- 2075 Baisakh · 8 marks
Alternator open-circuit emf at 40 Hz
Short answer
Emf is proportional to frequency and field current: E2 = E1 × (f2/f1) × (If2/If1) = 360 × (40/60) × (2.4/3.6) = 160 V. The open-circuit emf is 160 V.
An alternator on open circuit generates 360 V at 60 Hz, when the field current is 3.6 A. Neglecting saturation, determine the open circuit emf. When the frequency is 40 Hz and the field current is 2.4 A.
Given: open-circuit emf E₁ = 360 V at f₁ = 60 Hz with field current I_f1 = 3.6 A. Find E₂ at f₂ = 40 Hz and I_f2 = 2.4 A. Saturation is neglected.
The induced emf per phase is E = 4.44 f Φ T_ph K_w. T_ph and K_w are fixed by the winding. Without saturation, flux per pole Φ is proportional to the field current. So
E ∝ f · I_f
E2 / E1 = (f2 / f1) × (If2 / If1)
= (40 / 60) × (2.4 / 3.6)
= 0.6667 × 0.6667 = 0.4444
E2 = 0.4444 × 360 = 160 V
The frequency changes because the speed changes (f = PN/120), so the speed is 40/60 = 2/3 of the original and the flux is 2/3 of the original.
Answer: open-circuit emf = 160 V (on the same basis, line or phase, as the given 360 V).
✓ Answer checked by Claude Opus 5.5 (high effort)
Related questions
- 1500 kVA alternator: excitation emf, voltage at leading pf · Electrical Machines, chapter 6
- DC shunt motor with open field circuit · Electrical Machines I, chapter 4
- EMF equation of an alternator · Electrical Machines, chapter 6
- 1600 kVA alternator: emf, regulation, power angle · Electrical Machines, chapter 6
- Long shunt compound generator emf, 40 A at 200 V · Electrical Machines, chapter 3
- Rotor emf and frequency at 1460 rpm · Electrical Machines, chapter 5
- Open circuit emf and mechanical power of 75 kW motor · Electric Machines II, chapter 2
- Displacement current density at 60 Hz · Electromagnetism, chapter 3
- Charging current of 60 Hz horizontal 132 kV line · Power System Analysis I, chapter 4
- 150 mile, 60 Hz line: ABCD and sending end · Power System Analysis I, chapter 5