Pipe Networks. Problem Description Hardy-Cross Method Derivation Application Equivalent Resistance, K Example Problem. Problem Description. Network of pipes forming one or more closed loops Given Demands @ network nodes (junctions) d, L, pipe material, Temp, P @ one node Find
For Closed Loop:
*Schaumm’s Math Handbook for Binomial Expansion:
PA = 128 psi
f = 0.02
2. For each loop:
a) Assume discharge Qa and direction for each pipe. Apply Continuity at each node, Total inflow = Total Outflow. Clockwise positive.
b) Calculate equivalent resistance K for each pipe given L, d, pipe material and water temperature (Table 11.5).
c) Calculate hf=K Qan for each pipe. Retain sign from step (a) and compute sum for loop S hf.
d) Calculate hf / Qafor each pipe and sum for loop Shf/ Qa.
e) Calculate correction d =-S hf /(nShf/Qa). NOTE: For common members between 2 loops both corrections have to be made. As loop 1 member, d = d1 - d2. As loop 2 member, d = d2 - d1.
f) Apply correction to Qa, Qnew=Qa + d.
g) Repeat steps (c) to (f) until d becomes very small and S hf=0 in step (c).
h) Solve for pressure at each node using energy conservation.Hardy-Cross Method (Procedure)