Irrigation Pumping Plants

ByBlaine HansonUniversity 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? Will improving my pump’s

performance reduce my energy bill?

Basic Concepts

DefinitionEnergy = 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 costsReduce Input HorsepowerReduce Operating HoursReduce Unit Energy Cost

Improving Pumping Plant Efficiency

Adjust pump impellerRepair worn pumpReplace mismatched pumpConvert to an energy-efficient

electric motor

Shaft Frame Impeller Discharge Inlet

StuffingBox

BalanceLine

Volute WearingRings

Centrifugal or Booster Pump

Deep Well Turbine

Deep Well Turbine

SubmersiblePump

Terms

Total head or liftCapacityBrake horsepowerInput horsepowerOverall efficiency

Pumping Water Level

Pump

MotorDischarge Pressure Gauge

Discharge Pipe

Static or Standing Water Level

Pump Head

Ground Water

Ground Surface

Pumping Lift

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

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

Example

Pumping Lift = 100 feetDischarge Pressure = 10 psiDischarge Pressure Head = 10 psi x 2.31 = 23.1 feetTotal Head = 100 +23.1 = 123.1 feet

Intake Pressure

Discharge Pressure

Pump Discharge Pipe

Pump Intake

Total Head or Lift of Booster Pumps

Difference between pump intake pressure and pump discharge pressure

Multiply difference (psi) x 2.31 Example

o Intake pressure = 20 psio Discharge pressure = 60 psio Difference = 40 psio Total Head = 40 x 2.31 = 92.4 feet

Brake Horsepower = Shaft Horsepower of Motor or Engine

Input Horsepower = Power Demand of Motor or Engine

Overall Pumping Plant Efficiency =

Gallons per minute x Feet of Total Head

3960 x Input Horsepower

Pump Performance Curves

o Total Head or Lift - Capacityo Pump Efficiency - Capacityo Brakehorsepower - Capacityo Net Positive Suction Head -

Capacity (centrifugal pumps)

200 250 300 350 400

PUMP CAPACITY (gallons per minute)

0

50

100

150

200

250T

OTA

L H

EA

D (

feet

)

Worn Pump

New Pump

Total Head - Capacity

200 250 300 350 400

PUMP CAPACITY (gallons per minute)

0

10

20

30

40

50

60P

UM

PIN

G P

LA

NT

EF

FIC

IEN

CY

(%

)

Worn Pump New Pump

Efficiency - Capacity

0 100 200 300 400 500 600 700

CAPACITY (gpm)

0

2

4

6

8

10

BR

AK

E h

OR

SE

PO

WE

R

New Worn

Horsepower - Capacity

0 200 400 600 800 1000 1200 1400

PUMP CAPACITY (gpm)

0

20

40

60

80

100T

OTA

L H

EA

D (

fee

t)

0

20

40

60

80

100B

RA

KE

HO

RS

EP

OW

ER

Brake Horsepower

Total Head9

8.5

8

70

8082

8383

8280

83.5

How Do You Use Performance Curves?

o Selecting a new pumpo Evaluating an existing 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

A B C

Capacity (gpm) 940 940 940 Total Head per Stage (feet) 57 73 37 No. Stages 4 3 6 Actual Total Head (feet) 228 219 223 Pump Effi ciency (%) 84 69 76 Pump or Brake Horsepower 64 73 69 Annual Energy Cost ($) 7162 8225 7721

Selecting a New PumpDesign: Total Head = 228 feet, Capacity = 940 gpm

Common Causes of Poor Pumping Plant Performance

Wear (sand)Improperly matched pumpChanged pumping

conditionso Irrigation system changeso Ground water levels

Clogged impellerPoor suction conditionsThrottling the pump

Improving Pumping Plant Performance

Impeller Adjustment

Effect of Impeller Adjustment

Capacity (gpm)

Total Head (feet)

Overall Efficiency

(%)

Input Horsepower

Pump 1 Before 605 148 54 42 After 910 152 71 49 Pump 2 Before 708 181 59 55 After 789 206 63 65 Pump 3 Before 432 302 54 61 After 539 323 65 67 Pump 4 Before 616 488 57 133 After 796 489 68 144

Same Operating Time

Same Volume of Water

Pump 1 +16.7% - 22.8% Pump 2 +18.2% +5.0% Pump 3 +9.8% - 12.3% Pump 4 +8.3% - 16.8%

Effect of Impeller Adjustment on Energy Use

Repair Worn Pump

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%

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

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

A mismatched pump is one that is operating properly, but is not operating near its point of maximum efficiency

Capacity (gpm)

Effi

cie

ncy

(%)

00

Improperly Matched Pump

Matched Pump

Mismatched Pump

Pumping Plant Test Data Pumping Lift (feet) 113 Discharge Pressure (psi) 50 Total Head (feet) 228 Capacity (gpm) 940 I nput Horsepower 112 Overall Effi ciency (%) 48

Test 1 (Normal)

Test 2 Test 3

Capacity (gpm) 940 870 1060 Pressure (psi) 50 79 15 Pumping Lift (feet) 113 112 112 Total Head (feet) 228 295 147 IHP 112 112 104 Overall Efficiency (%) 48 57 38

Multiple Pump Tests

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

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?

Answer: Conduct a pumping plant test and evaluate the results using the manufacturer’s pump performance data

Pumping Lift

Pump Capacity

DischargePressure

FLOW METER

8 PIPE DIAMETERS DIAMETERS2 PIPE

FLOW

InputHorsepower

Is a pump worn or mismatched?

Multiple pump testsCompare pump test

data with manufacturer’s pump performance curves

0 100 200 300 400 500 600 700 800 900 1000 1100

PUMP CAPACITY (gpm)

0

50

100

150

200TOTAL HEAD (feet)

REPAIRED PUMPPumping Lift = 102 ftCapacity = 537 gpm

Input Horsepower = 28Overall Efficiency = 50%

Kwhr/af = 211

WORN PUMPPumping Lift = 45 ftCapacity = 624 gpm

Input Horsepower = 19Overall Efficiency = 39%

Kwhr/af = 123

LargeDifference

SmallDifference

2000 2400 2800 3200 3600

PUMP CAPACITY (gpm)

0

20

40

60

80

100TOTAL HEAD (feet)

1983 (64%)1984(54%)

1985 (62%)

2000 2400 2800 3200 3600

PUMP CAPACITY (gpm)

0

20

40

60

80

100TOTAL HEAD (feet)

1983 (64%)

1984 (66%)

1985 (55%)

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

Energy-efficient Electric Motors

Horsepower Standard Energy Efficient

10 86.5 91.7 20 86.5 93.0 50 90.2 94.5 75 90.2 95.0 100 91.7 95.8 125 91.7 96.2

Efficiencies of Standard and Energy-efficient Electric Motors

Variable Frequency Drives

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 demando A small reduction in pump rpm results in

a large reduction in the horsepower demand

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

Unthrottled Throttled VFD

Acres 80 50 50 Pressure (psi) 80 64 60 Capacity (gpm) 1,100 600 700 Input Horsepower 128 90 55 RPM 1770 1770 1345 Overall Efficiency (%) 40 24 44

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

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

Convert To Diesel 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%

Considerations

Brake Horsepower = Shaft HorsepowerEngines 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

Maximum horsepower Continuous horsepower

About ¾’s of the maximum horsepower

Derated for altitude, temperature, accessories, etc.

110

128

144

157167

173

1200 1400 1600 1800 2000 2200

ENGINE RPM

0

50

100

150

200B

RA

KE

HO

RS

EP

OW

ER

0.39

0.38

0.37 0.370.37

0.38

1200 1400 1600 1800 2000 2200

ENGINE RPM

0.30

0.32

0.34

0.36

0.38

0.40F

UE

L C

ON

SU

MP

TIO

N (

lb/b

hp

-hr)

33.2

34.2

35.1 35.134.7

33.9

1200 1400 1600 1800 2000 2200

ENGINE RPM

30

32

34

36

38

40

EN

GIN

E E

FF

ICIE

NC

Y (

%)

1400 1500 1600 1700 1800 1900 2000 2100 2200

RPM

0

20

40

60

80

100

120

140

160H

OR

SE

PO

WE

RPUMP HP CONTINUOUS ENGINE HP

Fuel Use Versus RPM RPM Pump Flow

Rate (gpm) Gallons of Diesel

per Hour Gallons of Water

per Gallon of Diesel

1500 1228 9 8187 1600 1731 11 9617 1700 2161 15 8644 1800 2486 19 8019

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

Electric Motor

Diesel Engine

Capital Cost $5,500 $11,500 $16,500 $16,500 Unit Energy Cost $0.14/kwhr $0.95/gal $0.95/gal $1.25/gal Total Cost ($/af) 60.5 37.8 39.9 48.5

Comparison of electric motor and diesel engine100 HP1,100 gpm2,000 hours per year

That’s All, Folks