Every pump industry professional faces the challenge of choosing the most efficient system for a particular application. This process is complicated by the uncertainties of efficiency regulations, mandates and parameters, especially when efficiencies alone do not tell the whole story. Sometimes the most efficient motor is not the best choice.
By using two simple calculations, pumping professionals can determine the best motor\'97as well as the best pump and drive—for each application.
To help both consumers and installers, nearly every product sold today features an energy cost or energy star rating. Industry standards and regulations developed by a coalition of industry experts help control these ratings, including multinational organizations such as The National Electrical Manufacturers Association (NEMA) and International Electrotechnical Commission (IEC).
Image 1. Analyze both the capital expenditure and the total energy cost of the pumping system to identify the motor that is truly the most efficient for an application. (Images and graphics courtesy of Franklin Electric)The goal of these organizations is to help individuals and power companies save money on energy and infrastructure costs by reducing confusion in methods of measurement and providing consistency in terminology and product labeling. This standardization minimizes incorrect measurements, misleading nomenclature and false competitive advantages. All reputable manufacturers support and collaborate with agencies like NEMA and IEC to maintain the professional standards of the organization. For this reason, users must ensure all of their products adhere to these standards.
Pump users often gravitate toward motors that report the highest efficiencies because they believe this decision will lead to future cost savings. This, however, is not always the case and is another reason end users must check the standards first. Companies that do not adhere to the standards could label their products as "super efficient," for example, or make similar false claims.
In many cases, the terms used on product labels are convoluted. For example, a company could invest in the original design of a 100-horsepower (HP) system, increase its efficiency by 16 percent and label it "super efficient." According to the standards, however, this improvement in design does not meet the criteria for premium efficiency. While the new product is 54 percent efficient compared with the previous product that was only 38 percent efficient, the minimum efficiency required for a premium label is 68 percent. Despite the fact that this product has been improved, it still does not meet the premium efficiency requirements. In this way, agencies and certifying bodies prevent manufacturers from making claims that do not meet the standards.
While understanding standard efficiencies is a vital part of choosing equipment, end users must go beyond efficiencies to select the ideal motor for an application. For example, submersible motors are generally 80 percent efficient, while similar surface motors are 90 percent efficient. Based on these numbers alone, one would assume that a submersible motor is less efficient. This, however, is not necessarily the case. Incorporating pump end efficiency in the assessment and evaluating horsepower requirements based on motor speed is vital.
Two simple calculations can help end users go beyond basic efficiencies to select the ideal motor for a particular pumping application.
Step 1. Measure WWE
The first step is to measure true efficiency by analyzing the complete wire to water efficiency (WWE) of the pumping system (see Equation 1).
WWE = [(motor efficiency) x (pump efficiency) x (control/drive efficiency)] x 100
Equation 1
One-hundred percent efficiency is impossible if any component within a system is less than 100 percent efficient. Consider the following examples.
Example 1. 50-HP, 2 Pole (3,600 RPM) pumping system:
The system has an 84 percent efficient motor, 45 percent efficient pump and 99 percent efficient control/drive. Its WWE is 37.42 percent (see Equation 2).
(0.84 motor x 0.45 pump x 0.99 control/drive) x 100 = 37.42% WWE
Equation 2
Example 2. 60-HP, 4 Pole (1,800 RPM) pumping system:
This system has a 90 percent efficient motor, 76 percent efficient pump and the same 99 percent efficient control/drive. Its WWE is 67.72 percent (see Equation 3).
(0.90 motor x 0.76 pump x 0.99 control/drive) x 100 = 67.72% WWE
Equation 3
Every motor and pump manufacturer is required to publish full load efficiency data for standard rated products. Be aware of data accuracy, and note that Web- or app-based sources for efficiency ratings are generally updated more frequently than printed documents are.
As the industry evolves, changes are inevitable. A subtle change has a multiplicative effect on total system efficiency. For this reason, end users should research the same efficiency data for pumps and controls/drives.
When choosing the proper motor efficiency, users should be aware that most motors will have a full load (FL) and a service factor (SF) efficiency rating, which will depend on the load point from the pump. When selecting the load point, reference the total HP requirements needed. If the HP is at the rated full load point, use FL efficiency. If the load point is at the SF rating, use SF efficiency. The correct efficiency rating is as important as the total system efficiency.
Most pumps will list a best efficiency point (BEP). To establish the pump efficiency, look at the head point on the efficiency curve (see Figure 1). Gather the control/drive details, and simply plug the data into the formula.
Figure 1. When analyzing pump efficiency, look at the head point of an efficiency curve rather than solely using the BEP that most pumps list.Analyzing these values closely is important because they represent a cost position. Looking only at total system product efficiency as noted above ignores the energy consumed by the system. Differences in speed and HP can result in higher energy costs, even when the efficiency ratings seem to indicate high efficiency.
Figure 2. The Application, Installation & Maintenance Manual (AIM) manual is one of the many places that you can obtain accurate motor efficiency information. Product efficiency information from online resources are typically more accurate because they can be updated more frequently.Users must also consider the needs of their applications. Some systems may run for only a couple of months out of the year, while others may run every day to supply water to a city/municipal system. For systems that will run for years, the cost of operation is paramount. While the energy cost for a continuous run system will be much higher than that of a short run, the savings calculation is similar. The amount of savings will be diminished by the amount of run time.
Step 2. Measure energy cost
The second calculation regarding efficiency determines energy cost. All power companies charge for kilowatt-hours (kW/HR). The rate varies by location and provider, so the following examples will use $0.20 per kW/HR. Regardless of whether an application is submersible or surface-mounted, motor speed and horsepower must be analyzed to determine kW/HR used. Whether a user is installing a submersible pumping system, a line shaft turbine or a centrifugal system, the calculations are the same. Note that the control/drive will have an efficiency point to include. This will provide the complete system cost. Examples 1 and 2 analyze energy cost using the same two pumping system examples from Step 1.
Example 1. 50-HP, 2 Pole (3,600 RPM) pumping system:
In the previous example, the system was 37.42 percent efficient. The motor consumes 49 kW/HR (50-HP motor from Figure 1) operating 24/7 for 30 days for a total of 35,280 kW consumed. At $0.20 per kW/HR, the power cost per month is $7,056 (see Equation 4).
49 x 24 x 30 = 35,280 kW/HR x $0.20 = $7,056
Equation 4
Example 2. 60-HP, 4 Pole (1,800 RPM) pumping system:
This system seemed to be the clear efficiency leader at 67.72 percent efficient. The more efficient motor in a line shaft turbine system is four-pole (slower speed) but requires a 60-HP motor to generate the same head and flow. Again, operating 24/7 for 30 days, this motor consumes 62 kW/HR and 44,640 kW per month for a cost of $8,928 (see Equation 5).
62 x 24 x 30 = 44,640 kW/HR x $0.20 = $8,928
Equation 5
Image 2. The submersible motor shown here may actually be a more efficient option than a surface motor, depending on the pump and drive/control it is paired with. The more efficient pumping system consumes more watts and costs more to operate, so it might not necessarily be the best choice for a continuous run application. When choosing a pumping system, determining motor and pump efficiencies is an important preliminary step. But when calculating the final system cost, users must include both the capital expenditure and the energy cost of the system. In many cases, the most efficient system may not be the most cost-effective system.
Measuring efficiency and energy cost offers more insight into total cost of ownership.
To learn more about industry standards, visit nema.org or iec.ch