8.7: Energy Consumption
- Page ID
- 44583
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Pumping water for irrigation consumes energy; it takes energy to lift water and it takes energy to pressurize water. Below we discuss ways to determine energy consumption so that irrigation managers can appreciate the energy costs of operating irrigation systems.
To analyze the rate of energy consumption, we will use what is called the Nebraska Pumping Plant Performance Criteria (Kranz et al., 2012a; Martin et al., 2017). given in Table 8.6. To illustrate how this table was developed, consider a 1.34 horsepower motor attached to an irrigation pump (one kilowatt is equivalent to 1.34 horsepower). Electric motors are not 100% efficient. For motors from 5 to 250 horsepower, the fully-loaded efficiency will range from 83 to 94%. The Nebraska Performance Criteria were developed assuming a motor efficiency of 88%. Thus, the power produced by the motor would be 0.88 times 1.34 horsepower or 1.18 horsepower. Therefore, 12% of the energy is lost due to the inefficiencies of the motor. The next step is to consider the energy that is transmitted from the motor to the pump. Many electric motors are directly connected to the pump and there is no energy loss in transmission. Thus, we would say that the drive efficiency is 100%. If a V-belt or right-angled gear drive is used to transmit the power from a motor or engine to the pump, energy is lost to heat in the drive. Typically, about 5% of the energy is lost between the motor and the pump if a gear drive or belt drive is used to transmit the power. With electric motors, the Nebraska Performance Criteria assumes a direct connection between the pump and the motor and thus a drive efficiency of 100%. Therefore, it is assumed that 1.18 horsepower is transferred to the shaft of the pump. The next step is to consider the efficiency of the pump. Nebraska Performance Criteria assumes a reasonable pump efficiency of 75%. Remember, as stated earlier, the peak efficiency of the pumps can vary from approximately 55 to 82%, depending upon the size and design of the pump. So, how much power is in the water leaving the pump? The power out of the pump will be equal to 1.18 horsepower going to the shaft of the pump times 0.75, which equals 0.885 water horsepower. Again, water horsepower is the power that is actually added to the water. Keep in mind now that we started with 1 kilowatt of power entering the motor. Thus, we have produced 0.885 water horsepower per kilowatt of input power.
Power is the rate of consuming energy. If power is multiplied by time, the result is the amount of energy consumed. Referring to Table 8.6, the Nebraska Pumping Plant Performance Criteria are expressed as an energy output power unit of energy input. If the water horsepower is multiplied by hours and the kilowatts by hours, the result is water horsepower hours and kilowatt hours, respectively. Thus, the Nebraska Pumping Plant Performance Criteria for electric powered pumping plants is 0.885 water horsepower hours per kilowatt hour.
The procedure that we just illustrated for developing the performance criteria for electric powered pumps was also followed for gasoline, diesel, natural gas, and propane. The only differences are how the units of energy are expressed and the fact that the drive efficiency of internal combustion engines is assumed to be 95%, because belt drives or right-angle gear drives are used. The Nebraska Pumping Plant Performance Criteria was developed with what are considered to be reasonable design objectives. We would expect well-designed and well-maintained pumping plants to perform at the level indicated. However, most pumping plants do not operate at this criteria. An index, called performance rating, is used to evaluate the performance and is calculated by:
\(\text{Performance Rating}=\dfrac{\text{Actual Performance}}{\text{Performance Criteria}}\)
|
Energy Source |
whp-hr/Unit of Energy[a] |
Energy Unit |
|---|---|---|
| Diesel | 12.5 | gallon |
| Propane | 6.89 | gallon |
| Natural gas | 61.7 | 1,000 ft3 (mcf) |
| Electricity | 0.885 | kW-hr |
| Gasoline | 8.66 | gallon |
| [a] whp-hr (water horsepower-hours)/unit of energy is the performance of the pumping plant as a complete unit—power unit, drive, and pump. The values are based on a field pump efficiency of 75% and natural gas energy content 925 btu/mcf. | ||
Given the following conditions, determine the performance rating of the irrigation pumping plant.
L = 100 ft
P = 50 psi
Q = 800 gpm
Measured diesel fuel consumption: 4 gal/hr
Find: Performance rating
Solution
\(TDH = 2.31 \dfrac{\text{ft}}{\text{psi}}(50 \text{ psi})+100 \text{ ft}=216 \text{ ft} \)
\(whp = \dfrac{(800 \text{ gpm})(216 \text{ ft})}{3960}= 44 \text{ hp} \)
\(\text{Performance}=\dfrac{44\text{ whp}}{4\text{ gal/hr}}=11\dfrac{\text{whp-hr}}{\text{gal}}\)
\(\text{Performance Criteria}=12.5 \dfrac{\text{whp-hr}}{\text{gal}} \)
\(\text{Performance Rating}=\dfrac{11\dfrac{\text{whp-hr}}{\text{gal}}}{12.5 \dfrac{\text{whp-hr}}{\text{gal}}}=0.88 \)
The example illustrates how pumping plants can be evaluated. By measuring the lift, discharge pressure, flow rate, and energy consumption, the actual performance of a system can be determined. This actual performance can then be compared to the Nebraska Pumping Plant Performance Criteria. To calculate the energy use rate per hour of an irrigation pump, use Equation 8.21
\(\text{Energy/hr} = \dfrac{whp}{(PC)(PR)}\)
where: PC = Nebraska Pumping Plant Performance Criteria and
PR = performance rating.
How much diesel fuel would be used per hour if a pumping plant is operating at 100% of the Nebraska Performance Criteria? Assume the same conditions as in Example 8.7.
If the pumping plant were operating at the criteria, the performance rating would be 1.
Solution
\(\text{Energy/hr}= \dfrac{44\text{ whp}}{\left(12.5 \dfrac{\text{whp-hr}}{\text{gal}} \right)(1.00)}=3.52\text{ gal/hr} \)
So, the pumping plant evaluated in Example 8.7 is using approximately one- half of a gallon per hour more diesel than needed according to the Nebraska Pumping Plant Performance Criteria.
Another equation that can be useful for determining the energy consumed per unit volume of water pumped is
\(E=\dfrac{TDH}{(8.75)(PC)(PR)} \)
where: E = energy consumed per ac-in of water.
For the same conditions given in Example 8.7, determine the energy required per acre-inch of water pumped.
Solution
\(E=\dfrac{216\text{ ft}}{(8.75)\left(12.5 \dfrac{\text{whp-hr}}{\text{gal}} \right)(0.88)}= 2.24\text{ gal/ac-in} \)
Example 8.9 shows that the diesel pumping plant that has a performance rating of 0.88 would consume about 2.24 gallons of diesel per acre-inch of water pumped. Now, what would the consumption rate be if the pumping plant performance were improved to 1? To find the answer, refer to Example 8.8. Example 8.8 shows that the pump would be consuming about 2 gallons of diesel per acre-inch if performing at the Nebraska Criteria. This is about 10% less energy per acre-inch than when it is performing at its current rating of 0.88.
Determine the energy consumption per acre-inch for the pump in Example 8.7 if the performance rating can be improved to 1.
Given: PR = 1
Find: E
Solution
\(E=\dfrac{216\text{ ft}}{(8.75)\left(12.5 \dfrac{\text{whp-hr}}{\text{gal}} \right)(1.00)}= 1.97\text{ gal/ac-in} \)
Equation 8.22 can be used by the irrigation manager to evaluate the costs of applying a known volume of water versus the expected return from that water. The equation can also be used to estimate the performance rating of an irrigation pumping plant if the manager knows the total dynamic head, the volume of water that is pumped in a year, and the energy consumed in that year. By using this equation, the manager can decide whether or not improvements need to be made to the irrigation pumping plant to improve its efficiency. The techniques for measuring water volumes were discussed in Chapter 3. Obviously, to determine total dynamic head, both lift and pump discharge pressure must be known. Investing in and maintaining accurate pressure gauges on an irrigation system is a must not only for energy management but also for managing and assessing the functionality of the irrigation system itself. For example, do the flow rate and system pressure agree with the original design? A distinction must be made here between pump discharge pressure, which as the name implies is measured immediately at the pump discharge, and system pressure which is the water pressure actually going into the irrigation system or mainline. Example 8.11 illustrates how Equation 8.22 can be used to assess energy management alternatives.
Suppose we have a sprinkler irrigation system that requires 800 gpm at 40 psi pressure. The pumping lift, L, is 143 feet. A vertical hollow-shaft electric motor powers the pump at 1770 rpm. A 5-stage 12 SKL pump (curve shown in Figure 8.11) is installed with 8.19-inch diameter impellers.
Given: Q = 800 gpm
Design discharge pressure = 40 psi
L = 143 feet at 800 gpm
Find: Does the pump match the pumping requirements, or is it oversized? If it does not match, what would be the correct impeller diameter? What effect would a throttling valve have on the energy consumption of this system?
Solution
Design TDH = 2.31 x 40 psi + 143 = 40 psi
Actual TDH produced = 5 stages x 57 ft/stage = 285 ft
So, the pump is oversized for this pumping condition. The correct impeller would deliver 47 feet per stage based on the following computation:
TDH/stage = 235/5 = 47 ft per stage
According to Figure 8.11, the correct diameter would have been 7.75 in.
Given the 8.19-in diameter impeller, a throttling valve would have to dissipate 50 ft of head, or 22 psi of pressure. Thus, using a throttling valve will result in a discharge pressure of 62 psi (between the pump and the valve) and 40 psi downstream of the valve. Assuming that the performance rating of this pump is 1, we can calculate how much extra energy is being consumed per ac-in of water:
At TDH = 235 ft, \(E=\dfrac{235\text{ ft}}{8.75)\times 0.885 \times 1}= 30.3\text{ kWh/ac-in} \)
At TDH = 285 ft, \(E=\dfrac{285\text{ ft}}{8.75)\times 0.885 \times 1}= 36.8\text{ kWh/ac-in} \)
Thus, the system should consume 30.3 kWh/ac-in with the proper impeller trim but instead, with the throttling valve, it is consuming 36.8 kWh/ac-in, 21% more energy than needed. This energy is being burned up or lost in the throttling valve.
In Example 8.11, the appropriate impeller diameter would have been 7.75 inches, but an 8.19-inch impeller was incorrectly installed. Once installed, it can be very expensive to make the impeller diameter change. A useful alternative would be to use a variable frequency drive (VFD) on the electric motor so that pump speed could be changed. The VFD is a motor controller which can be set to control the speed so that the desired pressure, 40 psi in our example, is maintained. In Example 8.11 the correct pump speed to obtain the 40 psi is 1670 rpm. This is based on application of the affinity laws discussed earlier. Variable frequency drives have many other useful applications in irrigation, especially where pumping conditions are not constant with time. A good example is variable rate irrigation, or a center pivot with a corner arm. A panel for a VFD motor controller is shown in Figure 8.16. Measuring pumping lift is probably the most difficult of all the measurements needed for evaluating the performance of the pumping plant. Permanent installation of air lines on the irrigation well can be a useful addition to the system. This method is discussed by USGS (2010). A detailed procedure for evaluating pumping plant performance is provided in Kranz et al. (2012b).
Figure 8.16. Panel for a variable frequency drive (VFD) electric motor controller.
