1. CEE 410 Hydraulic Engineering
14- Pipe Equivalency
Pipe Networks
Michael D. Doran, P.E. DEE
Professor of Practice

5. Understand concepts of “Equivalency” in piping systems.
Compute equivalent length of a pipe system.
Simplify a system using equivalency.
Understand how Continuity and Energy Principle are used to solve network problems.
Solve a simple network problems by hand.
Outcomes for today

6. L/D for valves and fittings is an ‘Equivalency’ concept

7. An ‘Equivalent’ pipe has the same headloss as a real system at the same flow.6 8

8. An ‘Equivalent’ pipe has the same headloss as a real system at the same flow.6 8
Equivalent Pipe

9. An ‘Equivalent’ pipe has the same headloss as a real system at the same flow.6 8 6
Simplifies calculations – especially if done by hand.

10. An ‘Equivalent’ pipe has the same headloss at the same flow.
From Hazen-Williams Equation (C=100):
Example 1
f for 12-inch pipe at 1,200 gpm is ft/100 ft
f for 8-inch pipe at at 1,200 gpm is 4.18 ft/100 ft

11. An ‘Equivalent’ pipe has the same headloss at the same flow.
From Hazen-Williams Equation (C=100):
Example 1
f for 12-inch pipe at 1,200 gpm is ft/100 ft
f for 8-inch pipe at at 1,200 gpm is 4.18 ft/100 ft
? How much 12-inch pipe is equivalent to 200 ft of 8-inch pipe?

12. From Hazen-Williams Equation (C=100):
Example 1
f for 12-inch pipe at 1,200 gpm is ft/100 ft
f for 8-inch pipe at at 1,200 gpm is 4.18 ft/100 ft
? How much 12-inch pipe is equivalent to 200 ft of 8-inch pipe?
L12 = L8 (ft 8-in) ft HL ft 12-in
100 ft 8-in ft HL

13. From Hazen-Williams Equation (C=100):
Example 1
f for 12-inch pipe at 1,200 gpm is ft/100 ft
f for 8-inch pipe at at 1,200 gpm is 4.18 ft/100 ft
? How much 12-inch pipe is equivalent to 200 ft of 8-inch pipe?
L12 = L8 (ft 8-in) ft HL ft 12-in
100 ft 8-in ft HL

14. From Hazen-Williams Equation (C=100):
Example 1
f for 12-inch pipe at 1,200 gpm is ft/100 ft
f for 8-inch pipe at at 1,200 gpm is 4.18 ft/100 ft
? How much 12-inch pipe is equivalent to 200 ft of 8-inch pipe?
L12 = 200 (ft 8-in) ft HL ft 12-in
100 ft 8-in ft HL
L12 = 1,440 ft

15. General expression for equivalency using Hazen-Williams (US units):
HL = 𝐿 100 (0.2083)( 100 𝐶 )1.85·Q1.85
D4.87

16. General expression for equivalency using Hazen-Williams (US units):
De4.87
D4.87

17. General expression for equivalency using Hazen-Williams (US units):
De4.87
D4.87
Le = Ce1.85·De4.87· Σ (LC-1.85D-4.87)

18. General expression for equivalency using Hazen-Williams (US units):
Le = Ce1.85·De4.87· Σ (LC-1.85D-4.87)
Le =Equivalent Pipe Length (ft)
Ce =Equivalent Pipe C
De =Equivalent Pipe Diameter (in)
L,C,D =Values for Individual Pipes in System

19. Example 2 – Let’s compute some Le’s using a table of C-1. 85D-4
Example 2 – Let’s compute some Le’s using a table of C-1.85D-4.87 values.
Le = Ce1.85·De4.87· Σ (LC-1.85D-4.87)
Pipe
Equivalent Pipe Le
500 ft 8-in
C = 100
12-in
C = 120
1,000 ft 12-in
C = 110
10-in
750 ft 60-in
48-in
2,000 ft 10-in
14-in

20. Example 2 – Let’s compute some Le’s using a table of C-1. 85D-4
Example 2 – Let’s compute some Le’s using a table of C-1.85D-4.87 values.
Le = Ce1.85·De4.87· Σ (LC-1.85D-4.87)
Pipe
Equivalent Pipe Le
500 ft 8-in
C = 100
12-in
C = 120
5,050 ft
1,000 ft 12-in
C = 110
10-in
345 ft
750 ft 60-in
48-in
215 ft
2,000 ft 10-in
14-in
8,630 ft

21. Example 3
Board example using capacity table and equivalent lengths.

23. Example 4 Q1 Q2 Le = 100 m; 0.76 m dia C = 100
WS = m
Pipes are submerged and flowing full.
WS = m
Le = 225 m; 0.76 m dia
C = 100 Q2

24. Example 4 Q1 Q2 Calculate Q1 and Q2. Le = 100 m; 0.76 m dia C = 100
WS = m
Pipes are submerged and flowing full.
WS = m
Le = 225 m; 0.76 m dia
C = 100 Q2
Calculate Q1 and Q2.

25. hA + PA/γ + VA2/2g = hB + PB/γ + VB2/2g + HL
Example 4 A B
hA + PA/γ + VA2/2g = hB + PB/γ + VB2/2g + HL

26. hA + PA/γ + VA2/2g = hB + PB/γ + VB2/2g + HL
Example 4 A B
hA + PA/γ + VA2/2g = hB + PB/γ + VB2/2g + HL
HL Line 1 = HL Line 2

27. (Minor + Major) HL1 = (Minor + Major) HL2
Example 4 A B
hA + PA/γ + VA2/2g = hB + PB/γ + VB2/2g + HL
HL Line 1 = HL Line 2
(Minor + Major) HL1 = (Minor + Major) HL2

28. (Minor + Major) HL1 = (Minor + Major) HL2
Example 4 A B
(Minor + Major) HL1 = (Minor + Major) HL2
HL = 1.1 Q L1(Q1)1.852
A12(64.4) C1.852D4.87
= 1.1 Q L2(Q1)1.852
A22(64.4) C1.852D4.87
= 0.75 m

29. HL, m
Q1 = 0.35 m/s
Q2 = 0.32 m/s
Q, m/s

32. Node
Loop

33. For Each Node:
ΣHL for Loop = 0
Qin = Qout

34. This loop is in balance for Q’s, but is it for HL?
Q = 200 gpm
HL = 20 ft
Q = 250 gpm
Q = 150 gpm B A
Q = 50 gpm
Q = 100 gpm
HL = 4 ft
HL = 15 ft C
Q = 50 gpm
This loop is in balance for Q’s, but is it for HL?
Both Q and HL need to be in balance!

35. Use ‘Successive Approximations’ to Balance
Q = 200 gpm
HL = 20 ft
Q = 250 gpm
Q = 150 gpm B A
Q = 50 gpm
Q = 100 gpm
HL = 4 ft
HL = 15 ft
AB=BC=AC=1,000 ft
D=4-in; C=100
Add K=2.0 Each Pipe C
Q = 50 gpm
Use ‘Successive Approximations’ to Balance

38. This loop is now in balance!
Q = 200 gpm
HL = 22 ft
Q = 250 gpm
Q = 135 gpm B A
Q = 65 gpm
Q = 115 gpm
HL = 6 ft
HL = 16 ft C
Q = 50 gpm
This loop is now in balance!

39. Using EPANET to Solve:
JU1
JU2
Pipe PI1
Pipe PI2
Pipe PI3
JU3

40. JU1
Pipe PI1
JU2
Pipe PI3
Pipe PI2
JU3

41. Understand concepts of “Equivalency” in piping systems.
Compute equivalent length of a pipe system.
Simplify a system using equivalency.
Understand how Continuity and Energy Principle are used to solve network problems.
Solve a simple network problems by hand.
Outcomes for today