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**Pipe Networks Problem Description Hardy-Cross Method**

Derivation Application Equivalent Resistance, K Example Problem

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**Problem Description Network of pipes forming one or more closed loops**

Given network nodes (junctions) d, L, pipe material, Temp, one node Find Discharge & flow direction for all pipes in network all nodes & HGL

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**Hardy-Cross Method (Derivation)**

For Closed Loop: * (11.18) *Schaumm’s Math Handbook for Binomial Expansion: n=2.0, Darcy-Weisbach n=1.85, Hazen-Williams

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**Equivalent Resistance, K**

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Example Problem PA = 128 psi f = 0.02

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**Hardy-Cross Method (Procedure)**

1. Divide network into number of closed loops. 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 / Qa for 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.

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Pipe Network Pipe network is analyzed based on: 1. Conservation of mass 2. Conservation of energy.

Pipe Network Pipe network is analyzed based on: 1. Conservation of mass 2. Conservation of energy.

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