his article examines some of the
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be hig
is wzth the Electrical E- Computer Engineerzng Department ut Clurkson U
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EE motors generally operate at a
higher speed (lower slip) thun SE
motors of the sume rutings.
horsepower (hp), and percentage load. As a consequence of this, the important question of downsizing of motors to operate closer to the smaller
motor's maximum efficiency is considered.
The article is organized as follows: The following section considers the impact of centrifugal
loads on the efficiency difference between EE and
SE motors. The third section considers the effects
of actual operating voltages on the performance of
EE and SE motors. The fourth section presents
manufacturers' data in graphical form for additional insight and examines the effects of downsizing of motors as a technique for improving
efficiency. The final section has the conclusions.
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495 I
490
485
480
475
.470 -
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End 9/29/93
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,
-Voltage
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hP
1
#of Poles
50
75
250
1 ip
89.5
91.7
1:
,
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,
.
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1
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492
490
488
486
484
482
480
478
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476t
200
400
600
800
Time Series Record
j - Voltaqe Q MCC
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-Voltaae
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1000 12dO
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Q Terminall
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Fig. 2.
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where they have the same speed. In payback calculations, the full load efficiency difference listed in
1 is often used, without regard to the difference in
the operating speed.
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__-
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1I;! Load Speed (rpm)
EE
~
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Full Load Speed Full Load
Diff. (rpm)
Eff. Diff.
3,530
1,760
1,175
1,765
1,175
3,560
1,775
1,185
3,560
1,775
1,185
0
0
2
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._
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r
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94.1
93.6
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8 MCC Y o l t a g e 8 Terminal
Table 1. Full Load Motor Data
EE
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Fig. 1 .
_-
_ _
Load ER.
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545PM
460 -4
4551,
.
. . , .
0 100 200 300 400 500 600 700
Time Series Record
Effect of Operating Speed Differentid
Losses in an induction motor are proportional to
slip. Hence, EE motors generally operate at a
higher speed (lower slip) than SE motors of the
same ratings. Table 1 gives the hp, pole number,
full load efficiency, and operating speed of SE and
EE motors for one motor manufacturer.
Generally, the EE motor speed is higher than
the SE motor speed, although there are some cases
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Effedive Full
Load Eff. Diff.
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2.90
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3.5
2.4
2.6
3.50
2.40
2.60
3,570
1,785
1,190
0
5
IO
3.5
3.3
4.0
3.50
2.50
1.63
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2
4
6
90.2
91.7
92.4
94.1
95
95
3,565
1,775
1,185
3,570
1,790
1,190
5
15
5
3.9
3.3
2.6
3.51
0.93
1.41
2
4
6
91.7
93
93
94.5
95.8
95.8
3,570
1,785
1,190
3,575
1,790
1,190
5
5
0
2.8
2.8
2.8
2.40
2.00
2.80
93
94.1
94.1
95
95.8
95.4
3,565
1,785
1,190
3,575
1,785
1,190
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0
0
2.0
1.7
1.3
1.21
1.70
1.30
93
93.6
94.3
95.4
96.2
95.8
3,570
1,780
1,190
3,575
1,790
1,190
5
10
0
2.4
2.6
1.5
2.00
1 .oo
1.50
/E€€ industry Applirotions Mogozine
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Jonuary/February
I997
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490
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2 485
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2 480
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470'45,PM'
'
, ,
0
100 200 300 400
'
1
500
600 700
Time Series Record
-Voltage
Q MCC
-Voltage
Q Termina
Fig. 3.
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4921
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100 200 300 400 500 600 700
Time Series Record
0
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M
-Voltage
MCC --Voltage 0 Terminal
~
Fzg. 4.
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0
100
300 400 500
Time Series Record
200
600
700
Fig. 5.
Consider an SE motor driving a pump, fan, or
centrifugal compressor load where the load torque
can be described as
TL = K
* o2
(1)
where T Lis the load torque, K is a constant, and ci)
is the actual operating speed Suppose the efficiency
of the SE motor is Eff (SE) and that ofthe EE motor
is Eff (EE). Suppose the speed of the SE motor is
(SE) and the speed of the EE motor is ci) (EE). Then
the load torque of the SE motor is
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TL(SE) = K
* CI)(SE)~
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/€€E lnduzfry Applitafions Magazine
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January/February 1997
(2)
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hP
#of Poles
3/4 Load Eff.
SE
20
EE
3/4 toad Speed (rpm)
3/4 Load Speed 3/4 Load
SE
Diff. (rpm)
EE
Eff. Diff.
Effective 3/4
Load EH. Diff.
2
4
6
91.6
92.1
89.0
93.8
93.9
93.4
3,548
1,770
1,181
3,555
1,774
1,181
7.5
3.75
0
2.2
1 .a
4.4
1.61
1.21
4.40
2
4
6
89.5
91.7
91.7
93.7
94.7
94.4
3,570
1,781
1,189
3,570
1,781
1,189
0
0
0
4.2
3.0
2.7
4.20
3.00
2.70
75
2
4
6
90.2
92.4
91.0
94.9
95.5
95.2
3,578
1,785
1,185
3,578
1,789
1,193
0
3.75
7.5
4.7
3.1
4.2
4.70
2.50
2.42
100
2
4
6
89.5
91.7
92.4
94.0
95.9
95.4
3,574
1,781
1,189
3,578
1,793
1,193
3.75
11.25
3.75
4.5
4.2
3.0
4.20
2.41
2.10
2
4
6
91.0
93.0
93.6
94.8
96.1
96.0
3,578
1,789
1,193
3,581
1,793
1,193
3.75
3.75
3.8
3.1
2.4
3.50
2.50
2.40
50
150
0
93.0
94.5
94.8
95.5
96.5
96.0
3,578
1,785
1,193
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3,574
1,789
1,193
250
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1,193
0
1.2
1.20
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",
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(7)
Since the system curve is normally steep around
the full load operating point, ignore the slight
increase in flow obtained from the EE motor which
is only linear in speed. That assumes that the only
usefud power is that generated by the SE motor.
Then the effective efficiency of the EE motor is
Table 1 summarizes this effective efficiency difference for several different motors from 20 hp-200
hp and for different pole numbers. It is clear that
the effective efficiency difference between the EE
motor and the SE motor reduces for high-speed
differences between the two motors. For example,
the 100 hp, &pole motor has a nominal full-load
efficiency difference of 3.3%, but because of the
large speed differential between the SE and EE
motors the effective efficiency difference is 0.93.
Payback calculations should be based on this lower
effective efficiency difference. Of course, for the
cases where the EE motor speed is identical to that
of the SE motor, the nominal efficiency difference
is the same as the effective energy difference. The
same procedure can be applied to find the effective
efficiency difference at part load as shown in Table
2 for 3/4 load and Table 3 for 112 load.
P~(SE) W(EE)~
--~
CO(SE)~
and the effective difference in efficiency is
O(SE)'
(8)
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Effects of Actual Operuthg Voltuges
Eff(EE)
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Motors are designed to operate from a 460 V
supply. Measurements in the petrochemical industry have revealed that the actual operating voltages
can be somewhat higher, as shown in Figs. 1-4 for
a 50 hp motor driving a pump, a 200 hp motor
driving a pump, a 100 hp motor driving a fan, and
a 30 hp motor driving an agitator. The corresponding loadings are shown in Figs. 5-8. These graphs
show both the motor control center voltage as well
as the actual motor terminal voltage obtained by
subtracting the line impedance drop. Coincidence
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I€€€Industry Applicafions Magazine January/February I997
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proper sizing of motors. These graphs indicate that
proper sizing will reduce efficiency if the motor
rating is 5 to 50 hp and is running around 75% of
its capacity. The oversized motor also runs cooler
than designed, thus increasing reliability and allowing some room for expansion, both ofwhich are
of prime concern in the petrochemical industry.
A,ny change in power factor is of concern when
replacing a standard-efficiency motor with an energy-efficient motor. Some manufacturers increase
the airgap of energy-efficient motors, which results
in reduced stray load loss at the expense of an
increase in magnetizing current and hence reduction in power factor. Thus EE motors quite frequently have a lower power factor than the
corresponding rated SE motor. Figs. 13 and 14
show the full load power factor vs. rated hp for SE
and IIE motors, respectively. Generally, 2-pole motors tend to have a better power factor than 4-pole
motors, which in turn tend to have a better power
factor than 6-pole motors. When calculating the
payback periods when retrofitting SE motors with
EE motors, it is necessary to include a calculation
on thle new VAR demand to ensure that additional
charges are not incurred, which may offset the
dollar savings achieved through reduced kWh consumption. Note that from the power company’s
point of view there is still a fuel savings even with
the reduced power factor, since fuel used is proportional to kWh and depends less so on VARs, the
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Any change in power factor is of
concern when rqlucing a
standurd-efficiency motor with un
energy-efficient motor.
only contribution here being increased transmission losses if the VARs are not generated locally.
Most of the rebate programs consider motors in
the 20-250 hp range. The case under consideration
is as follows: suppose a given SE motor is to be
retrofitted by an EE motor of a smaller size. What
should be the % loading for this to be economically
viable?
The procedure is as follows:
Suppose the current motor size is x hp, with the
loading at y%. Then the motor output is
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(10)
X*Y
bP
100
Suppose the efficiency of the motor at the y%
loading is h l ; then the input power equals
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x
* y * 0.00746 RW
(1 1)
rh
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20
50
2
4
6
90.9
91.9
89.0
92.9
93.6
93.3
3,565
1,780
2
4
6
87.5
91.0
91.0
2
4
6
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1,188
3,570
1,783
1,188
5
2.5
0
2.0
1.7
4.3
1.61
1.31
4.30
93.2
94.6
94.2
3,580
1,788
1,193
3,580
1,788
1,193
0
0
0
5.7
3.6
3.2
5.70
3.60
3.20
88.5
91.7
91.0
94.4
95.1
94.8
3,585
1,790
1,190
3,585
1,793
1,195
0
2.5
5
5.9
3.4
3.8
5.90
3.00
2.62
2
4
6
88.5
91.7
91.7
93.0
95.3
95.1
3,583
1,788
1,193
3,585
1,795
1,195
2.5
7.5
2.5
4.5
3.6
3.4
4.31
2.41
2.80
2
4
6
89.5
92.4
93.6
94.2
95.6
95.7
3,585
1,793
1,195
3,588
1,795
1,195
2.5
2.5
0
4.7
3.2
2.1
4.50
2.80
2.10
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2
4
6
91.0
94.1
94.5
93.7
96.0
96.0
3,583
1,793
1,195
3,588
1,793
1,195
5
2.7
2.31
0
1.9
0
1.5
1.90
1 .so
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2
4
6
92.4
94.5
94.8
94.5
96.1
95.8
3,585
1,790
1,195
3,588
1,795
1,195
2.5
5
0
2.1
1.6
1 .o
1.90
0.80
1 .oo
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75
100
150
200
250
IF€€IndustryApp/;ca!ions Magazine
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January/February I997
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96
>
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94
2
92
W
53
90
88
0
Rated load
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where
E =
F =
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(X
(X
*y
*y
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* 0.00746 * 12 * W )
* 0.00746 * s * t )
(k+ I
fg
+ h + b)
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= p years
1
[L- i)
* ( X * y * 0.00746) * (12 * w + J + t )
“rl
K+I+v+h+h
%
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years
0
12*w 12”w
___-__
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“r2
r12
n1
n2
(18)
i.e.,
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(20)
+ s*t
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From Equation (20), the following can be deduced:
The smaller the k + 1 + g + h + b, the lower the
payback period. The higher the output power, the
smaller the payback periods. The greater the differ-
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90% Voltage
110% Voltage
120%Voltage
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Starting and Max. Running Torque
Decrease 19%
increase21%
Increase44%
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SynchronousSpeed
No Change
No Change
No Change
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Percent Slip
increase23%
Decrease 17%
Decrease 30%
Full Load Speed
Decrease 1 -1/2%
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Increase 1 %
increase 1-1/2%
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Full Load - High Efficiency
Decrease 1-2 Points
Increase V2-1 Point
increase 1/2-1 Point
Decrease 1-4 Points
Small Increase
DEcrease 7-1 0 Points
Practically No Change
Increase 1-2Points
Practically No Change
Decrease 2-5Points
Decrease 1/2-2 Points
Increase 1-2Points
Increase2-4 Points
Decrease 1-2Points
Decrease 4-7 Points
Decrease7-20 Points
Decrease 14.18 Points
Full Load - High Efficiency
T-Line
increase 1 Point
increase 9-10 Points
Decrease3 Points
Decrease 10-15 Points
Decrease 5-15 Points
Decrease 10-30Points
3/4 Load - High Efficiency
Increase2-3 Points
increase 10-12 Points
Decrease4 Points
Decrease 10-15 Points
Decrease 10-30Points
Decrease 10-30Points
Increase4-5 Points
Increase 10-15 Points
Decrease5-6Points
Decrease 10-15 Points
Decrease 15-40Points
Decrease 10-30Points
increase 11%
Increase 3.6%
Decrease 7%
Increase2-1 1 %
Decrease 1 1 %
increase 15.35%
_
Increase 25%
....
...
.
....
.. ..... __
--...
Table 4. Effects of High or low Voltage on Induction Motors
OpercitingCharacteristic
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Effect of Voltage Change
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1-Line
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3/4 Load - High Efficiency
T-Line
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~~
1/2 Load - High Efficiency
T-Line
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Power Factor
T-Line
1/2 Luad - High Efficiency
1-Line
Full Lood Current - High Efficiency
T-Line
I Starting Current
Temperature Rise, Full Load
High Efficiency
1-Line
I Magnetic Noise, Any Load
, Decrease 10-12%
, Increase 10-12%
Increase23%
increase 6.12%
Decrease 14%
increase423%
I Decrease Slightly
I Increase Slightly
_
,
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Decrease 21%
increase 30.80%
I Noticeableincrease
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/E€€ Industry Applicaiians Magazine
Jonuary/February I 997
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Operation ut higher than rdted voltdge
tends t o reduce the ejjficiciency of SE
motors und can actually increase the
~ficiencyof E E motors.
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ence in the efficiencies, the lower the payback
period The higher the demand and energy charges,
the lower the payback period.
As an example, suppose
w = $14.70
t = 8,000 hours
s = 0.027 $ikwhr
x = 100 hp
y = 50%
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g = $115
h = $20
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= $92
hen the payback period for downsizing can be
calculated based on the efficiency difference as
follows:
The 100 hp SE motor at 50% loading could be
replaced with a 60 hp SE motor at a motor cost of
$1,304.00 Then ql for the 100 hp SE motor at
50% loading is 88.5% and q 2 for the 60 hp SE
motor at 83% loading is 89.0%. Then the total
costs to be recovered is $1,894.00 andp = 20 years,
which is clearly unacceptable. If the 100 hp SE
motor were loaded at 75%, it could be replaced
with a 75 hp SE motor at 100% at a motor cost of
$1,635 00 and a total installation cost of
$2,225 00. Here p = 5.5 years, which is alsoprobably unacceptable.
The 100 hp SE motor at 50% loading could be
replaced with a 60 hp EE motor at 83% loading,
say at a motor cost of $1,598.00 and a total cost of
$2,188 00. Then
for a 100 hp SE motor is
88 5%andq2fora60 hpEEmotorat83%loading
is 94.4%. Thus p from (20) = 1.76 years. However,
if the 100 hp SE motor were loaded at 7 5 % , then
the only option for downsizing is a 75 hp EE motor
at 100% loading at a motor cost of $1,962.00 and
a t o t a l c o s t o f $ 2 , 5 5 2 . 0 0 . H e r e ~=l 89.5%andq2
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I€€€ Industry ilpphcofions Muguzine Junour~i/FebruuryI997
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