Saving Energy with Variable Speed Drives
By Mark Gmitro
Product Manager, Drives
Baldor Electric Company
Drives provide additional savings by reducing the speed of the motor and therefore the horsepower required to operate.
Additional benefits include longer motor life, improved power factor, reduced noise and greater flexibility. The decision
to invest in energy efficient motors and drives will depend on the cost savings that can be achieved. Many variables will
impact the amount that can be saved, including the cost of electricity, annual operating hours, system duty cycle and
motor loading.
A simple energy savings calculation can be used providing a quick estimate of the payback that might be expected.
Information you will need to know:
HP
Hrs/Yr
$/kWh
SYSEFF
=
=
=
=
Motor full load horsepower
Number of hours per year in operation
Cost of electricity per kilowatt hour
System Efficiency
System Efficiency consists of the motor efficiency (published by manufacturer) multiplied by drive efficiency (a typical
value for variable speed drives is 98%).
First, calculate the kilowatts used by the system:
Kilowatts = Hp x 0.746 [1 / SYSEFF]
For our example, we will use a 100 Hp Premium Efficient Motor with an efficiency of 95.4% running a pump or fan
system:
Kilowatts = 100 Hp x 0.746 [1 / 0.954] = 78.2 Kilowatts
Next, calculate the energy cost in dollars for the motor to operate using this equation:
Energy Cost = Kilowatts x Hrs/Yr x $/kWh
We will assume annual operating hours of 3,000 and electricity cost of $0.10 per kilowatts:
Energy Cost = 78.2 kW x 3,000 Hrs/Yr x 0.10 $/kWh = $23,460 per year
The above assumes a motor running at full speed continuously. Pumps and Fans however spend 80% of there time
running at 60% speed and as a result of the affinity laws governing these systems only require 22% rated horsepower.
Using a variable speed drive (typical VFD drives are 98% efficient) to reduce speed we can calculate the cost of operation
based on partial load:
Kilowatts = 100 Hp x 22% Load x 0.746 [1 / (0.954 x 0.98) = 17.55 Kilowatts
Energy Cost at partial load = 17.55 Kilowatts x 3,000 hours/yr x 80% x 0.10 $/kWh = $4,212 per year
Energy Cost at full load = 79.79 kW x 3,000 hours/year x 20% x 0.10 $/kWh = $4,788 per year
Total Cost Operation = $4,212 + $4,788 = $9,000
Annual savings with VFD Operation = $23,460 - $9,000 = $14,460
Simply installing a variable speed drive will pay for itself in less than one year.
©Baldor Electric Company
Page 1 of 2
May 2009
These are estimated values. There are free energy savings software packages available that provide detailed information
about partial load system costs. The software can generate reports providing easy to understand information on actual
cost savings, one example appears below:
Theoretical Power (kW) consumed by Baldor Super-E
motor with and without the ASD
% Motor Load
Motor
20
30
40
50
60
70
80
90
100
49.3
52.4
55.5
57.9
61.8
64.9
69.6
73.5
78.2
®
Motor with
ASD
3.9
6.3
10.9
16.4
23.5
32.1
44.6
61.0
82.1
Kilowatt Hours (kwh) consumed by Baldor Super-E motor
with and without the ASD based on the load profile.
Hours
0
0
300
600
1200
450
300
150
0
3,000
Annual cost
at 0.10 $kWh
% Motor
Load
20
30
40
50
60
70
80
90
100
Total
0
0
16,656
34,719
74,131
29,207
20,879
11,026
0
186,617
Motor With
ASD
0
0
3,284
9,853
28,151
14,427
13,372
9,149
0
78,236
$18,662
$7,824
Motor
When looking at the actually duty cycle the real savings become $10,838, slightly more than the standard estimated
equations produce. Software packages produce reports useful when submitting a project for approval or capital
expenditure. The above example was produced using the Baldor BE$T Energy Savings Tool available at:
http://www.baldor.com/support/energy_savings.asp
©Baldor Electric Company
Page 2 of 2
May 2009