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Hardy Cross Method

Following is the procedure for the solution of pipe network problem.

1.    In this method, a trial distribution of discharge is made arbitrarily but in such a way that continuity equation is satisfied at node.

2.    Head loss in each pipe is calculated as hf = r Qn with assuming the value of Q.

3.    For each loop, the algebraic sum of head losses round each loop must be zero. i.e. the loss of head due to flow in clockwise direction must be equal to the loss of head due to flow in anticlockwise direction.

4.    For each loop, the quantity

Δ Q = - Σr Qn l Σr n-1l is calculated

for turbulent flow, the value of n = 2, hence above correction factor becomes,

Δ Q = -Σr Q2 / Σ2 r Q

5.    The value of Δ Q is calculated for all loops.

6.    Correction are now applied to each loop and to all loops. Clockwise direction is consider positive and anti-clockwise is negative.

7.    The procedure is repeated till Δ Q is very small.

8.    Some pipes which are common to two circuits, then two corrections are applied to these pipes.

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