PIPE NETWORKS AND THE HARDY CROSS METHOD  Virtually any collection of connected pipes can be considered a network Fluid Mechanics  Network analysis allows us to determine pressure drops, and flow rates within individual pipes and the network as a whole [ physical interpretation: what are we doing today? ]  Today’s practising fluids engineer would use software to perform network analysis, but, software and improper boundary conditions can often produce spurious results  It is therefore essential that we examine a technique to analyze networks “by hand” such that we can check the results of computational techniques

PIPE NETWORKS AND THE HARDY CROSS METHOD  (a) The first step in the Hardy Cross method is the assumption of initial flow in each pipe  (b) It is essential that continuity is satisfied at each node [ the essence of the Hardy Cross method ]  (c) Then we compute the head loss through each pipe (via Hazen Williams formula for head loss) Fluid Mechanics

PIPE NETWORKS AND THE HARDY CROSS METHOD  (d) Next, the head losses in each loop are summed, paying heed to the sign convention  (e) We recall that head loss between two joints is the same for each branch connecting the joints [ the essence of the Hardy Cross method ]  (f) The algebraic sum of losses in each loop must equal zero for the flow rates within the pipes to be correct, thus once the head loss sum in a loop is zero, the assumed flow rates are deemed correct and the problem has converged I Fluid Mechanics

PIPE NETWORKS AND THE HARDY CROSS METHOD  Our estimated initial distribution of flows is rarely correct, however with the application of the HC flow rate correction term, , we can converge on a proper solution  = -  (LH)/n  (LH/Q o ) [ the essence of the Hardy Cross method ] I - [1]  here,  = flow rate correction for a loop  (LH) = algebraic sum of head losses for all pipes in the loop n = an empirical constant that varies with the flow rate formula used (n=1.85 for Hazen Williams)  (LH/Q o ) = summation of lost head divided by the flow rate for each pipe in the loop Fluid Mechanics

PIPE NETWORKS AND THE HARDY CROSS METHOD  Let us consider the simple network shown below [ deriving Hardy Cross ]  We know that in the loop, the losses in parallel branches must be the same i.e., LH ABC = LH ADC or LH ABC – LH ADC = 0 - [2]  In order for us to use the relationship we write it in the form (work of HC) LH = kQ n - [3] Fluid Mechanics

PIPE NETWORKS AND THE HARDY CROSS METHOD [ deriving Hardy Cross ]  For Hazen Williams n=1.85, thus - [4]  Now, if we assume an initial flow Q o, we can express the correct flow as our guess plus a correction LH = kQ [5] Q = Q o +  Fluid Mechanics

PIPE NETWORKS AND THE HARDY CROSS METHOD [ deriving Hardy Cross ]  Now, if we invoke the binomial theorem, we can write - [6]  We can stop the expansion after the second term, following terms become negligible as delta is very small compared to Q o kQ 1.85 = k(Q o +  ) 1.85 = k(Q o Q o (1.85-1) ·  + …) Fluid Mechanics

PIPE NETWORKS AND THE HARDY CROSS METHOD [ deriving Hardy Cross ]  Now, let’s re-write [2], subbing in our binomial expression - [7]  or k(Q o Q o (0.85) ·  - k(Q o ′ Q o ′ (0.85) ·  ) = 0 k(Q o Q o ′ 1.85 ) k (Q o (0.85) - Q o ′ (0.85) ) ·  = 0 - [8] Fluid Mechanics

PIPE NETWORKS AND THE HARDY CROSS METHOD [ deriving Hardy Cross ]  Solve for  - [9]  or, more compactly  = k(Q o Q o ′ 1.85 ) / (1.85 k (Q o (0.85) - Q o ′ (0.85) )) = 0 - [10]  =  k(Q o 1.85 ) / (1.85  k (Q o (0.85) )) = Fluid Mechanics

PIPE NETWORKS AND THE HARDY CROSS METHOD [ deriving Hardy Cross ]  But we recall that kQ o 1.85 = LH and kQ o 0.85 = LH/Q o  so we rewrite [10] as  =  (LH) / (1.85  (LH/Q o )) - [11]  this correction is done for each loop in the network Fluid Mechanics