Irrigation Pumping Plants

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 • One kilowatt is 1.34 horsepower • Hours = 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. Centrifugal or Booster Pump Shaft Frame Impeller Discharge Inlet Stuffing Box Balance Line Volute Wearing Rings

6. Deep Well Turbine

7. Deep Well Turbine

8. Submersible Pump

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

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

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 • Intake pressure = 20 psi • Discharge pressure = 60 psi • Difference = 40 psi • Total 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 • Total Head or Lift - Capacity • Pump Efficiency - Capacity • Brakehorsepower - Capacity • Net Positive Suction Head - Capacity (centrifugal pumps)

19. Total Head - Capacity

20. Efficiency - Capacity

21. Horsepower - Capacity

23. How Do You Use Performance Curves? • Selecting a new pump • Evaluating 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 • Irrigation system changes • Ground 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. Before Pumping lift = 95 feet Capacity = 1552 gpm IHP = 83 Efficiency = 45% After Pumping lift = 118 feet Capacity = 2008 gpm IHP = 89 Efficiency = 67% Effect of Pump Repair

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

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. Discharge Pressure Pump Capacity