Synchronous Condensers

Synchronous Condensers    AU : April-97, Dec.-09, 17, May-11, 12

When synchronous motor is over excited it takes leading p.f. current. If synchronous motor is on no load, where load angle 8 is very small and it is over excited (E_b > V) then power factor angle increases almost upto 90°. And motor rims with almost zero leading power factor condition. This is shown in the phasor diagram Fig. 4.19.1.

This characteristics is similar to a normal capacitor which always takes leading power factor current. Hence over excited synchronous motor operating on no load condition is called as synchronous condenser or synchronous capacitor. This is the property due to which synchronous motor is used as a phase advancer or as power improvement device.

1. Disadvantages of Low Power Factor

In various industries, many machines are of induction motor type. The lighting and heating loads are supplied through transformers. The induction motors and transformers draw lagging current from the supply. Hence the overall power factor is very low and lagging in nature.

The power is given by,

P = V I cos ϕ  i.e. I = P / V cos ϕ       ... Single phase

The supply voltage is constant and hence for supplying a fixed power P, the current is inversely proportional to the p.f. cos ϕ . Let P = 5 kW is to be supplied with a voltage of 230 V then,

Case i) cos ϕ = 0.8, I - 27.17 A

Case ii) cos ϕ = 0.6, I = 36.23 A

Thus as p.f. decreases, becomes low, the current drawn from the supply increases to supply same power to the load. But if p.f. maintained high, the current drawn from supply is less.

The high current due to low p.f. has following disadvantages :

1. For higher current, conductor size required is more which increases the cost.

2. The p.f. is given by

cos ϕ  = Active power / Apparent power = Pin kW / S i.e. kVA rating

Thus for fixed active power P, low p.f. demands large kVA rating alternators and transformers. This increases the cost.

3. Large current means more copper losses and poor efficiency.

4. Large current causes large voltage drops in transmission lines, alternators and other equipments. This results into poor regulation. To compensate such drop extra equipment is necessary which further increases the cost.

Key Point : Hence power factor improvement is must in practice. Hence the supply authorities encourage consumers to improve the p.f

2. Use of Synchronous Condenser in Power Factor Improvement

The low power factor increases the cost of generation, distribution and transmission of the electrical energy. Hence such low power factor needs to be corrected. Such power factor correction is possible by connecting synchronous motor across the supply and operating it on no load with over excitation.

Now let V_ph is the voltage applied and Ilph is the current lagging V_ph by angle ϕ₁ This power factor ϕ₁ is very low, lagging.

The synchronous motor acting as a synchronous condenser is now connected across the same supply. This draws a leading current of I_2ph

The total current drawn from the supply is now phasor of Iph and I2ph. This total current IT now lags Vph by smaller angle due to which effective power factor gets improved. This is shown in the Fig. 4.19.2.

This is how the synchronous motor as a synchronous condenser is used to improve power factor of the combined load.

Example 4.19.1 A synchronous motor absorbing 75 kW is connected in parallel with a factory load of 300 kW having 0.9 lagging power factor. If the combined load has lagging power factor of 0.95, what is the value of leading kVAR supplied by the motor and what power factor is it working ?

Solution :

Load 1 = Industrial load, P₁ = 300 kW, cos ϕ₁ = 0.9, ϕ₁ = 25.8419°

The power triangle is shown in the Fig. 4.19.3 (a).

Q₁ = P₁ × tan ϕ₁ = 145.2966 kVAR

This is reactive power of load 1, lagging nature.

Load 2 = Synchronous motor, P₂ = 75 kW

The combined load active power is P_T = P₁ + P₂ = 375 kW

cos ϕ_T = 0.95, ϕ_T = 18.1948°

The power triangle is shown in the Fig. 4.19.3 (b).

Q_T = P_T × tan ϕ_T = 123.2565 kVAR

This is total reactive kVAR.

Q_T = Q₁ + Q₂        i.e. 123.2565 = 145.2966 + Q₂

Q₂ = - 22.04 kVAR        ... Negative indicates leading nature

This is kVAR supplied by synchronous motor and negative sign indicates its leading nature.

The power triangle for synchronous motor is shown in the Fig. 4.19.3 (c)

while overall diagram is shown in the Fig. 4.19.3 (d).

From Fig. 4.19.3 (c),

Examples for Practice

Example 4.19.2 An industrial load of 800 kW is operating at 0.6 lagging power factor. It is desired to improve the factor to 0.92 lagging by connecting a synchronous motor driving a load of 200 kW with an efficiency of 91 %. Determine kVA rating of the synchronous motor and the power factor at which it is operating

[Ans.: 669.356 kVA, 0.3283 leading]

Example 4.19.3 A synchronous motor improves the power factor of a load of 500 kW from 0.707 lagging to 0.95 lagging. Simultaneously the motor carries a load of 100 kW. Find : i) The leading kVAR supplied by the motor ii) kVA rating of a the motor iii) Power factor at which the motor operates.  UPTU : 2008-09

 [Ans. : 302.789 kVAR, 318.8753 kVA, 0.3136 leading]

Example 4.19.4 A factory has a total load of 1800 kW at a p.f. of 0.6 lagging. If it is desired to improve the p.f. to 0.95 lagging with the installation of a synchronous condenser, calculate

i) the kVA rating of the synchronous condenser.

ii) total kVA of the factory.

The condenser does not draw any active power.      VTU : March-03

[Ans. : - 1808.368 KVAR, 1894.7367 kVA]

 Example 4.19.5 A 440 volts, 50 Hz, 3 phase, star-connected circuit takes 35 Amps at a lagging p.f. of 0.8. A star connected synchronous motor is used to improve the power factor to unity. Calculate the kVA input, and the power factor of the synchronous motor when it is also supplying a load of 12 kW and has an efficiency of 85%.   VTU : Aug.-05

[Ans.: Negative sign indicates leading nature., 0.6615 leading, 21.341 kVA]

Review Questions

1. Write a note on synchronous condenser and its use.

2. What are the disadvantages of low p.f. ? How it can be improved using synchronous condenser ?

3. Explain how synchronous motor can be operated as a synchronous condensers. Draw a phasor diagram

AU : Dec.-09, 17, May-11, 12, Marks 12