Proof (r denotes k)

The following proof is taken from Hardy Cross's paper, “Analysis of flow in networks of conduits or conductors.”,^[1] and can be verified by National Programme on Technology Enhanced Learning Water and Wastewater Engineering page,^[4] and Fundamentals of Hydraulic Engineering Systems by Robert J. Houghtalen.^[5]

If the initial guess of flow rates in each pipe is correct, the change in head over a loop in the system, ${\displaystyle \Sigma rQ^{n}}$ would be equal to zero. However, if the initial guess is not correct, then the change in head will be non-zero and a change in flow, ${\displaystyle \Delta Q}$ must be applied. The new flow rate, ${\displaystyle Q=Q_{0}+\Delta Q}$ is the sum of the old flow rate and some change in flow rate such that the change in head over the loop is zero. The sum of the change in head over the new loop will then be ${\displaystyle \Sigma r(Q_{0}+\Delta Q)^{n}=0}$.

The value of ${\displaystyle \Sigma r(Q_{0}+\Delta Q)^{n}}$ can be approximated using the Taylor expansion.

${\displaystyle \Sigma r(Q_{0}+\Delta Q)^{n}=\Sigma r(Q_{0}^{n}+nQ_{0}^{n-1}\Delta Q+...)=0}$

For a small ${\displaystyle \Delta Q}$ compared to ${\displaystyle Q_{0}}$ the additional terms vanish, leaving:

${\displaystyle \Sigma r(Q_{0}^{n}+nQ_{0}^{n-1}\Delta Q)=0}$

And solving for ${\displaystyle \Delta Q}$

${\displaystyle \Sigma rQ_{0}^{n}=-\Sigma nrQ_{0}^{n-1}\Delta Q}$

${\displaystyle \Delta Q=-{\frac {\Sigma rQ_{0}^{n}}{\Sigma nrQ_{0}^{n-1}}}}$

The change in flow that will balance the head over the loop is approximated by ${\displaystyle \Delta Q=-{\frac {\Sigma rQ_{0}^{n}}{\Sigma nrQ_{0}^{n-1}}}}$. However, this is only an approximation due to the terms that were ignored from the Taylor expansion. The change in head over the loop may not be zero, but it will be smaller than the initial guess. Multiple iterations of finding a new ${\displaystyle \Delta Q}$ will approximate to the correct solution.^[1]

Process

The method is as follows:

1. Guess the flows in each pipe, making sure that the total in flow is equal to the total out flow at each junction. (The guess doesn't have to be good, but a good guess will reduce the time it takes to find the solution.)
2. Determine each closed loop in the system.
3. For each loop, determine the clockwise head losses and counter-clockwise head losses. Head loss in each pipe is calculated using ${\displaystyle h_{f}=rQ^{n}}$. Clockwise head losses are from flows in the clockwise direction and likewise for counter-clockwise.
4. Determine the total head loss in each loop, ${\displaystyle \Sigma rQ^{n}}$, by subtracting the counter-clockwise head loss from the clockwise head loss.
5. For each loop, find ${\displaystyle \Sigma nrQ^{n-1}}$ without reference to direction (all values should be positive).
6. The change in flow is equal to ${\displaystyle {\frac {\Sigma rQ^{n}}{\Sigma nrQ^{n-1}}}}$.
7. If the change in flow is positive, apply it to all pipes of the loop in the counter-clockwise direction. If the change in flow is negative, apply it to all pipes of the loop in the clockwise direction.
8. Continue from step 3 until the change in flow is within a satisfactory range.