October 27, 2008

Tom Dorn, UNL Extension Educator

The energy cost of operating a pumping plant depends on three variables:

• the amount of work output the pump is producing
• the efficiency of the pump
• the efficiency of the power unit

Work is being done by a pumping plant when it lifts water from a well or surface water source and/or pressurizes the water for delivery through a distribution system. Engineers report the useful work performed by a pumping plant as water horsepower hours (Whp-h) and the rate of useful work as water horsepower hours per hour (Whp-h/hour).

Nebraska Pumping Plant Research

The University of Nebraska has conducted hundreds of tests over the years on farmer-owned pumping plants. Based on these field tests and laboratory tests of engine efficiency, the University developed the Nebraska Pumping Plant Performance Criteria (NPPPC). Two measures of performance are given in the NPPPC. The first is the horsepower hours you should reasonably expect per unit of energy consumed by a power unit, measured as the work input (horsepower hours, hp-h) into the pump. The other represents the amount of work imparted to the water, whp-h you should reasonably expect per unit of energy consumed by the pumping plant as a whole unit.

The University of Nebraska has conducted several pumping plant efficiency studies over the years. A study was conducted in 1980-81 where 130 farmer-owned pumping plants were tested statewide. Some pumping plants were found to be very efficient, meeting or exceeding the NPPPC (criteria). About 85% of the pumping plants were found to be using more energy than called for by the criteria . When the performance ratings of all pumping plants tested were tallied, the average pumping plant in the study was found to be operating at only 77% of the NPPPC. In layman’s terms, they were only producing 77% of the useful work as they should have been for the amount of energy consumed. Stated differently, the average pumping plant was using 30% more fuel than if it were operating at the NPPPC (1/0.77 = 1.3) .

After analyzing the initial pumping plant tests to determine the cause for poor performance, 58% of the pumping plants tested were determined to potentially benefit from adjustments that could be made by the technician in the field. The technician adjusted the pump impellers when the pump test indicated a need. After the initial pumping plant test was run, the technician attempted to improve the air/fuel mixture setting and ignition timing on all spark ignition engines. If adjustments were made, a second pumping plant test was then conducted . Adjustments, either to the engine or pump or both, resulted in an average 14% reduction in energy costs.

A part of the pump test report given to the producer was a calculation of how much additional energy could be saved if repairs that go beyond simple adjustments were made, resulting in the pumping plant being made to operate at the NPPPC. The potential energy saved was then used to compute the economic feasibility of making repairs.

North Dakota Pumping Plant Research

Hla and Scherer, NDSU, conducted a pumping plant efficiency study in the year 2000 testing a total of 37 pumping plants. Unlike the Nebraska studies where all of the pumping plants tested were deep well turbine pumps, about half (18 of 37 pumps) were deep well turbine pumps pumping groundwater to center pivot systems. The other half (19 of 37 pumps) were centrifugal pumps with low lifts pumping from surface water sources and supplying water under low pressure to surface irrigation systems.

Twenty years elapsed between the Nebraska and North Dakota studies. Yet there is good agreement in the results when comparing the deep well turbine pump results. The performance ranges chosen by the authors to group the pumping plants differed somewhat between the two studies but overall the spread is quite comparable. This demonstrates the continuing importance of knowing the performance rating of individual pumping plants in order to identify the ones that are poor.

Causes for Poor Performance

The four main causes for poor pump performance are:

• Pump selections that are poorly matched to the job they are currently doing.
• Pumps that have worn impellers and/or damaged internal seals from pumping sand.
• Improperly adjusted pumps.
• Power units that are not operating efficiently.

Example 1: Effect of Changing the Pumping Conditions Without Changing the Pump

When his rain-fed field was first converted to irrigation, Pete wanted the pump to supply 1100 gallons per minute (gpm) and provide 10 PSI pressure at the pump discharge for the gated pipe distribution system. Based on the pumping water level of 160 feet when the well was test pumped at 1100 gpm, the well driller found a pump model that fit the conditions well and chose to install a three stage Western Land Roller 12BH pump. At 1760 rpm, this pump is able to deliver 1100 gpm while producing 62 feet of head per stage (total head of 186 feet). This is exactly the head needed to lift the water to the surface, overcome a small amount of friction loss in the pump column and produce 10 PSI at the discharge. From the published head-capacity curve, the estimated efficiency of the pump operating under these conditions was over 79% and the pump was operating near its best efficiency point. The pump is powered by a diesel engine through a gear head.

Pete replaced the gated pipe system with a center pivot system several years ago. The pivot is designed to deliver 750 gpm when the pressure gauge on top of the pivot reads 40 PSI (the pressure gauge at the pump discharge reads 45 PSI). No modifications have been made to the pumping plant or diesel engine except the pump speed had to be increased from 1760 rpm to 1900 rpm in order to create the necessary system pressure. The pumping water level is now 152 feet because Pete is pumping 350 gpm less water than with the gated pipe system. The operating point for the pump is now 750 gpm and 256 feet of head (85.3 feet if head per stage).

This pump, which was designed for a gated pipe system, can indeed be made to produce the head and capacity necessary for the center pivot, but what effect has it had on the operating efficiency of the pump?

Using affinity laws, we can extrapolate a new pump curve for 1900 rpm based on the published curve for 1760 rpm. The efficiency points transferred from the published curve show the efficiency of the new operating point to be 69% . When comparing the relative efficiency of the two operating conditions, the pump is producing (69%/79% = 87.3% ) as much useful work output per unit of energy consumed as it did when it was being used for the gated pipe operating conditions.

If a new pump with an impeller design better suited to the pumping conditions and able to match the original 79% efficiency would be installed, how much energy could be saved?

The expected energy consumption is calculated as the inverse of the efficiency ratio stated above (1/0.87 = 1.145). Conclusion: This pump is using 14.5 % more energy at 69% than it would if it were operating at 79%.

How can you know if a pumping plant is using more energy than it should?

It is difficult for producers to casually assess the relative efficiency of one pumping plant versus another because there are almost always differences in differences in lift, system pressure and acre-inches of water pumped per season between one pump and another. Even when the pumping plant is using 30% to 50% more energy than called for by the NPPPC, producers are often unaware their pumps are using more energy than they should.

The best way to assess the performance of a pumping plant is to conduct a short-term pumping plant test. Four measurements are taken using instruments: The flow rate in gallons per minute (gpm). The distance from the water level inside the well casing when the pump is operating to the pump discharge (lift). The system pressure, measured at the pump discharge (PSI). And the rate of fuel (energy) consumption (gallons/hour, kWh/hour, cubic feet of natural gas per hour).

Contact a reputable well driller and ask if they are equipped to run a short-term pumping plant test. Because the actual lift and pressure are precisely measured throughout the duration of the test, a short duration test provides the best data for determining the pumping plant efficiency and the performance rating as compared to the NPPPC.

An alternative to a short-duration pumping plant test is to use long term records to estimate pumping plant performance. The information required to estimate performance includes: total volume of water pumped over time (acre-inches), either calculated from the water meter totalizer or calculated from the flow rate (gpm) and the hours of operation; typical lift (pumping water level); average pressure at the pump discharge (psi) and energy consumed over the period corresponding to the water meter readings. Large changes in pressure and lift over the season may make the results less reliable. Shorter term evaluation periods may be better in these situations.

Example 2. Using Records to Estimate Long-Term Performance of a Typical Center Pivot System in Nebraska²

1. Test period = entire season
2. System = center pivot sprinkler covering 125 acres per circle
3. Lift, a.k.a. pumping water level (PWL) = 140 feet (measured while pump is running)
4. Pressure at the pump = 40 psi
5. Acre inches (Ac-in) of water pumped based on water meter readings = 1,250 ac-in
   (This is equal to 10 inches applied to 125 acres). To convert gallons to ac-in, divide by 27,154.
6. Total fuel used for test period = 3,450 gallons of diesel.
   Calculations:
   1. whp-h = acre-inches pumped x total head (ft)/8.75 Note: total head = lift + (PSI x 2.31 ft/PSI)
      = 1,250 ac-in x (140 ft + (40 PSI x 2.31 ft/PSI))/8.75
      = 1,250 x 232.4 /8.75
      = 33,200 whp-h
   2. Performance = whp-h/energy used for the test period = 33,200 whp-h/3,450 gallons
      = 9.623 whp-h/gallon
   3. Performance Rating = Performance /NPPPC (12.5 for diesel from Table 1) x100% = (9.623/12.5) x 100%
      = 77%
   4. Potential fuel savings = ((100% - %NPPPC)/100% x Energy used for test period = ((100% - 77%)/100% x 3,450 gallons of diesel
      = 0.23 x 3,450 gallons
      = 794 gallons of diesel

¹ Note: The 77% performance rating in this example matches the average performance found in the 1980-81 PUMP project in Nebraska. The lift, pressure and acre-inches applied are typical of many center pivot irrigated corn fields in Nebraska.
²An Excel worksheet has been developed to help with long-term performance assessments as illustrated in Example 2. The worksheet is available on the Irrigation page of the Lancaster County Extension Web site at lancaster.unl.edu/ag/crops/irrigate.shtml. Click on the Long_Term_Pump.xls. link at the bottom of the section titled What Can be Done About Irrigation Energy Bills or access the worksheet directly at http://lancaster.unl.edu/ag/crops/Long_Term_Pump.xls You can open this free worksheet and run it online in most popular Internet browsers or you can save it to your computer and open it using Microsoft Excel. The worksheet includes step-by-step instructions and three samples.

The 2009 UNL Extension Crop Budgets can be found on the Surviving High Input Costs in Crop Production Web site. The diesel price used in these budgets was assumed to be $4.00 per gallon. Using this price as the basis for analysis, the potential energy cost savings resulting from bringing this example pumping plant up to the NPPPC would be 794 gallons x $4.00 per gallon = $3,176 per year.

Computing the Feasibility of Repair or Replacement

Once we know the potential annual cost savings, we can calculate the economic feasibility of making repairs or installing any new components necessary to bring a pumping plant up to100% of the NPPPC. For example, let's assume a payback period of seven years with expenses financed at 7% interest with annual payments. The CRF for 7% interest and seven years is 0.1856 (Table 4). We can borrow $3,176 / 0.1856 = $17,112 and expect to make the annual principle and interest payments on the amortized loan using only the anticipated energy cost savings (assuming a constant diesel price of $4.00 per gallon).

Net Profit After Paying for Repairs

The useful life of irrigation pumps often exceed 20 years but in our analysis above, we have set a desired payback period from energy savings to only seven years. The energy savings after the investment in repair or replacement has paid for itself is added profit. If we assume diesel price continues to be $4.00 per gallon over the next 20 years, the net return, after the repairs have been paid for, would be 13 years x $3,176 / year or $41,288.

Average Net Profit Per Acre Per Year Over the Life of the Pump

If the $41,288 net gain is spread over the full 20-year expected service life of the pump, the annual net return above repair cost is $330 per acre ($41,288/125 acres). When broken down by year, the added profit would be $16.51 per acre per year for 20 years.

Note: Pump and engine adjustments may result in considerable savings without the expense of major repairs or replacement. This option should always be explored and adjustments made by a qualified technician before resorting to major repairs or replacement of components.