1. Simplifying Pipe Networks: Equivalent Pipes1 2 3 4 5 6 7 A B C D E F
Assume that the characteristics of all the pipes and the fluid are known, along with the flow rate entering at A. No other flow rates and no headlosses are given.
What do we know about the relative flow rates and headlosses in the various pipes?

2. By continuity: Q1=Q2=Q7=Q3+Q4+Q5, Q5=Q6A B C D E F
By continuity: Q1=Q2=Q7=Q3+Q4+Q5, Q5=Q6
Because total head (z+[p/g]+V2/2g) has the same value at a point, regardless of direction (it is a scalar, and isotropic):
hL,3=hL,4=hL,5+6
and (ShL)any loop=0

3. Flow Analysis in Systems with Multiple Pipes
Two levels of complexity
One inlet, one outlet, flow directions known
Multiple inlets/outlets, flow directions need to be determined
In both cases, key equations are the same
Continuity at junctions or nodes: total flow rate entering equals total flow rate leaving
Total head is well-defined at any point, so headloss between any two points is independent of path taken, and headloss around a loop is zero

4. 1 2 3 4 5 6 7 A B C D E F
IF the hL-Q relationship can be written as hL=KQn, then for m pipes connected in series, the total headloss when the flow rate is Q is:
A pipe with K=SKi will be hydraulically equivalent to the real pipes in series.

5. 1 2 3 4 5 6 7 A B C D E F
IF the hL-Q relationship can be written as hL=KQn, then for m pipes connected in parallel, the total flow rate when the headloss rate is hL is:
A pipe with K=(SKi-1/n)-n will be hydraulically equivalent to the real pipes in parallel.

6. Reduce that group of pipes to a single equivalent pipe. 1 2 3 4 5 6 7 A B C D E F
Approach
Find two points connected strictly by pipes-in-series or pipes-in-parallel.
Reduce that group of pipes to a single equivalent pipe.
Continue until the network has been simplified to the degree desired.

7. Example: Equivalent PipesB A C D E 1 2 3 4
Water
Pipe
Length (ft)
Diameter (in)
CHW 1
1200 8
110 2
800 6 3
900 4 90
500

8. Pipe l (ft) D (in) CHW K K- 1/n 1 1200 8 110 6.84 2 800 6 18.5 3 900 4B A C D E 1 2 3 4
Water
Pipe
l (ft)
D (in)
CHW K
K- 1/n 1
1200 8
110
6.84 2
800 6
18.5 3
900 4 90
217
500
16.8

9. Pipe l (ft) D (in) CHW K K- 1/n 1 1200 8 110 6.84 0.354 2 800 6 18.5B A C D E 1 2 3 4
Water
Pipe
l (ft)
D (in)
CHW K
K- 1/n 1
1200 8
110
6.84
0.354 2
800 6
18.5
0.207 3
900 4 90
217
0.0545
500
16.8
0.218
2+3
100
Chosen arbitrarily

10. Kequiv,parallel for pipes 2 and 3B A C D E 1 2 3 4
Water
Pipe
l (ft)
D (in)
CHW K
K- 1/n 1
1200 8
110
6.84
0.354 2
800 6
18.5
0.207 3
900 4 90
217
0.0545
500
16.8
0.218
2+3
100
12.0
0.261
Kequiv,parallel for pipes 2 and 3
SK- 1/n for pipes 2 and 3

11. Pipe l (ft) D (in) CHW K K- 1/n 1 1200 8 110 6.84 0.354 2 800 6 18.5B A C D E 1 2 3 4
Water
Pipe
l (ft)
D (in)
CHW K
K- 1/n 1
1200 8
110
6.84
0.354 2
800 6
18.5
0.207 3
900 4 90
217
0.0545
500
16.8
0.218
2+3
1765
100
12.0
0.261

12. Pipe l (ft) D (in) CHW K K- 1/n 1 1200 8 110 6.84 0.354 2 800 6 18.5B A C D E 1 2 3 4
Water
Pipe
l (ft)
D (in)
CHW K
K- 1/n 1
1200 8
110
6.84
0.354 2
800 6
18.5
0.207 3
900 4 90
217
0.0545
500
16.8
0.218
2+3
1765
100
12.0
0.261

13. Pipe l (ft) D (in) CHW K K- 1/n 1 1200 8 110 6.84 0.354 2 800 6 18.5B A C D E 1 2 3 4
Water
Pipe
l (ft)
D (in)
CHW K
K- 1/n 1
1200 8
110
6.84
0.354 2
800 6
18.5
0.207 3
900 4 90
217
0.0545
500
16.8
0.218
2+3
1765
100
12.0
0.261
1-4
5237
35.6

14. B A C D E 1 2 3 4
Water
Pipe
l (ft)
D (in)
CHW K
K- 1/n 1
1200 8
110
6.84
0.354 2
800 6
18.5
0.207 3
900 4 90
217
0.0545
500
16.8
0.218
2+3
1765
100
12.0
0.261
1-4
5237
35.6
The network is hydraulically equivalent to a single, 8-in pipe, 5237 ft long and with CHW=100

15. B A C D E 1 2 3 4
Water
Pipe
l (ft)
D (in)
CHW K
K- 1/n 1
1200 8
110
6.84
0.354 2
800 6
18.5
0.207 3
900 4 90
217
0.0545
500
16.8
0.218
2+3
12.0
0.261
1-4
5237
100
35.6
Note that the calculation of the length of the pipe equivalent to 2+3 was unnecessary if our only interest is in an equivalent pipe for the whole network