Branch diameters equal to a 450 mm direct pipe

Data. Main pipe AJ: 900 m, 450 mm. Branch JB: 1200 m to B, 36 m below A. Branch JC: 1500 m to C, 45 m below A. The branches carry equal discharges, $Q/2$ each; their sum $Q$ equals the discharge of a 450 mm pipe, 2100 m long, running directly from A to B (36 m head). Friction only, same $f$ everywhere. Use $h_f = k\dfrac{LQ^2}{D^5}$ with $k = \dfrac{8f}{\pi^2g}$ (the unknown $f$ and $Q$ cancel).

Step 1: total discharge from the direct pipe

$$36 = k\frac{2100\,Q^2}{(0.45)^5} \;\Rightarrow\; \frac{kQ^2}{(0.45)^5} = \frac{36}{2100} = 0.017143$$

That is, a 450 mm pipe carrying $Q$ loses 0.017143 m of head per metre of length.

Step 2: head loss in the main pipe AJ

It has the same diameter and carries the same $Q$:

$$h_{AJ} = 0.017143(900) = 15.43\ \text{m}$$

So the head at J is 15.43 m below the level of A.

Step 3: head available for each branch

• To B: $36 - 15.43 = 20.57$ m
• To C: $45 - 15.43 = 29.57$ m

Step 4: branch diameters (each carries $Q/2$)

$$h = k\frac{L(Q/2)^2}{D^5} \;\Rightarrow\; D^5 = \frac{kQ^2L}{4h} = \frac{(0.017143)(0.45)^5\,L}{4h}\quad\text{with } (0.45)^5 = 0.018453$$ $$D_B^5 = \frac{0.017143(0.018453)(1200)}{4(20.57)} \;\Rightarrow\; D_B = 0.3410\ \text{m} \approx 341\ \text{mm}$$ $$D_C^5 = \frac{0.017143(0.018453)(1500)}{4(29.57)} \;\Rightarrow\; D_C = 0.3316\ \text{m} \approx 332\ \text{mm}$$

Answer: The branch to B should be about 341 mm and the branch to C about 332 mm in diameter.