1. Irrigation Pumping Plants By Blaine Hanson University of California, Davis

2. Questions  How do pumps perform?  How can I select an efficient pump?  What causes a pump to become inefficient?  How can I determine my pump’s performance?  How can I improve my pump’s performance?  Will improving my pump’s performance reduce my energy bill?

3. Basic Concepts  Definition  Energy = kilowatt-hours oOne kilowatt is 1.34 horsepower oHours = operating time  Energy cost is based on kwhr consumed and unit energy cost ($/kwhr)  Reducing energy costs  Reduce Input Horsepower  Reduce Operating Hours  Reduce Unit Energy Cost

4. Improving Pumping Plant Efficiency  Adjust pump impeller  Repair worn pump  Replace mismatched pump  Convert to an energy-efficient electric motor

5. ShaftFrameImpellerDischargeInlet Stuffing Box Balance Line VoluteWearing Rings Centrifugal or Booster Pump

6. Deep Well Turbine

8. Submersible Pump

10. Terms  Total head or lift  Capacity  Brake horsepower  Input horsepower  Overall efficiency

11. Pumping Water Level Pump Motor Discharge Pressure Gauge Discharge Pipe Static or Standing Water Level Pump Head Ground Water Ground Surface Pumping Lift

12. Discharge Pressure Head  Height of a column of water that produces the desired pressure at its base  Discharge pressure head (feet) = discharge pressure (psi) x 2.31  Note: a change in elevation of 2.31 feet causes a pressure change of 1 psi

13. Total Head or Total Lift = Pumping Lift (feet) + Discharge Pressure Head (feet) Example Pumping Lift = 100 feet Discharge Pressure = 10 psi Discharge Pressure Head = 10 psi x 2.31 = 23.1 feet Total Head = 100 +23.1 = 123.1 feet

15. Total Head or Lift of Booster Pumps  Difference between pump intake pressure and pump discharge pressure  Multiply difference (psi) x 2.31  Example oIntake pressure = 20 psi oDischarge pressure = 60 psi oDifference = 40 psi oTotal Head = 40 x 2.31 = 92.4 feet

16. Brake Horsepower = Shaft Horsepower of Motor or Engine Input Horsepower = Power Demand of Motor or Engine

17. Overall Pumping Plant Efficiency = Gallons per minute x Feet of Total Head 3960 x Input Horsepower

18. Pump Performance Curves oTotal Head or Lift - Capacity oPump Efficiency - Capacity oBrakehorsepower - Capacity oNet Positive Suction Head - Capacity (centrifugal pumps)

19. Total Head - Capacity

20. Efficiency - Capacity

21. Horsepower - Capacity

23. How Do You Use Performance Curves? oSelecting a new pump oEvaluating an existing pump

24. Selecting an Efficient Pump Information needed –Flow rate (gallons per minute) –Total Head (feet) Consult pump catalogs provided by pump manufacturers to find a pump that will provide the desired flow rate and total head near the point of maximum efficiency

25. Selecting a New Pump Design: Total Head = 228 feet, Capacity = 940 gpm

26. Common Causes of Poor Pumping Plant Performance  Wear (sand)  Improperly matched pump  Changed pumping conditions oIrrigation system changes oGround water levels  Clogged impeller  Poor suction conditions  Throttling the pump

27. Improving Pumping Plant Performance

28. Impeller Adjustment

33. Effect of Impeller Adjustment

34. Effect of Impeller Adjustment on Energy Use

35. Repair Worn Pump

38. Effect of Pump Repair  Before  Pumping lift = 95 feet  Capacity = 1552 gpm  IHP = 83  Efficiency = 45%  After  Pumping lift = 118 feet  Capacity = 2008 gpm  IHP = 89  Efficiency = 67%

39. Summary of the Effect of Repairing Pumps  63 pump tests comparing pump performance before-and-after repair  Average percent increase in pump capacity – 41%  Average percent increase in total head – 0.5% (pumping lift only)  Average percent increase in pumping plant efficiency – 33%  IHP increased for 58% of the pumping plants. Average percent increase in input horsepower – 17%

40. Adjusting/Repairing Pumps  Adjustment/repair will increase pump capacity and total head  Adjustment/repair will increase input horsepower  Reduction in operating time is needed to realize any energy savings  More acres irrigated per set  Less time per set  Energy costs will increase if operating time is not reduced

41. Replace Mismatched Pump A mismatched pump is one that is operating properly, but is not operating near its point of maximum efficiency

42. Capacity (gpm) Efficiency (%) 0 0 Improperly Matched Pump Matched Pump

43. Mismatched Pump

44. Multiple Pump Tests

45. Replacing this pump with one operating at an overall efficiency of 60% would:  Reduce the input horsepower by 19%  Reduce the annual energy consumption by 34,000 Kwhr  Reduce the annual energy costs by $3,400 (annual operating time of 2000 hours and an energy cost of $0.10/kwhr)

46. Replacing a Mismatched Pump  Pumping plant efficiency will increase  Input horsepower demand will decrease  Energy savings will occur because of the reduced horsepower demand

47. How do I determine the condition of my pump? Answer: Conduct a pumping plant test and evaluate the results using the manufacturer’s pump performance data

48. Pumping Lift

49. Pump Capacity Discharge Pressure

51. Input Horsepower

52. Is a pump worn or mismatched?  Multiple pump tests  Compare pump test data with manufacturer’s pump performance curves

56. Recommended Corrective Action Eo greater than 60% - no corrective action  55% to 60% - consider adjusting impeller  50% to 55% - consider adjusting impeller; consider repairing or replacing pump if adjustment has no effect  Less than 50% - consider repairing or replacing pump

57. Energy-efficient Electric Motors

58. Efficiencies of Standard and Energy-efficient Electric Motors

59. Variable Frequency Drives

60. What is a Variable Frequency Drive?  Electronic device that changes the frequency of the power to an electric motor  Reducing the power frequency reduces the motor rpm  Reducing the motor rpm, and thus the pump rpm, decreases the pump horsepower demand oA small reduction in pump rpm results in a large reduction in the horsepower demand

62. When are Variable Frequency Drives Appropriate?  One pump is used to irrigate differently- sized fields. Pump capacity must be reduced for the smaller fields  Number of laterals changes during the field irrigation (odd shaped fields)  Fluctuating ground water levels  Fluctuating canal or ditch water levels

63. Centrifugal pump used to irrigate Both 80-and 50-acre fields

64. Note: Pumping plants should be operated at the reduced frequency for at least 1,000 hours per year to be economical

65. Convert To Diesel Engines

66. Options for Converting From Electric Motors to Engines &Direct drive (gear head) *Engine shaft to pump shaft efficiency = 98% &Diesel-generator *Engine shaft to pump shaft efficiency less than about 80%

67. Considerations  Brake Horsepower = Shaft Horsepower  Engines and motors are rated based on brake horsepower ( 100 HP electric motor provides the same horsepower as a 100 HP engine  Input horsepower of an engine is greater than that of an electric motor for the same brake horsepower

68. Engine Horsepower  Maximum horsepower  Continuous horsepower  About ¾’s of the maximum horsepower  Derated for altitude, temperature, accessories, etc.

70. 0.39 0.38 0.37 0.38 120014001600180020002200 ENGINE RPM 0.30 0.32 0.34 0.36 0.38 0.40 FUEL CONSUMPTION (lb/bhp-hr)

74. Electric Motors vs Diesel Engines: Which is the Best?  Unit energy cost  Capital costs, maintenance costs, etc  Hours of operation  Horsepower  Cost of pollution control devices for engines

75. Comparison of electric motor and diesel engine  100 HP  1,100 gpm  2,000 hours per year

77. That’s All, Folks