Transcript Irrigation Pumping Plants

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?
 Will improving my pump’s performance
reduce my energy bill?
Basic Concepts
 Definition
Energy = kilowatt-hours
o One kilowatt is 1.34 horsepower
o 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
Adjust pump impeller
Repair worn pump
Replace mismatched pump
Convert to an energy-efficient
electric motor
Centrifugal or Booster Pump
Shaft
Frame
Stuffing
Box
Impeller
Balance
Line
Discharge
Volute
Inlet
Wearing
Rings
Deep Well Turbine
Deep Well Turbine
Submersible
Pump
Terms
Total head or lift
Capacity
Brake horsepower
Input horsepower
Overall efficiency
Motor
Ground Surface
Discharge Pressure Gauge
Discharge Pipe
Pump Head
Static or Standing
Water Level
Pumping Lift
Ground Water
Pumping Water Level
Pump
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 feet
Discharge Pressure = 10 psi
Discharge Pressure Head = 10 psi x 2.31 = 23.1 feet
Total Head = 100 +23.1 = 123.1 feet
Discharge Pressure
Intake 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
o
o
o
Intake pressure = 20 psi
Discharge pressure = 60 psi
Difference = 40 psi
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
o
o
o
Total Head or Lift - Capacity
Pump Efficiency - Capacity
Brakehorsepower - Capacity
Net Positive Suction Head - Capacity
(centrifugal pumps)
Total Head - Capacity
250
Worn Pump
New Pump
TOTAL HEAD (feet)
200
150
100
50
0
200
250
300
350
PUMP CAPACITY (gallons per minute)
400
PUMPING PLANT EFFICIENCY (%)
Efficiency - Capacity
60
50
40
30
20
10
Worn Pump
New Pump
0
200
250
300
350
400
PUMP CAPACITY (gallons per minute)
Horsepower - Capacity
BRAKE hORSEPOWER
10
8
6
4
2
New
Worn
300
400
0
0
100
200
CAPACITY (gpm)
500
600
700
100
100
9
Total Head
80
70
8.5
80
82
8
60
83
83
60
82
80
40
40
83.5
Brake Horsepower
20
20
0
0
200
400
600
800
1000
PUMP CAPACITY (gpm)
1200
0
1400
BRAKE HORSEPOWER
TOTAL HEAD (feet)
80
How Do You Use Performance Curves?
o Selecting a new pump
o 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
Selecting a New Pump
Design: Total Head = 228 feet, Capacity = 940 gpm
A
Capacity (gpm)
Total Head per Stage (feet)
No. Stages
Actual Total Head (feet)
Pump Efficiency (%)
Pump or Brake Horsepower
Annual Energy Cost ($)
B
C
940
940
940
57
73
37
4
3
6
228
219
223
84
69
76
64
73
69
7162 8225 7721
Common Causes of Poor Pumping
Plant Performance
Wear (sand)
Improperly matched pump
Changed pumping conditions
o Irrigation system changes
o Ground water levels
Clogged impeller
Poor suction conditions
Throttling the pump
Improving Pumping Plant Performance
Impeller Adjustment
Effect of Impeller Adjustment
Capacity
(gpm)
Total
Overall
Input
Head Efficiency Horsepower
(feet)
(%)
Pump 1 Before
After
605
910
148
152
54
71
42
49
Pump 2 Before
After
708
789
181
206
59
63
55
65
Pump 3 Before
After
432
539
302
323
54
65
61
67
Pump 4 Before
After
616
796
488
489
57
68
133
144
Effect of Impeller
Adjustment on Energy Use
Pump
Pump
Pump
Pump
1
2
3
4
Same Operating Same Volume
Time
of Water
+16.7%
-22.8%
+18.2%
+5.0%
+9.8%
-12.3%
+8.3%
-16.8%
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
Efficiency (%)
Matched Pump
Improperly
Matched
Pump
0
0
Capacity (gpm)
Mismatched Pump
Pumping Plant Test Data
Pumping Lift (feet)
113
Discharge Pressure (psi) 50
Total Head (feet)
228
Capacity (gpm)
940
Input Horsepower
112
Overall Efficiency (%)
48
Multiple Pump Tests
Test 1 Test 2 Test 3
(Normal)
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
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
Discharge
Pressure
Pump
Capacity
8 PIPE DIAMETERS
2 PIPE
DIAMETERS
FLOW
FLOW METER
Input
Horsepower
Is a pump worn or mismatched?
Multiple pump tests
Compare pump test data
with manufacturer’s pump
performance curves
200
TOTAL HEAD (feet)
REPAIRED PUMP
Pumping Lift = 102 ft
Capacity = 537 gpm
Input Horsepow er = 28
Ov erall Efficiency = 50%
Kw hr/af = 211
150
Small
Difference
100
Large
Difference
WORN PUMP
Pumping Lift = 45 ft
Capacity = 624 gpm
Input Horsepow er = 19
Ov erall Efficiency = 39%
Kw hr/af = 123
50
0
0
100
200
300
400
500
600
700
PUMP CAPACITY (gpm)
800
900
1000
1100
100
TOTAL HEAD (feet)
80
60
1984(54%)
1983 (64%)
1985 (62%)
40
20
0
2000
2400
2800
PUMP CAPACITY (gpm)
3200
3600
100
TOTAL HEAD (feet)
80
1983 (64%)
1984 (66%)
60
1985 (55%)
40
20
0
2000
2400
2800
PUMP CAPACITY (gpm)
3200
3600
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
Efficiencies of Standard
and Energy-efficient
Electric Motors
Horsepower Standard
10
20
50
75
100
125
86.5
86.5
90.2
90.2
91.7
91.7
Energy
Efficient
91.7
93.0
94.5
95.0
95.8
96.2
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 demand
o 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 differentlysized 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
Centrifugal pump used to irrigate
Both 80-and 50-acre fields
Acres
Pressure (psi)
Capacity (gpm)
Input Horsepower
RPM
Overall Efficiency (%)
Unthrottled
80
80
1,100
128
1770
40
Throttled
50
64
600
90
1770
24
VFD
50
60
700
55
1345
44
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 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
Maximum horsepower
Continuous horsepower
 About ¾’s of the maximum horsepower
 Derated for altitude, temperature,
accessories, etc.
200
167
173
BRAKE HORSEPOWER
157
144
150
128
110
100
50
0
1200
1400
1600
1800
ENGINE RPM
2000
2200
FUEL CONSUMPTION (lb/bhp-hr)
0.40
0.39
0.38
0.38
0.38
0.37
0.37
1600
1800
0.37
0.36
0.34
0.32
0.30
1200
1400
ENGINE RPM
2000
2200
40
ENGINE EFFICIENCY (%)
38
36
35.1
35.1
34.7
34.2
33.9
34
33.2
32
30
1200
1400
1600
1800
ENGINE RPM
2000
2200
160
PUMP HP
CONTINUOUS ENGINE HP
140
HORSEPOWER
120
100
80
60
40
20
0
1400
1500
1600
1700
1800
RPM
1900
2000
2100
2200
RPM
1500
1600
1700
1800
Fuel Use Versus RPM
Pump Flow Gallons of Diesel
Rate (gpm)
per Hour
1228
1731
2161
2486
9
11
15
19
Gallons of Water
per Gallon of
Diesel
8187
9617
8644
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
Comparison of electric motor and
diesel engine
100 HP
1,100 gpm
2,000 hours per year
Capital Cost
Unit Energy Cost
Total Cost ($/af)
Electric
Motor
$5,500
$0.14/kwhr
60.5
Diesel Engine
$11,500
$0.95/gal
37.8
$16,500
$0.95/gal
39.9
$16,500
$1.25/gal
48.5
That’s All, Folks