1 / 13

7 Networks of Pipes

7 Networks of Pipes. multiple “inlets” and “outlets.”. use of node and loop equations. B C A G D F E. ∑h f = 0 , Q in = Q out , Darcy’s Formula or eq v . for h f. Steps of solution: Hardy Cross Method

lovie
Download Presentation

7 Networks of Pipes

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. 7 Networks of Pipes multiple “inlets” and “outlets.” use of node and loop equations

  2. BC AG D F E ∑hf = 0 , Qin = Qout , Darcy’s Formula or eqv. for hf

  3. Steps of solution:Hardy Cross Method • Assume best distribution of flow that satisfies the continuity • Compute hf = rQonin each pipe • Compute for each circuit • Evaluate • Compute Qrevised = Q +ΔQ • Repeat steps 1) – 5) beginning with revised Q until desired accuracy is obtained.(ΔQ<± 1) • r= RL/Dm; L is length of pipe and D is inside diameter in SI units. • R= 10.675/Cn ; R is resistance coefficient, n = 1.852 ≈2 and m = 4.8704 ≈5

  4. 50 20 A r=5 B r=2 r=1ΔQ2 r=1 7035 35 ΔQ1 Dr=4C 30 30 100 hf = rQon  Qrevised = Q +ΔQ Example 1: 15

  5. For the left side circuit ΔQ1can be estimated as follows: • A • r=1 • r=2 35 • 70 ΔQ1 • r=4 C • D 30 30 • 100 ΔQ1=

  6. For the right side circuit ΔQ2 can be estimated as follows: 20 15 50 A r=5 B r=1 r=1 35 ΔQ2 35 30 C ΔQ2=

  7. Hence the corrected flows in the individual pipes are: • ΔQ is added to Q in the clockwise flow • ΔQ is subtracted in the anticlockwise flow

  8. Repeating the same steps to calculate ΔQ1 and ΔQ2: • A • r=1 • r=2 17 • 57 ΔQ1 • r=4 C • D 43 30 • 100 ΔQ1=

  9. 20 20 50 A r=5 B r=1 r=1 17 ΔQ2 30 30 C ΔQ2=

  10. Hence the corrected flows in the individual pipes are: • ΔQ is added to Q in the clockwise flow • ΔQ is subtracted in the anticlockwise flow

  11. Repeating the same steps to calculate ΔQ1 and ΔQ2: 20 A 17 B 50 r=5 r=2r=121ΔQ2 r=1 58 ΔQ1 33 100 D 42 C 30 r=4 ΔQ1= ≈0 ΔQ2=

  12. Example 2: 3255 3 4 16 1316 ΔQ1 ΔQ2 4 10 4 2 3 2000’ , 8’’, 0.02 1500’, 6’’, 0.018 5 1000’ 600’ 1200’ 4’’, 0.02 6’’, 0.016 4’’, 0.016 3000, 6’, 0.015 10 2 n = 1.85

More Related