Parallel pipeline to increase discharge

The head available (reservoir to building) stays the same. Adding a parallel pipe over the first half reduces resistance there, so more flow passes. Darcy–Weisbach with friction factor $f$ is used: $h_f = \dfrac{fLV^2}{2gD} = \dfrac{8fLQ^2}{\pi^2 g D^5}$.

Step 1: Available head (existing pipe)

$$\begin{aligned} h &= \frac{8 \times 0.025 \times 2000 \times 0.09^2}{\pi^2 \times 9.81 \times 0.15^5} = 440.68\ \text{m} \end{aligned}$$

(This large value is simply the head the original pipe needs; it stays fixed.)

Step 2: New layout

           Pipe A (0.15 m, 1000 m), Q1
 Res ==+=================================+===== 0.15 m, 1000 m ====> Bldg
       +---------------------------------+        Q = 0.1125
           Pipe B (D2, 1000 m), Q2

New discharge $Q = 1.25 \times 0.09 = 0.1125$ m³/s.

Loss in the second (single) half:

$$h_2 = \frac{8 \times 0.025 \times 1000 \times 0.1125^2}{\pi^2 \times 9.81 \times 0.15^5} = 344.28\ \text{m}$$

Loss allowed in the parallel half:

$$h_p = 440.68 - 344.28 = 96.40\ \text{m}$$

Step 3: Flow in old pipe of the parallel portion

$$\begin{aligned} 96.40 &= \frac{8 \times 0.025 \times 1000\,Q_1^2}{\pi^2 \times 9.81 \times 0.15^5} \\ Q_1 &= 0.0595\ \text{m}^3/\text{s} \end{aligned}$$

So the new pipe must carry $Q_2 = 0.1125 - 0.0595 = 0.0530$ m³/s.

Step 4: Diameter of new pipe

Same $f$, same length and same head loss $h_p$, so $Q \propto D^{2.5}$:

$$\begin{aligned} D_2 &= 0.15\left(\frac{Q_2}{Q_1}\right)^{0.4} = 0.15\left(\frac{0.0530}{0.0595}\right)^{0.4} = 0.143\ \text{m} \end{aligned}$$

Answer: A parallel pipe of about 0.143 m diameter is needed. Recommend the next standard size, 150 mm, which gives slightly more than the required 25% increase.