Correct trial network flows and pressure at C

Data. Inflow 5 at A; outflows 2 at B, 2 at D, 1 at E. Trial flows and resistances ($h_f = rQ^2$, $n = 2$): AB 1 ($r$ = 40), AC 2 ($r$ = 10), AE 2 ($r$ = 30), BC 1 ($r$ = 20, flowing C to B), CD 1 ($r$ = 20), DE 1 ($r$ = 40, flowing E to D). Continuity check: A: $1 + 2 + 2 = 5$; B: $1 + 1 = 2$; C: $2 = 1 + 1$; D: $1 + 1 = 2$; E: $2 = 1 + 1$.

Loop I = A-B-C-A (AB +, CB $-$, CA $-$). Loop II = A-C-D-E-A (AC +, CD +, ED $-$, EA $-$).

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 | -20.00 | 160.00 | 0.125 | -100.00 | 280.00 | 0.357 | | 2 | -14.51 | 169.64 | 0.086 | -10.84 | 248.93 | 0.044 | | 3 | -1.82 | 172.22 | 0.011 | -3.90 | 243.73 | 0.016 |

The corrections fall below 0.001 after 14 iterations.

Correct distribution

Checks: A: $1.226 + 2.194 + 1.581 = 5.000$; B: $1.226 + 0.774 = 2.000$; D: $1.419 + 0.581 = 2.000$.

Pressure at C

The supply head at A from the 6 m tank is $H_A = 6$ m. The loss in AC is $h_{AC} = rQ^2 = 10(2.194)^2 = 48.12$ (in the units of the data). Taking A and C at the same level, the pressure head at C is

$$\frac{p_C}{\gamma} = H_A - h_{AC} = 6 - 48.12 = -42.12$$

With $r$ and $Q$ used exactly as given, the loss exceeds the 6 m of supply head, so the available pressure at C comes out negative, which is not physically possible. Either the 6 m tank is not enough for these resistances, or $r$ is on a scale in which $rQ^2$ is not in metres. The method is: available pressure head at C = 6 m $-$ $h_{AC}$ (in metres), and the pressure is $\gamma$ times this.

Answer: Correct discharges: AB = 1.23, CB = 0.77, AC = 2.19, CD = 1.42, ED = 0.58, AE = 1.58; $p_C/\gamma = 6 - h_{AC}$ (negative, $-42.1$, if the data are taken literally).