Loss of head (pressure)

According to Bernoulli's equation

$\frac{1}{2} \rho v^2 +\rho g h + p = const$ ,

where $v$ is the velocity of the fluid $\rho$ is the density $p$ is the pressure and $h$ is the elevation. This equation describes the conservation of energy for an ideal fluid, with no losses. First, let us see how high will the water go in this case.

We compare the top of the water tower ( $v=0, p=p_0, h=h_0$ ) with the exit of the nozzle ( $p=p_0, h=0$ ) and obtain $\rho g h_0 +p_0= \frac{1}{2} \rho v^2 + p_0$ . Here $p_0$ is the atmospheric pressure and $h_0$ is the height of the tower and $v$ is the velocity at the nozzle. The velocity is $v=\sqrt {2gh_0}$ . With this initial velocity the water will go to a maximum height of $h_0$ .

If the losses are proportional to the square of the velocity, each element in the transport line will contribute a term of $k_i v_i^2$ , where $k_i$ depends on the design of the element (valve, nozzle) and $v_i$ is the velocity of the flow in that element. Due to the equation of continuity (and to the fact that no one else is using the water) the velocity in each element can be expressed in terms of the velocity in the nozzle, $v_i=v A/A_i$ where $A$ and $A_i$ are the corresponding cross sections, and the loss is $k_i v_i^2=k_i (A/A_i)^2 v^2$ . Accordingly, the modified Bernoulli's equation, that includes the losses, can be written as

$\frac{1}{2} \rho v^2 +\rho g h + p +k_1\left(\frac{A}{A_1}\right)^2 v^2 +k_2\left(\frac{A}{A_2}\right)^2 v^2 +k_3\left(\frac{A}{A_3}\right)^2 v^2 + ... = const$ , or

$\frac{1}{2} \rho v^2 +\rho g h + p +k'v^2 = const$ or

$\frac{1}{2} (\rho+k') v^2 +\rho g h + p=const$

Here $k'=k_1\left(\frac{A}{A_1}\right)^2+k_2\left(\frac{A}{A_2}\right)^2+k_3\left(\frac{A}{A_3}\right)^2+...$ represents the sum of all the losses in the water line. This is a remarkable result, similar to Bernoulli's equation, except it looks like the water is more "dense" when it comes to the calculation of the kinetic energy and it has its original density for the calculation of potential energy.

Comparing the top of the water tower to the nozzle again yields a velocity of

$v'=\sqrt { 2gh_0 \frac{\rho}{\rho+k'}}$ ,

and maximum height of the outgoing water is

$h= \frac{\rho }{\rho+k'}h_0$

With the numbers in the problem $\frac{\rho }{\rho+k'}=\frac{h}{h_0}=\frac{1}{3}$ . If we are at 10m elevation the possible maximum height relative to that level is $h_0'=20$ m, and the actual height is $h'=20m/3=6.6m$ . Therefore relative to the original ground level the elevation will be 16.6 m