Purpose – The purpose of this advisory is to…
- Provide a brief description of how energy use and costs for pumping can be estimated
- Identify options for reducing energy use and cost
- Explain how a pump test is conducted and where to go to get a pump test
- Provide a calculator for estimating pumping costs and required power
INTRODUCTION
The imperative for improved water use efficiency in California will probably result in a significant amount of acreage being converted from surface irrigation systems to pressurized systems (sprinkler or micro). This Advisory explains how to estimate both energy requirements and costs for irrigation pumping. Sections are provided for both electric and diesel power. There is a short discussion on options for reducing energy use. Pump tests for determining pumping plant efficiency are explained. Definitions for some commonly used terms are provided. Immediately following is an energy use/costs calculator
IMPORTANT!! If you are one of those who will convert from a surface irrigation system to a pressurized system, make sure that your system designer is considering energy costs. This means that he is considering both the capital costs of the system as well as the annual operational costs.
Estimating Electric Energy Use for Irrigation
Estimating energy use for irrigation water pumping in any particular irrigation system can be complicated by many factors. For example, a basic question is what is the boundary of the subject irrigation system? That is, is the boundary set at the field where there may be a booster pump for a sprinkler system and the pipes and sprinklers of the system itself? Does the boundary include the well that supplies the water for the irrigation system? If the primary water supply is from an irrigation district through canals, are the pumps in the irrigation district included?
For analysis purposes, it is assumed that there is a single pumping plant supplying water to an irrigation system. This will provide a simple conceptual basis for identifying the potential options for energy savings. In this situation, the energy used for pumping water for irrigation can be calculated using four equations.
[1] kWh/yr = kWh/AF * AF/yr
where:
- kWh/yr = kiloWatt-hours usage by the irrigation system per year
- kWh/AF = kiloWatt-hours required to pump an acre-foot of water through the irrigation system
- AF/yr = acre-feet of water pumped through the system
And:
[2] kWh/AF = 1.0241 * TDH / OPE
where:
- kWh/AF = kiloWatt-hours required to pump an acre-foot of water through the irrigation system
- TDH = total dynamic head required by the system in feet (go to definition of TDH)
- OPE = overall pumping plant efficiency as a decimal (0 – 1.0)
And:
[3] AF/yr = Ac * (CL + (ETcyr – PPTeff) / ((1 – LR) * IE))
where:
- AF/yr = acre-feet of water pumped through the system
- Ac = acres of cropped land
- CL = conveyance system losses in acre-feet
- ETcyr = annual net crop water use in acre-feet/acre
- PPTeff = annual effective rainfall in acre-feet/acre
- LR = required leaching ratio to maintain a salt balance as a decimal
- IE = irrigation efficiency as a decimal
Equation [3] may also be written as
[3a] AF/yr = Ac * (ETcyr – PPTeff) / ((1 – LR) * CE * IE)
where:
- AF/yr = acre-feet of water pumped through the system
- Ac = acres of cropped land
- ETcyr = annual net crop water use in acre-feet/acre
- PPTeff = annual effective rainfall (rainfall that infiltrates and is stored for subsequent use by the crop) in acre-feet/acre
- LR = required leaching ratio to maintain a salt balance as a decimal
- IE = irrigation efficiency as a decimal
- CE = conveyance efficiency as a decimal
LR, the Leaching Ratio, is an indication as to how much applied irrigation water has to become deep percolation in order to leach excess salts from the root zone. The intent is to maintain a balance between the amount of salts being added to the root zone and the amount of salts being leached through the root zone. It should be understood that a salt balance (that is, the amount of salts applied with irrigation water equals the amount of salts leaving through deep percolation) can be maintained at any given level of salinity in the rootzone. However, rootzone salinity higher than some “threshold” level will result in yield declines. This threshold salinity is different for different crops and is probably different for different soil profiles and climates also.
Salinity theory is an inexact science at best. There have been several equations developed for determining the required LR. One of the most commonly used is:
[4] LR = ECi / (5 * ECe – ECi)
where:
- LR = required ratio of applied water that becomes deep percolation in order to maintain a salt balance in the effective root zone
- ECi = electrical conductivity of the irrigation water in deciSiemens/meter
- ECe = electrical conductivity of the saturated soil extract in deciSiemens/meter – usually set at the level that will just start to impact yields
Estimating Fuel Requirements for Irrigation
Estimating energy use for irrigation water pumping in any particular irrigation system can be complicated by many factors. For example, a basic question is what is the boundary of the subject irrigation system? That is, is the boundary set at the field where there may be a booster pump for a sprinkler system and the pipes and sprinklers of the system itself? Does the boundary include the well that supplies the water for the irrigation system? If the primary water supply is from an irrigation district through canals, are the pumps in the irrigation district included?
For analysis purposes, it is assumed that there is a single pumping plant supplying water to an irrigation system. This will provide a simple conceptual basis for identifying the potential options for energy savings. In this situation, the energy used for pumping water for irrigation can be calculated using four equations.
[5] Gal/yr = Hr/yr * BHP * Gal/BHP-Hr
where:
- Gal/yr = gallons of fuel required per season
- Hr/yr = the hours the pump runs each year
- BHP = the brakehorsepower requirements
- Gal/BHP-Hr = specific fuel consumption, gallons of fuel required per brakehorsepower-hour
and:
[6] Hr/yr = 5427 * (Cl + (Ac * (ETc – PPTeff) / ((1-LR) * IE))) / Q
where:
- Ac = acres of cropped land
- CL = conveyance system losses in acre-feet
- ETcyr = annual net crop water use in acre-feet/acre
- PPTeff = annual effective rainfall in acre-feet/acre
- LR = required leaching ratio to maintain a salt balance as a decimal
- IE = irrigation efficiency as a decimal
- Q = pump flow in gallons per minute
and:
[7] BHP = Q * TDH / 3960
where:
- BHP = required brake horsepower
- Q = pump flow in gallons per minute
- TDH = total dynamic head in system in feet
Please note again equation [4] as a common method for determining the leaching ratio, LR:
[4] LR = ECi / (5 * ECe – ECi)
where:
- LR = required ratio of applied water that becomes deep percolation in order to maintain a salt balance in the effective root zone
- ECi = electrical conductivity of the irrigation water in deciSiemens/meter
- ECe = electrical conductivity of the saturated soil extract in deciSiemens/meter – usually set at the level that will just start to impact yields
Options for Reducing Energy Use and Costs
Now, looking at equation [1] it is clear that to reduce energy use one must either reduce the kWh/AF or the AF/yr pumped. In other words you must reduce either the kiloWatt-hours required to pump each acre-foot of water or reduce the number of acre-feet pumped.
In regards to kWh/AF, equation [2] indicates that one must reduce TDH or increase OPE to reduce kWh/AF. In other words you need to reduce the pressure required by your irrigation system or improve the operating efficiency of the pump.
In regards to AF/year, equation [3] indicates that one must reduce Ac, CL, ETc, or LR, or increase PPTeff or IE. Taking these one-by-one:
- Ac – reduce acreage – you can always reduce planted acreage to reduce energy costs but this may not be a feasible option for many.
- CL – reduce conveyance system losses – compact or line earthen canals and reservoirs clean ditchbank and reservoir weeds; reduce the amount of time water is in the conveyance system (to reduce evaporative losses)
- ETc – reduce crop water use – in most situations this would mean changing crops although in some cases possibly a variety change may be warranted. There is much research and argument on-going currently regarding the split between the E (evaporation from soil and plant surfaces) and T (transpiration through the plant stomata) and the potential to reduce one or the other. Note that some forms of irrigation may reduce the E, notably sub-surface drip. However, sub-surface drip may act to increase plant vigor and thus increase the T.
- LR – reduce the required leaching ratio – this requires better quality water supplies or a more salt-tolerant crop (see equation [4]).
- IE – improve irrigation efficiency – this may be accomplished by changes in management or in the irrigation system. Note that changing an irrigation system from furrow to drip may actually increase energy use as the improvement in IE is offset by the required pumping pressure of the drip system. However, there have been projects where very deep well pumping was involved where this conversion did result in a net energy savings.
Note that the above (reducing Ac, Cl, ETc, and LR, or increasing IE) will also work to reduce the hours of operation per year as calculated by equation [6].
Equation [6] also involves the pump flow Q. It might be thought that increasing Q would decrease energy costs since it is in the denominator. But also note that Q is in the numerator of equation [7]. Thus, when equations [6] and [7] are inserted in equation [5], the pump flow cancels out. An increase in pump flow will decrease the required operating hours but will increase horsepower requirements. Thus, theoretically, Q does not impact energy use.
IMPORTANT!! However, it must be noted that increasing pump flow will tend to increase the total power bill for a number of reasons, including 1) larger pumps cost more and 2) larger pumps require more demand charges if electric.
Now there is one more thing that can be done to reduce energy costs, reduce the UNIT COST of energy. Consider the following:
- If your account is with one of the major utilities check with your account representative to see if you are on the right rate schedule. The correct rate schedule can make a huge difference in you average cost per kiloWatt-hour.
- If you are part of an aggregated group and are buying energy in bulk, make sure that your group’s representative has compared different bulk providers.
- Consider if you are part of an appropriate aggregate group.
- If buying in bulk check to see if a more accurate prediction of when and how much energy you will need can help in negotiating a lower price. There are consultants available who can help you develop a “load profile” for your pumps.
If you are using diesel, check that your economies of scale are correct. That is, do you have the appropriate amount of on-farm storage. Possibly with more storage you could negotiate a lower delivered price. Obviously the cost of increased storage must be compared to fuel cost savings..
What is a Pump Test?
A pump test is nothing more than a comparison between how much energy your pumping plant is using and how much it is producing.. The useable energy it produces is called water horsepower, the combination of water flow and pressure produced by the pumping plant. If an electric-powered plant you may have heard the term “wire-to-water” efficiency. Again, this is just indicating that the pump test is a comparison of how much electric energy is being consumed versus the combination of water flow and pressure produced.
Note the following relationships:
[8] WHP = Q * TDH / 3960
where:
- WHP = water horsepower produced by the pumping plant
- Q = water flow through the plant in gallons per minute
- TDH = total dynamic head of the pumping system in feet of water head
Note also that 1 horsepower = .746 kilowatts
Thus, when performing a pump test for an electric-powered plant, basically three variables need to be measured:
- The pump flow
- The total dynamic head of the pumping system
- The kilowatts input to the pumping plant
The Overall Plant Efficiency, OPE, is calculated as:
[9] OPE = WHP / HPin
where:
- OPE = overall plant efficiency as a decimal
- HP in = input horsepower
- WHP = water horsepower output
Note requirements for a valid pump test – Many pumps have been installed without flow meters. Pump testers can measure flow from a pump without a flow meter. They use a pitot tube inserted through a small hole in the discharge piping. However, there are some requirements. The diagram below explains the basics for a water well. The important factors are:
- The availability of the Sounding Tube – This allows the tester to measure the pumping water level.
- The access hole for the Pitot Tube which must have clear (no valves, turns, TEEs, etc.) piping for eight diameters upstream of the access hole and two diameters downstream. This ensures a smooth water flow for the pitot tube to measure.
If these requirements are not met then the pump test may not be accurate.
Another factor to consider when interpreting the results of a pump test, especially tests on water wells, is how representative was the operating condition. That is, when the pump test was performed, was the pump operating under normal conditions? Pumps that are operating outside of their normal conditions, with a water table that is lower than normal due to drought for example, will not be as efficient.
Pump tests may be available from your electric utility if one of their customers (or natural gas). Otherwise, it is best to contact your local pump dealer.
What is Total Dynamic Head (TDH)?
Total Dynamic Head, TDH, is the total head (another word for pressure) that the pumping system is working against. That is, it is the total head that the pump must overcome to move water. In reality this includes friction losses through the pump intake, in the pump column, and through the pump head. However, when testing pumps the only two factors measured are the the lift on the suction side of the pump and discharge head on the discharge side. Figure 2 is a schematic depiction of a typical well installation, showing the two components that would be measured, the Pumping Water Level and the Discharge Head. Thus, TDH for the purposes of a pump test would be the Pumping Water Level + Discharge Head. Many times, the Discharge Head will be read with a pressure gauge installed on the pump discharge piping.
It was just stated that the word “head” means the same as “pressure”. Most people are familiar with pressure gauges that read in units of pounds per square inch, psi. However, more often than not in water resources engineering we work with pressure in terms of “feet of water head”. The conversion rate is 1 pound-per-square-inch = 2.31 feet of water head at standard conditions.
What is Water Horsepower vs. Brake Horsepower vs. Overall Plant Efficiency?
Horsepower is a measure of the ability to do work over some period of time. That is, two pumps can lift 100 gallons of water 50 feet. The questions is, how long does it take to do this work? Note that in equation [8] for determining water horsepower, the water flow rate is involved, volume of water per unit time.
Most power sources convert one form of energy to another. In doing so there are losses. Thus, as in most things, the value of the horsepower rating depends on where you take the measurement.
For simplicity’s sake assume an engine coupled directly to a pump. The input horsepower to this pumping plant might actually be considered the energy value of the fuel consumption rate. The brake horsepower is the power at the driveshaft of the engine. It will be lower than the input power due to friction losses in the engine and the inability to fully utilize the power inherent in the fuel. The efficiency of the engine might be measured in terms of the Specific Fuel Consumption (BrakeHorsepower-Hours per Gallon of fuel in English units). The Specific Fuel Consumption and the brakehorsepower of the engine will change at different engine speeds.
The water horsepower is a result of the combination of the flow rate and total dynamic head from the pump. It will be lower than the brakehorsepower of the engine due to friction losses within the pump.
Overall Plant Efficiency (OPE) is a term peculiar to electric-powered pumping plants. It is a measure of how much input power, in terms of kiloWatts, is converted to water horsepower. Note that 1 horsepower equals .746 kiloWatts.