Hardy Cross correction derivation and network

Derivation of the correction $\Delta Q$ (Hardy Cross). Head loss in a pipe is $h_f = rQ^n$ (with $n = 2$ for the Darcy-Weisbach law, $r = \dfrac{8fL}{\pi^2gD^5}$). For a correct distribution the algebraic sum of head losses around every closed loop is zero (clockwise flows positive):

$$\sum rQ|Q|^{n-1} = 0$$

Take assumed discharges $Q_0$ which satisfy continuity at the nodes but not the loop condition. Let the true discharge be $Q = Q_0 + \Delta Q$, where the same correction $\Delta Q$ applies to every pipe of the loop (so continuity is preserved). Then

$$\sum r(Q_0 + \Delta Q)^n = 0 \;\Rightarrow\; \sum rQ_0^n + n\,\Delta Q\sum rQ_0^{n-1} + \frac{n(n-1)}{2}\Delta Q^2\sum rQ_0^{n-2} + \dots = 0$$

$\Delta Q$ is small compared with $Q_0$, so the terms in $\Delta Q^2$ and higher are neglected:

$$\Delta Q = -\frac{\sum rQ_0|Q_0|^{n-1}}{n\sum r|Q_0|^{n-1}} \;\xrightarrow{n = 2}\; \boxed{\Delta Q = -\frac{\sum rQ_0|Q_0|}{2\sum r|Q_0|}}$$

The numerator is the algebraic sum of head losses in the loop (signs by direction), the denominator is the sum of absolute values. Apply the correction to every pipe (add it to flows in the clockwise direction, subtract it from anticlockwise ones, and apply both loops' corrections to a common pipe), and repeat until $\Delta Q$ is negligible.

Network discharge through FG

Reading of the figure. The scan is cut off at the bottom, so the part below B and D is assumed to be: A (inlet) joined to B and C; B-D, C-D; D-E, D-G; E-F and G-F. Nodes B, E, F, G have the outflows shown (2, 2, 4, 1) and the inlet at A must therefore be $2 + 2 + 4 + 1 = 9$. Resistances $r$ ($h_f = rQ^2$) as in the table: AB 50, AC 50, BD 100, CD 50, DE 100, EF 100, FG 50, DG 50 (the $f$ and $D$ values are already contained in $r$). Flow units are those of the figure.

Initial distribution (satisfying continuity): AB 4, AC 5, BD 2, CD 5, DE 4, EF 2, GF 2 (G to F), DG 3.

Loop 1 (A-B-D-C, clockwise: AB, BD positive; CD, AC negative) and loop 2 (D-E-F-G, clockwise: DE, EF positive; GF as F to G, DG as G to D negative).

Correction in each iteration, $\Delta Q = -\dfrac{\sum rQ|Q|}{2\sum r|Q|}$:

| Iter | $\sum rQ|Q|$ (L1) | $2\sum r|Q|$ (L1) | $\Delta Q_1$ | $\sum rQ|Q|$ (L2) | $2\sum r|Q|$ (L2) | $\Delta Q_2$ | |---|---|---|---|---|---|---| | 0 | -1300.00 | 1800.00 | 0.722 | 1350.00 | 1700.00 | -0.794 | | 1 | 26.08 | 1872.22 | -0.014 | 63.06 | 1541.18 | -0.041 | | 2 | 0.01 | 1870.83 | -0.000 | 0.17 | 1532.99 | -0.000 |

After the third iteration the corrections are below 0.001. Final discharges:

Check at the nodes: D receives $2.708 + 4.292 = 7.000$ and sends $3.165 + 3.835 = 7.000$; F receives $1.165 + 2.835 = 4.000$, equal to its outflow of 4.

Answer: The discharge in link FG is $2.84$ units (flowing from G to F), $\approx 2.8$.