# Irrigation Chapter 13 - Energy Costs for Irrigation Pumping

Other Chapters in this home study course have discussed applying the correct amount of water to a crop.  This chapter will discuss the factors which affect the cost of pumping that water. Three general factors affect the energy cost of pumping a given quantity of water.  This chapter will be divided accordingly.  The three factors are:

I.   The horsepower imparted to the water by the pump.

II.  The efficiency of the pumping plant (the pump, power unit and right angle drive - a.k.a the gearhead, if used).

III. The cost of the energy used by the power unit.

Author: Thomas W. Dorn, University of Nebraska Lincoln Extension Educator, Lancaster County, NE.

## Horsepower

When water is moved through a plumbing system where the point of discharge is at a higher elevation than the point of origin, there must be an external source of motive power to move the water through the system. In irrigation, the source of motive power is a pumping plant.

Forcein the USA measured in pounds (lb), is applied to the water to overcome: (1) the pull of gravity when water is raised to the higher elevation, (2) the friction losses as the water moves through the plumbing system and (3) the pressure required to cause the water to move through the outlets in the distribution system.

Work is defined as force applied multiplied by distance. In the USA, work is measured in foot-pounds (ft-lb).

Power is the rate at which work is accomplished; it is expressed as work divided by time (foot-pounds per minute (ft-lb/min).  In the United States, the standard measure of power is horsepower.

Horsepower defined:  The classical definition of horsepower was developed by James Watt, the inventor of the steam engine, as a way to compare the power of his steam engines with the power of horses. After studying the power created by draft horses in coal mines and at the docks he determined the average full-sized draft horse was able to generate 33,000 ft-lb/min throughout a normal working day (with rest periods). For example, a horse walking 100 feet per minute would need to create 330 pounds of draft (pulling force) to generate one horsepower.

Water Horsepower: In irrigation, the rate useful work is performed by a pump is known as water horsepower (whp). The water horsepower to pump water is dependent on two factors, the pounds of water being pumped per minute and the height to which the water is raised (feet).

To calculate the water horsepower output of a pump, the volume of the water moved by the pump, (gallons per minute (gpm) and the total head (lift + pressure) the pump is generating must be known.  In the United States, we normally express water volume in gallons and one US gallon of water is assumed to weigh 8.33 pounds.

The horsepower imparted to the water (whp) can be calculated using Equation 13.1.

Equation 13.1    whp = gpm x 8.33 lb/gal x head (ft) / 33,000 ft-lb/min

or, since 33,000/8.33 = 3960, the equation is usually shortened to:

whp = gpm x head (ft) / 3960

## Horsepower: The relationship between psi and head

In pumping situations, some of the work performed by a pump is the work accomplished in the act of lifting the water from the pumping water level in the well to the surface.  The point of measurement at the surface is usually the center of the pump discharge head of the pump.

In pressurized irrigation systems, the pump must also overcome the pressure created by restrictions in the water distribution system which meters the flow of water.

The lift, measured in feet (ft) and the pressure, measured in pounds per square inch (psi), are both components of the head that the pump must overcome to move water through the pump and distribution system.  It is therefore useful to convert system pressure, measured in psi, to the equivalent lift in feet so both components can be combined into a common unit for computational purposes.

To understand the relationship between pressure and head, it may be useful to visualize a pipe oriented vertically, capped off on the bottom end and open on the top end. The interior of the pipe has a cross-sectional area of 1.0 square inch.

If one pound of water is poured into the pipe described, it would fill the pipe to a height of 27.72 inches (2.31 feet).  All of the one pound weight of the water is bearing down on the bottom of the pipe which has an area of one square inch. If a pressure gauge were attached to the bottom of the pipe it would register one pound per square inch (1 psi).

Likewise, if the vertical pipe (having a 1.0 square inch cross-sectional area) is filled to a height of 55.44 inches, (4.62 feet), it would contain two pounds of water and would be exerting 2 psi on the bottom of the pipe. Since fluids exert pressure in all directions, not just downward, the sides of the pipe would also be experiencing a pressure of 2 psi (at the very bottom of the pipe).

If a second pressure gauge were attached to the side of the pipe described in the paragraph above at the mid-point 2.31 feet below the top of the water in the pipe, that gauge would read only 1.0 psi because the portion of the water column above the gauge weighs only one pound and is acting over a cross-sectional area of one square inch.

Pressure within a water system is not dependent on the shape or cross-sectional area of the container; it is only dependent on the height of water above the point being measured (head).

Consider a hexagonal hollow column with a cross-sectional inside area of 5 square inches.  This column would contain 5 pounds of water for every 2.31 feet of height. The pressure at the bottom would still be 5 lbs / 5 in2. = 1 lb/in2 (1 psi) for every 2.31 feet of water in the column.

This relationship is expressed numerically by Equations 2  and 3.

Equation 13.2        head (ft) = psi x 2.31 ft/psi

This can be alternately expressed as:

Equation 13.3       psi = head (ft) / 2.31 ft/psi

## Example 13.1

If there was a square column with an inside dimension of 1.41 inches on a side (with a cross-sectional area of 2 square inches) that was filled to the 11.55 foot level,

A) What would the pressure be at the bottom of the column?

11.55 ft / 2.31 ft/psi = 5 psi

B) If you were to connect a 20 foot length of ½ inch pipe to the bottom of the container described above, orient the pipe so it is perfectly horizontal and add enough water to fill the horizontal pipe and the column so that the water level in the column is at the 11.55 foot level; what would the water pressure be, if measured at the far end of the ½ inch pipe?

The vertical distance between the end of the pipe and the top of the water column is still 11.55 feet.  Pressure at the end of the pipe would be:

11.55 ft/2.31 ft/psi = 5 psi

C)  If, instead of being level, the pipe attached to the bottom of the container angled downward so that the end of the pipe was 2.31 feet below the level of the bottom of the container, and sufficient water were added to fill the pipe and fill the column to the 11.55 foot level, what would the pressure be at the end of the pipe?

The elevation difference between the end of the pipe and the top of the water column is now 11.55 feet (column) + 2.31 feet (pipe) = 13.86 ft.

Using Equation 13.3, the pressure at the end of the pipe would be:

13.86 ft/2.31 ft/psi = 6 psi

## Horsepower: Measuring the lift component

Most people have observed that when a soda straw is placed in a glass of liquid, the fluid level in the straw will equalize with the level of the liquid in the glass. This holds true whether the straw is held vertically or allowed to rest at an angle in the glass. It also holds true if it were inserted to the bottom of the glass or held suspended within the water in the glass.

Liquids have no tensile strength (they can’t be pulled, only pushed). When you drink from the straw, you don’t pull the liquid out of the glass, instead, you create a partial vacuum which lowers the air pressure inside the straw and the atmospheric pressure acting on the surface of the liquid in the glass pushes the liquid up through the straw and into your mouth.

You used a small amount of diaphragm muscle power to create the partial vacuum which caused the liquid to rise in the straw. Stated differently, raising the water through the straw into your mouth required a small amount of work.

When you stopped creating the partial vacuum, (stopped producing work) the water quit flowing through the straw and the water level in the straw returned to the level in glass.

No external work is required to cause the water level inside the straw to equal the water level in the glass, even though the water level in the glass may be several inches above the lower end (the inlet) of the straw.

The phenomenon whereby water seeks its own level occurs whether one is observing a soda straw or a water pump and column.  The water level inside the pump column will equalize with the water level inside the well casing so long as both are open to atmospheric pressure.

Therefore, when a pump is used to lift water from the well, no matter how much column pipe is submerged in the water, the pump expends no work to lift the water from the inlet of the pump to the pumping water level in the well.

It does expend work to lift (push) the water up the pump column above the pumping water level in the well casing. The amount of work (ft-lb) performed to raise a given quantity of water to the surface is dependent on the distance from the pumping water level in the well to the pump discharge.

## Open Discharge Distribution Systems

In an unrestricted distribution system, such as a pump discharging from a delivery pipe into an open irrigation canal, fluid pressure at any point in the pipe system between the pump and the discharge will be dependent on the height differential between the point being measured and the elevation of the discharge point, just as described on the previous page. Equation 3 is used to calculate the pressure in psi at the point being measured.

## Horsepower: Pressurized distribution systems

In gated pipe systems on uneven ground, the operator generally partially closes the gates in the lower areas to equalize the output with the gates at higher elevations which we have shown would have lower pressure due to the elevation difference.

In sprinkler systems irrigation systems used today, water is discharged through precisely sized orifices to meter the flow. In these situations, a dynamic relationship develops between the flow rate (gpm) the pump can produce and the system pressure.

In practice, pressure builds within an irrigation distribution system until an equilibrium condition develops between the pump curve (head vs gpm pumped) and the system curve (pressure versus gpm output).

The useful work output of the system, the whp, is a function of the total system head the pump must overcome and the volume of water pumped.

In the normal irrigation pumping situation, the pump must overcome three components of head; the elevation difference between the pumping water level in the well and the pump discharge (Lift component), the elevation difference (if any) between the pump discharge and the pressure gauge registering the system pressure (Elevation component), and the system Pressure created within the irrigation system (measured with the pressure gauge.(In the USA, pressure is normally expressed as pounds per square inch (psi).

The total head is a summation of these three components and is expressed by Equation 13.4 below.

Equation 13.4    Total head (ft) = Lift (ft) + Elevation (ft) + (psi x 2.31 ft/psi)

Equation 13.4 ignores friction losses in the column and delivery pipe which may add a few feet of head which the pump must overcome and it ignores a very minor component of head associated with the velocity imparted to the water which is negligible and can be ignored in our discussion here.

## Example 13.2

Calculate the useful work (whp) expended by an irrigation pump that is pumping 800 gpm. The pumping water level is 110 feet below the pump discharge and the pressure gauge, located on top of the pivot point (12 feet above the pump discharge), reads 40 psi.

Step 1. Calculate total head using Equation 13.4:

Total head  = 110 ft. (lift) + 12 ft. (elevation) + (40 psi x 2.31 ft/psi) (pressure)  = 110 ft. + 12 ft. + 92.4 ft Total head = 214.4 ft.

Step 2. Calculate water horsepower using Equation 13.1

whp    =   gpm x head (ft) / 3960   =  800 x 214.4 / 3960   =   43.3 whp

## Efficiency of the Pump and Drive Unit

Efficiency is calculated by dividing the amount of useful work performed by a mechanical device by the amount of work that must be input into the system to cause it to operate. No device is 100% efficient. There are always losses due to friction, the creation of noise or heat; and internal losses, which will be discussed later.

All pumps used in irrigation will operate over a wide range of heads and capacities as can be seen in Figures 13.1 and 13.2. With all pump designs, operating at a given rpm, the lower the head (ft) the pump must overcome, the higher the output will be (gpm).

The best efficiency for a specific impeller design occurs at only one head versus capacity condition for a given rotational speed; with progressively lower efficiencies at incrementally higher capacities/lower heads or lower capacities/higher heads along the curve.

Notice the efficiency ratings along the pump curves in Figures 13.1 and 13.2. It is fairly apparent that the design engineer was attempting to optimize a  design for applications requiring 800 gpm at 1760 rpm (plus 100 or minus 200 gpm) when he designed Pump A and  Pump B was designed for applications requiring 1400 gpm at 1760 rpm (plus 100 or minus 200 gpm).

When selecting a pump model to install in the field, the goal is not only to select a design that can produce the desired volume while overcoming the anticipated head but to select a design that will do so while operating at the highest possible efficiency under the anticipated operating conditions.  Higher impeller efficiencies translate into lower input horsepower requirements which, in-turn, result in lower rates of energy consumption to pump a given quantity of water.

Two types of pumps are commonly used in irrigation applications, centrifugal pumps are often used to pump from a surface water source or shallow wells where the pump can be located a few feet above the water surface. They are also used as booster pumps to increase water pressure for end guns, etc.

Turbine pumps are used to pump water from deeper wells. Since most irrigation water in Nebraska comes from ground water sources, the large majority of irrigation pumps in Nebraska are turbine pumps.

The head and capacity output of both centrifugal and turbine pumps interact to produce a performance curve similar to those represented in Figures 13.1 and 13.2.  As can be seen, at a given rotational speed, the capacity (gpm) of any irrigation pump is dependent on the head the pump must overcome.  The lower the head produced, the greater the capacity and vice versa.

Every irrigation pump, will operate on a head/capacity curve. The manufacturers generally publish curves which represent the operating characteristics for a full diameter impeller when operating at several commonly used rotational speeds.

Manufacturers can also provide curves showing operation at a constant speed (usually the rpm of 3-phase hollow-shaft electrical motors (1760 - 1770 rpm) but have different “trims” Published curves will include a full diameter impeller and one or more curves representing impellers that have been milled to physically reduce their diameter.

At a given rpm, reducing the diameter of an impeller reduces its peripheral velocity and therefore reduces the centrifugal force at the outer rim of the impeller. This has the same effect on head and capacity as reducing the rotational speed (rpm) of a full-sized impeller.

To read the head and capacity a pump can develop at a given rpm, simply pick a point on the pump curve and draw horizontal and vertical lines and read the head from the vertical axis on the left margin and the capacity from the horizontal axis on the bottom.

Turbine pumps are designed to work as a multi-stage assembly.  It is not uncommon to find pump assemblies in the field with three to five or more stages bolted together in series.  The water passes from stage to stage through the assembly and each stage adds to the total head (pressure) the pump is creating. Pump installers pick an impeller model that will deliver the design output (gpm) when operating near the best efficiency point of the pump curve and which will produce the total head necessary with a given number stages.

Changes in the rpm of the pump will cause the pump to operate on a different curve as can be seen by the curves in Figures 13.1 and 13.2.  The published curves in the figures show the performance at 1760, 1460, and 1160 rpm but pumps can be made to operate at any rpm.  A qualified technician can determine the rpm necessary to produce the head per stage and gpm to match the design specifications.

If the power source will be an internal combustion engine, the rotational speed of the pump can be adjusted simply by changing engine rpm or by changing the gear ratio in the right-angle gear head.

If the power source is electricity, and a direct-coupled three phase hollow-shaft motor will be used as the power source, one or more impellers can be trimmed to physically reduce their diameter.

## Efficiency: The NPPPC

The University of Nebraska has developed the Nebraska Pumping Plant Performance Criteria (NPPPC) for deep well pumping plants. (Table 13.1)

The NPPPC assumes a pump efficiency of 75% in the field, including friction losses in the bearings supporting the lineshaft and bearings in the bowl assembly. Occasionally, a pump will exceed 75% effi­ciency in the field if it is an efficient pump design that is well matched to the job and properly adjusted.

In addition, certain efficiencies are assigned to the power units based on tests in the lab and in the field. Again, practical efficiencies are assumed, and can be exceeded with properly tuned power units that are well matched to the job. The criteria is a combination of the pump and power unit performance and is expressed as whp-h per unit of fuel.

 Energy Source hp.ha whp.hb   Unit of Energy Energy Unit of Energyc Diesel 16.66 12.5 gallon Gasoline 11.5 8.66 gallon Propane 9.20 6.89 gallon Natural gasd 82.20 61.7 1000 ft3 (MCF) Natural gase 8.89 6.67 Therm Electricityf 1.18 0.885 kW.h ahp.h (horsepower hours) is the work accomplished by the power unit with drive losses considered.bwhp.h (water horsepower hours) is the work accomplished by the pumping plant (power unit and pump) at the Nebraska Performance Criteria.cBased on 75% pump efficiencydAssumes an energy content of 925 BTU / cubic footeSome natural gas suppliers price their gas by the therm. A therm = 100,000 BTU.fAssumes 88 percent electric motor efficiency

## Efficiency: Uses of the NPPPC

The Nebraska Performance Criteria (NPPPC) can be used for many purposes.  The most obvious is to derive a performance rating for a pumping plant in the field.

## Example 13.3 — Performance Rating

A pumping plant performance test is run on a diesel-powered pumping plant and it is found to be producing 35.6 whp and is consuming 3.4 gallons of diesel per hour.

The performance of this pumping plant can be calculated by dividing the work output whp-h by the fuel consumption hour)

35.6 whp/3.4 gal/hr = 10.47 whp-h/gal.

The performance rating is calculated using Equation 13.5.

Equation 13.5     Performance. Rating = Actual Performance / NPPPC x 100%

For this pumping plant, the performance rating is:

Rating = 10.47 whp-h //12.5 whp-h /gal x 100%  =  83.8% of the NPPPC.

Interpretation: This pumping plant is only accomplishing 83.8% as much useful work as expected, given the amount of fuel it is using.

## Example 13.4 — Estimating Fuel Consumption

The NPPPC can also be used to estimate the fuel consumption of a pumping plant using Equation 6.

Equation 13.6     Fuel Use /hr = whp-h

Therefore, a diesel-powered pumping plant producing 35.6 whp should be using:

Fuel Use   =  35.6 whp / 12.5 whp-h / gal   =   2.85 gallons of diesel per hour.

## Example 13.5 — Calculating Possible Fuel Savings

Another very useful application of the NPPPC is to calculate possible fuel savings once the current performance of a pumping plant is known. Considering the situation described in Example 13.3 above, one could easily estimate the fuel savings possible as a result of bringing the pumping plant up to the performance criteria as a result of adjustment, repair or replacement.

Subtracting the fuel consumed at 100% of the criteria from the fuel consumed at 83.8% of the criteria, one finds that 3.4 gal/hr - 2.85 hr = 0.55 gal/hr of diesel could be saved if the pumping plant could be brought up to the NPPPC.

## Efficiency: Poor pumping plant performance

The Biological Systems Engineering department at the University of Nebraska has conducted extensive field studies over many years measuring the performance of pumping plants in the field. One result is the development of the Nebraska Pumping Plant Performance Criteria that is represented in Table 13.1.

While the criteria does represent a reasonable estimate of what one should expect from a properly designed and maintained pumping plant, the actual performance levels measured in the field varies greatly.

Table 13.2 is a summary of the results from a three-year study conducted in the early 1980's where 165 pumping plants were tested across the state of Nebraska.  As can be seen, while a fair number (15%) either met or exceeded the NPPPC, many others fell short.  A significant percentage of the pumping plants (7%) were found to be operating at less than 50% of the NPPPC (producing less than half the useful whp-h per unit of fuel as the NPPPC would call for).

An alternative way of stating this is those pumping plants that were operating at less than 50% of the NPPPC were using more than 1.0/0.50 = 2.0 times as much energy as they should have been for the work they were producing.

Overall, the average performance rating was 77% of the criteria (the average pumping plant was only producing 77% as much useful work as it should, given the amount of energy consumed.  Alternatively, this could be expressed as the average pumping plant was using 100%/77% = 1.3 times as much energy as it would have been if it were operating at the NPPPC.  (Notice that calculating possible energy savings is done by dividing 100% by the performance rating of the pumping plant (100/77), it is not calculated by subtracting 77% from 100%).

 Performance Rating, % of the NPPPC 100+ 90-100 75-89 50-74 49 or less Total Number of Tests 27 42 40 59 12 180 % of Total 15% 23% 22% 33% 7% 100% Average Performance Rating — 77%

## Efficiency: Effect of sand on pump wear

Every well pumps some sand along with the water. This is sand which makes its way through the gravel pack and well screen. Of course, some wells pump much more sand than others. Eventually, all pumps become worn by the sand in the water.

When the pump is running, nearly all of the sand in the water is carried along by the high velocity flow and passes through the impellers,  bowls and pump column and is expelled into the distribution system where most of it exits through the spray nozzles in sprinkler systems or gates in gated pipe systems.

Center pivots have a “sand trap” at the end of the mainline pipe. Irrigators clean the sand from the trap as a part of regular maintenance. The frequency depends on how much sand gets into the well. Sand collects at the far end of gated pipes as well, which is dealt with when the pipe is picked up at the end of the season. Water pumped for distribution through drip irrigation systems must pass through filters to trap the sand to avoid fouling the emitters.

Whenever the pump is shut down, any sand that happens to be entrained in the water at the time settles out and comes to rest when it can’t sink any further.

Sand that collects in the pump bowls collects in the wear ring area because that is the lowest area of the bowl/impeller assembly.  Sand particles that are small enough may slip into the gap between the skirt of the impeller and the wear ring area of the bowl. When the pump is restarted the sand caught between these two surfaces acts as an abrasive and wears away the metal surfaces on both the impeller and bowl. Over time, this widens the gap allowing coarser sand particles to fall into the seal area and wear process continues.

## Efficiency: The effect of wear on pump performance

We have shown that each stage in a turbine pump adds head (pressure) to the water. The water in each succeeding pump bowl up the line is under greater pressure than in the bowl below it.

There must be a good seal at the wear ring to keep higher pressure water in the bowl above from leaking into the lower pressure water in the bowl below. The seal is created by the close proximity between the skirt area of the rotating impeller and the stationary wear ring area of the pump bowl.

Wear in the seal area reduces head and capacity of the pump and reduces the efficiency in two ways.  A portion of the water that leaves the impeller and collects in the bowl is able to leak back through the worn seals and back into bowl below. The leakage enters the water below at the inlet of the impeller, disrupting the smooth flow of water passing through the impeller assembly. This water passes back into the impeller where it is pumped and re-pressurized - adding to the work performed on that portion of the total water output of the pump.

On most pump designs, some of the inefficiency caused by worn seals can be corrected by repositioning the impellers within the bowl assembly to establish a better seal between the impellers and the wear ring area of the bowl.

Figure 13.3 is an artist’s drawing of a cross-section of one stage of an impeller and bowl assembly depicting the seal created between the wear ring area of the bowl and the skirt of the impeller. The inset area is shown in Figure 13.4. The figure on the left represents a new pump with good seals. The figure in the middle shows a pump with worn surfaces in the seal area. The figure on the right shows a worn impeller following a pump adjustment to reposition the impeller to improve the seal and reduce some of the internal leakage.

CAUTION: Impeller adjustment must be done by a qualified person who understands the effect that hydraulic downthrust has on lineshaft elongation.  Improper adjustment can lead to the impeller grinding on the bowl assembly and may cause severe damage.

In the pumping plant test study mentioned earlier and summarized in Table 13.2, 69 of the 190 pumping plants were found to be operating within 10 percent of the Nebraska Pumping Plant Performance Criteria and therefore would not have benefitted from an impeller adjustment.  Impeller adjustments and sometimes engine adjustments were attempted on 111 of the remaining 121 pumping plants.  An average energy savings of 14% of the original consumption was accomplished simply by adjusting the impellers to achieve a better seal (and occasionally by adjusting the carburetor and ignition timing on spark ignition engines.)

## The Cost of the Energy Used by the Pumping Plant

Using the NPPPC from Table 13.1, it is possible to estimate the fuel an irrigation system should use and, therefore, the annual energy cost for pumping.

## Example 13.6 -- Estimating fuel (energy) consumption

We discussed a center pivot system in Example 13.2. We calculated the pump was producing 43.3 whp.

If we assume this pump was operated 1000 hours per year, the annual useful work accomplished by this pumping plant would be 43,300whp-h. If this pump is powered by a natural gas engine operating at 100% of the NPPPC, the fuel required to pump the water for this field can be estimated:

From Table 13.1, we know a natural gas pumping plant should produce 61.7 whp-h /MCF at 100% of the NPPPC.

Therefore, using Equation 13.6. we can estimate the annual fuel use:

43,300 whp-h / 61.7 whp-h /MCF  = 701.8 MCF per season

If natural gas is  \$12.30 / MCF, the annual cost of the fuel to pump the water would be:

701.8 MCF x \$12.30 / MCF = \$8632.   (Note: This estimate does not include the energy required to power the 480 volt generator to operate the center pivot system itself.)

## Cost: Comparing energy sources

Another purpose that can be made of the NPPPC is to compare one energy source to another.  By comparing the whp-h per unit of fuel stated in the NPPPC, one can compute how any two energy sources compare in terms of energy cost for irrigation pumping.

Since the NPPPC for all energy sources is stated in terms of whp-h per unit of fuel, you need only find the ratio of the whp-h output between two fuels to compare the cost of one source vs another source .

For example: if you know the cost for a gallon of Diesel, you can compute an equivalent energy cost for 1000 cubic feet of (MCF) of Natural Gas.

The NPPPC for Natural gas is 61.7 whp-h /MCF and the NPPPC for Diesel is 12.5  whp-h/gal. According to the NPPPC a  MCF of natural gas should produce 61.7 / 12.5 = 4.936 times as many whp-h as a gallon of Diesel.

Considering energy costs only, one could pay 4.936 times as much for 1000 cu-ft (MCF) of natural gas as for a gallon of diesel.  If the price ratio differs from this computed value, one can select the fuel source that would cost the least relative to the other.

 Equivalent Price Factors Diesel Natural Gas LP Gas Electricity Gasoline Diesel 1.0 4.936 0.551 0.071 0.693 Natural Gas 0.203 1.0 0.112 0.014 0.140 LP Gas 1.814 8.955 1.0 0.128 1.257 Electricity 14.124 69.718 7.785 1.0 9.785 Gasoline 1.443 7.125 0.796 0.102 1.0

Note: This table only compares energy prices and does not take into account differences in capital investment, expected useful life of the components, differences in labor requirements,  or any annual hookup charges, if any.

## Cost: How to use the equivalent price factors

When the price of one fuel source is known, an equivalent price for an alternate fuel source can be calculated using Table 13.3.

Find the fuel source with the known price along the left margin. Follow along the row of cells to the column corresponding to the second fuel source being considered.  The number in the table is an equivalent price factor.  Multiply the price per unit of the first fuel by the factor to obtain the equivalent price of the second fuel.

Example 13.7:.   If Diesel is selling for \$3.00 per gallon.

Using Table 13.3, calculate an equivalent price for Natural Gas, \$/MCF.

Procedure:  Find Diesel on the left margin of the table and follow across on that row to the column corresponding to Natural Gas.  The factor is 4.936.  The equivalent price for Natural Gas therefore would be \$3.00 X 4.936 = \$14.80 per MCF.

Example 13.8:

The most complicated situation to compare is when an energy source requires an annual hookup charge.  This example therefore will demonstrate how to handle hookup charges as well as the equivalent energy price factors from Table 13.3.

Assume an electrically powered irrigation pump with a 60 HP motor, producing 45 whp.  It typically is operated 878 hours per year to pump 12 inches of water on 130 acres.

Average electric rates are 6.4¢ per kW-hr plus an annual hookup charge of \$30 per horsepower.

What is the equivalent price per gallon of diesel, if the electric motor were replaced with a diesel engine to power this pumping plant?

Step 1. Calculate rate of energy consumption per hour.

At the NPPPC, the electric energy consumption would equal 45 whp / 0.885 whp-hr/kW-hr = 50.85 kW-hr per hour of operation.

Step 2.  Calculate annual energy consumption:

At 100% of the NPPPC, annual electrical energy use is therefore 878 hours x 50.85 kW-hr/hr = 44,644 kW-hr.

Step 3. Calculate annual electrical energy cost.

The annual cost of the Electricity is 44,644 kW-hr x \$0.064 / kW-hr = \$2857, plus the annual hookup charge of 60 hp x \$30/hp = \$1800.  Total energy cost therefore is \$2857 + \$1800 = \$4657.

Step 4.  Calculate average cost per unit of energy

Accounting for both the cost per kWh and the annual hookup charge, the actual price of the electricity consumed is \$4657 / 44,644 kW-hr = \$0.1043 / kW-hr.

Step 5.  Compare the prices of the alternative energy sources.

Using Table 13.3, find Electricity on the left margin and look across on the same row to the column corresponding to diesel.  The factor is 14.124.   Therefore the equivalent price of diesel would be \$0.1043 x 14.124 = \$1.47/gallon.

Interpretation: If diesel costs more than \$1.47 per gallon, the fuel cost will be higher for diesel than if no change were made and the electric motor were left in place. Presently, diesel fuel is higher than \$1.47 per gallon, therefore it would not be economically feasible to convert from electricity to diesel in this scenario.

## Summary

The energy cost for irrigating depends on many factors, including: Pumping the correct amount of water, the horsepower required to pump the water, the efficiency of the pumping plant, and the cost of the energy used to pump the water.

The power required to lift the water from its source and deliver it to the crop is a major factor and directly affects the amount of energy consumed for irrigation pumping.  The higher the total head, (lift plus pressure) the higher the horsepower requirements which translates into higher energy cost per acre-inch of water pumped.

The input power required is a function of the efficiency of the pump, power unit and drive unit.  If two different pumps were considered for an installation, picking one that can deliver the head and capacity desired more efficiently can result in considerable energy savings over the life of the pump.

Irrigation pumps experience wear. Keeping pumps in proper adjustment and/or replacing them when they become worn can save a considerable amount of energy and can keep pumping volume closer to design levels.

Each of the energy sources used for pumping irrigation water can be expected to produce a different amount of useful work, (water horsepower - hours (whp) per unit of fuel. One can use the Nebraska Pumping Plant Performance Criteria to estimate the annual cost of pumping for any energy source.  One can also compare relative prices between alternate energy sources and determine which would be most economical.