Irrigation pumping plants
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Irrigation Pumping Plants. By Blaine Hanson University of California, Davis. 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?

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Irrigation pumping plants

Irrigation Pumping Plants


Blaine Hanson

University of California, Davis



  • 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?

Basic concepts

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

Improving pumping plant efficiency

Improving Pumping Plant Efficiency

  • Adjust pump impeller

  • Repair worn pump

  • Replace mismatched pump

  • Convert to an energy-efficient electric motor

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Centrifugal or Booster Pump













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Deep Well Turbine

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Deep Well Turbine

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  • Total head or lift

  • Capacity

  • Brake horsepower

  • Input horsepower

  • Overall efficiency

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Discharge Pressure Gauge


Discharge Pipe

Pump Head

Ground Surface

Static or Standing

Water Level

Pumping Lift

Ground Water

Pumping Water Level


Discharge pressure head

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

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Total Head or Total Lift = Pumping Lift (feet) +

Discharge Pressure Head (feet)


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

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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

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Brake Horsepower = Shaft Horsepower of Motor or Engine

Input Horsepower = Power Demand of Motor or Engine

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Overall Pumping Plant Efficiency =

Gallons per minute x Feet of Total Head

3960 x Input Horsepower

Pump performance curves

Pump Performance Curves

  • Total Head or Lift - Capacity

  • Pump Efficiency - Capacity

  • Brakehorsepower - Capacity

  • Net Positive Suction Head - Capacity (centrifugal pumps)

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Total Head - Capacity

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Efficiency - Capacity

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Horsepower - Capacity

How do you use performance curves

How Do You Use Performance Curves?

  • Selecting a new pump

  • Evaluating an existing pump

Selecting an efficient pump

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

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Selecting a New Pump

Design: Total Head = 228 feet, Capacity = 940 gpm

Common causes of poor pumping plant performance

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

Improving pumping plant performance

Improving Pumping Plant Performance

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Impeller Adjustment

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Effect of Impeller Adjustment

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Effect of Impeller

Adjustment on Energy Use

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Repair Worn Pump

Effect of pump repair


Pumping lift = 95 feet

Capacity = 1552 gpm

IHP = 83

Efficiency = 45%


Pumping lift = 118 feet

Capacity = 2008 gpm

IHP = 89

Efficiency = 67%

Effect of Pump Repair

Summary of the effect of repairing pumps

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%

Adjusting repairing pumps

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

Replace mismatched pump

Replace Mismatched Pump

A mismatched pump is one that is

operating properly, but is not operating

near its point of maximum efficiency

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Matched Pump

Efficiency (%)






Capacity (gpm)

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Mismatched Pump

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Multiple Pump Tests

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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)

Replacing a mismatched pump

Replacing a Mismatched Pump

  • Pumping plant efficiency will increase

  • Input horsepower demand will decrease

  • Energy savings will occur because of the reduced horsepower demand

How do i determine the condition of my pump

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

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Is a pump worn or mismatched?

  • Multiple pump tests

  • Compare pump test data with manufacturer’s pump performance curves

Recommended corrective action

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

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Energy-efficient Electric Motors

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Efficiencies of Standard

and Energy-efficient

Electric Motors

Variable frequency drives

Variable Frequency Drives

What is a variable frequency drive

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

When are variable frequency drives appropriate

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

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Centrifugal pump used to irrigate

Both 80-and 50-acre fields

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Note: Pumping plants should be

operated at the reduced frequency

for at least 1,000 hours per year

to be economical

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Convert To Diesel Engines

Options for converting from electric motors to engines

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%



  • 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

Engine horsepower

Engine Horsepower

  • Maximum horsepower

  • Continuous horsepower

    • About ¾’s of the maximum horsepower

    • Derated for altitude, temperature, accessories, etc.

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Electric motors vs diesel engines which is the best

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

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  • Comparison of electric motor and

  • diesel engine

  • 100 HP

  • 1,100 gpm

  • 2,000 hours per year

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That’s All, Folks

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