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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

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Pipe Networks

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Pipe networks l.jpg

Pipe Networks

  • Problem Description

  • Hardy-Cross Method

    • Derivation

    • Application

  • Equivalent Resistance, K

  • Example Problem


Problem description l.jpg

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

    • Discharge & flow direction for all pipes in network

    • Pressure @ all nodes & HGL


Hardy cross method derivation l.jpg

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


Equivalent resistance k l.jpg

Equivalent Resistance, K


Example problem l.jpg

Example Problem

PA = 128 psi

f = 0.02


Hardy cross method procedure l.jpg

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 Shf/ Qa.  

e)  Calculate correction d =-S hf /(nShf/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)


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