To Pump or Not to Pump?

Applying Conservation of Energy in a suitable form for fluids systems (Energy per unit Weight):

$\textbf{\underline{Section 0 - 1}}$

$\displaystyle \frac{P_0}{\gamma} + z_0 + \alpha \frac{{v_0}^2}{2g} + H_p = \frac{P_1}{\gamma} + z_1 + \alpha \frac{{v_1}^2}{2g} + \sum{h_{l_{0-1}}}$

Begin by simplifying the expression and substitute in the relation for head loss in the problem statement, solve for $H_p$ : $P_0 = 0 \, \text{gage}, z_0 = 0, v_0 \approx 0, \alpha =1$

$\displaystyle H_p = \frac{P_1}{\gamma} + z_1 + \frac{{v_1}^2}{2g} + K_{0-1}\frac{{v_1}^2}{2g}$ ...EQ1

$\textbf{\underline{Section 1 - 2}}$

This is the same analysis as above applied between points 1 and 2 with: $H_p = 0, z_1 = z_2, P_2 = 0 \, \text{gage}$ . A goal is to eliminate $\frac{P_1}{\gamma}$ from EQ1, solving for it:

$\displaystyle \frac{P_1}{\gamma} = \frac{{v_2}^2}{2g}(1 + K_{0-1}) - \frac{{v_1}^2}{2g}$ ...EQ2

$\textbf{\underline{Section 1 - 3}}$

Same application as above between points 1 and 3 with, $H_p = 0, P_3 = 0 \, \text{gage}$ and the criteria $v_3 = 0$ , we find:

$\displaystyle \frac{P_1}{\gamma} = (z_3 - z_1) - \frac{{v_1}^2}{2g}$ ...EQ3

Now we can substite EQ3 into EQ1, eliminating $\frac{P_1}{\gamma}$ and simplify, finding:

$H_p = z_3 + K_{0-1}\frac{{v_1}^2}{2g}$ ...EQ1'

Assuming the velocity profile is approximately constant across the pipes, we can derrive the velocity of the flow in terms of the volumetric flow rate using the cross sectional area $A_1$ :

$\displaystyle Q_1 = v_1A_1$ ...EQ4

Rearrange EQ4 for $v_1$ and sub into EQ1':

$\displaystyle H_p = z_3 + K_{0-1}\frac{{Q_1}^2}{{A_1}^2 2g}$ ...EQ1''

Furthermore, we may apply continuity ( conservation of mass for incompressible flows) at point 1:

$\displaystyle Q_1 = Q_2 + Q_3$ ...EQ5

Letting $Q_3$ go to zero:

$\displaystyle Q_1 = Q_2$

It then follows that:

$\displaystyle H_p = z_3 + K_{0-1}\frac{{Q_2}^2}{{A_1}^2 2g}$ ...EQ1'''

Now, since Branch 1-2 & 1-3 are in parallel, they have a common pressure at point 1. We may equate EQ2 and EQ3 through the common pressure $P_1$ . ( $v_3 = 0$ already being accounted for in EQ3) we obtain ( solving for ${Q_2}^2$ in terms of $v_2, A_2$ ) :

$\displaystyle \frac{{v_2}^2}{2g}(1 + K_{0-1}) - \frac{{v_1}^2}{2g} = (z_3 - z_1) - \frac{{v_1}^2}{2g}$

$\displaystyle \rightarrow {Q_2}^2 = 2g{A_2}^2\frac{z_3-z_1}{1+K_{1-2}}$ ...EQ6

Finally, sub EQ6 into EQ1''', with $\frac{A_2}{A_1} = \left( \frac{D_2}{D_1} \right)^2$ :

$\displaystyle H_p = z_3 + \frac{K_{0-1}}{1+K_{1-2}}\left(\frac{D_2}{D_1}\right)^4(z_3 - z_1) = 50.5 \, [\text{ft}] + \frac{5}{1+3}\left(\frac{1 \, [\text{in}]}{1.5\, [\text{in}]}\right)^4 (50.5 \, [\text{ft}] - 10 \, [\text{ft}]) = 60.5 \, [\text{ft}]$

**Given a pump that outputs the required head, the system flow can be found from EQ6 to be approximately $62.5 \, [\text{gpm}]$ . Any further increase in head delivered by the pump and the criterion $v_3 = 0$ no longer holds. To determine head and flow state beyong $H_p$ a solution to a system of non-linear equations must be employed ( methods typically involve linearization and iteration)