Hardy Cross network with valve on diagonal

Data. Resistance $r$ in $h_f = rQ^2$ (discharges in l/min): AB 1, BC 2.5, CD 1.5, AD 2, DB 3. The valve (r = 1.5) sits in the diagonal, in series with DB, so the diagonal has $r_{BD} = 3 + 1.5 = 4.5$. Inflow 70 at A; outflows 30 at B and 40 at C. Sign rule: clockwise flow positive.

Initial distribution (continuity satisfied)

A to B 40, A to D 30; at B (out 30): B to C 10, B to D 0; at C (out 40): the 10 from B plus 30 from D (D to C 30).

Loop I = A-B-D-A, Loop II = B-C-D-B.

Hardy-Cross iterations

$\Delta Q = -\dfrac{\sum rQ|Q|}{2\sum r|Q|}$ for each loop, applied to all its pipes (the diagonal BD gets both corrections).

| 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 | -200.00 | 200.00 | 1.000 | -1100.00 | 140.00 | 7.857 | | 2 | -212.59 | 259.71 | 0.819 | 273.33 | 217.43 | -1.257 | | 3 | 57.52 | 239.40 | -0.240 | -29.55 | 196.23 | 0.151 | | 4 | -7.23 | 243.39 | 0.030 | 11.05 | 200.05 | -0.055 |

After about 10 iterations the corrections fall below 0.001.

Final distribution (l/min)

Checks: node A: $41.60 + 28.40 = 70.00$; node B: $41.60 + 5.10 = 30 + 16.70$ (46.70 = 46.70); node C: $16.70 + 23.30 = 40.00$.

Answer: AB = 41.6, AD = 28.4, BC = 16.7, CD = 23.3 (D to C) and the diagonal BD = 5.1 l/min (flowing D to B).