1. Irrigation Pumping PlantsBy
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. Pumping Lift 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 = = 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. Design: Total Head = 228 feet, Capacity = 940 gpm
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. Effect of Pump Repair Before After 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. Matched Pump
Efficiency (%)
Improperly
Matched
Pump
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

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
A 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. 0.40
0.39
0.38
0.38
0.38
0.37
0.37
0.37
0.36
FUEL CONSUMPTION (lb/bhp-hr)
0.34
0.32
0.30
1200
1400
1600
1800
2000
2200
ENGINE RPM

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