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

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

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

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

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

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.

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 0.581 ft/100 ft f for 8-inch pipe at at 1,200 gpm is 4.18 ft/100 ft

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 0.581 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?

From Hazen-Williams Equation (C=100): Example 1 f for 12-inch pipe at 1,200 gpm is 0.581 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 100 ft 12-in 100 ft 8-in ft HL

From Hazen-Williams Equation (C=100): Example 1 f for 12-inch pipe at 1,200 gpm is 0.581 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 100 ft 12-in 100 ft 8-in ft HL

From Hazen-Williams Equation (C=100): Example 1 f for 12-inch pipe at 1,200 gpm is 0.581 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) 4.18ft HL 100 ft 12-in 100 ft 8-in 0.581 ft HL L12 = 1,440 ft

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

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

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

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

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

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

Example 3 Board example using capacity table and equivalent lengths.

Example 4 Q1 Q2 Le = 100 m; 0.76 m dia C = 100 WS = 300.00 m Pipes are submerged and flowing full. WS = 299.25 m Le = 225 m; 0.76 m dia C = 100 Q2

Example 4 Q1 Q2 Calculate Q1 and Q2. Le = 100 m; 0.76 m dia C = 100 WS = 300.00 m Pipes are submerged and flowing full. WS = 299.25 m Le = 225 m; 0.76 m dia C = 100 Q2 Calculate Q1 and Q2.

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

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

(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

(Minor + Major) HL1 = (Minor + Major) HL2 Example 4 A B (Minor + Major) HL1 = (Minor + Major) HL2 HL = 1.1 Q12 + 10.7L1(Q1)1.852 A12(64.4) C1.852D4.87 = 1.1 Q22 + 10.7L2(Q1)1.852 A22(64.4) C1.852D4.87 = 0.75 m

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

Node Loop

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

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!

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

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!

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

JU1 Pipe PI1 JU2 Pipe PI3 Pipe PI2 JU3

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