Question

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

Short answer

The height of pressure head is$4.31\mathrm{m}$.

Step-by-step solution

Understanding about the Bernoulli’s equation

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.

Identification of given data

The speed of water is $v_1=9.2\mathrm{m}/\mathrm{s}$.

Determination of the pressure head

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 $\left(g=9.81\mathrm{m}/{\mathrm{s}}^2\right)$, $y_1$ 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.

$\begin{array}{c}\left(\frac{1}{2}\left(v_1^2\right)+g(0\mathrm{m})\right)=\left(\frac{1}{2}{(0\mathrm{m}/\mathrm{s})}^2+g{y}_2\right)\\ \frac{1}{2}\left(v_1^2\right)=g{y}_2\\ y_2=\frac{{v_1}^2}{2g}\end{array}$

After solving the above expression,

$\begin{array}{c}{y}_2=\frac{{(9.2\mathrm{m}/\mathrm{s})}^2}{2\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)}\\ =4.31\mathrm{m}\end{array}$

Thus, the required height of pressure head is $4.31\mathrm{m}$.