Page 136 - Electrician - TT (Volume 2)

ELECTRICIAN - CITS

Though the leakage fluxes are independent of saturation, they do depend upon the current and the phase angle
between the current and the terminal voltage `V’. These leakage fluxes induce a reactance voltage which is
ahead of the current by 90°. Normally the effect of leakage flux is termed as inductive reactance XL and as a
variable quantity. Sometimes the value XL is named as synchronous reactance to indicate that it refers to working
conditions.
Voltage drop due to armature reaction: The armature reaction in an alternator is similar to DC generators. But
the load power factor has considerable effect on the armature reaction in the alternators.

The effects of armature reaction have to be considered in three cases, i.e. when load power factor is
• unity
• zero lagging
• zero leading.
At unity P.F. the effect of armature reaction is only cross- magnetising. Hence there will be some distortion of the
magnetic field.
But in the case of zero lagging P.F. the effect of armature reaction will be de-magnetising. To compensate this
de-magnetising effect, the field excitation current needs to be increased.
On the other hand, the effect of armature reaction due to zero leading P.F. will be magnetising. To compensate the
increased induced emf, and to keep the constant value of the terminal voltage due to this additional magnetising
effect, the field excitation current has to be decreased.
Rating of alternators: As the power factor for a given capacity load determines the load current, and the
alternator’s capacity is decided on load current, the rating of the alternator is given in kVA or MVA rather than kW
or MW in which case the power factor also is to be indicated along with the wattage rating.
Example: A 3-phase, star-connected alternator supplies a load of 5 MW at P.F. 0.85 lagging and at a voltage of
11 kV. Its resistance is 0.2 ohm per phase and the synchronous reactance is 0.4 ohm per phase. Calculate the
line value of the emf generated.
Full load current = I = P
L
L
3 ECos
5 1000 1000 = 309 Amps.
3 11000 . 85
In star I = I
L P
IR drop = 309x0.2=61.8 V
a
Ix drop = 309 x 0.4 = 123.6 V
L
Terminal voltage (line)=11000 V
Terminal voltage(phase) 11000 = 6350V
3
Power factor = 0.85
Power factor angle = q = Cos –1(.85)
= Cos 31.8°
Sin q = 0.527.
Drawing the vector, as shown in Fig 2, with the above data, we have

2

E =(VCos +RI a ) ( VSin I X ) 2
L
o
2
. )
= (6350 . 0 85 +61.8) (6350 . 0527 1236 2
= 6468.787 volts.
Line voltage = 3 E P = 3 6469 11204 V

123

CITS : Power - Electrician & Wireman - Lesson 76-85