravitational acceleration \(\left( {g = 9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\), \({y_1}\) is the datum head \(\left( {{y_1} = 0} \right)\), \({y_2}\) is the required pressure head, \({v_2}\) if the final velocity of water when comes to rest \(\left( {{v_2} = 0} \right)\).

Substitute all the known values in the above expression.

\(\begin{array}{c}\left( {\frac{1}{2}\left( {v_1^2} \right) + g(0\;{\rm{m}})} \right) = \left( {\frac{1}{2}{{(0\;{\rm{m/s}})}^2} + g{y_2}} \right)\\\frac{1}{2}\left( {v_1^2} \right) = g{y_2}\\{y_2} = \frac{{{v_1}^2}}{{2g}}\end{array}\)

After solving the above expression,

\(\begin{array}{c}{y_2} = \frac{{{{(9.2\;{\rm{m}}/{\rm{s}})}^2}}}{{2\left( {9.81\;{\rm{m}}/{{\rm{s}}^2}} \right)}}\\ = 4.31\;\;{\rm{m}}\end{array}\)

Thus, the required height of pressure head is \(4.31\;\;{\rm{m}}\).