7 Networks of Pipes
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7 NETWORKS OF PIPES multiple “inlets” and “outlets.” use of node and loop equations
description
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 - PowerPoint PPT Presentation
Transcript of 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
∑hf = 0 , Qin = Qout , Darcy’s Formula or eqv. for hf
Steps of solution: Hardy Cross Method1)Assume best distribution of flow that
satisfies the continuity2)Compute hf = rQo
n in each pipe
3)Compute for each circuit4)Evaluate 5)Compute Qrevised = Q +ΔQ
6)Repeat steps 1) – 5) beginning with revised Q until desired accuracy is obtained.(ΔQ<± 1)
7)r= RL/Dm; L is length of pipe and D is inside diameter in SI units.
8)R= 10.675/Cn ; R is resistance coefficient, n = 1.852 ≈2 and m = 4.8704 ≈5
hf = rQon
Qrevised = Q +ΔQ15
Example 1:
Pipe rQon nrQo
n-1
DA 2x702 2x2x70
AC 1x352 2x1x35
DC -4x302 2x4x30
Sum 7425 590
For the left side circuit ΔQ1can be estimated as follows:
ΔQ1=
20 A
r=1
r=2 35 70 ΔQ1 r=4 C D 30 30 100
For the right side circuit ΔQ2 can be estimated as follows:
pipe hf=rQon hf/Q=nrQo
n-1
AB 5x152 2x5x15
BC -1x352 2x1x35
AC -1x352 2x1x35
Sum -1325 290
20 15 50 A r=5 B
r=1 r=1 35 ΔQ2 35
30 C
ΔQ2=
Hence the corrected flows in the individual pipes are:
Pipe Corrected flow
DA 70+(-13)=57
AC 35+(-13)-5=17
DC 30-(-13)=43
AB 15+5=20
BC 35-5=30
•ΔQ is added to Q in the clockwise flow
•ΔQ is subtracted in the anticlockwise flow
20 A
r=1
r=2 17 57 ΔQ1 r=4 C D 43 30 100
Repeating the same steps to calculate ΔQ1 and ΔQ2:
pipe rQon nrQo
n-1
DA 2x572 2x2x57
AC 1x172 2x1x17
DC -4x432 2x4x43
Sum -611 606
ΔQ1=
20 20 50 A r=5 B
r=1 r=1 17 ΔQ2 30
30 C
pipe rQon nrQo
n-1
AB 5x202 2x5x20
BC -1x302 2x1x30
AC -1x172 2x1x17
Sum 811 294
ΔQ2=
Pipe Corrected flow
DA 57+1=58
AC 17+1-(-3)=21
DC 43-1=42
AB 20+(-3)=17
BC 30-(-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
20 A 17 B 50 r=5
r=2 r=1 21 ΔQ2 r=1 58 ΔQ1 33
100 D 42 C 30 r=4
Repeating the same steps to calculate ΔQ1 and ΔQ2:
pipe rQon nrQo
n-1
DA 2x582 2x2x58
AC 1x212 2x1x21
DC -4x422 2x4x42
Sum 131 610
pipe rQon nrQo
n-1
AB 5x172 2x5x17
BC -1x332 2x1x33
AC -1x212 2x1x21
Sum -86 278
ΔQ1= ≈0
ΔQ2=
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 ΔQ1 ΔQ2
410 4 2
n = 1.85
Lp. no
P. no
D L K Q hl hl/Q Qrevised
1 1 4 1000 122.3 6 3365 561 6+(-2.25)=3.75
2 8 2000 7.65 3 58.4 19.5 3+(-2.25)=0.75
3 6 600 7.73 1 -7.73 7.73 1-(-2.25)+(-.48)=2.77
4 6 3000 36.2 4 -470 118 4-(-2.25)=6.25
∑hl =2946 n∑(hl/Q)=1307 ΔQ1 = -2.25
2 3 6 600 7.73 1 7.73 7.73 1-(-2.25)+(-.48)=2.77
5 6 1500 21.7 4 282 70.5 4+(-0.48)=3.52
6 4 1200 117.4 1 -117.4 117.4 1-(-0.48)=1.48
∑hl =172 n∑(hl/Q)=361 ΔQ2 = -0.48
