1 / 15

Problem 6.127 Network Flow

Problem 6.127 Network Flow. Scott Jewett BIEN 301 January 30, 2007. Problem Diagram. Horizontal Pipe Network. 2 ft 3 /s. D=8 in. f = .025 P A = 120 psi T= 20 °C. D. C. 3000 ft. D=6 in. D=3 in. D=9 in. A. B. 2 ft 3 /s. D=8 in. 4000 ft. Required.

Download Presentation

Problem 6.127 Network Flow

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. Problem 6.127Network Flow Scott Jewett BIEN 301 January 30, 2007

  2. Problem Diagram Horizontal Pipe Network 2 ft3/s D=8 in f = .025 PA= 120 psi T= 20°C D C 3000 ft D=6 in D=3 in D=9 in A B 2 ft3/s D=8 in 4000 ft

  3. Required • Determine the flow rate and direction in all the pipes • Determine the pressures at points B, C, and D.

  4. Assumptions • Liquid • Incompressible • Steady • Viscous

  5. Assumptions (cont.) • Flow directions • Loop directions Qcd C D L2 Qbc Qbd Qac L1 A B Qab

  6. Nodal Equations • Solve nodal equations Flow out - Flow in = 0 2ft3/s Qcd Node A: C D L2 Node C: Qbc Qbd Qac L1 Node B: A B 2ft3/s Qab

  7. Head loss • Use equation 6.10 to obtain head loss as a function of flow rate for each pipe

  8. Obtain five equations relating flow rate to head loss

  9. Loop Equations • Set up loop equations: Sum of head losses around loop = 0 Loop 1: Loop 2: Qcd C D If the flow is opposite the loop, then the head loss is negative. L2 Qcb Qbd Qac L1 A B Qab

  10. System of equations • Five equations, Five unknowns

  11. Solution • Solve using Mathcad or similar tool Qab = 1.187 ft3/s Qac = .813 ft3/s Qcb = .99 ft3/s Qcd = 1.803 ft3/s Qbd = .197 ft3/s

  12. Pressure Solution • Equation 6.8 relates pressure to head loss hf= (Pa-Pb)/(ρg)

  13. Pressure solution • Pb = Pa- ρghf(ab) Pb = 120 psi - ρg(19.116*(Qab)2) Pb = 108 psi • Pc = Pb - ρghf(cb) Pc = 102 psi • Pd = Pc - ρghf(cd) Pd = 74 psi

  14. Biomedical Application • Blood flow • Your body consists of blood vessels with varying: • Diameter • Friction • Height • All of these affect flow rate and pressure.

  15. Questions?

More Related