Effect of Change in Input or Mechanical Torque

Effect of Change in Input or Mechanical Torque AU : NOV.-04, May-04, 07, 13, Dec.-l0

For any alternator its driving torque can be changed by controlling the gate opening in case of hydrogenerators or by controlling the throttle opening in case of turbogenerators. Again we will consider two cases that are alternator with and without load respectively.

 

1. Alternator on No Load

Suppose that two alternators are running in parallel without any load on them. The excitations for two alternators are adjusted in such a way that the induced e.m.f.s. are equal in magnitude. The resultant voltage in the local circuit will be zero. With respect to external circuit the two e.m.f.s are in phase whereas in local circuit they are in opposition. 

Now the driving torque of alternator 1 is increased. This increment in torque will try to accelerate the alternator 1 and its induced e.m.f. E1 will lead e.m.f. E2. This will give rise to resultant voltage E_r. This voltage  circulates current IgY in local circuit which is given by

This current lags behind E_r by angle of approximately 90° if the resistances of the armatures of the two alternators are neglected. This is represented in following phasor diagram shown in Fig. 3.15.1.

This circulating current I_SY is almost in phase with E₁ and in phase opposition with E₂. Now here the synchronizing power will come into play. The alternator 1 produces a power E₁ I_SY cos ϕ₁ which is positive as ϕ₁ < 90° while alternator  2 generates a power E₂ I_SY cos ϕ₂ which is negative  as ϕ₂ < 90° Alternately we can say that alternator 1 experiences a generating action which will try to retard it and alternator 2 receives the power produced by alternator 1. Hence it will experience a motoring action which will tend to accelerate it. Thus there will be automatic synchronizing action which will retard the faster machine and accelerate the slower machine and synchronism is maintained.

It can be seen that the autosynchronizing action is on account of Z₁ and Z₂considered mainly reactive. If Z₁ and Z₂ are purely resistive then ISY will be in phase of E_r. Then power for both the machines is positive and both will experience generating action. So there would not be synchronizing power that will tend to accelerate the slower machine.

Key Point Thus reactance mainly causes auto synchronization but it is bad for voltage regulation.

 

2. Alternator on Load

Again we will consider two alternators which are loaded and running in parallel. The sharing of load between these alternators is governed by speed-load characteristics of their prime movers. In the Fig. 3.15.2 the two alternators are shown driven by prime movers 1 and 2. 

In Fig. 3.15.3 the lines 1 and 2 represent the speed load characteristics of prime movers 1 and 2 . For clarity and simplicity the slopes are exaggerated.

Horizontal line ab represents total load of 2P with load on each alternator as P. The frequency of bus bar is f.

Now if by govemer setting, the torque of prime mover 1 is increased, its speed will be increased which will shift its speed-load curve upwards. This is shown by dotted line 1'. The original operating points a and b are now shifted to c and d. This will give new operating conditions which  will increase load on alternator 1 from P to P₁ and decrease load on alternator 2 from P to P₂ with P₁ + P₂ = 2P. From the Fig. 3.15.3 it can be seen that frequency has increased from f to f'. Now, if it is desired to maintain the frequency constant then the input to prime mover 2 must be reduced which will shift its speed-load curve downward shown by dotted line 2'. The operating points c and d now shift to new points x and y. The horizontal line xy indicates that the load on alternator 1 is further increased from P₁ and P₁ and ₂that on alternator 2 is reduced from P to P₂ such that the relation P₁ + P₂ = 2P is maintained. Thus the load sharing between the alternators and the frequency can be controlled by changing the mechanical torque input to the alternators. By controlling the gate opening of water turbines or the throttle opening of steam turbines, the speed-load characteristics of prime movers can be shifted up and down.

 

3. Phasor Diagram

To consider what happens internally in the two alternators, let us consider the phasor diagram. 

The two alternators are running in parallel with their excitations constant. The armature currents I₁ and I₂ are also equal so that total load current is 2I₁ or 2I₂ . The terminal voltage V is constant. Each alternator is sharing a load equal to E₁V / X_s sin δ = E₂V / X_s sin δ = P.

Now when mechanical torque of alternator 1 is increased, its output will also increase. But E₁ V and X_s are constant. So to increase power output power angle must be increased from δ to δ₁, so new E₁, will be ahead of previous position. The alternator 1 shares greater load than P. Therefore for constant load of 2P the load on alternator 2 must be less than P. This will make new E₂ to fall back from its previous position. Due to the different positions of E₁, and E₂, resulting voltage AB appears in the local circuit which will send a circulating current I_SY lagging behind the voltage by 90°. This current I_SY must be added to I₁, and subtracted from I₂.

The alternator 1 carries increased current I₁ and alternator 2 carries decreased current I₂ but total load current remains same . The power factor of alternator 1 is improved from cos ϕ to cos ϕ₁ whereas it is reduced from cos ϕ to cos ϕ₂ for alternator 2. But the load power factor remains unaffected.

Thus increase in mechanical torque in case of alternator will increase armature current and improve the power factor. The alternator will share increased load whose driving torque is increased whereas the other alternator which is in parallel is relieved from the load whereas the reactive power distribution remains unaffected.

To consider the effect of change in input on corresponding power triangles of the two alternators we will assume that the two alternators are turbo alternators whose prime movers are supplied with steam.

Now the excitations for the two alternators are kept constant where steam supply i.e. power input to prime mover of alternator 1 is increased. The two alternators are running in synchronism. So machine 1 cannot overrun machine 2. The increased power input for alternator 1 makes it possible for carrying more load. This will make rotor for machine 1 advancing its angular position by an angle δ . 

The resultant e.m.f. E_r is produced in the local circuit which will set up a circulating current I_SY which lags E_r by 90° and almost in phase with E₁. The power per phase for alternator 1 is increased by an amount E₁ ISY whereas it is decreased by same amount for alternator 2. This current I_SY has no appreaciable reactive component and it will not disturb the reactive power distribution but active power output of alternator 1 will increase and that of 2 will decrease. This is shown in Fig. 3.15.5.

Key Point : The change in input to the prime mover will change the distribution of load between the alternators.

 

Example 3. 15. 1 A 3 phase, star connected alternator with R = 0.4 Q/ph and X = 6 Ω/phase delivers 300 Amps at a power factor of 0.8 lag to constant voltage, constant frequency 10 kV bus bar. If the steam supply is unchanged, find the percentage change in the induced e.m.f. necessary to raise the power factor to unity. Ignore changes in losses.

Solution :

Power developed in alternator = Power delivered to bus bar + I²R losses in armature

As rotational losses are assumed constant with steam supply unchanged, the power developed by alternator remains constant. With p.f. changed to unity, the armature current can be found as below.

Power developed in alternator = Power delivered to bus bar + I²R losses in armature

 

Example 3.15.2 Two identical 2000 kVA alternators operate in parallel. The governor of the prime mover of first machine is such that the frequency drops uniformly from 50 Hz on load to 48 Hz on full load. The corresponding uniform speed drop of the second machine is 50 Hz to 47.5 Hz. Find :

i) How will the two machines share a load of 3000 kW ?

ii) What is the maximum load of unity p.f. that can be delivered without overloading either machine ?

Solution :

The speed-load characteristics of the two alternators can be drawn as shown in Fig. 3.15.6.

Line PQ is drawn for machine 1 while line PR is drawn for machine 2. At any load the frequency of the two machines must be same. A line AB is drawn at a frequency x measured from point P. Total load at this frequency is given as 3000 kW.

AC + CB = 3000

Using the similarly of the triangles PAC and PQS

 

Example 3.15.3 Two exactly similar turbo alternators are rated at 25 MW each. They are running in parallel. The speed-load characteristics of the driving turbines are such that the frequency of alternator 1 drops uniformly from 50 Hz on no load to 48 Hz on full load and that of alternator 2 from 50 Hz to 48.5 Hz. How will the two machines share a load of 30 MW ? 

Solution :

Refer example 3.15.2 for the procedure and verify the answer as 12.85 MW and 17.15 MW.

 

Example 3.15.4  Two similar, 3 phase alternators work in parallel and deliver a total real power of 1800 kW at 11 kV and at 0.85 pf lagging to the load. Each alternator initially supplied half the load power. The excitation of the first alternator is then increased such that its line current becomes 60 A lagging. Find the line current delivered by the second alternator. Find the power factor of each alternator.

Solution : The load current is,

Wattfull or active component of current = I_L cos ϕ = 94.475 A

Wattless or reactive component of current = I_L sin ϕ = 58.55 A

Each alternator supplies half the load hence, current supplied by each machine = I_L / 2 = 55.5735 A. Since the steam supply of the first machine is not changed, the working components of both machines would remain the same as I_L cos ϕ  / 2 94.475 / 2 = 47.2375 A.

But the wattless or reactive component changes due to change in excitation. The line i.e. armature current of the first machine is changed from 55.5735 A to 60 A.

Wattless component of 1st machine = √60² - 47.2375²  = 36.9948 A

Wattless component of 2nd machine = 58.55 - 36.9948 = 21.555 A

Armature current of 2nd machine = √47.2375² + 21.555² = 51.923 A

The current diagrams are shown in the Fig. 3.15.7.

 

Examples for Practice

Example 3.15.5 Two identical 4000 kVA alternators operate in parallel. The govemer of the prime mover of first machine is such that the frequency drops uniformly from 50 Hz on load to 48 Hz on full load. The corresponding uniform speed drop of the second machine is 50 Hz to 47.5 Hz.

Find. : i) How will the two machines share a load of 6000 kW ?

ii) What is the maximum load of unity pf. that can be delivered without overloading either machine ?

[Ans.: 2667 kW, 3333 kW, 7200 kW]

Example 3.15.6 Two similar 6.6 kV, 3-phase generators are running in parallel at constant voltage and freuency bus bars. Each has an equivalent resistance and reactance of 0.05 ohms and 0.5 ohms respectively and supplies one half of a total load of 10000 kW at a lagging power factor of 0.8, the two machines being similarly excited, If the excitation of one machine is adjusted until the armature current is 438 A and the steam supply to the turbine remains unchanged, find the armature current, the e.m.f. and the power factor of the other alternator.

[Ans.: 769.21 A, 0.5686,7193.06 V]

Example 3.15.7 Two exactly similar turbo-alternators are rated 20 MW each. They are running in parallel. The speed-load characteristics of the driving turbines are such that the frequency of alternator 1 drop uniformly from 50 Hz on no-load to 48 Hz on full-load, and that of alternator 2 from 50 Hz to 48.5 Hz. How will the two machines share a load of 30 MW ? JNTU : Nov.-04

[Ans.: 12.85 MW, 17.15 MW]

Example 3.15.8 Two 750 kW alterantors operate in parallel. The speed regulation of one set is 100 % to 102 % from full-load to no-load and that of the other is 100 % to 104 %. How will the two alternators share a load of 1000 kW and at what load will one machine fail to supply any portion of the load ? JNTU : Nov.-04

[Ans.: 420 kW, 580 kW, 375 kW] 

Review Question

1. Dissuss the effect of change in the input power on the alternators running in parallel.  AU : May-07, Marks 8