Expression For Back E.M.F Or Induced E.M.F. Per Phase In S.M.
Case i)
Under excitation, Ebph < Vph .
Zs = Ra + j Xs = | Zs | ?? ?
? = tan-1(Xs/Ra)
ERph ^ Iaph = ?, Ia lags always by angle ?.
Vph = Phase voltage applied
ERph = Back e.m.f. induced per phase
ERph = Ia x Zs V … per phase
Let p.f. be cos?, lagging as under excited,
Vph ^ Iaph = ?
Phasor diagram is shown in the Fig. 1.
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| Fig. 1 Phasor diagram for under excited condition |
Applying cosine rule to ? OAB,
(Ebph)2 = (Vph)2 + (ERph)2 – 2Vph ERph x (Vph ^ ERph)
but Vph ^ ERph = x = ? – ?
(Ebph)2 = (Vph)2 + (ERph)2 – 2Vph ERph x (? – ?) ……(1)
where ERph = Iaph x Zs
Applying sine rule to ? OAB,
Ebph/sinx = ERph/sin?
So once Ebph is calculated, load angle ? can be determined by using sine rule.
Case ii) Over excitation, Ebph > Vph
p.f. is leading in nature.
ERph ^ Iaph = ?
Vph ^ Iaph = ?
The phasor diagram is shown in the Fig. 2.
Applying cosine rule to ? OAB,
(Ebph)2 = (Vph)2 + (ERph)2 – 2Vph ERph x cos(Vph ^ ERph)
Vph ^ ERph = ? + ?
... (Ebph)2 = (Vph)2 + (ERph)2 – 2 Vph ERph cos(? + ?) …….(3)
But ? + ? is generally greater than 90o
... cos (? + ?) becomes negative, hence for leading p.f., Ebph >Vph .
Applying sine rule to ? OAB,
Ebph/sin( ERph ^Vph) = ERph/sin?
Hence load angle ? can be calculated once Ebph is known.
Case iii) Critical excitation
In this case Ebph ? Vph, but p.f. of synchronous motor is unity.
... cos = 1 ... ? = 0o
i.e. Vph and Iaph are in phase
and ERph ^Iaph = ?
Phasor diagram is shown in the Fig. 3.
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| Fig. 3 Phasor diagram for unity p.f. condition |
Applying cosine rule to OAB,
(Ebph)2 = (Vph)2 + (ERph)2 – 2Vph ERph cos ? …………(5)
Applying sine rule to OAB,
Ebph/sin? = ERph/sin?
where ERph = Iaph x Zs V