Branch pipe diameters from a single feeder (f 0.025)

Data. Main pipe AJ: $L = 10$ km, $D = 0.75$ m, slope $2.2\times10^{-3}$. Branch JB: $L = 5.5$ km, slope $2.75\times10^{-3}$. Branch JC: $L = 3$ km, slope $3.2\times10^{-3}$. $Q_B = Q_C = 0.24$ m³/s, so the main pipe carries $Q = 0.48$ m³/s. $f = 0.025$. Only friction loss: $h_f = \dfrac{8fLQ^2}{\pi^2gD^5}$. Each reservoir level is 3 m above the end of its pipe (A: above the start of the main pipe; B and C: above the ends of the branches).

Elevations (datum: level of junction J)

• The main pipe falls $10\,000(2.2\times10^{-3}) = 22$ m from A to J, so the level of A is $22 + 3 = 25$ m above J.
• The branch to B falls $5500(2.75\times10^{-3}) = 15.125$ m, so the level of B is $-15.125 + 3 = -12.125$ m (12.125 m below J).
• The branch to C falls $3000(3.2\times10^{-3}) = 9.6$ m, so the level of C is $-9.6 + 3 = -6.6$ m.

Head loss in the main pipe and the head at J

$$h_{f,AJ} = \frac{8(0.025)(10\,000)(0.48)^2}{\pi^2(9.81)(0.75)^5} = 20.06\ \text{m}$$

Piezometric head at J (above J's level): $H_J = 25 - 20.06 = 4.94$ m.

Head available for each branch

• To B: $h_B = H_J - (-12.125) = 17.07$ m
• To C: $h_C = H_J - (-6.6) = 11.54$ m

Branch diameters

$$D = \left(\frac{8fLQ^2}{\pi^2g\,h_f}\right)^{1/5}$$ $$D_B = \left(\frac{8(0.025)(5500)(0.24)^2}{\pi^2(9.81)(17.07)}\right)^{1/5} = 0.5209\ \text{m} \approx 521\ \text{mm}$$ $$D_C = \left(\frac{8(0.025)(3000)(0.24)^2}{\pi^2(9.81)(11.54)}\right)^{1/5} = 0.4989\ \text{m} \approx 499\ \text{mm}$$

Answer: Diameter of the branch to B $\approx 521$ mm; diameter of the branch to C $\approx 499$ mm (head at J is 4.94 m above the junction level).