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You have probably noticed that there are several camps out there when it comes to centrifugal pump applications involving variable frequency drives (VFDs). One group believes that every pump should have one, while another thinks that they should be banned altogether. And, of course, there is the middle ground that says it depends upon the pump and application.

When we think of VFD control, the first thing that comes to mind is energy savings. There are, however, other benefits that can often justify the cost of this control technique, no matter whether there are energy savings.

For example, soft start and stop reduces mechanical and electrical stress while virtually eliminating water hammer. In some instances, this alone can double the life of a pump and motor. Single to three phase conversion and open delta current balancing are two other popular, non-energy related applications, and you still get soft start and stop as a bonus.

But if saving energy is the key purpose, it is very important that we understand how to compare energy consumption in the same application during constant and variable speed control. Let's take a look at a simple (water based) constant pressure / variable flow (cp/vf) application that uses the same pump and motor operating under both control schemes.

Motor, Pump and Total Efficiency

We will start by defining efficiency, since it is efficiency that will have a large influence on the amount of power consumed by either application. The efficiency of a typical machine is simply output power divided by input power.

If a motor consumes 1-kW of electrical power and produces 1-hp (0.746-kW), its efficiency is 74.6 percent. The same is true for the pump. Since 1-hp is equal to 33,000-ft-lb/min, the following equation will give us the theoretical horsepower required (or generated) at any point on a pump performance curve:

hp = (head in feet  x  flow in gpm  x  8.33 lb/gal) / 33,000 ft-lb/min

If you divide the theoretical horsepower by the actual horsepower measured during testing, the result is pump efficiency at that point. However, efficiency doesn't stop with the individual efficiencies. Anytime we couple a motor to a pump we have to look at the efficiency of both machines working together (total efficiency). For example, suppose we have an electric motor that is 90 percent efficient at its rated speed and directly couple it to a pump that is 80 percent efficient at its BEP. How do we calculate the total efficiency?

One approach would be to calculate the average of the two efficiencies. Another might be to assume that the system would operate at the lower of the two. Unfortunately, neither method is correct - the results would be quite a bit higher than the actual total efficiency. It turns out that the total, or system, efficiency is the product of the individual motor and pump efficiencies. In our example, the total system efficiency at BEP is 72 percent (0.8 x 0.9). At any point on a pump's performance curve, total efficiency will be the product of the motor and pump efficiencies at that specific operating point.

Another factor that can affect both motor and total efficiency is motor loading. The peak efficiency of a typical three phase motor occurs at about 75 percent to 80 percent of its rated load, but it will maintain a fairly flat efficiency curve between 60 percent and 100 percent when operating at rated speed. When loading drops below 60 percent, efficiency begins to drop, and will drop rapidly around 40 percent. This is not an issue in a typical, constant speed pump application as long as the horsepower required at minimum flow is at least 60 percent of rated motor horsepower.

Constant Speed

Figure 1 puts our discussion thus far into perspective. It shows a 4 x 4 x 7 end suction centrifugal pump with a manufacturer-approved flow range of 200-gpm to 800-gpm, operating in a constant pressure application.

Figure 1

A pressure reducing valve (PRV) is employed to maintain pressure at 100-ft across a flow range of 200-gpm to 700-gpm. The system curve (red) ignores any downstream changes due to friction and takes the form of a simple horizontal line. The blue data labels above the performance curve show the actual horsepower required at each major flow point, and are based upon the hydraulic efficiency (black data labels) at that same point.

As you can see, the system requires approximately 22.1-hp at 700-gpm and drops to about 17.7-hp at 200-gpm. This reduction in power is just what we would expect to see when a centrifugal pump is throttled at the discharge by a valve. Since the minimum load horsepower (17.7) is well above 60 percent, we can assume that motor efficiency (90 percent) remains relatively constant across the full range of flow.

What we have discussed so far is the horsepower required at each point - not the power required to produce that horsepower. The purple data labels below the system curve show the total efficiency at each flow point and assume a motor efficiency of 90 percent.

In order to obtain the actual kW consumed, we could recalculate the horsepower based on total efficiency and multiply by 746, or just use the following equation: [kW = (hp x 0.746) / motor efficiency]. This equation uses the original horsepower measurement and also takes into account motor efficiency. The red data labels show the power, in kW, consumed at each major flow point. As you can see, they range from 18.3-kW at 700-gpm down to 14.7-kW at 200-gpm.

Variable Speed Control

In this version of the application, the control valve is removed and a VFD is employed to control pump speed and system pressure across the same range of flow. When we introduce a VFD into the system we are essentially adding another machine to the equation, so we have to take a look at its effect on the rest of the system.

The VFD impacts the system in several different ways. The first is its own efficiency - how well it converts AC to DC and then pulses that DC to emulate an AC wave form. Energy losses are mostly in the form of heat, and if you check the rated efficiencies of several high quality drives you will find that they range from about 97 percent to 99 percent. For our comparison, we will take the middle ground and assume that the drive in our example is 98 percent efficient.

The second impact is due to the non-sinusoidal (pulsed) nature of the current supplied by the drive. This can give rise to harmonic losses in the motor and can result in an overall reduction in motor efficiency of about 1 percent. Since we will be using the same motor (90 percent efficiency) in this example, we will assume that harmonic losses will reduce its efficiency at rated speed and load to 89 percent.

Finally, reduced motor loading due to decreases in speed can also affect motor efficiency, although the decreases can be much smaller than those seen at fixed speeds. This quantity is a bit harder to define because it varies by motor size. For example, a typical 100-hp motor will experience a drop in efficiency of about 2.5 percent when operated at 30-Hz (1/2 speed). Smaller motors, especially those under 10-hp, will experience a larger drop.

From the information I have been able to obtain from several motor manufacturers, motors 10-hp and above under VFD control will experience no drop in efficiency down to 50 percent loading and only a small drop (1 percent) around 40 percent loading. In our example we will assume an additional 1 percent drop in motor efficiency (88 percent) at the two lowest load points. (This is a topic that should be addressed in greater detail by the motor and drive industry.)