1 / 6

Pipe Networks

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

saxon
Télécharger la présentation

Pipe Networks

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Pipe Networks • Problem Description • Hardy-Cross Method • Derivation • Application • Equivalent Resistance, K • Example Problem

  2. 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

  3. 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

  4. Equivalent Resistance, K

  5. Example Problem PA = 128 psi f = 0.02

  6. 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)

More Related