moter eficiency

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FACTA UNIVERSITATIS (NI S)

SER .: E LE C . ENERG . vol. 23, no. 3, December 2010, 333-344

Analysis of Induction Motor Efciency Class Change atPartial Load

Miloje M. Kosti c

Abstract: Motors with efciency high values can became inefcient then standardmotors when they are lightly loaded. The calculation procedure for the estimation of any part load efciency when two load points are given is presented in paper. Ef-ciency dependence is expressed in the form of = N k (1 N ), where is magni-tude k P = f 1(P0, P N , P / P N ), so that changes-deviations of efciency values comparedto rated value ( N ) would be explicitly expressed. Based on those deviations, energyefciency class changes of observed motors are directly identied with load change.Efciency curve for several motors, with typical shapes, are considered, especially forloads P / P N = 0.50 1.00. These results are shown justication claim that efciencyvalues at 50% and 75% load shall be stated in the documentation. The users can thenselect the motor best suited for their application.

Keywords: Induction motors, estimation, efciency, energy efciency class, partialload efciency, testing.

1 Introduction

M OST motor are over-dimensioned for safety reasons (and due to impossibilityof evaluating the load value) and because of standard power ratings. Thismeans that motors are usually used in the following range [1, 2]:

About 50% rated load, for small motors (to 10 kW),

50-75% rated load, for medium motors (11-30 kW), and

75-100% rated load, for larger motors (37-100 kW).

Manuscript received on August 5, 2009.The author is with the Institute of Electrical Engineering Nikola Tesla, Department of Power

Systems, Koste Glavinica 8a, 11000 Belgrade, Serbia (e-mail: [email protected] ).

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334 M. Kosti c:

By this reason, the topic of part load efciencies was considered in detail whenis presented and considered the IEC projects, IEC 60034-30 and IEC 60034-31[3, 4]. Some countries stated that they nominal efciency for a given class motorsbe reached at 75% or 100%. Other countries requested the nominal efciency to bereached at both 75% and 100% or even over the whole range from 60 to 100% [5].As Serbian member of IEC WG 31, which is performed IEC 60034-30 and IEC60034-31, Author these paper is presented a calculation procedure for the estima-tion of any part load efciency, when they for two load points are given [5, 6]. Thisprocedure is accepted in principle

Technically, the vast majority of motors (especially energy efcient motors)have at least the same if not a greater efciency at 75% compared to 100%. Withthe exception of very small motors (typically 1.1 kW and lower) it would actuallybe more expensive for a manufacturer to design a motor a greater efciency at 75%compared to 100%.

In order to simplify testing and harmonize with US procedures it was decided[3] to classify the motors according to their full load efciency only. However, from

16 June 2011, the information about part load efciencies (50% and 75%) shall bevisibly displayed on the technical documentation of motors [7]. The users can thenselect the motor best suited for their application.

If we consider the efciency curves shown in Fig. 1, we notice that the dif-ference in efciency between the motors varies with the load condition [6]. Thismeans that a motor with a high rated efciency does not guarantee a high efciencyat partial load. This is especially the case for motors with relatively high no-loadlosses - motors curve 1 and 2 in relation to motors curve 3.

2 Partial Load Efciency

The typical motor efciency dependence ( p) is given by following equations

=p

p + p0 + p Pn p2(1)

were are p = P / Pn-the p.u. load power, P0-no-load losses at U = U n in p.u. p0 =P0/ Pn and P Pn-load losses at rated load , in p.u. p Pn = P Pn / Pn

The difference in efciency between the motors varies with values of at ratio of no-load losses ( P0) to load losses in rated regime ( P Pn), the Fig.1

Total losses (a) and Efciency (b) dependencies, from output power in p.u., i.e.from ( P / Pn) values, varies for three (same) motors but with different number of turns per stator phase ( W s), -g.2 [8].


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Analysis of Induction Motor Efciency Class Change... 335

0 0.125 0.25 0.375 0.5 0.625 0.75 0.875 1 1.12570

72

74

76

78

80

82

84

86

88

90

P/P n

1 2 3

Fig. 1. Efciency curve = f (P / Pn), in relation to three P0/ P n values (in a per unitPn = 1): (P0/ P Pn )1=0.0618/0.0493=1.25, max = 112 for P / Pn = 112%, curve 1 ,(P0/ P Pn )2=0.0555/0.0555=1.00, max = n , for P = Pn = 100%, curve 2 , (P0/ P Pn )3=0.0400/0.0711=0.56, max = 75 for P = Pn = 75%, curve 3 .

Motor 1 (W s = 1.15W s3) with smallest light-load (and the greatest full-load) totallosses, Fig.2a, and with greatest light-load (and the smallest full-load) ef-ciency, Fig.2b.

Motor 3 (W s3 = W s) with smallest full-load (and greatest the light-load) totallosses, Fig.2a, and with greatest full-load (and the smallest light-load) ef-ciency, Fig.2b.

Motor 2 (W s2 = 1.09W s3 ) total losses and efciency dependencies are betweenMotor 1 and 3, Fig.2a and 2b.

3 Energy Efciency Motors and Partial Load Efciency

3.1 Efciency Classes of Three Phase Induction Motors

Three energy efciency classes are proposed, IE1, IE2 and IE3 [4]. The 60 Hzefciency values for class IE3 (premium) and IE2 (high) were taken from existingNEMA Premium and EPACT tables [9]. The 50 Hz values of class IE1 (standard)and IE2 (high) are similar to the existing CEMEP-EU e f f 2 and e f f 1 [10]. How-ever the values have been adjusted to take the different test procedure into account


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100

101

102

70

75

80

85

90

95

100

Motor nominal power [kW]

E f f i c i e n c y

[ % ]

IE1IE2IE3eff1eff2

eff3

Fig. 3. Comparative Survey of efciency rated values for classes IE1, IE2, IE3 [2] and eff1,eff2, eff3 [12].

Technically, the vast majority of motors (especially energy efcient motors)have at least the same if not a greater efciency at 75% compared to 100%. Withthe exception of very small motors (typically 1.1 kW and lower) it would actuallybe more expensive for a manufacturer to design a motor a lower efciency at 75%compared to 100% [14].

Testing of part load efciencies is a critical issue for manufacturers. In currentpractice all catalogue values for part load efciencies are taken from the test of fullload efciency. Therefore they are obtained at an operating temperature related tofull load instead of part loads. Since part load temperatures are generally lower, thepart load efciencies obtained by this practice are generally worse than in reality(so there is no problem for the user).

However when partly load efciencies became a mandatory requirement [7],the tests had to be carried out on the base of standard procedures [11, 15]. Thiswould require a signicant amount of extra work and cost, as it is shown by se-ries of IEEE 112A performance tests were conducted in 1999 and 2000 by MotorSystems Resource Facility (MSRF), [16]. That author suggests a similar series of performance tests for motor partial load efciency. Apart from part load there are


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338 M. Kosti c:

many other practical factors which inuence the actual efciency in the eld, likevoltage drops, ambient temperature deviations, harmonic voltage distortions, etc.Determination of stray-load losses is especially complicated [16].

This author presented a calculation procedure for the estimation of any partload efciency when two load points are given [6]. This procedure is accepted, inprinciple, and included into a new document Guide for energy efcient usage of electrical motors and variable speed drives [4].

3.2 Calculation of partial load efciency

It is usually that the efciency values:

4/ 4 = N i.e. at 100% rated load ( P / P N = 4/ 4), 3/ 4 i.e. at 75% rated load ( P / P N = 4/ 4),

be included in manufacturers information [13].If, the N and 3/ 4 (or P/ Pn ) values are given, it is able be determining of corre-

sponding total losses components, load losses component at rated load ( P N ), andno-load losses ( P0):

P Pn =P N P ,(P/ P N )

1 ( PP N )2

(3)

P0 = P N P ,(P/ Pn ) (4)

where is total losses at rated load ( P N ) calculated by equation (10)

P N =1 n

nP N (5)

On the base P0 and P Pn values, partial load demanded values P / Pn determine by

equation(1).By given procedure, on the base efciency values for Pn ( 4/ 4 = N ) and 3/ 4 Pn

( 3/ 4) for squirrel-cage motors 1LA and 1LG efciency classes e f f 1 and e f f 2[13], from 1.1kW to 200 kW, calculation of efciency values for 1/ 2 Pn ( 1/ 2) is per-formed. In such manner are obtained efciency dependencies from output power- (P / Pn).

3.3 Changes of motors efciency class at partial loads

Efciency dependence = i(P / Pn) is given in Fig.3, for four motors (7.5 kW,2 p = 2) with the same rated efciency value n = 90.1% (Fig. 4). The ef-ciency dependence differs more, the greater the difference in ratio values of P0/ P Pn .Namely, the smaller the losses ratio values P0/ P Pn :


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the bigger the maximum efciency values, i.e. 4max > 3max > 2max > 1max = n , and they are achieved at smaller values of relative loads P / Pn(Fig. 4), and

the lower the minimum load values at which efciency values i(P / Pn) n ,i.e. those motors may work in a wider load range with i(P / Pn) n (Fig. 4).

These given motors (7.5 kW, 2p=2) belongs to following efciency class [3]:These given motors (7.5 kW, 2 p = 2) belongs to following efciency class [3]:

Standard efciency class (IE1), if 86 .0% n < 88 .1%, High efciency class (IE2), if 88 .1% n < 90.1%, and Premium efciency class (IE3), if n 90 .1%.

As n = 90 .1% 90.1%, all considered motors corresponding efciency class IE3-Premim efciency class . These motors differ in values of efciency in the area of relative loads of motor P / Pn = 0 1.125, particularly in region P / Pn 0.50. This

is consequence mater of fact that those motors differ in values of at ratio of no-loadlosses ( P0) to load losses in rated regime ( P Pn ), Fig. 3.

It is interesting to determine the values of efciency at load P / Pn = 0.50 ( 50 ),and establish corresponding efciency class at given load (50%), for given motors1, 2, 3 and 4 (Fig. 4):

Motor 1 corresponding Standard efciency class IE1, as 1(0.5) = 87 .1%

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