In my previous post, I looked at using MotorMaster to take a life cycle cost perspective on the motor selection for a 5 hp chilled glycol pump serving an ice storage system. In doing that, I looked at upgrading from a Title 24/Energy Policy Act compliant motor to a premium efficiency motor as defined by NEMA. But, as you can see from the list below, there are a lot of choices, including motors with efficiencies that exceed the NEMA definition of premium efficiency.
The motors I contrasted previously are highlighted in green and yellow in the table above. In this post, I will take a look at some of the peak efficiency options, specifically, the ones highlighted in lavender and red.
Since the steps used to to generate the results with MotorMaster are identical to what I described in the last post, I will not repeat them here and have simply summarized the results in the table below which has been color coded to match the motor selections in the table above.
When you first look at the table, you may conclude that I made a mistake and imposed a slightly larger load on the motor that is color coded in lavender (96.5% of full load versus the 95% of full load number used for the other selections) causing the savings projected for that selection to be lower than it should be. After all, why wouldn’t two motors with virtually identical efficiencies deliver virtually identical savings?
The answer lies in the difference between rated motor speeds (1,750 rpm in one case and
1,760 rpm in the other) and the fact that the motor is driving a centrifugal machine, in this case, a pump which will by its nature, do more work if you run it at a higher speed, all other things being equal.
The rated speed for the squirrel cage motors we use on our HVAC equipment is the speed the motor will run at under the rated load. Specifically, in the table above, the number in the RPM column is the speed of the motor when 5 horsepower is being extracted fromthe output shaft.
As the motors unload, their speed will increase and approach the synchronous speed of the rotating field that exists in their stator windings. The rotational speed of the stator field is function of the number of motor poles and the distribution frequency of the utility system that is serving it (60 Hz in the USA and 50 Hz in most of the rest of the world).
If you want to see an example of this, Ryan Stroupe (a program manager that I teach with at the Pacific Energy Center) and I made a video clip that provides and overview of the pumps and ice tanks in the ice storage system at the PEC. At one point, the video captures the speed of the pump motor changing as we throttle and unthrottle the discharge valve. (Throttling and unthrottling the valve unloads and loads the motor because it reduces the flow and then allows it to increase again.)
The whole darn video is pretty exciting, but if you want the immediate gratification of seeing the speed change, then you should go to the 3 minute and 40 second point and start watching from there.
Returning to our discussion, all of the motors in the table are 4 pole motors with a synchronous speed of 1,800 rpm when operating on 60 Hz power system. The difference between the synchronous speed in the motor’s stator windings and the actual speed of the motor output shaft is called slip.
Slip is the mechanism by which the motor generates torque and is able to do work and will vary as a function of load. If there were no slip, a squirrel cage induction motor would not be able to generate torque.
The graph below illustrates this phenomenon using some test data I happen to have for a 25 hp motor.
The amount of slip required to generate a given amount of torque is a function of a number of motor design and operating parameters. Thus, different motors from different manufacturers will have different operating speeds at full load, as can be seen from the
The primary differences between the published data for the two motors I am contrasting in this discussion are:
- The one highlighted in red has a full load speed of 1,750 rpm, has a power factor of .80, and a list price of $657 where as,
- The one highlighted in lavender has a full load speed of 1,760 rpm, a power factor of .84, and a list price of $448.
- Both motors have the same published efficiency of 92.2%.
Given the better power factor and lower price of the motor highlighted in lavender, you would be tempted to purchase that one, anticipating the same energy savings as would be produced by the other for less capital cost with the added benefit of improved power factor.
So why is it that the seemingly more desirable motor does not save as much energy as the more expensive, lower power power factor motor? It’s because the torque produced by the motor with a higher rated speed will come into balance with the torque required by the pump at a slightly different operating point than the pump when it is equipped with a motor that has a lower rated speed.
As a result, the motor with the higher rated speed will tend to spin the pump’s impeller a bit faster than the one with the lower rated speed. If you spin the pump’s impeller faster, it does more work by moving a bit more water at a bit more of a pressure difference simply because that’s how a centrifugal machine works. The faster you spin the impeller (with-in the physical constraints of the design), the higher the centrifugal force generated will be, which translates into more energy transfer to the fluid stream to be converted to pressure as well as more fluid moving through the impeller.
Think of it this way. The pump in our example requires 4.75 bhp to provide design performance in terms of flow and head. So, it would not fully load a 5 hp motor rated for 1,750 rpm at full load and the motor would require a little less slip to do the job. As a result, the torque required by the pump would come into balance with the torque developed by the motor at a speed somewhere above 1,750 rpm (remember, as an induction motor unloads, it tends to speed up). For the sake of discussion, lets say that the motor and pump come into balance at 1,755 rpm.
By come into balance, I mean that the torque extracted from the shaft by the pump impeller as it spins exactly equals the torque developed by the motor. They both will vary with speed, but as the pump speeds up, it will require more torque. In contrast, as the motor speeds up, it will develop less torque.
If we remove the old motor that is rated 1,750 rpm at 5 hp and replace it with a new motor that is rated 1,760 rpm at 5 hp, we now have a motor installed that, by virtue of its design will spin above 1,760 rpm to deliver less than 5 bhp. The specific relationship would be described by its speed vs. torque curve and its speed vs. load curve.
But, if we begin to spin the pump faster than the 1,755 rpm that was assoicated with delivering the required performance with the motor rated at 1,750 rpm, it will start to use more power because it will be doing more work. Specifically, the torque will vary with the square of the speed change.
Bottom line is we have a machine that delivers less torque as it speeds up driving a different machine that requires more torque as it speeds up and eventually, the two will come into balance.
This is one of those things where a picture is probably worth at least a thousand words (or at least 363 words, which is how many I have used since the phrase think of it this way …).
This image is a graph of the speed vs. torque curve for a 1,750 rpm motor and a 1,760 rpm motor along with the speed vs. torque requirement for the pump. Where the pump curve crosses the motor curves is the point where things will come into balance. It’s the same concept as ploting a system curve on a pump curve to determin where the pump will operate in a given system.
To really see the answer, you need to zoom into the curve up where everything converges, which is exactly what the inset graph does for the area inside the tiny red square in the lower right corner of the main graph. As you can see, when the new motor is installed, if nothing was changed, the pumping system operating point would shift to where the pump used about 14.34 lb-ft of torque instead of 14.22 lb-ft.
That translates to a bhp increase of 0.06 bhp or 0.05 kw. Not much but still an increase that would cost $61 per year in energy for a constant volume system running round the clock in California, where electricity can cost $0.15 per kWh. In fact, if you had changed the motor to gain an increase in efficiency of 1.5%, the increased operating cost would have eradicated the anticipated savings and the measurement and verification folks would not be pleased if you were doing the work under an energy efficiency program that paid incentives based on measured savings. Neither would the Owner when they didn’t get the anticipated incentive money.
MotorMaster accommodates this phenomenon by providing a “Centrifugal Load” check-box (see the red arrow in the figure below) that, when selected, adjusts the projected performance of the new motor based on its speed relative to the original motor.
If one recognizes this during the design process and selects the pump based on the actual motor speed, not a nominal speed, then you can have the best of both worlds. But, if the situation is a retrofit application, where the energy baseline has already been established by the pump running with a motor at 1,750 rpm, then replacing the motor with a new motor that has a higher efficiency rating but operates at a higher speed may not generate the anticipated savings if the impact of the speed change has not been taken into account and accommodated.
To accommodate the extra speed you, you might consider trimming the pump impeller slightly to modify its performance or even install a variable speed drive to slightly slow the pump down and thus ensure that the new motor would only have as much power
extracted from it as the old motor did, and thus generate the intended energy savings.
But, there is no free lunch in the world of physics; the impeller trim would likely reduce the efficiency of the pump slightly and the drive would introduce efficiency losses of its own. So, the minor adjustments you might consider to keep all things equal would actually not keep all things equal and you still would not see the anticipated savings even
though the modifications set the pump up to deliver the same head and flow (water horsepower) as it was delivering before the motor swap.
After considering all of this, you would likely conclude that you will be better off buying the more expensive motor and realizing the maximum energy savings benefit, assuming
there is no financial benefit associated with the improved power factor and you don’t need a bit more capacity out of the pump. (Bear in mind that the motor is only a small component of the over-all facility power factor and just because you make an improvement in a small motor does not mean there will be a commensurate change in the over-all facility power factor.)
The following table contrasts the future value of the energy savings benefits of the three options I’ve explored relative to the baseline.
Interestingly enough, the motor with the best simple payback also delivers the least savings because it sacrifices some of the energy it saves to the pumping system due to its higher operating speed. In contrast, the $157 premium paid for the red best efficiency selection pays for itself many times over during its lifetime especially when one recognizes that energy savings generally also translate to lower power plant emmissions. In addition, it is likely that energy prices will increase over the motor’s operating life and the annual savings and future value will increase as well.
The bottom line is that subtle differences between motors can make a big difference in the results you get when you apply them to a given load. If you don’t recognize that up front, then you may not achieve your goal, be it energy savings, power factor improvement, or both. MotorMaster can be a very useful tool to help you take everything into account and make the right decision when contemplating a motor replacement or substitution.
Senior Engineer – Facility Dynamics Engineering