Flow distribution in five-pipe network

Data. 100 units enter at A; 20 leave at B, 50 at C and 30 at D. Pipes: AB (300 m, 200 mm, f 0.02), BC (250 m, 150 mm, 0.03), CD (300 m, 100 mm, 0.02), AD (250 m, 150 mm, 0.03), AC (500 m, 100 mm, 0.025). $h_f = kQ^2$ with $k = \dfrac{8fL}{\pi^2gD^5}$ (flow in m³/s units; the final distribution depends only on the ratios of $k$, so it applies to any consistent flow unit).

Initial distribution (continuity): A to B 30, A to C 30, A to D 40; B to C 10; D to C 10 (check: B: $30 = 20 + 10$; D: $40 = 30 + 10$; C: $30 + 10 + 10 = 50$).

Loop I = A-B-C-A (AB +, BC +, CA $-$). Loop II = A-C-D-A (AC +, CD $-$ as DC, DA $-$).

Hardy-Cross iterations

| Iter | $\sum rQ|Q|$ (I) | $2\sum r|Q|$ (I) | $\Delta Q_I$ | $\sum rQ|Q|$ (II) | $2\sum r|Q|$ (II) | $\Delta Q_{II}$ | |---|---|---|---|---|---|---| | 1 | -90744818.41 | 6453183.05 | 14.062 | 74940519.31 | 7841390.76 | -9.557 | | 2 | 3527371.24 | 1847342.04 | -1.909 | -34798261.81 | 4066053.33 | 8.558 | | 3 | -22562136.76 | 3972538.03 | 5.680 | 9604960.48 | 5240079.52 | -1.833 |

Converged after 68 iterations (corrections below $10^{-9}$).

Final distribution

Checks: A: $52.44 + 11.15 + 36.40 = 100$; B: $52.44 = 20 + 32.44$; D: $36.40 = 30 + 6.40$; C: $32.44 + 11.15 + 6.40 = 50$.

Answer: AB = 52.4, BC = 32.4, AC = 11.2, AD = 36.4 and DC = 6.4 units (all in the directions shown).