1. { CEE 410 Hydraulic Engineering 2- Economic Selection Michael D. Doran, P.E. DEE Professor of Practice Risk and Reliability Friction Losses in Pipes

2. Outcomes for today 1.Understand basics of economic selection (from several options) 2.Understand some basic concepts of risk and reliability 3.Understand ‘data’ vs ‘knowledge’ as potential reasons for failure 4.Understand basic concepts for computing friction losses in piping systems 1 1 May need to complete next Tues/Weds.

3. Economic Selection Economic Selection is a rigorous method for evaluating and selecting best solution among alternatives. Monetary Cost Non-Monetary Factors

4. Monetary Cost Life-Cycle Costs Present Value Costs Equivalent Annual Costs Monetary Cost Analysis incorporates projections of time value of money.

5. Monetary Cost - Symbols n = year(s) i = discount rate P = Present Value (PV) of money F = future value (at n) of money A = annual amount of money G = annual gradient amount g = decimal fraction for annual increase to A

6. Present Value of an Option Initial Capital (constr. + other) Cost +PV of Future Capital Costs +PV of Uniform Annual Costs (e.g., O&M) +PV of Gradient Annual Costs +PV of Geometric Annual Costs -PV of Salvage (Value at end of Life-Cycle) =Total PV (as Cost)

7. Present Value of an Option 5 10 15 n

8. Present Value of an Option 5 10 15 n P

9. Present Value of an Option 5 10 15 n F5F5 F 10 F 15 P

10. Present Value of an Option 5 10 15 n F5F5 F 10 F 15 A P

11. Present Value of an Option 5 10 15 n F5F5 F 10 F 15 A G P

12. Present Value of an Option 5 10 15 n F5F5 F 10 F 15 A G P AgAg

13. Present Value of an Option 5 10 15 n F5F5 F 10 F 15 A G P FSFS AgAg

15. PV Formulae and Calcs We will spend some time in class next Wednesday if needed working with PV calculations. See the posted example.

16. Non-monetary Factors How affordable? How reliable? How easy to operate & maintain? How much odor and visual pollution potential? How expandable? How upgradable? How green?

17. Decision Matrix Approach FactorScoreWeightGrade Monetary Cost Non-Monetary 1 Non-Monetary 2 Non-Monetary 3 Non-Monetary 4 …… Summation

18. Risk and Reliability http://abclocal.go.com/kgo/story?section =news/local/peninsula&id=7630626

19. Some Examples Pumping Station design error (static head of discharge point) Wastewater Treatment Plant hydraulic profile calculation error Water Treatment Plant design error for draining of Membrane Module tanks

20. Examples of Risk Inadequate DataInadequate Knowledge gal/capita; peak demands Industrial/commercial Qs Water and wastewater quality Soil/geological conditions Groundwater conditions Existing Infrastructure conditions Actual growth/development that will occur Future trends in water use Future chemical conditions Actual Service life of old and new infrastructure Long-term performance factors for infrastructure Future costs for labor, energy, etc.

21. Examples of Risk Inadequate DataInadequate Knowledge gal/capita; peak demands Industrial/commercial Qs Water and wastewater quality Soil/geological conditions Groundwater conditions Existing Infrastructure conditions Actual growth/development that will occur Future trends in water use Future chemical conditions Actual Service life of old and new infrastructure Long-term performance factors for infrastructure Future costs for labor, energy, etc. And, of course, Human Error

22. Improving Reliability Conservative hydraulic design Provide standby equipment Design for pipeline trench support and restraint at bends Careful material selection Choose pumps for future larger motor and impeller Size structures for future larger equipment Design with future connections in mind Etc.

23. Reducing Risk Collect more data ($/benefit) Use best quality materials and equipment ($/benefit) Overdesign system ($/benefit) Careful material selection Choose pumps for future larger motor and impeller Size structures for future larger equipment Design with future connections in mind Etc.

24. Reducing Risk Build strong client relationship Discuss risk factors with client Agreements should define how risk is to be shared Use appropriate agreement language to avoid responsibility for items you (as engineer) cannot control Carry adequate Professional Liability Insurance Inform client and work immediately to resolve problems when they arise

25. Pipe Friction Some factors affecting pipe friction H L Velocity (H L ~ v 2 ) Temperature/viscosity of bulk fluid Roughness of pipe interior Deposits, growths Misalignments/offsets with age

26. 3 Normal Methods of Calc MethodApplication Darcy- Weisbach General Hazen- Williams Full Pipe under pressure Water-Wastewater 40 – 75 deg F Manning’sGravity flow Partially full pipe Water-Wastewater Normal water-wastewater temp

27. Pipe Friction Relationships H L ~ L H L ~ 1/D H L ~ v 2 /2g H L ~ Re

28. Darcy-Weisbach Equation ffriction factor (unitless) Lpipe length (ft) Dpipe diameter (ft) vvelocity (ft/s) ggravitational constant (ft/s 2 ) H L ~ L H L ~ 1/D H L ~ v 2 /2g H L ~ Re US Units

29. Darcy-Weisbach Equation ffriction factor (unitless) Lpipe length (m) Dpipe diameter (m) vvelocity (m/s) ggravitational constant (9.81 m/s 2 ) H L ~ L H L ~ 1/D H L ~ v 2 /2g H L ~ Re SI Units

30. US units: ft, ft 2, ft 3, ft/s, ν in ft 2 /s

31. The flow is: laminar when Re < 2300 transient when 2300 < Re < 4000 turbulent when 4000 < Re

32. ε/D Relative Roughness Moody Diagram

33. ε Values

34. Darcy Weisbach Procedure 1.Determine viscosity and density of fluid at temperature. 2.For V, ν and D compute the Re. 3.Determine the Relative Roughness for your pipe material and diameter. 4.Read f from Moody Diagram (carefully) 5.Compute HL from Darcy-Weisbach Eqn

35. Example 1 (US) – what is the friction loss (ft) for 1,000 ft of 6.0 inch (0.50 ft) ID pipe with a flow of 50% ethylene glycol and 50% water of 300 gal/min (0.67 ft 3 /s) if the fluid is at 160 deg F? Data: ν = 10 x 10 -6 ft 2 /s (from handbook) ε = 102 x 10 -3 inches (from handbook)

36. Example 1 (US) – what is the friction loss (ft) for 1,000 ft of 6.0 inch (0.50 ft) ID pipe with a flow of 50% ethylene glycol and 50% water of 300 gal/min (0.67 ft 3 /s) if the fluid is at 160 deg F? Data: ν = 10 x 10 -6 ft 2 /s (from handbook) ε = 102 x 10 -3 inches (from handbook) Re = QD/νA = 0.67 ft 3 (0.50)ft(s) s(10 x 10 -6 ) ft 2 (π0.25 2 )ft 2 = 170 x 10 3 Flow is Turbulent (Re>4,000)

37. Example 1 (US) – what is the friction loss (ft) for 1,000 ft of 6.0 inch (0.50 ft) ID pipe with a flow of 50% ethylene glycol and 50% water of 300 gal/min (0.67 ft 3 /s) if the fluid is at 160 deg F? Data: ν = 10 x 10 -6 ft 2 /s (from handbook) ε = 102 x 10 -3 inches (from handbook) Re = 170 x 10 3

38. Example 1 (US) – what is the friction loss (ft) for 1,000 ft of 6.0 inch (0.50 ft) ID pipe with a flow of 50% ethylene glycol and 50% water of 300 gal/min (0.67 ft 3 /s) if the fluid is at 160 deg F? Data: ν = 10 x 10 -6 ft 2 /s (from handbook) ε = 102 x 10 -3 inches (from handbook) Re = 170 x 10 3 ε/D = 102 x 10 -3 /6.0 = 0.017 (Relative Roughness)

39. f = 0.047

40. ffriction factor (unitless) Lpipe length (ft) Dpipe diameter (ft) vvelocity (ft/s) ggravitational constant (ft/s 2 ) H L = 0.047(1,000 ft)(0.67/π·0.25 2 ) 2 ft 2 (s 2 ) (0.50 ft) (s 2 )(2)(32.2)ft H L = 5.0 ft

42. Do not tear out your hair 1.You need to use D-W except when you are working with water at ~40-75 deg F. 2.There are computational aids that will help you in that case. 3.For most situations you can use the simpler Hazen-Williams or Manning’s Equations.

43. Hazen-Williams Equation Use for 40-75 deg F, full pressure pipe (US) Water and Wastewater. H L = 1,043·Q 1.85 ·(C) -1.85 ·(D -4.87 ) H L in ft/100 ft of pipe L is pipe length in ft C is H-W Friction Factor (normally 100 for DI) Q is flow in gal/min D is diameter in inches

44. Hazen-Williams Equation Use for 4-25 deg C, full pressure pipe (SI) Water and Wastewater. H L = 10.67·Q 1.85 ·C -1.85 ·D -4.87 H L in m/m of pipe C is H-W Friction Factor (normally 100 for DI) Q is flow in m 3 /s D is diameter in m

45. Example 2 (US) - 1,000 ft of 12-inch ID pipe has a flow rate of 1,800 gal/min of water at 65 deg F. If the “C” factor is 100, what is the H L due to friction? = 1.22 ft/100 ft  1.22(10)ft = 12.2 ft H L = 1,043·Q 1.85 ·(C) -1.85 ·(D -4.87 ) ft/100 ft = 1,043 (1,800 1.85 )(100 -1.85 )(12 -4.87 ) ft/100 ft

46. Example 3 (SI) - 305 m of 0.30 m ID pipe has a flow rate of 0.12 m 3 /s of water at 18 deg C. If the “C” factor is 100, what is the H L due to friction? H L = 10.67·Q 1.85 ·C -1.85 ·D -4.87 m/m = 10.67(0.12 1.85 )100 -1.85 (0.30 -4.87 ) m/m = 0.014 m/m  0.014(305)m = 4.3 m

47. Manning’s Equation Use for gravity flow pipe.(US) v = 1.486 · R 0.67 ·S 0.50 n Q = v is velocity (ft/s) n is Manning’s Roughness Coefficient R is Hydraulic Radius in ft (A/P w ) S is slope (ft/ft) Q is flow rate (ft 3 /s) 1.486·A · R 0.67 ·S 0.50 n

48. Hydraulic Radius R = Area/Wetted Perimiter For full circular pipe: R = πD 2 · 1= D/4 4 πD h w For rectangular pipe not full: R = wh/(w+2h)

49. Manning’s Equation Use for gravity flow pipe.(SI) v = 1.00 · R 0.67 ·S 0.50 n Q = v is velocity (m/s) n is Manning’s Roughness Coefficient R is Hydraulic Radius in m (A/P w ) S is slope (m/m) Q is flow rate (m 3 /s) A · R 0.67 ·S 0.50 n

50. Example 4 (US) – Use Manning’s equation with n=0.015 to determine the necessary slope S for a pipe flowing full. The pipe has an ID of 1.0 ft and is to convey a flow of 2.3 ft 3 /s. Q = 1.486·A · R 0.67 ·S 0.50 n n·Q= S 0.50 1.486(A)R 0.67 S = n·Q 1.486(A)R 0.67 2 0.015(2.3) 1.486(π)(0.50 2 )(1.0/4) 0.67 = 2 S = 0.0055 ft/ft

51. Outcomes for today 1.Understand basics of economic selection (from several options) 2.Understand some basic concepts of risk and reliability 3.Understand ‘data’ vs ‘knowledge’ as potential reasons for failure 4.Understand basic concepts for computing friction losses in piping systems 1 1 May need to complete next Tues/Weds.