BALANCED OPERATION OF THREE PHASE SYNCHRONOUS MOTORS CONNECTED TO A SINGLE PHASE SUPPLY

BALANCED OPERATION OF THREE PHASE SYNCHRONOUS MOTORS CONNECTED TO A SINGLE PHASE SUPPLY A. I. Alolah Elect. Eng. Dept., College Of Engineering, King S...
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BALANCED OPERATION OF THREE PHASE SYNCHRONOUS MOTORS CONNECTED TO A SINGLE PHASE SUPPLY A. I. Alolah Elect. Eng. Dept., College Of Engineering, King Saud University, P.O.Box 800, Riyadh 11421, Saudi Arabia Abstract Balanced operation of three phase s ynchronous motors has been dealt with extensively in the literature. However, unbalanced operation has not received as much attention. One of the interesting unbalance modes of these motors is their operation from a single phase supply. In this paper the possibility of operating three phase synchronous motors from a single phase supply through the use of a two element-phase balancer is explored. The values of the reactances of the phase balancer to cause complete balance, together with their variations with load have been presented. Furthermore, the paper includes two methods to obtain positive as well as negative sequence impedances of the motor. Symbols V, I V, I R, X Z,Y z, y δ, θ γ j, a

phasor value of voltage and current. rms value of voltage and current. resistance and reactance. impedance and admittance. Z and Y. load angle and angle of Z. Shift angle between Vand I 1∠90 o and 1∠120 o.

Subscripts a, b, c p, n, 0

stator phases positive, negative and zero sequence.

motors, when fed from a single-phase supply are well documented in the literature [3-16]. Recently, starting and steady state performance of three-phase synchronous motor when fed from a single phase supply has been introduced using a single capacitor as a phase balancer [17-19]. This paper deals with the steady state operation of three phase synchronous motor when operated from single-phase supply through a two element-phase balancer as shown in Fig.1. These two elements are made passive and reactive to avoid any extra losses to the system. The values of the reactances to cause complete balance between the stator voltages are studied. The variation of these values with the load angle and the ratio of the stator indu ced emf to the supply voltage, is shown. Furthermore, the paper includes two tests to obtain the negative sequence impedance of the motor. II. Mathematical Modeling And Analysis The system under study comprises a star connected three phase synchronous motor fed from a single phase ac supply as shown in Fig.1. Synchronous motor with cylindrical rotor is only considered in this paper. A. Voltage equations

I. Introduction Three phase synchronous motors have been used for long time in industrial applications requiring constant speed operation in addition to power systems for improving the voltage profile of the power grid by absorbing the excessive reactive power available at selected locations. Balanced operation of these motors has been dealt with extensively in the literature [1-2]. However, no similar interest was paid to the unbalanced operation of such motors. One of the interesting unbalanced modes of three phase ac motors is their operation from a single phase supply. Such a situation can be dictated by the occurrence of an open circuited single line fault or in remote areas, where only a single phase supply is available. Nearly all steady state and transient performance of three phase induction and reluctance

The loop voltage and current equations for the motor circuit of Fig.1 are: Vi Va + Vb = 0 Va V1 Vc = 0 Vc V2 Vb = 0 Ia + Ib + Ic = 0 I1 I2 - Ic= 0 V1 =+ j I1 X1 V2 =+ j I2 X2

(1) (2) (3) (4) (5) (6) (7)

According to the three phase symmetrical components, stator voltages and currents are:    

ga   1   gb  =  a 2 g c   a

1 a a2

1  1 1 

    

gp  gn  g 0 

(8)

Ii

Ia

Zp

Va

X1

V1

Vi Vb

Ic

Solving eqns. (13) and (14) for X1 and X2 yields:

X2 Fig.1 Connection of three phase motor to single phase supply using two element phase balancer.

where g stands for voltage or current. As can be realized from eqns. (4) and (8), zero sequence current and consequently voltage are zero. Moreover and as shown in Fig.2, sequence voltages and currents are related as follows: Ip = (Vp Ef )/ Zp = (Vp Ef )Yp

(9)

where Ef is the stator internal induced voltage. In = Vn / Zn = Vn Yn

(10)

Solving equations (1) - (10) yields: Vp = Vn

[ Vi D p + X 1 X2 Yp E f ]

[V =

D

2 i D n + a X 1 X 2 Yp E f

In

Fig.2 Positive and negative sequence equivalent circuits of cylindrical rotor three phase synchronous motor.

Vc V2

Ef Ip

I1

Ib I2

Vp

Zn

Vn

]

D

(11) (12)

− 3 ( R 2p + X 2p ) 3 k 2 R p + k4 X p

and X 2 =

6 ( R 2p + X 2p ) k1 k4 − k 2 k3

Positive value of X1 or X2 means inductive reactance while negative value means capacitive reactance. C. Sequence Impedances Evaluation of Vp and Vn in addition to the value of the phase balancer to cause complete balanced operation requires determination of both the positive and negative sequence impedances of the motor. Positive, negative an d zero sequence impedances of any machine are the impedances offered by that machine to the flow of positive, negative and zero sequence currents, respectively. As shown above, the zero sequence component is zero. Accordingly, only positive and negative sequence impedances are required for the analysis. These impedances can be obtained as discussed below. C.1. Positive Impedance Positive sequence impedance (Zp) is equal to the input impedance of the motor under normal balanced threephase operation. From eqn. (9), Zp is:

where D, Dp and Dn are as given in the Appendix. B. Balanced Operation

Zp = ( Vi Ef )/ Ia

Under this mode, perfect balanced operation of the motor can be ensured when Vn is zero. Equating real and imaginary parts of eqn. (12) to zero yields: 2 X1 X p + k1 X 2 + k2 X1 X 2 = 0

(13)

2 3 X1 R p − k3 X 2 − k4 X1 X 2 = 0

(14)

where k factors are also given in the Appendix.

X1 =

Zp can be determined from open circuit and short circuit characteristics obtained when the machine is run as a generator while Rp is obtained from the dc test. C.2. Negative Impedance Different methods may be practically applied to obtain the negative sequence impedance (Zn) of synchronous machine. Two of these methods are explained below.

2

p.u

1.6

Ef

fitted curve expermintal

Ef ,

V

- - - -

U

1

Ia , p.u

0.8

Ia

I

A 0.0 0.0

Fig.3 Connection of indirect test for determing Zn of cylindrical rotor three phase synchronous motor. The machine is run as a generator.

Direct Method: The negative impedance of the three phase synchronous machine can be directly obtained by applying a three phase balanced supply to the stator while the rotor is driven at synchronous speed but in an opposite direction to that of the stator field. The input impedance under this condition is equal to the negative impedance of the machine, which can be determined by measuring the input voltage, current and power. The field winding under this test is short circuited [20]. Indirect Method: The negative impedance of the three phase synchronous machine can be indirectly obtained by operating the machine as a generator while the stator is connected as shown in Fig.3. Under this test, Zn is given by [21]: zn = U / ( I

3 ) , R n = z n sin β and X n = z n cos β

where β = cos-1 ( P / U I ) . U, I and P are respectively, the measured output voltage, current and power as shown in Fig.3. III. Results

if

0.8

, p.u

0 1.6

Fig.4 Open circuit and short circuit characteristics of the machine under study

Zn = 0.125 + j 0.36 p.u ( direct test) Zn = 0.122 + j 0.367 p.u (indirect test). This motor is investigated to determine the values of the phase balancer to yield zero negative sequence voltage when the motor is fed from a single phase supply. Figs. 5 shows the variation of X1 and X2 versus the load angle δ, when ef = 1/√3. It can be realized that X1 is equal to X2 in magnitude but X1 is a capacitive while X2 is an inductive, for the whole range of loading. Fig.6 shows the variation of the shift angle (γp) between Vp and Ip versus ef for different values of δ. At ef less than 0.6, γp is positive while it is negative for ef greater than 0.6. Fig.7 illustrates the variations of X1 and X2 against ef for different values of δ. It is interesting to note the following: γp > +60° + 60° < +60°

γp >- 60° - 60° < -60°

X1 inductive open circuit capacitive

X2 inductive open circuit

capacitive

0

300

Zp = 0.09 + j 2.7 p.u

p.u

p.u

X1

ef= 0.57 7

-150

150

X2

X1

The motor under study is a three phase 0.8 kW, 380 V, 1.5 A, 4-pole, 60 Hz, synchronous motor. Rp of the machine is obtained by a dc test on the stator winding. Open circuit and short circuit characteristics of this machine are shown in Fig.4. Zp is obtained from this figure. Direct and indirect tests as mentioned above, are carried out to determine the negative sequence impedance (Zn) of the machine. These experimental tests yielded:

X2 -300 -50

-25

δ

Fig.5. Variation of X1 and X2 versus δ at ef =1/√3.

0 0

90

From the investigations carried out in this paper, the following can be concluded:

60

v) For γp > +60° X1 is inductive then changes to capacitive for γp < +60°. X2 is inductive for γp > -60 ° then changes to capacitive for γp < -60°.

δ= −10 δ= −30

-60 -90 0.10

1.00

Fig.6. Variation of the angle γp versus ef for different values of δ. 10000

1800

δ= −5 0 0

-10000 X2 X1 -2 0000 0.10

0. 55

-1800 1.00

ef

1 800

C., Concordia,"Synchronous Machines", Book, John Wiley, New York, 1951.

39000

[2]

M., Sarma,"Synchronous Machines", Book, and Breach Science Publishers, New York, 1979.

[3]

S. Murthy, G. Berg, B. Singh and J. Singh,"Transient Analysis of a Three-Phase Induction Motor with Single-Phase Supply ", IEEE Trans., Vol.PAS-102(1), 1983, p.28.

15000

A.L. Mohamadin, A.A. Al-Ohaly and A.H. AlBahrani,"On the Choice of Phase-Balancer Capacitance for Induction Motors Fed from Single Phase Supply", IEEE Trans., Vol.EC -2, No.3, Sept. 1987, pp.458-464.

-9000 0 .10

Gordon

δ= −10

p.u

X2

p.u

[1]

ef

0.55

vi) X1 and X2 is open circuited when γp is +60 ° and -60 °, respectively. V. References

δ= −5

p.u

iv) The value of X1 and X2, to achieve perfect balanced operation varies with the load, the ratio between the stator induced emf and supply voltage and the shift angle between Vp and Ip, (γp).

-30

δ= −15

X2

iii) The reactance of the two elements (X1 and X2) can be selected to achieve perfect balanced operation of the motor.

0

p.u

ii) These two elements are made passive and reactive to avoid any extra losses to the system.

30

X1

i) The operation of three phase synchronous motors from a single phase supply is possible through the use of a two element-phase balancer.

angle bet. Ip & Vp

IV. Conclusions

R. Habbermann, "Single Phase Operation of Three Phase Induction Motor Connected to a Single-Phase Supply systems", AIEE Trans., Vol.PAS, 1954, pp.833837.

X1

- 1800 1.0 0

4000 X1

p.u

X2 -100

0

X2

[6]

R. Jha and C Jha,"Operation of Three-Phase Induction Motor Connected to a Single-Phase Supply System", Journal of the Institution of Engineers (India), Vol.58, pt EL6, 1978, p.339.

ef

p.u

[5]

0.5 5

0

X1

[4]

X1

X2

0

δ= −15

-200 -4000 0.10 0.55 ef 1.00 Fig.7 Variations of X1 and X2 against ef for different values of δ.

-4

5000

[13] A.I. Alolah and M.A. Badr, "Starting of Three Phase p.u

p.u

δ= −30

Reluctance Motor Connected to Single Phase Supply", IEEE Trans., Vol.EC -7(2), 1992, pp.295-301.

[14] A.I. Alolah, "Balanced Operation of Three Phase

X2 -3500

X1

X2

-7

Reluctance Motors Connected to Single Phase Supplies", IEEE and Middle East Technical University 7th MELECON 94 Conference, Antalya, Turkey, April, 1994, pp.813-816.

X1

[15] M.A. Badr and A.I. Alolah, "Transient Analysis of -10 0.10

0.55

ef

-12000 1.00

Three Phase Reluctance Motors Fed From A Single Phase Supply", IEE Proc. on Electric Power Applications, Vol.142(2), 1995, pp. 104-112.

17000

[16] A.I. Alolah, A.M. Alsalloum and M..A. Badr, "Effect of

-4

p.u

δ= −35

p.u

X1

-7

X2

7000

X1

X2

-10 0.10

0.55

ef

-3000 1.00

Machine and Balancer Parameters on the Pulling Into Step of Three-Phase Reluctance Motors Fed From A Single Phase Supply", European Transactions on Electrical Power Engineering (ETEP), Germany, Vol.5(3), 1995, pp.207-214.

[17] A.I. Alolah, "Operation of Three Phase Synchronous Motors Fed From A Single Phase Supply", IEEE and Middle East Technical University7th MELECON 94 Conference, Antalya, Turkey, April 1994, pp.12571278.

[18] M.A. Badr and A.I. Alolah, "Starting Transients of [7]

[8]

[9]

J. Brown and C Jha, "The Starting of a Three-Phase Induction Motor Connected to a Single-Phase Supply System", Proc. of IEE, pt A, Vol 106, 1959, p 183. Danials and B Belly, "A Locus Diagram to Determine the Starting Performance of a Three-Phase Induction Motor Connected to a Single-Phase Supply ", Proc. of IEE, Vol 108, 1961, pp.244-249. M.A. Badr, A.I. Alolah and M.A. Abdel-halim, "A Capacitor Start Three Phase Induction Motor", IEEE Trans., Vol.EC -10(4), 1995, pp.675-680.

Three Phase Synchronous Motors Connected to a Single Phase Supply ", IEEE Trans., Vol.EC-10(1), 1995, pp.48-55.

[19] A.I. Alolah, "Steady State Operation of Three-Phase Synchrouns Motor from a Single-Phase Proc. of ICEM, Espoo, Finland, August 2000.

[20] Ivanov, "Electrical Machines edition, Translated Moscow, 1988.

[11] C., Tindall and W. Monteith, "Balanced Operation of Three-Phase Induction Motor Connected to SinglePhase Supplies", Proc. of IEE, Vol 123(6), 1976, pp.517-522.

[12] A.I. Alolah, "Steady State Operation of Three Phase Reluctance Motor from Single Phase Supply", Journal of Institution of Engineers (India), Vol.70, Pt.EL(5), 1989, pp.157-161.

Part 2", Book, 2nd Russian, Mir Publishers,

from

[21] "Westinghouse

Electrical Transmission and Distribution", Reference Book, 4th edition, USA, 1964.

[10] A.A. Al-Ohaly, A.L. Mohamadin and A.H. Bahrani, "Effect of Phase-Balancer Capacitance on the Dynamic Behavior of a Three-Phase Induction Motor Operated from Single-Phase Supply", Journal of Engineering Sciences, King Saud University, Vol.15(1), 1989.

Supply",

VI. Appendix D = − 3 j ( X1 + X2 ) + X1 X2 ( Yp + Yn ) 1 D p = − jX1 + e − j150 X 2 + X1 X2 Yn e − j 30 3 D n = − jX 1 + e − j 30 X 2 +

1 3

X1 X 2Y p e j 30

k1 = 3 R p + X p

k2 =1 − 3 e f cos δ + e f sin δ

k3 = 3 X p − 3 R p

k4 =1 − 3 e f cos δ − 3 e f sin δ

ef = | Ef / Vi |