Question
Download Solution PDFCalculate the line value of induced emf of a 10-pole, 3-phase, 60 Hz star-connected alternator with 60 slots and 4 conductors per slot. The value of the pitch factor is 0.966, the distribution factor is=0.966, the flux per pole is 0.12 Wb and it is sinusoidally distributed.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept
The per phase induced emf in an alternator is given by:
\(E_p=4.44fϕ K_pK_dT_{ph}\)
where, f = Frequency
ϕ = Flux per pole
Tph = Turns per phase
Kp = Pitch factor = \(cos({\beta\over 2})\)
Kd = Distribution factor = \({sin({m\beta\over 2})\over m \space sin({\beta \over 2})}\)
Calculation
Given, Kp = Kd = 0.966
ϕ = 0.12 Wb
Total conductor = 60 × 4 = 240
We know that 2 conductors = 1 turn
∴ 240 conductors = 120 turn
Tph = 120/3 = 40
\(E_p=4.44× 60× 0.12× 0.966× 0.966× 40\)
Ep = 1193.24 V
The line value of induced emf is given by:
\(E_L=√{3}E_p\)
EL = √3 × 1193.24 = 2066.76 V
Last updated on May 29, 2025
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