Hardy Cross method

The Hardy Cross method assumes that the flow going in and out of the system is known and that the pipe length, diameter, roughness and other key characteristics are also known or can be assumed.[1] The method also assumes that the relation between flow rate and head loss is known, but the method does not require any particular relation to be used.[1]

In the case of water flow through pipes, a number of methods have been developed to determine the relationship between head loss and flow. The Hardy Cross method allows for any of these relationships to be used.

The general relationship between head loss and flow is:

${\displaystyle h_{f}=k\cdot Q^{n}}$

where k is the head loss per unit flow and n is the flow exponent. In most design situations the values that make up k, such as pipe length, diameter, and roughness, are taken to be known or assumed and thus the value of k can be determined for each pipe in the network. The values that make up k and the value of n change depending on the relation used to determine head loss. However, all relations are compatible with the Hardy Cross method.[4]

Head loss equation	Relation	k	n
Hazen-Williams equation	${\displaystyle h_{f}=L\cdot {\frac {10.67\quad Q^{1.85}}{C^{1.85}\quad d^{4.87}}}}$	${\displaystyle L\cdot {\frac {10.67}{C^{1.85}\quad d^{4.87}}}}$	1.85
Darcy-Weisbach equation	${\displaystyle h_{f}={\frac {8fLQ^{2}}{g\pi ^{2}d^{5}}}}$	${\displaystyle L\cdot {\frac {8f}{g\pi ^{2}d^{5}}}}$	2

It is also worth noting that the Hardy Cross method can be used to solve simple circuits and other flow like situations. In the case of simple circuits,

${\displaystyle V=K\cdot I}$

is equivalent to

${\displaystyle h_{f}=k\cdot Q^{n}}$.

By setting the coefficient k to K, the flow rate Q to I and the exponent n to 1, the Hardy Cross method can be used to solve a simple circuit. However, because the relation between the voltage drop and current is linear, the Hardy Cross method is not necessary and the circuit can be solved using non-iterative methods.

The method of balancing heads uses an initial guess that satisfies continuity of flow at each junction and then balances the flows until continuity of potential is also achieved over each loop in the system.[1]

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]

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 the 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.

The method of balancing flows uses an initial guess that satisfies continuity of potential over each loop and then balances the flows until continuity of flow is also achieved at each junction.

The Hardy Cross method is useful because it relies on only simple mathematics, circumventing the need to solve a system of equations. Without the Hardy Cross methods, engineers would have to solve complex systems of equations with variable exponents that cannot easily be solved by hand.

The Hardy Cross method iteratively corrects for the mistakes in the initial guess used to solve the problem.[1] Subsequent mistakes in calculation are also iteratively corrected. If the method is followed correctly, the proper flow in each pipe can still be found if small mathematical errors are consistently made in the process. As long as the last few iterations are done with attention to detail, the solution will still be correct. In fact, it is possible to intentionally leave off decimals in the early iterations of the method to run the calculations faster.