Synchronizing Power

Synchronizing Power  AU : Nov.-96, Oct.-97, 98, April-2000, 01, 04, May-01, 03, 08

Consider an alternator connected to infinite bus bar. Let V be the bus bar voltage and E be the e.m.f. induced in the alternator. The excitation of the alternator is adjusted in such a way that E and V are equal in magnitude.

In the local circuit the two voltages E and V are in phase opposition while in    the external circuit they are in the same phase. This is represented in the Fig. 3.9.1.

Consider the alternator to be on no load. If by some means power input to machine is decreased and its induced e.m.f. E will then lag behind V by say angle 28. Due to this difference, E and V will not remain in exact phase opposition but will give rise to resultant e.m.f. Er. This Er will act in the local circuit and a synchronizing current ISY will start flowing in the local circuit. The synchronizing current is given by, 

I_SY = E_r/Z_s

I_SY is lagging behind Er by an angle θ given by,

θ = tan^-1(X_s/R)

R is very very small it can be neglected.

θ ≈ 90°

The angle 2δ is very very small and θ is approximately equal to 90° so the synchronizing current IgY is almost in phase with V and in phase opposition with E. So infinite bus bar will deliver some power to the alternator. As current in the local circuit is always opposing to induced e.m.f. E, the alternator will act as a synchronous motor. Thus synchronizing torque will be developed which will try to accelerate the machine. Thus the angle 2 δ will go on decreasing and resultant e.m.f. Er also goes on decreasing. Finally the two e.m.f.s E and V will again be in phase opposition and the machine will now act as an alternator in synchronism with bus bar.

Thus the power which automatically comes into play and accelerates the machine which was retarding and decelerates the machine which tries to accelerate is called synchronizing power. This power will keep the machine in step with infinite bus bar.

 

1. Expression for Synchronizing Power (P_SY)

Consider an alternator which is operating at an power angle δ        i.e. E leads V by an angle δ         

Let power input of this alternator be increased suddenly so that      it will now operate at a new power angle given by δ + δ '. So the    synchronizing emf E_SY will come into play and sends a circulating current given by I_SY = E_SY / Z_S. This current produces synchronizing power. Now we will derive the expression for synchronizing    power per phase.

Before increasing the input of alternator, the power input Pil is given by,

P_il = E/Z_S [E cos θ – V cos (θ + δ)

When power angle δ has changed to δ ± δ ' (positive sign indicates acceleration while negative sign indicates deceleration) the power input P_i2 is given by,

P_i2 = E / Z_S [E cos θ – V cos (θ+ δ ± δ)]

The difference between these two power is nothing but synchronizing power P_SY.

The above expression is per phase power. Therefore for the machine having 'm' phases the synchronizing power is given by,

P_SY = m E I δ' 

 

2. Synchronizing Torque (T_SY)

The synchronizing torque T_SY per phase in N-m when two alternators are connected in parallel is given by,

When alternator is connected to infinite bus bars then T_SY is given as

Here N_s = Synchronous speed = 120f/P

 

3. Effect of Load on Synchronizing Power

When the alternator is loaded then the expression for P_SY is given as,

P_SY = ɑ E V /X_s

Here V is bus bar voltage.

E is alternator induced e.m.f. per phase.

E = V + I Z_s

The corresponding phasor diagram is shown in the Fig.3.9.4

From phasor diagram,

 

Example 3.9.1 A 700 KVA, 11 kV, 4 pole,3 phase star connected alternator has percentage resistance and reactance as 1.5 and 14 respectively. Determine synchronizing power per mechanical degree of displacement at full load 0.8 p.f. lagging.

Solution:

Using the values, E_ph = 6991.1104 volts.

The phasor diagram is as shown in Fig. 3.9.5 Power angle, δ can be calculated as follows.

 

Example 3.9.2 A 3 MVA, 6 pole alternator runs at 1000 r.p.m. on 3.3 kV bus bars. The synchronous reactance is 25 %. Calculate the synchronizing power and torque per mechanical degree of displacement when the alternator is supplying full load at 0.8 pf. lag. AU : Nov.-96, April-2000, Marks 8

 

Example 3.9.3 A 5 MVA, 10kV, 1500 r.p.m. 3-phase, 50 Hz, 4 pole alternator is operating on infinite bus bar. Find the synchronizing power per mechanical degree of angular displacement under no load condition X_s = 20 %

Solution :

Example 3.9.4 A 5000 kVA, 10000 V, 1500 r.p.m., 50 Hz alternator runs in parallel with other machines. Its synchronous reactance is 20 %. Find the (1) no load (2) full load at power factor 0.8 lagging, synchronizing power per unit, mechanical angle of phase displacement and calculate the synchronizing torque if the mechanical displacement is 0.5°. AU : May-08, Marks 12

Solution :

This is synchronizing torque for 1° mechanical displacement .

For 0.5° mechanical displacement,

T_SY = 0.5 × 5555.55 = 2777.77 Nm

ii) Full load, cos ϕ = 0.8 lagging, ϕ = cos^-1 0.8 = 36.869° 

I = 288.675 ∠ 0° A, V = 5773.5 ∠ 36.869° V      ... I lags V

V = 4618.8542 + j 3464.02 V

Thus E leads I by 44.99° and V leads I by 36.869°

The angle δ is the angle between E and V hence δ = 44.99° - 36.869° = 8.12°

For 0.5° mechanical displacement,

T_SY = 0.5 ×  6221.048 = 3110.524 Nm

 

Example 3.9.5 A 3 phase, 50 Hz, 2 pole alternator is excited to generate the bus bar voltages of 11 kV at no load, Calculate the synchronizing power per degree of mechanical displacement of the rotor. The machine is star connected and the short circuit current for this excitation is 1200 amperes. Neglect armature winding resistance.

Solution :

E = 11 kV (line), P = 2, I_sc = 1200 A

 

Examples for Practice

Example 3.9.6 A 10 MVA 3 phase alternator has an equivalent short circuit reactance 20 %. Calculate the synchronizing power of the armature per mechanical degree of phase displacement when running in parallel on a 10,000 V, 50 Hz bus bars at 1,500 r.p.m.

[Ans.: 1745.325 kW]

Example 3.9.7 A 700 kVA, 11 kV, 4 pole, 3 phase star connected alternator has percentage resistance and reactance as 1.5 and 14 respectively. Determine synchronizing power per mechanical degree of displacement at full load 0.8 p.f. lagging.       

[Ans.: 191.24 kW, 1217.47 N-m]

Example 3.9.8 Calculate the synchronizing power in kW for 1 mechanical degree of displacement at full load 0.8 power factor lagging for 3 phase, 2000 kVA, 6600 V, 50 Hz, 12 pole synchronous machine having a synchronous reactance of 25 % and negligible resistance.

[Ans.: 961.48 kW]

Example 3.9.9 A 5 MVA, 11 kV, 50 Hz, 4-pole, star-connected synchronous generator with synchronous reactance of 0.7 p.u. is connected to an infinite bus. Find synchronizing power and the corresponding torque per unit of mechanical angle displacement -

i) At no load and ii) At full load of 0.8 p.f. lagging.

[Ans.: i) 1587.3 N-m, ii) 2253.494 N-m]

Example 3.9.10 A 6.6 kV, 3 MVA, 3 phase, 50 Hz, 8 pole alternator has synchronous reactance of 2.9 ohm and negligible resistance. When it is operating on infinite bus, calculate the synchronizing power and the corresponding torque per mechanical degree of phase displacement at

i) No load ii) Full load 0.85 pf lag.

[Ans.: 1048.64 kW, 13.35 × 10³ N-m, 1158 kW, 14.74 × 10³ N-m]

Example 3.9.11 A 2000 kVA, 3 phase, 8 pole star connected synchronous generator runs on 6000 volts, 50 Hz infinite bus bars. Find the synchronizing power and synchronizing torque per mechanical degree of displacement for full load at 0.8 p.f. lagging. The resistance of generator is 0.01 p.u. and synchronous reactance is 1.2 p.u.

[Ans.: 4.77°, 1255.6 kW, 15986.78 Nm]

Example 3.9.12 A 2 MVA, 3-phase, 8 pole, 750 rpm, sychronous generator is operating on 6000 V bus bars. The synchronous reactance is 6 Q/phase. Find the synchronizing power and torque per mechanical degree at full load and 0.8 pflag.

[Ans.: 502.248 kW, 6394.82 Nm] 

Example 3.9.13 A 2 pole, 50 Hz, 3 phase, turbo alternator is excited to generate the bus bar voltage of 11 kV on no load. The machine is star connected and short circuit for this excitation is 1000 A. Calculate the synchronizing power per degree of mechanical displacement of the rotor and the corresponding synchronizing torque.

[Ans.: 332.52 kW, 1058.44 N-m]

Example 3.9.14 A 3 MVA, 6 pole alternator runs at 1000 r.p.m. in parallel with other machine on 3.3 kV bus bars. The synchronous reactance is 20 %. Calculate the synchronizing power per mechanical degree of displacement and the corresponding synchronizing torque.

[Ans.: 261.797 kW, 7500 N-m]

Review Question

1. Derive the expression for the synchronizing power when an alternator is connected to infinite bus bar.