Payback Analysis for Variable Frequency Drives
Copyright © 2003 Francis J. Martino
The use of variable frequency drives on variable torque loads for
the purpose of gaining energy savings is a common application.
Calculation or estimation of energy savings is needed to justify
expenditures for the drives.
In the State of Connecticut, nominal power cost is 8.5 cents per kWH.
The utility will pay for the drive but not for the installation.
A 10 HP, 460 volt drive with line reactor will cost about $1300.
Installation time, materials and start-up will cost $500 or more.
In this example, operation will be 2000 hours per year, which is the
minimum acceptable operating time to the utility. The motor in this
example will be driving a centrifugal pump.
When operating across-the-line, the cost of power consumption of the
motor will be 10 HP x 746 Watts/HP x 2000 Hours x $ .085 per kWH, or,
$1268 per year.
We now add a variable frequency drive that will, according to the affinity
curve, save 50% energy by allowing a 20% reduction in flow. Savings will
be $ 634 per year.
Since the affinity curve neglects the static head of the system as seen
by the pump, most applications will save between 1/2 and 1/3 of what is
determined by the affinity curve calculation. Thus, the savings on the
present application will be $317 to $211 per year. See Elusive Energy Savings:
Centrifugal Pumps and Variable Speed Drives.
With the higher system head and a savings of $211 per year, the payback
to the customer is $500 installation cost/$211 = 2.37 years.
From past experience, small drives typically fail in eight years and are
replaced rather than repaired. The larger horsepower drives last longer by
virtue of repairs that often require replacement circuit boards and other
components. After bearing the cost of two or three replacement boards in
any one drive, the user often realizes that the drive should have been
The drive in our example fails after eight years. The user has gained a
(8 years minus the 2.37 year payback period) x $211 per year, or, 1188.
The user then replaces the drive for $1300 plus a $500 installation
cost. Payback on the replacement drive is now $1800 /$211 per year, or,
8.53 years. The drive then fails in eight years after giving a savings of
8 x $211 = $1688. Net loss on the second drive is $1800 - $1688 = $112.
At the end of the sixteen years, the user has saved $1188 - $112 = $1076
on a total investment of $500 + $1800, or, $2300. Return on the
investment averages 2.9% per year.
Perhaps the user, upon failure of the first drive, should replace the
drive with a full voltage starter and keep the $1188 savings on the initial
$500 investment, a return of 238% in eight years, or, 29.7% per year.
If the pump is pumping waste water from a sufficiently large holding
tank then down-time during installation of the drive doesn’t matter. If the
pump is pumping a process fluid and installation down-time causes loss of
production, then the user may have lost the $1188 saving on the day the
first drive was installed.
The question is not one of payback on a drive but, rather, one of
spending limited budget funds on capital projects that would raise the
overall facility production-to-kWH ratio, which is the only true measure
Power Quality and Drives LLC