Energy Use/Costs for Pumping 
In This Advisory 
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.

Energy Use/Cost Calculator
IMPORTANT!! Water Horsepower is not the horsepower rating of the electric motor (or engine). It is the output power of a
pump measured by the flow and total dynamic head generated. Specifying motor or engine sizes should only be done by competent pump engineers.

Instructions 
IMPORTANT!! Choose the units and energy type first. 
To Calculate Flow knowing Total Dynamic Head and Water Horsepower
a. Enter 2. and 3. below
b. Press button marked "FLOW", read at 1.

To Calculate Total Dynamic Head knowing Flow and Water Horsepower
a. Enter 1. and 3. below
b. Press button marked "HEAD", read at 2.

To Calculate Water Horsepower knowing Flow and Total Dynamic Head
a. Enter 1. and 2. below
b. Press button marked "WHP", read at 3.

To Calculate Energy Use and Costs per Unit Water
a. Enter either 1. and 2., or 3 only below
b. Enter 4. if electricpowered, 5. if fuel
c. Enter 6. if known
d. Press button marked "USE / COST", read at 9. and 10.

To Calculate Annual/Seasonal Energy Use and Costs
a. Enter either 1. and 2., or 3 only below
b. Enter 4. if electricpowered, 5. if fuel
c. Enter 6. if known
d. Enter either 1. and 7., or 8. only
e. Press button marked "USE / COST", read at 9., 10., 11., and 12.



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 = kiloWatthours usage by the irrigation system per year
kWh/AF = kiloWatthours required to pump an acrefoot of water through the irrigation system
AF/yr = acrefeet of water pumped through the system
And:
[2] kWh/AF = 1.0241 * TDH / OPE
where:
kWh/AF = kiloWatthours required to pump an acrefoot 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 = acrefeet of water pumped through the system
Ac = acres of cropped land CL = conveyance system losses in acrefeet
ETcyr = annual net crop water use in acrefeet/acre
PPTeff = annual effective rainfall in acrefeet/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 = acrefeet of water pumped through the system p
Ac = acres of cropped land ETcyr = annual net crop water use in acrefeet/acre
PPTeff = annual effective rainfall (rainfall that infiltrates and is stored for subsequent use by the crop) in acrefeet/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/BHPHr
where:
Gal/yr = gallons of fuel required per season
Hr/yr = the hours the pump runs each year
BHP = the brakehorsepower requirements
Gal/BHPHr = specific fuel consumption, gallons of fuel required per brakehorsepowerhour
and:
[6] Hr/yr = 5427 * (Cl + (Ac * (ETc  PPTeff) / ((1LR) * IE))) / Q
where:
Ac = acres of cropped land CL = conveyance system losses in acrefeet
ETcyr = annual net crop water use in acrefeet/acre
PPTeff = annual effective rainfall in acrefeet/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 (go to definition of TDH)
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 kiloWatthours required to pump each acrefoot of water or reduce the number of acrefeet 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 onebyone:

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 ongoing 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 subsurface drip.
However, subsurface 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 salttolerant 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 kiloWatthour.

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 onfarm 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 electricpowered plant you may have
heard the term "wiretowater" 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 (definition of TDH)
Note also that 1 horsepower = .746 kilowatts
Thus, when performing a pump test for an electricpowered 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. 
Figure 1  Schematic of water well pumphead showing requirements for a valid pump test (courtesy PG&E).

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.

FIGURE 2  Schematic of a water well showing different components of Total Dynamic Head (courtesy PG&E)

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 poundpersquareinch = 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 (BrakeHorsepowerHours 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 electricpowered 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.



