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UPPCL JE Previous Paper 7 (Held On 25 November 2019 Shift 2)

Option 1 : Load angle is equal to internal angle

ST 1: General Knowledge

5609

20 Questions
20 Marks
20 Mins

__Mechanical power developed by synchronous motor:__

The mechanical power developed by the synchronous motor is given by

\({{\bf{P}}_{{\bf{mech}}}} = \frac{{{\bf{EV}}}}{{{Z_s}}}\;{\bf{cos}}\left( {{\bf{θ }} - {\bf{δ }}} \right) - \frac{{{{\bf{E}}^2}}}{{{{\bf{Z}}_{\bf{s}}}}}\;{\bf{cos}}\;{\bf{θ }}\)

Where

E is the excitation voltage line to line

V is the terminal voltage line to line

Z_{s} is the synchronous impedance

θ is the internal angle and δ is the load angle

**Condition for maximum mechanical power developed:**

For maximum power developed \(\frac{{{\bf{d}}{{\bf{P}}_{{\bf{mech}}}}}}{{{\bf{dδ }}}} = 0\)

⇒ **\(\frac{{{\bf{EV}}}}{{{Z_s}}}\;{\bf{sin}}\left( {{\bf{θ }} - {\bf{δ }}} \right) - 0 = 0\)**

⇒ sin (θ - δ) = 0

⇒** θ = δ **

**∴ The maximum power developed by the synchronous motor when the load angle is equal to the internal angle.**