Page 184 - Electrician - TT (Volume 1)

ELECTRICIAN - CITS

Instantaneous values of voltage,

EA=EmSin wt.
EB=EmSin (wt-90).
As the two windings are wound on the same armature and have same number of turns, the generated EMF in the
two coils have the same maximum value, frequency and wave length.
The voltage between two phases is the vector sum of voltage generated in two windings.
AB²=OA²+OB²
OR

ER²= EA²+ EB²
EA and EB are phase voltages and are equal
Therefore
ER²=E²+E²
ER²=2E²
ER=√2E

• This means that the line voltage in two phase system is √ 2 times phase voltage.
Power derivation in star connection
The total power in the circuit is quite logically,the sum of the powers in the three phases. For balanced load power
consumed in each phase is same
Total power = 3 × power in each phase
3 × V I cosØ
ph ph
For star connection V = V /√3 ;I = I L
ph
ph
L
P = 3 × V /√3 × I cosØ
L L
=√ 3V I cosØ
L L
Ø is the phase difference between a phase voltage and corresponding phase current.
Note that in either case star or delta ,the expression for the total power is same provided that the system is
balanced
Related problems
1.A three phase 415 V A.C induction motor requires a current of 50 A . Calculate its H.P at a P.F of 0.8 and
efficiency 90 %
Solution
Line voltage V = 415 V
L
Line current I = 50A
L
Power factor cosØ = 0.8
Efficiency of motor = 90%=0.9
As it is 3 phase motor :
Input power= √ 3.VL .IL.cosØ
= 1.732 x 415 x 50 x 0.8 watts
Out put of motor = input to motor x efficiency
= 1.732 x 415 x 50 x 0.8 x 0.9 watts
= 1.732 x 415 x 50 x 0.8 x 0.9 / 735.5 ….1 H,P = 735.5 watts
= 35.18 H.P

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CITS : Power - Electrician & Wireman - Lesson 26-29