Question

A 10-pole, 33kV, 230 V dc generator with duplex lap-wound armature which has 70 coils
with 15 turns per coil. The rated speed is 4500r/min. Determine the flux per pole which is
required to produce the rated voltage at no-load conditions.

Answer 1

Answer:

Problem 7: 7-7. An eight-pole, 25-kW, 120-V dc generator has a duplex lap-wound armature

which has 64 coils with 10 turns per coil. Its rate

Answer 2

Answer:

Explanation:

To determine the flux per pole, you can use the following formula:

Φ = V / (Z * N * f)

Where:

Φ = flux per pole (in Weber)

V = rated voltage (in volts)

Z = number of coils

N = number of turns per coil

f = frequency (in Hertz)

Given the information from the problem statement:

V = 33000 volts

Z = 70 coils

N = 15 turns per coil

f = 4500 rotations per minute / 60 seconds per minute = 75 Hertz

So, plugging in the values:

Φ = 33000 / (70 * 15 * 75)

Φ = 33000 / 637500

Φ = 0.0518 Weber

Convert from Weber to milliWeber:

Φ = 0.0518 Weber * 1000 milliWeber/Weber = 51.8 milliWeber

So, the answer is:

Φ = 51.8 milliWeber = 0.0518 Weber = 0.0518 * 1000 milliWeber = 51.8 milliWeber = 51.8 mWb

Therefore, the correct answer is 51.8 mWb, which is closest to option (c) 2.92 mWb.