1. 7 NETWORKS OF PIPES multiple “inlets” and “outlets.” use of node and loop equations

2. B C A G D F E ∑h f = 0, Q in = Q out, Darcy’s Formula or eqv. for h f

3. Steps of solution: Hardy Cross Method 1)Assume best distribution of flow that satisfies the continuity 2)Compute h f = rQ o n in each pipe  3)Compute for each circuit 4)Evaluate 5)Compute Q revised = Q +ΔQ 6)Repeat steps 1) – 5) beginning with revised Q until desired accuracy is obtained.(ΔQ<± 1) 7)r= RL/D m ; L is length of pipe and D is inside diameter in SI units. 8)R= 10.675/C n ; R is resistance coefficient, n = 1.852 ≈2 and m = 4.8704 ≈5

4. h f = rQ o n  50 A r=5 B r=2 r=1 ΔQ 2 r=1 70 35 35 ΔQ 1 D r=4 C 30 20 100 30 Q revised = Q +ΔQ 15 Example 1:

5. PiperQ o n nrQ o n-1 DA2x70 2 2x2x70 AC1x35 2 2x1x35 DC-4x30 2 2x4x30 Sum7425590 For the left side circuit ΔQ 1 can be estimated as follows: ΔQ 1 = 20 A r=1 r=2 35 70 ΔQ 1 r=4 C D 30 30 100

6. For the right side circuit ΔQ 2 can be estimated as follows: pipeh f =rQ o n h f /Q=nrQ o n-1 AB5x15 2 2x5x15 BC-1x35 2 2x1x35 AC-1x35 2 2x1x35 Sum-1325290 20 15 50 A r=5 B r=1 r=1 35 ΔQ 2 35 30 C ΔQ 2 =

7. Hence the corrected flows in the individual pipes are: PipeCorrected flow DA70+(-13)=57 AC35+(-13)-5=17 DC30-(-13)=43 AB15+5=20 BC35-5=30 ΔQ is added to Q in the clockwise flow ΔQ is subtracted in the anticlockwise flow

8. 20 A r=1 r=2 17 57 ΔQ 1 r=4 C D 43 30 100 Repeating the same steps to calculate ΔQ 1 and ΔQ 2 : piperQ o n nrQ o n-1 DA2x57 2 2x2x57 AC1x17 2 2x1x17 DC-4x43 2 2x4x43 Sum-611606 ΔQ 1 =

9. 20 20 50 A r=5 B r=1 r=1 17 ΔQ 2 30 30 C piperQ o n nrQ o n-1 AB5x20 2 2x5x20 BC-1x30 2 2x1x30 AC-1x17 2 2x1x17 Sum811294 ΔQ 2 =

10. PipeCorrected flow DA57+1=58 AC17+1-(-3)=21 DC43-1=42 AB20+(-3)=17 BC30-(-3)=33 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. 20 A 17 B 50 r=5 r=2 r=1 21 ΔQ 2 r=1 58 ΔQ 1 33 100 D 42 C 30 r=4 Repeating the same steps to calculate ΔQ 1 and ΔQ 2 : piperQ o n nrQ o n-1 DA2x58 2 2x2x58 AC1x21 2 2x1x21 DC-4x42 2 2x4x42 Sum131610 piperQ o n nrQ o n-1 AB5x17 2 2x5x17 BC-1x33 2 2x1x33 AC-1x21 2 2x1x21 Sum-86278 ΔQ 1 = ≈0 ΔQ 2 =

12. 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 Example 2: 3 2 5 5 3 4 1 6 1 3 1 6 ΔQ 1 ΔQ 2 4 10 4 2 n = 1.85

13. Lp. no P. no DLKQhlhl h l /Q Q revised 1141000122.3633655616+(-2.25)=3.75 2820007.65358.419.53+(-2.25)=0.75 366007.731-7.737.73 1-(-2.25)+(-.48)=2.77 46300036.24-470118 4-(-2.25)=6.25 ∑h l =2946n∑(h l /Q)=1307  ΔQ 1 = -2.25 2366007.731 1-(-2.25)+(-.48)=2.77 56150021.7428270.54+(-0.48)=3.52 641200117.41-117.4117.41-(-0.48)=1.48 ∑h l =172n∑(h l /Q)=361  ΔQ 2 = -0.48